Guest post by Willis Eschenbach
There is a lot of misinformation floating around the web about how the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, a “mirror” that reflects part of the heat back to Earth, or “a pane of glass” that somehow keeps energy from escaping. It is none of these things.
A planetary “greenhouse” is a curiosity, a trick of nature. It works solely because although a sphere only has one side, a shell has two sides. The trick has nothing to do with greenhouse gases. It does not require an atmosphere. In fact, a planetary greenhouse can be built entirely of steel. A thought experiment shows how a steel greenhouse would work.
Before we start, however, a digression regarding temperature. The radiation emitted by a blackbody varies with the fourth power of the temperature. As a result, for a blackbody, we can measure the temperature in units of radiation, which are watts per square meter (W/m2). For objects with temperatures found on the Earth, this radiation is in the range called “longwave” or “infrared” radiation. See the Appendix for the formula that relates temperature to radiation.
This means that we can denote the temperature of a blackbody using W/m2 as well as the traditional measures (Fahrenheit, Celsius, Kelvin). The advantage is that while temperature (degrees) is not conserved, energy (W/m2) is conserved. So we can check to see if energy lost is equal to energy gained, since energy is neither being created nor destroyed by the climate.
For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square meter of surface area. How is it heated from the interior? Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.
The planet is in interstellar space, with no atmosphere and no nearby stars. The equilibrium surface temperature of this planet is, of course, 235 W/m2. To maintain the equilibrium temperature, it constantly radiates this amount of energy out to space. Coincidentally, this is the amount of solar radiation that makes it past the clouds to warm the Earth. If we convert 235 W/m2 to one of our normal temperature scales, it is -19 Celsius (C), or -3° Fahrenheit (F), or 254 Kelvins (K). It’s a possible temperature that the Earth might have if there were no greenhouse effect. That is to say … cold.
Now imagine that the planet gets completely surrounded by a thin black steel shell, located a few thousand meters above the surface, as shown in a cutaway view in the picture above, and in Figure 1 below. What happens to the surface temperature of the planet? (To simplify calculations, we can assume the shell has the same outer surface area as the surface. For an earth-sized planet with a shell two kilometers above the surface everywhere, the difference in area is only six-hundredths of one percent. This assumption makes no difference to the argument presented.)
In order to maintain its thermal equilibrium, including the new shell, the whole system must still radiate 235 W/m2 out to space. To do this, the steel shell must warm until it is radiating at 235 watts per square meter. Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet. The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell. This will warm the planetary surface until it reaches a temperature where it radiates at 470 watts per square meter. In vacuum conditions as described, this would be a perfect greenhouse, with no losses of any kind. Figure 1 shows how it works.
Figure 1. Building a steel greenhouse. (A) Planet without greenhouse. Surface temperature is 235 W/m2 heated from the interior. (B) Planet surrounded by steel greenhouse shell. Shell radiates the same amount to space as the planet without the shell, 235 W/m2. It radiates the same inward, which warms the planet to 470 W/m2 (29 C, 83°F, 302 K). [Clarification added] Note that the distance from the shell to the planet is greatly exaggerated. In reality, it is close enough to the planet that the difference in the areas can be neglected in practice.
The trick can be repeated again, by surrounding the planet and the shell with a second outer shell. In this two shell case, the planetary surface temperature (in W/m2) will be three times the initial surface temperature.
In nature, planets have atmospheric shells, of course, rather than steel shells. The trick works the same way, however. Rather than being warmed from the inside, the Earth receives 235 W/m2 from the sun. Solar radiation passes through the atmosphere. Outgoing longwave radiation is absorbed by the atmosphere, just as it is absorbed by the steel shell shown in Fig. 1.
So that’s the trick of the greenhouse. It has nothing to do with blankets, mirrors, or greenhouse gases. It works even when it is built out of steel. It depends on the fact that a shell radiates in two directions, inwards and outwards. This radiation keeps the planet warmer than it would be without the shell.
