# Trenberth's Energy Budget



## IanC (Oct 23, 2011)

how many of you think it is a deception to have 390 (W/m2) going up and 324 coming down? while it is true in an absolute way it disguises the fact that only 66 w/m2 is available as the difference between the temps of the surface and the lower atmosphere. out of that 66, 40 leaves directly with only 26 W/m2 warming the lower atmosphere.

so 26 W/m2 is the greenhouse effect and I have never seen any estimates higher than 26% for CO2's share of the GHE so that means ~8 W/m2 or less for the total amount of CO2. the absorption of CO2 is logrithmic so that means we are into the ninth doubling (256-512) so every doubling is ~(8W/m2)/8 doublings= 1 W/m2 or less.

I find it hard to believe that something that is responsible for ~1 W/m2 is the grand control knob of the climate. it is an amount that is lost in the error bars of measurement. but that is just my opinion but now you know why I think that way.


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## code1211 (Oct 23, 2011)

IanC said:


> how many of you think it is a deception to have 390 (W/m2) going up and 324 coming down? while it is true in an absolute way it disguises the fact that only 66 w/m2 is available as the difference between the temps of the surface and the lower atmosphere. out of that 66, 40 leaves directly with only 26 W/m2 warming the lower atmosphere.
> 
> so 26 W/m2 is the greenhouse effect and I have never seen any estimates higher than 26% for CO2's share of the GHE so that means ~8 W/m2 or less for the total amount of CO2. the absorption of CO2 is logrithmic so that means we are into the ninth doubling (256-512) so every doubling is ~(8W/m2)/8 doublings= 1 W/m2 or less.
> 
> I find it hard to believe that something that is responsible for ~1 W/m2 is the grand control knob of the climate. it is an amount that is lost in the error bars of measurement. but that is just my opinion but now you know why I think that way.





Yeah!  What HE said!.


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## Mr. H. (Oct 23, 2011)

Looks like a coronary bypass.


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## Mr. H. (Oct 23, 2011)

Aine said:


> Reflective clothing can include high visibility vests, jackets or simple armbands. These can be equipped with reflective strips and even flashing LED lights for super visibility.It is safe and save energy



That would be a handsome accompaniment to your foil hat.


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## IanC (Oct 24, 2011)

IanC said:


> how many of you think it is a deception to have 390 (W/m2) going up and 324 coming down? while it is true in an absolute way it disguises the fact that only 66 w/m2 is available as the difference between the temps of the surface and the lower atmosphere. out of that 66, 40 leaves directly with only 26 W/m2 warming the lower atmosphere.
> 
> so 26 W/m2 is the greenhouse effect and I have never seen any estimates higher than 26% for CO2's share of the GHE so that means ~8 W/m2 or less for the total amount of CO2. the absorption of CO2 is logrithmic so that means we are into the ninth doubling (256-512) so every doubling is ~(8W/m2)/8 doublings= 1 W/m2 or less.
> 
> I find it hard to believe that something that is responsible for ~1 W/m2 is the grand control knob of the climate. it is an amount that is lost in the error bars of measurement. but that is just my opinion but now you know why I think that way.



does everybody understand the conceptual difference between the 168 W/m2 coming from the sun that is available to directly heat an object, and the 324 W/m2 backradiation that can only counteract part of the 390 W/m2 upward radiation from the surface? on a cloudy day you feel immediate warmth when the sun peeks through.


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## IanC (Oct 24, 2011)

I forgot to note that the Sun's 168 W/m2 is mostly higher energy photons than the radiation from and back to the earth, which would also affect why it seems warm when the sun comes out


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## wirebender (Oct 24, 2011)

IanC said:


> I forgot to note that the Sun's 168 W/m2 is mostly higher energy photons than the radiation from and back to the earth, which would also affect why it seems warm when the sun comes out



There is no energy coming back to the earth.

Second Law of Thermodynamics:  It is *not possible *for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will *not flow *spontaneously from a low temperature object to a higher temperature object. 

Second Law of Thermodynamics

_Clausius statementGerman scientist Rudolf Clausius is credited with the first formulation of the second law, now known as the Clausius statement:

No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.

Spontaneously, heat cannot flow from cold regions to hot regions without external work being performed on the system, which is evident from ordinary experience of refrigeration, for example. In a refrigerator, heat flows from cold to hot, but only when forced by an external agent, a compressor._


Which part of "not possible" and "will not" is it that you guys don't understand.  "Net" energy flow is a fiction, a fabrication, a concoction, a fable, and a falsehood invented by people to whom the actual second law of thermodynamics is an inconvenience.  

Of course, if you can show some observable, repeatable experimental evidence that proves otherwise, I would be very interested in seeing it as it would be the basis for a damned profitable perpetual motion machine.

BTW, absorption and emission by CO2 molecules does not constitute work.


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## flacaltenn (Oct 24, 2011)

IanC said:


> how many of you think it is a deception to have 390 (W/m2) going up and 324 coming down? while it is true in an absolute way it disguises the fact that only 66 w/m2 is available as the difference between the temps of the surface and the lower atmosphere. out of that 66, 40 leaves directly with only 26 W/m2 warming the lower atmosphere.
> 
> so 26 W/m2 is the greenhouse effect and I have never seen any estimates higher than 26% for CO2's share of the GHE so that means ~8 W/m2 or less for the total amount of CO2. the absorption of CO2 is logrithmic so that means we are into the ninth doubling (256-512) so every doubling is ~(8W/m2)/8 doublings= 1 W/m2 or less.
> 
> I find it hard to believe that something that is responsible for ~1 W/m2 is the grand control knob of the climate. it is an amount that is lost in the error bars of measurement. but that is just my opinion but now you know why I think that way.



I've stared at this before -- and no it's not illogical to have 390 going out with 324 bouncing back. That's the heat loss rate of the earth that HAS to happen at night when the left side of the diagram is not operating. Futhermore, we know that the heat loss at night would be mitigated by cloud and atmospheric conditions. So as far as your calculations for CO2 contribution -- you could be right. 

Part of the confusion is that the Trenberth numbers are for heat FLOW RATES, not energy. The earth and atmosphere is a heat battery that gets charged by the left side rates and discharged by the right side rates. So "average rates" aren't very meaningful unless you consider the CAPACITY of the various components to store energy. Add water vapor and greenhouse gases and that CAPACITY increases in the atmosphere. So even if the rates didn't change (they would) -- the ability to coast thru a colder night is increased by the storage capacity. Can't really analyze the dynamics of temperature change without the time and capacity factors. 

2nd -- the right side of the chart is happening all day long. It's just easier to see it's sum contribution by considering it at nighttime. Whereas the left side is operating at VARYING efficiency all day long. So the rate of heat charge to replace loss at night has to EXCEED the sum discharge rate on the right side. That's why when you find the net discharge to be 66w/m2, -- it's no surprise that the net CHARGE rate is closer to 168w/m2. That's MORE then double the net discharge -- which makes perfect sense because of "daytime variability" of power coming in. 

So yeah --- after staring at it a couple times, it passes the first level reasoning that at least I can apply to it...


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## flacaltenn (Oct 24, 2011)

One thing that could be looked at more BTW --- is that we are concentrating on the energy balance at the earths surface.. That is just ONE of the TWO major heat storage systems in the diagram. Even if we could affirm "balance" at the surface of the earth, there is still "balance" in the temperature of the atmosphere which has DIFFERENT flow rates associated. Could be that more energy is being stored in the atmosphere than actually at the surface. Thus modifying the "back radiation" value. In fact, it's a little flaky that I see about 200w/m2 going OUT of the atmosphere, but LESS than that coming in.. That can't proceed that way for very long -- can it... !!!

But I thought that they've scratched their heads about not finding the predicted warming at higher altitude that the GHG folks expected..


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## wirebender (Oct 24, 2011)

flacaltenn said:


> I've stared at this before -- and no it's not illogical to have 390 going out with 324 bouncing back. That's the heat loss rate of the earth that HAS to happen at night when the left side of the diagram is not operating. Futhermore, we know that the heat loss at night would be mitigated by cloud and atmospheric conditions. So as far as your calculations for CO2 contribution -- you could be right.



That budget isn't based on a model that has a left and a right side.  That model is based on a flat earth that is receiving sunlight 24 hours a day at a rate of 1/4 of actual.  There is no dark side of the earth associated with that model and as a result, a greenhouse effect is necessary to explain the observed temperature.

  If you model the earth as a sphere that is being radiated across 180 degrees of its surface and dark across 180 degrees of its surface, no greenhouse effect is necessary to account for the observed temperature of the earth.


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## flacaltenn (Oct 24, 2011)

wirebender said:


> flacaltenn said:
> 
> 
> > I've stared at this before -- and no it's not illogical to have 390 going out with 324 bouncing back. That's the heat loss rate of the earth that HAS to happen at night when the left side of the diagram is not operating. Futhermore, we know that the heat loss at night would be mitigated by cloud and atmospheric conditions. So as far as your calculations for CO2 contribution -- you could be right.
> ...



With that kind of Sesame Street assumption, it's a wonder that any flow rate numbers would come close to balancing long term. Where does one find all these "assumptions"???


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## IanC (Oct 25, 2011)

flacaltenn said:


> One thing that could be looked at more BTW --- is that we are concentrating on the energy balance at the earths surface.. That is just ONE of the TWO major heat storage systems in the diagram. Even if we could affirm "balance" at the surface of the earth, there is still "balance" in the temperature of the atmosphere which has DIFFERENT flow rates associated. Could be that more energy is being stored in the atmosphere than actually at the surface. Thus modifying the "back radiation" value. In fact, it's a little flaky that I see about 200w/m2 going OUT of the atmosphere, but LESS than that coming in.. That can't proceed that way for very long -- can it... !!!
> 
> But I thought that they've scratched their heads about not finding the predicted warming at higher altitude that the GHG folks expected..




it looks like all the numbers add up. but there is a discrepancy between the satellite surface temps and ground based surface temps. the whole Urban Heat Island effect is a bit of a conumbrum, as are other land use issues. but at least factors other than CO2 are getting looked at more in depth now.


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## IanC (Oct 25, 2011)

wirebender said:


> IanC said:
> 
> 
> > I forgot to note that the Sun's 168 W/m2 is mostly higher energy photons than the radiation from and back to the earth, which would also affect why it seems warm when the sun comes out
> ...







> Since the temperature of an object is a function of both energy gain AND energy loss, the temperature of the plate (or anything else) can be raised in 2 basic ways: (1) increase the rate of energy gain, or (2) decrease the rate of energy loss. The temperature of everything is determined by energy flows in and out, and one needs to know both to determine whether the temperature will go up or down. This is a consequence of the 1st Law of Thermodynamics involving conservation of energy.


Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still « Roy Spencer, Ph. D.


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## wirebender (Oct 25, 2011)

flacaltenn said:


> wirebender said:
> 
> 
> > flacaltenn said:
> ...



They aren't assumptions.  Trenberth et al treat the earth as a blackbody and divide incoming energy from the sun in the form of /4 which is fine if you have an actual black body that is radiating the same amount of energy from any point.  You can lay such a body out as a flat surface and accurately represent radiating energy.  The earth, however is not a self illuminating black body that can be accurately represented as flat.   The earth is an illuminated sphere that is radiated on one side and dark on the other.  Incoming energy must be divided in the form of /2, not /4.


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## wirebender (Oct 25, 2011)

IanC said:


> > Since the temperature of an object is a function of both energy gain AND energy loss, the temperature of the plate (or anything else) can be raised in 2 basic ways: (1) increase the rate of energy gain, or (2) decrease the rate of energy loss. The temperature of everything is determined by energy flows in and out, and one needs to know both to determine whether the temperature will go up or down. This is a consequence of the 1st Law of Thermodynamics involving conservation of energy.
> 
> 
> Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still « Roy Spencer, Ph. D.



Are you really presenting a failed thought experiment as evidence of two way energy flows between warm and cool objects?  Really Ian?


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## IanC (Oct 25, 2011)

thought experiments using basic principles are the best way to understand anything without getting bogged down in extraneous complexities.

why do you consider it a failed experiment?


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## IanC (Oct 25, 2011)

wirebender said:


> flacaltenn said:
> 
> 
> > wirebender said:
> ...




while I appreciate the need to be as accurate as possible in calculating how much energy is coming in, what real difference does it make? the energy in must match the energy out. surely no one actually thinks the estimates for the various pathways in and out are fixed in stone, they are just a model to get a better understanding of what may be going on.


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## Iridescence (Oct 25, 2011)

IanC said:


> how many of you think it is a deception to have 390 (W/m2) going up and 324 coming down? while it is true in an absolute way it disguises the fact that only 66 w/m2 is available as the difference between the temps of the surface and the lower atmosphere. out of that 66, 40 leaves directly with only 26 W/m2 warming the lower atmosphere.
> 
> so 26 W/m2 is the greenhouse effect and I have never seen any estimates higher than 26% for CO2's share of the GHE so that means ~8 W/m2 or less for the total amount of CO2. the absorption of CO2 is logrithmic so that means we are into the ninth doubling (256-512) so every doubling is ~(8W/m2)/8 doublings= 1 W/m2 or less.
> 
> I find it hard to believe that something that is responsible for ~1 W/m2 is the grand control knob of the climate. it is an amount that is lost in the error bars of measurement. but that is just my opinion but now you know why I think that way.



This is very interesting... and reminds me a bit of a book that I may pick up, eventually... Scott Mackay, Award Winning Mystery Writer Official Website *Phytosphere*. Otherwise...  Looks too incomplete, what else is missing?


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## wirebender (Oct 25, 2011)

IanC said:


> thought experiments using basic principles are the best way to understand anything without getting bogged down in extraneous complexities.
> 
> why do you consider it a failed experiment?



It was a violation of the 2nd law of thermodynamics.  A poster named Gord, among others, disproved spencer's claims and one couldn't help but notice that spencer didn't attempt a rebuttal. In the end, he was unable to adequately defend his experiment.   Also, the experiment takes place in a perfect vaccum (as opposed to the open atmosphere) and that, in and of itself takes it out of the realm of reality and makes it a pointless exercise.


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## wirebender (Oct 25, 2011)

IanC said:


> while I appreciate the need to be as accurate as possible in calculating how much energy is coming in, what real difference does it make? the energy in must match the energy out. surely no one actually thinks the estimates for the various pathways in and out are fixed in stone, they are just a model to get a better understanding of what may be going on.



That budget is not based on anything even remotely resembling reality.  How then, do you suppose it represents anything even resembling reality?  As has been pointed out, if the earth is modeled as a rotating sphere being irradiated across 180 degrees of its surface, and dark across 180 degrees of its surface, a fictitious greenhouse effect is not necessary to account for the observed temperatures on the earth.  That is a pretty "real" difference if you ask me.

And what sort of understanding do you suppose is to be gleaned from a model that doesn't even approach reality?


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## gslack (Oct 26, 2011)

wirebender said:


> IanC said:
> 
> 
> > thought experiments using basic principles are the best way to understand anything without getting bogged down in extraneous complexities.
> ...



As A non-scientist. I see one problem.. Maybe its my nature to notice the obvious... But the second plate shouldn't cause the first plate to get hotter, the heat source sets the temperature of it. The second plate is an insulator forcing more heat retention in the first.. The first plate will get hotter because less or slower heat loss causing a build up of heat energy. But if the first plate is at maximum temperature the heat source produces it will not get hotter than its source.

its a logic test... he is associating heat retention due to an added insulator with an increase in heat energy. Maybe I dont have all the terms or scientific jargon to give this a proper explanation, but in essence this is a logic test. All he did was switch a concept and use misleading terms to confound the simplicity of it all. Everyone was arguing his methods and convection, etc. but no one asked the simple thing..

The temp of the first plate will not go beyond that allowed by its heat source. Adding the second plate slows the heat loss, but that does not constitute more heat due to the second plate.

WTH? is it really that easy to distract smart people?


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## IanC (Oct 26, 2011)

gslack said:


> wirebender said:
> 
> 
> > IanC said:
> ...



well, you are right that the experiment is a test of analytical skills. unfortunately you fail again. 

the conditions are that the bars are in a cooled thermos with a vacuum to represent outer space. the bars arent touching so there is no conduction, there is no air so there is no convection. that leaves only radiation. heat loss through radiation is proportional to difference in temperature between the radiator and the absorber. 

both bars will be warmer than if only one bar was present. the extra heat comes from the area of radiation blocked from escaping to 'outer space' from both bars. so in reality 'outer space is cooler by the same amount that is retained by the bars.


the earth's surface and atmosphere are much more complicated, with conduction convection and water-based effects coming into play. so what? the question was 'can something cooler make a warm thing warmer'. the answer to that is yes, depending on the conditions.


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## IanC (Oct 26, 2011)

wirebender said:


> IanC said:
> 
> 
> > while I appreciate the need to be as accurate as possible in calculating how much energy is coming in, what real difference does it make? the energy in must match the energy out. surely no one actually thinks the estimates for the various pathways in and out are fixed in stone, they are just a model to get a better understanding of what may be going on.
> ...




well, I dont really disagree with you. but most of the inputs and outputs are still there even if you dont agree with the actual figures for them. I am all for better models but I cant see just throwing out the existing model until it is supplanted with a better one.


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## wirebender (Oct 26, 2011)

gslack said:


> wirebender said:
> 
> 
> > IanC said:
> ...



Precisely.  The energy input to the heated plate establishes the maximum temperature and to get even the smallest fraction of one degree above that is to create energy and that simply can not happen.


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## wirebender (Oct 26, 2011)

IanC said:


> well, you are right that the experiment is a test of analytical skills. unfortunately you fail again.



Well someone failed, unfortunately for you, since you believe yourself to be the smartest guy in the room you don't realize that it was you.



IanC said:


> the conditions are that the bars are in a cooled thermos with a vacuum to represent outer space. the bars arent touching so there is no conduction, there is no air so there is no convection. that leaves only radiation. heat loss through radiation is proportional to difference in temperature between the radiator and the absorber.



It doesn't matter whether it was in a vaccum or not Ian.  The second law and the law of conservation of energy are laws of nature, not laws of systems or laws of vacuums, or laws of anything else you care to name.  They apply everywhere.



IanC said:


> both bars will be warmer than if only one bar was present. the extra heat comes from the area of radiation blocked from escaping to 'outer space' from both bars. so in reality 'outer space is cooler by the same amount that is retained by the bars.



The electrical energy coming into the heated bar establishes the maximum temperature.  It can not go higher than that because to do so would be to create energy.  You were shown mathematically that it can not happen and instead of grasping what you saw, you reverted to your faith that cool objects can heat warmer objects.

Here is the proof that was shown to you Ian.  What precisely do you find to be wrong with the math?

_Re: Vacuum Chamber with plates.

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts

(***That&#8217;s ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)

The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2
&#8212;&#8212;&#8211;
If another identical &#8220;non-heated&#8221; and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:

The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.

The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.

The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

And the equilibrium temperature for both plates will be 284.69 K or 53 deg F.
&#8212;&#8212;&#8212;&#8212;-
*NOTE:

This is simple effect is demonstrated every day and is happening in your Computer right now.

It&#8217;s called a &#8220;Heat Sink&#8221; and is used to cool the microprocessor in your computer by increasing the radiating surface area of the microprocessor*._

Here was the mathematical proof that in order for the heated bar to reach a temperature of 160, additional energy would have to be created.

_PROOF:
160 deg F = 344.11 K and that means it would require:
P = (5.67X10^-8) X 1m^2 X (344.11 K)^4 = 795 Watts!

795 Watts exceeds the TOTAL Energy available which is 744.95 Watts.

An obvious violation of The Law of Conservation of Energy._

You were shown that in order to get the heated plate to 160 degrees, an additional 50.05 watts above and beyond what is available is required.  Where do you believe that additional energy came from?  The plate that had no power source?  Well of course you believe that is where it came from because you also beleive that the atmosphere, which is not an energy source, provides energy to the surface of the earth.



IanC said:


> the earth's surface and atmosphere are much more complicated, with conduction convection and water-based effects coming into play. so what?



Hell yeah it is more complicated but you are fooled by a very simple thought experiment in which it is claimed that by placing a non powered bar next to a powered bar you can somehow coax 6.7 more watts out of the system than is going in.  If that is possible, why aren't we doing it on a large scale and making use of that excess energy?  Someone should be making billions off of the process.



IanC said:


> The question was 'can something cooler make a warm thing warmer'. the answer to that is yes, depending on the conditions



The answer is no Ian.  You accept that it can happen on faith, not on the basis of any physical law and there is no condition that we can create in which the second law of thermodynamics or the law of conservation of energy is violated.  Hell Ian, that experiment violates the 2nd law of thermodynamics even if you throw in the fictitious "net" energy flow allowing energy from cool objects to be absorbed by warm objects but the net result is that more energy is radiated to the cooler object because the net result of that experiment was that the heated bar actually got hotter.


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## wirebender (Oct 26, 2011)

IanC said:


> well, I dont really disagree with you. but most of the inputs and outputs are still there even if you dont agree with the actual figures for them. I am all for better models but I cant see just throwing out the existing model until it is supplanted with a better one.



The inputs are there, but the outputs are the result of a falsehood.  They are based on something other than reality so why on earth should they be used for anything other than perhaps a bird cage liner?

Why keep a model that is patently and obviously wrong simply because the climate change community won't accept a more realistic model?  We both know the answer to that question Ian.  It is because a more realistic model doesn't require an ad hoc greenhouse effect to explain the temperature of the earth and without an ad hoc greenhouse effect, there is no climate alarmism and without climate alarmism, the grant money will dry up faster than a puddle of piss in Death Valley.


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## IanC (Oct 26, 2011)

wirebender said:


> IanC said:
> 
> 
> > well, you are right that the experiment is a test of analytical skills. unfortunately you fail again.
> ...



I just went back and looked at Spencer's article again. assuming has made an ass out of me. I assumed that the second bar was also heated but to a lower temp. I cannot believe Spencer could make such a fundemental error. perhaps edthecynic is right about him.

yah what a goof that Spencer is!!!! now if he had said the second bar totally enveloped the first bar it would be _obvious_ that the inside bar would be warmer because of the thermal impedance of the outer bar which would be much cooler than 150 degrees because of the greater surface area to radiate from. the inside heating element would be hotter too because of the insulating property of the outside bar not allowing it to shed heat as quickly as before.

if there was a hole in the outside bar the inside bar would radiate out at a more powerful intensity because the inside was hotter but soon would cool a bit because of the extra radiation allowed to escape. and the outside bar would be a little cooler because some of the heat didnt have to travel through it to escape.

and if you made the hole a little bigger more heat would escape but the inside block would still be hotter than 150 degrees because of the insulating property of the outside bar.

hey! wait a minute!!! if you keep shrinking the outside bar until it only insulates on one side of the inside bar it still adds thermal impedance and causes the heated bar to be warmer!

the heated bar has less surface area available to transfer heat to the cold outer space therefore the remaining surface area is hotter to shed the same amount of available heat.


wirebender dont quit your day job because you wouldnt make it as a physicist. retard, did you really think a noted scientist and physicist would make stupid mistakes and leave them uncorrected?


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## wirebender (Oct 26, 2011)

IanC said:


> I just went back and looked at Spencer's article again. assuming has made an ass out of me. I assumed that the second bar was also heated but to a lower temp. I cannot believe Spencer could make such a fundemental error. perhaps edthecynic is right about him.



It goes much deeper than that Ian.  The entire climate change community, right down to the textbooks used to teach climate science are using a form of the Stefan-Boltzman Law that allows two way energy exchange between warm and cool objects.  The version of SB that they are using is not, and never has been used or supported in the physics literature.  As I said, "net" is a concoction.  It is another example of ad hoc pseudoscience by the climate science community to address an inconvenient fact.



IanC said:


> yah what a goof that Spencer is!!!! now if he had said the second bar totally enveloped the first bar it would be _obvious_ that the inside bar would be warmer because of the thermal impedance of the outer bar which would be much cooler than 150 degrees because of the greater surface area to radiate from. the inside heating element would be hotter too because of the insulating property of the outside bar not allowing it to shed heat as quickly as before.



Insulation can not make a thing get hotter Ian.  Insulation can only slow down the rate at which it cools.  The amount of power going into the heated bar establishes the maximum temperature possible.  You can not coax even the tiniest fraction of a watt beyond that no matter how much you reflect, insultate, or surround.  To do so would be a violation of the Law of Conservation of Energy.



IanC said:


> and if you made the hole a little bigger more heat would escape but the inside block would still be hotter than 150 degrees because of the insulating property of the outside bar.



You could never get that heated bar above 150 unless you bring in more electricity.

Look again Ian:

_P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts_



IanC said:


> hey! wait a minute!!! if you keep shrinking the outside bar until it only insulates on one side of the inside bar it still adds thermal impedance and causes the heated bar to be warmer!



The only way to make the heated bar warmer than 150 is to either add more electricity or reduce the size of the heated bar.  



IanC said:


> the heated bar has less surface area available to transfer heat to the cold outer space therefore the remaining surface area is hotter to shed the same amount of available heat.



No Ian, if you reduce the size of the non heated bar, the only thing you will achieve is to reach equilibrium faster at which point, the two can be considered to be a single radiating unit.



IanC said:


> wirebender dont quit your day job because you wouldnt make it as a physicist. retard, did you really think a noted scientist and physicist would make stupid mistakes and leave them uncorrected?



His whole exercise is flawed Ian because he is assuming wrongly that a cool object can radiate energy to a warmer object in violation of the 2nd law of thermodynamics.  I asked before, what is wrong with the math that proves the experiment wrong.  I note that neither Dr. Spencer, nor you have any answer.

By the way, noted physicists make stupid mistakes and leave them uncorrected all the time.  It would be nothing new.


----------



## gslack (Oct 26, 2011)

IanC said:


> gslack said:
> 
> 
> > wirebender said:
> ...



Ian I have been as tolerant as I am going to be with you lately considering your behavior and showing not only your true lack of understanding scientifically, but also your complete and total inability to see just how obvious your bullshit has become lately... Do not condescend on me, you lack the ability to do so...

Ian in all of your BULLSHIT above how does that change anything I said? The point still remains the first plate nearest the heat source cannot reach a temperature above that of the heat source. Be it in a vacuum or anywhere else, the point will still remain the same..

Also why add a second bar? We only have one sun... 

See Ian I know you can't grasp this but what he did was over-complicate the system with variables and factors that do nothing to further the original concept. All they do is confound it and draw otherwise highly intelligent people into arguing over the variables instead of noticing the simple thing staring them in the face...

The second plate does not make the first plate warmer than its source will allow. That would be creating energy from nothing, but what it DOES do is slow the heat loss of the first plate, allowing for more build up of energy in the first plate. That first plate CANNOT and WILL NOT reach a temperature higher than that of its source without more work applied to it somehow. If the plate were flammable and caught fire then it could get warmer than its original source, but than again that would change the source from the original element to the damn fire consuming it, and that rule would still apply but to the new source.

Now you want to prove that there is energy creation by adding that second plate in his thought experiment please be my guest. But try not to get bogged down in all the extra inconsequential variables he added.. it makes you look like a google scientist with a thesaurus and a "how to sound like you know what you are talking about even if you don't" book....


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## konradv (Oct 26, 2011)

Forget the trolls, Ian.  You and I may not agree, but at least we live on the same planet.


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## gslack (Oct 26, 2011)

konradv said:


> Forget the trolls, Ian.  You and I may not agree, but at least we live on the same planet.



And thats Melmack right???


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## IanC (Oct 26, 2011)

man you are dense.

the heating coil inside the bar is hotter than 150 degrees. insulating any part of the heated bar  subtracts part of the surface area that can radiate heat away at the full 150-(-273) difference. the bar will be different temperatures on different parts but the overall outside temperature average will be higher.  your exaggeratedly accurate calculation does not take temperature gradients of the bar or surface area x temperature differential differences into account. as usual you either emphasize extraneous details or ignore important ones. get a brain, or at least use the one you have


----------



## IanC (Oct 26, 2011)

gslack said:


> IanC said:
> 
> 
> > gslack said:
> ...



as usual you dont have a clue. what temperature do you think the heating coil is at if the outside of the bar is only 150 degrees while it is radiating heat away as fast as it can?

dont you guys _ever_ try to think things through?


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## gslack (Oct 26, 2011)

IanC said:


> man you are dense.
> 
> the heating coil inside the bar is hotter than 150 degrees. insulating any part of the heated bar  subtracts part of the surface area that can radiate heat away at the full 150-(-273) difference. the bar will be different temperatures on different parts but the overall outside temperature average will be higher.  your exaggeratedly accurate calculation does not take temperature gradients of the bar or surface area x temperature differential differences into account. as usual you either emphasize extraneous details or ignore important ones. get a brain, or at least use the one you have



I didn't say anything about insulating the bar Ian... I said the second plate acts as an insulator for the first plate... Once again you try and make an obtuse argument and imply I said something I did not...

If you cannot debate honestly, then just ignore my posts...


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## gslack (Oct 26, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC said:
> ...



There ya go Ian confound it to cover your bullshit again...

Stop focusing on the bar, you and I both know what I said was the second plate acts as an insulator for the first plate.. Not the bar or heat source, but the first plate that receives the radiant heat from the heat source... Now if his experiment only uses the heat source and one plate, than it is even more ridiculous because it does not show anything that remotely relates to his point.. Using his example the sun should get hotter simply because our planet is orbiting it...

Your continued pretense and obfuscation shows once again that you cannot argue what i said... Fact is you know I am right, but rather than admit anybody is right besides yourself, you have to try and make the argument something else...

Now argue my point as it is, or ignore me.. But deliberate obfuscation will not hide you backed the wrong horse again...


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## wirebender (Oct 26, 2011)

IanC said:


> man you are dense.



You are the one that is dense Ian.



IanC said:


> the heating coil inside the bar is hotter than 150 degrees. insulating any part of the heated bar  subtracts part of the surface area that can radiate heat away at the full 150-(-273) difference. the bar will be different temperatures on different parts but the overall outside temperature average will be higher.  your exaggeratedly accurate calculation does not take temperature gradients of the bar or surface area x temperature differential differences into account. as usual you either emphasize extraneous details or ignore important ones. get a brain, or at least use the one you have



Sorry Ian, but you were given the equations that showed how much energy was required to raise the temperature of the bar to X degrees.  Adding another bar doesn't change that.  Adding 50 bars won't change that.  You have been fooled and just don't have the intellectual wattage to see it.  Sorry.


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## wirebender (Oct 26, 2011)

gslack said:


> I didn't say anything about insulating the bar Ian... I said the second plate acts as an insulator for the first plate... Once again you try and make an obtuse argument and imply I said something I did not...
> 
> If you cannot debate honestly, then just ignore my posts...



Actually gslack, the second plate acts as a heat sink for the first plate.  When they reach equilibrium, they act as a single radiating source at a lower temperature than the heated bar alone would be.   As with all heat sinks, it doesn't serve to make the powered radiating source hotter, but to allow it to bleed off energy more quickly.  It can radiate energy faster, and by doing so, will never reach a higher temperature.  Ian is so concerned with being right that he can't see the observable proof in the real world that the system described doesn't act as he imagines it.

Ian apparently beleives that the heat sinks in his computer serve to heat up his processor rather than cool it down.


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## IanC (Oct 26, 2011)

wirebender thinks computers are cooled by radiation rather than conduction and convection.


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## gslack (Oct 26, 2011)

wirebender said:


> gslack said:
> 
> 
> > I didn't say anything about insulating the bar Ian... I said the second plate acts as an insulator for the first plate... Once again you try and make an obtuse argument and imply I said something I did not...
> ...




Thank You! I didn't get the heat sink connection at the time, but YES exactly. Ironic I just had to replace a GPU fan and heatsink on this very PC I am using just the other day.

Either way as you said, it will not serve the purposes of generating energy (in this instance heat) out of nothing. 

If this concept worked as he described we could power the world with a simple heat engine. And if the photonic-vibrational ? energy generation properties of CO2 worked as claimed we could slap a tank of CO2 onto a light bulb make a 200% effective machine...


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## gslack (Oct 26, 2011)

IanC said:


> wirebender thinks computers are cooled by radiation rather than conduction and convection.



Why don't you for once actually debate what he said? What are you that afraid of him or that afraid of being wrong? Seriously, you know what he said and what he meant, why not for once show some integrity and address his points honestly?

He did not argue the way computer processors (computers are entire device) or chips are cooled using heat sinks.. He simply stated how the bar in the experiment acts like one. 

the heat sink allows heat from a small, hot, perhaps less heat conductive surface or object, to dissipate and spread out over a broader often more heat conductive object or surface.

And actually cooling a CPU or chip uses conduction, convection, and radiation Ian...

Conduction happens when heat sink is in physical contact with the CPU.

Convection occurs when the air inside the computer case interacts with the heat sink.

Radiation when the heat dissipates up from the cpu across the fins of the heat sink and radiates out into the air.

The primary would be convection in cases of higher temperatures, and a mixture of the other two in the case of lower temps. Good rule of thumb for cooling a chip, given a choice between a bigger fan or a bigger heat sink, go with the heat sink, its cheaper and requires no energy to do its job...

Now please debate HIS points as they are and not what you try and make them out to be..


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## wirebender (Oct 27, 2011)

IanC said:


> wirebender thinks computers are cooled by radiation rather than conduction and convection.



Ian, it doesn't matter whether the means of moving the heat is radiation, convection, or conduction, the laws of nature still apply.   Whether it is in a vaccum or not doesn't allow it to break the 2nd law of thermodynamics.  Because it is in the open atmosphere, the heat sink and the heat source will never reach equilibrium and become as a single radiating source as would happen in the vaccum of spencer's imagination.  

Once more Ian, here is the math that proves you wrong.  Do tell exactly what you believe is wrong with it.  As you can see, the bottom line is both plates ending up at 53 degrees F; considerably cooler than the plates in spencer's imagination.  Which physical law do you believe is being misapplied there, or what mathematical error do you believe has been made.  The laws of nature are called the laws of nature for a reason Ian and not the laws of systems, or laws of vacuum vs laws of open space.  You can not break the laws of nature and create energy simply because you suck all the air out of a chamber.  If you could, our energy woes would be a thing of the past.

You have been willingly duped because your faith is stronger than your analytical mind.  You want to believe that bar will get warmer because you are emotionally attached to the idea that CO2 can somehow cause the atmosphere to heat up.  Give up the belief that the cooler bar will somehow make the heated bar warmer and you have to give up your fantasy about CO2.

_Re: Vacuum Chamber with plates.

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts

(***That&#8217;s ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)

The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2
&#8212;&#8212;&#8211;
If another identical &#8220;non-heated&#8221; and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:

The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.

The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.

The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

And the equilibrium temperature for both plates will be 284.69 K or 53 deg F._


The only way to make that heated bar warmer Ian, is to either pump in more electricity through the wire or to make it smaller so that its radiating surface is smaller.  Placing a heat sink next to it serves to increase the radiating surface thus bleeding heat off faster and making it cooler.  You can't increase the amount of energy in the system by increasing the radiating surface; that only serves to make the system cooler and it doesn't matter a whit whether it is happening in the open atmosphere or in a vaccum chamber.


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## wirebender (Oct 27, 2011)

gslack said:


> Now please debate HIS points as they are and not what you try and make them out to be..



I am afraid that is never going to happen gslack.  Remember the stink he made over me not directing him to the math I did on the board earlier relating to the same argument except the material was CO2 and earth instead of a powered plate and a non powered plate as if he would destroy my argument if he could just get to that math.  Well, there is some math right here, of a more simple nature as it doesn't involve time and distance and yet, I don't see him doing anything but dodging it.  Later, after sufficient time has passed I will wager that if this thread comes up, he will ask for directions back here and pretend that if he could just see the math, he could destroy the argument.

It is clear by his avoidance of this relatively simple math that he is not the student of physics that he claims to be.  The straight forward formulas above are used day in and day out by engineers of all stripes who must deal with heat and they accurately predict the temperature materials will reach given the amount of energy going in, the amount of radiating surface, and the nature of the materials.  If they didn't work, every project would be guesswork and you can bet that there is no guesswork involved in designs intended to bleed off heat.

I doubt that you find many designs out in the real world that have to handle energy created by the heat sink.  Can you imagine suddenly having to deal with energy that you weren't getting from the electric company or your own generators by simply adding to your radiating surface?    Add some extra radiating surface, create some free energy.  We could build radiators the size of sky scrapers, plug them into a couple of AAA's and power entire cities.

Perpetual motion incarnate; and Ian believes in it.


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## gslack (Oct 27, 2011)

wirebender said:


> gslack said:
> 
> 
> > Now please debate HIS points as they are and not what you try and make them out to be..
> ...



Going by their hypothesis, we could put a low pressure tank of CO2 as a addon to an electric generator and BAM! turn 10 generators into 1000...


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## IanC (Oct 27, 2011)

> Re: Vacuum Chamber with plates.
> 
> First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.
> 
> ...



this is the thought experiment that wirebender and gslack want me to respond to. so I will



> Vacuum Chamber with plates.


I am assuming that the chamber is cooled to some specific constant temperature



> First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.
> The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.


the heater inside the plate is hotter than the outside of the plate because of the thermal impedance of the material of the plate. the outside plate temp is determined on how fast the internal energy can be shed to the vacuum of constant temperature. any change to the power source, thermal impedance of the plate material or temperature of the vacuum will change the equilibrium (changing the shape of the plate would also affect the temp if it changed the surface area).



> Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2
> Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:
> P = e*BC*A*T^4
> Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K
> ...


your definition, live with it



> If another identical &#8220;non-heated&#8221; and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:
> The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2


at the beginning the second plate (if at the same temp as the vacuum) will make little difference. but as the second plate heats up it will start to radiate in all directions, including back at the heated plate. this changes the ability of the heated plate to shed energy. the heated plate still wants to get rid of the same amount of heat therefore more heat will be dispersed through the area that can escape to the vacuum. increased heat means increased temperature.



> We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.
> The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.
> The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2



this is where your example really goes off the rails, hahahahaha. what sort of gullible idiot would think that both plates would be at the same temperature at equilibrium? hahahahaha. what kind of moron would believe that the areas exposed to the cold vacuum loses heat at the same rate as the areas next to warmer surfaces? could you imbeciles get more things wrong?


it is easy to see that the heated plate must warm up to compensate for the loss of exposure to the cooled vacuum due to the thermal impedance of the second plate, and that all four surfaces of the two plates (I am ignoring the edges) have different temperatures. the heated inside facing area is the warmest, the heated outside facing area is next, the inside facing unheated area is third and the unheated outside facing area is the coolest of them all.


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## gslack (Oct 27, 2011)

HAAHAHAHAHAHAHAHAHAHAHAHAAHAHAHAAAAAAA!

Ian you really honestly, and truly believe that adding a heat sink alone (no other work added) to a hot object will make that object get hotter??? Seriously???

Ian I think you have some fundamental flaw in your logic... Follow me for a second here..

Remember these in your claimed studies Ian?

1. You cannot win.

2. You cannot break even.

3. You cannot get out of the game.

A british scientist coined the phrases above, can't remember his name off-hand.. Do you know what the mean? they are a very simple way to remember the first 3 laws of thermodynamics.. Notice I said the first 3 laws... Thats important cause there is another one that came along later..

1. means you can't get something for nothing. conservation of energy...

2. Means in this instance heat cannot flow from colder body to a warmer body on its own. Or energy exhibits entropy... meaning it flows away from its source..

3. means absolute zero (molecules cease to move) cannot be reached. 

and the newest one..pay very close attention here..

4. The Zeroth law.. Its an observation where when two objects are separately in thermodynamic equilibrium with a third object, they are in equilibrium with each other. basically how thermometers work..

Now your contention breaks 1, 2 , and 4 ...Like it or not, believe it or not its how it is regardless.. So your argument would require first that energy out would have to be greater than energy in, second that heat would have to suddenly ignore its entropic nature, and third it would have to ignore equilibrium...


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## IanC (Oct 27, 2011)

gslack said:


> HAAHAHAHAHAHAHAHAHAHAHAHAAHAHAHAAAAAAA!
> 
> Ian you really honestly, and truly believe that adding a heat sink alone (no     	other work added) to a hot object will make that object get hotter??? 	Seriously???
> 
> ...



wow! a coherent post with a minimum of ad hom. good job.

my explanation doesnt involve any extra heat, just a redistirbution. the same 	amount of power goes in and the same amount of radiation goes out. lets 	look at another scenario. a symetrical object has a heater embedded in it but 	instead of being in the centre it is offset half way to the surface on one of the 	planes. the escaping energy will be larger on the side that is closer to the 	surface and less on the other side. hopefully you can agree with that.

next, cut the object in half in the same plane that the the heater was offset 	but keep the two pieces close but not touching. is there now a difference in 	heat flow? yes the heat flow by conduction is stopped and the heated side 	will increase its heat flow by that much and the non-heated side will 	decrease its heat flow by that much. after equilibrium has been reached we 	are now at the two body state in Spencer's and wire bender's examples, 	although coming from a different direction. 

what is the temperature of the exposed sides of the heated block (plate, 	hemishere, etc)? I dont know exactly but I do know that there is more heat 	radiating from the exposed sides than is flowing through the previously cut 	side. when we remove the unheated half what happens? all the sides have 	open access to the cold vacuum and radiate equally. obviously the power of 	the heater divided by six sides is less than (five sides and one side with less 	than unity heat flow). therefore the single heated piece is colder than when 	there were two pieces in the vacuum.


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## wirebender (Oct 27, 2011)

IanC said:


> this is the thought experiment that wirebender and gslack want me to respond to. so I will



Dishonesty right off the bat Ian.  You are the one who brought spencer's failed thought experiment into the discussion



IanC said:


> I am assuming that the chamber is cooled to some specific constant temperature



Assuming?  You brought the thought experiment into the discussion, are you saying now that you don't even know the parameters spencer specified?



IanC said:


> the heater inside the plate is hotter than the outside of the plate because of the thermal impedance of the material of the plate.



That makes absolutely no difference Ian.  The temperature of the heated bar is determined by the electricity going into the heating element and the surface area of the bar.  That's it.  You might get a different temperature if you use different materials but that isn't part of the experiment.  The bar reaches its maximum temperature based on the amount of power going in and the amount of surface area available to radiate off the heat.



IanC said:


> the outside plate temp is determined on how fast the internal energy can be shed to the vacuum of constant temperature. any change to the power source, thermal impedance of the plate material or temperature of the vacuum will change the equilibrium (changing the shape of the plate would also affect the temp if it changed the surface area).



None of those things change during the course of the experiment.




IanC said:


> your definition, live with it



Spoken like a man who doesn't have a clue as to what he is looking at.  If there is a problem with the equation, then state it.  If there is no problem with the equation, then you have lost the discussion.  Simple as that.  Clearly, you don't grasp the math and are arguing based on an article of faith.



IanC said:


> at the beginning the second plate (if at the same temp as the vacuum) will make little difference. but as the second plate heats up it will start to radiate in all directions, including back at the heated plate.



No, Ian, it will not radiate back to the heated plate. See the second law of thermodynamics.  _It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object. _

I have asked before and will ask again, which part of NOT POSSIBLE and WILL NOT is it that you don't grasp.  There is nothing in the second law about two way energy flow with a net result.  The statement is pretty straight forward.  It IS NOT POSSIBLE for heat to flow from a cooler body to a warmer body without work having been done to accomplish the flow.  Sorry Ian, but that is a law of nature and it defeats your argument hands down.



IanC said:


> this changes the ability of the heated plate to shed energy. the heated plate still wants to get rid of the same amount of heat therefore more heat will be dispersed through the area that can escape to the vacuum. increased heat means increased temperature.



It acts as a heat sink Ian.  Do you know what the term means and implies?  No more heat is going to go through the heated plate than at the onset of the experiment.  The only thing that has happened is that by adding a second bar, you have inserted a heat sink into the equation which increases the radiating surface area which in turn will lower the temperature of the heated bar.  Adding a heat sink does not increase the amount of energy being fed into the system.  The amount of energy remains constant in the experiment.



IanC said:


> this is where your example really goes off the rails, hahahahaha. what sort of gullible idiot would think that both plates would be at the same temperature at equilibrium? hahahahaha. what kind of moron would believe that the areas exposed to the cold vacuum loses heat at the same rate as the areas next to warmer surfaces? could you imbeciles get more things wrong?



Again, you fail to recognize the implications of the parameters that govern the experiment.  It is in a vacuum Ian so of course they reach equilibrium, or within a fraction of a degree of equilibrium.  Equilibrium would never be reached in the open atmosphere, but in a cooled vaccum, equilibrium would be readily reached.   As to different surfaces being different temperatures; again, maybe a fraction of a degree but the bars are in a cooled vacuum.

Either point out a problem with the equations in question Ian, or forfiet the argument because if the equations are correct, then the results of the experiment are as the equations predict.  The numbers don't lie Ian while faith will lead you off in all sorts of directions.


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## wirebender (Oct 27, 2011)

gslack said:


> HAAHAHAHAHAHAHAHAHAHAHAHAAHAHAHAAAAAAA!
> 
> Ian you really honestly, and truly believe that adding a heat sink alone (no other work added) to a hot object will make that object get hotter??? Seriously???
> 
> ...



What amazes me is his complete disregard for the laws of physics.  Faith is a hell of a thing, isn't it?  I can easily imagine his eyes sparkling with a zealot's zeal as he thinks up even more convoluted thought experiments that also can not break the laws of physics.


----------



## IanC (Oct 27, 2011)

wirebender said:


> IanC said:
> 
> 
> > this is the thought experiment that wirebender and gslack want me to respond to. so I will
> ...





> your experiment's equations are wrong in many areas. the most startling is that the author thought that the two plates would be the same temperature at equilibrium and that he thinks that heat loss to a cold vacuum is the same as to a less cold object between the heated plate and the vacuum. how could anyone not recognize that the areas facing each other between two close objects is not able to shed radiation as easily as free access to cold vacuum.




next time could you specifically state which experiment you are discussing


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## gslack (Oct 27, 2011)

IanC said:


> gslack said:
> 
> 
> > HAAHAHAHAHAHAHAHAHAHAHAHAAHAHAHAAAAAAA!
> ...



Ian, that is perhaps the single best example of Bullshit walking I have seen.. Seriously you couldn't defend the concepts and claims in the first experiment so you make up another one that makes absolutely no attempt to address anything in the first experiment....WTH?

What kind of posturing, preening wannabe peacock have you turned into? What in the hell does that experiment show other than heat transfer is stronger the closer an object is to its heat source... Seriously you just took a shit load of time to come up with a few paragrapsh that not only does not address the problems in the first experiment, but do nothing but affirm that which we already know about heat transfer relative to the distance from its heat source..

WOW! You really are pulling shit out of your ass and shouting Eureka!


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## gslack (Oct 27, 2011)

IanC said:


> wirebender said:
> 
> 
> > IanC said:
> ...



Playing obtuse again Ian? Shameless... The experiment has been the one you brought into the discussion from the start, you brought in another experiment to try and bolster the previous one which did nothing but state the obvious and uncontested in this discussion... in BOTH cases the experiments where brought in by you.. 

Deliberate obfuscation makes you look like a desperate fake trying to save face...


----------



## IanC (Oct 28, 2011)

gslack- why did you edit out almost all of my answers to wirebender in that last quote? are you taking dishonesty tips from the SkepticalScience blog?


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## wirebender (Oct 28, 2011)

IanC said:


> I brought up Spencer's, you brought up an example from [Gord] and then I was accused of ducking the issue of the plates



It is clear by now that you don't grasp the math even though it is little more than algebra but DAMN MAN, get with the program.  The math, courtesy of Gord, is addressing spencer's experiment, and yes, you ducked the issue of the plates.  It is abundantly clear now why you ducked.



IanC said:


> your experiment did not define the temperature size or shape of the containment vessel



Once more Ian, get with the program.  I have no experiment.  The math you were presented with was addressing spencer's thought experiment as he defined the parameters.  Geez guy, is it all that far over your head?  I am embarassed for you again over that stink you made regarding the math that I did here that you couldn't find.  You acted as if you could destroy my argument if you could just see the math.  Now here you are looking at an experiment and don't even recognize the same experiment when it is stated in mathematical terms.    The math regarding spencers experiment is simpler than what I did as it doesn't involve distance or time.  Come on guy.



IanC said:


> every object warmer than 0K radiates. the direction of heat flow is determined by the difference of temperature between the two objects being considered. why do you deny these basic physics facts?



I understand it perfectly Ian and have been stating exactly that for lo these hundreds of posts.  You have just agreed with my statements and apparently fail to comprehend the implications of what you just said.

Every object in spencer's experiment is radiating, even the chamber because he set the parameters at 0 degrees F.  The direction of heat flow is away from the heated plate.  The heated plate is radiating an EM field and because it is warmer than every other object in the experiment, it determines the direction of propagation and that direction is away from itself.  No other object there can radiate energy towards the warmed plate.  

Now apply what you just said to the earth and atmosphere and see how wrong you have been all this time.  You stated correctly that the direction of heat flow is determined by the difference of temperature between the two objects being considered.  The earth is warmer than the atmosphere and therefore the direction of heat flow is away from the earth.  The atmosphere can not radiate a fraction of a watt of energy towards the earth because as the 2nd law of thermodynamics states, it IS NOT POSSIBLE for heat to flow from a cool object to a warmer object without some work having been performed to make it happen.  

The cool bar can not radiate energy to the warmer bar and the cool atmosphere can not radiate energy to the warmer earth.

Why is it that you can state the obvious but can not make the connection to the real world?



IanC said:


> next you will be telling us that objects stop radiating if they are next to something the same temperature. you are a loon



It is called equilibrium Ian and is a well known and long understood phenomenon.  Two objects at the same temperature radiate as a single object.  The do not warm each other.



IanC said:


> the same amount of heat is being shed from the heater, just as I have been saying all along. what part of 'heat flow depends on the difference in temperature' dont you understand?



Of course the same amount of heat is being shed Ian but it is being shed faster because of the heat sink and because of that, its temperature is going to drop.  Are you really this far behind the curve?  



IanC said:


> somehow you think a heated plate and an unheated plate in a cold vacuum are going to be within a fraction of a degree with each other? are you high or something??



Not high Ian, just doing the math.  The math was done for you but clearly it is over your head so you don't grasp it.  Hell, you didn't even realize that the math you were given was a mathematical statement about your damned experiment.

The bars are in a vacuum Ian, of course they will reach equilibrium or very close to it.  As I so patiently explained to you before, there is no conduction or convection allowed in the experiment because the bars are in a vacuum.  Conduction and convection are far and away more efficient means of transporting heat but neither are available for the bars.  They must radiate into the vacuum and as a result, will reach equilibrium quite quickly.  

Of course they could never achieve equilibrium in the open atmosphere because of convection and conduction.  Do you believe that the fins on your heat sink ever even approaches the temperature of the processor in your computer?  Do you think that the radiator in your car ever approaches the temperature of your engine block?  It is because of convection and conduction.  If you put your processor and its heat sink into a vacuum, they will achieve equilibrium or very close to it.  

I don't know if you could find the info today, but there used to be volumes on the topic of vacuum tubes for electronics and one of the major problems with them was heat and the fact that everying inside the tube was operating at, or very near the same temperature.



IanC said:


> your experiment's equations are wrong in many areas. the most startling is that the author thought that the two plates would be the same temperature at equilibrium and that he thinks that heat loss to a cold vacuum is the same as to a less cold object between the heated plate and the vacuum. how could anyone not recognize that the areas facing each other between two close objects is not able to shed radiation as easily as free access to cold vacuum.



So here is where the rubber meets the road.  State specifically what you believe is wrong with the equations.  They are standard derivations of the Stefan-Boltzman Law for blackbody radiation.  They are used hundreds of thousands of times per day by engineers of all sorts who have to deal with transporting heat away from whatever it is that they are working on.  Those equations are used for everyting from determining what sort of heat sink is necessary to keep your computer processor cool to the size and shape of the radiator in your car.

If you can't state what is wrong with the equations (here is a hint, nothing is wrong with the equations) you have effectively lost the argument.  I couldn't help but note that even dr spencer didn't try to argue with those equations.  The fact that dr spencer didn't catch on to the fact that he was merely adding a heat sink to his heated bar and understand the consequences of that says a great deal about the way an academic's mind works vs that of an engineer who actually has to use the math in the real world and be able to predict whtih high accuracy what is going to happen, temperature wise, when a million dollars is spent to produce his design.  If those formulas didn't work, nothing would ever be built.

So tell me Ian, what is wrong with the math?



IanC said:


> next time could you specifically state which experiment you are discussing



And one final time Ian, get with the program.  This whole thing has been only about spencer's failed thought experiment.  The fact that you didn't recognize the math as a statement about the experiment that spencer proposed exposes you as a fraud Ian.  Perhaps it would be best to simply bow out now and avoid further public humiliation.  The fact that it went that far over your head says all that need be said about your scientific "skills".


----------



## wirebender (Oct 28, 2011)

IanC said:


> gslack- why did you edit out almost all of my answers to wirebender in that last quote? are you taking dishonesty tips from the SkepticalScience blog?



For the simple reason that you didn't say anything Ian.  Flailing about and grasping for straws and generally demonstrating that the topic is over your head doesn't constitute any sort of an argument at all.


----------



## IanC (Oct 28, 2011)

wirebender- this is what you posted


> Re: Vacuum Chamber with plates.
> 
> First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.
> 
> ...



is it a different experiment by [Gord] or just his description of Spencer's example? you didnt specify.

it doesnt really matter because Spencer's description is correct and your copied description is incorrect. 

you still havent described the mechanism by which you think both plates (or bars) will come to equilibrium at the same temperature even though one is heated and the other is not.

you still havent explained why you think the heat loss in the direction of the second bar is the same as the heat loss in the direction of the cooled vacuum, even though there is an obvious difference in temperatures.

the math in your example is not only wrong but farfetched. no wonder you refuse to bump your post where you 'proved that photons disappear in space, and did the math'. obviously it was just as infintile and incorrect as your 'proof' that Spencer was wrong.


----------



## wirebender (Oct 28, 2011)

IanC said:


> is it a different experiment by [Gord] or just his description of Spencer's example? you didnt specify.



Sorry Ian, this is just to far over your head.  If you can't look at the math and determine which damned experiment it relates to, you really don't have anything to add to the discussion.  You have become rocks.  Congratulations.



