(1)I dont know if the effect of CO2 is vanishingly small but I certainly agree with you that it is impossible to separate out from the noise of natural variability.
(2)back to the shell problem...equilibrium will converge to 235W/m2 outside radiation just like before the shell was in place. until equilibrium is reached the deficit of outside radiation will charge the heat sink in the planet.
let me ask you a question. would the inside of the planet be warmer if the planet was made out of asbestos or out of steel? Joe wants to argue the question as a strawman mathematical problem but it is really just an insulated heat sink. the surface has been moved out to the shell, which must radiate 235 out but because it is separated by a vacuum that defeats conduction it is also radiating 235 in. therefore it needs a 470 source.
Joe, and others, have mangled the problem by not including the energy needed to charge the heat sink. in the beginning almost all of the source will be used to warm the planet, only as the planet starts to get much warmer does the shell receive enough radiation to be able to shed anywheres close to 235W/m2. if the energy source from the core stopped it would take a long time for the radiation to space to decrease as the heat sink would be giving up its energy.
1.) In any scientific trace analysis the detection limit standard is set at 2 X the background "noise"
2.) Don`t forget that a warmer body also radiates heat at a higher rate.
Heat Transfer, Heat Transfer Coefficient | [email protected]
There is no heat to be gained by "backradiation" from a colder body...period !!!Stefan Boltzmann Law
The emissive power of a (black) body is proportional to the forth power of the absolute temperature
Even building contractors are aware of that:
5-Step Heat Loss Calculation
In this example the "greenhouse" was at 72 F and "outer space" was only 5 F cooler. Even though you can see how much higher the heat loss is if you increase the "greenhouse" temperature by only 1 F while leaving the "outer space" the same...but now it`s 6 F lower than the "greenhouse"Assuming that T1 is 72F and T2 is 5F, Delta T = 72F - (-5F) = 72F + 5F = 77F
Follow the steps 1 through 4 to calculate heat loss separately for windows, doors, and ceiling.
Door Heat Loss = 0.49 x 24sq ft x 77F = 906 BTUH
(U-value is based on assuming a solid wood door)
Window Heat Loss = 0.65 x 14sq ft x 77F = 701 BTUH
(U-value is based on assuming a double-panel window)
Ceiling Heat Loss = 0.05 x 352sq ft x 77F = 1355 BTUH
(U-value is based on assuming a 6 fiberglass insulation. Ceiling surface is 22ft x 16ft)
Note
The above online calculator is for house wives and works in the normal room temperature range when it is expressed in deg Fahrenheit...by numbers coincidence .
That`s why the web site says
And they do it using (T1)^4 - (T2)^4 in degrees Kelvin where T1 is ALWAYS the HIGHER temperatureFor actual calculations, contact your contractor or system designer.
and add this:
Or measure it using thermal imaging...Heat loss through roofs should be added 15% extra because of radiation to space. (2) can be modified to:
H = 1.15 A U (ti - to)
There is no heat to be gained by "backradiation" from a colder body...period !!!
But according to Roy Spencer`s "thought experiment" where he has a colder body warming a hotter one , the hotter one looses even less heat after it got warmer.
There is no heat to be gained by "backradiation" from a colder body...period !!!
The only way that could happen if Roy manages to heat outer space to get a lower (T1)^4 - (T2)^4 temperature differential
It works both ways...not just from air+CO2 back down, but up an out also.
Up and out outpaces the air+CO2 even without using convection
because there is a temperature gradient of 2 F (drop) per 1000 feet...called the "standard lapse rate".
I`ve been around this bend with the AGW fanatics in this forum before,...
It never occurred to Boltzmann that some day there would be people who are too dumb to realize that in his equation T1^4 - T2^4 T1 was always the higher temperature.
Now we got "climatologists" who put the lower Temperature where T1 is and generate "back radiation energy".
I explained all that over a year ago and even plotted the function T1^4 - T2^2 showing the direction of heat transfer if T1 < T2 = blue line
and if T1 > T2 = green line
http://www.onlinefunctiongrapher.com/?f=-1*x^4|x^4&xMin=-10.03603&xMax=10.03603&yMin=-15723.02&yMax=16343.02
There is no heat to be gained by "backradiation" from a colder body...period !!!
The red roof thermal image from the outside follows the green function and looks red
The same roof would look blue with thermal imaging because on the inside it follows the blue function
These pictures are hard to find because nobody wants to crawl in the attic lay on his back in a bunch of fiberglass and point a camera at the inside of a roof.
Before Christmas I uploaded a video showing what a thermistor, well insulated from the ambient inside a 6 inch reflector telescope registers at the focal point when you point it at something cold...because I did not have a thermal imaging cam at hand.
There is no heat to be gained by "backradiation" from a colder body...period !!!
Poophead, RollUnder and Saigon replied "Lolololoh"
but in this video you can see what a surface looks like with thermal imaging when it`s loosing heat....blue
The same area would look red outside
Obeying the direction of heat transfer from hot to cold just like in the graph above when T1 is either higher or lower than T2
There is no heat to be gained by "backradiation" from a colder body...period !!!
Check out the video at 4:19
To see the effect of moisture evaporative cooling
The thermal imaging shows a temperature drop between 2 and 5 degrees Celsius..!!!.Something climatologists prefer not to discuss
SSDD and I mentioned this many times...
Poophead the fake physicist who kept spelling it "physisist" finally shut up, but Saigon and UndertheRoll replied: "Lolololoh"
If a building contractor/engineer would do math like climatologists do it he would be out of business or loose his license in no time.
This is why we got over 31 000 and counting engineers and scientists signing this petition:
Global Warming Petition Project
31,487 American scientists have signed this petition,
including 9,029 with PhDs
There is no heat to be gained by "backradiation" from a colder body...period !!!
If you got a degree in science I urge you to sign as well
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