Now, it is tempting to think that we could model the Earth in this manner, as a sphere surrounded by a single shell. This is called a “two-layer” model, with the two layers being the surface and the atmospheric shell. In fact, a number of simplified climate models have been built in this way. Unnoticed by their programmers, however, is that is not physically possible to model the Earth as a two-layer system. Figure 2 shows why. Note that in all cases, the system has to remain in thermal equilibrium. This means that the surface must radiate as much as it absorbs, and the shell must also radiate as much as it absorbs. In addition, radiation from the shell to space (upwelling longwave radiation or ULR) must equal radiation from the shell to the Earth (downwelling longwave radiation, or DLR)
Figure 2. Single-shell (“two-layer”) greenhouse system, including various losses. S is the sun, E is the Earth, and G is the atmospheric greenhouse shell around the Earth. The height of the shell is greatly exaggerated; in reality the shell is so close to the Earth that they have about the same area, and thus the small difference in area can be neglected. Fig. 2(a) shows a perfect greenhouse. W is the total watts/m2 available to the greenhouse system after albedo. Fig. 2(b) is the same as Fig. 2(a) plus radiation losses Lr which pass through the atmosphere. Fig. 2(c) is the same as Fig. 2(b), plus the effect of absorption losses La. Fig. 2(d) is the same as Fig. 2(c), plus the effect of thermal losses Lt.
Figure 2(a) shows the same situation as Figure 1(B), which is a perfect planetary greenhouse. In this case, however, it is heated by an amount of energy “W”, which is coming from the sun. The planet receives solar radiation in the amount of “W” from the sun, and longwave radiation “W” from the atmospheric shell. The surface temperature is thus 2W. All energy flows are in Watts/square metre (W/m2).
Figure 2(b) adds two losses. The first is the reflection from the Earth’s albedo (Wo – W). This is energy which never enters the system and is reflected back into space. We are still getting the same energy entering the system (W). The second loss Lr is from radiation which goes from the surface to space through the “atmospheric window”. Because of the second loss, the surface of the Earth does not get as warm as in a perfect system. In a perfect system, the temperature of the surface is 2W. But including the radiation loss Lr, the surface temperature drops to 2W – 2Lr.
Figure 2(c) adds another loss La, which is the solar radiation which is absorbed by the shell. This cuts the radiation hitting the surface down to W – La. Including this loss, the surface temperature is 2W – 2Lr – La.
Figure 2(d) adds the final loss. This is Lt, the thermal loss. It is the sensible energy (energy you can feel) and latent energy (evaporation) which is transported from the surface to the shell by convection. Including all of these losses, the surface temperature of the Earth is 2W – 2Lr – La – Lt.
Now, why can’t the Earth be modeled in this manner? A look at the size of the various losses shows why. Here is the canonical global energy budget, called the “Kiehl/Trenberth” budget, or the K/T budget.
Figure 3. The Kiehl/Trenberth Global Energy Budget. This is a “two layer” representation, with the surface and the atmosphere being the two layers. Lr, the radiation loss, is the 40 W/m2 labeled “Atmospheric Window”. La, the absorption loss, is the 67 W/m2 labelled “Absorbed by Atmosphere”. Lt, the thermal loss, is 102 W/m2. This is the sum of the 24 W/m2 labelled “Thermals” and the 78 W/m2 labelled “Evapo-transpiration”. W, the energy from the sun, is the incoming solar radiation of 342 W/m2 less the 107 W/m2 that is reflected by the surface and the clouds. This means that W is 235 W/m2. SOURCE
What’s wrong with this picture? Note that the temperature of the Earth is 390 W/m2, labeled as “Surface Radiation”. This is 15 C, or 59°F. But from Fig. 2(d), we know that the surface temperature of a greenhouse system with a single shell, as shown in the drawing, is 2W (470) – 2Lr (80) – La (67) – Lt (102) = 221 W/m2. This is far below the known surface temperature of the Earth. In other words, a single shell greenhouse system simply isn’t efficient enough to give a surface temperature which is warm enough to allow for the known losses.
So where is the problem with the K/T budget diagram? The hidden fault is that the upward radiation from the atmospheric layer does not equal the downward radiation. There is 195 W/m2 going to space from the atmospheric shell, and 324 W/m2 going down to the surface.
In order to get enough energy to allow for the known losses, the simplest model requires two atmospheric shells. A perfect greenhouse with two shells would give a surface temperature of 3W, or 705 W/m2. This is enough to allow for the known losses, and still give a surface temperature which matches that of the Earth. Figure 4 shows one such possible system.