IanC said:


> you still havent described the mechanism by which you think both plates (or bars) will come to equilibrium at the same temperature even though one is heated and the other is not.



I have described it in detail Ian.  Sorry it is so far over your head.  I can't help you any further.



IanC said:


> you still havent explained why you think the heat loss in the direction of the second bar is the same as the heat loss in the direction of the cooled vacuum, even though there is an obvious difference in temperatures.



I have described that in detail as well Ian.  Once again, sorry this is so far over your head.  You had me fooled at one time into thinking that you had a clue.  It is obvious now that you don't.



IanC said:


> the math in your example is not only wrong but farfetched.



The Stefan-Boltzman law dealing with black body radiation is far fetched?  Interesting Ian.  You should run along now.  At this point all you can do is further embarass yourself.  Maybe rocks will come along and give you some sugar though, or wave his pom poms.



IanC said:


> no wonder you refuse to bump your post where you 'proved that photons disappear in space, and did the math'. obviously it was just as infintile and incorrect as your 'proof' that Spencer was wrong.



I am laughing at you Ian. 

By the way, it has also become abundantly and undeniably clear that you don't have the slightest notion of what a photon is and don't have even an inkling of how particle wave duality might relate to the topic at hand.  

I am sorry for you Ian.  At one time, you were viewed as a respected, reasonably intelligent member of this board.  You have been amazingly successful at completely tearing that reputation to ribbons.


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## gslack (Oct 28, 2011)

IanC said:


> gslack- why did you edit out almost all of my answers to wirebender in that last quote? are you taking dishonesty tips from the SkepticalScience blog?



Ian I did not edit out anything in your post.. The board software does that to reserve space in quoted posts. It always does this when you quote from a post with more and a few quoted sections in it..

For example your post I responded to, had multiple quoted lines in it. I am not sure the cut off number in the software as I am not a site admin, but in cases where the post has multiple quotes in it, even if they are just a few lines each, it will automatically cut off some in any reply with quotes to that post.

If you don't like it, join the club I have complained about it before, but seeing as i don't run the site or pay for the bandwidth I really have no grounds to dictate policy. Its happened to me numerous times for even fewer quoted lines...

if it bothers you can copy the entire posts text and paste into the quick reply box at the bottom of the page. The board will assume it is a new post and not edit it. Well as long as the admins don't see it as a repeated post..

Again, editing posts is something I do not do.. Matter of fact if you look you can see I have take offense when it is done to me by many posters here in the past.


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## IanC (Oct 28, 2011)

wirebender said-





> Now apply what you just said to the earth and atmosphere and see how wrong you have been all this time. You stated correctly that the direction of heat flow is determined by the difference of temperature between the two objects being considered. The earth is warmer than the atmosphere and therefore the direction of heat flow is away from the earth. The atmosphere can not radiate a fraction of a watt of energy towards the earth because as the 2nd law of thermodynamics states, it IS NOT POSSIBLE for heat to flow from a cool object to a warmer object without some work having been performed to make it happen.
> 
> The cool bar can not radiate energy to the warmer bar and the cool atmosphere can not radiate energy to the warmer earth.
> 
> ...



this is the crux of the problem. wirebender does not understand the difference between radiation and heat flow. radiation is a function of the temperature of the radiating object (in very simple terms) and heat flow is the measurement of the size and direction of net radiation.

two objects of the same temperature both radiate at the other but the radiation of one replaces the radiation of the other and therefore there is no net heat transfer. if there is a difference in temperature between the two objects the warmer one radiates more and is slightly cooled while the cooler object still radiates towards the warmer one but not enough to fully replace the warmer object's radiation. 

an object releases the same amount of radiation whether it is in cold outer space or next to a blast furnace. the heat flow is simply a matter of whether the incoming radiation is lesser or greater than the outgoing radiation. there is no magical disappearance of photons, they are always there and it is only the net amount of photons in a direction that counts.


----------



## gslack (Oct 28, 2011)

IanC said:


> wirebender- this is what you posted
> 
> 
> > Re: Vacuum Chamber with plates.
> ...



I just want to address a few things directly...

_"you still havent described the mechanism by which you think both plates (or bars) will come to equilibrium at the same temperature even though one is heated and the other is not."_

Its due to the laws of thermodynamics Ian... Particularly the last one I mentioned in a previous post.. The zeroth law... I suggest you read it again..

here is a better explanation than I gave I believe... 

Zeroth Law, Thermal Equilibrium and Temperature



> Zeroth Law
> 
> The zeroth law is a consequence of thermal equilibrium and allows us to conclude that temperature is a well-defined physical quantity. The zeroth law of thermodynamics states:
> 
> ...



Now if you are somehow changing the distance of the plates from one another that would be effected by that new distance. I was under the assumption the plates were placed on top of one another. This was implied throughout the experiment I believe. if they are indeed in contact with one another the outside temps only effect the outside surface areas of the plates. Of course the closer to the heat source the warmer it will be no one made any claim to the contrary. however, the assumed term "all things being equal" will apply here. 

The entire experiment is a thought experiment. The physical reality is its not even a good example of how a real physical experiment would be. First there is always a difference in the temperature of any heated surface the farther it gets from its point of contact with the heat source. Second standing plates are not a moving atmosphere with all the variables it entails. Third, the vacuum is only applicable to the relation of the Sun to the atmospheric surface. Once it hits that atmosphere the vacuum is no longer a factor. The earth sits in a vacuum, but the atmosphere is not a vacuum, and that is the part that is in contention in greenhouse effect theory. That would make the entire premise of the experiment pointless. 

However you brought this experiment into the discussion and professed its accuracy. So we tried to show how it was not an accurate example of how the "greenhouse effect" works. You vehemently defended it and still do so now. This experiment forces many of the natural finer points and details to be ignored.

Now if you want us to ignore the finer points and details and allow this experiment to go through to its claimed result, you cannot very well fall back on those finer points to defend it when it falls flat.. The experiment assumes "all things being equal" from the start, that is the only way it can work even in a thought experiment. So either all things are indeed equal in it or all things are as they would be in the real world, either way it falls flat. But you must stick to one or the other, switching between them only confounds the thing and makes it impossible..


----------



## IanC (Oct 28, 2011)

IanC said:


> wirebender said:
> 
> 
> > IanC said:
> ...



test

my apologies. I did not realize the message board did that.


----------



## gslack (Oct 28, 2011)

I see your test had the same result...

EDIT: Maybe next time you should test it BEFORE neg repping me and making the false claim...


----------



## wirebender (Oct 28, 2011)

IanC said:


> this is the crux of the problem. wirebender does not understand the difference between radiation and heat flow. radiation is a function of the temperature of the radiating object (in very simple terms) and heat flow is the measurement of the size and direction of net radiation.



The crux Ian, is that this is all just way to far over your head for you to discuss.  When you said that the Stefan-Boltzman equations for dealing with blackbody radiation were far fetched and incorrect, you lost any remnant of credibility you might have retained. 

Get a clue Ian, it isn't to late.


----------



## IanC (Oct 28, 2011)

gslack said:


> IanC said:
> 
> 
> > wirebender- this is what you posted
> ...



I agree with the zeroith law but it does not apply to one object that is being heated and another one that is only absorbing the radiation from the first. the two objects are not the same temperature even if they are in an equilibrium of sorts.

if you are talking about Spencer's experiment the diagram specifically shows separation between the two bars, to stop conduction. and states a vacuum, to stop convection. he was specifically dealing with the radiation properties of objects. and he successfully showed that adding an object between the source of heat and the cold outside would raise the temperature of the radiating source.

an analogy is a water source (electricity) going to a connection with multiple hoses and sprinklers (the heated bar). if you put a crimp into one of the hoses (second bar which hampers the ability to radiate to the cold outside) then the water pressure backs up and the remaining sprinklers are more forceful (the heated bar is warmer). the whole thought experiment is logical and easy to understand if you just think about it.


----------



## IanC (Oct 28, 2011)

gslack said:


> I see your test had the same result...
> 
> EDIT: Maybe next time you should test it BEFORE neg repping me and making the false claim...



I'll pos rep you when I am able. again, my apologies


----------



## IanC (Oct 28, 2011)

gslack said- 





> However you brought this experiment into the discussion and professed its accuracy. So we tried to show how it was not an accurate example of how the "greenhouse effect" works. You vehemently defended it and still do so now. This experiment forces many of the natural finer points and details to be ignored.



Spencer's experiment is not an explanation of the greenhouse effect. it is an explanation of why the greenhouse effect is not in violation of thermodynamic laws, specifically 'a cooler body cant warm a warmer body'.


----------



## wirebender (Oct 28, 2011)

IanC said:


> I agree with the zeroith law but it does not apply to one object that is being heated and another one that is only absorbing the radiation from the first. the two objects are not the same temperature even if they are in an equilibrium of sorts.



Mighty caucasian of you even though you clearly don't grasp its inplications.



IanC said:


> he was specifically dealing with the radiation properties of objects. and he successfully showed that adding an object between the source of heat and the cold outside would raise the temperature of the radiating source.



The only thing spencer showed is that he doesn't recognize a heat sink when he imagines it.



IanC said:


> an analogy is a water source (electricity) going to a connection with multiple hoses and sprinklers (the heated bar). if you put a crimp into one of the hoses (second bar which hampers the ability to radiate to the cold outside) then the water pressure backs up and the remaining sprinklers are more forceful (the heated bar is warmer). the whole thought experiment is logical and easy to understand if you just think about it.



The math is logical and easy to understand, your interpretation of the experiment and end result is twisted in tangled up knots.


----------



## wirebender (Oct 28, 2011)

IanC said:


> Spencer's experiment is not an explanation of the greenhouse effect. it is an explanation of why the greenhouse effect is not in violation of thermodynamic laws, specifically 'a cooler body cant warm a warmer body'.



The experiment was a failure precisely because it violates both the 2nd law of thermodynamics and the law of conservation of energy.  You are hopeless Ian.


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## IanC (Oct 28, 2011)

wirebender said:


> IanC said:
> 
> 
> > I agree with the zeroith law but it does not apply to one object that is being heated and another one that is only absorbing the radiation from the first. the two objects are not the same temperature even if they are in an equilibrium of sorts.
> ...



from a description of heat sinks-


> The heat transfer from the heatsink is mediated by two effects: convection via the coolant, and thermal radiation.
> 
> Heat transfer by radiation is a function of both the heat sink temperature, and the temperature of the surroundings that the heat sink is optically coupled with. *When both of these temperatures are on the order of 0 °C to 100 °C, the contribution of radiation compared to convection is generally small, and this factor is often neglected.* In this case, finned heat sinks operating in either natural-convection or forced-flow will not be effected significantly by surface emissivity.
> 
> ...



please explain to me how you think the two bars will equilibrate to the same temperature in Spencer's experiment even though one is heated and the other is not.

you have said many foolish things in the past but that is right up there with saying that a blanket make you colder.


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## gslack (Oct 28, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC said:
> ...



Well I am glad you agree with the zeroth law.. With it being a natural law you really little choice in the matter it simply is regardless of your agreement with it....

Ian will you *PLEASE* stop with the nonsense about "if you are talking about spencer's experiment" ... There wouldn't be either experiment in this discussion if you hadn't brought them here.. Dude pick one already! Your first one or your second one matters little to me but PICK ONE! You brought them both into the discussion for pete's sake...

Now after you said all that about which experiment, you went and did this again...

_"an analogy is a water source (electricity) going to a connection with multiple hoses and sprinklers (the heated bar). if you put a crimp into one of the hoses (second bar which hampers the ability to radiate to the cold outside) then the water pressure backs up and the remaining sprinklers are more forceful (the heated bar is warmer). the whole thought experiment is logical and easy to understand if you just think about it."_

Ian WTH??? You just brought in another experiment using a completely different set of variables and using completely different methodology.. My god man are you aware that you have been asking us which experiment we are talking about in each post, when you were the one who brought all of them in? look if you can't focus on one why bring other ones in? YOU confounded this conversation just like you have done in the previous ones.. you did it because you cannot make the claims hold water, yet you feel you must.. 

I am sorry Ian... The theory has holes. its a fact. Scientists are human first. They can and do make mistakes, and with their brilliance and egos its very difficult to get many of them to admit them especially when they are this big.. They will do what they have to, to protect their pride, position and esteem, as well their research they may have spent their lives on. Not to mention colleges that support them, and students hoping to go into the same field one day. That is an insane amount of pressure to be not only right about this, but insure the science and investments in it...

Now your new experiment above....

Dude electricity, is not heat... And water is a liquid. Come on Ian you damn well know better.... that was just a bunch unrelated nonsense to further confound the discussion.. Are you really wanting to prove something here or just wanting to give the appearance you proved something here?

EDIT: sorry forgot one thing..

You said... _"the diagram specifically shows separation between the two bars, to stop conduction. and states a vacuum, to stop convection. he was specifically dealing with the radiation properties of objects. and he successfully showed that adding an object between the source of heat and the cold outside would raise the temperature of the radiating source."_

Ian first and foremost, the experiment was in fact an effort by spencer to show how a cold object can make a hot object hotter in some circumstances, its entire point was to show how this can happen in the case of greenhouse effect in the atmosphere... 

Now please remember that first part especially.. I will state it again....*"the experiment was in fact an effort by spencer to show how a cold object can make a hot object hotter in some circumstances"* Okay that was what spencer made a point of early in that article you cited.....

Now in spencers experiment he mentions a plate and an electric heater, then he mentions another plate but his picture shows a bar... See the problems yet? Later on he mentions bars, then plates, and comments on them back and forth making the entire thing very hard to follow fro a true skeptical mind... if you go in there with the preconceived notion you are going to get a logical experiment from a solid scientist with no ambiguity or deception, you will more times than not accept what he shows as true and accurate. However clear your mind and treat him and his post as that of a person with the same strengths and weaknesses as any other human, and look it over with the desire to know the truth whatever it may be, you will end up with questions that the experiment cannot address..

I ask you once more to go and read it through very carefully, as will I... When I am done I will come clean with you.. If I find hes right I will say so... But I ask you to do the same and without all the confounding and nonsense.. No more double talk, no more BS obfuscation just flat out let the chips fall where they may truth, no matter who gets their pride hurt or what we find... Agreed?

I hope so, either way I am going to have to switch to my laptop for a bit and do some work, Then I will go over that experiment with all my natural anal-retentiveness and either show how it is wrong or how it is right to the best of my ability..


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## IanC (Oct 29, 2011)

gslack- I linked Spencer's article, then wirebender said it was debunked by some poster named Gord. wire then put up a comment about plates instead of bars that had bogus math. it may or may not be about Spencer's thought experiment but you guys werent explicit even when I asked for clarification. if it was about Spencer then it was hopelessly flawed.

so far I have tried to explain the actual physical characteristics of the experiment. when that failed I gave you a simple analogy that anyone older than eight could understand.

I have tried in many ways to point out where you and wirebender are confused but all you do is repeat that I am wrong with no counter explanation nor any response to my explicit simple questions. 

for instance, I would be very interested in hearing why you think a heated object will be the same temperature as an unheated object when both are losing heat to a cooled vacuum container. will you at least attempt to answer that question which is the basis of your supposed debunking of Spencer?


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## gslack (Oct 29, 2011)

Okay I am back.. this is going to take a bit so stay the course...

Forget plates, it only confounds his experiment and confuses the point.. Stick with bars...

Okay he says a cooling a cylindrical cooling chamber (he says round but I am sure he meant cylindrical) and he puts a bar in it that is being heated by an electric heater. In this case lets say the bar itself is a heating element as I feel it was implied from the picture. he goes on to say the chamber is in a vacuum and cooled. he does not specify the temperature it is cooled to. And that bar is heated to a constant temperature of 150 degrees F... Okay first part done.. we following each other here?

Then he adds a second bar next to the first, he does not specify exact distance. And contends the second bar through radiation from the first bar will warm up to a temp he chose of 100 F (equilibrium). he then contends the introduction of that second bar will cause the first heated bar to get even hotter... Agreed? okay..

Problems...

1. First and foremost spencer calls the bars plates, and then bars repeatedly especially after he introduces the second one. Perhaps this was an error, but the fact is it makes the experiment harder to follow and as a scientist he should know better.. Proper, established terms for respective elements in an experiment even a thought experiment is paramount.. Its enough to get the entire thing dismissed off-hand on any serious scientific level..

2. He does not give a temperature for the cooling or anything that would give an idea of it. he also fails to mention a material for the chamber. These two factors alone make the experiment dubious, but added to the already mentioned problems above, it only furthers the pointlessness of it all...

3. He mentions the second bar acts as an insulator and slows the heat loss of the first bar causing it to get hotter. This may seem plausible but he fails to realize one simple fact.. the second bar is not in equilibrium with the first bars temp. bar #1 150 F, bar #2 100 F...its in equilibrium with the amount of radiative energy received from the first bar comparative to distance, and whatever the temperature of the cooled chamber is and the various energy transfer rates the conditions allow. Again we do not know this as he did not tell us materials used in the chamber or the temp it is cooled to.. If Place my hand next a heat source the closer I get the warmer it will be, and if its colder out the closer i can get to it before I get burned. So there is a difference relative to distance and temperature regardless of a vacuum or atmosphere. This again makes the experiment a failure...

4. he makes the claim several times in the experiment as well as responses to posters, that because the first bar is constantly receiving new energy from the wires and constantly radiating new heat, then so is the second bar and that  allows the second law of thermodynamics to be a non-factor or however he tried to explain it (he gets very vague on that and says it doesn't violate the rule when it clearly does). Now we both know that either he breaks the law with this or he doesn't. Breaking it and claiming he doesn't is just retarded.. the experiment breaks the law period.. if the amount of energy it takes to make the first bar 150 F is a constant flow of new energy, than that is what it takes to make that bar 150F. And the amount of radiative energy from that bar it takes to make the second bar 100F is again that amount constantly. OF course its new energy it makes no difference the amount of heat generated from that energy will still be 150F, new or old, or stored or whatever, the point is the heat it gives off. the second law tells us energy flows in one direction, and that direction is from the source outward. So long as the second bars temp is below that of the first that energy flow will go from the hotter to the colder period no ifs ands or buts about it. if there is any incidental extra warming its due to proximity and reflection.

5. he makes it very clear that this experiment was to show how this type colder to warmer reaction can work in our atmosphere. That is shown from the beginning and at the end. So yes it is an attempt to show how one aspect of greenhouse effect works. Using completely unrelated and unrealistic means and methods...

Basically his contention falls flat on many levels but the one that bothers me the most is the fact he says one thing then shows another, claims one thing and shows another.. 

That and the very simple factor that the heat source is the first bar which would represent the sun. meaning the second would have to represent the earth. So does our presence make the sun hotter? Don't think so, and that is the entire point... Nothing in his experiment has an electro-magnetic field, nothing orbits around the heat source, there is nothing representing the atmosphere, there is no telling what he cools the chamber too or the materials used.

its nonsense designed to confound a simple thing and I am now officially disgusted with Spencer... From reading his responses I see very clearly is not a scientist as much as he is an egotist... 

Solid objects are not gases... The Stefan-Boltzmann Law was not written for gases, and he relies on its principles to make his experiment and apply it to greenhouse theory.. Spencer is desperate to ride his gravy train till he can't any longer... 

Earth's atmosphere may be more efficient at releasing energy to space than climate models indicate, satellite data suggest 

_"*Earth's Atmosphere May Be More Efficient at Releasing Energy to Space Than Climate Models Indicate, Satellite Data Suggest*

ScienceDaily (July 29, 2011)  Data from NASA's Terra satellite suggests that when the climate warms, Earth's atmosphere is apparently more efficient at releasing energy to space than models used to forecast climate change may indicate, according to a new study."_

Spencer's own words.._""The satellite observations suggest there is much more energy lost to space during and after warming than the climate models show," Spencer said. "There is a huge discrepancy between the data and the forecasts that is especially big over the oceans."_

Yeah hes a sell out... he claims one thing but then when its found to be inaccurate, he doesn't come out and say so anywhere he doesn't have to... on his blog its business as usual after all he has books to sell and money to make. But in the science sites very few people read it's another story...


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## wirebender (Oct 29, 2011)

IanC said:


> please explain to me how you think the two bars will equilibrate to the same temperature in Spencer's experiment even though one is heated and the other is not.



Because they are in a vacuum Ian where conduction and convection into the atmosphere are eliminated.  Radiation is the only means of bleeding off heat and the bars (plates;same thing) are in close proximity.  They will achieve equilibrium or very close to it.

As I have already told you, volumes have been written about the issue of heat within vacuum tubes from the old days when everything used them.  In a vacuum, materials heated and passive reach an equilibrium temperature very quickly.  Great pains had to be taken in the design of vacuum tube components using materials of widely varied absorptivity and emissivity, and wild heat sink designs were fabricated precisely to keep the internal components of the vacuum tube from reaching thermal equilibrium.  It doesn't happen in the open atmosphere because convection and conduction are in operation to carry energy away from the heat sink.  Not so in a vacuum.  

Face it Ian, math and the laws of physics are stronger than your faith in a mythical greenhouse effect.

The math is correct and is a correct useage of the Stefan-Boltzman law dealing with blackbody radiation.  You are accepting your intuition over hard mathematical evidence that has been proven reliable for over a hundred years.  

What else need be said?



IanC said:


> you have said many foolish things in the past but that is right up there with saying that a blanket make you colder.



Funny you should mention that Ian, and state it with your usual confidence as if it were a statement of fact.  Sorry Ian, the second law of thermodynamics defeats you again.  The fact is, that a blanket does reduce your surface temeprature.  Once again, your intuition, belief, faith, or whatever intermal system your intellect runs on, has lead you off in the wrong direction.  The fact is, Ian, that putting a blanket over your body does make you colder just as the second law of thermodynamics predicts.  

Your problem Ian, is that you believe you are smarter than the laws of physics and the math that proves those laws.  You aren't.

_"Human Body Emission

As all matter, the human body radiates some of a person's energy away as infrared light.

The net power radiated is the difference between the power emitted and the power absorbed:






Applying the Stefan Boltzman Law






The total surface area of an adult is about 2 m², and the mid- and far-infrared emissivity of skin and most clothing is near unity, as it is for most nonmetallic surfaces.[ Skin temperature is about 33 °C, but clothing reduces the surface temperature to about 28 °C when the ambient temperature is 20 °C. Hence, the net radiative heat loss is about




_"

If you put a 20C blanket (which is colder) on a warmer 33C body, the surface temperature is going to reduce to about 28C.  Heat flowed from the warmer body to the cooler blanket just as the 2nd law of thermodynamics predicts.

It is true that the blanket will trap warm air between the body and itself, but that heat will not increase the temperature of the body.

This buisness is a matter of the laws of physics and the mathematics that prove them Ian.  Not, as you seem to believe, a matter of faith.  You pit yourself against the laws of nature and you will lose every time.


----------



## wirebender (Oct 29, 2011)

gslack said:


> Well I am glad you agree with the zeroth law.. With it being a natural law you really little choice in the matter it simply is regardless of your agreement with it....



As with all the other laws of physics that he so blythely disregards in favor of his intuition, or faith, or hope, or fealty or whatever failed intellectual method he employs. (is shuck and jive considered to be an intellectual method?)  I don't know about you, but I am about finished with him.  The math that has been presented stands unassailed by any proof to the contrary and my bet is that the math I just gave him regarding the fact that a blanket does in fact cool you down will remain unchallenged by anything more than a verbal objection and perhaps an ad hominim delivered in frustration; and the laws of physics remain standing, unscathed.  That being the case, I must assume that our side has prevailed.  Congratulations.  (high five, knucle bump,..... whatever)

More patience than he deserves has been lavished upon him in an effort to explain what is actually going on and the failures of trenberth's energy budget, but as they say, you can lead a horse to water but you can't make him cook cornbread once he is finished drinking.


----------



## IanC (Oct 29, 2011)

wirebender said:


> IanC said:
> 
> 
> > please explain to me how you think the two bars will equilibrate to the same temperature in Spencer's experiment even though one is heated and the other is not.
> ...



wirebender- your link supports my position! skin radiates more when it is exposed. when an insulator is put between skin and a colder ambient temperature there is less heat loss. surely you didnt think that the skin underneath the clothing has cooled? are you really that stupid? 

your link even spells it out. net power radiated equals power emitted less power absorbed. but there wasnt the faintest hint of photons magically being extinguished.

again. please explain how a heated object and an unheated object will equilibrate at the same temperature when the heated object is the only source of heat in an open system. just describe the heat flow from one to the other so we can understand this amazingly counterintuative feat of physics.

gslack- I havent had time to go over S's article but I will.


----------



## wirebender (Oct 29, 2011)

IanC said:


> wirebender- your link supports my position! skin radiates more when it is exposed. when an insulator is put between skin and a colder ambient temperature there is less heat loss. surely you didnt think that the skin underneath the clothing has cooled? are you really that stupid?



Of course the skin underneath the clothing has cooled Ian.  The math is right there.  Skin temperature drops from 33 degrees C to 28 degrees C just as the second law of thermodynamics predicts.  Geez Ian, you get dumber every time you post.  Which part of this exactly is it so difficult for you to grasp:

 Skin temperature is about 33 °C, but *clothing reduces the surface temperature to about 28 °C *when the ambient temperature is 20 °C. 

The skin gets cooler exactly as the 2nd law of thermodynamics predicts.  Your instincts are wrong Ian and clearly you don't grasp the math so I suppose you are doomed to be wrong every time you engage the topic.



IanC said:


> again. please explain how a heated object and an unheated object will equilibrate at the same temperature when the heated object is the only source of heat in an open system. just describe the heat flow from one to the other so we can understand this amazingly counterintuative feat of physics.



It has been explained to you over and over Ian and the math has been done.  You just aren't bright enough to grasp the explanation much less grasp the math.


----------



## wirebender (Oct 29, 2011)

wirebender said:


> ... my bet is that the math I just gave him regarding the fact that a blanket does in fact cool you down will remain unchallenged by anything more than a verbal objection and perhaps an ad hominim delivered in frustration;



And Ian's response is....drumroll........



			
				IanC said:
			
		

> surely you didnt think that the skin underneath the clothing has cooled? are you really that stupid?




Ding ding ding ding.....we have a winner!!!  Once more, when the math is staring him right in the face, he doesn't get it.


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## IanC (Oct 29, 2011)

wirebender you have mixed apples with oranges again. the skin under clothing is NOT the surface. dont you ever actually think about the physical processes going on? the skin under the clothing is warmer than exposed skin, that is the main point in wearing them. if clothing made people colder no one in temperate or cold climates would go to the bother.

if you still think clothes or blankets make you colder please explain the mechanism of how an insulator increases heat flow rather than reducing it.


----------



## gslack (Oct 29, 2011)

Ian, the skin is the radiator, the clothing or blanket if it is colder than the skin at the time it is placed in contact with it will indeed make the skin overall cooler for a time. Once the the inside of the blanket reaches equilibrium with the skin its in contact with, then the situation will change of course. But at the point a cooler body (blanket) first makes contact with a warmer body (skin) the skin will get cooler until equilibrium is reached.

I think Wirebender is referring to that, and you are most likely referring to after equilibrium is reached. When broken down fully you will find they are two different situations both mathematically and naturally...

A good example, is a cold pillow.. I like to turn my pillow over at night so I have a cooler side on my face until I fall asleep. I can feel it making my skin cooler at first, then once equilibrium is established, it of course causes my skin to get warmer on the side touching that same pillow. I turn that same pillow over and its cold on that side, and again it cools my skin for a time...


----------



## IanC (Oct 29, 2011)

gslack said:


> Ian, the skin is the radiator, the clothing or blanket if it is colder than the skin at the time it is placed in contact with it will indeed make the skin overall cooler for a time. Once the the inside of the blanket reaches equilibrium with the skin its in contact with, then the situation will change of course. But at the point a cooler body (blanket) first makes contact with a warmer body (skin) the skin will get cooler until equilibrium is reached.
> 
> I think Wirebender is referring to that, and you are most likely referring to after equilibrium is reached. When broken down fully you will find they are two different situations both mathematically and naturally...
> 
> A good example, is a cold pillow.. I like to turn my pillow over at night so I have a cooler side on my face until I fall asleep. I can feel it making my skin cooler at first, then once equilibrium is established, it of course causes my skin to get warmer on the side touching that same pillow. I turn that same pillow over and its cold on that side, and again it cools my skin for a time...



certainly I can agree with that point. but the cool side of the pillow is mostly a combination of conductance and heat sink. I thought we were discussing radiation at equilibrium. you may even be correct at the boundary of skin and clothes because of the better thermal capacity of conduction over radiation. but a clothed body loses less heat than an uncovered one so the body will not need to burn as much food to keep at its desired temperature of 37C internally.


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## IanC (Oct 29, 2011)

gslack- go back and reread my description of the two bars together. there would be a temperature gradient because the heater was offset. once the bars were separated the gradient would be greatly increased because of the loss of conductance. there is no way the seperate bars would be the same temp, and there is no way that the originally closed faces would send all or even most of their radiation to the cold exterior.


----------



## gslack (Oct 29, 2011)

IanC said:


> gslack said:
> 
> 
> > Ian, the skin is the radiator, the clothing or blanket if it is colder than the skin at the time it is placed in contact with it will indeed make the skin overall cooler for a time. Once the the inside of the blanket reaches equilibrium with the skin its in contact with, then the situation will change of course. But at the point a cooler body (blanket) first makes contact with a warmer body (skin) the skin will get cooler until equilibrium is reached.
> ...



Ian you are trying to nitpick this until you can walk away thinking you are right.. I will not play a semantics tit-for-tat game with you any longer.. The point remains the fact the colder blanket will cool the warmer skin for a time, regardless of the means the end result is the same... Jesus dude what the hell is wrong with you?

I said I believe the two of you were referring to separate points in the blanket/skin heat interaction. He was referring to point of contact which from the wording of your post I can hardly blame him, thats what I took it to imply. And you are referring to equilibrium state..

on a side note... I an I think you are confused regarding equilibrium and what it means in cases like spencers experiment and our discussions of it.. the best definition I have seen is this one : *Equilibrium (&#8652 is the condition of a system in which competing influences are balanced. *

 This does not mean both bars will reach the same temperature. What it means is they will be in balance as far as energy in, out, and with limitations inherent with environment, conditions, materials, et al..

Equilibrium in the experiment and our discussion of it have been in reference to the energy not the heat produced from that energy or the temperatures of the bars..

If you read spencers experiment more carefully you will see *HE* says they reach equilibrium when the first bar is at 150 F and the second bar is at 100 F...

Ian I seriously don't think you want to find truth here, I think you just want to walk away feeling you are right no matter what... Your arguments are all too often misusing terms and conditions, you seem to not have many of the core knowledge any form of physical science study would give you.. You didn't know the two-slit experiment, you didn't know that a photon was EM radiation, and know you show you do not what equilibrium ins systems means...

I am growing tired of this game Ian, I really am..


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## IanC (Oct 29, 2011)

here is a relevent part of wirebender's 'math' -


> If another identical &#8220;non-heated&#8221; and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:
> 
> The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2
> 
> ...



here is a diagram of the end stage equilibrium from Spencer's article-






the second unheated plate obviously cannot be as warm as the heated one. personally I think the inside faces of the two plates will be slightly warmer than the outside faces but this is a simple diagram. 

it is also obvious from the diagram that the energy flow between the two inside faces will involve less net power transfer than the faces that lose heat to the outside because the temperature difference is less. this makes a mockery of wirebender's equal radiation from all surfaces EVEN if he could come up with an explanation for both bars being the same temperature.

the total power lost to the outside is the same in both parts of the experiment. only the amounts radiating from the different faces change. the amount of radiation from the heated inside face is less effective because of the back radiation from the unheated block. the difference backs up, and raises the temperature in the heated block which in turn raises the amount of radiation released by all of the faces of the heated block until equilibrium is again reached.


----------



## gslack (Oct 29, 2011)

Ian you left out the pertinent part at the end...

_"And the *equilibrium temperature* for both plates will be 284.69 K or 53 deg F.
&#8212;&#8212;&#8212;&#8212;-
NOTE:

This is simple effect is demonstrated every day and is happening in your Computer right now.

It&#8217;s called a &#8220;Heat Sink&#8221; and is used to cool the microprocessor in your computer by increasing the radiating surface area of the microprocessor.

Here was the mathematical proof that in order for the heated bar to reach a temperature of 160, additional energy would have to be created.

PROOF:
160 deg F = 344.11 K and that means it would require:
P = (5.67X10^-8) X 1m^2 X (344.11 K)^4 = 795 Watts!

795 Watts exceeds the TOTAL Energy available which is 744.95 Watts.

An obvious violation of The Law of Conservation of Energy.

You were shown that in order to get the heated plate to 160 degrees, an additional 50.05 watts above and beyond what is available is required. Where do you believe that additional energy came from? The plate that had no power source? Well of course you believe that is where it came from because you also beleive that the atmosphere, which is not an energy source, provides energy to the surface of the earth."_

now again this is due to the parameters described in spencers experiment... Are the bars touching or not? if not what is the distance between them? What is the temperature in the chamber? All of these things are factors that spencer ( i think deliberately ) either makes vague, or fails to clarify.. If the bars are touching the math is true, if not then there are other variables that cannot be quantified due to missing information.

One problem is, is the first bar the heating element or a bar sitting on the heating element? he eludes to one then the other several times, making assumptions all too frequent. If indeed the bar is a separate entity from the heating element, than we must ask if it is sitting on that element or sitting above, below, or to the left or right of it. And then we must ask how far in any direction..

IF the heating element and bar are separate entities than the equation is correct, if not than its a different equation.

personally from reading it I took the first bar to be the heating element since the drawing shows wires going to it. And if that is the case then you cannot consider the first bar anything but the heat source. Which makes any attempt to show a cold object making a warm object warmer, dubious in this experiment. 

This experiment is an experiment used to show the zeroth law at work. but he took out one of the bars.

A common way to describe zeroth law is to use the three identical bars concept. bar A, bar B, and Bar C. If A and C are in equilibrium with each other than object B must also be in equilibrium with both A and C. this of course will vary depending on materials, conditions, et al.. Remember it assumes "all things being equal".. Spencer removed one of the bars and tried to use incidental heat gain from kinetic energy transfer as a way to explain the problem with cold to hot transfer in greenhouse theory..

I don't know any other way to put it Ian, frankly I am learning a great deal of this as I go and the continued side arguing over variables and non-factors is giving me a migraine now.. Fact is as spencer relates the experiment its not applicable in real world atmospheric heat transfer no matter how he tries to BS it ..

This may help...
Methods of Heat Transfer

1. 1. Consider Object A which has a temperature of 65°C and Object B which has a temperature of 15°C. The two objects are placed next to each other and the little bangers begin colliding. Will any of the collisions result in the transfer of energy from Object B to Object A? Explain.


Answer: Most certainly yes.

The average kinetic energy of the particles in Object A is greater than the average kinetic energy of the particles in Object B. But there is a range of speeds and thus of kinetic energy in both objects. As such, there will be some highly energetic particles in Object B and some very non-energetic particles in Object A. When this combination of particles encounter a collision, there will a transfer of energy across the boundary from Object B (the colder object) to Object A (the hotter object). This is just one collision. Since majority of collisions result from the more energetic particles of Object A with less energetic particles of collision B, there will be a net kinetic energy transfer from Object A to Object B.

2. Suppose that Object A and Object B (from the previous problem) have reached a thermal equilibrium. Do the particles of the two objects still collide with each other? If so, do any of the collisions result in the transfer of energy between the two objects? Explain.

Answer:

The collisions will still take place because the particles are still moving. Just because the temperatures are the same doesn't mean the collisions will stop. The fact that the temperature is identical means that the average kinetic energy of all the particles is the same for both objects. As such, there will be just as much energy transferred from Object B to Object A as there is energy transferred in the opposite direction. When the effect of these collisions is averaged, there is no net energy transfer. This explains why the temperature of the two objects remains the same. Thermal equilibrium persists.


So according to that physics online classroom you are both right and both wrong, just as spencer is neither right nor wrong.. The transfer is not as spencer describes it, and it does violate the second law the way he describes it. Again he took a variable in a controlled environment and tried to use it to save a severe flaw in a completely different theory.

Until spencer stops playing his little mind games like these he can kiss my ass.. He is a  charlatan plain and simple.. he knew damn good and well taking a demonstration of the zeroth law and changing the elements involved would lead to mass misunderstandings and confusion on this, he knew this and did so intentionally to confound it even further...

Thats why I know AGW theory is BS...If it were so undeniably true why do they have to  pull shit like this all the time? 

 In my 20+ years as a data analyst both for the military and for others i have found one undeniable and unquestionable fact.. If there is a lot of deliberate BS in a simple thing, some one is lying! Either at the source or in transit someone has planted BS in the beans... The truth is never so complicated Ian, trust me on this... You don't have to sell the truth, it sells itself. 

if spencer were being truthful he wouldn't soil it with this nonsense... Again his experiment is designed to confound and confuse not to shed light or explain... He achieved his goal apparently...


----------



## IanC (Oct 29, 2011)

gslack- the difference is that one block was heated and they werent touching. try again.


----------



## IanC (Oct 29, 2011)

Spencer is against CO2 as a major factor in the warming of the earth over the last 200 years by the way. he just isnt willing to discount the real effects that CO2 plays.


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## gslack (Oct 29, 2011)

IanC said:


> gslack- the difference is that one block was heated and they werent touching. try again.



Ian thats it.... That was a deliberate and unjustified dismissal of all that I said and showed you.. If there was any doubt you are just another fake googleing terms to give a false impression of knowledge your response above to my post should kill it...

I have been spending more time on this than I should have and for that I get that from you after the shit you have pulled with me alone?

FU Ian... I am done from here on out you get my disdain, you do not dismiss work I have done without even reading the damn thing again and get to play nice with me again.. thats too many times I gave you a chance....

Now you imbecilic internet fraud, if you had an ounce of integrity in you, you would know i made a point in that post you responded to, to give both scenarios a fair explanation.. You can't even be honest at all can you fraud....


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## gslack (Oct 29, 2011)

IanC said:


> Spencer is against CO2 as a major factor in the warming of the earth over the last 200 years by the way. he just isnt willing to discount the real effects that CO2 plays.



he's a douchebag and a charlatan. he says one thing and then another. he has books to sell even the article you linked is a book advertisement.. So no thanks, he's a sell out..

you are dishonest, as well as reactionary.. And your inane BS posturing has gotten on my last nerve.. If you had half the experience and knowledge you claim to have in this, you would have known about the two-slit experiment, what an EM field is, what a photon is, and that spencers experiment was a deliberate whitewash using only part of an original and accurate experiment to show the zeroth law in action.. 

I call them as I see them Ian, and you are full of shit... You know less about physics than my 15 year old son... get a new persona this ones blown his cover fake..


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## IanC (Oct 29, 2011)

you were just like wirebender. mixing apples with oranges. why are you pissed with me for pointing out that the scenario described in your link was different from what we were talking about?


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## IanC (Oct 29, 2011)

gslack said:


> IanC said:
> 
> 
> > Spencer is against CO2 as a major factor in the warming of the earth over the last 200 years by the way. he just isnt willing to discount the real effects that CO2 plays.
> ...



I obviously know about more photon paradoxes than you. both you and wirebender have some strange view of the thermodynamic laws that only hold true for the two of you. I am more than willing to discuss any of these thing because I actually find them interesting. So far the only thing you have shown me is that the message board has an odd way of dealing with unidentified quotes. which I am still sorry about BTW.

I am anything but dishonest. if I make a mistake I own up to it, and learn from it. I would be pleased if you if you found something that taught me something. so far I am still waiting though. on the other hand I have been trying to show you two guys a more realistic way of looking at physics and you refuse to even consider it. your loss.


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## gslack (Oct 29, 2011)

IanC said:


> you were just like wirebender. mixing apples with oranges. why are you pissed with me for pointing out that the scenario described in your link was different from what we were talking about?



No weasel I am pissed at you for being a deliberate asshole repeatedly.. the fact is the link was to show the fact that spencers experiment was ambiguous in its pretense and how that is so.. The rest of the post (80%) was dismissed by without even reading it..

Don't try and lie by reading it now, you dismissed it without reading again, you do this continually and then try and bullshit your way out of it.. If you had read it you would have commented on something more than just the reference link to show how spencer made the confusion work to his benefit. There was a great deal more to my post, I put time into it I know...

So go take a flying leap you fake...


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## IanC (Oct 29, 2011)

gslack said:


> IanC said:
> 
> 
> > Spencer is against CO2 as a major factor in the warming of the earth over the last 200 years by the way. he just isnt willing to discount the real effects that CO2 plays.
> ...



I'm curious. what's the original and accurate experiment?


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## gslack (Oct 29, 2011)

Google it yourself fraud... Here's a hint: Theres reference to it one of the places I linked to previously... if you had read them you would know...

I don't do research for fakes..


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## IanC (Oct 29, 2011)

gslack said:


> IanC said:
> 
> 
> > Spencer is against CO2 as a major factor in the warming of the earth over the last 200 years by the way. he just isnt willing to discount the real effects that CO2 plays.
> ...



I'm curious. what was the original and accurate experiment?


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## IanC (Oct 29, 2011)

sorry double post. internet glitch. I see you didnt want to answer the question.  it seems you are guilty of many of the things that you accuse me of. projection perhaps? hahaha


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## gslack (Oct 29, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC said:
> ...



I told you already, now go fetch asshole...


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## IanC (Oct 30, 2011)

just like wirebender declaring to everybody that he provided the proof and did the math. in a past post. that he refuses to show anybody. just like the worst of climate scientists.

I dont really care, I was just curious. you two guys seem to think huffing and puffing and bluster is the same as describing your side of the story. I cant really judge the merits of your ideas if you refuse to produce them.


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## gslack (Oct 30, 2011)

IanC said:


> just like wirebender declaring to everybody that he provided the proof and did the math. in a past post. that he refuses to show anybody. just like the worst of climate scientists.
> 
> I dont really care, I was just curious. you two guys seem to think huffing and puffing and bluster is the same as describing your side of the story. I cant really judge the merits of your ideas if you refuse to produce them.



Stop posturing for the board asshole, want to talk to me than try honesty for once...

You are a fake, we both now it so knock off the bullshit.. You have no intention of examining the truth here all you want to do is save your internet identity by confounding the argument every time you get tested... FU phony, I do not dance for you no mater what you say here.. 

You are caught again showing you know nothing you claim you know, and actually understand even less... You are about as much a science expert as any other internet googlologist... Pathetic..


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## IanC (Oct 30, 2011)

gslack said:


> IanC said:
> 
> 
> > just like wirebender declaring to everybody that he provided the proof and did the math. in a past post. that he refuses to show anybody. just like the worst of climate scientists.
> ...



hahaha. as usual you type a lot of words....but no substance.


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## gslack (Oct 30, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC said:
> ...



Stop posturing you fake... You couldn't argue my post so turned into the real you again same as always. Just a little punk playing smart-guy on the net. One of many nothing more..

Why didn't you respond to my post fairly then fraud? You couldn't and right now you are googling your ass of to find some terms you throw in here to save face... Same thing you did last time. 

I just bet you make another thread where you google up someones study and try to comment on it again. Yeah you did that last time and it was a pathetic display of "see im still smart".. The truth is you aren't smart, you're just a fake with a little more patience with character creation than most of your ilk.. Same animal just a slightly better act..

So go make your ego stroking thread now and play smart for the forum. I know you for a fake and so does wiredbender, soon your BS will show with the rest of the forum.. People aren't fooled by a posturing buffoon for too long they will catch on...


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## wirebender (Oct 30, 2011)

gslack said:


> I think Wirebender is referring to that, and you are most likely referring to after equilibrium is reached. When broken down fully you will find they are two different situations both mathematically and naturally...



No offense gslack,  but I am talking about the skin touching the blanket.  Because the blanket is in open air and able to bleed off heat into the atmosphere, it will never reach equilibrium with the skin.  A couple of Christmases ago, my kids pitched in together and got me a handy dandy infrared imager.  This thing has a sensitivity of 0.15 C.  I love toys like that.

The first time someone told me that a blanket will cool you down, my first instinct told me that it just wasn't so.  Having a handy dandy infrared imager, I put the claim to the test.  Sure enough, the skin in contact with the blanket is initially, and remains cooler than the skin not in contact with the blanket.  It doesn't matter how long you leave that blanket on, the skin in contact with the blanket remains cooler than the skin not in contact with the blanket. 

Those laws of physics are something.  And interestingly enough, if you add blankets, the skin in contact with the blanket cools a little bit more.  Not a lot,  but it does get cooler.  More blankets = more absorptivity and you can't generate more heat than you can generate so the skin in contact with the blanket remains cooler than the skin not in contact with the blanket.  Lift the blanket and in about 15 - 30 seconds, your skin goes back to its normal temperature.  Of course, your skin temperature varies across  your body but it goes back to normal.

When you look at the void between you and the bed (the tented area) you can see the radiation spread out and the temperature will eventually get relatively close, but never in equilibrium with your body.  Even after hours in the bed, even with two bodies, the air space never gets to body temperature no matter how many blankets you have.


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## wirebender (Oct 30, 2011)

IanC said:


> gslack- the difference is that one block was heated and they werent touching. try again.



They were in a cooled vacuum Ian.  Sorry that the implications of that escape you, but then the implications of damned near everything that doesn't jibe with your faith escapes you, or you escape them.  Either way, the result is the same.

Here are a couple of hints for you:
Vacuum
Heat sink
Radiating as a single body


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## wirebender (Oct 30, 2011)

IanC said:


> Spencer is against CO2 as a major factor in the warming of the earth over the last 200 years by the way. he just isnt willing to discount the real effects that CO2 plays.



CO2 is no factor because it can not radiate IR back to the surface of the earth.


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## wirebender (Oct 30, 2011)

IanC said:


> I obviously know about more photon paradoxes than you. both you and wirebender have some strange view of the thermodynamic laws that only hold true for the two of you.



Ian, you clearly don't know the first thing about photons.  You were, and apparently remain completely unaware that photons are nothing more than tiny bits of EM energy and forget about wave particle duality, that topic may as well be in another galaxy for you for all your apparent understanding.  You picture them as little free agents zipping about the universe and because of that picture that is apparently indellibly scribed in your mind, you will never understand what is happening here.  You drag in virtual photons as if they had anything to do with this topic completely unaware that virtual photons are nothing more than a means to describe an energy transfer between electrons and protons or neutrons, two particles that have absolutely nothing to do with EM fiields outside the confines of individual atoms.  You are, in short, clueless Ian and pretending that you know something that everyone else is missing just brings your ignorance of the subject into high relief.


----------



## wirebender (Oct 30, 2011)

IanC said:


> just like wirebender declaring to everybody that he provided the proof and did the math. in a past post. that he refuses to show anybody. just like the worst of climate scientists.



You have some math above Ian which is little more than algebra and it is obviously over your head.  You look at it and may as well be looking at ancient aramaic for all the sense it makes to you.  You claimed that the Stefan-Boltzman equations were far fetched and had no place in an experiment dealing explicitly with blackbody radiation.  The math I refered to in the previous thread involved time and distance.  Hell Ian, you can't even talk intelligently about math involving nothing more than radiation and you completely failed to even address the proof provided that additional energy would be required to raise the temperature of that heated bar.  In short Ian, you are a fake and have proved it beyond any reasonable doubt.



IanC said:


> I dont really care, I was just curious. you two guys seem to think huffing and puffing and bluster is the same as describing your side of the story. I cant really judge the merits of your ideas if you refuse to produce them.



In case you haven't noticed, and it is clear that you haven't, it is you, Ian, who is huffing and puffing and blustering.  I showed you the math to prove spencer's conclusions wrong and the math that proves that his claims violate the law of conservation of energy.  You merely declared that the math was wrong and the Stefan-Boltzman equations which deal explicitly with blackbody radiation are not applicable in an experiment about blackbody radiation.  You didn't prove the equations wrong, or offer up what you believe to be corrections of the equations or even suggest a different derivation of the Stefan-Boltzman equation (which would be flawed by the way), you just pronounced them wrong and expected that to mean something.  It is you who is all bluster Ian, I laid out the math for you and it meant about as much to you as it would mean to one of my donkeys.


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## wirebender (Oct 30, 2011)

IanC said:


> hahaha. as usual you type a lot of words....but no substance.



The substance is in this thread Ian.  The math is there.  Prove it wrong.  Prove that the Stefan-Boltzman equations are inappropriate for an experiment about blackbody radiation.  Prove that the law of conservation of energy was not broken in order for the experiment to end as spencer claimed.  Prove that the second law of thermodynamics was not broken.

Prove it mathematically Ian.  That is where the substance is.  You have the math that supports my postion.  I have provided substance.  Thus far, you have provided notthing but hot air.


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## wirebender (Oct 30, 2011)

One other thing Ian, just to further highlight the inherent weakness of your position.  You claim adamantly that the two bars could never reach equilibrium and yet, you believe the cooler bar can somehow make the heated bar hotter.  I would ask for an explanation for that bit of magic, but frankly, I am no longer interested in your fantasies.


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## IanC (Oct 30, 2011)

ok, lets describe the situation

I am going to assume that the two bars are separated by the minimum distance that stops conduction but will allow virtually all the radiation from the two inside faces to only pass back and forth between each other. there is no convection in a vacuum.

first bar, heated.   assume symetrical with height twice the width and a surface area of 1m2. that means there are 4 faces of 1/5 m2 and two faces 1/10 m2. the heater is symetrically imbedded and we will ignore the differences at the corners. your math says that it radiates 750 W/m2 so I will go with that but drop the units from now on. I will also combine the top and bottom face together and consider them one face for ease of calculating, five equal faces.

original conditions are at 150C with 750W divided by five faces = 150W each

after the second bar is added the heated bar is now 160C and I will make up a figure of 800W for ease of calculation. 800 divided by 5 faces is 160W. 

the question is where did the extra 50W come from. 

the 50W came back from the second bar. the inside face of the second bar is being heated by the 160W of the inside face of the heated bar but 50W is being radiated back. net power = power emitted - power absorbed. 110W= 160W- 50W. lets go back and add up the heated bar. 160+160+160+(80+80)+110=750W. exactly the power output we expected.

now lets look at the other bar. it has no internal heating, just the 160W radiation from the inside face of the heated bar. unlike the heated bar, the second bar is not the same temperature at all points but has a temperature gradient going from the side receiving radiation to the opposite face. the inside face would be by far the warmest, capable of radiating 50W. the opposite side radiates the least, perhaps 10W. the other faces radiate the remaining 100W, 33W per full face, but starting at 50W and dwindling to 10W at the edge farthest away from the heated bar.

now lets add up the radiation escaping to the cooled vacuum container. 

   from the heated bar. four full faces  (1+1+1+1/2+1/2) x 160W= 640W
            the inside face does not escape but goes into the unheated bar

   from the unheated bar. three full faces (1+1+1/2+1/2) at 33W= 100W
                                                                                one face at 10W
              the inside face does not escape but goes into the heated bar

grand total of escaping heat is 640W+100W+10W= 750W. the same power going out as the power going in.

the temperature of the heated bar is relatively easy to give a uniform temperature to, indeed we specified that it started at 150C and ended at 160C. the unheated bar is not a uniform temperature. I dont know why Spencer assigned it a temperature of 100C. 

this is only my mental picture of what happens in the thought experiment. I am not a physicist so I am likely to be wrong in some places. but dont just say it violates some law, point out where you think my radiation picture is wrong and describe how you think the heat is flowing.


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## wirebender (Oct 30, 2011)

IanC said:


> original conditions are at 150C with 750W divided by five faces = 150W each
> 
> after the second bar is added the heated bar is now 160C and I will make up a figure of 800W for ease of calculation. 800 divided by 5 faces is 160W.
> 
> the question is where did the extra 50W come from.



Hell Ian, why not make it 20 faces, or 200 faces.  Maybe you can make it 1000 faces and power a whole damned neighborhood on 750 watts worth of electricity.  What is sad is that you actually believe that crap.

You were given the math that reflects the reality of the experiment Ian.  Sorry you don't get it.  If you believe you can multiply energy by adding more faces to your radiator, then get yourself some investors (surely there are goobs out there who will believe you) and go into the power business.  You will make billions and be adored by the masses if it works.

More likely, the goobs you con out of their money will break your kneecaps when they realize that they have been had by a huckster.


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## IanC (Oct 30, 2011)

wirebender said:


> gslack said:
> 
> 
> > I think Wirebender is referring to that, and you are most likely referring to after equilibrium is reached. When broken down fully you will find they are two different situations both mathematically and naturally...
> ...



sounds like a cool toy to play with and an interesting experiment.
did you try measuring the temperature of your feet, one under the blanket and one exposed? I wonder how deep in the skin are the body's temp sensors that regulate blood flow and hence temperature control. as you say it is very counterintuative that exposed skin would be warmer while losing more energy than covered skin which loses less heat but is cooler. I suppose it has something to do with the greater ability of conduction to move heat compared to radiation. and the functioning of the body's thermostatic control somehow reacting to heat loss rather than just temperature. 

thanks for an interesting topic.


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## wirebender (Oct 30, 2011)

IanC said:


> sounds like a cool toy to play with and an interesting experiment.
> did you try measuring the temperature of your feet, one under the blanket and one exposed? I wonder how deep in the skin are the body's temp sensors that regulate blood flow and hence temperature control. as you say it is very counterintuative that exposed skin would be warmer while losing more energy than covered skin which loses less heat but is cooler. I suppose it has something to do with the greater ability of conduction to move heat compared to radiation. and the functioning of the body's thermostatic control somehow reacting to heat loss rather than just temperature.
> 
> thanks for an interesting topic.



It is a cool toy and I have tried damned near every permutation of heat loss that I can think of.  Hell, I even went next door and aimed it at my neighbor's siamese cats.  Its true that the dark areas of a siamese cat's body are slightly cooler than the light areas.  Just barely within the range of my imager, but cooler none the less.  Extremities are cooler than the core of the body.

Another interesting thing is to point it at the items in the refrigerator and the freezer.  You get zilch.  That tells me that even though the items in the cold are radiating, they are not emitting to the imager which is warmer.  I believe I suggested that you try it yourself.

