Trenberth's Energy Budget

[gslack]

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...

 

[IanC]

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

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

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.

 

[gslack]

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..

 

[IanC]

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.

 

[gslack]

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..

 

[wirebender]

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.

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’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. ".

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’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.

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’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’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’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’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.

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.

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’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.

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.

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 “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

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.

 

[wirebender]

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?

 

[wirebender]

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.

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.

 

[IanC]

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

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

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

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.

 

[wirebender]

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?

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.

 

[IanC]

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?

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.

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).

 

[IanC]

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.

 

[gslack]

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

 

[wirebender]

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.

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?

 

[wirebender]

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

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.

 

[wirebender]

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.

 

[IanC]

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.

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?

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?

 

[IanC]

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.

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

 

[gslack]

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..