Figure 4. Simplest greenhouse global energy budget capable of representing the Earth. Note that all major flows in the K/T energy budget have been maintained. There are two shells, which must be physically separate. These are the lower troposphere and the lowest stratosphere. They are separated by the tropopause.
This budget fulfills all of the requirements for thermal equilibrium. The same amount is radiated upwards and downwards by each layer. The amount absorbed by each part of the system equals the amount radiated. Further verification of this energy budget is the 147 W/m2 emission from just above the tropopause. This converts to a Celsius temperature of about -50 C, which is a typical temperature for the lowest part of the stratosphere.
I have written a simplified Excel radiation/evaporation/convection model which encapsulates the above system. It is available here. Click on the “File” menu on the webpage and select “Download”.
I invite people to play with the model. It is what I term a “Tinkertoy” model, which is what I call the simplest possible model that can represent a particular reality. One of the interesting results from the model is that there can be very little thermal loss between the inner and the outer atmospheric layers. If there is more than a trivial amount of leakage, the surface cools significantly. Among the other insights yielded by the model is that a change equivalent to a doubling of CO2 (an increase of 3.7 W/m2 downwelling radiation at the top of the atmosphere) can be canceled by a 1% increase in the upper and lower cloud reflections.
Experiment with the values, it is an interesting insight into the energy flows in the simplest possible climate model that can represent the Earth’s greenhouse system.
APPENDIX
The formula that relates the temperature to radiation is called the “Stefan-Bolzmann” equation:
R = sigma * epsilon * T^4
where r = radiation (W/m2)
sigma = the Stefan-Boltzmann constant, 5.67 * 10^-8
epsilon = the emissivity of the body, which for a blackbody = 1
T^4 = absolute temperature in Kelvins (which is Celsius plus 273.15) raised to the fourth power
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THANK You John A. Nice explanation! Couldn’t have said it better myself.
Gary says:
Actually, with no albedo (or absorption by the atmosphere), the amount of solar radiation absorbed by the earth would be ~342 W/m^2. And, in reality, the earth would still have an albedo without an atmosphere of ~0.09, so in fact without an atmosphere, the amount absorbed would be ~310 W/m^2. This would yield a surface temperature of ~272 K. When you compare this with the actual surface temperature of ~288 K and the temperature in absence of the greenhouse effect but no change in albedo of ~255 K, what we can say is the follows: The greenhouse effect due to all the greenhouse gases (water vapor, clouds, and the long-lived GHGs like CO2 and CH4) raises the temperature of the Earth by an amount of ~33 K (which is 288K – 255K); the albedo due to cloud reduces the temperature by ~17 K (which is 272 K – 255 K); the net effect of both the GHGs and the cloud albedo is ~16 K (which is 288K – 272K).
Joel Shore (14:17:35) :
Joel, always a pleasure to hear from you, thanks for the physicist’s vote of confidence.
You may be correct, but it is drawn as a shell model, where the atmosphere is considered as a single entity. My point remains — to get enough energy from a greenhouse system to model the earth, you need two physically separated shells. If there is thermal loss and air mixing between the shells, you won’t get the energy you need to explain the Earth’s temperature. The system, as my Tinkertoy model shows, is very sensitive to thermal loss between the shells. That’s the part that the K/T analysis glosses over.
You are generally correct. If the albedo goes from 30% to 31%, it has increased by 3.3%. And clouds are only part of the equation as you say. However, as my analysis shows, in the Pacific Ocean the change in clouds is responsible for a 60 W/m change between 10:30 and 11:30 each day …
And you are correct that when clouds come over, they warm as well as cool. However, the true change is masked by the use of averages. For example, in the tropics the mornings are generally clear, warming the earth when it is cool. After about 11:00, clouds form and cool the earth when it is warm. Since it is the tropics, the solar radiation is on the order of a kilowatt per square metre, and the afternoon change in net radiation is quite large. In addition, since over the tropics the humidity is quite high, adding clouds doesn’t change the downwelling radiation as much as it would in the desert. Finally, during the night the clouds disperse, allowing increased outgoing radiation.
The net result of the timing of these processes is critical. If we swapped day clouds and night clouds, the average cloudiness would be the same … but the result in terms of the change in net radiation would be huge.