Exposed skin feels cooler than clothed skin because the clothes hold dead air which is warmer than the air outside.  When you look at a clothed person with the imager, it is interesting to note how little of a person's clothing actually touches their body.  Look at your own clothes, about the only place they touch you is across your shoulders, down your back to a point just below your shoulder blades, your waist, and your thighs.  You are warming dead air in more places than you are losing surface heat due to the clothing touching your body.  Make yourself aware of that and then go outside and stand still for a while.  Without looking you can touch the places where your clothing is touching your body because those areas will be a bit cooler than where your clothing is not touching.

Really Ian, the math doesn't lie while your intuition will lie to you in a new york second.


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## gslack (Oct 30, 2011)

wirebender said:


> gslack said:
> 
> 
> > I think Wirebender is referring to that, and you are most likely referring to after equilibrium is reached. When broken down fully you will find they are two different situations both mathematically and naturally...
> ...



Agreed that was what I assumed. When the cooler blanket touches the warm skin it will make the warmer skin cooler for a time.. Sorry if it became fuddled and unclear... After all the back and forth with the little idiot trying to play smart I most likely got it mucked up.. Sorry for that..

I am sorry wirebender but I have run out of patience with that POS posturing fake.. he has no desire to prove anything or show truth, all he cares about is trying to convince others hes not losing and a real science "expert".. 

I ask you honestly, can you point to one thing he has put up here of his own creation from his own mind and in his own words, that can give credence to his claim of higher knowledge on this or any other physical science discipline? I have been going over a lot of posts here and frankly all I can find is him sitting there claiming he knows this or that and never once has he been able to back it up when he has been tested..

he is a fake plain and simple, he has shown it over and again, and his lack of knowledge of the very basics of physical science tell me he is not even well read enough to make a good case in his defense. His last action where he obviously couldn't follow my posts contentions, and decided to dismiss it rather than try told me exactly what we were dealing with.. he doesn't think on multiple levels, he doesn't question that which is established, he does not feel he has to read anything he doesn't want to, he already assumes he knows the answer to every possibility, and on top of all of this he reacts like a child when he can't keep up...

he is the worst kind of internet fake.. he pretends neutrality while all the while he views everyone here with disdain. he assumes he is the smartest guy in every room, when in reality he has never tested his own mental abilities.. He is one minute insulting me and dismissing my points as so much nonsense for no good reason, and the next he is trying to pretend he is the peaceful one just here to understand something... he is useless...

There isn't a genuine bone in him. So I am done being decent with the POS.. From now on I treat him as I treat konradv, because as far as I have seen they are two of the same animal...


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## wirebender (Oct 30, 2011)

gslack said:


> I ask you honestly, can you point to one thing he has put up here of his own creation from his own mind and in his own words, that can give credence to his claim of higher knowledge on this or any other physical science discipline?



Truthfully?  No.  I have done the math on this board in public and brought math that I agree with here in defense of my position.  To date, he has done nothing but claim that it isn't right, or not appropriate but remains completely unable to say what is wrong.  He put himself completely in the barking moonbat camp with is suggestion that you could get more wattage out than is going in if you simply add some sides.  Add some sides.  Is that, or is that not the craziest thing you ever heard.  Hell, lets do the experiment and add a thousand sides, we might be able to add enough sides to get a megawatt out of that initial 750 watts.

If it works, I can buy that 200' super yacht I always wanted.  What will you do with your billions?  Imagine, changing 750 watts into a megawatt or more, just by adding some more sides.  If the laws of nature were that easy to break, it makes you wonder why some genius hasn't already done it.  Hell, if you could eek out the 50 watts Ian suggested, you could make a profit.  750 watts in, 800 out.  106.7% efficiency.  You could earn a fine living for yourself if it were actually possible.  What is the average efficiency of nuclear reactors"  Somewhere between 35 and 50%?  Maybe Ian should tell them that all they need to do is add some sides and they to can get 106% or more out of those fuel rods.

What gets me is that he stated that you could get the extra 50 watts out with a straight face.  He actually believed it because spencer said so.  Surely a great physicist like spencer couldn't make a mistake; especially one so basic as to confuse a heat sink with an energy multiplier.


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## gslack (Oct 30, 2011)

wirebender said:


> gslack said:
> 
> 
> > I ask you honestly, can you point to one thing he has put up here of his own creation from his own mind and in his own words, that can give credence to his claim of higher knowledge on this or any other physical science discipline?
> ...



A couple of years ago I would have considered what spencer did as a mistake, but now? No I think hes doing it deliberately and doing so to get his money out of this.. he has written multiple books on this, and since his admission (under duress) of the data showing less actual warming than they previously thought, I believe he has to either ride it till he can't any longer or confound the science as best he can until it can be salvaged..

Spencer used a commonly referenced thought experiment regarding zeroth law application, removed one of the parameters and used the incidental kinetic energy from the objects proximity to its heat source to try and justify greenhouse theory... he is far too educated and experienced to not know both the original thought experiment, and the often overlooked kinetic energy from a heat source placed in proximity to a warming or warmed object. As they heat up the molecules move more, and that movement is energy too. it has to go somewhere and not all of it is converted to heat as he would have us believe.

Technically the second bar will get a very little bit hotter but thats due to proximity and incidental kinetic energy inherent in two heated stationary solid materials. make those objects gases that move and release their kinetic energy in the process of their movement around a chamber, and then lets see that theory in action... It will fail, and he has to know this..

I think spencer did this and crap just like it to cause mass confusion and confound the discussions in variables and parameters that have no bearing in greenhouse theory.. he has a position, a job, a career, and people who depend on this being right. And add in all those books to sell... he is stuck, and hes milking it for all he can...

And yes, I want to take my idea before about adding a tank of CO2 to an electric generator. We can turn 1 generator into 100...


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## IanC (Oct 31, 2011)

wirebender said:


> IanC said:
> 
> 
> > original conditions are at 150C with 750W divided by five faces = 150W each
> ...



are you denying that a rectangular block with height=2, width=1, depth=1 does not have six faces, four equal 2x1 faces and two 1x1 faces(these two half size faces add up to the area of a larger face). 

we know only one of the larger faces is interacting with the second bar and all the other sides are interacting with the cooled outside. surely you understood this? from my post-


> now lets add up the radiation escaping to the cooled vacuum container.
> 
> from the heated bar. four full faces (1+1+1+1/2+1/2) x 160W= 640W
> the inside face does not escape but goes into the unheated bar
> ...



where is the multiplying effect that you are accusing me of?


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## gslack (Oct 31, 2011)

IanC said:


> wirebender said:
> 
> 
> > IanC said:
> ...



More BS from you.... Ian just ONCE why don't you man the F*** up and respond to what HE DOES SAY instead making shit up or arguing something HE DIDN'T SAY... Man could you get any more of a chicken shit without having feathers?

Man up wuss..


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## IanC (Oct 31, 2011)

you guys really are mental midgets. of all the things that you could question about my simple description of where and how the heat flows in Spencer's thought experiment you pick on my choice of dividing the surface areas of the blocks into whether they radiate into cold space or back into a block!

I expected you to complain that the radiation back into the heated block would cause a temperature gradient there, just like in the unheated block. it would, except there is already a temperature gradient from the imbedded heater that would totally swamp the insignificant effect.

or you could have complained that the radiation/backradiation was an infinite series. it is but the end result is different by a negligible amount.

you could have complained that the corners of the blocks would be cooler than the interior of the face area because they have ability to shed radiation in more planes. true but I defined them as equally radiating at the start to escape from the fantastically difficult math involved.

obviously you two dont do any thinking for yourselves. you take someone else's statements, repeat them ad nauseum, and insult people who challenge their veracity. the problem is-- you guys take true statements but then attach them to situations that dont apply.

the zeroith law doesnt apply to equilibriums where one of the objects is heated by external power.

the 2nd law doesnt state that it is impossible for a cooler object to radiate into a warmer object. it states that the balance of radiation will flow from warmer to cooler.

I look forward to your next round of bluster and ad homs. I just wish that you would add a little insight into the topic of discussion once in a while.

BTW- I'm still waiting for you to prove that photons are extinguished in space without the presense of matter....and do the math of course. hahahaha


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## IanC (Oct 31, 2011)

gslack said:


> IanC said:
> 
> 
> > wirebender said:
> ...



wirebender said-





> If you believe you can multiply energy by adding more faces to your radiator, then get yourself some investors (surely there are goobs out there who will believe you) and go into the power business. You will make billions and be adored by the masses if it works.



and I responded. 

one of the difficulties I have in discussing things with you two guys is that you imbed your questions in reams of insults. I usually ignore them.

or how about you gslack. how many times have you accused me of making perpetual motion machines? or getting energy for free?

you never quote me, you just state that I said it. when I ask you about it you just go on to the next insult.

I am more than willing to answer questions but I do expect some compliance in returning the favour.


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## wirebender (Oct 31, 2011)

IanC said:


> are you denying that a rectangular block with height=2, width=1, depth=1 does not have six faces, four equal 2x1 faces and two 1x1 faces(these two half size faces add up to the area of a larger face).



The number of faces is completely irrelavent Ian and in a cooled vacuum, it doesn't matter which direction the faces are pointing or what they are pointing at.  Sorry you fail to get it.  More importantly, however, a heat sink does not, and can not heat up the radiator it is serving to help cool.


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## gslack (Oct 31, 2011)

IanC said:


> you guys really are mental midgets. of all the things that you could question about my simple description of where and how the heat flows in Spencer's thought experiment you pick on my choice of dividing the surface areas of the blocks into whether they radiate into cold space or back into a block!
> 
> I expected you to complain that the radiation back into the heated block would cause a temperature gradient there, just like in the unheated block. it would, except there is already a temperature gradient from the imbedded heater that would totally swamp the insignificant effect.
> 
> ...



Ian you have just described exactly what you have been doing here all along, and the ridiculous part is you really honestly think you can do crap like that and come back in the next day and claim we did it and its all good..

yeah Ian you fool somebody I am sure....

Please Ian can you point to YOUR original work here? I seemed to have missed it... I just explained my points and positions to you over several posts, you couldn't respond to them adequately, you knew this and decided to pretend some higher intellect, insult me and run away like a coward until you could google some more terms...

You look ridiculous dude... I almost feel sorry for you..Almost...


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## IanC (Oct 31, 2011)

wirebender said:


> IanC said:
> 
> 
> > are you denying that a rectangular block with height=2, width=1, depth=1 does not have six faces, four equal 2x1 faces and two 1x1 faces(these two half size faces add up to the area of a larger face).
> ...



the number of faces is irrelevent. what is relevent is whether they are radiating to the cold space or a warmer object. net power=power emitted-power absorbed. my post made that abundantly clear.


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## IanC (Nov 1, 2011)

wirebender said:


> No offense gslack,  but I am talking about the skin touching the blanket.  Because the blanket is in open air and able to bleed off heat into the atmosphere, it will never reach equilibrium with the skin...
> ....When you look at the void between you and the bed (the tented area) you can see the radiation spread out and the temperature will eventually get relatively close, but never in equilibrium with your body.  Even after hours in the bed, even with two bodies, the air space never gets to body temperature no matter how many blankets you have.



so you did your own experiment and found that objects around a heated object never reach equilibrium at the same temperature, even with conduction and convection added to the arguably weakest form of energy transfer, radiation.

yet you mock me for saying the same thing. 

explain in your own words how both the heated and unheated bars in Spencer's experiment will end up at the same temperature. and while you're at it explain how all the surfaces will radiate away the same amount of energy even though some are radiating out to the cold and some to the warm.


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## wirebender (Nov 1, 2011)

IanC said:


> wirebender said:
> 
> 
> > No offense gslack,  but I am talking about the skin touching the blanket.  Because the blanket is in open air and able to bleed off heat into the atmosphere, it will never reach equilibrium with the skin...
> ...



My experiment happened in the open atmosphere.  Spencer's experiment happened in a cooled vacuum and only involved what you clearly stated was the weakest form of energy transfer.  Sorry that you are unable to grasp the difference; and terribly sorry that you beleive the two to be the "same thing".


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## IanC (Nov 1, 2011)

wirebender said:


> IanC said:
> 
> 
> > wirebender said:
> ...



OK, explain what difference the open atmosphere vs vacuum is making regarding the equilibrium.

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change. in the original the bars were side by side with no specified distance between them. lets rotate the bars 90 degrees and keep them the same distance apart. now it is just two of the smaller faces available to transfer radiation into the unheated bar. from your original 'math' this will make no difference even though there is half the area to transfer heat and lead to equilibrium. not to mention an extra 10% of the surface area radiating directly into the cold outside. Am I correct in my assumption that you are sticking with your original 'math'? the bars will come to equilibrium at an equal temperature?

hahahaha, ludicrous


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## gslack (Nov 1, 2011)

IanC said:


> wirebender said:
> 
> 
> > IanC said:
> ...



*Example #1* *Changing the experiment or parameters to try and make it work.*



			
				IanC said:
			
		

> because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change.



Changing the experiment again? Its what you have done from the start and do so every time you cannot accept what it tells you..

*Example #2 **Doing the same thing over and again while expecting a different outcome, and pretending to get one regardless.*



			
				IanC said:
			
		

> in the original the bars were side by side with no specified distance between them. lets rotate the bars 90 degrees and keep them the same distance apart. now it is just two of the smaller faces available to transfer radiation into the unheated bar.



Ian, try and get your head around this... The size of the face close to the heat source changes nothing in this experiment. Shrinking the radiator and receptor surface area size will also shrink the radiating surface area and thus shrinking the amount of radiating heat from that vector to the receptor. Ian...seriously...

you keep changing the parameters to make it work and it still falls flat. I explained it, wire explained it, hell wire showed you the math, I explained how spencer took an existing experiment to show zeroth law in action and removed a parameter to make an unclear and deceptive claim about energy transfer in the atmosphere, and you still try and pretend we are both wrong... Why is that Ian? Can't accept you are wrong? yeah...

Ian you BS is tiring, you make less and less sense and your arguments are getting ignorant now...


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## wirebender (Nov 1, 2011)

gslack said:


> because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change.



Hard to pin down?  What the hell are you talking about.  I gave you the expanation in the most explicit language possible; mathematics.  Sorry you don't speak that language.  That being said, it has been explained to you in verbage as well and you didn't get that either.  The bottom line, Ian, is that the topic is apparently so far over your head that it can't be explained to you.


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## IanC (Nov 1, 2011)

wirebender said:


> gslack said:
> 
> 
> > because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change.
> ...



your math is highly suspect. if shrinking the area available for heat flow and expanding the area available for heat loss doesnt effect the final answer then your modelling of the problem is wrong. its as simple as that.


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## gslack (Nov 1, 2011)

IanC said:


> wirebender said:
> 
> 
> > gslack said:
> ...



Nice try with the slight of hand there Ian but you said to turn BOTH objects to the smaller areas face one another at the same distance... In effect you shrank BOTH surface areas that transfer the heat... Increasing the distance from the heat will change the entire experiment again.. The original experiment was not to see if altering the heated and receiving areas geometry relative to position and vector would alter the result, and attempting to do so is a deliberate distraction from the original...

Stop trying to confound it Ian....


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## IanC (Nov 1, 2011)

gslack said:


> IanC said:
> 
> 
> > wirebender said:
> ...





> Ian, try and get your head around this... The size of the face close to the heat source changes nothing in this experiment. Shrinking the radiator and receptor surface area size will also shrink the radiating surface area and thus shrinking the amount of radiating heat from that vector to the receptor. Ian...seriously...



hahahaha. what a bunch of nonsense. worthy of Cliff Clavin! explain how less available heat absorbance and more available heat shedding adds up to the same result. add up the warming surface area in both cases, the cooling surface area in both cases. 50% less warming, 10% more cooling. yet you two guys say there is no change. what a couple of retards


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## gslack (Nov 1, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC said:
> ...



Please show where I said "no change"... yeah, i said changes nothing in the experiment.. The experiment was not and is not an attempt to show changes in vector or geometry relative to heat surface area and rate of heat loss... 

you tried again to alter the topic to cover your ass Ian.. Now you can stop lying anytime...


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## IanC (Nov 1, 2011)

wirebender's 'math' gives the same answer regardless of orientation. the orientation obviously makes a difference in heat flow. therefore wirebender's math is incorrect.

can I make it any simpler?


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## wirebender (Nov 2, 2011)

IanC said:


> your math is highly suspect. if shrinking the area available for heat flow and expanding the area available for heat loss doesnt effect the final answer then your modelling of the problem is wrong. its as simple as that.



You lose again Ian.  I already told you that the only way to increase the temperature of the heated bar was to either decrease its area or increase the power going to it.  Clearly adding the heat sink added to the radiating area and the final result was both radiating at a considerably lower temperature than the heated bar alone.

If the math is wrong, then prove the math is wrong.  Or you might man up and simply admit that you don't know whether the math is wrong.

By the way Ian, it isn't "my" math and it isn't gord's math either.  It is Jozef Stefan's math and it has stood for a very long time and my bet is that you are not going to tear it down either.


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## wirebender (Nov 2, 2011)

IanC said:


> hahahaha. what a bunch of nonsense. worthy of Cliff Clavin! explain how less available heat absorbance and more available heat shedding adds up to the same result. add up the warming surface area in both cases, the cooling surface area in both cases. 50% less warming, 10% more cooling. yet you two guys say there is no change. what a couple of retards



Prove the Stefan-Boltzman equation wrong Ian or prove that it isn't applicable in problems of radiation.


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## wirebender (Nov 2, 2011)

IanC said:


> wirebender's 'math' gives the same answer regardless of orientation. the orientation obviously makes a difference in heat flow. therefore wirebender's math is incorrect.
> 
> can I make it any simpler?



Wirebender doesn't have any math.  Neither does gord.  It is a straight forward application of the Stefan-Bolzman law to a problem of radiation.  The answer is what it is Ian.  Either prove the Stefan-Boltzman law wrong or prove that it isn't applicable to radiating bodies and we have something to talk about. 

Going on and on about the math being wrong, or far fetched isn't getting the job done Ian.  Either you can prove it wrong, in the language of mathematics, or you can't.  My bet is that you can't because it isn't wrong.  Since you can't prove it wrong, then you might consider admitting that you are working from a position of faith, or instinct, or wishfullness or something other than a position of hard, unblinking science and a willingness to accept the laws of physics no matter what the little guy behind your eyes is screaming to the contrary.

Here is the mathematical sequence again Ian.  You tell me where it goes off the tracks and show me the math to prove it.  By the way, this sequence addresses spencer's thought experiment.  He didn't specifiy the emissivity of the plates so it was assumed to be 1 and he didn't specify the surface area of the plates so they were assumed to be one square meter.  Those numbers were chosen for convenience.  You can change the emissivity if you like and you can change the surface area if you like.  The final answer will be different in accordance with the changes you make but the bottom line will remain the same.  The heated plate and the cooled plate situated in a cooled vaccum will attain an equilibrium temperature considerably lower than the heated plate in the chamber alone.

Last time, you got off the tracks by assuming that because the heated plate was cooled by the addition of the other plate that perhaps more energy was flowing through the heated plate via the power cord attached to the heated plate.  If more energy is coming in, then spencer is still wrong because the cooled plate didn't warm the heated plate, the additional energy input did. Remember the proof that to raise the temperature of the heated plate the wattage going in would have to increase from 744.95 to 795 watts?  The energy input remains the same.  Spencer is trying (and failing) to show that backradiation is possible in the atmosphere.  Letting more power flow into the bar because it now has essentially more surface area makes it behave in a manner other than as a blackbody and the whole experiment is geared as an attempt to prove something about blackbody radiation.

_Re: Vacuum Chamber with plates.

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefans constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts

(***Thats ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)

The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2
 
 If another identical non-heated and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:

The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.

The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.

The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

And the equilibrium temperature for both plates will be 284.69 K or 53 deg _

You can either prove the math wrong or you can't Ian.  I am betting that you can't in which case, the discussion has been over for some time and you merely didn't realize that you lost.  The orientation of the plates doesn't matter because there is no such thing as backradiation.  The failure of the experiment was due to the fact that spencer ignored the 2nd law of thermodynamics and the law of conservation of energy and assumed that backradiation is possible.  Tell me Ian, do you also believe in back conduction, or back convection?  how might they work?  If you believe backradiation is possible, you must also believe back convection and back conduction are possible.  how me the math that supports such nonsense in the physics literature.


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## IanC (Nov 2, 2011)

you really dont think these things through, do you wirebender. I will ignore your strawman defense of accusing me of trying to prove thermodynamics wrong. in fact they are what I use to imagine the energy flow in these type of problems.

lets try again to make you 'see the light' (bad pun)

both bars are at 150C and are in separate vacuum containers. what happens with respect to time and equilibrium? the heated bar continues to receive 750W therefore continues to radiate 750W @ 150C with an area of 1m2. assuming a regular rectangular shape and even heat distribution that means four faces radiating 150 W each and 2 faces at 75w.   total =750w= 150+150+150+150+75+75
no change with respect to time for the heated bar.

the unheated bar starts at 150C and also radiates 750W with six faces adding up to 1m2, same as the heated bar. because the unheated bar is losing heat but not receiving any replacement heat the temperature will drop, exponentially until it is essentially the same as the cooled container.

next variation- the unheated block somehow magically recieves the 150w of radiation from one of the large sides of the heated block. this heat is evenly distributed so the six sides of the now heated block would radiate at 30+30+30+30+15+15=150w

next variation- the unheated block somehow magically receives the 75w of radiation from on e of the small sides of the heated block, evenly distributed. 15+15+15+15+7.5+7.5=75w

your 'math' of what is happening in the thought experiment is already smashed to bits but lets keep going.

next variation-if the heated bar and the unheated bar were placed in the cooled vacuum container close enough to totally exchange the radiation from the two inside faces but not touching so there is no conduction what would happen? the heated face would radiate 150w to the unheated block but get 30w back. because there is thermal impedance the heat would build up at both faces but I will ignore that as an unnecessary complication. the 30w back radiation is not heat flowing into the warmer block, it is just part of the calculation. net Power= emitted Power- absorbed Power. the 30w back radiation needs to be equally divided between all the faces of the heated block.....

sorry, to be continued when I have time.


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## gslack (Nov 2, 2011)

Ian if you can prove his math wrong as it is, why do you keep making all these variations?Why all the complication if it really is so simple to show it wrong just do it already...

Jesus man, you are like a bullshit fountain... Show his math wrong already! Damn man..

EDIT: Ian his post was not a strawman, you ARE trying to claim thermodynamics laws wrong in your continued pretense by altering the original experiment enough you can somehow make the laws different. Strawmen have been your entire MO in all of this. Everytime you bring a variation or a different experiment or change the parameters like you just did, its a strawman.. Wow man you going to wake up today sometime?


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## wirebender (Nov 2, 2011)

IanC said:


> you really dont think these things through, do you wirebender. I will ignore your strawman defense of accusing me of trying to prove thermodynamics wrong. in fact they are what I use to imagine the energy flow in these type of problems.



There is your first problem Ian.  You are imagining rather than doing the math.  I have done the math and like it or not, it trumps whatever imaginings you care to do.  The math trumped spencer on his own blog because he made no effort to address it.  He simply sat quitely hoping that it would go away.  It didn't.

You have the math concerning radiation and your "sides" fantasy has no place in it.   Of course if you like, feel free to show me the Stefan-Boltzman derivitive that shows an increase of energy as the result of adding sides.  Do be sure to square that increase in energy with the law of conservation of energy.


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## wirebender (Nov 2, 2011)

gslack said:


> Jesus man, you are like a bullshit fountain... Show his math wrong already! Damn man..



I am afraid that isn't in the cards.  Imagine, proving the Stefan-Boltzman law wrong by simply adding sides.  Who would have thought?  What sort of thinking leads one to actually believe such gibberish.


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## IanC (Nov 3, 2011)

IanC said:


> you really dont think these things through, do you wirebender. I will ignore your strawman defense of accusing me of trying to prove thermodynamics wrong. in fact they are what I use to imagine the energy flow in these type of problems.
> 
> lets try again to make you 'see the light' (bad pun)
> 
> ...



continued...

so in the variation with the large (0.2m2) heated face radiating 150w into the large (0.2m2) unheated we have a choice of bookkeeping methods. I prefer to imagine 150w out of the heated bar and 30w back from the second bar, for a net exchange of 120w. wirebender is vehemently against any mention of back radiation so we can instead imagine only 120w out of the heated face and 0w out of the unheated face. either way works for calculating, I just find it easier to visualize with back radiation.

what are the energy balances now that the two blocks are together? the heater is still inputting 750w into the heated bar but now one of the large faces is only outputting 120w instead of 150w, meaning 30w is not radiated away but is used to _heat up_ the bar until the faces are radiating 6w hotter for the large faces and 3w hotter for the smaller faces. 156+156+156+78+78+[(156-30)the heated face radiating into the second bar]= 750w. the faces of the heated bar are emitting more power which means they are warmer than before. _the temperature of the heated bar has increased!!!!_

what about the unheated bar? it has received 120w through the inside face and also nullified an extra 30w of power. I prefer to imagine it as absorbing 150w and reradiating 30w back at the source but it is a matter of preference. so the remaining 0.8m2 is radiating 120w. 30+30+30+15+15+0(for the inside face)=120w. of course this is vastly oversimplified. the heat flow is coming in through the side instead of from the centre so the inside face would be much warmer than the face opposite it. and the extra 6w from the heated bar after it equilibrates for the nullified 30w would slightly heat the unheated bar which would cause more back radiation, oops I mean power nullification and warm up the heated bar by a smidge, etc.

it is easy to see that the heated bar will *always* be warmer than the unheated bar. it is also easy to see that any thing warmer than the temperature of the cooled container will impede the loss of heat from the heated bar which will force the heated bar to warm up until the increased flow from the unimpeded sides balances the power in=power out.

I have asked wirebender to explain how the two bars will be exactly the same temperature at equilibrium--no answer. his 'math' states that they are, therefore it is incorrect.

perhaps he can explain how a heat source can bleed away the same amount of energy when it has become more insulated without becoming warmer itself. but I doubt it.

personally I think the small amount of temperature increase of the earth's surface caused by CO2 is mitigated by increased water evaporation which causes rising air and clouds that release latent heat are much more efficient at transferring energy than radiation. just look at the tropics and the afternoon thunderstorms that cool the surface in minutes. but I dont make up the inputs for the computer models, or decide the various values of Trenberth's energy pathway graph (which has changed considerably over the years, I wonder if _this_ one is right?).

anyways wirebender. two simple questions for you. you've ducked in the past and I'm sure you will duck now. explain how the bars come to equilibrium at the same temperature when one is heated and the other is not. explain how a heated insulated object can come to equilibrium without warming up.


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## gslack (Nov 3, 2011)

Ian why are you confounding this even more? Why not stick with the original thought experiment and prove it right, instead of continually altering it? That ones already established here, its been written out, and yet you continue to claim its right but refuse to use it to show its right...

If wire wants to continue playing your silly deflection its on him, but your "physics" persona is done for.. You know about as much physics as my 13 year old...

BTW Ian, remember the chamber, and the vacuum.. And thats all the help im giving you..


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## wirebender (Nov 3, 2011)

IanC said:


> I have asked wirebender to explain how the two bars will be exactly the same temperature at equilibrium--no answer. his 'math' states that they are, therefore it is incorrect.



Incorrect based on what Ian?  Your intuition?  The math is correct but do feel free to prove the Stefan-Boltzman Law wrong.  

I answered your question in the most explicit language possible.  Sorry you don't speak it.  That being the case, you really are in no position to question the answer.

You have lost the point, you have lost the discussion and have lost any respect that I may have had for you.  Why continue?


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## gslack (Nov 3, 2011)

Right now I am heading to the patent office to secure my rights to the CO2 molecular vibration infinity engine... Drive... Thing...Whatever it is it will work I just know it...


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## IanC (Nov 4, 2011)

gslack said:


> Ian why are you confounding this even more? Why not stick with the original thought experiment and prove it right, instead of continually altering it? That ones already established here, its been written out, and yet you continue to claim its right but refuse to use it to show its right...
> 
> If wire wants to continue playing your silly deflection its on him, but your "physics" persona is done for.. You know about as much physics as my 13 year old...
> 
> BTW Ian, remember the chamber, and the vacuum.. And thats all the help im giving you..




the reason I keep making small alterations to the thought experiment is to imagine what different results will occur. 

now I will try to imagine a scenario where you two could be correct. 

what are the absolute basics of the experiment? a heater producing 750w and a cooled container capable of recieving 750w without changing temperature. is this the point where your thinking stops. the trivial position of 750w in then out? any material inside the container is meaningless, only the input and output?

the bare heater radiating directly to the cold walls is the lowest temperature possible. once the bar is placed around it, insulating it, the temperature of the heater must increase to compensate for the shrinking of the temperature differential. the bar itself has a temperature gradient from the heater to the outside surface which can then radiate to the cooled wall. wirebender's math has already broken down, there has been an increase in the temperature of the heater due to insulation and there is a temperature gradient in the bar. but the input still equals the output.

adding a second bar only increases the insulation. which increases the the temperature of the first bar which then increases the temperature of the heater. but input still equals the output. 750w.

is it possible that wirebender and gslack are only considering the final outcome? that because the input and output match exactly that there is no information worth knowing about the temperature gradients? is this what gslacks's reference to the zeroith law and wirebender's proclaimations of Stefan-Boltzman have been about? that just because the output is equal to the input then we shouldnt worry about all that fiddily stuff in between?

even then wirebender's math is wrong because...


> If another identical non-heated and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:
> 
> The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2
> 
> ...


wirebender doesnt take the radiating abilities of the different areas into account. hell, if you push the two bars together you have lost 20% of the area (10% if end to end). that would magically increase the bars temperature dramatically wouldnt it? and you dont even have to consider the change by adding conduction to the mix because wirebender thinks they are both the same temperature anyways!

wirebender is a fool to invoke general formulas when the results of those formulas are not the answer we are looking for. in Spencer's thought experiment we are looking to find  the outcome of adding a second bar. in reality the second bar is only insulation. insulation causes the heat source to become hotter so that it can force the energy across the insulation. therefore the the first bar is forced to become warmer by the insulating properties of the second bar. the heater is forced to warm up because of the insulating properties of the first bar. the output remains the same as the input, although there has been some energy stored as a heat sink. Spencer is correct on the basics although his labelled temperatures of 150 and 160 degrees for the first bar and 100 degrees for the second bar seem out of proportion to me, and for the second bar especially the temperature would vary according to distance from the heat source.

I have looked at, and described this problem from many different angles. from bare basics to more complex. to do the actual derivation of temperature gradients on each bar would take extremely complicated math but I am not interested in specific numbers, only general principles. but not so simple that only the input and output are noted.

wirebender foolishly uses thermodynamic laws as a hammer to examine a soap bubble and loses all the detail in the problem. but then turns around and adds in exaggeratedly precise numbers at the end.


> wirebender stated- equilibrium temperature for both plates will be 284.69 K


which is wrong not only because he has failed to calculate the effective surface area of the masses but because there is a temperature gradient on and between both bars. the only part of his calculations which is right is the part where he states the power input equals the power output. and even that is only true after equilibrium has been reached.


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## gslack (Nov 4, 2011)

Ian once again its in a vacuum and in an enclosed space. If you want to use the argument you are using now, than perhaps you should remember original claims made in the original spencer experiment about heat transfer in a vacuum (particularly your previous defense of his experiment)BEFORE you go any further... 

You keep changing the parameters because in each new variation you run out of quick, lazy, and half-read/understood responses you can throw in. You don't care about knowing, you care about the appearance of knowing and that's the problem. You can't fake this, there is no way you can pretend a real scientific understanding of something when its truly tested, when you aren't even willing to take the time and read or study it. And the fact is you haven't, and further you have no desire to learn it, all you want to do is appear the victor or somehow smarter..

Seriously Ian go study the basic natural laws at least... They cannot be broken in the physical world we see and interact with directly. Even for spencer..


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## IanC (Nov 4, 2011)

where did your comment go wirebender? 

anyways, so why dont you start cranking out the math.

first define the heater and show what the inside gradient to the outside is. once you have found the outside temps then show how much radiation will go into the second bar. then mathematically show how the heat will cross the second bar and come out of the surfaces. for bonus points you could show the radiation difference from the container side closest to the heated bar compared to the container side directly opposite across the second bar.


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## gslack (Nov 4, 2011)

IanC said:


> where did your comment go wirebender?
> 
> anyways, so why dont you start cranking out the math.
> 
> first define the heater and show what the inside gradient to the outside is. once you have found the outside temps then show how much radiation will go into the second bar. then mathematically show how the heat will cross the second bar and come out of the surfaces. for bonus points you could show the radiation difference from the container side closest to the heated bar compared to the container side directly opposite across the second bar.



he is tired of your silly ass for one, and hes not here...

Spencer should define the heater. Matter of fact he should have provided all of that you just asked for... he didn't and you are trying to defend it and get everyone else to do the work for you. You change the experiments, change the variables, change the questions, change anything you can all in an attempt to prove yourself right.. You do all of that but refuse to actually DO the work to either prove it right or show it wrong.

he isn't going to do any more work for you, its wasted on you. You don't care about the truth, you care about being right..


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## wirebender (Nov 4, 2011)

IanC said:


> the reason I keep making small alterations to the thought experiment is to imagine what different results will occur.



Ian, you can imagine 10,000 different experiments but till you learn and apply the laws of physics, your end results will be wrong.



IanC said:


> now I will try to imagine a scenario where you two could be correct.
> 
> what are the absolute basics of the experiment? a heater producing 750w and a cooled container capable of recieving 750w without changing temperature. is this the point where your thinking stops. the trivial position of 750w in then out? any material inside the container is meaningless, only the input and output?



No need Ian, we are already correct and I am just about done trying to imagine ways of getting you to see what you so adamantly refuse to see.

First off, in spencer's experiment the bar was radiating 744.95wm2.  Here is the initial calculations again just to refresh your memory as if that would do any good.

_Re: Vacuum Chamber with plates.

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.6 K.

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.6K)^4 = 744.9 Watts_

You are imagining parameters that don't exist in spencer's experiment right off the bat Ian.  I don't know where you get the idea that the container is capable of only receiving 750 watts per square meter  (wm2) without changing temperature.  Spencer states quite clearly that the chamber represents the space around the earth.  He said:  "One way to demonstrate the concept is with the following thought experiment, which* I will model roughly after the Earth suspended in the cold of outer space.* ". 



IanC said:


> the bare heater radiating directly to the cold walls is the lowest temperature possible. once the bar is placed around it, insulating it, the temperature of the heater must increase to compensate for the shrinking of the temperature differential.



You have missed the train right off the bat Ian and at this point, I call bullshit on your claims of having taken physics.  I believed for a while that you might have actually taken physics but not done so well at it, but with that statement, you have proved that you don't know physics, and haven't the slightest clue as to the physical laws that make the science possible.

To claim that the bare heater radiating into the cold walls is the lowest temperature possible is about the stupidest thing you have said to date Ian.  In short, you have proven beyond any reasonable doubt that you don't have a clue.

You are simply wrong to assume that the temperature of the heating element must rise.  It can only rise if more electricity is delivered to it and that is not one of the parameters of spencer's experiment.  Imagine the electricity being delivered to the element via a rheostat if it helps you.  The electricity coming into the heating element is a constant and therefore it can not rise.

Let me walk you through this a step at a time and after this if you don't get it, I am finished with you.  

To follow your "most basic" senario, lets strip away the bar from the heating element.  We now have just a bare element radiating out into the vacuum.  For convenience lets assume the heating element has an emissivity of 1 just like the bar and lets assume that it is still radiating 744.9wm2 but since it is just a heating element that was inside the bar, lets make it a quarter of a square meter.

So we have a heating element radiating that has a radiating surface of .25 square meters, an emmisivity of 1, and it is radiating 744.9wm2.  You claim that it is at the lowest temperature possible.  Laughable Ian, and f'ing pittiful.

Apply the Stefan-Boltzman Law to what we know

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

744.9=5.67 X 10^-8 X .25 m^2 X(478.79K)^4 

Remove the plate and expose a bare heating element with a surface area of .25 meters and it will radiate 744.95wm2 at a temperature of 402 degrees F.  Thats 252degrees hotter than the 1 square meter plate radiating the same wattage per square meter.



IanC said:


> once the bar is placed around it, insulating it, the temperature of the heater must increase to compensate for the shrinking of the temperature differential.



Again, Ian, your intuition has taken you into the exact wrong direction.  Again, you aren't getting more electricity from the power cord.  The amount of power coming into the heater element remains constant througout the experiment.   Imbed that heating element into a plate that is 1 square meter and you are not going to insulate it, you are going to  increase the radiating surface area.  Assume the bar radiating 744.9wm2, and the emissivity is still 1 and the temperature will change.

Apply the Stefan-Boltzman Law:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

744.9=5.67 X 10^-8 X 1 m^2 X(338.6K)^4  

Increase the area of the radiator from .25 square meters to 1 square meter, keep emitting the same wattage per square meter and the temperature will drop from 478.8K to 338.6K or from about 402F to 150F.  

Just for fun, lets increase the temperature of the bar so that it is at the same temperature as the bare heating element.  Apply the Stefan-Boltzman Law

P = e*BC*A*T^4 

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

2982.37=5.67 X 10^-8 X 1 m^2 X(478.86K)^4 

As you can see Ian, to keep the bar just at the same temperature as the element, it will have to be radiating 2982.37wm2.  Where do you suppose all that energy is going to come from?    

Suppose we want the heating element to be radiating at 150 degrees.  Apply the Stefan-Boltzman law:

P = e*BC*A*T^4 

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

186.24=5.67 X 10^-8 X .25m^2 X(338.6)^4 

If the heating element is at 150 degrees it is only radiating 186.24wm2 compared to the 744.9wm2 required to raise the bar to 150 degrees.


If we have the heating element radiating 186.24wm2 at 150 degrees and imbed it in a 1 square meter bar and change nothing else, apply the Stefan-Boltzman Law:

P = e*BC*A*T^4 

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

186.24=5.67 X 10^-8 X 1m^2 X(239.4)^4 

The bar is now at 28.75 degrees F.





IanC said:


> the bar itself has a temperature gradient from the heater to the outside surface which can then radiate to the cooled wall. wirebender's math has already broken down, there has been an increase in the temperature of the heater due to insulation and there is a temperature gradient in the bar. but the input still equals the output.



The Stefan-Boltzman law and the accompanying math says that the temperature drops when you imbed the heating element into the bar and thus increase the surface area.  



IanC said:


> adding a second bar only increases the insulation. which increases the the temperature of the first bar which then increases the temperature of the heater. but input still equals the output. 750w.



Imbedding the bare heating element into the bar didn't constitute an insulator Ian, it increased the radiating surface.  It was a heat sink.  Adding another heat sink to the equation further lowers the temperature of the first bar.  Apply the Stefan-Boltzman Law

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan&#8217;s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

744.9=5.67 X 10^-8 X 2 m^2 X(284.69K)^4 

Double the radiating area and the temperature will reach equilibrium between the two bars at about 53F.  

You clearly don't understand the difference between an insulator and a heat sink Ian which tells me that you know exactly squat about physics.



IanC said:


> is it possible that wirebender and gslack are only considering the final outcome?



It is possible, and more than probable that Ian simply doesn't know jack.  You are wrong Ian.  The math says you are and I have gone through it in the simplest terms I can.  If you still don't get it, I am sorry for you.  You should never have engaged this discussion because you simply don't know enough.  You lack the requsite knowledge of physics and the physical laws that govern the science.  You are operating on intuition and it has led you off in the wrong direction because you lack the fundamental knowledge base.




IanC said:


> even then wirebender's math is wrong because...



My math is correct Ian.  Your intuition is wrong.  You reamin unable to prove me wrong because I quite simply am not wrong.



> If another identical &#8220;non-heated&#8221; and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:
> 
> The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2
> 
> ...





IanC said:


> wirebender is a fool to invoke general formulas when the results of those formulas are not the answer we are looking for.



You are the idiot Ian and I have proven it beyond any reasonable doubt.  Try this one last time to see what you are claiming.  If you can't get it, then you really must be an idiot.  You claim that by putting a bar around the heating element, you will increase the temperature without any additional power coming in through the power cord in direct oppostion to the Stefan-Boltzman law.  

Imagine you have a lighted flashlight filament in a vacuum.  It is buring at about 5000 degrees.  Do you believe that you could encase that filament in a metal bar and raise the temperature to above 5000 degrees without increasing the power coming in from the battery?  If you do Ian, then you are an idiot.


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## wirebender (Nov 4, 2011)

IanC said:


> where did your comment go wirebender?
> 
> anyways, so why dont you start cranking out the math.
> 
> first define the heater and show what the inside gradient to the outside is. once you have found the outside temps then show how much radiation will go into the second bar. then mathematically show how the heat will cross the second bar and come out of the surfaces. for bonus points you could show the radiation difference from the container side closest to the heated bar compared to the container side directly opposite across the second bar.



I have cranked out the math and once again, it proves you wrong.  What could possibly make you so stupid Ian?


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## wirebender (Nov 4, 2011)

gslack said:


> he is tired of your silly ass for one, and hes not here...



Yeah, I am about tired of him.  At least rocks and konradv know when they have lost and run away for a while before they start making the same stupid claims over again.  That puts Ian a step below those two.  How sad for him.



gslack said:


> he isn't going to do any more work for you, its wasted on you. You don't care about the truth, you care about being right..



I did it one more time accounting for his bare heating element.  That he thought a bare element would radiate at a cooler temperature than an element that is buried in a bar of metal larger than itself is just plain stupid.  There just isn't a better adjective unless you want to use something more colorful like moronic, insensate, or even "out to lunch".  And you are right, he isn't going to do the math because it is all just way over his head.  Maybe he will add some more sides trying to prove the Stefan-Boltzman Law wrong again.


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## IanC (Nov 6, 2011)

sorry for the delay. sometimes reality interfers with the internet.

as usual wirebender doesnt think these things through, he confounds one thing for another.

P = e*BC*A*T^4       what is this formula describing? it is describing the power emitted by a body at a certain temperature. in a system where there is no back radiation. in other words a container at zero degrees Kelvin.
wirebender said-





> Apply the Stefan-Boltzman Law to what we know
> 
> P = e*BC*A*T^4
> 
> ...



I agree that the heated bar in a zero degree Kelvin container would be 479K. his math is correct to this point.

but is the container 0K? no it is not. it is zero degrees farenheit = -18 degrees celsius = 255 degrees Kelvin. if the heated bar was receiving no outside power source it would be at the same temperature as the container, 255K. plugging that number into P = e*BC*A*T^4 =60w. it is already radiating 60 w before we add the 750w input. once we add the 750w and 60w for a total of 810w we can use the formula to work out the temperature of the heater. simple math calculations hold no interest to me so I will let you work out for your self how much warmer it is than 479K. but it is, and has to be warmer than 479K.











therefore P(emitted)= P(net)+P(absorbed)   are you following so far?

everytime you put matter in between the source of radiation(the heater) and the final radiation absorber(the 255K container) you have slowed down the transmission of power from the speed of light and created a heat sink that warms up. this heat sink radiates back at the source, reducing the abilty of the heat source to effectively shed heat.
     P(emitted)= P(net)+P(absorbed)     again. the emitted power is going up because of the absorbed power is increasing yet the _net_ power is staying the same. the heater has to keep getting hotter to force the energy across the heat sink until it finally can radiate unimpeded at the speed of light into the container. if you turn off the 750w input all the stored heat will bleed away and the unpowered heater and heat sink will come to equilibrium radiating at 255K just like the container.

wirebender likes to exclude the details in his models. in this case he ignored that the container was warmer than absolute zero, and that the heat sink would impede the flow of energy by changing the temperature differential. the heater must become warmer to force the energy across the matter. any addition to the heat sink around the heater will increase the necessary Power(emitted) and hence the temperature to balance the formula






why do you think delicate electronics burn out when you drop a pile of laundry on them? because the wires get colder?


----------



## IanC (Nov 6, 2011)

why do clothes keep you warmer when it is cold outside?





it has to do with the term {T^4- T(0)^4}

if you decrease the temperature differential then the net power goes down. clothing mechanically slows the speed of heat loss from the speed of light for straight radiation to a combination of radiation, conduction and convection. the last two are much slower, that is why there is a heat sink that builds up overall temperature in the clothing but the outside of the clothes is always cooler than the inside (if the ambient temp is lower than body temp). your body feels warmer because it doesnt have to produce as much heat to keep the internal temp at 37C.


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## wirebender (Nov 6, 2011)

IanC said:


> as usual wirebender doesnt think these things through, he confounds one thing for another.
> 
> P = e*BC*A*T^4       what is this formula describing? it is describing the power emitted by a body at a certain temperature. in a system where there is no back radiation. in other words a container at zero degrees Kelvin.



There is no such thing as backradiation Ian, any more than there is back conduction or back convection.  Backradiation defies the 2nd law of thermodynamics.  Which derivitive of the SB law do you believe suggests the existence of backradiation and can you show even one shred of observed, repeatable, experimental evidence to support the claim?



IanC said:


> I agree that the heated bar in a zero degree Kelvin container would be 479K. his math is correct to this point.



My math is correct from beginning to end Ian.


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## IanC (Nov 6, 2011)

wirebender said:


> IanC said:
> 
> 
> > as usual wirebender doesnt think these things through, he confounds one thing for another.
> ...



all of the SB derivatives incorporate backradiation except the ones that deal with a single radiating body.




the T^4 - T^4 tem is the backradiation.





the (minus Power absorbed) term is the backradiation.

are you quibbling over some wirebender imposed definition of the term back radiation? do you deny that the temperature of the object being heated is not an integral part of the equation?

and your math is incorrect as early as when you stated the heater's temperature was 479K in the vacuum container kept at 255K. and then proceeded to become more garbled by your incorrect simplifications.

do you agree or disagree that the heater's temperature is warmer than 479K when placed in the 255K container? if you disagree show me how it can be the same temperature even though the P(absorbed) term went from k(0^4) to k(255^4).


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## IanC (Nov 6, 2011)

interesting description there wirebender....no back conduction!

does that mean that no molecules are allowed to jiggle in the opposite direction of the heat flow? is it a physical impossibility? is it like the radiation that magically disappears (somewhere, you wont be specific) after being emitted from a CO2 molecule towards the earth? hahahaha.


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## gslack (Nov 7, 2011)

Ian, If you poke a hole in the sidewall of your cars tire, can you place your hand a few centimeters away and force the air back in it? or is the energy from the expelling air to great to allow the air to go back whence it came?

if you point a flashlight at a mirror will it make that flashlight brighter? or will the illumination make the mirror and environment brighter?

You can virtually take any type of real world application and find the same type of examples. Every single one in the natural world without help to achieve a difference will result in the stronger energy winning over the weaker energy.

Yet somehow you seem to think its not going to apply in spencers experiment or in your variations... WOW... I have no words man... Good luck with your patent on your endless heat engine bud...LOL


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## wirebender (Nov 7, 2011)

IanC said:


> all of the SB derivatives incorporate backradiation except the ones that deal with a single radiating body.
> 
> 
> 
> ...



Sorry Ian, but they do not.  The equation p=sigma (T^4 - T^4) most certainly does not describe backradiation.  That equation accurately describes the SB law as proven by SB. ( T^4 - T^4) describes radiation from a blackbody at temperature T^4 into a background at temperature T^4.  There is nothing like backradiation in that equation.

If you want an equation incorporating backradiation, then it must take the form P=(sigma T^4)-(sigma T^4).  

Algebraically the two look similar but they describe physics that are very different.  One describes the SB law as derived and proven by SB and the other describes a non physical event that has never been seen to happen in nature or the laboratory.

p=sigma (T^4 - T^4) is the actual SB law and it describes radiation from the blackbody to the background, period.

P=(sigma T^4)-(sigma T^4) on the other hand describes (sigma T^4) radiation from the blackbody to the background minus (sigma T^4) radiation from the background to the blackbody.

The SB law in the form p=sigma (T^4 - T^4) has appeared in physics texts since the late 1800's.  The law in the form P=(sigma T^4)-(sigma T^4) on the other hand only appears in atmospheric physics textbooks and is a new phenomenon.  If you are interested in verifying that bit of information, here is a short list of physics texts that teach SB as derived by SB.  Interestingly enough, I can't find a single physics text printed before 1980 that teaches the corrupt SB law.  

Radiative Transfer by Chandrasekhar
An Introduction to Radiative Transferby Peraiah
Radiative Heat Transfer by Modest

And here are a few that teach the corrupted SB law that describes backradiation.

An Introduction to Atmospheric Physics by Fleagle-Businger
Radiative Transfer in the Atmosphere and Ocean by Thomas-Stamnes
An Introduction to Atmospheric Physic by Andrews
A First Course in Atmospheric Radiation by Petty

Interesting to note who is teaching backradiation as reality and why, isn't it?


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## wirebender (Nov 7, 2011)

gslack said:


> Ian, If you poke a hole in the sidewall of your cars tire, can you place your hand a few centimeters away and force the air back in it? or is the energy from the expelling air to great to allow the air to go back whence it came?
> 
> if you point a flashlight at a mirror will it make that flashlight brighter? or will the illumination make the mirror and environment brighter?
> 
> ...



He is never going to get it because it would call his faith into question.  He proved pretty clearly above that the math is way over his head.  Showing an equation that clearly and obviously describes a one way energy transfer and claiming that it showed backradiation.

He has drunk the kookaid and is a believer.  Nothing is going to ever convince him otherwise; especially the math because he can't do the math.  He is in the same boat as rocks, and konradv et al.  He simply listens to and beleives those scientists who are politically palatable to him.  He doesn't believe the raving barking moonbats that konradv and rocks listen to, but he will believe someone who leans a little more to the right.  He obviously can't do the math and discover anything for himself so he looks for someone who he believes he can trust.  Clearly, spencer is someone who he believes he can trust but the math that spencer is using is the same math that trenberth and mann et al are using.


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## wirebender (Nov 7, 2011)

IanC said:


> interesting description there wirebender....no back conduction!
> 
> does that mean that no molecules are allowed to jiggle in the opposite direction of the heat flow? is it a physical impossibility? is it like the radiation that magically disappears (somewhere, you wont be specific) after being emitted from a CO2 molecule towards the earth? hahahaha.



You are proof of Clark's adage that any sufficiently advanced technology is indistinguishable from magic.  In this case, the math (which isn't that tough by the way) is so far over your head that you seem to be able to only refer to what it describes as "magic".  Not surprising, but really, really sad.


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## IanC (Nov 7, 2011)

wirebender said:


> IanC said:
> 
> 
> > all of the SB derivatives incorporate backradiation except the ones that deal with a single radiating body.
> ...



wow!!!

first wirebender overthrows known physics by stating photons magically disappear in open space by a method only known by him and now....he has decided that the associative law of mathematics only holds true when he want it to.

Ax(B-C) no longer equals (AxB)-(AxC)  who knew we had been mistaken for thousands of years?

sarc off

quit dodging the question. why did you pretend the calculated temperature for the heater radiating into a container at 0 degrees Kelvin was the same as the heater radiating into a container at 255 degrees Kelvin?

this is first and worst of the mistakes in your 'math'. fix this deception and we can proceed to the next mistake.

will you answer this simple question or will you duck it again?


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## IanC (Nov 7, 2011)

wirebender said:


> IanC said:
> 
> 
> > interesting description there wirebender....no back conduction!
> ...



the laws of thermodynamics deal with the net results of countless interactions. individual interactions are not held to the general rule. eg radioactivity can be calculated to great precision but you cant calculate which atom will decay


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## gslack (Nov 7, 2011)

Ian you dodged my last post to you...

YOU are the one making exceptions to the rules to suit your claim.. I asked you categorically how it is that the claims you make do not exist anywhere else in the real world, and you now claim someone else is bending the laws to suit..

Ian are you going to stop posturing anytime soon or are you going to continue being obtuse? I just want to know before going any further if you will be honest or try and grasp any further discussion... If this is going to continue to be a "im right cause I can't be wrong" discussion I am done..


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## IanC (Nov 7, 2011)

gslack said:


> Ian, If you poke a hole in the sidewall of your cars tire, can you place your hand a few centimeters away and force the air back in it? or is the energy from the expelling air to great to allow the air to go back whence it came?
> 
> if you point a flashlight at a mirror will it make that flashlight brighter? or will the illumination make the mirror and environment brighter?
> 
> ...



what am I dodging here?

if the ambient air pressure surrounding the tire was increased would that slow the flow of air out of the tire? yes. you didnt explain how a hand  away from the hole was supposed to effect the situation. why would I respond to a nonsense scenario?

again, what is your flashlight comment supposed to show? are you saying that it is impossible for a photon to bounce back from the mirror and be absorbed by the filament if it can find a molecule or atom that is in a state capable of absorbing that photon?it certainly can. thermodynamics and entrope state that you cannot make perfect mirrors so that ALL the energy is returned to the source. it does not state that NO energy can return.

accusing me of building a perpetual motion machine or a free energy machine is strictly an 'ad hominem' attack, totally concocted in your imagination with no reference to what I have said.


you have agreed that wirebender's math on Spencer's experiment is correct. perhaps you can answer the simple question that wirebender refuses to address....

why does wirebender use the calculated temperature for the heated bar in a zero degree container, in an experiment that specifies a container at 255 degrees Kelvin?







how can the term (T^4- 0^4) equal (T^4- 255^4)?


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## wirebender (Nov 7, 2011)

IanC said:


> wow!!!
> 
> first wirebender overthrows known physics by stating photons magically disappear in open space by a method only known by him and now....he has decided that the associative law of mathematics only holds true when he want it to.



You believe the laws of physics obey the laws of mathematics?  Interesting.  Clearly you don't grasp the physics when you claimed that p=sigma (T^4 - T^4) incorporated backradiation.  You are out of your depth and clearly in over your head.  I am not and have not dodged anything.  I have been as patient as I can with you and am finished with you till you can show me observable, repeatable evidence of backradiation either in a laboratory or in nature.


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## IanC (Nov 7, 2011)

wirebender said:


> IanC said:
> 
> 
> > wow!!!
> ...



in other words you are cowardly ducking my question.

you proudly proclaimed that you had done the math and it proved you right. now that I showed your math incorrect you dont want to talk about it anymore.

apparently you no longer believe in the Stefan-Boltzmann equations. at least not now that they show you to be mistaken.


you know, its not that big a deal to be wrong once in a while. I dont see why you are taking it so personally. you cribbed someone else's mathematical 'debunking' of Spencer's thought experiment and you really, really wanted it to be true. so what if it wasnt correct? does it surprise you that a noted physicist had a deeper understanding of physics that some poster named Gord?


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## wirebender (Nov 7, 2011)

IanC said:


> in other words you are cowardly ducking my question.



I am not ducking anything Ian.



IanC said:


> you proudly proclaimed that you had done the math and it proved you right. now that I showed your math incorrect you dont want to talk about it anymore.



The only thing you showed is that the math is over your head in your claim that P=sigma(T^4 - T^4) was an expression of backradiation.



IanC said:


> apparently you no longer believe in the Stefan-Boltzmann equations. at least not now that they show you to be mistaken.



That you actually believe that you have shown me to be mistaken is nothing but more evidence that you don't grasp the math.  Why do you believe that the physics texts taught p=sigma (T^4 - T^4) for a hundred years and then suddenly, when backradiation was required to support AGW alarmism, the climate science physics books started teaching P=(sigma T^4)-(sigma T^4)?

And the fact that you don't recognize that the two equations are describing very different physics is further evidence that the math is over your head.



IanC said:


> you know, its not that big a deal to be wrong once in a while.



No, it isn't.  In this case, however, I am not wrong as evidenced by your failure to demonstrate any observable, repeatable evidence of backradiation in the laboratory or nature.  You should take your own homilies to heart.




IanC said:


> I dont see why you are taking it so personally. you cribbed someone else's mathematical 'debunking' of Spencer's thought experiment and you really, really wanted it to be true. so what if it wasnt correct? does it surprise you that a noted physicist had a deeper understanding of physics that some poster named Gord?



I cribbed nothing.  I used very similar mathematics earlier which were also so far over your head that you had nothing to say when you saw them before.  Further, I am the one who pointed out your error in claiming that p=sigma (T^4 - T^4) was an expression of backradiation and had to show you what the equation looks like if you want to claim backradiation.  It is you, ian, who doesn't grasp the math.

I also find it interesting how you so often appeal to the board while I, on the other hand, talk directly to you.  You do realize don't you that appealing to others is a subconscious admission that you lack any confidence in your argument and are looking for approval from anyone whether they have a clue or not?


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## gslack (Nov 7, 2011)

IanC said:


> gslack said:
> 
> 
> > Ian, If you poke a hole in the sidewall of your cars tire, can you place your hand a few centimeters away and force the air back in it? or is the energy from the expelling air to great to allow the air to go back whence it came?
> ...



LOL, you posturing phony! 

I asked you:

_"Ian, If you poke a hole in the sidewall of your cars tire, can you place your hand a few centimeters away and force the air back in it? or is the energy from the expelling air to great to allow the air to go back whence it came?"_

You responded with:

_*"if the ambient air pressure surrounding the tire was increased would that slow the flow of air out of the tire? yes.* you didnt explain how a hand  away from the hole was supposed to effect the situation."_

Right there the underlined part you blatantly twisted my question. Why? If it is so nonsensical why not answer it as it was? Thats been your MO since this thing began, and even from the previous arguments you did this.. If its such a nonsensical question why did you have to alter it and answer your own question?  Its simple Ian, you are a fake and a fraud and only care to save your BS identity...

The answer to my question as you well now was that the air would not go back into the tire whence it came simply by placing your hand close to it, or anything else for that matter. And why? Because the energy coming from the tire will beat the energy from the air reflecting off your hand every time. Simple yet you couldn't give an honest answer to an honest question, you had to alter the question again....

Ian you don't understand entropy any more than duality, or any other concept here. You are a fake, and a fraud and not only is painfully obvious, but very sad to watch...

Ian Spencer relied on people to be too-pig headed to look into entropy and understand it. People just like you...


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## IanC (Nov 7, 2011)

how _desparate_ you must be to be reduced to arguing that rearranging terms in a formula is making a case. 

you still havent answered the question. if your math is correct, why do you get the same temperature for the heater whether it is radiating into a zero degree container or a 255 degree container? simple question, asked many times, always ducked.

do you need the formula? you have posted it many times but I can repost it for you if you want. its the same one that you complain about if someone rearranges the terms. you must have had a rough time in math and calculus if you are uncomfortable with rearranging terms.

will you answer my question this time or are you going to duck again?


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## IanC (Nov 7, 2011)

gslack said:


> IanC said:
> 
> 
> > gslack said:
> ...



gslack--how does having your hand in a specific place force air to do anything? explain the concept that your ridiculous example is supposed to be illustrating.


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## gslack (Nov 7, 2011)

IanC said:


> *how desparate you must be to be reduced to arguing that rearranging terms in a formula is making a case*.
> 
> you still havent answered the question. if your math is correct, why do you get the same temperature for the heater whether it is radiating into a zero degree container or a 255 degree container? simple question, asked many times, always ducked.
> 
> ...



Ian you seriously do not get the difference in his and yours?



All of your posturing, all of your BS, all of your googled terms and big talk and it can all be boiled down to that statement I made bold...

LOL, okay Ian. Mr physics..


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## gslack (Nov 7, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC said:
> ...





Im done Ian, you are an idiot, you think linearly and one dimensional.. Ian you I think you are deliberately trying to be obtuse now, you know full and well the implications of energy in my question, just as you know full and wll why you cannot answer it honestly.. Just like you couldn't answer my question about the flashlight honestly or any other question you have been asked... If you answer them honestly you know full and well what that means in your argument.. It means the laws are being bent to support your claim...

Ian you aren't the physics expert you have pretended to be here, I know it, wire knows it, and so do you.. You had a long run of bullshitting with googled terms and pretense, but you just like many other internet fakes got full of yourself and forgot it was bullshit... You bought into your online character and overstepped you bullshit safe zone.. We are now into areas you cannot bullshit or fake..

If you knew the math it would show, just as it shows you don't. And if you actually understood the concepts, again it would show. But so far all that shows is you do not even care to understand any of it, all you care about is saving face...

Pretty pathetic Ian..... You should be ashamed of yourself lying to people for no reason other than appear better or more knowledgeable in a simple web forum.. What possesses people to act this way I have cannot imagine...


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## IanC (Nov 7, 2011)

I see that you have no coherent explanation for your hand/ slashed tire scenario. if you cant figure out what you're trying to say why do you think anyone else should? 

and still no response from tweedledum and tweedledumber about the first mathematical mistake in wirebender's 'debunking' of Spencer. have you two conceeded? ungraciously I might add.


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## gslack (Nov 7, 2011)

Ian quit being a douchebag already.. grow up man.. You are a fake its just ignorant now..

Ian can you in the very least explain what a true "black body" is in terms of this discussion in your own words?  

No altering the question, no bullshit just answer it on your own words..

I will wait.. Any idea how long its going to take you to google an obscure answer you feel safe enough to post in here? I mean I have work to do and kids so try and not take a couple of days again...


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## IanC (Nov 8, 2011)

more attempts at deflection?

you guys asked me to prove your math wrong. I accept your challenge. back up your case or admit defeat


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## gslack (Nov 8, 2011)

IanC said:


> more attempts at delection?
> 
> you guys asked me to prove your math wrong. I accept your challenge. back up your case or admit defeat



No douchebag you claimed we were wrong, no one asked you to prove anything.. Don't you dare even try to make that lie stick you incessant little twerp.... YOU DID ALL OF THIS YOU IGNORANT FOOL!!!

No deflection except all the deflections you have done till this point.. I asked you another simple freaking question regarding basic thermodynamics and AGAIN you refuse to answer.. 

How many basics of physics can you not have any knowledge of BEFORE you come clean or at least let it rest? How many basics of physical science have been unable to explain here now Ian? 

Lets review....

1.* You didn't know anything about the double-slit experiment in either form I posted examples of, and you even make the implication they were nonsense...* They ARE one of the base experiments used in beginning physics classes all over the world and have been for many many years...  You not only didn't recognize it, you didn't even deem it a viable experiment at all.. According to whom? The rest of the physics education community or you?

2. You do not understand wave-particle duality and when asked to explain your understanding of its principles you again flatly refused to do so. In fact likening the implications to "magic" at one point.

3. You did not recognize implications in swapping the order of variables in an equation.. its an equation Ian, a mathematical formula the sequence is part and parcel to the formula.. Try baking a cake with that claim once Ian.. tell us how it works out for you...

4. You didn't recognize Fermat's last theorem... Dude seriously? Any physicist or student with any desire to find truth especially BEFORE the "proof" has in the very least spent some time trying to figure it out themselves.. jesus man..

5. You have altered our direct questions to you at every turn, rather than answering what we ask, you try and answer what you wish or want us to ask..

6. When asked even some of the most basic questions you divert, posture and preen like a peacock and google up new terms and altered versions of the original experiment all in the attempt to appear smarter or knowledgeable. And then try and claim we are the ones doing it...

7. Every time Ian does the math you cry for on here you run away for a day or more and then come back with multiple posts in an attempt to bury it under ever more complications that are not only unnecessary and do nothing to counter his math, but only show how little of this you actually do understand.

8. Now you cannot explain backradiation, or back radiation (they are indeed the same thing I have seen both in various physics sites lately) in your own words as pertains to this discussion.. The entire concept of backradiation is the very core of your claims here Ian, if you do understand what you are arguing for or against, you should in the very least be able to explain its core principle in your own words....

There are more, I am just too tired to bother with your childish ignorant ass anymore.. You are a fake, and whats worse a fake who believes his own bullshit online identity now... Pathetic, truly pathetic... You have gone from a bullshit "expert" to an obvious fake and all because you bought into your own lies so much and for so long you forgot where the lies stopped and truth began...

Nice work phony! Take a bow!


----------



## wirebender (Nov 8, 2011)

IanC said:


> how _desparate_ you must be to be reduced to arguing that rearranging terms in a formula is making a case.



Ian, talking to you is like talking to a slightly retarded child.  You have to have every single thing explained to you in detail and even after you get the explanation, you still don't get it.  

You might start by showing me in the physics literature where it is permissable to randomly apply the distributive property (or any property for that matter) of algebra for phisics problems.  To the best of my knowledge, in physics, if you are going to apply an algebraic property then the property must be assigned a physical meaning.  In this case it hasn't or if it has, then feel free to point me to the text in the physics literature in which it has.  If you had ever taken physics, you would know that you can't simply apply random mathematical properties to equations dealing with physical laws and then say look, they get the same answer so they must be the same.  It doesn't work that way in physics.  

When you apply the distributive property to the SB law, you describe an entirely different set of physics.  The SB law is expressed as p=sigma (T^4 - T^4) which describes an emitter at one temperature emitting into a background at a lower temperature.  

When you apply the distributive property to the equation, you are applying the SB law twice which describes a very different thing.  By using the distributive property you are describing (sigma T^4) the emitter at its temperature radiating into a background at a lower temperature minus the background (sigma T^4) which, according to SB is also radiating into a background at a lower temperature.  Obviously this isn't what is happening.  It is a violation of the 2nd law of thermodynamics.  You have the background which is at the lower temperature radiating back to the emitter but to do so you must apply the SB law a second time and claim that now the emitter is at the lower temperature even though we both know that the emitter is at a higher temperature.  The answer is the same but one is describing a phyisical process which is observable and repeatable and the other describes a fictional process (backradiation) invented to support AGW alarmism which is not observable and exists neither in nature nor in the laboratory.

I am sorry Ian, but you have made it more than clear that you simply don't posess the requsite math skills, analytical skills, or knowledge of the physical laws to adequately discuss this topic.

If you believe there exists observable, repeatable proof of backradiation, then present it.  Failing to do that loses you the argument as my argument stands.



IanC said:


> you still havent answered the question. if your math is correct, why do you get the same temperature for the heater whether it is radiating into a zero degree container or a 255 degree container? simple question, asked many times, always ducked.



I answered your question Ian.  As I have said, you lack the requsite math and analytical skills to understand what you are being told.  Did, or did not spencer state that his experiment was to represent the bars as an analog to earth sitting in space?  He said:



			
				roy spencer said:
			
		

> One way to demonstrate the concept is with the following thought experiment, which I will model roughly after the Earth suspended in the cold of outer space.



Now Ian, do you think that spencer doesn't realize that space is represented at 0 K or do you think that perhaps spencer doesn't know that space isn't at 0 F?  He said cold space and that being the case I am supposing that setting the temp at 0 F was an oversight on his part.

I have answered your question Ian, ad nauseum.  Again, sorry that you lack the requsite skills to even realize that you have been answered, much less understand the answer you were given.  

When you can show me in the physics literature that it is permissable to apply algebraic properties to physics problems without having first assigned a physical meaning to the property, or you can show observable, repeatable evidence of backradiation, we will have something to talk about.  Till that time, you are wrong, and you have been proven wrong whether or not you posess the math skills necessary to realize it.

Further posts on your part will be answered with the paragraph above.  You know and I know, (or maybe you really don't know) that no such carpet permission exists in physics to apply algebraic properties in physics and by now it is obvious that you know that no evidence of backradiation exists or you would have already posted a link to it.


----------



## gslack (Nov 8, 2011)

I wish this board software allowed the use of mathematical code in posts. It would make writing equations out a breeze. you could simply place <code> brackets and then use {%RHO} or {%EPSILON} and it would show as P or E.. 

Trying to show correct mathematical symbols in standard text using a keyboard is a pain in the ass.. I appreciate what ya did, because I do understand what a nuisance it is and what kind of crap idiots like Ian can try and imply from it.. Actual greek symbols, not the closest fit as written on here would eliminate a lot of his excuse making and wiggle room. He knew this and thats why he keeps asking for "your math" again and again. he's hoping to appear the victor by making it too much a pain in the ass to do it again and again..

BTW, I don't know if you are aware but (check your messages in about 2 minutes)


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## wirebender (Nov 8, 2011)

gslack said:


> I wish this board software allowed the use of mathematical code in posts. It would make writing equations out a breeze. you could simply place <code> brackets and then use {%RHO} or {%EPSILON} and it would show as P or E..
> 
> Trying to show correct mathematical symbols in standard text using a keyboard is a pain in the ass.. I appreciate what ya did, because I do understand what a nuisance it is and what kind of crap idiots like Ian can try and imply from it.. Actual greek symbols, not the closest fit as written on here would eliminate a lot of his excuse making and wiggle room. He knew this and thats why he keeps asking for "your math" again and again. he's hoping to appear the victor by making it too much a pain in the ass to do it again and again..
> 
> BTW, I don't know if you are aware but (check your messages in about 2 minutes)



It wouldn't matter if he and I were at a blackboard and had 50 math and physics texts to reference.  He isn't working from a position of math or physics.  He has proven beyond any reasonable doubt that the math and physics are beyond him already with his repeated very basic errors.  He is operating from a position of faith and simply will not see anything that challenges the faith he has placed in those climate scientists he trusts (spencer).  

He also gives his lack of confidence away in his constant appeals to the board rather than direct conversation with us.  He regularly appeals to the board for some support when he doesn't know what he his talking about.


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## IanC (Nov 8, 2011)

gslack said:


> IanC said:
> 
> 
> > more attempts at delection?
> ...



1. only in your head do I not know about duality paradoxes. when I brought up a different paradox that was pertinent to the conversation you ignored it and claimed some experiment with mechanical waves in oil was important although pretty much everyone besides you knows that light waves and mechanical waves are fundementally different. 

2. the only magic I have mentioned is wirebender's supposed explanation of how IR photons magically disappear between the CO2 molecule and the earth's surface. actually it wasnt an explanation it was only an assertion because he refused to show where or how it happened.

3. mathematical formulae are rearranged to solve the equation for the unknown quantity. some higher order functions that use matrices and vectors need to be done in a specific order but the S-B equation we are discussing is not one of them.

4. Fermat's equation is a fine historic question that took hundreds of years to solve. but it only deals with Number Theory and has no relevance to physics that I know of. my favourite historical question relates to the density of prime numbers. does the value converge to a specific number and is that number in the same family as _pi_ or _e_. interesting

5. usually when I rephrase your questions it is to bring the conversation back to the topic at hand. you lay down red herrings that I refuse to follow whereas you and wirebender refuse to answer specific, defined questions that are integral to the problem being discussed.

6. actually I google very little. perhaps I should put up more links. I suppose you confuse  my describing the physical aspects of a problem as preening but I find the only way to increase my understanding is by actually thinking about the problem rather than just proclaiming your position and doggedly refusing to entertain any further discussion.

7. I assume you meant to say wirebender rather than Ian. I respond when I have the time and have something to say. I prefer not to lay down page after page of _ad hominems_ while adding little to the discussion.

8. I will continue to use the term back radiation simply because it pisses off the two of you. I dont really care what you want to call it. but the radiation going in the opposite direction of the net flow of energy is there. that is why it is taken into account when solving the S-B equations. no amount of hysteria on your part changes the reality of both objects radiating.


when are you guys going to own up to your mistake of using zero degrees Kelvin as the temperature instead of 255K when the thought experiment specifically defined the temperature of the container? I want to move on to the next mistake.


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## IanC (Nov 8, 2011)

wirebender said:


> IanC said:
> 
> 
> > how _desparate_ you must be to be reduced to arguing that rearranging terms in a formula is making a case.
> ...



does all this dancing around the issue mean that you are finally admitting that you used the wrong temperature for the container? the thought experiment specifically defined it as zero degrees farenheit (255K) and you used (0K).  it really is that simple. answer yes or no.

I want to get on to the next mistake but it is impossible until you fix the first mistake and correct your figures downhill from the first mistake. will you correct your figures or not?

if you refuse to fix your math then at least be man enough to admit that you have not proved Spencer wrong. show some integrity, you have been caught making a simple mistake. own up to it and move on


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## gslack (Nov 8, 2011)

More bullshit how surprising....


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## IanC (Nov 8, 2011)

gslack said:


> More bullshit how surprising....



as usual, _ad homs_ are your favourite response. 

when are you going to admit that you guys made a mistake at the very beginning of the 'math' that supposedly debunked Spencer.

I responded to your eight points (of mostly ad homeninum) and yet you refuse to respond to my specific question asked eight times (or more).


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## polarbear (Nov 8, 2011)

IanC said:


> gslack said:
> 
> 
> > More bullshit how surprising....
> ...



Hi IanC, I`m trying not to take "sides" in what started out to be a rather amusing science/technical argument, but has rapidly become an emotionally charged argument.

Remember,...some months ago we had the same discussion about 'back radiation"...?

I`ll give You my take (again) , *but please don`t take it the wrong way* when I tell You right up front that a so called *"energy budget"* and well what should we call it ...?...is not  a "temperature budget".

But in "climatology" they still get away claiming just that !
One of the examples cited here by You was Roy W. Spencer`s  publication.


> *Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still*


And I`ld like to point out a few things in the assertions he,...and the entire "climatology" community for that matter continue  to make
concerning "back-radiation" , energy & increased temperature :

Roy fails to distinguish in his "experiment" between reflectivity  end emissitivity.
Fact of the matter is, that the higher the emissitivity the lower the reflectivity.

here is a Wiki quote concerning the difference between these  rather simple principles:
Heat transfer - Wikipedia, the free encyclopedia

Note to "OldRocks" & his flock...:
Before I get laughed at for using words like "emissitivity" please direct the ridicule @ Wiki & not me...



> The radiation of such perfect emitters is called black-body radiation. The ratio of any body's emission relative to that of a black body is the body's emissivity, so that a black body has an emissivity of unity. Absorptivity, reflectivity,  and emissivity of all bodies are dependent on the wavelength of the  radiation. The temperature determines the wavelength distribution of the  electromagnetic radiation


The better the ability of  a material is to absorb heat, the  better  of an emitter it will be ( black body)
and since any body which  has a higher emissitivity has a  reflectivity of  1-emissitivity is is a poorer reflector, not even Spencer can have
it both ways *as he is* using it to make his his so called  "experiment" plausible.


In his "experiment" which  he never did carry out,  in other words it is only an example 
with no actual observations or measurements  he (Spencer) states :
"Again, the reason the heated plate became even hotter is that the second plate  has, in effect, &#8220;insulated&#8221; the first plate from its cold surroundings, keeping  it warmer than if the second plate was not there. "


Since he put the word INSULATE inside quotes ...as in  "insulated" ...his rhetoric side stepped how  different the heat transfer heat mechanism in which "insulation" plays a role is from  the radiative heat (budget)  he wants to con You into.
Heat or thermal insulation has nothing to do with radiative heat transfer whatsoever, but has everything to do with heat flow.


*But in his vacuum chamber example heat flow (which an insulator impedes) is  ruled out.*
  Yet to  make his case he is  using the words "*insulated* ....from it`s cold surroundings...keeping it warmer as if the second plate was not there"
Then he goes on to use this assertion  that the second plate if it were improved to  surround  the heat emitting plate as an "insulator"...would raise the temperature even more..
*Then  he continues with an entirely different concept, namely  heat radiation   to make the claim  that  the temperature of the first plate would therefore increase..*.
*in his vacuum chamber example....in which an insulator and how that affects  heat transfer is not even an issue.*

*But the only way the second so called colder plate could  alter the energy budget of  the heated plate in his example , is by
how good of a reflector the second , the so called colder plate is.*

If  it is not a good reflector, then it may as well not even exist.
If it is a good reflector, then  it can not be a good absorber at the same time...
that would violate not just one, but an entire hierarchy of Physics laws.

And with the black body  so called "back radiation" he is also shit out of luck...
since a black body  absorbs  radiation but can`t  reflect it  back at the same
frequency,...(_*only at a lower frequency = colder.!!!*_)  since it certainly is not a reflector.

This secondary radiation, 1.rst of all does not just go inward as a reflector would do it,
but radiates Infrared  at a much longer wavelength,   in all directions and diminishing with the square of the distance.

So this "insulation" (effect) as he puts it one way first to borrow the effect he needs so badly,...he    then goes on  to use that as a "back-radiation" effect
With this trick he is trying to wiggle off the facts  that  the longer the wavelength, the  lower  the temperature.

Yes the overall *ENERGY*  of  what a black body radiates  out is the same as how much *energy*
it absorbed, but  it  does so at a much lower *TEMPERATURE*...since that`s happening at
a l*ower frequency,* ...there are of course many equations attesting to that, but  any thermal imaging instrumentation clearly shows that too.

In essence this "expert" has been stating, that a cooler body can heat a hotter body to a  higher temperature
by willfully or skillfully confusing the casual reader what the difference between
ENERGY and  TEMPERATURE  is.

He is using an *"energy budget*" as he calls it to make a case for a *higher temperature*...


I`ve been around the same bend with You almost a year ago and then I used the example...
to highlight the difference between (heat) *ENERGY* and  *TEMPERATURE*
that  there is no way that  the *temperature* of 1 gram of water  at 40 deg C
can be raised by even 100 kilos of  water which is at 39 degrees.
even though the (heat) *ENERGY* of the 100 kilos @ 39 degrees is 100 000 times more
than the 1 degree hotter  water.

But according to Spencer, his daughter Virginia the realtor & others,...they claim they can do so by surrounding the hotter water with colder water and thus trap more heat...imagine that


None of  the GW literature in circulation ever made the claim, that  CO2 can raise the energy ...
even they are aware how ridiculous it would be to claim that, ...*mostly because  what is generally associated with the word energy..*
Once You claim that the total energy has been raised people get out a sharp pencil and check up on Your math
  So in the end they all do say it raises the *TEMPERATURE*...and they all make their case
by  exploiting the fact, *that a lot of folks confuse energy & temperature*
along with heat flow and heat radiation....and how wave mechanics work when
You mix a lower frequency  with a higher one...also with electromagnetic waves
there is no way  to achieve a higher frequency, or as it were with Infrared  a consequently
higher temperature...








Yes You can raise the* ENERGY *with carefully controlled dimensions, symmetry and when all conditions are right to avoid wave cancellation are add overall ENERGY by mixing waves.
...*but  no way can  you raise the frequency...which = proportional to temperature in the Infrared Spectrum.... and do so with a frequency which is lower *
If You have any doubts about what I am saying here please do consult
"Wiki" what the difference between heat radiative transfer and heat conduction is..:
Heat transfer - Wikipedia, the free encyclopedia


And if  You don`t have an oscilloscope and have never seen or are not familiar with wave mechanics or with super imposed waves,this might be helpful...: 
Superposition of Waves






I hate to put it in these terms, but the examples like the one with the hand near the tire with a leak were in fact legitimate examples, although I would have used something different, like :

The Spencer "experiment" which he never actually did carry out concludes in effect, that a star can get hotter if it has... or more exactly phrased because it has some planets around it...and would be somewhat "cooler" if there were no other (cooler) "bodies" in the vicinity...

I`ld like to witness the roar of laughter that assertion would cause in the Astrophysics community...
I`m quite certain that Spencer`s so called experiment and conclusions would not be accepted  no matter how many other non-denier- "scientists" agree,...including his daughter who as he said has no clue about physics agree with him..:



> *Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still*
> 
> Even my oldest daughter, a realtor who has an aversion to things scientific, got  the right answer when I used this example on her.


*Can`t You see how absurd the "Spencer experiment" or for that matter this so called GW "energy budget" is...????????

*I`ll give You another example...You might be too young tube familiar with the vacuum tubes we had before semi-conductor technology.
However what was called a "triode tube" was exactly what Spencer is trying to conjure up in your mind.
There was an (electron) emittor, just like in Spencer`s example..it was a glow-filament typically running on 5 Volts ac & about 500 milli-amps depending on it`s purpose. This was surrounded by the "grid" which modulated the electron flow from the heated emitter to the anode.

See if You can find somebody else who still knows about these vacuum tubes....*I`m sure he would also tell You, that Spencer is full of it* and tell You that a "triode" with the same heated cathode and the grid that surrounded it got hotter than what was used then as a rectifier tube with the same "heater" ratings...which has no "colder" grid surrounding the heated filament...


And if You really want to find out the truth, it`s not all that hard to find an amateur radio buff who collects vintage equipment...and then You can see for Yourself...!!!!!!!
*P.S. back then there were even "pentodes" with 5 grids and they also did not get any hotter than a triode or a rectifier tube with the same ratings and currents.*...as Spencer and his agreeing idiot daughter "Virginia" would have it
Touching these tubes & feeling the temperature worked almost as well as the "tube testers" they had in any drug-store when TV`s still worked with vacuum tube technology...so anyone who is familiar with these would raise eyebrows when he reads that bullshit crap that Spencer is claiming...
but hey....I know...Spencers "example" sounds a lot more impressive than my drug-store vacuum tube examples:


> Imagine a heated plate in a cooled vacuum chamber, as in the first illustration,  below. These chambers are used to test *instruments and satellites that will be  flown in space.*


And with the likes of "OldRocks" that really matters...


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## wirebender (Nov 8, 2011)

Thanks polarbear, you are clearly more well versed in the subject than I and I doubt that I could ever have explained where Ian was missing the boat as well given 50 more pages.  Of course, he isn't going to believe you either.

Tell me something, in your opinion, am I on the right track with the two versions of the SB law.  Does the original indeed describe a one way energy flow from the blackbody to the background and the other with the distributive property applied describe a one way energy flow from the blackbody to the background minus a one way energy flow in which the background is called the blackbody and the actual blackbody is called the background?


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## IanC (Nov 8, 2011)

hey polar bear, good to hear from you again.

there are lots of oversimplifications in Spencer's example and I have no problem with people pointing them out. 

but it seems obvious to me that every time you raise the ambient temperature, or reduce the radiating effectiveness of the body, then the original heat source must operate at a higher temperature to pump the power across the temperature differential. its as simple as that. the input and output remain the same, the individual temperatures of radiating faces change. a point or sphere is the most effective shape for a radiator. every deformation of shape or addition of mass between the radiator and the radiation acceptor decreases the efficiency of energy shedding, therefore neccessitating an increase in the temperature of the heat source to compensate. in the case of adding mass, the complexity of the calculations go up tremendously because of the formation of a heat reservoir and the addition of conduction into the mix. while the input and output of power remain the same, temperature differentials form, inefficiencies form, heater temp goes up.


----------



## polarbear (Nov 8, 2011)

IanC said:


> hey polar bear, good to hear from you again.
> 
> there are lots of oversimplifications in Spencer's example and I have no problem with people pointing them out.


 1.) @wirebender...I have not had a chance yet to read the whole thread  who said what ...but I`m going to...this is a very interesting  discussion...and I`ll try answer that to the best of my ability...

 I`ll start out by saying that in Physics text  can be expressed as equations, and  anyone who truly understands an equation has no trouble to substitute  the equation with words..., that anybody can understand..
That is especially true with all the German Physics literature...it is a lot easier to understand than reading the same statements in English.
Many therefore prefer to study Physics and certain Math in German although that is not their mother tongue.

 I could do this either way, in Text form or with equations...I do have all the necessary symbols in my alternate keyboard tables...which do You prefer..??

 back to You Ian
 Great, I was quite sure that You had no problem with that, but my  problem regarding Spencer is not so much the simplification, as the  conclusion he draws from it.
 As a matter of fact in experimental physics, the simpler You keep the  experiment the better, as long as the simple experiment is not used  afterwards to walk over and railroad parameters that the simple  experiment did not include.
 I`m not quite sure what You mean by:


> but it seems obvious to me that every time you raise the ambient  temperature, or reduce the radiating effectiveness of the body, then the  original heat source must operate at a higher temperature to pump the  power across the temperature differential.


 So I have no choice to make assumptions what You meant, and if I assumed  wrong, and there is more than one way I could do that,... don`t worry,  come back  and re-elaborate,...I promise I won`t link You to my cussing  elephant URL in retaliation.

   It`s not that obvious to me how  I should understand *exactly*  what You are saying with the multiple statements in this sentence ...the color coding is what I do to aid me when I am trying to sort out things like that..
 So let`s look at the red first, I have a problem with the word "must" here, and we have to  agree on some common terminology especially settle which is the effect  and which was the cause.
 Surely You intended the original heat source  as the cause.
 Well if is the cause then the word "must" has to be stricken. In physics  the "cause" must not anything...the only thing that "must" follow  is the  effect that`s being caused, and that would be to raise the ambient  temperature.

 So are You trying to address how this effect may be impeded with the words "reduce the radiating effectiveness of the body" in the same sentence..?


 I am trying as hard as I can not to put words in Your mouth, that You did not intend to say, 
 So now the "reduced radiating effectiveness"  ...in the context with the red and the blue part of this sentence, is  that  a second variable You introduced as a cause with an effect on the  original heat source?

 You did say:
 "or reduce the radiating effectiveness of the body, then the  original heat source must operate at a higher temperature to pump the  power across the temperature differential"

 We have to clear a few things up here and the first thing is the word  choice & the "pump" word analogy. If You employ these analogies with  a body that radiates heat You will fall in a trap of totally erroneous  thinking how heat RADIATION really works....and that will be the exact  same trap, that Spencer has lured not just You, but many others   into as well...and that is to state that if another object is placed  next to the body that RADIATES heat, that now somehow the "radiating effectiveness of the body"  as You put it has been changed.
 And now it ( *the effect*) should be " obvious to me that every time you raise the ambient  temperature," .....

 Is that what You are asking me?
 If so, then as before my answer is no...again.
 No matter what You place next to the radiating body, it has no bearing on the energy rate or the temperature at which that body *radiates *energy....
The only parameter that may be effected is not the radiating body, but the spread of the radiation if you place an obstruction into the field..
and in no way will that cause an overall temperature increase


 If You place a *reflector *next to this body...and remember what a reflector is...!...The less electromagnetic radiation a material *absorbs* the better of a reflector it is...and yes a good *reflector *will reflect heat radiation back to the radiating body at the *same frequency and consequently at the same temperature at which the radiating body has sent it out...*

 Will that raise the temperature of the *radiating* body...?
 No it won`t, it can`t ...and  how could it do a silly thing like that...

 It`s no accident that Spencer resorted to the word "insulate" with his  example because now we have left the realm of heat radiation and entered  into heat conduction.
 If You *insulate* a heat source then indeed the *heat flow* is impeded.
 But then that would also have the opposite effect on what You called "the ambient  temperature"
 In that case "*ambient*"  would be whatever *is not an integral part of the heat source and the ambient would receive less heat .*

 The heat source may or may not increase in temperature...that all depends what Your choice of heat source was in the example.

 I quote Spencer now:


> Let&#8217;s heat the plate continuously with electricity.


 Yes very convenient and also very crafty.
 Because that is a heat source where the temperature of the *heat source* will go up if you *INSULATE* it...but even a housewife wearing oven mittens knows, that if you surround such a heat source with a heat *INSULATOR *that the ambient, (her fingers, which are not an integral part of the heat source ) don`t get as hot as they would without the heat *INSULATOR*.



Holding Spencers EXACT words up against him, I would be entitled to accuse him of having said that in so many words in his example that the heat source + fingers ... in the long run *would get hotte*r _*as they would without this second "shielding" body*_,...In essence that is exactly the claim made how CO2 has supposedly raised our planet`s temperature..but I`m not as much into hair splitting as "climatologists" are with their critics..Would I, then I would get a lecture on the finer nuances  between heat conduction and heat radiation..
But in the Spencer example these 2 totally different concepts are amalgamated by skilfully  using ambiguities just as a politician rather than a scientist would do it..like placing quotation marks around the word "insulate". 

I`ll ask You this question now;

If You put a cardboard in front of Your car radiator and the engine overheated...would  you believe Spencer if he tells you, that since your motor + what was in the radiator got hotter, that the air around the radiator also got hotter than it would have without the cardboard..????


Or would You rather believe an engineer, any of which I`m certain would tell You, that the cardboard reduced the amount of heat the radiator transferred to the air was to little...Had I said "plugged" furnace air filter instead of radiator, You`ld have no trouble figuring out why the air did in fact *did not get warmer*


 It gets really obvious once You walk away from Spencer`s electric hot plate and I am sure that it was no accident why he chose
that as a heat source in his example
 You can electrically heat a tiny filament wire, like in a flash light bulb to white  hot temperatures...with a thin Wolfram Metal wire that would be >  1000 deg Celsius...and only feed in energy at a rate of a few hundred  milli-watts.

 Or in a thicker wire filament, like inside a cloth dryer with as much as  500 or more watts get around 450 deg Celsius...depending at what rate  the heater can dissipate the heat...
 You could possibly surround it in a parking lot with as many "colder" cars as You want...the heater wires won`t get any hotter...
 However if you wrap it with asbestos or ceramic *insulation*, yes the wire will get up to temperatures hot enough to melt.
 But lets assume You limit the amount of current to a point just below  where the wire would melt and keep it there...and observe the  temperature at the surface of the insulator...
 If You leave all other parameters the same...long enough...then the  insulator will eventually be at the same maximum temperature the heater  could achieve with these current settings in *a vacuum,*...and would  radiate the * exact same thermal signature it would have**without the insulator * with temperature calibrated imaging...

 So now we got that out of the way, I hope..

 That leaves us and led us to the hugest discrepancy in Spencer`s "logic"...
 And that was projecting what he believes might happen when as he said "Let&#8217;s heat the plate continuously with electricity. "...
 and then use what he thought would happen...because he never even  bothered trying it and afterwards uses the parameters of this example to  make silly statements like :


> Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still


 by asserting that his surrounded electric hot-plate heat loss rate theory can be applied to :


> reduce the radiating effectiveness of the body, then the  original heat source must operate at a higher temperature to pump the  power across the temperature differential


 I have no idea why so many people who should know better get so mystified by Infrared Heat radiation,...
 The laws governing this process is in no way different from any other electromagnetic radiation.
 The higher the frequency the higher is the quantitative energy of a wave..
 I don`t care what You place in and around any such radiation source, be  that Ultraviolet, visible or infrared, there is no way You can change  the wavelength of the source just by parking another object next to it.
 Or in other words *cause* "then the  original heat source must operate at a higher temperature"...Then  this effect should also be observed at other wavelength and higher  frequencies...like Your car headlights are turning blue because a train  went by as you stopped at the crossing.

 I know *EXACTLY* what kind of bullshit Spencer has said...he leaves no other interpretation...

 But I can`t say I`m *exactly *sure what to make out of:



> but it seems obvious to me that every time you raise the ambient  temperature, or reduce the radiating effectiveness of the body, then the  original heat source must operate at a higher temperature to pump the  power across the temperature differential.


 If I got it wrong...well nobody is perfect...and don`t forget my primary  language is German, not English... and the German language is quite  exact,leaves really no  room for ambiguities and as a consequence it can  only be interpreted one way...much like mathematical equations...
 I`m not as good in guessing when I have to guess what people meant to say  

 My wife is ...!!!...and always expects the same near psychic abilities from me.
 Guessing wrong always  gets me into a lot of shit...

 Like she would say "honey aren`t You thirsty ?"
 "No" I replied and drove right by a truck stop...
 See how I fucked up...?
 What she really meant to tell me as politely as possible that she wanted  me to pull in at the "Flying J" and buy her a diet coke...

P.s.:
There is one more thing You said, that I feel compelled to address before I hit the hay..:


> a point or sphere is the most effective shape for a radiator. every  deformation of shape or addition of mass between the radiator and the  radiation acceptor decreases the efficiency of energy shedding,  therefore neccessitating an increase in the temperature of the heat  source to compensate.


No, a sphere is actually not so good...it has the least surface area per mass of any other 3-d geometry shape You care to choose and with a "radiator" You want the exact opposite...You want as much surface area as possible...the rougher the better...
And that applies not only just for heat exchangers and heat conduction, but also for heat radiation....
That has been well tested and tried...ask any engineer

And last not least but actually the most important part of what You wrote was:


> in the case of adding mass, the complexity of the calculations go up  tremendously because of the formation of a heat reservoir and the  addition of conduction into the mix


And I whole heartedly agree with You on that one..
So now let`s go back a few months ...I`ll try and find the post later, where I showed that on any graph that "climatologists" have published so far , they 
show a _*simple milk-maid math linear relation ship*_ between the ppm CO2 (the absorber) and the ambient temperature..
 a LINAER RELATIONSHIP, ....while at least one of the prime functions, the % absorption CO2 exhibits are already logarithmic functions !!!!
Even if there was a relationship between ppm CO2 as the cause and observed temperature increase as effect, that You would have to go down to at least the 3rd or 4th derivative function to get a linear correlation once mass & convective heat transfer is incorporated...
here again "climatologists" themselves don`t either seem to know the difference between absorbance and absorption or are secure in the knowledge that no decision maker at the political level knows the difference.

In parting company for today I would like to suggest You carry out an experiment which reveals a lot more than Spencer`s so called example which he never even bothered trying...because I know had he tried it he would have never published this crap.

It`s very easy to find 2 light bulb versions of the same outside dimensions and wattage,...
use one version with clear glass and as a comparison the same make bulb but with a milky glass.
Turn them both on and compare the glass temperature...
According to Spencer and "climate science" there better be a huge difference !


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## gslack (Nov 8, 2011)

Don't know if this was already touched upon previously, but I have a problem with the term "black body/blackbody) in reference to earth. The earth is not a blackbody by either absorpivity or emissivity . Its a graybody, and to me this makes the entire contention of greenhouse theory as its claimed dubious.. 

I don't know but it would seem to me any use of complex mathematical principles or physical laws has to start with accurate descriptive terms...


----------



## wirebender (Nov 9, 2011)

gslack said:


> Don't know if this was already touched upon previously, but I have a problem with the term "black body/blackbody) in reference to earth. The earth is not a blackbody by either absorpivity or emissivity . Its a graybody, and to me this makes the entire contention of greenhouse theory as its claimed dubious..
> 
> I don't know but it would seem to me any use of complex mathematical principles or physical laws has to start with accurate descriptive terms...



True.  If you ask me, right there is where the whole climate science model of the earth's energy budget goes off the rails.  trenberth treats the earth as a blackbody when he divides the incoming energy by 4 and in effect, turns the earth into a flat plate receiving energy 24 hours a day at a rate of 1/4 of what it actually receives.  You can divide energy by 4 with a blackbody (star for example) and represent it pretty accurately as a flat disk because no matter where you look at a star from, it appears roughly the same.  The earth can't be accurately represented as a flat disk because it isn't a 3D self illuminating body.  It is a 3D body being illuminated across 180 degrees of its surface at any given time and the amount of radiation it is receiving across that 180 degrees varies because it is a sphere.  Across that 180 degrees along any lattitude, there is only a short time where it receives the maximum energy from the sun.  The rest of the day, the energy is either increasing or decreasing, and you must also account for increasing and decreasing amounts of energy at any given time along the lines of longitude.

trenberth's energy budget doesn't represent anything like the actual conditions found on earth and therefore doesn't describe anything like the energy budget of the earth and it is not an acceptable excuse to use it because climate science has produced nothing better.


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## wirebender (Nov 9, 2011)

polarbear said:


> 1.) @wirebender...I have not had a chance yet to read the whole thread  who said what ...but I`m going to...this is a very interesting  discussion...and I`ll try answer that to the best of my ability...
> 
> I`ll start out by saying that in Physics text  can be expressed as equations, and  anyone who truly understands an equation has no trouble to substitute  the equation with words..., that anybody can understand..
> That is especially true with all the German Physics literature...it is a lot easier to understand than reading the same statements in English.
> ...



Thanks.  Given a choice, I would prefer an explanation in words.  I took a good deal of math, chemistry and physics in college, but when I decided to go into medicine, most of that fell by the wayside.  I can do the math, but it is an endeavor every time I do.  I don't make my day to day living with the math; I bend wires (orthodontics).

I do remember though from physics class that we weren't allowed to just use algebraic properties with physics problems because we found the problem easier to work out as a result of applying the property.  When you alter the equation, even when the answer is the same, you have altered the physical meaning of what is happening.  

Specifically, the claim that  p=sigma (T^4 - T^4) represents the same physical processes as  P=(sigma T^4)-(sigma T^4).  To my understanding, the first represents the SB law as published by SB, that being a blackbody at a warmer temperature radiating into a colder background (no backradiation); whle the second represents the net difference between a blackbody at a warmer temperature radiating into a colder background and a colder background which is represented to be a blackbody radiating into a warmer blackbody which is represented as a colder background.  The answer is the same, but the physical processes they describe seem to me to be very different.  One representing reality and the other representing a means of promoting AGW alarmism.  

If you don't have backradiation, you don't have AGW alarmism.


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## gslack (Nov 9, 2011)

wirebender said:


> gslack said:
> 
> 
> > Don't know if this was already touched upon previously, but I have a problem with the term "black body/blackbody) in reference to earth. The earth is not a blackbody by either absorpivity or emissivity . Its a graybody, and to me this makes the entire contention of greenhouse theory as its claimed dubious..
> ...



Agreed and as you stated, energy (light) the earth receives from the sun is almost always in a state of increasing (heading towards noon or day) or decreasing (heading towards night). Also the curvature of the planet allows for far more energy to "slip off" (for lack of a better term) the atmospheric surface the closer it gets to the poles. Also the electro-magnetic field of the Earth is often treated as little consequence in climate research, as well as our distance from the sun at any point in time..

You take those obvious problems, add them to others that are even less understood or in the least less regarded; the procession of the equinoxes and our solar system tracking the galaxy itself, the fact as our sun is less active its electro-magnetic field weakens which allows more cosmic radiation into our planet, variations in our proximity to other radiating or reflecting bodies, and many other things the "science" seems to be able to disregard when it sees fit.

If i were to go at my job with the same single-mindedness and already set culprit of climate science, I would not have been doing my 20 years now thats for sure. I would most likely have been standing tall in proceedings for it..


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## wirebender (Nov 9, 2011)

I believe the day is coming where climate science (or at least its practitioners and financial supporters) are going to become responsible in a monetary way for their predictions of woe.  Vast quantities of money are being spent based on thier very flawed science and eventually, someone is gonig to win a lawsuit against some part of the climate science establishment which will start an avalance of lawsuits based on the existing legal precedent.

It will be bad for science in the short term, but in the long term maybe it will serve as a warning to those who get the idea of promoting snake oil for fun and profit.


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## gslack (Nov 9, 2011)

wirebender said:


> I believe the day is coming where climate science (or at least its practitioners and financial supporters) are going to become responsible in a monetary way for their predictions of woe.  Vast quantities of money are being spent based on thier very flawed science and eventually, someone is gonig to win a lawsuit against some part of the climate science establishment which will start an avalance of lawsuits based on the existing legal precedent.
> 
> It will be bad for science in the short term, but in the long term maybe it will serve as a warning to those who get the idea of promoting snake oil for fun and profit.



Just to clarify the bigger problems..

Took my oldest daughter over to College of Wooster for her campus sleepover last weekend. They of course had all their majors and minors information and all kinds of info on their various popular courses all laid out in snazzy little papers and pamphlets. I was browsing them looking for classes she could sit in on the next day particularly psychology (her choice of field right now) and in the science department I found one called "climate change"..

Now you and I as well as any other person with sense realizes "climate change" is not a real course of study by any measure. Its not a real science. You aren't going to become a climatologist by taking climate change, you will not have the math to even be a meteorologist. in fact I would consider the course a waste of money in terms of its usefulness and real world application. But there it is and you can minor in it.. A minor in climate change is less useful than a minor in crocheting or basket weaving. At least in those you have a sell-able skill after.

That is what this has turned into already. They are married to this thing and no way they can gracefully bow out now without massive repercussions across the board. The UN pushed governments into this kind of half-cocked "science" through their BS "reports", sustainable development incentives and  all kinds of other programs and sanctions through themselves directly or the world bank, or any of the other agencies they have at their disposal. Now the world has colleges teaching crap like "climate change" to the masses preparing them for nothing.

Personally I think a scientist who willfully deceives masses like this should be held accountable. Holding the UN or governments accountable will amount to nothing we know this. But you remove the safety net for these scientists and things will change..


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## wirebender (Nov 9, 2011)

gslack said:


> wirebender said:
> 
> 
> > I believe the day is coming where climate science (or at least its practitioners and financial supporters) are going to become responsible in a monetary way for their predictions of woe.  Vast quantities of money are being spent based on thier very flawed science and eventually, someone is gonig to win a lawsuit against some part of the climate science establishment which will start an avalance of lawsuits based on the existing legal precedent.
> ...



Come on gslack, a degree in climate change, or even a minor would be at least as useful to earning a living as a degree in Medieval french poetry.

Personally, I belive basketweaving 1 and 2 would be more profitable in the long run.  My wife just dropped $375 for a hand made basket to set on the flagstone at the fireplace to keep pine cones in.


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## gslack (Nov 9, 2011)

wirebender said:


> gslack said:
> 
> 
> > wirebender said:
> ...



Damn, I knew I should have taken basket weaving over Tae Kwon Do. . I help teach kali/Escrima/ Arnis 2 nights a week down the road from the house, and that 90 bucks a month per student (about 20 of them IF they remember to pay) divided among 3 people and insurance, renting a place, utilities, licenses et al. is really lucrative...NOT!

I wonder if the pottery class is till taking apps at the community college...


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## wirebender (Nov 9, 2011)

gslack said:


> I wonder if the pottery class is till taking apps at the community college...



Don't even get me started on pottery.  We passed through Albuquerque a few years ago and my wife went certifiably insane over some of the native american pottery in that area.  Don't ask what that cost me.  Apparently pottery is a much more lucrative business if you happen to be native american.


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## gslack (Nov 9, 2011)

wirebender said:


> gslack said:
> 
> 
> > I wonder if the pottery class is till taking apps at the community college...
> ...



yeah i think it runs on the same mysterious principles that make my wife think that "Hecho en Mexico" is synonymous with quality. And if its at a flea market, swap meet, or an antique show its automatically a good deal even if its not...


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## IanC (Nov 9, 2011)

> but it seems obvious to me that every time you raise the ambient temperature, or reduce the radiating effectiveness of the body, then the original heat source must operate at a higher temperature to pump the power across the temperature differential.
> So I have no choice to make assumptions what You meant, and if I assumed wrong, and there is more than one way I could do that,... don`t worry, come back and re-elaborate,...I promise I won`t link You to my cussing elephant URL in retaliation.
> 
> It`s not that obvious to me how I should understand exactly what You are saying with the multiple statements in this sentence ...the color coding is what I do to aid me when I am trying to sort out things like that..
> ...



what I mean by the ambient temperature is the defined vacuum container that is kept at 0F (255K).  this was the first mistake in wirebender's 'debunking'. as the temperature of the container increases, so does the starting temperature of the heater which leads to a higher end point temperature once the 750w is applied. carried to an extreme, you could warm the container and the unplugged heater until the heater was radiating 750w. when you plug in the heater the extra 750wmust escape and the only way to do that is by increasing the temperature of the heater so that it can force energy past the (back)radiation of the container. as the ambient radiation of the container goes up there must be an increase in the operating temperature of the heater to shed the energy. is that more clear?

in a similar fashion, when you put matter between the radiating source and the side of the container where the energy escapes, you have slowed the escape of energy from the speed of light to something less. the radiation then warms that matter which in turn decreases the the temperature differential which in turn lowers the rate of energy shedding. the only way for the radiator to continue shedding the necessary amount of heat is to increase its temperature to make up for the difference.

your reference to reflectors is off topic because wirebender already defined the emmissivity as unity therefore there is no reflection.

your reference to cold water not being able to warm up warmer water is also a non sequitur. the electricity going into the heater is not limited to producing radiation of a specific wavelength and is capable of being used in a range of outcomes until disruption occurs at very high temperatures.

I must admit I lost interest in your long and rambling post so if you could stay on point and limit yourself to just a few topics when responding to me it would be appreciated.


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## gslack (Nov 9, 2011)

WOW Ian... Just wow... You are one egotistical man, and with no good reason for it.. It certainly doesn't show in your knowledge here and most assuredly not in your logic..

3 people now are in agreement to some degree on this at least, and rather than face the possibility you may be wrong, you decided to insult the calm and relaxed voice of reason.. Seriously Ian, grow up..


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## IanC (Nov 9, 2011)

gslack said:


> WOW Ian... Just wow... You are one egotistical man, and with no good reason for it.. It certainly doesn't show in your knowledge here and most assuredly not in your logic..
> 
> 3 people now are in agreement to some degree on this at least, and rather than face the possibility you may be wrong, you decided to insult the calm and relaxed voice of reason.. Seriously Ian, grow up..



polar bear is the calm and relaxed voice of reason? hahaha, even polar bear wouldnt agree with that. polar bear is a brusk, insensitive, longwinded blowhard who is smart enough and educated enough to have his own opinions and back them up. that's why I like him, or at least find him entertaining. at least for a paragraph or two, before my eyes glaze over.


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## IanC (Nov 9, 2011)

by the way gslack, you still havent responded to the question of why you support wirebender's 'math' even though he used the wrong temperature in the S-B equation. why do you think others should respond to your questions when you wont reply to other people's questions?


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## wirebender (Nov 9, 2011)

IanC said:


> what I mean by the ambient temperature is the defined vacuum container that is kept at 0F (255K).  this was the first mistake in wirebender's 'debunking'.



Wirebender made no mistakes Ian.  If you could do the math, you might be able to check but clearly you can't.  There are a few things that you apparently don't know about the SB law.  The first is that the SB law calculates the radiation emitting from any surface that is above absolute zero.  It does so even when the background is not at absolute zero.  As I said, if you could do the math, you could check the math and see for yourself that I made no mistake.


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## wirebender (Nov 9, 2011)

gslack said:


> WOW Ian... Just wow... You are one egotistical man, and with no good reason for it.. It certainly doesn't show in your knowledge here and most assuredly not in your logic..
> 
> 3 people now are in agreement to some degree on this at least, and rather than face the possibility you may be wrong, you decided to insult the calm and relaxed voice of reason.. Seriously Ian, grow up..



He is one of the faithful and there isn't much you can say to someone after they have drunk the kookaid.


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## gslack (Nov 9, 2011)

IanC said:


> by the way gslack, you still havent responded to the question of why you support wirebender's 'math' even though he used the wrong temperature in the S-B equation. why do you think others should respond to your questions when you wont reply to other people's questions?



Ian you haven't responded honestly to any of my questions, so go shit in your hat..

You spent days being a posturing imbecile and now you try and pretend you want an answer from me?  

A post or two ago I was too dumb to talk too anymore, now you want my opinion.. I have an opinion for ya, You are a douchebag and a internet fake.. how's that?


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## gslack (Nov 9, 2011)

IanC said:


> gslack said:
> 
> 
> > WOW Ian... Just wow... You are one egotistical man, and with no good reason for it.. It certainly doesn't show in your knowledge here and most assuredly not in your logic..
> ...



You really do think you are the smartest guy around don't you.... Wow its not your online persona you really do believe yourself so much smarter than everyone else... WOW!!!!!

How exactly did you come this hypothesis "oh great internet brain"? Seriously dude what gave you the impression you are so brilliant? I don't see it here, and from your lack of knowledge shown repeatedly its not in your education.. So please enlighten us all as to how you reached the conclusion you were so brilliant?


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## wirebender (Nov 9, 2011)

gslack said:


> You really do think you are the smartest guy around don't you.... Wow its not your online persona you really do believe yourself so much smarter than everyone else... WOW!!!!!
> 
> How exactly did you come this hypothesis "oh great internet brain"? Seriously dude what gave you the impression you are so brilliant? I don't see it here, and from your lack of knowledge shown repeatedly its not in your education.. So please enlighten us all as to how you reached the conclusion you were so brilliant?



Maybe he sleeps at Holiday Inn Express a lot.


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## IanC (Nov 9, 2011)

wirebender said:


> IanC said:
> 
> 
> > what I mean by the ambient temperature is the defined vacuum container that is kept at 0F (255K).  this was the first mistake in wirebender's 'debunking'.
> ...









I did do the math. you used T^4 - 0^4 for temperature radiating into space. but the experiment defined the receiving container as 255K not 0K. (T^4 -0) does not equal (T^4 - 255^4). it works out to about a ten degree difference.

why cant you just admit to making a simple mistake?


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## polarbear (Nov 9, 2011)

wirebender said:


> polarbear said:
> 
> 
> > 1.) @wirebender...I have not had a chance yet to read the whole thread  who said what ...but I`m going to...this is a very interesting  discussion...and I`ll try answer that to the best of my ability...
> ...