As a result, the use of averages in these matters can be very misleading. That’s why it’s a Tinkertoy model.
Your comments are always welcome. They are clear, to the point, and backed by the science.
w.
The validity of this model is demonstrated in a practical
application. The inside of a Dewar flask (thermos bottle) used to store liquid gases contains not just a vacuum between the inner and outer surfaces, but many layers of plastic film that is metalized on both sides. This reflects the radiation as in the model, and greatly (order of magnitude) reduces the net energy flow to the stored gas.
Willis:
Willis,
You site a solar heater as an example, but how does this prove your point? They say the solar heater can get up to 100C at mid-day, no clouds.
Ok, well the the sun in mid-day, no clouds can deliver 1366 w/m^2. Put that into Stephan-Boltzmann equation and you get a temperature of around 120C for an ideal blackbody. 120 is greater than 100. I’m missing how your example proves me wrong.
Anton Eagle (14:12:49) :
Um, based on what I’ve read from you, you’re the one in need of reason lessons.
For the record, the shell has twice the area because there is both and inside and an outside, both of which have approximately the same surface area as the planet itself. Well, there is about a 0.2% error, a point Willis has made clear on more than one occasion, yet you repeatedly don’t get the point.
Actually, you can do this at home with a dichroic mirror. Phil. often talks about doing just such an experiment. Once you wrap your head around the transient state in which , it is fairly easy to understand the equilibrium result.
Mark
Can someone fix the tags in my previous comment. I think I didn’t close the first set of blockquotes correctly.
Thanks
Mark T.
No… again… the shell does not necessarily have twice the area. It COULD have twice the area if its very close to the surface (give or take a percent or two like you said)… but the model he proposes does not in any way require the shell to be close to the surface… so it could just as easily be 10,000 miles above the surface without changing his model in anyway whatsoever. If the shell can be some arbitrary distance above the surface, then it can have any arbitrary area that is greater than twice the area of the planet. If you disagree, then show me where his model requires the shell to be at any particular distance from the planet.
There are many contradictions here.
Here’s another. Let’s accept for the moment his model. Then, lets add another shell just one inch further out than the first shell. According to the author, and his model, this second shell would cause a further increase in temp. Okay, so then add 8 more, each 1 inch further out from the previous. Now, we have a huge increase in temp (according to this flawed model).
However, how are 10 shells all 1 inch appart any different than 1 shell that is just 10 times as thick? And, since the thickness of the shell does not affect his (flawed) radiative balance, this results in the contractory situation where 10 shells 1 inch apart cause a much larger temp increase than a single shell exactly as thick as the sum of the ten shells. Huh? It makes no sense! He has created some kind of weird thermodynamic perpetual motion machine involving flawed radiative balances.
Stop focusing on the mathematics of the W/m2 accounting, and focus on the basic principles. If you do so, its easy to see that its nonsense.
-a
Anthony, anybody,
Mercy KIll this silly thing. Its embarassing.
The thought experiment raises a question: How is a temperature gradient created in a vaccuum – 470 Wm-2 at the planetary surface to 235 Wm-2 at the inside surface of the shell?
Willis Eschenbach (14:08:30) :
Its agreed that at 15C and 27C, a human is warmer than his surroundings and is there for unlikly to absorb heat from surrounding objects.
try on a google search (Cambridge journals) enter
“Description of a human direct calorimeter” and you should be able to read the measured radiation in w/m2 a huma produces.
for average human values in various situations on human heat exchange with the environment:
http://personal.cityu.edu.hk/~bsapplec/heat.htm
for various measurements of energy required to heat room:
http://www.energyinst.org.uk/content/files/4g.pdf
cites 70-100w/m2
i can’t readily find the value for the energy required to boil sunflower oil, as i’m not sure its on the internet, but it omes under the category of experiments in food engineering, so tomorrow i’ll find the source
John A (14:59:54) :
Willis:
You are making a simple mistake. You are confusing net warming with the energy interchange between two objects.
You are correct that net heat flows from warm to cold. But that says nothing about the individual flows.
This is the mistake. You agree, I hope, that both planets radiate energy. The warmer planet radiates more than the cooler planet. As a result, an energy exchange occurs in both directions. And you are right that the net flow goes from the warmer to the cooler planet.