GW also uses back-radiation where nothing is missing from convective heat transfer, which is a massive amount...
But as You noticed they are  also using the same proportionality factor (sigma) of the ideal black body to make the case for CO2 as a black body
radiator....
a black body can absorb light energy at all wavelength, that`s why it is called a "black body"

CO2 does not even qualify as a so called "grey body"
the Bolzmann  law   quantifies "j"  as a  THEORETIC thermodynamic Temperature

"Thermo-dynamic temperature"  has the dimension of  watts per square meter AT a GIVEN Temperature and the Bolzman law states it the original German text, that a black body, which absorbed X amount of watts in the form of radiation  will re-radiate the amount (j)  again at   so many watts per
square meter...
The original Bolzmann equation therefore expresses the ENERGY FLUX DENSITY, which  is a far cry from  the ACTUAL TEMPERATURE  which has the single dimension deg K...and even more of a far cry is  to claim as
GW alarmists do,   that CO2 acts as if it ACTUALLY had been IRRADIATED by  a black body which by an actual black body that,...   as their latest claim would have it  *ACTUALLY at 288 deg Kelvin.
* 

First of all if  anyone wants to go from a flux density which is watts / m^2 at any temperature to express it as an actual temperature, this calculation cannot possibly done,  unless we also assign* on what receiving  mass* the (j) watts per square meter is acting.
If we use for an area, let`s say 1 m^2, that still leaves us with a number that expresses Watts.
Which is a POWER UNIT,...the same as calories per time is....
So to get rid of the  dimension for time let`s say we choose 1 second, now we have an energy unit as in (j) Watt seconds  or (j) calories...
But in order to be able to calculate the EFFECT, by how much the ACTUAL TEMPERATURE  (j) Watt-seconds would be raised per one second there is no way around specifying on what mass these (j)  Watt-seconds
have been exerted.

And that is the only way  ACTUAL TEMPERATURE  could be calculated from Bolzmanns THERMODYNAMIC TEMPERATURE equivalent energy flux.

And again that would assume that the other mass is absorbing  *THE ENTIRE *Bolzmann energy flux...
*And only an ideal black body could do that*

"climatology" does make this (blatant) assumption and then goes on to _run this second "black body" the CO2 yet again through the same Bolzmann equation and feed energy back to the first original radiating body_
without assigning a value for the mass yet again and try tell You, that this will now heat this radiation body to an even higher temperature as it could have achieved without the second body, which has received only a 
portion of the energy  from the first radiator ...

So they have a lot more watts (= Energy per time)  *and have it generated* by another *passive * body, yielding in total now more than  
  the original watts (Energy per time) coming in from  the sun which has fed the entire system.

I don`t care who is using what kind of rhetoric, but neither Bolzmann, Planck or any other Physisist would try tell You, that You can take X amount of Energy (watt-second) and simply by placing another object (...which does not* GENERATE *any *ADDITIONAL* energy whatsoever...)
that some how You can now have X+x /y energy  ...
or X+x /y  power as in watts or calories per second.

Then every engineer to date has been stupid...we could have squeezed X+1 times x/y  Energy out of every fuel known to man, had we only increased the mass  of the apparatus,  that burns these fuels.

Why stop with one single passive receiver- or so called "back radiator"...lets use 10 of them,...then we can apply the Bolzmann and the Planck equations not just twice as these quacks are, but 10 times...
Then we boost the energy  with 10 more "back radiation" steps to to a new X which in their logic is now  X= X + x/y +  (x/y)^2 + (x/y)^3....and so on till  + (x/y)^10 
and all we need to start with is just X watt-seconds.

This is the very same 'logic" that novices to higher math fall for,  Aristotle has used it to test the intellect of his pupils in a very similar mathematical manner with this statement:
Although the sprinter can run faster than the turtle he can never overtake the turtle...
because while he is running x amount of time, covering  x times his speed (v1) the distance d1, but the turtle has meanwhile moved  x  times its speed (v2) =d2.
So now the distance separating them is not just the original distance d-d1 but is d-d1+d2.
This will add to the time when the sprinter can overtake the turtle the time
delta t = d2/ (v1-v2)....
but after the sprinter has run for delta t, the turtle yet again has  added the "extra distance" delta t *  v2 =d3 to the distance
and so on...all the way to 1/infinity....
And that is exactly the same joke "climatology" has performed with the "extra" -"back-radiation" energy.

Yes I know in the strange world of finance "math"  this is routinely done and You can "generate"
million$  more starting with way less $$ without even spending any energy as in "work"..simply by using compound interest $$$%%%%  which like "back radiation" is not just an interest percentage of the original principal amount of $ they started with but is the principal + p% and then
doing it again  ...and now they have $$$=  principal$ +p% +  (Princ.$ + the new $+p%)* p%...and so on and on...and by repeating this step over and over again they "generated" many million$ from next to nothing
without any additional input...so they claim...but we all know by
now what really does happen with this "magic money" generator...the same as any other perpetual motion machine that have supposedly been invented...they all stall after having
consumed the original energy,...and then it`s up to us tax payers to fork over our money which we earned with real & hard work to "re-stimulate"
this magic  interest "back-radiation"  money machine , ...and are supposed to believe, that if we participate we`ll get more out of it as we put in to begin with...
Hey You could even apply the same "black body" and "back-radiation" analogy  to the way  banks should work in theory.

By the way no matter where You look but especially so at "wikipedia"...every definition of almost every law of Physics that may be applied to "climatology" has been shit and 
pissed over by "climatologists" adding their own assertions...
Example:

Stefan
*Stefan&#8211;Boltzmann law*


From Wikipedia, the free encyclopedia
"Since the *emissivity with greenhouse effect* (weighted more in the longer wavelengths where the Earth radiates) is *reduced more than the absorptivity *(weighted more in the shorter wavelengths of the Sun's radiation) is reduced, the equilibrium temperature is higher than the simple black-body calculation estimates. As a result, the Earth's actual average surface temperature is about 288 K (14 °C), which is higher than the 255 K effective temperature, and even higher than the 279 K temperature that a black body would have."

No math, no measurements, no proof, nothing more than an assertion and a very ridiculous one at that...


and that was written by  P. K. Das, 

retired Metereologist @ the U of Nairobi (retired)
in a publication ", _The Earth's Changing Climate"_ 
  before that he was a meteorologist @ the meteorological department of India...
 a bureaucrat.

The problem is, that people who`s major is not in physics will interpretate that as a valid amendment to Staphan Bolzmann`s laws
because Wikipedia lists his garbage on the same page.

No matter where You look, these people have been pissing ink by the barrel fulls all over the internet and
it would take a lifetime and the entire physics community to go through all the red tape involved with editing
invalid claims like these from Wikipedia.
These bastards are taking full advantage secure in the knowledge that most people don`t have their own
books or have access to a University library  where they could read up on Bolzmann without these  garbage attachments   disguised as amendments which "climatologists" have added almost everywhere You look.


@ IanC...
No I don`t cite  "my own opinion" as You accuse me of how I "back up my own claims"...
I do`nt need the internet for this either, but I do have shelves full of Physics & Chemistry books that I needed for my work and I do have access to my Universities Library even though I am now retired.

Come see me and I`ll show You what`s really in the book that for example publishes the original German Text, say regarding Stephan Bolzmann`s laws.
So, but when a "Meteorologist" who worked as a civil servant in India adds his crap opinion and that is now being used as a reference by all the rest of this "climatology" crowd to back their rather strange opinion You accept that as a "proven fact of science".
But hey...I don`t care, stay the course...and let these clowns "educate" You,...if You so prefer...
I doubt it though, that this way You would pass even an entry exam in Physics...maybe at some holy cow town U in India where You can tip them like a a waitress expects a tip and then get passed the exam... or maybe in "climatology"...
I passed my exams probably long before You were even born and then went on to teach Physics & Chemistry at several U`s in North America...but I never ever taught my OPINIONS to anybody.


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## IanC (Nov 9, 2011)

gslack said:


> IanC said:
> 
> 
> > gslack said:
> ...



please show where I have claimed to be brilliant. wirebender put out a call to prove his math wrong. I have and yet you two refuse to admit it.


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## IanC (Nov 9, 2011)

> @ IanC...
> No I don`t cite "my own opinion" as You accuse me of how I "back up my own claims"...
> I do`nt need the internet for this either, but I do have shelves full of Physics & Chemistry books that I needed for my work and I do have access to my Universities Library even though I am now retired.



I am too terse and you are too verbose. what I meant by your own opinions is that you synthesize your understanding of a process from basics rather than accepting them whole cloth from someone else. I like your abilty to analyze things coming from different directions rather than simply accepting authority.


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## wirebender (Nov 9, 2011)

IanC said:


> wirebender said:
> 
> 
> > IanC said:
> ...



Here is a newsflash Ian, you still haven't proven me wrong.  I doubt very seriously that you even understand what you have written, but it does not prove me wrong.


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## IanC (Nov 9, 2011)

> GW also uses back-radiation where nothing is missing from convective heat transfer, which is a massive amount...
> But as You noticed they are also using the same proportionality factor (sigma) of the ideal black body to make the case for CO2 as a black body
> radiator....
> a black body can absorb light energy at all wavelength, that`s why it is called a "black body"
> ...



I am not supporting GW or AGW except to say that extra CO2 does send some radiation back at the earth. I am not trying to quantify other than to say it is one process occuring in the atmosphere that adds to energy retention by choking off energy loss  along one pathway.

I am not supporting the specific calculations done for Trenberth's energy budget other than to say the pathways identified are indeed pathways for energy to come in and go out.

I am not claiming the earth is a perfect blackbody or that the S-B laws can be used in any fashion other than to give a general idea of what may be going on.


my dog in this fight is to prick the balloon of pomposity that wirebender carries around when he shouts 'here is my position, and the math to prove it'. but when push comes to shove he chickens out. he refuses to show the post where is math is, now that he finally provided some math that he copied from someone else to debunk Spencer he refuses to admit a simple mathematical mistake at the very beginning.

as far as Spencer's experiment goes, I have problems with it as well. and I have pointed them out. but wirebender's 'math' is completely out to lunch. 


oh, and one of your comments intrigued me. if you did shine an airplane or train light at a car headlight and waited for it to come to equilibrium are you sure that no change would occur? I think it would draw less power rather than change emmission values but it could be both.


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## wirebender (Nov 9, 2011)

IanC said:


> > @ IanC...
> > No I don`t cite "my own opinion" as You accuse me of how I "back up my own claims"...
> > I do`nt need the internet for this either, but I do have shelves full of Physics & Chemistry books that I needed for my work and I do have access to my Universities Library even though I am now retired.
> 
> ...



Spoken by the guy who blindly accepts authority.


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## gslack (Nov 9, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC said:
> ...



I never said you claimed it, or even said it.. I said you think it, and that is plain as day. it shows in your behavior and actions all too clearly. You have done nothing, I repeat nothing but talk shit and post googled terms. You pretend you know, but your behavior, manner, lack of knowledge, lack of everything except your pathetic big talk and deliberate attempts to confound everything you cannot disprove with a google search, tell us all the exact opposite..

Come on douchebag, you think are a genius and whats more THE GENIUS.. You show it in the way you respond to anyone who dares to show any mathematical ability or knowledge and express opinions or facts that may differ from your own. You sit back call them wrong and cite googled terms and definitions you do not understand while pretending some greater knowledge and they're so far beneath you.. 

You do it all the time tool, you did it with wire, then me, and now polarbear who was actually trying to clarify what he felt (by giving you the benefit of the doubt) was a simple misconception of a few of the finer points in spencers experiment. he wasn't rude, dismissive, or belittling you in any way shape or form but you had to be rude to try and cover your lack of knowledge..

What you think your act is new? HAHAHAHAHA! Dude your act is as old as web forums. Gimme a break there are so many like you on these kinds of places, every forum has at least a few of you. You talk shit and pretend to be so much above everyone else usually your types are in some kind scientific field... Coincidence? I think not.. Dr Gregg, was supposed to be a chemist and as it soon turned out he was a fraud. That was just a year ago on this very forum. We have several others trying the same act, all of it bullshit and all of it may differ slightly but still the same act.

You choose science because in your vanity you believe you can out think anyone even though you have no reason to believe it. You obviously haven't had the education for it, you most assuredly don't have the ability to think on more than one level at a time, and as far as I can see you just unjustifiably deem yourself better my right or by intellect.

You have been dismissive of everyone who dares to disagree with you. Whats that tell you Ian.. Wire, me, and now polarbear all of whom you have agreed with at times, and even shown respect for. Respect until anyone of us dare to challenge your superiority then you become a rude, arrogant, egomaniac who will do or say anything to protect his pride.. A pride which YOU have not earned. You bullshitted your way into it and did it so long you forget the reality.

Reality check Ian..

1. You are no more a physics expert or even competent student than I am a one-armed lion tamer..

2. Being the smartest guy in a room full of idiots makes you the smartest idiot, not a genius. 

3. You cannot fake real knowledge. Every time you have to go into depth or detail you will ALWAYS do or say something that a perceptive person who is not easily intimidated by titles (real, implied or imagined) or claims of brilliance you have yet to show, will call you on it. Just like I did, I tested you multiple times and you failed every single one miserably..

4. There is a little white spot on top of chicken shit, and despite its pure white appearance, that part is chicken shit too. That means no matter how much you sling or how good you think your bullshit is, its still limited by the simple fact it is bullshit, and susceptible to the weakness of its originator. Meaning YOU are the weakest link in your fake internet persona.  only bullshit requires a salesman, the truth just is and will stand on its own. You try to hard Ian, bullshitters always do.

5. You are internet phony nothing more. You play a part online, and thats the extent of your brilliance. And being weasel enough to talk in circles and dismiss everyone off-hand is not a win by any standard other than your own.

Ian truth is I would bet there are at least 50 people on this board I have talked to that can think rings around you. At least 20 with far superior applicable mathematical knowledge than you, and at least as many who can dance and posture better than you do. you aren't even a good second place in any area i can think of here. Except maybe being a total douchebag... Well no there are better ones than you at that too..


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## IanC (Nov 10, 2011)

gslack said:


> IanC said:
> 
> 
> > gslack said:
> ...



wow! what a rant!

just out of curiosity, what would make you put as much time and effort into discussing the topics at hand as you do in typing out _ad hom_ smears?


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## wirebender (Nov 10, 2011)

IanC said:


> I am not supporting GW or AGW except to say that extra CO2 does send some radiation back at the earth. I am not trying to quantify other than to say it is one process occuring in the atmosphere that adds to energy retention by choking off energy loss  along one pathway.



Of course you are Ian.  To accept backradiation is to accept AGW alarmism because without backradiation, there is no AGW alarmism.  In order to use the SB law to prove backradiation you must call CO2 a blackbody radiating to the earth and in order to do that you must call the earth a cooler background than CO2.

That is what I mean when I say that you don't even begin to understand the math or the ramifications of your claims.  You rely on your instinct and your political leanings to determine whom you should trust.



IanC said:


> I am not supporting the specific calculations done for Trenberth's energy budget other than to say the pathways identified are indeed pathways for energy to come in and go out.



Those "pathways" Ian, are EM vectors and there is a very specific set of physics and mathematics that govern operations with them and there is no "pathway" by which CO2 can radiate back to the warmer earth.



IanC said:


> my dog in this fight is to prick the balloon of pomposity that wirebender carries around when he shouts 'here is my position, and the math to prove it'. but when push comes to shove he chickens out. he refuses to show the post where is math is, now that he finally provided some math that he copied from someone else to debunk Spencer he refuses to admit a simple mathematical mistake at the very beginning.



On top of all else, now you have become a liar ian.  Congratulations.  I brought that math from spencers experiment because it effectively proved him wrong and unlike you, I recognized it as proving him wrong.  Had I done the calculations myself, the answers would have been the same.  That's the thing about math Ian, when you do it right, everyone gets the same answers.  As to the math I did earlier regarding EM fields, again, you were in the conversation and had no comment whatsoever on the math other than the sort of hit and run snide remarks that rocks makes about things he doesn't begin to understand.  If you are interested, I suppose you could find it but since you are apparently so brilliant it involved the subtraction of EM fields.  If you could do the math and understand what it means, you wouldn't be claiming that CO2 radiates an EM field back to the surface of the earth.  In fact, you would reject the notion of backradiation entirely.  You don't, so you accept backradiation and even tried to defend back convection and back conduction.



IanC said:


> oh, and one of your comments intrigued me. if you did shine an airplane or train light at a car headlight and waited for it to come to equilibrium are you sure that no change would occur? I think it would draw less power rather than change emmission values but it could be both.



No light from the more cooler fillament would ever reach the warmer fillament.  If it did then the warmer fillament would absorb it and in turn emit more light, some of which the cooler bulb would absorb and re emit which the warmer bulb would then absorb and become brighter and on and on in an endless perpetual motion feedback loop.  When you break the laws of thermodynamics Ian, you must be willing to accept and prove the consequences.  The consequences in this case is an endless feedback loop that would result in each light eventually emitting an infinite amount of radiation.  You know and I know that it would never happen but you, unlike I am not willing to break the laws of thermodynamics just a little in order to try and prove a point.

By the way, does a heated bar in a 0F chamber in any way accurately represent earth in cold space?  That was spencer's goal and he made so many mistakes in his thought experiment that calling OK OF would be the least of them.  If you redo the calculations at 0F, the cooler bar is still not going to cause the heated bar to get warmer.  If pointing out want amounts to be a spelling error on spencer's part gives you a sense of victory,  you truely have my deepest pity.


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## IanC (Nov 10, 2011)

an interesting excerpt out of Pseudo Critical Thinking in the Educational Establ via Judy Curry's blog



> Unfortunately, there is not simply good and bad thinking in the world, both easily recognized as such. There is also bad thinking that appears to be good and therefore wrongfully, sometimes disastrously, used as the basis of very important decisions. Very often this &#8220;bad thinking&#8221; is defended and &#8220;rationalized&#8221; in a highly sophisticated fashion. However flawed, it successfully counterfeits good thinking, and otherwise intelligent people are taken in. Such thinking is found in every dimension of human life and in every dimension it does harm; in every dimension it works against human well-being.
> 
> &#8232;&#8232;Sometimes when people think poorly, they do so out of simple ignorance. They are making mistakes, they don&#8217;t know they are making mistakes, but they would willingly correct their mistakes if they were pointed out to them. Often mistakes in thinking are quite humble.
> 
> ...



reasoned judgment; neither fact alone nor opinion alone.

pseudoscience can make a good case for itself if allowed to cherrypick what facts it wants presented, and when it can establish the opinion among many that it is reasonable and accepted. but real science must always be open to questions, no matter how awkward or uncomfortable. climate science is rotted through and through with pseudoscience. sometimes its the actual data that is screwed up, more often it is the faulty conclusions drawn up and presented as fact.


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## IanC (Nov 10, 2011)

and just for wirebender and gslack. from the same blog post at Disinformation and pseudo critical thinking | Climate Etc.



> 1. Hear no evil, see no evil, speak no evil.  Regardless of what you know, don&#8217;t discuss it &#8212; especially if you are a public figure, news anchor,  etc.  If it&#8217;s not reported, it didn&#8217;t happen,  and you never have to deal with the issues.
> 2. Become incredulous and indignant.
> 3. Create rumor mongers.  Avoid discussing issues by describing all charges, regardless of venue or evidence, as mere rumors and wild accusations.
> 4. Use a straw man.
> ...



I think wirebender and gslack are especially talented in performing 4,5,7,9,17 and 18.

but they are best at just _not answering the question!_


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## wirebender (Nov 10, 2011)

IanC said:


> and just for wirebender and gslack. from the same blog post at Disinformation and pseudo critical thinking | Climate Etc.
> 
> 
> 
> ...



Funny Ian, you were the one who was constantly changing the parameters of spencer's experiment or inventing new experiments all together.  The very definition of a strawman and 9 describes every bit of your argument up to and including this last post.  Your constant altering of the parameters of spencer's experiment and the "new" experiments you injected were also examples of changing the subject.  Clearly, you are doing nothing more than attempting to blame your opponent for what you are doing.

Your constant appeals to the board are most telling Ian.  If you honestly believed you were right, you would not need to appeal to the board for approval.  If you want rocks to come around and give you some more sugar out on the public board, why not just send him an IM and ask him to come around.  I am sure that he would jump at the opportunity to praise anyone who was arguing with me.  Appealing to the board is one of your most obvious "tells" ian, when you are unsure of yourself, you appeal to the board.  It has been obvious through this whole discussion that you are unsure of yourself and equally obvious that you were out of your depth.


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## wirebender (Nov 10, 2011)

IanC said:


> an interesting excerpt out of Pseudo Critical Thinking in the Educational Establ via Judy Curry's blog
> 
> 
> 
> ...



Interesting that you don't recognize yourself in the quote you brought here ian.


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## gslack (Nov 10, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC said:
> ...



Ian that was just 15 minutes.. Actual thought occurred as written.. Impressed aren't you... Yeah I get that a lot. See that's how it works when you actually DO SOMETHING besides sit on your ass calling yourself a genius..

Ian everything in that regarded what we have seen from you here in this thread.. Ad hom is vital in this instance because your character and moral fiber are in question due to your actions.. You sir are a liar and fake and that is fact... The words _Ad Hominem  _ literally translate to _"to the man"_. So when you start showing character and stop lying, weaseling, waffling, obfuscating and generally acting like a child, you cry about ad hom attacks.. 

You going address anything I brought up in the last 20 posts honestly? Nope didn't think so. until you do you are bringing your character into this discussion.

Now go cry punk..


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## gslack (Nov 10, 2011)

I find it all too telling that after I put my own thoughts and words on Ian's behavior down to type, he goes and googles up a list in response... Not his own words, not his own thoughts, but someone else's... And he is supposed to be so much more educated...

Lets see here.. Me, not a physicist, or expert and teaching myself as I go. Ian, supposedly educated in physics and from his big talk we would have to assume he must claim to be some kind of expert in the field. I use my own words, write my own thoughts and write on her what i think and why or how I come to my conclusions. While Ian, googles up terms and what other people think..

Hmm...Maybe logic and reason have changed since last I checked but I am pretty sure the first requirement in any scientific study would be to think... Ian must have never had to show his work...Ever..


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## wirebender (Nov 10, 2011)

And then accuses me of "cribbing".


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## polarbear (Nov 10, 2011)

wirebender said:


> IanC said:
> 
> 
> > please explain to me how you think the two bars will equilibrate to the same temperature in Spencer's experiment even though one is heated and the other is not.
> ...




I had to read all these posts to find where this perversion of the Stephan Bolzmann equation came from...the one You asked me about...:






And it turns out it came from Wikipedia...I only use "Wiki" because I don`t have the time to scan in a page of a real physics book and upload it.

Amazing how "climatology" can transmute this into the kind of total crap, like "back radiation"...and then have a second & cooler body "radiate energy" back to the hotter body where it got the energy from and then they go on and claim that this will now raise the TEMPERATURE of the original emitter.

By now everyone is familiar with the dimensions used in the Bolzmann equation, ...?

So now let`s go to the altered Bolzmann equation, the way "climatology" is trying to exploit it and to the question You (wirebender) asked me:



> Specifically, the claim that p=sigma (T^4 - T^4) represents the same physical processes as P=(sigma T^4)-(sigma T^4)


Stephan Bolzmann never made this claim..here again somebody felt free to ride Stephan Bolzmann`s coat tails and added this amendment into the Wikipedia pages.

Here is Bolzmann`s original law exactly as You would find it in the textbooks:





This equation expresses the energy flux in watts per square meter at a given temperature...

But in no way is it valid to do this form of this law for such a thing as "BACK-RADIATION":


> p=sigma (T1^4 - T2^4)


(I had to change Your "T`s" to T1 and T2)

Why not....?
Because that would also state that if both radiators were at the same temperature no matter how hot that the outcome would be 0....as in zero watts/m^2...and that would apply over the entire system.
And in no way can anybody say that there would be a zero energy flux condition between say 2 suns that are equally hot...



Obviously that paradox does not just happen when T1= T2, but over the entire range as T1 and T2 diverge in magnitude.

*I kept saying that no second body, no matter what has the ability to alter the heat radiation energy flux of any other one...*


The author of this equation that was published in Wikipedia





*Also never even claimed this either, because he is expressing the heat INSULATION effect of a heat INSULATOR and nothing else*.

*He is not talking in any way shape or form about "back-radiation"*

And what he is saying about *HEAT INSULATORS* is EXACTLY RIGHT...
Also the second temperature the "T0" to the 4.th power anywhere in physics has the meaning 0 degrees Kelvin...and if the author of this equation meant anything else with it, that would be nonsense, since if the T is already expressed in Kelvin, then there is no need to do this subtraction.
*That subtraction has to be done only if You are expressing T in degree Celsius*

I am sure that who ever posted it @ Wikipedia never even dreamed what "climatologists" would do with it...
They are not even adhering to his equation either, because they don`t
*do a subtraction with T1 and T2 they ADD them*, and have used the "T0" as a second so called "cooler black body" assigning a Temperature > ) degrees Kelvin for their back radiation lunacy.
So now poor Bolzmann who must be turning over in his grave by now has been ab-used to "T1^4-T0^4 and T0 has been assigned "climatology" values way above zero degrees Kelvin



*Like I already told You. IanC...be careful what kind of "physics lessons" You get in "climatology"...or when You start quoting equations that You don`t fully understand.

*


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## gslack (Nov 10, 2011)

I just recently discovered its actually called *Stefan-Boltzmann* two names as in two people. A Jozef Stefan and one Ludwig Boltzmann both were credited for it. As it turns out Jozef Stefan deduced it from work Ludwig Boltzmann was doing on some theories from even another guy named Tyndall.. I guess they figured Steffan-Boltzmann-Tyndall would have been too long... LOL, just to clarify I was under the assumption it was a Stephen Boltzmann (one guy) and in my defense I have noticed its written like that a lot on the net.. But its actually two guys work..

BTW, if you guys haven't played around a new LaTEX or similar equation editor yet, its a blast. And I highly recommend trying one out.. It allows to write out equations that will look on type as it would if written out by hand (albeit neater).. So basically you can show actual 10 to the 4th in much like you would writing it on a black board using a simple code and the program will write a graphic in real time for it. Really nice.. It makes me even look a bit smarter LOL.


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## IanC (Nov 11, 2011)

wirebender said:


> As to the math I did earlier regarding EM fields, again, you were in the conversation and had no comment whatsoever on the math other than the sort of hit and run snide remarks that rocks makes about things he doesn't begin to understand.  If you are interested, I suppose you could find it but since you are apparently so brilliant it involved the subtraction of EM fields.  If you could do the math and understand what it means, you wouldn't be claiming that CO2 radiates an EM field back to the surface of the earth.  In fact, you would reject the notion of backradiation entirely.  You don't, so you accept backradiation and even tried to defend back convection and back conduction.



I couldnt care less whether you are unwilling to present your 'masterpiece'. just stop daring people to find the fault in your math, when you are unwilling to produce it.

where to start...hmmm....

a CO2 molecule absorbs an IR photon, vibrates for a while, then usually emits the same energy IR photon in a random direction, sometimes back towards earth. four events, emmission and absorption of two photons. radiation and backradiation. this is where you hold your breath until your face goes red and then blurt out 2nd Law!!! and gslack calls me an idiot for describing a perpetual motion machine. 

but as I told you before, the system has changed. everytime a photon is emitted or absorbed a minute amount of momentum is transferred. these crumbs are the stuff of entrope. there is no prohibition for even the same two molecules to pass the photon back and forth, other than the fantastically long odds of it happening.

your conception of EM fields is distorted. you can measure the amount of radiation coming from one object and compare it to the radiation from a second object, then calculate the net flow of energy but you are crazy to think that photons that make up the fields fight each other to the death in open space. both fields radiate, all photons reach their target, the net energy exchanged matches the subtraction of the fields.

to reiterate- excited CO2 molecules dont give a flying fuck about EM field, they only care about shedding the extra energy. and some of that energy goes back towards earth. you can argue defintion of terms but most people just call that backradiation because it imparts more information about where the energy originated and where it went.


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## IanC (Nov 11, 2011)

polarbear said:


> wirebender said:
> 
> 
> > IanC said:
> ...



it was wirebender's link and I was more than happy to use it because it shows he calculated incorrectly.

previously you muddied the waters by bringing up reflection. wirebender defined the emissivity as 1, no reflection. I dont particularly want to get into discussions of planetary systems of energy transport until this simple thought experiment of Spencer's has run it course.

are you just arguing the definition of 'backradiation'? are you saying that there is no difference in the amount of net energy transferred per metre squared when the hot object is radiating into zero Kelvin space or into a close object at many hundreds Kelvin? if there is a difference, where does that difference escape from? 750w is being pumped in, it needs to get out somehow. my though was that the temperature of the heater had to increase to force the energy out past the blockage and to increase flow in areas that were left unblocked. you seem to disagree so perhaps you can clear up my confusion as to how and where the energy escapes.

Im not trying to be combatitive, I would just like to know where my thinking is incorrect.


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## wirebender (Nov 11, 2011)

IanC said:


> it was wirebender's link and I was more than happy to use it because it shows he calculated incorrectly.



Why do you lie so ian?  Do you think that people aren't bright enough to catch you at it?  The link to wiki was regarding your claim that putting clothing over your skin would cause your skin to warm and for that purpose, the math was dead on correct.  The wiki article was not in response to spencer's failed experiment.  Regarding spencers experiment, I didn't calculate incorrectly.  Spencer made a typo and I corrected for it.  He said his model was based on the earth sitting in cold space.  If the vacuum chamber was above 0K, then the chamber itself was radiating and was therefore another energy source.  He made no mention of a second energy source.  Congratulations on finding yet another error in spencers failed experiment.



IanC said:


> are you just arguing the definition of 'backradiation'? are you saying that there is no difference in the amount of net energy transferred per metre squared when the hot object is radiating into zero Kelvin space or into a close object at many hundreds Kelvin?



Apply the laws of physics ian, geez guy.  Do you not understand the relationship between absorptivity and emissivity?  It doesn't matter what the temperature of the background is so long as it is cooler than the emitting object.  If the background is hotter than the object, then the object is not radiating anything as that would violate the 2nd law of thermodynamics.  It is absorbing from the wamer emitter.




IanC said:


> if there is a difference, where does that difference escape from? 750w is being pumped in, it needs to get out somehow.



Where do you get the idea that 750 is being pumped in?  The experiment doesn't say anything like that.  The heated bar is emitting 750 to be at the temperature that spencer specified.  That doesn't say anything about how much electricity is flowing in.  The only parameter regarding how much electricity was flowing through the electrical cord was that it didn't change.  We don't know how much electricity was flowing in, we only know the amount of wattage the bar was emitting in order to be at the specified temperature.  

The more you talk ian, the more clear it becomes that you don't really get any of this.  



IanC said:


> my though was that the temperature of the heater had to increase to force the energy out past the blockage and to increase flow in areas that were left unblocked. you seem to disagree so perhaps you can clear up my confusion as to how and where the energy escapes.



What "blockage"  The heater simply absorbs and emits.  It doesn't decide to warm up or cool down based on the surroundings.  It is receiving an unknown quantity of electricity and simply emits.  It can't warm up unless A) the amount of electricity coming in increases, or B) the temperature of the background increased to a temperature higher than itself.  If the powered bar increases its temperature without one of those two things happening, then you have created energy which is a violation of the law of conservation of energy.



IanC said:


> Im not trying to be combatitive, I would just like to know where my thinking is incorrect.



Start at the very beginning.  That is where you went off the rails.  Rather than do the math and accept what it tells you, you tried to reason your way around the applicable physical laws in order to preserve what you believe regarding CO2's non existent ability to radiate energy back to the surface of the earth where it is then absorbed.


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## gslack (Nov 11, 2011)

Ian you are full of it.. Do you really think you are fooling anyone now? Grow up man, you aren't physicist, you aren't a chemist, dude you aren't even a good student...

Wire was right, you tried to imply (weakly because you are that type of a weasel) that there was such a thing as back conduction and back convection... You didn't do it outright, what you did was your weasel act where give little snigger or make some comment that implies it. One that you can pretend you didn't mean like it read later if it turns out you are wrong.. Again showing how little you actually know...

Keep on going ego boy, EVERYONE must be wrong because you are right because.... Yeah...


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## gslack (Nov 11, 2011)

Ian can you explain how it is you cannot define in your own words a single concept I have asked you regarding even the most basic principles in physical science? You didn't recognize any of the standard beginning experiments, you didn't recognize Fermat's last Theorem, and you have likened the role of Quantum mechanics in this as an esoteric theory and likened it to "magic". 

Where in the hell did you study physics? Melmack university? Is there some other form of physical science that avoids actual math somewhere we aren't aware of?


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## polarbear (Nov 11, 2011)

IanC said:


> I dont particularly want to get into discussions of planetary systems of energy transport until this simple thought experiment of Spencer's has run it course.
> 
> are you just arguing the definition of 'backradiation'? are you saying that there is no difference in the amount of net energy transferred per metre squared when the hot object is radiating into zero Kelvin space or into a close object at many hundreds Kelvin? if there is a difference, where does that difference escape from? _*750w is being pumped in, it needs to get out somehow.*_ my though was _*that the temperature of the heater had to increase *_to force the energy out past the *blockage* and to increase flow in areas that were left unblocked. you seem to disagree so perhaps you can clear up my confusion as to how and where the energy escapes.
> 
> Im not trying to be combatitive, I would just like to know where my thinking is incorrect.




Being combative against  another person  has no place in a  physics discussion anyway. 
Some people, mostly climatologists resort to that  when someone points out the flaws in their theory.

Perhaps the best place to start clearing up the confusion is for You to consider asking yourself the same questions as
any physisist would ask...

1.) Where  is the EXTRA energy supposed to be coming from, that is supposed to increase the temperature of this heater?
It does take EXTRA energy to raise the temperature of a Mass M1 (using Your words, "the heater") from  T1 to T2...

2.) Using Your words, "the blockage" as the second mass M2 which has to be heated in this example from a portion
of the radiation it has received from M1.

3.) Using Your 750 *watts *that M1 has absorbed and is re-emitting   a portion of the 750 watts is now absorbed by M2
How large this portion is depends on the mass, the spacial arrangement and the optical properties of M2 at the part of the involved infrared spectrum...but since You are free to choose anything You want for the mass and also for the spacial arrangement You are also free to assign any portion You want for  750/n .th  Watts M2  is absorbing.


Your final question was:
"*so  perhaps **you can clear up my confusion as to how and where the energy  escapes."

* L_e_t`s  start by answering first  by how much "escape" *energy * ...as You call it we do have in total.

Your example starts out with 750 *Watts,* which by the way is not *energy*, but is a *power unit*. (energy per time)
You keep using the word "*energy*" instead of the word "*power*" so then , by all means we`ll discuss this subject using *energy *instead of power

*So the 1 number  priority is to  change that to the proper dimension and use Watt-seconds  and stop mixing up apples and oranges...!!*

Now we can begin.
After the time interval Tsec1 the mass M1 has been heated with *750 Watts X Tsec1---->    to Watt-seconds Energy...*
That is all the *ENERGY* M1 contains and has to dissipate or that has to "escape" as You put it.

Of these *750 Watt-secs  ENERGY *the mass M2 or the 'blockage" as You call it , depending on the properties I mentioned before is going to
absorb a portion of these 750 *Watt secs.*...You are free to choose a simple number for reasons of clarity...and
so am I...so if You don`t object, I choose  1/10 th.  

So then, after the time interval  Tsec1   Your "heater" the mass M2  has already allowed  75 Watt seconds to "escape" over to
the "blockage" which now contains these 75 Watt seconds.

But in no way will that now raise the *TOTAL ENERGY * of this system to more than *750 Watt seconds *which  have to "escape"


Regardless if You use * POWER * as in Watts instead of *ENERGY* as in  Watt seconds that applies.


No matter which number You want to choose as a time interval if You meant to ask :
"*so  perhaps **you can clear up my confusion as to how and where the POWER   escapes."

*The conversion to POWER Units as in Watts  that have to "escape" involves nothing more than a simple division of the above Numbers that represented *ENERGY *by whatever You want to choose as a *time interval for Tsec1....*

It also won`t matter if You use calories as energy, or calories per second as power or whichever You prefer, the outcome is the same
  Surely You are not of the opinion, that  M2, ...Your so called "blockage" will  add  *POWER **to Your original 750 Watt heater *....*and do so with some of the 75 Watts  it  scavenged  from  from the heater in the first place.*


So which part of  my "opinion" as You called it  confuses You...?
If  You have scientific evidence You would like to present, that my "opinion" is wrong then  I encourage You...don`t  just to talk about it, go ahead and use it,  because  such scientific evidence could  revolutionize all prior technology known to man...and You could get filthy rich.

But in real science & technology You would have to build a working model, even one as simple "Spencer...Yes Virginia" example would suffice as long as You demonstrate that it really does as advertized.

Nobody ever got a patent just on his say so and no matter how many titles he included in his name... or perhaps because "Vriginia" the realtor daughter agreed.
But "climatologists" like Spencer (Al Gore, M.Mann etc.) get to side step this pesty burden of proof altogether and still get filthy rich,...
Because there are all these dummies who believe quacks like that and support the politics that funds the $$$ that flow into these pockets.

A cynic basing people who do not agree with the likes of Spencer might reply, that this revolutionary technology already exists...
we call it "insulation" and use it in building construction.

Then again these cynics have forgotten, that in this example as with all the other ones they do use, not a single one has ever managed to prove that CO2 is an "insulator".
..having failed to do so ..."climatology" has been using the same method over and over again...
and that is to present  "back-radiation" HYPOTHESIS` (in plural...such as Spencer`s) as "proof" for their original CO2 caused heat increase "HYPOTHESIS".

Tell me now, how exactly has Spencer`s example  demonstrated the validity of "back-radiation"...

No, aside from the utter math & physics blunders...it`s not just the definition I do not accept. 
I *also *do not accept the claim that *he has proven anything *with his "Yes Virginia" example...*which exists only on paper.*

Also any "computer modelling" as a substitute for the failure to demonstrate experimentally is not far above from anything else, that is just on paper either..!!!

....and neither would anyone who is involved in R&D and has to prove the validity of statements with a working experiment, that actually demonstrates that the hypothesis was correct with observable results.


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## polarbear (Nov 12, 2011)

"Climatology" is using  "* radiative forcing of 1.66 W/m*2*, " *for this so called "back radiation" energy budget
to make the case for man made global warming.

Then they go on and use "models" like Spencer`s "example"  to "explain" how this back-radiation  allegedly adds heat energy to the overall system...the energy that they claim would have "escaped", if the CO2 were not there

I can also use  an "energy budget"  which involves CO2  and it is one that "climatology" chooses to ignore with their "earth=black body energy budget simulation".

It goes like this:
The Mauna Lua Observatory tells us, that at any time during the plant growth season the average molar ppm CO2 is 5 ppm lower....







and Wikipedia will also tell You that this is due to seasonal plant growth & bio synthesis.
By that they refer to this equation, *but do not a single "climatologist" wants to specify the energy which is required *for this bio chemical reaction...well then I shall  specify it:

Bond Energy


*C6H12O6 + 6O2 -> 6CO2 + 6 H2O, &#916;  G = -686 kcal.

Before You go on a "left spin" here, be aware that each and every proper chemical reaction is using "<-->" and not just "->"...
which of course also applies for the delta Energy requirement when the reaction goes from CO2 to **C6H12O6 + 6O2 in the opposite reaction, as
in the bio-synthesis of Carbohydrates from CO2, while generating Oxygen
** so the 686 k-cals per 6 Moles CO2 converted to a carbohydrate monomer DOES APPLY Exactly as above...You can look that up in any half decent chemistry book...!!!
And again, NO, that is not just my personal opinion, as You so often classify my statements

*So now with that cleared up...(I hope) we can proceed:

The Molar Volume at "Standard Atm"=  22 414 cm^3. 
so 1 m^3 containing  5 molar ppm CO2 corresponds to    44.6 * 5 * 10^(-6)  Moles  per cubic meter 

Each and every second we add CO2 and for the seasonal  level when bio-synthesis applies the molar ppm is by 5 ppm lower than without the bio-synthesis
when generating Carbohydrates + Oxygen from CO2
as shown in the equation above

The energy budget per cubic meter for this reaction is:
(44.6 * 5 * 10^(-6)  * 686 )/ 6 =  25.5 calories 

which converts to *106 watt seconds* required by bio-synthesis to lower the CO2 level in each 1 cubic meter  with this bio-chemical reaction.
......and each cubic meter  of course has a base area of 1 square meter....
...and as You surely know 106 watt seconds per 1 second *= then a power consumption of 106 watts per square meter*

"Climatology" does pay some *qualitative * homage  to the CO2 which is consumed by bio synthesis,...but nothing more beyond,...and God forbid to how many watts per square meter this amounts to








Not a single "Climatologist"  wants to show you the math  for the energy this bio synthesis requires in terms of Watts per square meters, allowing a comparison  with the alleged  *1.66 W/m*2* "back radiative forcing"* 
which is supposed to eventually drown us all by melting the polar ice caps.


I would also like to draw Your attention to the popular terminology CO2 = "Greenhouse Gas".
Well CO2 is in fact used in Greenhouses,....but it is used to accelerate  this very bio-synthesis equation we have been using...and certainly not for the purpose of raising the temperature.

Thus it is also valid to make the claim, that outside a "Greenhouse" CO2 will also accelerate plant growth ....Yet there is not a single "computer simulation"  that incorporates not just the *present 106 Watt/m^2 requirement*, but also how this number would increase,...*as it must in an actual Greenhouse when You increase CO2 content and therefore the rate of plant growth.*


So IanC, You have confronted me with this so called "Spencer Model" and "Trenberth`s Energy budget".
And now I am confronting You with a bio-synthesis energy budget.

I am waiting for Your (non-combative) rebuttal...
Before You say again, something like  *C6H12O6 + 6O2 -> 6CO2 + 6 H2O, &#916;  G = -686 kcal.  *is my "personal opinion", go & Google,...and then do the rest of the math yourself


----------



## IanC (Nov 13, 2011)

wirebender said:


> IanC said:
> 
> 
> > it was wirebender's link and I was more than happy to use it because it shows he calculated incorrectly.
> ...



you provided the link. I used it to show your math was wrong. how is that a lie? why did you post the link if you dont agree with it?

Spencer made no typo. he defined the container temperature as zero degrees farenheit and acknowledged that it radiates as this diagram from his article clearly states






that said, I disagree with the size of some of the arrows in the diagram.


----------



## IanC (Nov 13, 2011)

wirebender said:


> IanC said:
> 
> 
> > are you just arguing the definition of 'backradiation'? are you saying that there is no difference in the amount of net energy transferred per metre squared when the hot object is radiating into zero Kelvin space or into a close object at many hundreds Kelvin?
> ...



I wish _you_ would apply the laws of physics. just because there can only be heat flow in one direction that does not mean there is radiation in only one direction.






when you plug higher numbers into the equation for Tnaught, then the power is obviously being reduced.


because the board was down last night I browsed around for a while. I came across the writings of Claes Johnson and immediately recognized all of your talking points, and even your arguing style. you, wirebender, are nothing but a poser. stealing someone else's ideas and promoting them as your own without giving credit to the source. plagarism.

I dont have any major problems with Johnson's work on radiation because it doesnt actually change any of the previous work, it only reframes the descriptions in a way that makes no difference to calculating the forces. even he admits there is radiation impacting on the warmer surface which is absorbed and reradiated immediately, 'resonated reemission' or something like that. akin to reflection. but he doesnt make the erroneous statement that radiation from the cooler source isnt part of the total radiation from the warmer source.

an analogy....a payphone (remember them?) requires 50cents a minute to call someone. you put two quarters in, but one quarter drops into the coin return every time even though it was counted as paid. every minute afterwards you have to put another two quarters in, one from your pocket and one from the coin return. the net cash flow is going into the phone, some is coming back, thus diminishing the actual cost of the phone call. you guys are arguing that there is no back radiation in the same way as you could argue that the pay phone wasnt giving back any money because no quarters went back into your pocket. but you are paying less for the same effect.


----------



## IanC (Nov 13, 2011)

wirebender said:


> Where do you get the idea that 750 is being pumped in?  The experiment doesn't say anything like that.  The heated bar is emitting 750 to be at the temperature that spencer specified.  That doesn't say anything about how much electricity is flowing in.  The only parameter regarding how much electricity was flowing through the electrical cord was that it didn't change.  We don't know how much electricity was flowing in, we only know the amount of wattage the bar was emitting in order to be at the specified temperature.
> 
> The more you talk ian, the more clear it becomes that you don't really get any of this.



I am so happy that polar bear jumped into this thread. now you are actually saying things I can argue rather than just reading through streams of _ad homs_ that add nothing to the debate.

I dont care how much electricity is going into the heater. as long as it stays the same. it takes roughly 750w of power to make the heated bar radiate at 150F when it is by itself so let that be the input. the input will always be 750w, therefore the outside container will always be receiving 750w once equilibrium has taken place. are you OK with that? input equals output.

for the sake of simplicity we assume that all the surfaces of the heater are radiating equally, at least until the second bar is introduced. are you OK with that?

let us define the area of the face radiating into the second bar as 0.2m2 out of a total area of 1.0m2. that means 150w out of 750w will radiate into the second bar, given that emissivity =1 and the bars are very close to each other (but not touching so there is no conduction). are you OK with these simplifications of conditions for ease and clarity of sorting out where the power goes?

once the second bar is placed into the chamber it will absorb the 150w of power and warm up until equilibrium is reached. the face absorbing the radiation from the heater will be warmer than the opposite face which is emitting radiation into the cool container. are you OK with this? do you have some real world explanation of how there would not be a temperature differential between these two opposite sides, and to a lesser extent across the other faces as well.

Spencer stated that the second bar was 100F at equilibrium. I do not believe this to be true for the _average_ temperature but it could be true for the one face that is receiving radiation from the heated bar. the other faces would be less.

according to




the power of the 100F face would be roughly 75w. I am ignoring the radiation from the container because we ignored it when defining the power of the heated bar as well. the precise numbers are not overly important here because we are just getting an idea of the relative strengths of radiation. 

now let us look at the net radiation between the heated face radiating into the closest unheated face. I prefer to use 




150w -75w =75w   I find it easier to keep track that way.

but you demand that it be done as




= 75w     it doesnt matter mathematically. the final answer is the same.

the inside face of the heated bar is now only radiating 75w by your bookkeeping. the inside of the unheated bar is now radiating 0w by your bookkeeping. but the rest of the unheated bar was only radiating 75w before so when you add it up 75w input equals 75 w output. (remember the original 150w input was divided into 75w backradiation and 75w left over for the rest of the bar. wirebender wants us to use 75w from the heated inside face and no backradiation from the unheated inside face. much more confusing, I know, but it still adds up correctly)

the net power of the heated inside face has been reduced from 150w to just 75w. but the heater is still inserting 750w into the first bar! where does the extra 75w go? it must go somewhere! we cant just ignore the discrepancy. there are only two options.

option one- the heater is thermostatically controlled to only draw enough power to warm the first bar to 150F, which is now 750w -75w = 675w. this is plausible. but what are the implications? you have _cooled the container by 75w!_ it is no longer getting 750w it is only getting 675w! cooling the container while keeping the bar the same temperature is still increasing the the temperature differential!

option two- the heater still produces 750w of power and the 75w radiation shortfall must be made up by dividing the extra 75 watts amongst the total area of the heated bar. by definition, raising the radiation of the surface is raising the temperature.

Spencer is correct.


----------



## IanC (Nov 13, 2011)

polarbear said:


> Then again these cynics have forgotten, that in this example as with all the other ones they do use, not a single one has ever managed to prove that CO2 is an "insulator".
> ..having failed to do so ..."climatology" has been using the same method over and over again...



please excuse my use of power and energy incorrectly. I have always tried to impress upon my sons the need for using units when solving physics problems as a good way of avoiding simple mistakes. unfortunately I no longer solve physics problems and I tend to try using different words when describing things. no excuse, simply imprecise terminology for the sake of readability which in fact had the reverse effect.

as to the quote....I find it hard to believe that you think CO2 has no potential to slow the escape of energy from the earth's surface.

the earth emits 15 micron radiation that can interact with CO2. there are 3 scenarios that are important to this discussion-

1. the photon simply escapes into space at the speed of light carrying the energy with it.
2. the photon is absorbed by a CO2 molecule, which then reemits a similar photon which escapes into space. the energy is still gone but it has taken much longer to escape.
3. the photon is absorbed by a CO2 molecule, which then reemits a similar photon back towards earth. the energy does not escape and the time has been used up that could have been spent losing the radiation. a decrease in the intensity of energy shedding.

by what standard of 'insulation' do #2 and #3 not meet the criteria?


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## wirebender (Nov 13, 2011)

IanC said:


> you provided the link. I used it to show your math was wrong. how is that a lie? why did you post the link if you dont agree with it?



I provided the link in response to your claim that putting a blanket over your skin caused your skin to warm ian, not in response to spencers failed experiment.



IanC said:


> Spencer made no typo. he defined the container temperature as zero degrees farenheit and acknowledged that it radiates as this diagram from his article clearly states



Of course he did.  He said that he was modelling his experiment after earth sitting in cold space.  Is earth sitting in space at 0F?


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## IanC (Nov 13, 2011)

wirebender said:


> IanC said:
> 
> 
> > you provided the link. I used it to show your math was wrong. how is that a lie? why did you post the link if you dont agree with it?
> ...








Spencer specifically stated the temperature, he showed the radiation, and explained the real world type of equipment where you could actually run the experiment. although he did not. if he _had_ run the experiment would you have complained because the cosmic background radiation means the temperature of space around here is 3 degrees K? grow up, you got caught again making a false statement, live with it.

as to you complaining about me using your own link to show that you are wrong....either something is true or it isnt. you dont like those formulae, so does that mean you are retracting your claim about blankets making a body colder?


----------



## wirebender (Nov 13, 2011)

IanC said:


> I wish _you_ would apply the laws of physics. just because there can only be heat flow in one direction that does not mean there is radiation in only one direction.



The warmer object does not absorb radation from the cooler object ian.  If it did, it would become warmer and then radiate more which the cooler object would then backradiate and you would get an endless feedback loop in which both would end up emitting infinite radiation.  Perpetual motion, ian, it isn't happening.



IanC said:


> when you plug higher numbers into the equation for Tnaught, then the power is obviously being reduced.



And you still don't get it.  If you raise the temperature of the background, it takes less power for the blackbody to radiate at the same temperature, but if you raise the background temperature and keep the power coming in through the electric cord constant, you will increase the temperature of the bar.  I have told you before ian, the only way to raise the temperature of that bar is to either increase the amount of energy coming in through that electric cord, or raise the temperature of the background.

Funny that you don't grasp even that basic concept and still have the nerve to call me a poser.



IanC said:


> because the board was down last night I browsed around for a while. I came across the writings of Claes Johnson and immediately recognized all of your talking points, and even your arguing style. you, wirebender, are nothing but a poser. stealing someone else's ideas and promoting them as your own without giving credit to the source. plagarism.



I read Claes Johnson, and Joseph Postma, and Hans Schreuder, and Will Alexander, and Joeseph Reyden, and Harry Huffman, and numerous others who apply the actual laws of physics and show their math to reach their conclusions.  I have stolen nothing from anyone although I have gained a great deal of knowledge through reading people who actually do the math to prove their conclusions. I never claimed to be a mathematics professor, or a scientist.  I am an orthodontist who happens to grasp the math.  I don't feel constrained to give credit to everyone who has ever shaped my thinking for my entire life and clearly you don't either.   By the way, the only math I have used in this discussion has been the Stefan Boltzman equations and I believe I have given them credit.



IanC said:


> I dont have any major problems with Johnson's work on radiation because it doesnt actually change any of the previous work, it only reframes the descriptions in a way that makes no difference to calculating the forces. even he admits there is radiation impacting on the warmer surface which is absorbed and reradiated immediately, 'resonated reemission' or something like that. akin to reflection. but he doesnt make the erroneous statement that radiation from the cooler source isnt part of the total radiation from the warmer source.



And you fail to understand his work as well.  If you grasped it, you would know that he makes no consession to mythical backradiation.



IanC said:


> an analogy....a payphone (remember them?) requires 50cents a minute to call someone. you put two quarters in, but one quarter drops into the coin return every time even though it was counted as paid. every minute afterwards you have to put another two quarters in, one from your pocket and one from the coin return. the net cash flow is going into the phone, some is coming back, thus diminishing the actual cost of the phone call. you guys are arguing that there is no back radiation in the same way as you could argue that the pay phone wasnt giving back any money because no quarters went back into your pocket. but you are paying less for the same effect.



There is no backradiation ian.  Feel free to do the math and prove it if you believe it exists.


----------



## wirebender (Nov 13, 2011)

IanC said:


> wirebender said:
> 
> 
> > Where do you get the idea that 750 is being pumped in?  The experiment doesn't say anything like that.  The heated bar is emitting 750 to be at the temperature that spencer specified.  That doesn't say anything about how much electricity is flowing in.  The only parameter regarding how much electricity was flowing through the electrical cord was that it didn't change.  We don't know how much electricity was flowing in, we only know the amount of wattage the bar was emitting in order to be at the specified temperature.
> ...



And you still don't get it and I doubt that you ever will.  You can't get around the laws of physics ian.  If you think you can, by all means build the prototypes, get yourself a patent and become a bazillionaire.  Here is a hint for you though, don't invest your life savings unless you believe you can develop a taste for cat food in your old age.


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## IanC (Nov 13, 2011)

where exactly do you think I am gaining or losing energy?

you are the one who is saying that physical objects that must be radiating by existing at a temperature greater than 0K, are not radiating if an object near by is warmer by any fraction of a degree. and I am not sure because you dont explain yourself but I think that you might perhaps think that that secondary warmer object _still radiates at full strength_ for its temperature. 

please give a description of two blocks close to each other, with a temperature difference of say, 10 degrees Kelvin. describe what amount of radiation +/or heat is exchanged. assume a vacuum for simplicity.


----------



## polarbear (Nov 13, 2011)

IanC said:


> polarbear said:
> 
> 
> > Then again these cynics have forgotten, that in this example as with all the other ones they do use, not a single one has ever managed to prove that CO2 is an "insulator".
> ...




*"please excuse my use of power and energy incorrectly."*....no apology necessary and I have no intention to harp on it,...other than reiterating that
in this case, the difference is crucial.  It was an oversight, than can happen to the best of us, so don`t worry too much about it...Okay...!!! 
But I do have to bring it up again, not for the purpose of rubbing Your face into it , but because I have to in order to respond to the
3 points You raised here

Power is energy  per time( as in divided by)  and  energy is power multiplied *by the time frame in question.
*and  I know that You don`t need a lecture using  numerical examples to show the difference between a division and a multiplication .

Let me expand the meaning of this last sentence in 2 directions.
1.) I am not trying to lecture You nor do I view this as a contest who is smarter on the grounds who defends what.
But more importantly it is 2.)  the vastly different conclusions that can be drawn  if this un-intentional oversight
is allowed to run the course through the statements 1,2,3

So now let me clear up this uncertainty You have, where I stand (or any other Phys/Chem professional)..:
"I find it hard to believe that you think CO2 has no  potential *to slow the escape of energy* from the earth's surface."

CO2 can and does absorb  ENERGY  on several bands in the IR spectrum...nobody disputes that.
The entire field of  Spectroscopy in analytical Chemistry & Physics relies on the fact that the presence of a specific substance can be detected by the *ENERGY* *absorption* at very specific wavelengths.

However we never had and to the best of my knowledge still don`t have any such analytical spectroscopic instrument
which would be able to measure what You  (& all of  "climatology") keeps claiming....this mysterious *"POWER" increase...
*
If this *"POWER" increase* is enough to warm the globe we (anybody who ever did IR spectro-analysis) should have noticed that a long time ago.

This is seriously SENSITIVE INSTRUMENTATION,...in case You did not know that...and if there would indeed be a POWER increase with an increase of ppm CO2, *there is no way, that would have gone un-noticed *

Nobody (You included) is entitled  to run through Your 1.,2,3 statements without specifying the time frame in statement
number 2:  " the* energy* does not escape and the*  time* has been used up..."

*So exactly how much EXTRA time are we talking about here.*...?

In so many words You and all of "climatology" is stating that if You spread out an original * ENERGY *amount of  A watt seconds
over a *longer time*, that You can somehow wind up  with an *ENERGY INCREASE *of A ..+X *Watt*s (as in *POWER*)  _*using an increase of TIME*_

So that rules out ENERGY all together, because You know how wrong  that is..right.?..it`s *even going in the opposite direction *of these claims.
Same *ENERGY* divided by *MORE TIME* = a *DECREASE in POWER, not an INCREASE*...and You did say, that You knew that and simply overlooked it

Yet You and "climatologists" begin this argument :
"the *energy* is still gone but it  has taken *much longer* to escape.
*And in the end wind up with the EXTRA POWER this "longer time"  generates in a universe, ....where "climatology", not nature rules *

And then "climatology" makes the pseudo-logic leap from energy to power without ever even bothering to quantify  the *time* other by simply stating  that it would take *"much longer"* .


You and all the others are in effect claiming  that the CO2  in the path of an infrared beam  is reducing the speed of light of  using Your words "* the photon simply escapes into space at the speed of light"*


Really...?   So* by how much* does CO2 then reduce this speed of light as it must if , using Your words again  " similar photon which escapes into space. the energy is still gone but it  has taken *much longer* to escape."

And lastly:
"by what standard of 'insulation' do #2 and #3 not meet the criteria?"

well if You can`t quantify the time or any detectable difference in the speed of light then You have not shown anything at all.

And as far as the leap back to the 'insulation' escape hatch when the "back-radiation"  theory "logic" starts to collapse....even I am growing tired of re-explaining the difference between insulation and this "back-radiation"

There are tables upon tables of insulation properties for just about any substance that has insulation properties.
It`s quite easy to measure and engineers call it the "R-" factor.
What has been the problem preventing the heavily $$$$ funded "climatologists" to present the rest of science with a table that shows an
*actually measured increase* in the "R" factor for CO2, specifically how much that is per ppm increase of CO2 in the air...????????????

*That will never happen...and You know that deep down as well...*

What will happen and always has happened (You just now, did it again )... is the usual 2 way use of the escape hatch , in this case from the "insulation" argument back to the "back-radiation" ....the alter universe of "climatology" *in which the dimension TIME does not exist.

*In this alter universe "climatologists" can generate more power from less energy and do so simply by extending the time dimension,...or letting this dimension dis-appear altogether ...but as You said that was a simple error and an oversight...

That leaves now (again) the original problem, to explain how  the "slowing down photons" with CO2 can GENERATE any extra POWER...You know, the *POWER* in *Watts *per square meter that *CO2 is supposed to add* in all these "climatology" computer models that are supposed to melt all these cubic miles of ice at the poles within a certain *TIME*..


----------



## wirebender (Nov 13, 2011)

IanC said:


> where exactly do you think I am gaining or losing energy?
> 
> you are the one who is saying that physical objects that must be radiating by existing at a temperature greater than 0K, are not radiating if an object near by is warmer by any fraction of a degree. and I am not sure because you dont explain yourself but I think that you might perhaps think that that secondary warmer object _still radiates at full strength_ for its temperature.
> 
> please give a description of two blocks close to each other, with a temperature difference of say, 10 degrees Kelvin. describe what amount of radiation +/or heat is exchanged. assume a vacuum for simplicity.



ian, you can formulate a million pertubations of that experiment but you will never raise the temperature of the heated bar unless you either reduce the dimensions of the heated bar, increase the power going into the heated bar from the electrical cord, or raise the temperature of the background.  I used 0K for my calculations because spencer said that he wanted to simulate the earth in cold space.  If you begin with 0F,  instead of 0K then the heated bar is radiating 340.3 wm2 in order to be 150 degrees F.  Initially, you will get different numbers but you still won't get the second bar to raise the temperature of the heated bar by even the tineist fraction of a watt per square meter unless you reduce the size of the heated bar, increase the amount of electricity coming into the heated bar from the wall socket, or raise the temperature of the background.  Spencer's chamber is cooled to 0 so its temperature isn't rising as the heater radiates.   To do so would violate the 2nd law of thermodynamics which states:

It is *not possible *for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy *will not flow spontaneously *from a low temperature object to a higher temperature object. 

I don't know what it is about the phrases not possible and will not flow spontaneously that is so difficult for you to understand but it is what the second law of thermodynamics states and you simply can't get around it.  You can play word games and build thought experiments till the cows come home but the fact remains that you can not build a machine that does what you claim and not even the tiniest shred of observable, repeatable evidence exists for the existence of backradiation.

You can point a parabolic reflector into open sky (away from the sun) on a clear sunny day and the temperature in that reflector will decrease.  If backradiation existed as you claim, the temperature in that reflector would not decrease.  Or perhaps you might like to explain how some mysterious "cold" backradiation might heat the earth.


----------



## gslack (Nov 13, 2011)

I love how Ian tries to make it seem all the discussion on our part was ad hom... He started insulting everyone who disagreed with his view on this from the very start. All I did in the beginning was verify some things wirebender had said as being factual and he went off following me around and harassing my every post for days after. But it was us...sure..

Ian you implied EM fields were some made up esoteric theory.. I showed this was not the case, and from there you went ape shit.. Do we have to start linking to every one of your tantrums to get some honesty out of you?


----------



## wirebender (Nov 13, 2011)

gslack said:


> I love how Ian tries to make it seem all the discussion on our part was ad hom... He started insulting everyone who disagreed with his view on this from the very start. All I did in the beginning was verify some things wirebender had said as being factual and he went off following me around and harassing my every post for days after. But it was us...sure..
> 
> Ian you implied EM fields were some made up esoteric theory.. I showed this was not the case, and from there you went ape shit.. Do we have to start linking to every one of your tantrums to get some honesty out of you?



Better make sure someone famous hasn't already said ape shit or he will start calling you a plagerist also.  Imagine, plagerizing the laws of physics.


----------



## IanC (Nov 13, 2011)

polarbear said:


> IanC said:
> 
> 
> > polarbear said:
> ...



the CO2 is decreasing the power of out going radiation. any slowing down of the radiation is diminishing it. less is less, I dont have to put a numerical figure to it. less than the speed of light therefore it is escaping slower.

the incoming radiation from the sun warms the planet, the reduction of heat loss from radiation to space increases the net warming from the sun. it couldnt be a simpler concept.


----------



## IanC (Nov 13, 2011)

wirebender said:


> IanC said:
> 
> 
> > where exactly do you think I am gaining or losing energy?
> ...



no heat is flowing from cold to hot. the electricity is warming the heater. the second bar insulates the first bar making the electrical power more efficient at raising the temperature of the heated bar. 

are you saying the electricity cant warm the heater any hotter or are you saying the second bar cant insulate the first bar?


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## Rozman (Nov 13, 2011)

IanC said:


> IanC said:
> 
> 
> > how many of you think it is a deception to have 390 (W/m2) going up and 324 coming down? while it is true in an absolute way it disguises the fact that only 66 w/m2 is available as the difference between the temps of the surface and the lower atmosphere. out of that 66, 40 leaves directly with only 26 W/m2 warming the lower atmosphere.
> ...





> *how many of you think it is a deception*




Actually Sheldon from The big Bang Theory TV show on CBS covered this very topic in depth in season one of their show....

Who do you think you're dealing with here... a bunch of rubes?


----------



## gslack (Nov 13, 2011)

Ian fix your quote... its gives a false impression of the conversation here..

The one above where you quoted polarbear, is missing a tag making it seem his quote is yours... Fix that please and try and refrain from editing peoples quoted posts, that way this sort of thing is avoided..


----------



## IanC (Nov 14, 2011)

wirebender said:


> IanC said:
> 
> 
> > where exactly do you think I am gaining or losing energy?
> ...



wirebender- you have a distorted view of thermo dynamics.

why does a blanket make you feel warmer? because it decreases heat loss. does the blanket actually heat your body? no, your body heats itself by processing food. less food needs to be 'burned' to keep your internal temperature stable when you use clothing or blankets to decrease the heat loss to the environment.

in Spencer's experiment, does the second bar actually heat the heated bar? no, the electrical power going into the heater heats the bar, the second bar only insulates which means the rate of heat loss to the environment is decreased. if the same amount of power goes into the heated bar then the temperature will be higher when it is insulated than when it was not. but it is the electrical power that heats it. have you been quibbling over some strange semantics? Spencer said the first bar will become warmer if a second bar is placed next to it. are you denying that, or are you just denying that heat energy is coming from the second bar? you have built a strawman by saying anyone has claimed heat was flowing into the first bar from the second. the claim is that the second bar insulated the first causing the heater to be more efficient in raising the temperature.

the same holds true for the earth/atmosphere CO2/greenhouse argument. no one has said that the atmosphere is warming the earth's surface (except under rare inversion conditions). the claim is that CO2 is decreasing heat loss by radiation which means the sun's incoming radiation will warm the surface more than when there was less insulation.

no one claims that insulation actually 'heats' anything. it slows heat loss, making it easier to achieve a higher temperature or maintain a temperature with less inputted energy. I have said this to you over and over again. apparently there really are deniers. you are just as foolish in your ways as the crackpot warmists are in their claims of certain doom. there should be at least three categories. Deniers, Warmists and Realists.


----------



## wirebender (Nov 14, 2011)

IanC said:


> no heat is flowing from cold to hot. the electricity is warming the heater. the second bar insulates the first bar making the electrical power more efficient at raising the temperature of the heated bar.



The second bar is a heat sink ian, not an insulator.  I am not sure why you don't seem to be able to understand the difference between an insulator and a heat sink and what each actually does.  Here are a couple of definitions.  By the way, I got them from a dictionary and aren't the product of entirely original breakthrough thinking on my part.

_insulator - A material or an object that does not easily allow heat, electricity, light, or sound to pass through it. Air, cloth and rubber are good electrical insulators; feathers and wool make good thermal insulators. 

heat sink - a metallic heat exchanger designed to absorb and dissipate excess heat from one of the devices, as a transistor or resistor, in a circuit.  A metal plate specially designed to conduct and radiate heat from an electrical component._

Now you tell me how you believe the second bar to be acting as an insulator ian, and show me the math or describe the physical law by which you make the claim.



IanC said:


> are you saying the electricity cant warm the heater any hotter or are you saying the second bar cant insulate the first bar?



I am saying both.  You have a fixed amount of power going into the bar and you have a fixed background temperature.  Therefore, the only way to raise the temperature of the heated bar is to reduce its surface area.   

The second bar is a heat sink, not an insulator so no, it can't cause the temperature of the first bar to increase, it effectively gives the second bar mor radiating surface so it serves to lower the temperature of the heated bar.


----------



## wirebender (Nov 14, 2011)

IanC said:


> wirebender- you have a distorted view of thermo dynamics.



ian, my view of thermodynamics is fine and the math supports my view.  In fact, I hold my view because the math supports it.  If the math supported your view, I would support your view and subscribe to backradiation.



IanC said:


> why does a blanket make you feel warmer? because it decreases heat loss. does the blanket actually heat your body? no, your body heats itself by processing food. less food needs to be 'burned' to keep your internal temperature stable when you use clothing or blankets to decrease the heat loss to the environment.



There is yet another mistake on your part ian.  Why does a blanket make you "feel" warmer?  It doesn't matter a whit what you feel ian, it doesn't actually make you warmer.  The blanket traps dead air which you feel around you but it does not make you warmer.  If you wrapped a blanket tightly around yourself so that no dead air space were trapped, it would serve as a heat sink and drop your body temperature.  

The second bar in the experiment is not acting as a blanket however, it is acting as a heat sink.



IanC said:


> in Spencer's experiment, does the second bar actually heat the heated bar? no, the electrical power going into the heater heats the bar, the second bar only insulates which means the rate of heat loss to the environment is decreased.



The second bar is a heat sink ian, not an insulator.  I really don't see why that is so difficult for you to grasp.  There is no real area for confusion between heat sinks and insulators.  The second bar is identical to the first bar except it has no heating element inside.  Where does this insulating property you are so adamant about come from.  Which physical law are you depending on to support the claim that it is a heat sink.  I am using the Stefan-Boltzman law to support my claim that it is a heat sink.

The second bar effectively adds surface area to the first bar.  Adding surface area will cause a temperature decrease.  



IanC said:


> if the same amount of power goes into the heated bar then the temperature will be higher when it is insulated than when it was not.



But the second bar is a heat sink, not an insulator.  If you wrapped fiberglass insulation around the heated bar, you could certainly maintain a higher temperature with less electricity or generate a higher temperature with the same amount of electricity but that isn't what the second bar is doing.  

The claim in that experiment is that backradiation from the second bar is causing the heated bar to be warmer and it simply isn't happening.  Heat sinks don't cause the radiator to increase in temperature.



IanC said:


> but it is the electrical power that heats it. have you been quibbling over some strange semantics?



I have been arguing ofer your gross misunderstanding of what is happening in the experiment.  The difference between a heat sink and an insulator is not a matter of semantics.  They are two different things and facilitate two entirely different reactions.  The second bar, identical to the first can not be construed to be an insulator.  That is like saying that if you put a house identical to your own next to your own home, the second house will act as an insulator.  Do you really believe your neighbors house keeps your own house warmer?




IanC said:


> Spencer said the first bar will become warmer if a second bar is placed next to it. are you denying that,



Yes I am denying that.  Spencer clearly doesn't grasp the difference between an insulator and a heat sink if he believes anything like what you are describing is happening.  Again, does your neighbor's house insulate your house?  




IanC said:


> or are you just denying that heat energy is coming from the second bar?



Spencer is claiming backradiation from the 2nd bar is warming the first bar.  If you don't get that, then you have once again, entirely missed the boat and don't even know what you are arguing about.



IanC said:


> you have built a strawman by saying anyone has claimed heat was flowing into the first bar from the second. the claim is that the second bar insulated the first causing the heater to be more efficient in raising the temperature.



ian, you really don't get it do you?  Look at the arrows on spencers diagarams.  He is claiming backradiation from the second bar is causing the first bar to be warmer.  



IanC said:


> the same holds true for the earth/atmosphere CO2/greenhouse argument.



Sorry ian, that is another argument you have entirely missed the boat on.  



IanC said:


> no one has said that the atmosphere is warming the earth's surface (except under rare inversion conditions). the claim is that CO2 is decreasing heat loss by radiation which means the sun's incoming radiation will warm the surface more than when there was less insulation.[/quote
> 
> Of course they are.  Look back to the beginning of this thread at trenberth's diagram.  That tan bar on the far right hand side of the diagram pointing towards the earth with the number 324 labeled backradiation and the words underneath saying absorbed by the surface is most certainly making the claim that 324 watts per meter squared are being radiated from the atmosphere and are being absorbed by the surface of the earth and by being absorbed is serving to warm the surface allowing it to radiate more energy than it receives from the sun.
> 
> ...


----------



## polarbear (Nov 14, 2011)

IanC said:


> polarbear said:
> 
> 
> > IanC said:
> ...


----------



## wirebender (Nov 14, 2011)

Polarbear, you are going to run afoul of ian here because he views photons as free agents that go zipping about the universe in more or less straight lines till such time as they run into some sort of solid matter.  

He doesn't recognize a photon as the smallest possible unit of energy in an electromagnetic field which can exhaust itself in opposition to an EM field of greater magnitude propagated from the opposite direction.  He seems to believe that when EM fields cancel, the photons that made up those fields continue to zip about the universe till they hit something even though the EM fields which they made up effectively cancelled each other out or were diminished to the point of non existence via the subtraction of EM fields.

And don't even bother mentioning particle-wave duality or the fact that phenomena like interference and cancelling of EM fields can only be explained if one assumes EM fields to be waves rather than separate photons zipping about the universe till they run into some sort of solid matter.


----------



## polarbear (Nov 14, 2011)

wirebender said:


> Polarbear, you are going to run afoul of ian here because he views photons as free agents that go zipping about the universe in more or less straight lines till such time as they run into some sort of solid matter.
> 
> He doesn't recognize a photon as the smallest possible unit of energy in an electromagnetic field which can exhaust itself in opposition to an EM field of greater magnitude propagated from the opposite direction.  He seems to believe that when EM fields cancel, the photons that made up those fields continue to zip about the universe till they hit something even though the EM fields which they made up effectively cancelled each other out or were diminished to the point of non existence via the subtraction of EM fields.
> 
> And don't even bother mentioning particle-wave duality or the fact that phenomena like interference and cancelling of EM fields can only be explained if one assumes EM fields to be waves rather than separate photons zipping about the universe till they run into some sort of solid matter.



That may be so, but then again this is more or less how the strange universe which is ruled by "climatology-physics" photons are supposed to behave...and if IanC is restricting his reading material to this kind of "simple matter" physics to defend Trenberth`s "energy budget" or Spencer`s example, which is a pseudo-logic classic  then  I won`t fault him (personally) .

What worries me is not so much what IanC is reading or what he subscribes to,...what really worries me a lot, is ...:
Are the schools preaching this kind of nonsense in today`s physics classes also..????...or is it just in the Internet...?
And is IanC a typical product of this Orwellian "new-speak" physics...?

Unlike some of the mud flinging trolls( You know ...like what`s his name Thunderfart or something like that)... You, Gslack, Westwall, Skooker-bill & I had run into in some of the other threads, I can`t see any such malice or ill will in IanC, ...he is convinced he is believing and doing the right thing by continuing to defend this hypothesis which does not have a leg to stand on in the so called "hard sciences".

As far as I can see I have answered all the questions he asked,...
but I have yet to get an answer to any of the questions I have asked him.

But be that as it may up to this point I have enjoyed this discussion...
Thanks for letting me in on it guys...!!
And my best Greetings to all of You..


----------



## gslack (Nov 14, 2011)

Frankly my faith in humanities future has been lessened lately.. one can now minor in "climate change" from virtually any liberal arts college. That And the complete ignorance of the general populace regarding money and finance, tell me that there will be a reckoning all too soon.. One that we as a whole are ill-prepared for.  

Imagine a biblical apocalypse and we have scientists like spencer and so many others to turn to... Shudders me to my soul...


----------



## IanC (Nov 14, 2011)

polarbear said:


> IanC said:
> 
> 
> > polarbear said:
> ...


----------



## gslack (Nov 14, 2011)

Ian backtrack much? Gotta love revisionist history especially when its recent enough to verify.. Or dispute... I loved this line...

_*"I am simply describing how CO2 slows the shedding of heat from the earth's surface."*_



no ian what ya did was call wire everything from wrong with no proof, to outright calling him a crackpot using esoteric claims, likened EM fields to made-up nonsense, and even harassed me simply because I proved you were dead wrong on that... Hell man you followed me around for 3 days harrassing my posts over it like a child..

YOU DIRECTLY ATTACKED WIREBENDER FROM THE START!! 

Sorry but your behavior has been inexcusable, and until there is an apology and acceptance of your responsibility in this you can spare your act....


----------



## wirebender (Nov 15, 2011)

polarbear said:


> What worries me is not so much what IanC is reading or what he subscribes to,...what really worries me a lot, is ...:
> Are the schools preaching this kind of nonsense in today`s physics classes also..????...or is it just in the Internet...?
> And is IanC a typical product of this Orwellian "new-speak" physics...?



I can't speak to most of it, but the corrupted version of the SB equations where the SB law is applied twice, once from the emitter to the background and again from the background to the emitter is being taught in climate science physics classes.  On an earlier post in this thread, I provided ian a list of physics books in which the corrupted version of SB was taught and was not taught.  Classical physics where the practitioners are likely to move on to engineering, astrophysics, etc are taught the actual SB law while climate physics teaches the corrupted version of SB.


----------



## wirebender (Nov 15, 2011)

IanC said:


> I am simply describing how CO2 slows the shedding of heat from the earth's surface.



Like the difference between heat sinks and insulators, you also don't seem to grasp what "slowing down" means.  IR radiates away from the earth at, or very near the speed of light and doesn't slow down and none of it is beamed back towards the earth by CO2 molecules. 



IanC said:


> my concern is that if we deny obviously true effects that reasonable people will ignore us on other topics as well.



Take your own advice ian, you are denying the laws of physics in an effort to support an invention of climate pseudoscience that is necessary to promote climate alarmism.  Which law of physics is it that you believe supports the notion of back radiation?



IanC said:


> why should warmists believe your criticisms of real problems like climate feedback numbers when you disavow even simple physical concepts?



It is you who is denying simple physical concepts ian.  Hell you just applied the workings of an insulator to a heat sink in order to support the crackpot idea of backradiation.

This may come as a surprise to you ian, but you are not the voice of reason here.  You are the barking moonbat who supports the corruption of physical laws for pete's sake in an effort to rationalize a hypothesis that simply isn't supported by hard science.  You readily stand by even when you are shown how a physical law has been corrupted and continue to support claims which have not, can not, and never will be substantiated either mathematically or by repeatable, observable, experimentation in the lab or in nature.


----------



## polarbear (Nov 15, 2011)

IanC said:


> all of what you said *simply* reinforces my statements.
> 
> the earth wants to shed heat by radiation.
> 
> ...


 


So You have finally committed Yourself to specify where this *"extra time"* is supposed to come from...the *"longer time"* that the 1.7 or so watts are spending in the 
 strange universe of "climatology".
 True, the speed of light is proportionally reduced by the optical  density of the transparent gas a beam of light has to traverse.
 This is also observed as the "refractive index"...which also goes up as the density of this gas increased.

 Finally we get to do the  math and see as a *number *what You say is gained as "*extra time*"..instead of using _*"much longer" *_etc.
 The much longer is  in the case of air at standard pressure & temperature as opposed to a vacuum :

   (1/299792368)-(1/ 299792458)=  ~ 1 * 10^(-15) * "extra time*" in seconds for each meter
 I really don`t care how many energy (photon) quant`s You want to use, use as many as You like to get Your back-radiation *POWER flow* of  say 1.7 Watt. and please do apply  this "*extra time*" how "*much longe*r" these climatology photons are spending because of the "CO2 obstructions"

 So IanC, do tell us how much  *ENERGY* that amounts to ... these extra 1 *10^(-15) seconds
 (....ENERGY expressed as Watt seconds...)
 If You won`t tell then I shall. 
 It comes out to a miniscule 1.7 * 10^(-15)  Watt seconds  of heat *ENERGY *
 So, Wirebender, Gslack, I`m sure they would...but _I shall ..._decide to let You have say a  couple of hundred watts worth` of "back-radiation" for an "extra time" of 0.0000000000000015 seconds in your explanation how "climatologists"  can measure with thermometers the so called extra heat ENERGY increase You  could
 achieve with that...
"climatologists" keep claiming they have measured this extra CO2 caused temperature increase,...and that they did so with thermometers...or do You argue that...???

 And now to Your other statement:


> if you were correct that it* is 10^24 then it  only strengthens my case*. the *longer *the photon is held in the  atmosphere the _*more inefficiency in heat loss*_.


 Why do I have to continue to give You free tutoring in physics....in this case what is what in an absorption spectrum..
 I linked You to some literature and You could have educated Yourself in that regard...
 Obviously You did not...

 Absorption is measured by  a spectro-photometer which sends a beam of light from a source through air
 containing CO2 in our case.  
 Nearly 100 % of the light would reach the detector if there is nothing in the straight line path from the source to the detector.
 So now we put air containing 380 ppm molar CO2 in this path.
 The detector will now experience a drop in received energy, the dark absorption lines in the picture + the explanation I have linked You to.
 Had You studied up on this subject then You should not have any trouble at all understanding that
 the more molecules are encountered that can absorb this light on the* straight path *from the source to the detector,....the more
 of Your mechanical "climatology photons' have strayed from the straight pass from the source to detector and went off
 in all other possible directions.

 They don`t spend any "extra time" in there which the detector experiences as a drop in energy transmission.
 And  they did not "back-radiate" to the source either.

 Because on the way back these ever so mechanical  Ian photons encounter the same problem as they had
 on their path to the detector.

 And You  come back and say, the higher the number of encounters as can be calculated using the Loschmitt number
 the better for "back-radiation" or using Your words it "strengthens Your case".

 I noticed You use the word *"simple"* quite often...and that is the problem.
 You should start doing some serious,...not just* simple * thinking and reading before You make any more statements like this 

 Also have You ever given it a simple thought what your climatology  photons have to pass through on their way back, as opposed to the way .

 On the way back  the air + CO2 gets increasingly denser and is as You keep putting it "obstructs" IR radiation 
 while on the way out the air gets progressively less dense and the number of CO2 molecules that are
 obstructing Your climatology photons get less and less.

 Lets not even get into that  air which has been warmed plus the CO2 in it which is supposed to "back-radiate"
 starts climbing at rates which can be in excess of 3000 feet per minute...adding this distance  (=*~ 17 additional meters each second*) the
 "climatology photons" would have to traverse on their way back to earth.

 IanC...Do You understand how a diode or a "rectifier" works..???
 Well if somebody would have to design a "rectifier" or a diode for  electro magnetic (heat) radiation
 which  transmits stored heat  quite well in one direction, away from earth... and  impedes the "back-radiation" path..
 a more efficient design than how our atmosphere and applying the laws of thermodynamics  work
 does to date not exist.

 In the absence of gravity on this "climatology" model-planet,  "warm" air+CO2 would just sit and hover where they were and be a  drag on photons preventing the "escape of heat" but add heat "back-radiating" heat, now with no obstruction and all day long over the same short distance that gravity/convection in reality does add in addition to everything else I have pointed out to You.
 But  in Your *"simple"* concept which You insist we should accept as a  proof of concept for global warming the planet earth is a simple  "black-body", has no gravity  and  "climatology photons", 
 CO2 has a lot of "drag" on departing photons, slowing these down, 
 ... but the same "drag" does not apply on the "back-radiation" way back down...

 IanC  in essence Your "climatology photons" behave like rocks falling back to earth.

 Say where exactly did You get this :
 "the estimate for the free path of an IR photon is less than 10  interactions, I believe"

 Let me guess, that`s 10 climatology-photon "interactions" with the  380 molar ppm CO2
 Not only does light no longer exist as electro-magnetic radiation ,  equally spreading in all directions and therefore diminishing in  intensity by the square of the distance...it also has to hit something  or have some sort of "interaction" ....say like the 10 You estimated
 before light can diminish in intensity...

 Now You have finally given us an insight into Your simple "climatology"  universe...where the photons fly around like little space ships and  hit  or miss  CO2 molecules that behave like little planets.


 That is alarmingly similar to how simple teenagers  would imagine it after smoking a few too
 many  "joints"....after that everything,....also whatever little brains remain... gets rather *simple*


----------



## polarbear (Nov 15, 2011)

gslack said:


> Ian backtrack much? Gotta love revisionist history especially when its recent enough to verify.. Or dispute... I loved this line...
> 
> _*"I am simply describing how CO2 slows the shedding of heat from the earth's surface."*_
> 
> ...



Yes by a whooping 1 *10^(-15) seconds
But what caught my attention here is the little bomb You dropped here:


> Gotta love revisionist history especially when its recent enough to verify.. Or dispute... I loved this line...


I love it too...!!!
Especially the history that has been verified by none other than the honorable Kamerad Stalin...
The same Stalin who`s word we are supposed to take that under all that steel re-enforced concrete are some of the 6 million Jews us Germans have "zyklon-B`d" to death and which then vanished...

He says, that could be verified by anyone who would be able to dig through all this concrete...which he says the Soviets poured there so that the dead will not be disturbed:




But since ground penetrating radar we have in addition to all this concrete a #130 "holocaust denial" paragraph in Germany, which earns You a 10 year prison sentence if You would pry in any way shape or form to check if Stalin was a liar.
Same goes for anyone who dares to actually analyze for HCN residues in the ruins that are claimed to have been "gas chambers"...
At least one of these, the holocaust "gas chamber" + all the "ovens" icons in Auschwitz has been "re-constructed" by the Soviets from an air-raid shelter,...The Russians + the Polish Auschwitz curator Dr.Piper  finally admitted that on camera to David Cole... that this "re-construction" magic was performed during the time while nobody from the West was allowed to inspect anything inside Soviet occupied territory...

Well I am not claiming one thing or the other, but as far as "verification" is concerned "climatology" is inching ever closer to the same methodology as "holocaust verification research" ...
as long as there are enough fools who believe something then it must be true.


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## IanC (Nov 16, 2011)

thats odd. where did your last post go wirebender? do you mind putting that link up again, my phone doesnt have a history on the browser.

its an interesting idea but I am not really sure that it is fair to compare emissivity of CO2 over a large range when we are interested in IR that the earth gives off. CO2 has a couple of fat bands right in the middle of earth's radiation, with the 15 micron band practically to itself.

judging from the big bite taken out of the 15 micron portion of TOA outward radiation it appears that CO2 does a pretty good job of absorbing that band of radiation. actually I would like to read a paper on how much more energy can even be stopped because it looks like most already has been. I suppose that is why we are in the low slope portion of the log scale for doubling CO2 concentration.


----------



## polarbear (Nov 16, 2011)

IanC said:


> thats odd. where did your last post go wirebender? do you mind putting that link up again, my phone doesnt have a history on the browser.
> 
> its an interesting idea but I am not really sure that it is fair to compare emissivity of CO2 over a large range when we are interested in IR that the earth gives off. CO2 has a couple of fat bands right in the middle of earth's radiation, with the 15 micron band practically to itself.
> 
> judging from the big bite taken out of the 15 micron portion of TOA outward radiation it appears that CO2 does a pretty good job of absorbing that band of radiation. actually I would like to read a paper on how much more energy can even be stopped because it looks like most already has been. I suppose that is why we are in the low slope portion of the log scale for doubling CO2 concentration.



Well if You want to read a paper on it here it is, these are ACTUAL MEASUREMENTS,...:
The Climate Catastrophe - A Spectroscopic Artifact



> [FONT=Arial, Geneva]A 10 cm glass cylinder (150 cm[SIZE=-2]3[/SIZE], with IR-transparent window) was filled with synthetic CO[SIZE=-2]2[/SIZE]-free and vapour-free air. Then a microlitre syringe was used to add CO[SIZE=-2]2[/SIZE] so that the concentration was 357 ppm (concentration in 1993). Moreover 2.6% water vapour was added. Applying the IR beam source (a so-called *Globar *, an electrically heated silicon carbide bar at 1000 to 1200 degC and an adjustable interference filter) on one side, the absorption spectrum arriving at the other end was recorded. Then CO[SIZE=-2]2[/SIZE] was added to make 714 ppm. The equipment was an *FTIR* spectrometer *"Bruker IFS 48"* coupled to a PC. The program *OPUS* was used as analyzing software. A zero bias measurement was made to be subtracted later.[/FONT]





and here are the results:




> [FONT=Arial, Geneva]Crucial is the *relative increment of greenhouse effect *. This is equal to the difference between the sum of slope integrals for 714 and 357 ppm, related to the total integral for 357 ppm. Considering the n[SIZE=-2]3[/SIZE] band alone (as IPCC does) we get[/FONT]
> [FONT=Arial, Geneva](9.79*[SIZE=+1]*[/SIZE]*10[SIZE=-2]-4[/SIZE] cm[SIZE=-2]-1[/SIZE] - 1.11*[SIZE=+1]*[/SIZE]*10[SIZE=-2]-4[/SIZE] cm[SIZE=-2]-1[/SIZE]) / 0.5171 cm[SIZE=-2]-1[/SIZE] = 0.17 %[/FONT]
> 
> 
> ...


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## IanC (Nov 16, 2011)

wirebender said:


> IanC said:
> 
> 
> > I am simply describing how CO2 slows the shedding of heat from the earth's surface.
> ...



it obviously untrue that none of the radiation makes it back to the surface. a great many CO2 molecules near the surface absorb a photon and then randomly reemit it, sometimes directly back to the surface. it is inconceivable that none of the photons reach the surface. photons dont cancel out. calculations can derive net flows but no photons actually disappear because of calculations.

heat sinks operate by conduction and convection. the radiation aspect is so low it is usually ignored. without conduction and convection heat sinks have insulating properties because they decrease the efficiency of radiation loss by lowering the temperature differential.


----------



## IanC (Nov 16, 2011)

polarbear said:


> IanC said:
> 
> 
> > thats odd. where did your last post go wirebender? do you mind putting that link up again, my phone doesnt have a history on the browser.
> ...



thanks for the link polarbear. the whole idea of absorbtion to extinction was my first impression before I started looking into the whole global warming thing. I assumed total absorbance of 15 microns, with enough  swappings to convert the energy into photons that can escape through atmospheric windows. there is not enough information available, and it seems that most scientists on either side of the debate find an increase of ~1C for doubling of CO2. perhaps they are all idiots who refuse to listen to the visionaries like Hug, or perhaps Hug is looking at a restricted range that gives a false impression of the overall picture. I dont know, I am not a physicist. I do know that CO2 absorbs and slows down radiation from the earth from escaping therefore it warms the earth. I do not believe the positive feedback numbers though, I have a strong impression that they are physically improbable.


----------



## wirebender (Nov 16, 2011)

IanC said:


> it obviously untrue that none of the radiation makes it back to the surface.



Of course it is true.  How do you suppose a massless photon which is nothing more than a bit of energy manages to get to the surface of the earth against the current, so to speak, of the EM field emitted by the earth which is hundreds of orders of magnitude more powerful?  



IanC said:


> a great many molecules near the surface absorb a photon and then randomly reemit it, sometimes directly back to the surface.



None make it to the surface ian, even one being absorbed by the surface of the earth woudl violate the second law of thermodynamics which states explicitly that it is not possible for heat to spontaneously flow from cool objects to warm objects without some work having been done to accomplish the task.  Absorption and emission do not constitute work.



IanC said:


> it is inconceivable that none of the photons reach the surface. photons dont cancel out. calculations can derive net flows but no photons actually disappear because of calculations.



If you believe the 2nd law of thermodynamics to be true, it is perfectly concievable.  What is inconcievable is that someone would believe that the 2nd law of thermodynamics is being broken by a trace gas in the atmosphere.  



IanC said:


> heat sinks operate by conduction and convection.



Really?  Your lack of knowledge just keeps expressing itself more clearly as you go on ian.  Perhaps you should read a bit on the topic of vacuum tubes, specifically internal heat sinks.  If heat sinks operate only by conduction and convection, some electrical engineers wasted a great deal of money putting heat sinks into vacuum tubes.  Heat sinks work via conduction, convection, and radiation ian.  How might they know the difference?



IanC said:


> the radiation aspect is so low it is usually ignored. without conduction and convection heat sinks have insulating properties because they decrease the efficiency of radiation loss by lowering the temperature differential.



Sorry ian, but you couldn't be more wrong.  Again, refer to internal heat sinks in the design and manufacture of vacuum tubes. For that matter look at space itself.  Space acts as a heat sink for the earth and radiation is how heat finally gets into cold space.  Conduction and convection are the primary means of moving heat to the upper atmosphere, but from there it radiates into the heat sink of space.   You have, however, hit on the reason that the two bars will eventually reach thermal equilibrium or very close to it and radiate as a single object.  I don't guess you will grasp that either.


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## wirebender (Nov 16, 2011)

IanC said:


> I do know that CO2 absorbs and slows down radiation from the earth from escaping therefore it warms the earth. I do not believe the positive feedback numbers though, I have a strong impression that they are physically improbable.



I believe that you genuinely believe you know that to be true, but alas, it just doesn't happen ian.  

Tell me ian, which law(s) of physics do you beleive support and predict such an idea?


----------



## polarbear (Nov 16, 2011)

IanC said:


> thanks for the link polarbear. the whole idea of absorbtion to  extinction was my first impression before I started looking into the  whole global warming thing. I assumed total absorbance of 15 microns,  with enough  swappings to convert the energy into photons that can  escape through atmospheric windows. there is not enough information  available, and it seems that most scientists on either side of the  debate find an increase of ~1C for doubling of CO2. *perhaps they are  all idiots who refuse to listen to the visionaries like Hug, or perhaps  Hug is looking at a restricted range that gives a false impression of  the overall picture. I dont know, I am not a physicist.* I do know  that CO2 absorbs and slows down radiation from the earth from escaping  therefore it warms the earth. I do not believe the positive feedback  numbers though, I have a strong impression that they are physically  improbable.


 

 No, they are not all idiots and neither are You.
 But they all have one thing in common, You included..and how could I possibly exclude myself....and that is that  each and every one has "overlooked" something,...not just You, but Yes  Heinz Hug also.
 But then the only process which does not "overlook" anything would be a  thoroughly and COMPLETE representation of each and every process that  affects this so called "energy budget"...
 And as it becomes very very clear, this is a far cry from the *"simple"* examples, like Spencer`s and Trenberth`s.

 All people like Heinz (Hug), many others and I are doing is pointing out  what has been "overlooked" in these "simple" examples that conclude  with some rather non-sensible if not outright ridiculous results in  absolute numerical terms.

 No matter who does what calculation, nobody would be able to proceed  even on the first step without Beer Lambert`s laws of absorption.






 So let`s take a look how many ways You can violate this law with  "overlooking" something, before we start going beyond it and go on to  "black bodies"...:

Beer



> *Prerequisites*
> 
> There are at least six conditions that need to be fulfilled in order for Beer&#8217;s law to be valid. These are:
> 
> ...


 Well as You can see it will be rather hard to find anybody who was  innocent,...and has not violated at least one if not all of #1 to #  6,...
 Would You not agree...?


 Wirebender, I hope he does not mind discussing his private message,  which was a Link to an extremely interesting paper which I would like to  mention in context with the above. Here is the Link he sent me:

Mean Free Path Length of Photons in the Earth's Atmosphere

 This publication has more or less the same numbers in it when we  discussed how due to the wave properties of a photon, the "absorption"  of energy as observed by Beer Lambert is really a matter of how much of  the incident light as mentioned in condition # 4 above has been diverted  from the "*radiation must consist of parallel rays'
*
 I did stress this point and have stated that each time there was a  "photon encounter"...in Your examples that each and every time the  energy quantum (the photon) is no longer on this path and thus manifests  itself as a "darkened" absorption band in the spectrum.


 The paper which is doing it the other way around...calculating a "mean  free path" using the molecular cross section and # of molecules per  volume, is thus considering the remaining transmittance of IR, which is  the residual after the absorption...

 The author of this paper is by no standard an idiot...
 Yet he "overlooked" some pretty important things also...

 I have nothing at all against "examples"...no matter, even if an example  has been using photons like Newtonian particles with a momentum,  velocity or directional / a "resting" Mass....and is ignoring the wave  duality properties altogether....I`m still willing to apply the math  walking through such examples..

 But then I should also be allowed the use of examples...
 Although the author of this paper has performed proper math,.. it is not entirely free of oversights.
 Consider this question:
 If a photon has to traverse a gas in which the sum of all absorbing  molecular cross sections is say X...or using the "free path" Y= Total  area-X  ...
 Do they have the same number of "encounters" as You put it,.... if these absorbing CO2 Molecules stand still...as opposed to Molecules in "molecular motion"..???
 Or are there more of these "photon-CO2" encounters if we consider the well known fact that *stored heat *is in the final analysis increased Molecular motion...in random directions

 So does the wind shield of a parked (CO2 molecule" car "encounter" as many photon "snow-flakes" as it would when it moves.

 Would it encounter more snow flakes if the (molecular motion) speed  increases...or does none of that matter and it stays the same as when  the (molecular) car is parked...and it`s only a matter of cross section  area which "blocks" or "catches" the photon (snow flakes)...

But by not considering this rather important detail, ...in no way would this make an idiot out of the author of this paper...

 The main thrust of his publication was to compare the magnitudes of the  CO2 absorption process with the rest of the absorption process not  attributable to CO2 and he came to this conclusion:



> [FONT=Georgia, Times, serif]_Conclusions_
> [/FONT]
> [FONT=Georgia, Times, serif]
> [/FONT]
> ...


And what he said + factoring in what I added, that would not just apply for "departing" photons, but more so to "back-radiation" photons...
_because on the way back down the CO2 molecular motion & temperature increases  progressively _...
and for the so called "obstructed" way out...or the "escape" path, I think You called, *the exact opposite applies*


 I would also like to add, that it would be a monumental task to express  in mathematical terms,...that is to say, perform a "climate energy  budget" calculation which has not "overlooked" anything or has  considered each and every process that in reality is involved...

 That would not be a "simple" thing, like "climatology" would like to  have it with how "simple" it is to proof their hypothesis   and these  "simple" examples

 But back to You...You are on the right track now....
 please do keep reading and discussing what You have been reading.

 Thank You for starting this thread...
 It motivated me to think about physics while I am re-painting the  interior walls of my house...instead of letting TV news background noise  entertain me

 And Thanks to You also,  wirebender for that link..
It was by far more interesting than this morning`s daily news papers






[FONT=Georgia, Times, serif]
[/FONT]


----------



## IanC (Nov 17, 2011)

wirebender said:


> IanC said:
> 
> 
> > a great many molecules near the surface absorb a photon and then randomly reemit it, sometimes directly back to the surface.
> ...



the 2nd law states that there can be no _net_ flow of heat from colder to warmer. the cancelling out of radiation flow going in both directions is a mathematical calculation, not a physical obliteration of photons. and it certainly doesnt happen without the presence of matter. you have put up zero evidence of this magical destructions of photons. you wont even pinpoint _where_ this is supposed to happen. I have asked you dozens of times whether it happens at the surface, CO2 molecule or just in space but you refuse to give an answer.

the closest I have come to finding something that supports your view is Claes Johnson's discussion on how he thinks the back radiation is reversed at the surface, much like reflection. no cancelling out, and the 'resonated reemission' is part of the radiation calculated for the earth's surface, not some extra bit. this calculation works out to exactly the same numbers as the 'corrupted version' of the S-B formula.

one of the biggest problems in physics is being able to resolve what happens on a micro scale to the reality of the macro scale. you are trying to switch back and forth from large to small scale. heat flow only goes in one direction but radiation flow goes in both. an excited molecule doesnt check to see if it is allowed to emit energy by radiation, it just emits. the photon doesnt care which direction it is heading in, it just goes. the photon doesnt care whether it has enough energy to warm a body or just cancel out radiation in the other direction leading to a 'net flow'. CO2 doesnt care if it is a good emitter and absorber at certain wavelengths and a poor one at others, it just is what it is regardless.

do I like the surface radiation and atmospheric back radiation components in Trenberth's diagram? no, I think it should just show ~60W/m2 going up, perhaps with a side diagram showing the the calculation of radiation in both directions. do I think Trenberth has the right figures for the energy coming in and going out? I dont think it is exactly right but it is probably reasonably close.

CO2 stops most of the 15 micron radiation from leaving directly. H2O stops many other bands and also carries latent heat aloft, water droplets reflect microwaves, and a hundred other things happen. perhaps the CO2 portion will be reassigned a different number in the future, maybe not. But there is a physical explanation for what physicists say is happening. until there is proof that that explanation is wrong, and accepted by more than a few fringe crackpots, I will continue to believe that CO2 is a factor in atmospheric processes but I will also continue to be sceptical of the high feedback values which lead to global warming alarmism.

wirebender, care to explain where and how all these photons magically disappear? in the real world and not in some calculation?


----------



## IanC (Nov 18, 2011)

I finally found a nice diagram of blackbody radiation for two temperatures in the range of earth/atmosphere interactions. do you notice how they almost perfectly overlap? how can a body tell the difference between a photon radiated from 10C or at -10C? how can a 0C body only accept radiation from 10C but refuse the same photon from -10C? 

not only did I find the diagram but the article that goes along with it is outstanding.

The Amazing Case of &#8220;Back Radiation&#8221; &#8211; Part Three « The Science of Doom






the upwards and downwards radiation was measured...






and compared to the theoretical radiation that should have been emitted that day for the daily temperatures.


----------



## gslack (Nov 18, 2011)

Ian please stop using science of doom as a reference.. he is a 7th grade teacher who was let go from his job as a PR man for the NOAA.. And he routinely does the same crap spencer does using half-science mixed with speculation and twisted logic..

Seriously, He is a PR man cashing in..


----------



## polarbear (Nov 18, 2011)

IanC said:


> The Amazing Case of &#8220;Back Radiation&#8221; &#8211; Part Three « The Science of Doom
> 
> 
> 
> ...



When academics discuss a subject they can afford to use "simple examples" and can even afford to "overlook" things while experimenting.
Engineers may not have the amount of academic schooling to win such purely academic discussions, but on the
other hand they cannot afford any oversights or justify their decisions, should they result in a disaster with "simple examples" that in theory sounded "logical" or that an "experiment showed"...or that "almost all agreed that blah blah" etc etc.
Engineers  do not experiment, lives depend on it that they don`t .
So we have to be dead sure,...just boasting that the silenced critics had no counter argument never got any fuck up engineer off the hook


 While purely academic people do concern themselves with the laws of physics, they are not involved in applying these
to a real world scenario which is the realm of engineering..

 Engineers  must understand the laws of physics not just in the abstract, but also all the implications that go with these.
To carry out their task they do use certain principles and the numbers that go with it all the time on an almost daily basis....

 I have been an academic first and turned engineer later...so  I would be the last one to ridicule one or the other
but I do know the limitations of  any person who chose to be ONLY either one or the other.

 So the first thing I noticed in Trenberth`s energy Budget and all the other "examples" was this gaping hole which
remained when trying to bridge the gap from paper based theory to a real world scenario:

 Engineers measure Heat conductivity...
"Standard" air as it is ( the exact CO2 was not specified) has a heat conductivity of  0.025 Watts per meter per degree K
That was most likely measured in this century...so let`s plug in 370 ppm (Molar) CO2

 pure CO2 has under the same conditions a heat conductivity of  14.65 milli watts per meter and per deg Kelvin temperature
difference between the "hotter" and the "colder" .


So we can calculate that the loss in heat conductivity the climatologists like to refer to as the insulation effect of
CO2...but since they want to avoid going into heat convection etc they prefer to put the this word "insulation"
inside  quote marks and keep on talking about "black body back radiation" that has manifested itself as an
"insulation effect"...."blanketing the earth" and so on...


Well when engineers do measure the heat conductivity of a gas such as CO2, they do not install an optical
barrier which blocks heat radiation, be that back or forward radiation...
Engineers measure the entire heat energy flow in watts that can pass through CO2.

 So let`s do the math for how much a gas that has a heat conductivity of 14.65 milli watts / (m*K)
can reduce the heat conductivity of air with no CO2

 We can also say that A (Air) +  370 ppm C (CO2) = 0.025 

 that  A without any CO2  has a thermal conductivity of

  A = 0.025 -(0.0145*370*10^-6) =0.024994635.

In the final analysis over the distance of 1 meter and over a Temp differential of 1 Kelvin  the heat conductivity dropped by only  0.005365 milli-watts
  which is the amount of "extra heat energy" per second  the 370 molar ppm CO2 can "insulate."

And I do want to re-iterate that  in all the actual measurements for heat conductivity the radiation component is not blocked out
but is included...
The only thing that is excluded as much as it is possible in these measurements is convection.
And that`s the way we design the insulation we do use...in the so called "Fiber glass insulation"
the role of the fiberglass is to suppress convection, the air it traps is the insulator...
and don`t even think for one second that aside from common air no other gas has not thoroughly been tested.
They have been, especially in the design of "thermal windows"...
Last year I made a joke that window any manufacturer knows a lot more about CO2 and "back-radiation"
than any of these climatologist...the sad part is, that it was`nt a joke

So my argument against academics like Trenberth is not just purely academic, ...

Let me put this in other terms.
Suppose an engineer is commissioned to design a greenhouse for very expensive & exotic plants and
uses the climatology "back-radiation" numbers + the "back radiation examples" .

Then trying to use the extra CO2 "back radiation-("insulation") effect" to save fossil fuel , or to save energy  in a general sense this engineer  under designed the Greenhouse heat requirements
by the wattages climatology keeps coming up with.
There is no way this engineer would get another job, no matter how often he quotes Trenberth or anybody else.


----------



## IanC (Nov 18, 2011)

gslack said:


> Ian please stop using science of doom as a reference.. he is a 7th grade teacher who was let go from his job as a PR man for the NOAA.. And he routinely does the same crap spencer does using half-science mixed with speculation and twisted logic..
> 
> Seriously, He is a PR man cashing in..



as usual you add nothing to the conversation. the only thing different is that you insult someone other than me.

do you have some sort of explanation as to how photons of the same wavelength and energy are sometimes absorbed and sometimes not, depending on the temperature of the object radiating them? 

it sure makes things a lot easier to imagine with both objects radiating towards each other and heat flowing in the direction of net radiation.


----------



## IanC (Nov 18, 2011)

polarbear said:


> IanC said:
> 
> 
> > The Amazing Case of Back Radiation  Part Three « The Science of Doom
> ...



I am sorry but I dont quite understand what I am supposed to take away from your comment. CO2 is a greybody absorber and emitter. are you saying that we should ignore its good emission/absorbance at 15 microns because of its poor emission/absorbance at other wavelengths? the 15 micron band is where it has an effect because other substances in the atmosphere dont react strongly there. the TOA emission measured from satellites show that the 15 micron band has been strongly reduced by passing through the atmosphere. obviously the 15 micron band has been transformed into some other type/wavelength of radiation before the energy escaped.


----------



## polarbear (Nov 18, 2011)

IanC said:


> I finally found a nice diagram of blackbody radiation for two temperatures in the range of earth/atmosphere interactions. do you notice how they almost perfectly overlap? how can a body tell the difference between a photon radiated from 10C or at -10C? how can a 0C body only accept radiation from 10C but refuse the same photon from -10C?
> 
> not only did I find the diagram but the article that goes along with it is outstanding.
> 
> The Amazing Case of &#8220;Back Radiation&#8221; &#8211; Part Three « The Science of Doom



Well IanC I have had the time to read this publication and the only thing I can find which is outstanding is how ignorant of physics the author is.



> Notice the similarity between the 10°C and the -10°C radiation curves.
> Think of individual photons as anonymous &#8211; a 10&#956;m photon from a  2,000K source has exactly the same energy as a 10&#956;m photon from a 200K  source. No one can tell them apart.


So not only are all thermodynamic laws "imaginary", but so is Wien`s displacement law:






where _&#955;_max is the peak *wavelength*, _T_ is the absolute temperature of the black body, and _b_ is a constant of proportionality called _Wien's displacement constant_, equal to 2.8977685(51)×10&#8722;3 m·K (2002 CODATA recommended value).

This diagram where this guy says :


> Notice the similarity between the 10°C and the -10°C radiation curves.
> Alert readers who have pieced together these basics will already be  able to see why the imaginary second law is not the real second law.


You can see Wien`s and the other "imaginary laws"...
did You not notice how the +10 T total emitted energy peak has been shifted by ~ 3[microns] to the left...the shorter wavelength..!!!

I won`t fault You if You did not spot it, especially not when You allowed this physics tourist guide to give you a tour like he does.


Even "Wikipedia" would have been a better tourist guide, at least they point it out to their physics tourist visitors coming to their web site:

Wien's displacement law - Wikipedia, the free encyclopedia


> *Wien's law is not obvious in the picture*,  because the total emission includes a geometrical factor of 1/_&#955;_2 which counts the number of fourier modes of wavelength _&#955;_, and a second factor of 1/_&#955;_2 to convert intensities per-unit-frequency to intensities per-unit-wavelength


So since as this goof ball claims these are just "imaginary laws" all of this must also be merely an illusion:

Near, Mid and Far-Infrared



> The *wavelength* at which an object radiates most intensely *depends on its temperature.* In general, as the temperature of an object cools, it shows up more prominently at farther infrared wavelengths. This means that some infrared wavelengths are better suited for studying  certain objects than others.
> 
> As we move from the *near-infrared* into *mid and far-infrared *regions of the *spectrum*, some celestial objects *will appear while others will disappear from view.  *


And on top of that this I am searching for a proper word here...has the nerve to use the "Wien down-shift" on solar radiation to the wave length where CO2 can "absorb"...actually You should know by now, that is really just a dispersal in other directions rather than having "absorbed" energy and storing it as heat...but on the way up he is trying to tell you that there is no "Wien up-shift" when a body gets hotter.

And thus according to him :


> Notice that there is *no dependence on the temperature*  of the source. Think of individual photons as anonymous &#8211; a 10&#956;m photon  from a 2,000K source has exactly the same energy as a 10&#956;m photon from a  200K source.



And goes on to use this as "proof" that the photons from a "colder body" can heat a "hotter body"
I live in Canada, and I wish I could heat my house by hanging a few large ice cubes outside my windows.

So what would be the proper word (???) to attribute to somebody who publishes garbage like he does...I let You pick it


----------



## IanC (Nov 18, 2011)

> You can see Wien`s and the other "imaginary laws"...
> did You not notice how the +10 T total emitted energy peak has been shifted by ~ 3[microns] to the left...the shorter wavelength..!!!



why wouldnt the peak be more intense and at a higher wavelength for a warmer temp? I dont follow your reasoning.

the point that the author was making is that there is a spread of wavelengths for any temperature and that the overlap between close temps is very close to complete. there is no way to tell whether any individual photon is from the warmer or cooler object.

the flawed 2nd law-


> It (the flawed theory) goes like this:
> 
> 1.the atmosphere emits &#8220;a photon&#8221;
> 2.the photon reaches the surface of the earth
> ...



and the point of the two curves- 


> Alert readers who have pieced together these basics will already be able to see why the imaginary second law is not the real second law.
> 
> If a 0°C surface can absorb radiation from 10°C radiation, it must be able to absorb radiation from -10°C radiation. And yet this would violate the imaginary second law of thermodynamics.



edit- I meant to say higher energy wavelength, which means a shorter wavelength, shifted to the left on the graph


----------



## IanC (Nov 18, 2011)

and just to be clear, he is not saying the 2nd law of thermodynamics is imaginary. he is saying that some people's interpretation of the 2nd law is imaginary. I believe he is referring to Claes Johnson, who has been parroted in an even crazier form by wirebender who claims the photons are not just bounced but actually obliterated in space without the need of matter to be present.


----------



## polarbear (Nov 18, 2011)

IanC said:


> > You can see Wien`s and the other "imaginary laws"...
> > did You not notice how the +10 T total emitted energy peak has been  shifted by ~ 3[microns] to the left...the shorter wavelength..!!!
> 
> 
> ...


 

 Look again closely at the diagram. The peak where the highest output from the "hotter body" is, *has been shifted to the shorter wave-length.*.. by about 3 microns.
*And the peak is higher at this higher frequency..!!!*

 If You can`t understand how an IDEAL "black body" radiates then read up  on the theory again. A "black body" is assumed to consist of a material  that can in THEORY oscillate over the entire frequency range...but it is  the AMPLITUDE You get at a higher frequency that makes the difference.

 You fell for it again...!
  That tells me that You still don`t understand Plancks`s law, because You  are trying to do so studying only the kind of "material" that comes  from sources that probably don`t understand it either.

  Here, read some more about it:

Planck's law - Wikipedia, the free encyclopedia



> In terms of frequency (&#957 or wavelength (_&#955;_), Planck's law is written:
> 
> 
> 
> ...


 Look again at this diagram






 This guy has sold You  the con, that there is no frequency shift and  that "photons" coming from a colder source are indistinguishable from  colder source photons...

Even  in his diagram where he collapsed the scale to mask the frequency shift You can still see it
 I don`t have the time right now to go back and read his silly web page..but as far as I remember he used these words:
 "photons coming from a 2000 K source are the same as photons coming from a 200 K source"...

 How exactly has he established this astonishing feat...
 On that web-page there is a picture of a calibrated thermal imaging  device which for all intents and purposes is just a stage prop.
 He is not MEASURING with that anything we did not know, or which shows  that photons from a "colder source" can add heat energy to a hotter  source.
 The only thing they have measured in this farmer`s field was incident  and outgoing IR...which is specific to the temperature, they did not  even bother to record and what kind of dirt & plants the thermal  detector was looking at...NOTHING ELSE...and NOTHING NEW...



> If a 0°C surface can absorb radiation from 10°C radiation, it  must be able to absorb radiation from -10°C radiation.


 What kind of reasoning is that..???
 of course a 0°C surface  can absorb heat energy from a + 10 °C source..
 How do You get from that to :


> ...it must be able to absorb radiation from -10°C radiation.


 and  have allowed yet again allowed some quack to present photons to You  as if they were particles that You could somehow accumulate and save  like pebbles, till You have enough "photon/heat energy" from the colder  source to raise the temp. of the hotter source where these photons came  from....and he is trying to use a Planck "black-body radiation" versus  Temperature graph trying to establish this...

You also continue to foist the burden to de-bunk this kind of crap on me...!
Why don`t You study up on the physics which are involved ;

Second law of thermodynamics - Wikipedia, the free encyclopedia



> *Clausius statement*
> 
> German scientist Rudolf Clausius is credited with the first formulation of the second law, now known as the _Clausius statement_:[4]
> _No process is possible whose sole result is the transfer of heat  from a body of lower temperature to a body of higher temperature._[note 1] Spontaneously, heat cannot flow from cold regions to hot regions  without external work being performed on the system, which is evident  from ordinary experience of refrigeration, for example. In a  refrigerator, heat flows from cold to hot, but only when forced by an  external agent, a compressor.


So that settles it for *HEAT FLOW.*..if You want to *rule out* heat flow, and want to show me an experiment + the  measurement that makes the case for *back-radiation *then You have to come up with something better than this silly stage prop setup:





Because the first thing You`ld have to do is to *disallow HEAT FLOW*, and *disallow any contact of the air above the farmer`s field and the ground below.*


So then let`s go from HEAT FLOW to radiation:



> *Kelvin statement*
> 
> Lord Kelvin expressed the second law in another form. The Kelvin statement expresses it as follows:[4]
> _No process is possible in which the sole result is the absorption  of heat from a reservoir and its complete conversion into work._ This means it is impossible to extract energy by heat from a  high-temperature energy source and then convert all of the energy into  work. At least some of the energy must be passed on to heat a  low-temperature energy sink. Thus, a heat engine with 100% efficiency is  thermodynamically impossible. *This also means that it is impossible to  build solar panels that generate electricity solely from the infrared  band* of the electromagnetic spectrum without consideration of the  temperature on the other side of the panel


But if these climatology photons, the  ones Your physics mentor claims are no different if they come from a "hot" source or a "cold source" then these should work just fine on solar panels..
*Unitarity (physics)*




> In quantum field theory one usually uses a mathematical description which includes unphysical fundamental particles, *such as a longitudinal photons. **These particles must not appear as the end-states of a scattering process.*


*And as You should know by now *that these IR light (photons) are not really absorbed by a transparent medium like air or CO2 and heat this medium, they are *merely SCATTERED as an electro-magnetic wave* in all possible directions...and are as a consequence "missing" at the detector which is in line of sight with the source-emitter..

But You have not been reading any of this and keep coming back with yet another photon "back radiation" re-appearance example which are in gross violation of all quantum physics...and as an end product "add heat" from a colder receiver to the hotter source...

There are no "simple examples" to help explain quantum physics to somebody that has only a spotty knowledge of physics.

But that should not be viewed as some sort of insult,..after all You were not the author of the web-page that published this nonsense
The Amazing Case of &#8220;Back Radiation&#8221; &#8211; Part Three « The Science of Doom

and all the other physics laws including quantum physics laws are "imaginary laws" I think are the words he used...
But never mind that,...look at this spin doctor twist quantum physics with curves that to him look perfectly "similar"




> Notice that there is *no dependence on the temperature*  of the source. Think of individual photons as anonymous &#8211; a 10&#956;m photon  from a 2,000K source has exactly the same energy as a 10&#956;m photon from a  200K source.
> Notice the *similarity *between the 10°C and the -10°C radiation curves.
> 
> 
> ...


and then uses this as some sort of "proof"...You did not notice this at all..?

How about what he says here..::



> If you can afford an FT-IR to do a spectral analysis you will be able to  prove your theory beyond a shadow of doubt &#8211; as the spectrum will have  those characteristic CO2, O3 and water vapor peaks that were shown in  DLR spectra in Part Two.





> That he did not even bother to do a *spectral analysis*, which could have singled out his "DLR" (Downward *Longwave *Radiation)  if it existed in the *entire IR frequency range *his  Hollywood like stage prop setup was looking at.
> 
> Where exactly did he measure anything that excluded any of the incoming solar radiation and the wavelength associated with his so called "DLR measurement"
> 
> ...


----------



## IanC (Nov 19, 2011)

this graph shows that practically all of the radiation from both the +10C and -10C bodies is radiated from 4-50 microns. the average wavelength for the warmer body is shifted towards 4 microns side(more energetic), and the warmer body is radiating more at every wavelength. 

the type of radiation is almost exactly the same! 4,5,6,7....49,50. both bodies emit the same types of radiation! the warmer body just radiates more at every wavelength. if both bodies radiate a portion of their energy at 12 microns how is a 0C body supposed to know that it can accept the 12 micron photon from the warmer body but not the 12 micron photon from the colder body? there are no tags on radiation saying what temperature body they were emitted from!

the warmer body gives off MORE radiation, and at a slightly higher average energy levels. the graphs for any temperature have the same shape (within reason). the reason why heat always flows from warm to cold is because there is always MORE radiation coming from the warmer body. the warmer body always has the same amount of radiation as the colder body (both in wavelength and intensity) PLUS the extra radiation above the colder body's graph( which is shifted to a higher average energy=lower wavelength). 

the 2nd law is derived from how bodies radiate. a body's radiation is not derived by the 2nd law


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## IanC (Nov 19, 2011)

polarbear said:


> The Amazing Case of Back Radiation  Part Three « The Science of Doom
> 
> and all the other physics laws including quantum physics laws are "imaginary laws" I think are the words he used...
> But never mind that,...look at this spin doctor twist quantum physics with curves that to him look perfectly "similar"



the only thing he called 'imaginary' was the faulty interpretation of the 2nd law. but I will go back and reread it to make sure.


----------



## IanC (Nov 19, 2011)

> In this article we will consider what happens when this radiation reaches the ground. The reason we want to consider it is because so many people are confused about &#8220;back radiation&#8221; and have become convinced that either it doesn&#8217;t exist &#8211; covered in the previous two parts &#8211; or it can&#8217;t actually have any effect on the temperature of the earth&#8217;s surface.
> 
> The major reason that people give for thinking that DLR can&#8217;t affect the temperature is (a mistaken understanding of) the second law of thermodynamics, and they might say something like:
> 
> ...



his reason for writing the article. with a scanned in page of a physics text.



> One commenter on an earlier post asked this question:
> 
> But if at the surface the temperature is higher than in the atmospheric source then might the molecules which might have absorbed such a photon be in fact unavailable because they have already moved to a higher energy configuration due to thermal collisions in the material which contains them?
> 
> ...



first utterance of imaginary---in reference to a distorted view of the 2nd law.



> Of course, radiation from different temperature sources do have significant differences &#8211; in aggregate. What most, or all, believers in the imaginary second law of thermodynamics haven&#8217;t appreciated is how similar different temperature Planck curves can be:



second use of imaginary---again in reference to the distorted definition of the 2nd law. 



> Notice the similarity between the 10°C and the -10°C radiation curves.
> 
> Alert readers who have pieced together these basics will already be able to see why the imaginary second law is not the real second law.
> 
> If a 0°C surface can absorb radiation from 10°C radiation, it must be able to absorb radiation from -10°C radiation. And yet this would violate the imaginary second law of thermodynamics.



third and fourth use of 'imaginary'----again in reference to the mistaken interpretation of the 2nd law.



> So from basic physics and basic material properties it should be clear that radiation from a colder surface cannot be all reflected while at the same time radiation from a warmer surface is absorbed.
> 
> And if any radiation is absorbed it must change the surface temperature and therefore violate the (imaginary) second law of thermodynamics.
> 
> You have to ditch something. I would recommend ditching the imaginary second law of thermodynamics. But you can choose &#8211; instead you could ditch the first law of thermodynamics, or the basic equation for the energy of a photon (make up your own), or invent some new surface properties.



fifth and sixth utterances. still aimed at the twisted version of the 2nd law that wirebender espouses.

that seems to be all of them.


----------



## polarbear (Nov 19, 2011)

IanC said:


> > In this article we will consider what happens when this radiation reaches the ground. The reason we want to consider it is because so many people are confused about back radiation and have become convinced that either it doesnt exist  covered in the previous two parts  or it cant actually have any effect on the temperature of the earths surface.
> >
> > The major reason that people give for thinking that DLR cant affect the temperature is (a mistaken understanding of) the second law of thermodynamics, and they might say something like:
> >
> ...



It`s a little bit taxing for me to explain the same thing over and over again, what the 
difference between Energy and Power is...
Yet here we are again...
In addition to that it`s my turn to stand vigil at my daughters bed-side in the ICU
She has been scheduled for an operation to stop severe internal bleedings

But I`m  going to humor You in the short time I have before I have to leave my house.
I wish You`ld read a real physics book instead of  me having to read it for You.

Perhaps You don`t have any such book, ...and  in my hurry, for now  this is just  a little substitute I could find for You in the 
Internet 

Quantum energy



> In Planck's assumption, radiant energy is emitted in small bursts, known as "quanta". Each of the bursts called a "quantum" has energy E that depends on the frequency f of the electromagnetic radiation by the equation:
> 
> 
> In classical physics, energy of electromagnetic (EM) radiation was thought to be absorbed or emitted continuously. It wasn't until late 1900 the German scientist Max Planck (1858-1947) made a radical assumption in explaining the black body radiation spectrum, the idea of discrete energy arose.
> ...


>>>>>>
Well  they are using skin as the body that absorbs these photons which according to Your climatology physics mentor
are all the same,...no matter how hot or cold the source was that emitted them...and there are much better examples
but right now I`m a little pressed for time and my thoughts are with my daughter...as I already told You in a private message..

But if You still don`t understand the difference between the* frequency dependent energy quantum *of  light and
and the power, no matter if expressed as watts or luminosity ....You will never  be able to understand it
reading the kind of crap like the the stuff You Google for on the Internet.


I wish I had the time to go yet again through every idiotic statement the author of  the publication You cited.
"photons" are all the same no matter how hot or cold the emitting source was....and uses Planck`s curves
to sell You this lunacy...
all the while Planck was trying to explain why they are not..:

Quantum - Wikipedia, the free encyclopedia


> The concept of quantization of radiation was discovered in 1900 by Max  Planck, who had been trying to understand the emission of radiation from heated objects, known as black body radiation.
> By assuming that energy can only be absorbed or released in tiny,  differential, discrete packets he called "bundles" or "energy  elements,",[8]
> Planck accounted for the fact that certain *objects change colour when heated*


>>>>>>
Since as You and  the "climatologist" keep claiming that all photons are the same, then
it should also not be possible to observe a "color"...which is a frequncey change...which is
also a quantum energy change of the light emitted by any such body as the temperature goes up.

So if You wish to ignore the quantum nature of a photon , how do photons get absorbed by anything...if not by  orbital jump
of an electron to a higher energy level...(speak potential)

The difference in the energy level  of the orbital jump shows up as a higher frequency  of the light that
has been emitted.
But once again I remind You, that  any such photon that has initiated the jump must have
a sufficiently high  ENERGY  QUANTUM to do so.

But  You still believe, "all photons" are the same and then *You wind up in a bind*
*and I am supposed to explain to You* why a photon that packs an energy quantum which is
too low can not  move the electron orbital  to the higher orbital ...

*Show me where exactly in physics does it say that many more low energy quantums *
*can do the same thing as a (short wave) photon high energy quantum....*

If that was so then  the entire field of  atomic absorption spectroscopy would seize to exist.
Any atom in the ground state could get it`s electrones "pumped up" to a higher orbital and
absorb light, no matter what the wavelength...

If photons were "all the same" as this author is claiming then it would not be necessary to
supply  photons  at very specific wavelength that pack the sufficiently high energy where light can be
absorbed.....

And no, You can`t  add these "photons which are all the same"   and pack them up as a new and higher photon
ENERGY quantum...*.if You could then You could also convert low energy Infrared to a high energy X-ray quantum.*

And You continue to fall for  the same con these cliamtologists "explain" photons *while totally ignoring ALL OF QUANTUM PHYSICS..*
They do just that....they are accumulating enough  photons as low energy watt seconds  till in their
strange minds they have according to Planck enough ENERGY in watt seconds  to have a "black-body" emitt light at  a shorter wavelength
i.e....then go on  claim  that  they have managed to heat a hotter body with  a colder one using the
low  frequency "back-radiation"...


Unfortunately I have to go to the hospital now and leave it as that.

And whatever differences You have with wirebender You have to sort out with wirebender.
I don`t have the time to do this.


----------



## gslack (Nov 20, 2011)

Not sure if this may help with the concept overall but I will try once more..

Ian remember spencers experiment?

He mentions the plate being heated to 150 F?

Okay are we to assume he means the heat source is through a thermostat which keeps enough energy (electricity) flowing in it to keep it at that temp? or it is limited to enough energy required to make that plate that temperature? See the difference? With a thermostat to regulate it we are observing the temperature and not necessarily the energy directly just the result. The amount of energy it would take could change with environment, variables, insulators, heat sinks, et al. with the concern being the temperature. However if we were wanting to maintain the same energy input that would cause the plate to reach 150F with the set parameters and environment at the beginning any insulating body would require less energy to create 150F temperature in that plate. 

So which are we concerned with the energy input to output vs back from the second plate or the heat?

One big problem I see with it is he mixes the two concepts giving the impression they are one and the same, When they are not.In his experiment Energy = cause, heat = effect... Changing parameters like adding another body in close enough proximity to allow reflected heat to factor into it (as he claims) would have to change the energy input from the source if we were concerned with maintaining 150F temperature, but he claims the source stays the same.. In which way does it stay the same? Input energy or output heat?

he does not specify that I read anyway, anything like that other then the 150F for the first plate. So to me even IF he could cause the temperature the first plate to rise (which i think not) it wouldn't be applicable to atmospheric activity on our planet anyway.

Even IF backradiation existed, it would amount to virtually no significance when you realize the loss of energy in the work involved at each stage of transfer. We have to take into account all other manner in which heat is transferred in the atmosphere and the surface, and the constant flow of the formless gasses due to their temperature, wind, pressure, concentration, and cloud cover...

The problem here is too many educated people (along with uneducated) that have focused on one aspect so intently they forget the bigger picture and fuss inconsequential factors that do nothing when taken into proper context with the whole.. They would love to see the forest but they have the damn tree in their way...


----------



## IanC (Nov 20, 2011)

gslack said:


> Not sure if this may help with the concept overall but I will try once more..
> 
> Ian remember spencers experiment?
> 
> ...



a bit late to finally start thinking about the problem isnt it? oh well, better late than never.

as I previously pointed out....input equals output. the two options are as follows

1. the input of electricity remains at 750w and the heated bar increases temperature when the second bar is placed next to it.

2. the electricity is decreased to keep the heated bar at 150F, by 75w the last time I explained it, and the container is cooled because it is only receiving 675w instead of 750w.

(inputs, outputs and temps at equilibrium)


----------



## IanC (Nov 20, 2011)

polarbear said:


> IanC said:
> 
> 
> > > In this article we will consider what happens when this radiation reaches the ground. The reason we want to consider it is because so many people are confused about back radiation and have become convinced that either it doesnt exist  covered in the previous two parts  or it cant actually have any effect on the temperature of the earths surface.
> ...



I dont think you are reading what SOD or I am writing.

the graphs of the two blackbodies, one at 10C and one at -10C, have exactly the same range of possible wavelengths for radiation, with the possible exception of very rare high energy photons at the flat area of the curve past the left side of the graph.

the graph of the 10C blackbody ENTIRELY contains the graph of the -10C blackbody! every photon made by the -10C body can be made by the 10C body. likewise every photon made by the 10C body can be made by the -10C body. the area from the top of the -10C curve to the 10C curve is the positive difference in radiation between the two, and it is all in the previous range of wavelengths. get it now?

both bodies produce the same types of radiation but the warmer one produces more radiation and at a higher average energy (shorter wavelength).

back radiation cancels out all the area under the -10C curve, and the area between the curves is the extra energy that is used to transfer heat. because a warmer object always radiates more than a cooler one, heat always flows from the warmer to the cooler. the graphs make this obvious.

your complaint about lower energy photons being unable to do specific types of work is a non sequitur. the only time when a warmer object wouldnt have the right amount of radiation to cancel out the cooler object's radiation is when one or the other is not a blackbody and the curve would be a different shape. 

the earth's surface is quite close to a blackbody because it has many types of compounds that absorb and emit over a large range of wavelengths. the atmosphere is less of a blackbody because it has less constituents therefore less ability to absorb and emit over a wide range. CO2 by itself has a limited range of emission and absorbance.

the earth can absorb pretty much any type of radiation, the atmosphere less so, CO2 limited to specific bands. 

one thing that I find interesting but I have never heard anyone talk about is what happens when the temp of the earth goes up or down. the 15 micron band is to the right of the highest intensity radiation. if the temp goes up then the 15 micron band will increase in absolute number but decrease in proportion to the overall output. this seems like a negative feedback to me. I have no idea if this would really affect things much but it seems like it would be a factor that should be compensated for. earth surface temps range from minus 60C to plus 40C.


----------



## wirebender (Nov 20, 2011)

IanC said:


> the 2nd law states that there can be no _net_ flow of heat from colder to warmer.



ian, the second law of thermodynamics says no such thing.  I have posted it for you numerous times and there is nothing whatsoever there about "net" energy flow.  Again, here is the 2nd law of thermodynamics:

"It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object. "

"Second Law of Thermodynamics: It is *not possible *for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will *not flow spontaneously *from a low temperature object to a higher temperature object. "

So tell me ian, where do you see anything there about "net" energy flow?  You don't.  The second law doesn't say anything about "net" energy flow but somewhere along the way, you have bought a bill of goods so thoroughly that you can apparently look at the statement above and see the word "net" in there somewhere.  The idea of "net" energy flow is in all likelyhood an artifact that resulted from the corruption of the Stefan-Boltzman law which you have also bought into.  The argument in support of backradiation is circular in nature ian, sorry that you can't see it.  

The idea of backradiation is supported by the corrupted version of the SB law which is supported by a corrupted statment on the 2ndl law of thermodynamics which the corrupted SB law introduces.  You have bought into pseudoscience ian, not actual science.  The second law of thermodynamics as stated above is science and I am sorry to tell you, but it doesn't support your view of the physical universe.



IanC said:


> the cancelling out of radiation flow going in both directions is a mathematical calculation, not a physical obliteration of photons. and it certainly doesnt happen without the presence of matter.



Again ian, sorry that your very immature view of physics doesn't allow you to see what is actually happening.  The cancelling out of radiation flow via the subtraction of EM fields is a very real phenomenon that is calculated, and put into actual physical practice by electrical engineers every day.  Your view of the behavior of photons, on the other hand, is really nothing more than a mathematical artifiact and the result of a corruption of one of the laws of physics to boot.



IanC said:


> you have put up zero evidence of this magical destructions of photons. you wont even pinpoint _where_ this is supposed to happen. I have asked you dozens of times whether it happens at the surface, CO2 molecule or just in space but you refuse to give an answer.



Because it isn't magical ian.  Apparently it is so far over your head (which is sad) that it only appears to be magic to you.  When you subtract EM fields ian, what do you suppose you are subtracting?  When the magnitude of an EM field diminishes, what do you suppose it has lost in order for it to be diminished?  In order for you to grasp this ian, you are going to have to embrace particle-wave duality and until you do, this is going to look like magic to you and you are going to continue to misunderstand.



IanC said:


> the closest I have come to finding something that supports your view is Claes Johnson's discussion on how he thinks the back radiation is reversed at the surface, much like reflection. no cancelling out, and the 'resonated reemission' is part of the radiation calculated for the earth's surface, not some extra bit. this calculation works out to exactly the same numbers as the 'corrupted version' of the S-B formula.



Johnson says no such thing.  Again, your immature view of photons can not be applied to reality.  Till you accept particle-wave duality, this will continue to escape you.  As to the answer being the same with the corrupted version of the SB law, again, you fail to see the issue.  The corrupted version of the SB law describes a physical process that is not happening.  It describes a violation of the 2nd law of thermodynamics in order to support a non physical phenomenon which allows a corruption of the 2nd law of thermodynamics all for the purpose of promoting AGW alarmism.

Without backradiation, there is no AGW alarmism.  Without corruptig the SB law, there can be no support of backradiation which in turn allows you to claim that the 2nd law of thermodynamics is describing net flows which, in reality, it is not. 

It is unfortunate that the level of public education has dipped to the level that allows so many people to be taken in by that very bit of sleight of hand ian; and not just you, but scientists who should know better and whole generations are being taught that very bit of rubbish as if it were true.



IanC said:


> one of the biggest problems in physics is being able to resolve what happens on a micro scale to the reality of the macro scale.



Your biggest problem with regard to what is happening here is your inability to understand and grasp wave particle duality.  Till you get that ian, this will continue to go right over your head.  Phenomena like EM interference, and cancellation can only be explained by looking at EM fields as waves. The phenomena are real and measurable, but not explainable if you assume EM fields are made up of discrete particles.  Your thinking along this line is very immature and as a result, to rigid for you to actually grasp what is going on.  You are stuck with particles and particles don't explain what is happening.



IanC said:


> you are trying to switch back and forth from large to small scale.



No I am not ian, I have a grasp of particle wave duality and therefore am not confined within your small mental box trying to apply particle physics to a physical phenomenon that particle physics can not, and does not try to explain.  



IanC said:


> heat flow only goes in one direction but radiation flow goes in both.



Radiation can go in both directions if the frequencies are different, but we aren't talking about different frequencies.  The frequency of IR changes as it passes through a CO2 molecule but it does not change enough to get outside of the broad band of IR being emitted by the surface of the earth.  Cancellation, interference, and subtraction of EM fields is a very real, observable phenomenon ian and it can't be explained rationally by viewing EM fields as being made up of discrete particles.  You have to look at EM fields as waves in order to explain what is happening in the real world.  Your thinking isn't mature enough for that at this point so you are left calling it magic.



IanC said:


> an excited molecule doesnt check to see if it is allowed to emit energy by radiation, it just emits. the photon doesnt care which direction it is heading in, it just goes. the photon doesnt care whether it has enough energy to warm a body or just cancel out radiation in the other direction leading to a 'net flow'. CO2 doesnt care if it is a good emitter and absorber at certain wavelengths and a poor one at others, it just is what it is regardless.



Yes, it just goes ian, but it goes in the direction of propagation of the EM field which has the larger magnitude.  If it is in opposition to that field, it is expended (subtracted).  Again, when you subtract EM fields, what do you think you are subtracting?  The explanation on spencer's page by gord was ok.  The block of wood being pushed on in two directions doesn't know, or care which direction it is supposed to go.  It just moves in the direction of greater force.  The photon can not move in a direction opposite to the direction from which the EM field with the larger magnitude was propagated.



IanC said:


> do I like the surface radiation and atmospheric back radiation components in Trenberth's diagram? no, I think it should just show ~60W/m2 going up, perhaps with a side diagram showing the the calculation of radiation in both directions. do I think Trenberth has the right figures for the energy coming in and going out? I dont think it is exactly right but it is probably reasonably close.



At this point ian, I really don't care what you like and don't like and don't even take such things in consideration because you have demonstrated beyond even the smallest reasonable doubt that you simply don't get it.  This whole subject is beyond you and as a result, you are operating from a position of faith, not any inherent knoledge.  When you do bring bits of science forward in an attempt to support your position, it is clear that you don't understand what you are saying.



IanC said:


> Until there is proof that that explanation is wrong, and accepted by more than a few fringe crackpots, I will continue to believe that CO2 is a factor in atmospheric processes but I will also continue to be sceptical of the high feedback values which lead to global warming alarmism.



There is proof that the explanation is wrong and hard, repeatable, observable proof that backradiation from the atmosphere to the surface of the earth does not exist.  You can prove it yourself in your own back yard.  I have done it in my back yard.  If you would like plans for the experiment, I will gladly provide them to you.

You make a parabolic dish and point it into the sky on a sunny day.  If you point it at the sun, you get a great deal of heat very quickly.  But if you point it towards clear sky, even on a bright sunny day, the temperature at the bottom of the dish drops below the abient temperature.  If backradiation were "beaming"back down to the surface of the earth and warming the surface, then there is no physical way the temperature in that dish could drop below the ambient temperature.  The dish is behaving precisely as the second law of thermodynamics predicts.  Heat is flowing from warm to cool and nothing is moving in the opposite direction.  At night, you can cause ice to form in that dish if the ambient temperature is around 45 degrees.  Even lower if the humidity is lower.

There is hard, observable, repeatable evidence that the claim of backradiation is wrong ian but like the scientists who have a financial stake in AGW alarmism, my bet is that you wll ignore what you can see in favor of faith.



IanC said:


> wirebender, care to explain where and how all these photons magically disappear? in the real world and not in some calculation?



First, there is no magic.  It only appears to be magic to you because it is beyond your understanding.  Do you deny that EM fields interfere with, and cancel each other out and this phenomenon is measurable?  If you do, then you are well and truely full of BS.  Since the phenomenon of cancellation and interference happens, and one EM field can be reduced in magnitude by another and can be measured, what do you suppose is being subtracted?  The answer is photons ian.   It looks like magic to you because your thinking hasn't matured to encompass particle wave duality.  The answer lies there and till you can grasp it, this will remain over your head.

You are the one who is engaging in magical thinking ian.  Cancellation and interference are real, measurable phenomenon that happen without the presence of matter and can not be explained by the view that EM fields are composed of particles and yet, you continue to hold the rigid belief that EM fields are particles even though phenomena are happening that can't be explained by your belief.


----------



## wirebender (Nov 20, 2011)

gslack said:


> Ian please stop using science of doom as a reference.. he is a 7th grade teacher who was let go from his job as a PR man for the NOAA.. And he routinely does the same crap spencer does using half-science mixed with speculation and twisted logic..
> 
> Seriously, He is a PR man cashing in..



Science of doom as a scientific reference.  You may as well reference Archie comics.


----------



## wirebender (Nov 20, 2011)

IanC said:


> gslack said:
> 
> 
> > Ian please stop using science of doom as a reference.. he is a 7th grade teacher who was let go from his job as a PR man for the NOAA.. And he routinely does the same crap spencer does using half-science mixed with speculation and twisted logic..
> ...



Ian, it has been explained to you over and over.  The fact that the explanation is over your head or in opposition to your faith, or both does not change the fact that it has been explained.


----------



## wirebender (Nov 20, 2011)

IanC said:


> the TOA emission measured from satellites show that the 15 micron band has been strongly reduced by passing through the atmosphere. obviously the 15 micron band has been transformed into some other type/wavelength of radiation before the energy escaped.



That is simply not true.  You have bought another bill of goods from someone who didn't show you the whole picture and you either didn't grasp that you weren't being shown the whole picture or you didn't care because the partial picture meshed wtih your belief.  Here is the whole picture:

Here is an overlay of snapshots of outgoing long wave radiation taken in 1970 by the sattellite IRIS and in 1997 by the sattellite IMG in 1997. Both snapshots were taken over the central pacific at the same time of the year and under the same conditions. 






The X axis of the graph indicates wavelengths. The wavelengths that CO2 absorbs, remember are 2.7, 4.3, and 15 micrometers. The light colored line is the IRIS data collected in 1970 and the darker line is the IMG data from 1997.  Where do you see evidence of a reduction of outgoing LWR in the 15 micrometer band?

 The next two images were taken by IRIS in 1970 and TES in 2006 respectively. In these graphs, the black line represents the actual measurement taken by the sattellite, the red line represents what the climate models predict and the blue line represents the difference between the model data and the actual data.








Feel free to print out the two graphs and overlay them. You will find that the black lines (actual measured data) are identical indicating this time, that there is no difference between outgoing longwave radiation in the CO2 absorption spectrum between 1970 and 2006.  The reduction you claim is output from the models as indicated by the blue line, not any actual decrease.  You are accepting the output from a model as if it were actual data.


----------



## wirebender (Nov 20, 2011)

IanC said:


> and just to be clear, he is not saying the 2nd law of thermodynamics is imaginary. he is saying that some people's interpretation of the 2nd law is imaginary. I believe he is referring to Claes Johnson, who has been parroted in an even crazier form by wirebender who claims the photons are not just bounced but actually obliterated in space without the need of matter to be present.



Once more ian, when you subtract EM fields, what do you believe you are subtracting?


----------



## wirebender (Nov 20, 2011)

IanC said:


> the only thing he called 'imaginary' was the faulty interpretation of the 2nd law. but I will go back and reread it to make sure.



The second law is a statement ian, it doesn't require interpretation and it states pretty clearly that heat can not flow from cold to warm without some work having been done to accomplish the task and absorption and emission do not constitute work.


----------



## wirebender (Nov 20, 2011)

IanC said:


> fifth and sixth utterances. still aimed at the twisted version of the 2nd law that wirebender espouses.



Ian, this is the version of the second law of thermodynamics upon which I base my position:

"Second Law of Thermodynamics: It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object. "

It is the only version of the 2nd law I have ever referenced because it is, in fact, THE second law of thermodynamics.  Any deviation from that law, as stated above would render a twisted version of the 2nd law of thermodynamics.  Now ian, it is you who believes that heat can flow from cold to warm and you who believes in net flows in opposition to what the 2nd law actually says, so it is you, ian, who is operating from a twisted version of the 2nd law of thermodynamics.


----------



## polarbear (Nov 20, 2011)

wirebender said:


> gslack said:
> 
> 
> > Ian please stop using science of doom as a reference.. he is a 7th grade teacher who was let go from his job as a PR man for the NOAA.. And he routinely does the same crap spencer does using half-science mixed with speculation and twisted logic..
> ...



Hi guys, right now some other relatives are staying with my daughter @ the ICU, giving me a lunch break @ home.
So I looked in here while I`m waiting for my dinner...

I can`t believe what I am reading here...seldom have I seen statements as ridiculous. For example what IanC has written in post # #283 



> every photon made by the -10C body can be made by the 10C body. likewise  every photon made by the 10C body can be made by the -10C body



I beginning to wonder if Ian or this 7th grade teacher who he keeps quoting ever seen a spectroscope...

Lets see if a "body" at a lower "black-body" temperature can make every photon a hotter one can make...:








2,7003,300 K Incandescent light bulb...:







1,850 K Candle flame,...:








It is futile to discuss physics with somebody who continues to use the same absurd kindergarten interpretations of what a photon is...

and this is also the reason why so few physics professionals bother with crap like "trenberth`s energy budget"....heating a hotter black body with a colder one etc etc...

This garbage is written specifically for people below a certain IQ and with only limited education....
So how could it possibly be debunked if the people who are falling for it lack the education to understand the explanation...
You might as well try convincing some jungle tribe that it was`nt their chicken sacrifices that caused the rain which put out the forest fire.

Anyway IanC, it started out as an interesting discussion like some TV show with an interesting title to catch one`s attention.. and a few minutes into the show they are "explaining" paranormal activity and "negative energy"....which they say nobody else (conventional science) has  explained ...so it must be true... I have seen enough,.... and know when to pull the plug


----------



## wirebender (Nov 20, 2011)

You are right polarbear.  Till ian's thinking matures he isn't going to be equipped to even glimps what is happening here, much less understand it.  He is stuck on photons as discrete particles and till he gets past that, he will remain behind the curve.

By the way, I hope nothing to serious is going on with your daughter.


----------



## gslack (Nov 20, 2011)

IanC said:


> gslack said:
> 
> 
> > Not sure if this may help with the concept overall but I will try once more..
> ...



Douchebag, you keep insulting me and condescending on me , and then turn around and try to play friendly? FUCK YOU ASSHOLE!!!

Don't you even think for one minute you have the mental, physical, or even educational muscle to condescend on me you little weaseling internet wannabe..

You don't know squat, you have been playing scientists so long you started to believe it.. You didn't recognize any of the most basic principles any of us have brought forth all you have done is google terms and other peoples work or explanations, you haven't expressed a single original thought or anything in your own words other simply calling us wrong and insulting me..

GO shit in your hat you fraud.. You are done playing fake scientists with me, and I am done being even remotely tolerant of your immature bullshit any longer.. You just insulted me for the last time lowlife...

Now I told you from the get go the basic issues in the experiment, you fraud. You went and made up new parameters and new examples to defend it and confound the topic..

BTW, douchebag, your response did not address the point I made.. All you did again was ramble bullshit and talk in a circle..

Point remains: Temperature and power are not the same thing.. IF you limit or increase power to maintain the temperature you change the output of the power source, and that makes the entire experiment pointless in its claim..

Catching on yet dumbass? Changing power input to make a set temp is not what the sun does asshole.. Temperature is a measure of heat, energy is what causes that heat you incompetent imbecile.. They are not one and the same.. Its the same problem you have with polarbear trying to show you the your mistake with work in his posts..

You incompetent, incoherent, fraudulent, posturing phony, you cannot even keep two concepts separate and have the nerve to try and condescend on me? 

Go and play scientists with people dumber than yourself all you want, but do not try your act with me asshole..


----------



## IanC (Nov 21, 2011)

wirebender said:


> IanC said:
> 
> 
> > wirebender, care to explain where and how all these photons magically disappear? in the real world and not in some calculation?
> ...



you keep saying that _I_ am the one with faulty understanding of the duality of photons but it is _you_ that is confused on this topic and others.

opposing photon waveforms do show interference patterns according to how they match up in orientation. here is a diagram of two waves that add together to form a standing wave that is twice the amplitude of the original waves. by changing the phase you could get the waves to cancel out but a straight flat line isnt very informative








> .....
> 4. a standing wave does not tranfer energy(its two components however, do transfer energy in their respective directions)



Google Image Result for http://www.a-levelphysicstutor.com/images/waves/statw-formation.jpg

you are mistaking interference patterns for actual change in the photons. the photons go through the area of waveform interference and emerge totally unchanged and in the same direction as before. they continue on their path until they are absorbed by a particle of matter that is capable of interacting with them.


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## IanC (Nov 21, 2011)

gslack said:


> IanC said:
> 
> 
> > gslack said:
> ...



back to ranting and _ad homs_? not surprising. I am sorry that you feel I am condescending to you but that is only your opinion of what I have said, probably born out of your own fear that you are at a severe disadvantage in a debate over scientific concepts.

I am somewhat amused that you think Spencer would change the power input into the heated bar according to a thermostat set for 150F when the whole concept of the experiment is to find the change in temperature when the second bar is added. that is just as ridiculous as wirebender changing the temperature of the container even though Spencer specifically defined it as 0F. and then you guys pretend that _Spencer_ made the mistake!?!?

anyways, back to your comment. as I previously explained to you, the bar temp goes up if the power input remains the same. the container temp is cooled (at least less energy is needed to keep it at 0F than when the heater was receiving full power) when the power is reduced to keep the bar at 150F. do you understand the concept of input=output? I think it is funny that you insulted me for bringing up this scenario previously (and explaining it), while you think you are brilliant and original for bringing it up yourself! I suppose you didnt understand it the first time and you probably dont understand it now. hahahahahaha


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## wirebender (Nov 21, 2011)

IanC said:


> you are mistaking interference patterns for actual change in the photons. the photons go through the area of waveform interference and emerge totally unchanged and in the same direction as before. they continue on their path until they are absorbed by a particle of matter that is capable of interacting with them.



I am not mistaking anything ian.  As I have told you before, if two EM fields are propagated in the same direction, even a weaker field will add to the magnitude of the other.  Interesting that you pick an example of two waves that are propagated in the same direction in an effort to prove that they do not reduce each other if propagated from different directions.

Face it ian, you are wrong and till you learn a bit more or have some sort of revelation, you will continue to be wrong.


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## IanC (Nov 21, 2011)

polarbear said:


> I can`t believe what I am reading here...seldom have I seen statements as ridiculous. For example what IanC has written in post # #283
> 
> 
> 
> ...



ho hum, more _ad homs_

your pretty pictures seem to be incomplete.







> Figure 1. A blackbody radiation curve at T=2800K, which is a typical operating temperature of a 100W incandescent bulb. Approximately 88% of the light is emitted in the infrared region. BB: blackbody.



this graph comes from a paper on how to make more efficient lightbulbs by sequestering the wasted IR.



> Recycling processes have previously been developed in the form of reflecting envelopes using either a dielectric metal film stack1,2 or a dielectric multi-layered film.3 However, for both structures the reflectance in the near-infrared region is not high enough to bounce back all the infrared light. To overcome these limitations, we employed a two-dimensional metallic photonic band gap (PBG) filter architecture to enclose the incandescent filament. The filter acts as a perfect transmitter for the useful visible light and a perfect reflector for the undesirable infrared light. *The reflected light is re-absorbed which, in turn, helps to heat up the filament. This infrared recycling process has two major energy consequences. First, it reduces the amount of electricity required to maintain a hot filament and thus improves electric-to-optical conversion efficiency*. Second, it reduces the thermal radiation of the bulb as infrared photons cannot escape.



hmmmm.......that seems to go totally against wirebenders 'imaginary' 2nd law of thermodynamics! intersting paper. while the glass bulb of an ordinary light cuts off a lot of the infrared, the coating in frosted ones also cuts down on IR transmission. test it yourself if you have clear bulbs in your bathrooms.

I couldnt find the same flourescent light as in polarbear's picture. perhaps the unlabelled y axis is part of the problem.






flourescent lights do not emit like a blackbody. the curve is a different shape because it uses specific atomic bands of emission.






sodium lights are even less like a blackbody


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## gslack (Nov 21, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC said:
> ...



IanC you have just shown how immature and ignorant you are..



> back to ranting and _ad homs_? not surprising. I am sorry that you feel I am condescending to you but that is only your opinion of what I have said, probably born out of your own fear that you are at a severe disadvantage in a debate over scientific concepts.



No you little lowlife you have gone back and forth insulting and the pretending we insult you over and again for weeks now. You were condescending again and the fact you try and deny this is all the more pathetic..



> *I am somewhat amused that you think Spencer would change the power input into the heated bar according to a thermostat set for 150F when the whole concept of the experiment is to find the change in temperature when the second bar is added.* that is just as ridiculous as wirebender changing the temperature of the container even though Spencer specifically defined it as 0F. and then you guys pretend that _Spencer_ made the mistake!?!?



Your words your previous post to me: 

_"the electricity is decreased to keep the heated bar at 150F, by 75w the last time I explained it, and the container is cooled because it is only receiving 675w instead of 750w."_

Your words above:

*I am somewhat amused that you think Spencer would change the power input into the heated bar according to a thermostat set for 150F when the whole concept of the experiment is to find the change in temperature when the second bar is added.*

Recognize them? You better dumbass they are your words...

So...whats your excuse this time? You just said one thing and then tried to make it seem ridiculous I would think that one thing..So, which is this time dancing bear does he change the input level or not? Or does he change it then not change to suit your silly posturing ass?

 Got an explanation didpshit? NO?????? Didn't think you would.. Now dance boy!



> anyways, back to your comment. as I previously explained to you, the bar temp goes up if the power input remains the same.



Really? Hows that work?  What planet does something get hotter with less energy input? 



> the container temp is cooled (at least less energy is needed to keep it at 0F than when the heater was receiving full power) when the power is reduced to keep the bar at 150F.



Yes yes and what exactly does this address regarding my post or contention, other than the fact your are once again talking out of your ass? nice way to dodge and divert dancing bear..



> do you understand the concept of input=output? I think it is funny that you insulted me for bringing up this scenario previously (and explaining it), while you think you are brilliant and original for bringing it up yourself! I suppose you didnt understand it the first time and you probably dont understand it now. hahahahahaha



Yes I do it seems you don't though because you made a contention then denied making it or was it I made it then you denied I made it? Does it matter? After all you knew that too didnt ya phony.. Sure ya did because you know everything don't ya... You even talked about this before but didn't if its wrong... Sure pal sure..

IanC, you want to explain to me how it is a self proclaimed expert on this didn't recognize fermats theorem? How about the fact you didn't know the two-slit experiment? Remember what you told me about that experiment genius? Oh I sure do.. YOU told me you can't learn about physics watching waves in a liquid... LOL..

Now dance, and squirm and google up terms to try and save face phony, but from here on out I will make sure I remind you how, when and where you got outed as a fake..


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## IanC (Nov 21, 2011)

wirebender said:


> IanC said:
> 
> 
> > you are mistaking interference patterns for actual change in the photons. the photons go through the area of waveform interference and emerge totally unchanged and in the same direction as before. they continue on their path until they are absorbed by a particle of matter that is capable of interacting with them.
> ...



geez, you are a total train wreck. you will say anything wont you?

here you go, I should have put this one up instead anyways.






from the same article that wirebender said the waves were in the same direction. I would paste the text but it doesnt seem to work.


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## wirebender (Nov 21, 2011)

Sorry ian, you are so caught up in trying to prove that you are right that you simply can't see how wrong you are.  I hope for you that you don't end up living your entire life like.

We are done.  I proved my point on the first try and my proof stands.  You, on the other hand have failed at every turn for the simple reason that you are wrong.  Enough said.  Life is to short to waste it on people who can't, or more importantly won't learn.


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## gslack (Nov 21, 2011)

Notice the little punk avoids the post reminding him of his outing.. What a coward...


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## polarbear (Nov 21, 2011)

IanC said:


> *your pretty pictures seem to be incomplete.*



First You tried to make Your case by confusing POWER and ENERGY,...then after I pointed this out to You You said it was an
 oversight..
 Then You tried to make the same erroneous case again using the "extra time" a photon  spends in Your phantasy climatology world,..and then I did the math for You and the "extra time"  was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
 try to use to get the "extra Energy"...and when You realized, that Your "extra Spencer`s example Power"  evaporated by a factor of 10^(-15)
 You come back here claiming that all photons are the same...we all know where and from who You got that one from...

 So now You are claiming that these "all the same photons" each having been emitted by a radiation source say at 10 C....
 that 10 of these photons stack up to a new and higher temperature of  10 times X ("climatology photon  degrees")...

 You got that nonsense from this idiot who told You that  the photons coming from a 2000 degree black body are the same
 as 10  photons coming from a 200 degree source and therefore that enough these are able to raise the temperature of a 2000 K black body

Then I pointed out the spectral energy shift an increased temperature causes (Wien`s law)

And now You claim that I don`t understand the laws of thermo-dynamcis, Plank`s, Wien`s and Kirchoff`s laws...( but You do..??? )

Let`s take a look at Kirchoffs laws again:

Kirchhoff's law of thermal radiation - Wikipedia, the free encyclopedia



> >>>if we consider an ideal situation in which an enclosure with _perfectly reflecting_ walls contains radiation with a certain amount of energy, then at equilibrium, this "photon gas" will have a Planck distribution of energies.[1] This will be true even though the walls are perfectly reflecting due to the very small amount of interaction between the photons themselves. The equilibration process will take a considerable amount of time, but the distribution of energies and radiation density will ultimately approach a Planck distribution. The next step is to realize that, as a result of the second law of thermodynamics, _any_ enclosure at thermal equilibrium must also have a Planck distribution of radiation.[1] If this were not true, then we could bring that system in contact with the above ideal system, both at the same temperature, and by connecting them through an optical filter, we can have a net amount of radiation pass from one body to the other. For example, suppose in the second system, the density of photons at narrow frequency band around wavelength &#955; were higher than that of a black body at that temperature. If a filter that passed only that frequency band was inserted in an opening that connected the two bodies, then there would be a net transfer of photons, and their energy, from the second system to the first.* This is in violation of the second law of thermodynamics, which states that there can be no net transfer of energy between two bodies at the same temperature*.<<<<


So let`s see when You start violating it...shall we...:
  You say that  there is an energy transfer from a -10 C body to a +10 C.
So does this energy transfer increase when this colder energy donor gets colder...or does it increase
when the colder energy donor is warmer, say -9 C instead of  -10 C.

  Let`s see how  fucked up You get with Your  climatology energy transfer from a colder to a hotter body as we increase the temp. of the colder body..
How far can You go up using this idiotic climatology  cold====> hot  "energy transfer"
without violating all the existing laws of thermodynamics...

  Can You go up from -10C on the colder body to 9.9999 C  ....?
Remember Your theory has it that  there are now even more photons coming from the
"colder body" as did when it was -10 C....

  So, are You disputing all laws of thermodynamics...or are You claiming
that You understand these, but I don`t...?



 So who`s picture is not complete...Your`s or mine.
If You want to play quantum physics with me make sure You are playing with a full deck Ian


----------



## IanC (Nov 21, 2011)

gslack said:


> IanC you have just shown how immature and ignorant you are..
> 
> 
> 
> ...



you said- 





> Really? Hows that work?  What planet does something get hotter with less energy input?



you guys were the ones who insisted that the bar would not heat up. I simply told you what would happen if the bar was kept at the same temperature. in that case the power input would have to be lowered, which in turn would cool the container by the same amount that the input was lowered.  to answer your question that was so badly phrased. in Spencer's experiment, I showed that the electrical power would have to be decreased by 75w to stay the same temperature, therefore if you decrease it by any value from 1-74w the bar will still increase in temperature. increased temp with decreased power when the second bar is placed next to the heated bar. on this planet or any other.


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## polarbear (Nov 22, 2011)

IanC said:


> ho hum, more _ad homs_
> 
> your pretty pictures seem to be incomplete.
> 
> ...




Just a minute here IanC...
*Where exactly did I say that sodium lights are a black body...???


*Let me remind You that it was You and Your 7th grade "teacher" that said :


> all photons are the same...
> Nobody can tell them apart
> photons coming from a 2000 K source are the same as photons coming from a 200 K source


Let me get this straight,...and now You are saying that we have to distinguish between "light bulb photons" and "Spencer`s metal bar example photons"...?

Are You in fact getting Your physics ideas  from Archie comics books like wirebender remarked?

And by the way I did not post the light spectrum emitted by a sodium light, had I You should have seen several very bright yellow lines.

The 2 picture composite unfortunately had on the right side a fluorescent light...but I posted it as it was @ upload.wikimedia.org/wikipedia/commons/b/b0/Spectral_Power_Distributions.png  because the left side has a 2800 K incandescent light...








what was I supposed to do here...use MSPAINT and clip this picture, then upload it to photobucket and put another url between  the {img} tags so that You can`t nit-pick
And why would that bother You if I did not bother to waste even more time responding to these outlandish claims You continue to make here...after all according to You all photons are supposed to be the same, so why would "fluorescent light photons" not fit into Your "climatology billiard ball photon" scheme...?
Having a little trouble finding those in a -10 C Planck curve have You...according to you and the crap You quote they should be there..should they not..?


I also noticed that You rather play word games with wirebender  than answering my question about the  energy flow that You say flows from the cooler body to the hotter one...
Does in Your world of climatology  flow more energy from the cooler (-10C) body to the hotter (+10C)  one if the cooler body is  only 0 C,...and even more if it is +5C...and more again if it is 9.99999 C...??

Then it follows according to your "climatology physics" that even more energy is transferred if this second body is @ +10 C....

...which as You should know by now, if You read a real physics book is in fact in gross violation of Kirchoff`s law..You know that pesty physics law, that Your grade 7 teacher, who turned "climatologist" called "imaginary"...

well You directed this @ wirebender, not me,...but I can`t resist:



> I showed that the electrical power would  have to be decreased by 75w to stay the same temperature, therefore if  you decrease it by any value from 1-74w the bar will still increase in  temperature. *increased temp with decreased power when the second bar is  placed next to the heated bar. on this planet or any other. *                    __________________
> "Nothing shows a lack of mathematical understanding more than an exaggeratedly accurate calculation.&#8221;
> Carl Friedrich Gauss (1777-1855)


Tell me now how did You "calculate" the power to increase the temperature of the second bar...if You don`t even know the mass or the distance of the second bar....and exactly what temperature did You get for either bar in this calculation You pretend to have done...

In Missouri they say "show me...!"

Let me ask You another question about Your 2-bar calculation, regarding this:


> *increased temp with decreased power when the second bar is  placed next to the heated bar. on this planet or any other. *


Okay, let`s use a mass of 1 kg, the unspecified metal of your "metal bar" shall be lead as in Pb with an oxidized surface...looks pretty black...
the distance between the 2 is 10 cm,...and each cylindrical bar is say 5 cm thick...one bar = @ +10 C, the other @ -10 C

Allright IanC...do the math...because ""Nothing shows a lack of mathematical understanding more than faking it"

Oh and please do tell me also what happens when the bars mover ever closer...
and what happens when they make contact...does the -10 C bar still heat the +10 C bar..???
as "contact each other" lets use the distance at which they would bridge to make an electrical contact at 1 Volt...
According to You it still should...after all  atoms can`t really touch each other...or do "climatology atoms" touch each other ...?

Hey lets use Carbon instead of Lead...Carbon rods are as close as You can get to a "black body"...and to make it even simpler let`s use square graphite rods, 10 cm long and 1 cm sides...and if You look it up the atom/atom distance for Carbon is  1.42 Ångstroms...

So Your "climatology back-radiation photons" coming from the -10 C rod flat surface should have no problem "heating" the + 10 C carbon rod flat surface...as opposed to doing so over a distance of 10 cm..

There is nothing unfair about this problem...I have seen You write the word "simple" many many times...and after all the back-radiation coming from the CO2 in the air that climatologists  claim heats our planet does not come from a distant bar in a vacuum...it comes from the air, that undeniably touches the "balck body climatology earth"...so do the math and use the Avogadro gas constant & the Loschmidt number to calculate the Molecular distance at - 10C and 760 Torr ...and use that distance 
If You can show, as Your "climatology" explains it with their "simple examples..." like Spencer etc.. that -10 C air can "warm" +10 C water over this very short "back-radiation" distance, then all the ice we see on our lakes must be just as imaginary as Kirchoff`s law the way established physics "mis-understood" it to this day...to quote Your source


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## IanC (Nov 22, 2011)

polarbear said:


> Then You tried to make the same erroneous case again using the "extra time" a photon  spends in Your phantasy climatology world,..and then I did the math for You and the "extra time"  was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
> try to use to get the "extra Energy"...and when You realized, that Your "extra Spencer`s example Power"  evaporated by a factor of 10^(-15)




when you jumped into the thread I was excited that there was finally someone who was going to discuss ideas without all the distortion and unfounded denial of wirebender and gslack. unfortunately that is not the case. I dont know if it because you couldnt be bothered to read anything I write or if you just like being an ignorant asshole like those other two.

with respect to the extra time for escape of radiation- I gave three basic options to consider. 1.the photon escapes at the speed of light 2.the photon is absorbed and reemitted by the CO2 molecule and escapes, at less than the speed of light 3. the photon is absorbed and reemitted towards the surface and does not escape.

the first option leads to the coldest surface because the shedding of radiation is the most efficient.
the second option slows the escape of radiation, making the surface warmer. you are implying that is the only mechanism for warming. I am stating that at the very least there is some warming. you agree that it takes time to absorb and reemit the radiation therefore you agree that CO2 warms the earth. the amount doesnt matter in this scenario, just the concept.
the third option stops the loss of radiation completely and returns the energy to the surface. obviously this is where the bulk of the inefficiency of radiation shedding due to CO2 occurs. 

I made a point of not putting values on any of the three options, nor did I bring up any of the numerous complexities of multiple absorption/emission, transforming to different wavelengths etc. I brought up real world mechanisms that warm the surface by reducing the radiation loss. I have no problem with you pointing out that the second option doesnt account for much warming, I do have a problem with you implying that it was the only option and that you have 'refuted' my case somehow.


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## IanC (Nov 22, 2011)

polar bear said-





> Just a minute here IanC...
> *Where exactly did I say that sodium lights are a black body...???*


*

you put up a pretty picture of visible light from an incandescent bulb. unfortunately it cut off the rest of the radiation, more than 80% from the blackbody curve.

flourescent lights and sodium lights are not blackbody curves.*


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## IanC (Nov 22, 2011)

polar bear said-





> Just a minute here IanC...
> *Where exactly did I say that sodium lights are a black body...???*


*

you put up a pretty picture of visible light from an incandescent bulb. unfortunately it cut off the rest of the radiation, more than 80% from the blackbody curve.

flourescent lights and sodium lights are not blackbody curves.*


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## polarbear (Nov 22, 2011)

IanC said:


> when you jumped into the thread I was excited that there was finally someone who was going to discuss ideas without all the distortion and unfounded denial of wirebender and gslack. unfortunately that is not the case. I dont know if it because you couldnt be bothered to read anything I write or if you just like being an* ignorant asshole* like those other two.
> 
> 
> 
> ...



So You won`t or can`t do the math, ...?
And You "excused" Yourself from not doing the math by saying:


> *I made a point of not putting values ..
> 
> *


and claim:


> * I brought up real world mechanisms *that warm the surface by reducing the radiation loss.


Well haven`t You noticed, that in the real world this second "Spencer example balck body imaginary bar" is air containing CO2 and that they *do make a point to put numerical values on this "extra heat"...and post these all over the internet and news papers*

Not just with Your "Trenberth`s energy budget" but with every temperature graph with which "climatologists" claim that the temperature increase is caused by 380 ppm CO2...

I`m not asking You to do the math for an entire energy budget...
Do the math then for 2 bars in a chamber, just like Spencer`s "simple example" but use 2 carbon bars, 10 X 1 X 1 cm at a distance where they could make low voltage electrical contact and 1 bar is at +10 C, the other one is -10 C...

If You know math & physics then You should not have any problems with that...

Because any Kindergarten kid could figure out that if You don`t supply any more (extra) heat other than what You have in calories or watt seconds in these 2 bars *that they will both stabilize  at 0 C

Now please do  use Your "climatology math & physics" for Your "climatology photons" and  assign numbers and do the math...and tell us what temperature You come up with

*Unless of course You want to continue to dodge this simple problem with "intelligent" answers like (me) :





> being an* ignorant asshole* like those other two.


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## wirebender (Nov 22, 2011)

polarbear said:


> *Okay, let`s use a mass of 1 kg, the unspecified metal of your "metal bar" shall be lead as in Pb with an oxidized surface...looks pretty black...
> the distance between the 2 is 10 cm,...and each cylindrical bar is say 5 cm thick...one bar = @ +10 C, the other @ -10 C*
> 
> Allright IanC...do the math...because ""Nothing shows a lack of mathematical understanding more than faking it"
> ...



You had me holding my breath there for a second when you laid out the parameters using cylindrical bars.  Ian seems to believe that adding sides to one bar or the other, or both will cause the cooler bar to warm the heated bar.  I didn't know how he would handle a cylindrical bar.

Don't expect any set of rational calculations that actually incorporate any known physical law.  The one pseudomathematical response he has gave me in this whole long discussion appeared to suggest that if you increase the number of faces an object was radiating from, you could increase the amount of power it was radiating.  I don't know which physical law that is based on but I wish I did.  Add enough faces and you could power a city from the output of a AAA battery.

For that matter, if you could harness the energy created by backradiation, you could power the earth.  I never could get ian to name a physical law that supported and predicted backradiation.  It is a matter of faith for him.  Those free agent photons are just zipping around the universe carrying their power till they run into something solid and phenomena like interference and cancellation which result in the subtraction of EM fields don't actually decrease the strength of the fields by reducing the number of photons that make up the fields.  

Apparently the fields are diminished or cancelled, but either retain the same number of photons or the number of photons by which the field was reduced break ranks with the rest of the photons in the field and continue on their merry way across the universe till they hit a brick wall or some other suitably solid bit of matter.  

In ian's eyes, the subtraction of EM fields is magical.  With that, I believe I have said enough.  Good luck, but I don't think you will get anywhere.  What you are doing is far more complicated than what I was doing and he didn't even get what I was doing.  He thought it was a description of a magical process.  

You guys have fun and let me know if he manages to disprove the SB law.  Till that happens, my proof that the heated bar will not be further warmed by the addition of a second unheated bar stands.


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## gslack (Nov 22, 2011)

IanC said:


> gslack said:
> 
> 
> > IanC you have just shown how immature and ignorant you are..
> ...



IanC do you really think you can talk in a circle and I won't call you on it?

You didn't respond to my question, you talked in a circle and pretended to...

You do that a lot dancing bear....

You are such a fake ..


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## konradv (Nov 22, 2011)

gslack said:


> You are such a fake ..



Expert on fake, eh?  I guess you go with what you know!!!


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## gslack (Nov 22, 2011)

konradv said:


> gslack said:
> 
> 
> > You are such a fake ..
> ...



Sorry junior this thread has an age limit.. Now run along and play with the other kids..


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## konradv (Nov 22, 2011)

gslack said:


> konradv said:
> 
> 
> > gslack said:
> ...



Fake, fake, fake, fake fake.  Nothing you post makes much sense.  Your only value is as an object of ridicule.


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## wirebender (Nov 22, 2011)

konradv said:


> gslack said:
> 
> 
> > konradv said:
> ...



Do you want to prove that mathematically or are you just here to wave the pom poms for ian?


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## westwall (Nov 22, 2011)

IanC said:


> polarbear said:
> 
> 
> > Then You tried to make the same erroneous case again using the "extra time" a photon  spends in Your phantasy climatology world,..and then I did the math for You and the "extra time"  was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
> ...







Ian,  I have been reading this and the various other threads for awhile now and the one thing that has become very plain is you are terribly ignorant of real world physics and the mathematical skills one needs to express the world of physics.  I suggest you get yourself a CRC and a good book on math and study, study, study.

You are far, far out of your depth here and need a good deal of study to catch up.


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## gslack (Nov 22, 2011)

konradv said:


> gslack said:
> 
> 
> > konradv said:
> ...



Come back with one of your socks.. You know one of them that's older..or something...


----------



## polarbear (Nov 22, 2011)

westwall said:


> IanC said:
> 
> 
> > polarbear said:
> ...




Hi there old friend...!!! How are You today...??...By the way guys, my daughter was released from the hospital now...*and we thank You all f*or Your concern and well wishing
@ Westwall...:
Have You noticed how the "lefties" are joining in, just as soon as You corner a pro-warmer or the author he quoted with the math..:



> konradv
> post number 312
> Quote: Originally Posted by *gslack*
> 
> ...


Well he did not direct that at me, but at wirebender and Gslack...
All that`s missing now is "OldRocks" and his sidekick "Thunderfart"

But You know I don`t really blame IanC so much as the authors who spread this kind of garbage and have the nerve to pass themselves off as scientists...IanC does not really do that...:
The Amazing Case of &#8220;Back Radiation&#8221; &#8211; Part Three « The Science of Doom


> The major reason that people give for thinking that DLR can&#8217;t affect  the temperature is (a mistaken understanding of) the second law of  thermodynamics, and they might say something like:A colder atmosphere can&#8217;t heat a warmer surface​There are *semantics *which can confuse those less familiar with thermal radiation.
> 
> If we consider the specific terminology of heat we can all agree and  say that heat flows from the warmer to the colder. In the case of  radiation, this means that more is emitted by the hotter surface (and  absorbed by the colder surface) than the reverse.
> However, what many people have come to believe is that the colder surface can have *no effect at all*  on the hotter surface. This is clearly wrong.  And just to try and  avoid upsetting the purists but without making the terminology too  obscure I will say that the radiation from the colder surface _can have an effect_ on the warmer surface and_ can change the temperature _of the warmer surface.
> ...


This guy can`t do the math any more than IanC...actually I think if IanC had really tried to do the math as I asked him now how many times...he would have done better than this fraud.
Anybody can make a mistake, make a typo or fuck up by accident with a decimal place...it can happen to all of us...*but I`m certain this is not the case *in this "science of doom back-radiation"...
It seem s to happen everywhere You look in "climatology"...we all remember how they fucked up with their Himalayan glacier predictions and demanded they should be cut some slack for this decimal place slip-up...but they have been doing so all over in their statistics,....everywhere You look...especially in their hockey stick "average temperature graphs " and just a little while ago again with the amount of ice in the arctic...*it`s a pattern*, they have these "accidental/mistakenly" placed decimal points, which produces numbers which they then all use and publish,....knowing full well that not very often somebody checks up on that...and when It is discovered then it was merely a slip of the pen..and the next fraud is already concocted again, using the same method

But let`s see how many SERIOUS mistakes this copy&paste "author" has made in just these few lines...
If You actually bother doing the math for his example You get for a 10 &#956;m  wavelength an  energy quantum of:
*1.98 ^ (-21) *

and he not only came up *with a number 10 times higher than what it really comes* out at if You do the math for E =  hc/&#955;, he commits much more gross errors than that...totally disregarding what the "h" in the equation hc/&#955; really represents:


So what`s the dimensional difference You may ask Ian...well let`s take a look, it happens to be the difference what an Energy quantum really is, as opposed to the "photon energy" in "all photons are the same"....(as stipulated in "climatology physics")..:
Joule-second - Wikipedia, the free encyclopedia


> The* joule-second *is a unit equal to a joule multiplied by a second, *used to measure *action* or *angular momentum. The joule-second is the unit used for Planck's constant.
> In SI base units, the joule-second is
> 
> 
> ...


And the guy that said :


> You can find this in any basic textbook and even in Wikipedia. *So, for example, the energy of a 10&#956;m photon = 2 x 10*&#8722;20* J.*


has been using photons as if they were little pebbles that You can add up into a bigger one...

I am 100 % certain that this fraud has no idea what an angular momentum means in quantum (wave) mechanics..or that he would be able to comprehend it...

Maybe You will IanC, if You let me explain it to You:

Any rotating mass has an angular momentum which resists a change...
it is a product of angular speed in radians per second and a mass...in classical mechanical physics..
In wave quantum physics the "angular speed" is the FREQUENCY of the wave ...which resists a change in frequency and orientation the same way a mechanical angular momentum does...it is also the reason why the speed of light is constant unless You change the medium,...and then it only changes very very little..the energy quantum of a wave at any frequency is a discrete quantity ...You cannot take a fraction of this quantity, ... somehow divide it and use these fractional components  *or other whole photons* *of a lower energy t*o add up and  to merge these to a new photon that carries a higher energy quantum and thus increase the angular speed or the frequency.







So who is the asshole here...this usurp who pretends  to  understand science or the guys who quoted and have been relying on him...:
The Amazing Case of &#8220;Back Radiation&#8221; &#8211; Part Three « The Science of Doom


> There are *semantics *which can confuse those less familiar with thermal radiation.
> 
> If we consider the specific terminology of heat we can all agree and  say that heat flows from the warmer to the colder. In the case of  radiation, this means that more is emitted by the hotter surface (and  absorbed by the colder surface) than the reverse.
> 
> ...


At least IanC never *PRETENDED* to have done his own math...he actually told me, that he did not want to assign any specific numbers when I pressed him for it...

I have been trying in vain to explain the most elemental parts of quantum physics to IanC, like the wave duality of a photon,...and that no not all photons are the same as this quack keeps saying...and I did already mention to IanC, that a photon has no "resting mass" and what in quantum physics the energy of a wave is 

Unfortunately he keeps going back to the same fodder troth...
In the future I suggest to You Ian, to keep an eye out who Your PC is connecting to when You go Googling for science, ...
and if it connects to something like "wordpress.com"....which is a favorite server for climatology propaganda, like what You have been defending here...don`t expect to find reliable  information on a server like that,... as You would in a real & serious physics book for quantum physics.
and by the way what this "scientist has told You here was also utter bullshit:


> *a 10&#956;m photon  from a 2,000K source has exactly the same energy as a 10&#956;m photon from a  200K source.*


the Energy for a photon at  a  10 &#956;m  wavelength =1.98  * 10^ (-21) 
and for a photon where the Planck curve peaks according to the Wien`s displacement law is at a wavelength *of ~ 1.2 **&#956;m *...
which has a quantum angular momentum energy of *~ 2 * 10^(-20) 
which is in turn what the author of "the science of doom" had for :
*


> *So, for example, the energy of a 10&#956;m photon = 2 x 10*&#8722;20* J*


See it was not just a simple oversight like You made, when You confused POWER and ENERGY...
He never did do his own math...he did a copy&paste...of somebody else`s numbers but he did a typical "OldRocks" and a copy&paste on another "climatology science" garbage page and also confused the numbers what the Planck equation has for *1 **&#956;m  and 10 **&#956;m*...!!!

If that doesn`t expose him as a fraud, what will it take to (finally) open Your eyes InaC....??
Wirebender told Yu more than once that this "physics expert" is nothing more than a (fired) grade 7 school teacher, who is now cashing in as a "climatology" P.R. asshole.

And what I really would like to see is a "climatology" equation for "back-radiation" this time let`s settle on graphite rods, 10 by 1 by 1 cm with a 0.5 cm groove milled in the 2 sides of these rods as they face each other...now You have a perfect cavity and with graphite the best thing You can get as a "black body"..:
Black body - Wikipedia, the free encyclopedia


> Blackbody radiation becomes a visible glow of light if the temperature of the object is high enough. The Draper point is the temperature at which all solids glow a dim red, about 798 K.[4][5] At 1000 K, the opening in the oven looks red; at 6000 K, it looks white. No matter how the oven is constructed, or of what material,
> A closed box of graphite walls at a constant temperature with a small hole on one side produces a good approximation to ideal blackbody radiation emanating from the opening


One rod is at +10 C, the other at -10 C...then we move them together till the rods make contact and the optical cavity is 10 by 1 cm with the 2 channels milled 0.5 cm deep.
I can assure You IanC, that both rods if they have the same mass, will settle at 0 deg C and *that You will not wind up with the +10 C rod being heated* by "back-radiation" or any other means to more than +10 from the -10 C rod...
Has it not occurred to You, that if indeed the colder rod "back-radiates heat energy" then it just got also even colder...now it will emit less and less high quantum energy photons as this "climatology" Kirchoff`s laws defiant heat gain of the hot rod getting even hotter is supposed to progress the way these quacks are describing it...even when the 2 rods don`t touch with nothing but a channel for space between the 2,... *there is no way You can gain energy from the colder rod...*and the colder rod will gain energy from the hotter rod till You have a perfect thermodynamic equilibrium, and at the same time and *only then the energy distribution as shown in Planck`s curves*..*.and it will be exactly the same in both rods*.

If You were to heat either rod, You would not even have an energy distribution over the spectral range as in Planck`s curves till the new & extra heat energy You are feeding to one of these rods has been distributed over the entire mass of the black body...*and that takes a lot longer than You think.!!!!*
 

You can get graphite rods from almost any welding supply store, scrape off the copper coating, square them off with a file and scratch or file a grove into each rod....
In Germany we say "studieren macht schlau, studieren und probieren noch schlauer"
{ study can educate, to study and to try it out yourself makes You even more educated and smarter}
That`s why in real science we don`t just use "examples" that exist only on paper, like "Spencers example" with his 2 rods, no mass, distances  or temperature specified...and he managed to convince You that the hotter rod can get even hotter from the "back-radiation heat energy" that is supposed to come from the colder rod...in real science we have labs and confirm with experiments..
So go & buy 2 graphite rods, heat one and not the other one, use some thin wire to bind them together and then suspend them in a $20 thermos bottle ..





Well I had enough of this...I`m renovating my house at the same time as I`m reading & writing here...the putty is dry..I can start sanding and painting the next wall now...
Have a good time guys and good bye again for quite some time


----------



## IanC (Nov 23, 2011)

polarbear said:


> You come back here claiming that all photons are the same...we all know where and from who You got that one from...
> 
> So now You are claiming that these "all the same photons" each having been emitted by a radiation source say at 10 C....
> that 10 of these photons stack up to a new and higher temperature of  10 times X ("climatology photon  degrees")...
> ...



this is the type of nonsense that pisses me off. you read the article so you know he didnt say any of those things. 

1. he said a photon of a specific wavelength is exactly the same whether it is emitted from a cooler body or a warmer one. do you disagree? do you think there is some sort of brand labels on photons with warmer surfaces turning their noses up at Levis and only accepting Calvin Kleins?

2. he said that a warmer blackbody emits a set of radiation that totally contains the set of radiation from the lower blackbody. do you disagree? every Planck curve diagram for blackbodies that I have seen shows the lower curve totally under the higher curve, do you have some evidence otherwise?

3. the radiation graph of the lower temp blackbody is perfectly matched by the warmer body. these are the two sets of radiation involved with backradiation. each body is radiating into the other body this set of radiation, so there is no gain and no loss. these two set exactly cancel each other out but that does not mean the radiation magically disappears. it is still there, going back and forth. just like if there were two bodies at the same temperature, the radiation would cancel out. this is where you obviously disagree but where is your proof that the radiation of these two equal sets disappears?

4. the left over part of the warmer Planck curve, above the balanced out lower portion, is the reason why heat always flows from hot to cold. the warmer body ALWAYS radiates more, as can easily be seen in the graph of +10C and -10C. or in any other Planck curve for blackbodies of differing temperatures. this excess set of radiation is considerably shifted to lower wavelength which is more energetic.

5. the last concept he brought up is more counterintuative. the range of wavelengths radiated by the two bodies only 20C different is _almost_ exactly the same. the only difference is at the far left, probably off the graph. the warmer body can give off a rare photon of higher energy that the cooler body cannot. but 99+% of the range is identical although the amount of radiation at each wavelength is always higher for the warmer body and the proportional excess continues to get higher as you move left into higher energy photons. when you compare larger temperature differences it is much easier to see how the average wavelength and intensity deliver more energy.






please note that the area of back radiation is the area under the pink curve which perfectly balances out radiation going back and forth between the two bodies. the area between the pink and blue curves is the radiation capable of transfering heat that only goes in one direction. a visual desciption of how and why the 2nd law of thermodynamics works.






please note that even with a 2000K difference in temperature the range of wavelengths for emitted photons almost completely overlaps. there is no difference in a 500nm photon emitted from a 3500, 4500 or 5500K source. there is no difference in a 15micron photon emitted from a 263, 283 or 5500K source.


----------



## IanC (Nov 23, 2011)

http://spie.org/documents/Newsroom/Imported/1199/1199_4082_0_2008-07-08.pdf


> Incandescent bulbs emit light in a manner closely resembling
> Plank&#8217;s law of blackbody radiation. The law describes how a
> body capable of absorbing all radiation contacting it (a blackbody)
> will emit at a given range of wavelengths dependant on its
> ...





> The filter acts as a perfect transmitter for the useful visible
> light and a perfect reflector for the undesirable infrared light.
> The reflected light is re-absorbed which, in turn, helps to heat
> up the filament. This infrared recycling process has two major
> ...








first off, this paper demolishes wirebender's claim that no photons can be reflected back at the filament of a flashlight.

second, it demolishes the claim that back radiation can have no effect on the warmer body. if the 'generic' heat is not needed to produce the IR because it is already there, then that heat is freed up to produce higher energy radiation.


to answer polarbear's request for an explanation of my figures for Spencer's thought experiment-

I used wirebender's value of 1 for emissivity. I used wirebender's value of 1m2 area for each bar. I used wirebender's calculation of ~750w for 150F heated bar. I constructed a rectanglular block of proportions 2H,1D,1W for ease of calculations. I am also ignoring the temperature differential due to the shape. the face of the heated bar which will be towards the second bar is therefore 0.2m2 and radiates 150w.

the second bar is placed in the container. I have placed it as close to the first bar as possible to reduce stray radiation but not touching so there is no conduction. Spencer said the second bar reached 100F which works out to ~75w for 0.2m2 but did not point out that there would be a temperature differential across the second bar because the radiation is only coming in through the inside face.

the heated bar radiates into the second bar until equilibrium is reached and the inside face of the second bar is 100F and therefore radiating 75w. it is time to recalculate the heated bar. it is recieving 750w from the heater, and 75w from the second bar for a total of 825w. that means the 0.2m2 inside heated face is now radiating 165w. 

time to check the inputs and outputs. the two inside faces are not exposed to the container so they do not output. 825w times the 0.8m2 effective radiating surface of the heated block is 660w. what about the second bar? it has 165w coming in, and 75w going back for a net gain of 90w. that 90w is unevenly divided between the 0.8m2 remaing of effective radiating surface. 660w and 90w =750w.  input equals output.

I have not done the calculation for the temperature of 825w but I bet it is ~160F

my model is vastly oversimplified and only meant to give a rough idea of how the heat flows through the system. but it is vastly more sophisticated than wirebender's claim that both bars will be the same temperature with the full 2.0m2 area radiating equally.


----------



## gslack (Nov 23, 2011)

IanC I can't wait to see them destroy that bit of garbage you just posted as science.. 

The "paper" you cited is not a scientific paper at all.. Its sales pitch.. Its selling a conceptual (at least from the way it reads) invention where as they invention attempts to recycle emitted IR into more incandescent visible light using nanostructures... As in a machine doing more work... HAHAHAHAHAHAHAA!

Notice the way it works at all? No of course not all you did was shout eureka I have done it and go with it... Again... LOL it recycles the heat from the light source through a system designed specifically for the task and does so by redirection.. Jesus IanC...WOW...

And one thing you didn't see because once again you didn't read it fully... They say this...

_"Photon recycling via a metallic PBG filter is a promising new
route to creating a &#8216;cool&#8217; light bulb. Our next step is to study a
cylindrical filter geometry that is comparable to the commonly
used tungsten-filament configuration."_

Ah they say "is promising" as in maybe, or could or theoretically by their calculations should.. So... Yeah...

They also say this: 

"To illustrate the validity of our approach, we have employed
an ideal system that has a spherical blackbody filament enclosed
by the filter: see Figure 2(a). "

So they are using and "ideal system" "that has a spherical blackbody filament"... So then its not even based on anything real is it.. A perfect blackbody? Got one lying around do ya? LOL...

I will stop there because I really want to see wire and polarbear trash this further...LOL


----------



## wirebender (Nov 23, 2011)

IanC said:


> first off, this paper demolishes wirebender's claim that no photons can be reflected back at the filament of a flashlight.



It doesn't touch my argument ian, but I suppose you can't be faulted to much for wishing that it did.  



IanC said:


> second, it demolishes the claim that back radiation can have no effect on the warmer body. if the 'generic' heat is not needed to produce the IR because it is already there, then that heat is freed up to produce higher energy radiation.



It demolishes the 2nd law of thermodynamics?  Prove it ian.  Lets see your math.  For that matter, lets see "your particular version" of the 2nd law of thermodynamics which speaks to net energy flows.  

Face it ian, you were wrong from the start, you never had a chance.  The laws of physics and nature were against you.  As westwall said, you are out of your depth and in order to even begin to grasp how far from the facts you are, you are going to have to get a handle on what a photon actually is.  Till you understand that, the rest is going to look like (to quote you) magic.


----------



## wirebender (Nov 23, 2011)

gslack said:


> Notice the way it works at all? No of course not all you did was shout eureka I have done it and go with it... Again... LOL it recycles the heat from the light source through a system designed specifically for the task and does so by *redirection*.. Jesus IanC...WOW...



Redirection.  That's right up ian's alley.

I am going to bow to polarbear here gslack.  He clearly makes his living with mathematics every day.  I know enough to prove ian wrong (simple laws of physics) but polarbear is in a different league.  I don't have time to wade through enough physics texts to speak at polarbear's level even if I were inclined to do so.  I would have to give up half my practice and 2/3 of my family life to manage the time it would require.

I am done with ian.  At this point, I don't think that it is a matter of him being unable to see that he has been wrong, it is a deliberate unwillingness to see how far out on a limb he is or even admit it as a possibility.   The rather rudimentary work I brought dealing with just the SB laws was enough to prove him wrong, polarbear is decorating the cake, gift wrapping it with hand painted wrapping paper and covering the box with elaborate hand tied bows.

ian was asked to do the simple math with a couple of carbon bars and is apparently unable to even do that but still maintains that he is right and we are wrong.  He has the faith of a zealot and no amount of actual science can ever penetrate that sort of quasireligious fervor.

Heretics and infidels can never be right even when they are right.


----------



## polarbear (Nov 23, 2011)

Okay IanC, I`ll write one more post in this quite interesting thread You started here, but I really don`t want to go into "He says she said" arguments,....but the author of the publication You quote did say
that photons are all the same and nobody can tell them apart...and from Your reasoning I could see that maybe not now, but
at some time You did understand photons like that...
If You very carefully scrutinize the way this author is leading the readers to such conclusions, then You can see
where he made the leap that a  10&#956;m  photon coming from a 200 K source is the same as a 10&#956;m  photon coming from
a 2000 K source  to  " *ALL PHOTONS ARE THE SAME*...NOBODY CAN TELL THEM APART"...

And he needs You to go along with him on this leap in order to conjure up a situation where as he says...
"something has to give"....which then turns out to be Kirchoff`s law...because in the end he wants to make a statement that a colder body can heat a hotter one with these "all the same photons" via back-radiation.

You  and many others have been vulnerable to this brain hack, because this assertion exploits a gap
in specialized physics schooling....if somebody chose to study for another profession...
 and a gap, *that many physics books have also *when it comes to explaining the nature of electro-magnetic radiation
and TEMPERATURE...

I shall try and close this gap and explain it in less abstract terms ,  because a gap in understanding is quite similar to a security gap and this brain hack is quite similar to a cyber hack...so let`s try fix it...:

In the end we are talking about* TEMPERATURE,* so let`s start here.
What exactly is it...?...You have matter consisting of atoms, if You talk about a pure element or
matter consisting of molecules if You are talking about something other than an element...
Lets choose molecules, because this second cooler "black-body" are gasses, and the bone of contention is CO2.

The earth`s surface, the second and "hotter black body"...let`s also stick with molecules here....just for the sake of some oversight and to avoid confusion
So what exactly is going on here when Molecules get "hotter"...?
Well first off they don`t stand still, they can only stand still @ zero motion if they are at a TEMPERATURE of
absolute Zero (0 Kelvin)...
So any *temperature increase* must be able to accelerate the average speed of this molecular motion.
when that happened  a hotter gas either expands or if You don`t let it,  will exert more pressure.
In the case of a solid the same is true except it`s from a bit more difficult to impossible...to dis-allow expansion.

Now in this hotter state both Molecules and Atoms have electrons in their outer shell
in orbitals that want to fall back into the least deformed state, ...and the positive charge of the nucleus
in the center is responsible and is the cause for this  "fall back"....which then emits photons

As  the *speeds* ( speak TEMPERATURE) of  the individual molecules increases the more deformed, these electron orbitals will get....just like cars would behave in collisions...only difference is that  Molecules don`t have  hard metal  shells like colliding cars, but  deformable electrons that repell  and deform other
electron shells in other molecules  that come too close....

So now that we  have an understanding of temperature...let`s think about it, if  a molecule at a lower speed can "bump" another molecule which is also at a lower speed so that this molecule attains a higher speed...( is getting hotter )....or can a slow speed molecule bump or "rear end "
a fast molecule to make it even go faster...?
*And if it could have then we would indeed have to ditch Kirchoff`s 2.nd law

*So now that leaves us having to examine if "back-radiation"  from a cooler (speak slower) molecule can accelerate the molecular motion (speak temperature) of  a  hotter molecule.

To resolve this we have to take a look  how  heat generates radiation.
As we do, let us not forget, that a soon as a "hotter"...(speak faster) molecule emits a photon, which is now the energy carrier for the energy that this previously "hotter" (faster) molecule has just given up energy ...as the deformed orbital relaxed back to the lower energy orbital state.

And when this photon that thermal imaging/spectroscopy   sees as a  *light equivalent ENERGY* of   
the *heat (motion) energy  as You measure it with a thermometer *
is converted back to heat .....as is the case  when it is absorbed by another * COOLER molecule

*It can only do so much  "electron car bumper" damage as the energy  equivalent that generated this photon ...and this energy amount is called a wave quantum, .....in terms of our phyisics here ..
this frequency dependent quantum  now has to distort the  "electron car bumper"...the orbital of a molecule  that is absorbing this photon radiation ...and in the end cause this "bumped" molecule to increase in speed ,
...no matter if that speed is molecular motion or increased molecular bond vibrations

Although photons move at the speed of light, remember I told You over and over again, they do not have a "resting mass"...and You can`t envision them as a projectile at the speed of light and having a mass....had they then such light speed fast photon masses coming from a colder source
could accelerate the molecular motion of a hotter molecule ....and we would be in
the "science of doom" and  the back-radiation from colder bodies  could heat hotter bodies and Kirchoff`s laws would have to be revised...pending of course an actual experiment that proves this assertion..

There have been  countless experiments that  have shown  how our current understanding of wave quantum mechanics is correct.
For example a "smart climatologist" (which is an oxy-moron)  might claim
that  10&#956;m  photon coming from a 200 K  can be seen as a  2000 K source 1&#956;m  photon by  molecules that streak with enough speed toward  the  "cooler"  back-radiation source.

But that won`t help either, because the observed Doppler shift  due to the constant speed of light + the addition of the object speed* can not add up to more than the speed of light.

*Could this be done, then the "Science of doom" would have also just managed to
"create"  a  1&#956;m  photon  from a 10&#956;m  photon  that  came from a 200 K source
instead of a 2000 K source...

It is not the photon that has been changed...the only thing that changed is what is
being observed as You move into the opposite direction of  the  10&#956;m  wave propagation

So You see, no matter how You try and arrange these  photon absorptions by directional changes of the molecules (or Atoms)  there is no way a 10&#956;m  can  effect the same as a  1&#956;m  photon..*even  when they  do appear to be the same  *because of the Doppler shift...

I hope You can understand now why photons are no exception to Kirchoff`s second law.
Also I did not read EVERYTHING in this thread,  because I rather read EVERYTHING
in the phyisics books on my shelves...


I hope after reading this You have an understanding that the "*heat radiation*" You
can observe as photons is *just a snapshot of a process*  when the *heat *as You measure
it with a *thermometer* in degrees Kelvin , (the speed of *molecular motion*)  has been  converted from  an energy that can be measured in *velocity (speed) and mass *to an energy form * where the mass has disappeared* , the speed (c)  *remains constant*  as is no longer variable as earlier in this process..and energy  must now be expressed in terms of  the quantum a  1&#956;m   or a  10&#956;m  long wave can possibly carry with it.

And when this photon is absorbed and  shows up again as heat the way a
thermometer can measure it  can  only  have  so much of a "speed effect" on
the existing mass that absorbed this photon...and no more "speed" as
the process that generated this photon had...

So there is no way a  slower -10 C  system   can make the faster +10 C system even faster in any way...and that includes the transition from  heat motion  to photon and then back again

And also let`s be clear about this "slowing the loss of heat"....
*This is not the issue here, the issue was that "climatology" claims *
*that a colder body can HEAT a  hotter one.*


I don`t know if wirebender or somebody else made a mistake...and if indeed
he made one...I can assure You that his mistake is in no comparison as the
mistakes in the "science of doom"...
I also make mistakes, but that does in no way get subtracted from what I do
know about physics...the way a cynic would like to discredit in science as if it were politics
But You see, this is exactly how this lefty debating tactic wants to work it..
Somebody made a mistake and now they want to use that like a debit card
and a bank account..and  subtract  from the knowledge of  physics brain cells account

I have no problem with the mistakes I can spot in the "science of doom" and
the "colder body photon back-radiation can heat a hotter body"   publications
that are all over the internet lately...each is just a copy of another...including
the mistakes in the math & the numbers have been copied...that`s what
"climatology" calles "peer review"...!!!!!.....*.But I do have a problem when*
*this crap is being used to  do brain hacking attacks on people who just **had to study something other than quantum physics to make an honest  living..



*


----------



## polarbear (Nov 23, 2011)

@ IanC
Now after having read what I just posted before, perhaps You will read this text with a whole new visual ability[SIZE=+2]
and please pay extra attention to the words I highlighted in red.!!!! :
[/SIZE]Photon - Wikipedia, the free encyclopedia



> The modern concept of the photon was developed gradually by Albert Einstein to explain experimental observations that did not fit the classical wave model  of light. In particular, the photon model accounted for the frequency  dependence of light's energy, and explained the ability of matter and radiation to be in thermal equilibrium. It also accounted for anomalous observations, including the properties of black body radiation, that other physicists, most notably Max Planck*, had sought to explain using *_semiclassical models_*, in which light is still described by *Maxwell's equations*,  but the material objects that emit and absorb light *are quantized.  Although these semiclassical models contributed to the development of  quantum mechanics, further experiments[2][3] *validated Einstein's hypothesis that light itself is quantized*; the quanta of light are photons.In the Standard Model of particle physics, *photons are described as a necessary consequence of physical laws* having a certain symmetry at every point in spacetime. The intrinsic properties of photons, such as charge, mass and spin, are determined by the properties of this gauge symmetry.
> *The *neutrino theory of light*, which attempts to describe the photon as a composite structure, has been unsuccessful so far.*


So, please don`t cement Your feet in the position, that "climatology" tries to convince You what a photon is..even Max Planck does not have it right, as has been proven by Einstein..
As for these  few short wavelength photons that a colder body seemingly emitted,...You know the few that appear under the Gaussian statistical curve of Planck`s distribution...let me remind You again that also in a "black-body" the molecules or atoms that emitted these "hotter" photons are in motion...randomly in all direction...
So from any angle You would have atoms or molecules moving in this direction at the thermal "speed"...the TEMPERATURE these black body molecules are...

And now You get a Doppler shifted photon *that appears to have a shorter wavelength.*..and these are the few ones appearing under the Planck distribution curve.
A Doppler shifted photon is not the same as a photon that has a shorter wavelength...without having been Doppler shifted...
Modern physics takes all that into account in the ongoing experimentation at the CERN Hadron collider...
Unfortunately it is not possible to accelerate photons  as You can nuclear particles , but You can accelerate  particles and then have these emit light later, You collide them and observe every photon energy quantum and particle fragment that has been generated...
and in the case of the photons the *Doppler shift is taken into account..*
In Your Planck`s distribution curves for various temperatures *it is not taken into account...
*Remember that, the next time You quote somebody who is trying to pull the wool over Your eyes again,...with that a 10 &#956;m photon coming from a 200 K source is the same as one coming from a 2000 K source and then leaps from there to how a thermodynamical slower 200 K system can speed up a thermodynamical faster 2000 K system if You use photons as (back) radiation instead of thermal heat exchange via heat flow...which certainly will never happen from colder to hotter..So to make the case for this imaginary man made CO2 based dooms day pseudo science these "climatology photons" that fit into these assertions had been created with nothing more than word-games and number-less "examples" that exist only on paper and only in the minds of people that may have heard of Einstein, but never studied, pondered and fully understood what he actually wrote


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## James_Of_Gala (Nov 24, 2011)

Shall we start with arithmetic

6.6^(-34) * 3^8  * 10^5 = 19.8^(-21) = 1.98^(-20)

any comments?


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## polarbear (Nov 25, 2011)

James_Of_Gala said:


> Shall we start with arithmetic
> 
> 6.6^(-34) * 3^8  * 10^5 = 19.8^(-21) = 1.98^(-20)
> 
> any comments?




6.6^(-34) * 3^8  * 10^5 = *4.33026 ^(-25)* not *1.98^(-20)*

any comments?


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## James_Of_Gala (Nov 25, 2011)

Er OK let's use a less abbreviated notation

6.6*10^(-34) * 3*10^8 * 10^5 = 1.98*10^(-20) 

comments?


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## polarbear (Nov 25, 2011)

James_Of_Gala said:


> Er OK let's use a less abbreviated notation
> 
> 6.6*10^(-34) * 3*10^8 * 10^5 = 1.98*10^(-20)
> 
> comments?



Hey 3*10^8 is not the same as 3^8
3^8 =6561
and 3*10^8=300000000

And You did say 3^8 not 3 *10^8


James_Of_Gala said:


> Shall we start with arithmetic
> 
> 6.6^(-34) * 3^8  * 10^5 = 19.8^(-21) = 1.98^(-20)
> 
> any comments?



We all make  mistakes...I`m no exception either, and I also know *where I made it*..
I have a "quick launch" for windows cale.exe...when I`m too lazy or too tired...I use it
But...!!! take a look at it how small the * and the - sign is...
they both look like dots to me unless I wear my reading glasses

anyway for  6.6*10^(-34) *10^5 I don`t need glasses or a calculator.. I can read my screens large type font okay...that`s 6.6*10^(-29)
and with my glasses on if I multiply that by the 3^8 (=6561) as You did have it,... : 
*6561 * 6.6 * (10^(-29)) = 4.33026 × 10^(-25)*

I know full well You are not a dummy, and I also know that You meant 3*10^8 instead of 3^8...because the speed of light c is certainly not 3^8 meters per second...
But as far as I`m concerned we are even now,...I f`d up not seeing one decimal point in the correct position without my glasses and I also know that 19.8 *10(-21) is the same as 1.98 *(-20)
But I think You would agree that *Your "c" = 3^8 *missed the mark by quite a bit more...
Imagine what that would be like, if the speed of light would be that slow, 6.5 km per second..why worry about any kind of radiation.. 

I`m not trying to pull Your leg or call You down...because I know full well how easily something like that  can happen...

For example how often does that happen...: You write a capitol letter and Your left hand finger is still on the shift key...and oops you typed a "("  instead of the number 9
On top of that more often than not I loose everything I type in here quite often because our server here keeps dropping the connections...and as of tomorrow where we live there is no more high speed server...no more e-mail...no more Internet period....
We are way out of range from all the other wireless servers and the only other option would be to open a new account and use the telephone & a 50Kb slooooow modem...
That`s Okay...I rather live way out in the country, no cable TV and such than in the city,...the only thing that`s faster there is the Internet but certainly not the traffic.!!
So have fun without me


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## IanC (Nov 25, 2011)

polarbear- I enjoyed your bumpercar analogy.

I have a specific request for information, if you know where to find it. Conduction is the amalgam of energy transfer by physical and radiation means. do you know of a source that gives the relative efficiencies of conduction at a boundary and radiation at a boundary, so that you could subtract the radiation portion of conduction and get a rough idea of the kinetic energy portion of conduction?

I still have a problem with you claiming that SOD said a 200K source could warm a 2000K body, he didnt. nor did he say that 10 micron photons were being tranfered into 1 micron photons. he was describing two bodies of relatively equal temperatures, and how naturally occuring higher energy photons from the cooler surface would impact the local area of the warmer surface. 

your other point was that a low energy photon couldnt make a molecule go faster, which is true, but it can alter the direction of the molecule by adding a sideways component. I also doubt that the relatavistic events that you brought into the conversation are a meaningful factor in the low energy/low temperature scenarios that we are discussing.


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## James_Of_Gala (Nov 25, 2011)

OK so we agree what the energy of a 10 micron photon is, now that I have sorted out my nottaion.

That's a minor point though. What I really wanted was to ask a simple question. Supposing you have a single photon, just one photon all by itself, how do you know the temperature of the source from which it originated? (assuming it did originate by a temperature dependent mechanism, as it may not have)


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## IanC (Nov 26, 2011)

there are no labels on photons to identify their source. and photons do not disappear until they interact with matter.


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## gslack (Nov 28, 2011)

This will be my last post on this topic...

Its called entropy fellas.. It does not require the energy to know whence it came or the temperature from its origin. Energy flows out from its source it doesn't decide to do this after some thought, nor does it check to see if its starting point is hotter than its destination. Just as the destination does not check the temperature of incoming energy sources to be sure its indeed correct in accepting it.

The claim you two are making is just nonsensical... Does a golf ball check the air and ground before or after a club hits it to see if it can overcome the resistance and forces acting against it? Does it check the force of that club hitting it before leaving and then decide how far to go? Or does the action of the club hitting and your decision in swinging that club make those things happen? I don't think the ball had much of a choice in the matter do you?

By the thinking showed by you two (or you alone, lets be honest here) we have to be ready to accept the "feelings" or "thoughts" of matter or energy in all things.. Nice try.. Not really, more like nice try at BS...

Why don't you go and google up proof of back radiation and see what you find. You won't find any proof of it, you will find a few papers claiming it possible in perfect conditions (that do not exist outside of theory) and only mathematically. However you will find many papers showing it doesn't happen naturally and they will be based on real world observation and experiments.


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## IanC (Nov 28, 2011)

interesting. if you substitute a CO2 molecule for the golfer then you are supporting my position. the photon doesnt care where it came from or where it is going, it just goes. and sometimes it goes to the earth's surface and adds to the pool of energy. it doesnt heat the surface because the net flow is outwards but it does slow the flow of outward radiation by cancelling out (edit-the net effect of) a simular outbound photon.


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## wirebender (Nov 28, 2011)

IanC said:


> interesting. if you substitute a CO2 molecule for the golfer then you are supporting my position. the photon doesnt care where it came from or where it is going, it just goes. and sometimes it goes to the earth's surface and adds to the pool of energy. it doesnt heat the surface because the net flow is outwards but it does slow the flow of outward radiation by cancelling out (edit-the net effect of) a simular outbound photon.



The ball moves in the direction propagated by the greater force just as the IR emitted by the CO2 molecule.  It is subtracted from the EM field emitted by the earth and as such, diminishes the field by that amount, but alas, never reaches the surface.


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## IanC (Nov 28, 2011)

wirebender said:


> IanC said:
> 
> 
> > interesting. if you substitute a CO2 molecule for the golfer then you are supporting my position. the photon doesnt care where it came from or where it is going, it just goes. and sometimes it goes to the earth's surface and adds to the pool of energy. it doesnt heat the surface because the net flow is outwards but it does slow the flow of outward radiation by cancelling out (edit-the net effect of) a simular outbound photon.
> ...



you seem quite certain of this and I have no doubt that you believe it. all I am asking for is some independent confirmation and an explanation of how it happens and where it happens. I have already asked whether it happens inside the CO2 molecule, enroute to the surface, or at the surface but you have refused to give any details of this process that seems to contradict the known rules of physics. could you just give a few more details so that we mere mortals might learn at the feet of the master?


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## wirebender (Nov 29, 2011)

IanC said:


> you seem quite certain of this and I have no doubt that you believe it. all I am asking for is some independent confirmation and an explanation of how it happens and where it happens. I have already asked whether it happens inside the CO2 molecule, enroute to the surface, or at the surface but you have refused to give any details of this process that seems to contradict the known rules of physics. could you just give a few more details so that we mere mortals might learn at the feet of the master?



ian, I have answered all your questions ad nauseum.  As I have said, you won't find much information on the how because the how is not understood.  The fact that opposing EM fields reduce each other's magnitude however is well known, observable, and repeatable science.  The energy can not be destroyed but is gone none the less.  Since the energy is composed of its carrier, ie photons, it is the number of photons that is being diminished.  

As I have said before, till you get off the mistaken idea of photons as independent discrete free agent particles and accept and embrace wave particle duality, you aren't going to be able to understand this and it will remain a mystery to you but it will allow you to hold to your belief that somehow the 2nd law of thermodynamics is being violated by individual photons emitted by CO2 molecules moving against the overwhelming magnitude of the EM field radiated away from the surface of the earth. 

I have told you that it does not happen inside the molecule which is why your incessant harping on virtual photons is useless as virtual photons are confned to the boundries of the atom or molecule.  

I will not go over this with you any further.  Each and every one of your questions has been answered over and over.  Asking again is not going to change the answers.  If you want to review them, you are welcome to go back over this thread and the other threads in which this topic has been discussed.  You will never win a discussion ian, by going over and over points that you have already lost.

By the way, subtraction of EM fields is supported and predicted by the 2nd law of thermodynamics.  The 2nd law states explicitly that it is not possible for energy to flow from cool objects to warm objects.  Regardless of of your belief in net flows, the 2nd law doesn't say anything about net flows.  It states that energy will not flow from cool to warm.  The subtraction of EM fields resulting in no energy flow from cool to warm is precisely in accord with the laws of physics.

It is your terribly flawed understanding of the laws of physics and your gross misunderstanding of what photons are and your refusal to accept wave particle duality that leaves you in the dark ian and that fact is not going to change till you change.

I have asked you over and over which law of physics you believe supports your beliefs regarding energy transfer and you can not state one since none either support or predict what you believe happens.  Let me know when you get something new or come around to actually being willing to accept what is happening with regard to the tiny EM field radiated by a CO2 molecule vs the overwhelming EM field radiated by the surface of the earth.


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## IanC (Nov 29, 2011)

wirebender said:


> IanC said:
> 
> 
> > you seem quite certain of this and I have no doubt that you believe it. all I am asking for is some independent confirmation and an explanation of how it happens and where it happens. I have already asked whether it happens inside the CO2 molecule, enroute to the surface, or at the surface but you have refused to give any details of this process that seems to contradict the known rules of physics. could you just give a few more details so that we mere mortals might learn at the feet of the master?
> ...



you have not answered my questions, you have ducked them ad nauseum. at least this time you admitted that you have no explanation for how or where the photons disappear.

lets move on to the next question then. you have repeatedly stated that if even one photon returned to earth that would be a violation of the second law. this is one of your fallacies that I have called you on before. microscopic vs macroscopic. where did you learn that the second law was in effect for single particles rather than for macroscopic systems made up of vast numbers of particles and positions? its not in any physics text I have read. the arrow of time, entropy only works with systems not individual particles or photons. there is no prefered position for one molecule of dye in a container of liquid, and there is no law that prohibits a large amount of dye from all being in one portion of the container either, except for the fantastically long odds of it happening.

the photons emitted from the earth have next to zero effect on the photons emitted from the atmosphere. all of the photons reach their destination, whether it is the surface, open space or a particle in the atmosphere. the cancelling out in the Stefan-Boltzmann equation is simply a mathematical calculation. there is simply a one-to-one replacement of the radiation from the cooler body. you subtract the data set of radiation from the cooler body with an identical set from the warmer body, and the radiation left over is the net flow. no photons were destroyed magically in their flight through space to their destination. 
the second law is true because there is always more radiation produced by a warmer body. this is a trivial result that can easily be seen by comparing Planck curves for bodies of differing temperatures. 

virtual photons carry the energy in electric and magnetic fields. if they find a particle to interact with then they become real, otherwise they are reabsorbed. where would the energy come from that is necessary to send photons in every direction and at all times if the unused ones continued to exist? I am OK if you just want to say that we dont understand electric and magnetic fields except to calculate the force but if you want to explain how it physically happens then you need virtual photons. that is the difference between reactive and radiative photons. radiative photons are paid in advance, reactive virtual photons are only paid if something buys them during the short term 'sale'.


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## wirebender (Nov 29, 2011)

I have explained to you ad nauseum ian and have ducked nothing.  We are finished.  My proofs stand and you have failed at every attempt to prove a point.  Repeating the same thing over and over to you is both boring and pointless.

By the way, virtual photons only exist within the confines of an atom or molecule.  They carry EM fields from electrons to protons or neutrons.  That is it.  They have no place in discussions on the topic of EM fields outside atoms or molecules.  It doesn't matter at this point.  I consign you to your faith.


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## IanC (Dec 2, 2011)

wirebender said:


> I have explained to you ad nauseum ian and have ducked nothing.  We are finished.  My proofs stand and you have failed at every attempt to prove a point.  Repeating the same thing over and over to you is both boring and pointless.
> 
> By the way, virtual photons only exist within the confines of an atom or molecule.  They carry EM fields from electrons to protons or neutrons.  That is it.  They have no place in discussions on the topic of EM fields outside atoms or molecules.  It doesn't matter at this point.  I consign you to your faith.



as you wish. I still laugh that you refused to disclose your 'proof'. do you still claim that it is back in some unnamed thread on some unnamed page? you admitted that you knew were it was but I imagine it wasnt quite the artistic piece of work you thought it was back when you dared everyone to prove your math wrong. hahahaha.


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## IanC (Dec 6, 2011)

wirebender said:


> ian, I have answered all your questions ad nauseum.  As I have said, you won't find much information on the how because the how is not understood.  The fact that opposing EM fields reduce each other's magnitude however is well known, observable, and repeatable science.  The energy can not be destroyed but is gone none the less.  Since the energy is composed of its carrier, ie photons, it is the number of photons that is being diminished.



is this the _magnum opus_ that you have been talking about? the grand explanation?hahahahaha. the how is not understood? hahahahaha


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## OohPooPahDoo (Dec 6, 2011)

IanC said:


> how many of you think it is a deception to have 390 (W/m2) going up and 324 coming down?



Is that not the case?



> while it is true in an absolute way it disguises the fact that only 66 w/m2 is available as the difference between the temps of the surface and the lower atmosphere. out of that 66, 40 leaves directly with only 26 W/m2 warming the lower atmosphere.


LOL! It only 'disguises' that fact to imbeciles that can't do subtraction. You've got to be kidding me, right?




> so 26 W/m2 is the greenhouse effect


NO. The "greenhouse" effect refers to the gross trapping of energy by the atmosphere - not the net difference. Without it we'd all freeze to death.



> and I have never seen any estimates higher than 26% for CO2's share of the GHE so that means ~8 W/m2 or less for the total amount of CO2. the absorption of CO2 is logrithmic so that means we are into the ninth doubling (256-512) so every doubling is ~(8W/m2)/8 doublings= 1 W/m2 or less.



Wow you've said a lot in a little amount of space, care to back it up and explain yourself a bit more?




> I find it hard to believe that something that is responsible for ~1 W/m2 is the grand control knob of the climate. it is an amount that is lost in the error bars of measurement. but that is just my opinion but now you know why I think that way.



Whether or not 1 W/m2 - wherever you got that number - is within the error bars isn't really a matter of opinion.


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## IanC (Dec 6, 2011)

all of the numbers are approximations taken from Trenberth's graph and public sources.

most sources say that we are 33C warmer because of the green house effect. and most sources say that CO2 is responsible for 1/4 or less of that effect. therefor 8 doublings of CO2 is responsible for ~8C. ~1C per doubling. another way of calculating and another result showing the minor result from CO2.

we are just about 400ppm CO2 now. that means ~1C increase from now when we get to about 800ppm. 

is that the sort of  answer you were looking for?


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## OohPooPahDoo (Dec 6, 2011)

IanC said:


> all of the numbers are approximations taken from Trenberth's graph and public sources.
> 
> most sources say that we are 33C warmer because of the green house effect. and most sources say that CO2 is responsible for 1/4 or less of that effect.* therefor 8 doublings of CO2 is responsible for ~8C. ~1C per doubling. another way of calculating and another result showing the minor result from CO2.*
> 
> ...




stop at the bold highlighted letters and explain further where you are getting your logic.


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## IanC (Dec 6, 2011)

1,2,4,8,16,32,64,128,256,512. eight doublings, working on the ninth

33C times 1/4 = ~8C

~1C per doubling of CO2

CO2 has a warming effect of about 8C on the climate, about 1C per doubling. if we add another 395ppm then it should be about 1C warmer than now.


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## OohPooPahDoo (Dec 6, 2011)

IanC said:


> 1,2,4,8,16,32,64,128,256,512. eight doublings, working on the ninth
> 
> 33C times 1/4 = ~8C
> 
> ...


According to who?


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## IanC (Dec 6, 2011)

OohPooPahDoo said:


> IanC said:
> 
> 
> > 1,2,4,8,16,32,64,128,256,512. eight doublings, working on the ninth
> ...



which part?   the 30 odd degree increase for the GHE? the roughly 25 percent allocation to CO2? the number of doublings? the 1C increase for the next 400ppm?

I must admit I have taken the numbers from authorities but they are similar from both sides of the debate. is there something in particular that you dont agree with?


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## OohPooPahDoo (Dec 6, 2011)

IanC said:


> OohPooPahDoo said:
> 
> 
> > IanC said:
> ...




The part where you assume a function that relates warming due to CO2 to the amount of Co2. You seem to be suggesting warming due to Co2 is always exactly proportional to the base 2 logarithm of the quantity of Co2 in  the atmosphere - yet provide no source for that information.


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## IanC (Dec 6, 2011)

OohPooPahDoo said:


> IanC said:
> 
> 
> > OohPooPahDoo said:
> ...



are you suggesting that it is not logarithmic? I have put up the graph a few times in the past. if you want to disagree find one for CO2 that shows a different relationship.


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## OohPooPahDoo (Dec 6, 2011)

IanC said:


> OohPooPahDoo said:
> 
> 
> > IanC said:
> ...



No, I'm asking how you know its exactly proportional to a base 2 logarithm always? I'm not disagreeing. I'm asking you to justify your statements.


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## IanC (Dec 20, 2011)

> JoNova- Greenhouse gases don&#8217;t  warm the planet but they do create a situation where the Sun can warm the world to a temperature above where it previously was without the gases


So what is the Second Darn Law? « JoNova: Science, carbon, climate and tax


an interesting blog entry with participation by some of the 'Slaying the Skydragon' principals. it really is a matter of how you frame the question and how you define the system.



> In the end this boils down to the energy flows in a whole system, and people who think that the addition of greenhouse gases to our atmosphere won&#8217;t warm the planet (due to a conflict with the Second Law) are forgetting that the system includes the sun, and the extra energy flowing in continuously is what drives the system. CO2 may have little effect on Planet Earth, but that&#8217;s for other reasons. You can&#8217;t just make a comment about &#8220;the Earth-Atmosphere&#8221; part of the system and ignore the {sun}



and for extra things to think about, Monckton has another interesting piece-



> The posting begins by making the common error of assuming that a blackbody cannot have an albedo. Of course it can. The Stefan-Boltzmann equation accounts for albedo in the simplest possible way: by simply taking it that the fraction of incident radiation that is reflected away by the albedo of the Earth plays no part in the radiative transfer at the characteristic-emission surface. Here is how it&#8217;s done.
> 
> The characteristic-emission surface of the Earth is not the surface we stand on. It is about 5 km up in the troposphere, varying quite a bit with latitude. At that surface, by definition, incoming and outgoing radiative fluxes balance, and there are no non-radiative fluxes as there are at the Earth&#8217;s surface.


Monckton on blackbody radiation « JoNova: Science, carbon, climate and tax


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