But that means nothing about the individual flows. If you agree that both planets radiate, what do you think is happening to the radiation from the cooler planet that hits the warmer planet? It has to warm the warmer planet … but not as much as the warmer planet is cooled by its own radiation.
I’m not. There is nothing unphysical about radiation. All objects radiate. Their radiation adds energy to whatever object it hits. Do you seriously believe that when radiation hits an object, it first checks the object’s temperature to see whether to add energy to that object? It adds the energy regardless of the temperature of the object it strikes. This makes the object warmer than it would be without that radiation.
Lets try to keep a clear accounting of what parameters affect this model, and what parameters do not affect this model.
First. This model is not dependent in any way on the distance of the shell from the planet. No matter what the distance, according to the model, the shell has to radiate 235 out to space, and then of course also must radiate 235 back to the planet. According to the model, this must be so if the shell is only 1 foot off the ground or 10,000 mile up. Again… according to this model.
The above fact inevitibly leads to the fact that the area of the shell is not a factor (that alone should raise some eyebrows). And thus, all attempts at carefully accounting for the radiative balance by stating that the shell is twice the area of the planet are nonsensical, because it doesn’t have to be twice the area… it could easily be much more without changing the basic premise of the model.
Second. This model does not in anyway depend on the thickness of the shell. Again, no matter how thick the shell… according to the model… the shell must radiate 235 out to space, and 235 back to the planet.
As I posted above, this leads to the non-sensical result that one thick shell results in a totally different radiative balance than 10 thin shells.
All you folks arguing that the model is valid are doing so by performing accounting on the radiative balance (235 out = 470 out – 235 in, etc.). But, by doing so you are accepting the basic premise of the model, and you can’t do that, because the basic premise is flawed because it inevitibly leads to contradictions.
Look, I’m not saying that there isn’t some way to make a model of the atmosphere that can be represented by a shell. I’m just saying that this model isn’t it. It violates basic priniciples of thermodynamics.
If you can’t resolve the above stated two contradictions, then you can’t defend the model. Period.
Ian Schumacher (15:03:31) :
Your numbers are wrong. You have given 1366 W/m2, which is the strength of the sun at the top of the atmosphere. The max sun on the earth’s surface (clear tropical noon) is about a kilowatt per square metre, because even on a clear day there is absorption in the atmosphere. See here for actual measurements. The maximum they found was 1044 W/mw, and rarely exceeded a kilowatt. So we can take a kilowatt as a practical maximum.
By Stefan-Bolzmann this is only 90C for a kilowatt, and 95C for the absolute maximum measured.
So if you think that you can boil water (100C) simply by sitting it in the noonday sun, you need more assistance than I can give you. Do you really think that people put the vacuum shell around the heater pipes for fun?
If the two planets were at the same temperature they would be in equilibrium and wouldn’t radiate to each other. If terrestrial amtter is in equilibrium with the atmosphere then earth matter isn’t going to radiate much. Oceans might throw heat into the air or the sun heats oceans although at the optimum, a safe estimate would be to say that earth re-radiates around 5% of its heat, and not the 117% that is suggested that it has to get rid of in order to be in equilibrium
Physics 101 for black body radiation (epsilon =1):
Object A and B are in a vacuum and transfer heat to each other by black body radiation. Object A is at temperature T(a) and radiates a flux of R(a). Object B is at temperature T(b) and radiates a flux of R(b).
The gross radiative fluxes are:
R(a) = sigma*T(a)^4
R(b) = sigma*T(a)^4
The net radiative fluxes are:
For body A
=R(a) – R(b)
= sigma*T(a)^4 – sigma*T(b)^4
= sigma*(T(a)^4 – T(b)^4)
For body B
=R(b) – R(a)
= sigma*T(b)^4 – sigma*T(a)^4
= sigma*(T(b)^4 – T(a)^4)
All fair and above board. I can’t make it any simpler. So can we please stop arguing that this model contradicts basic physics because it doesn’t.
P Wilson (15:59:21) :
Let me stop you there, because everything that follows flows from this misconception.
I say again, people are confusing net heat flow with individual heat flows. Conduction is different from radiation. When a warm object touches a cool object, there is only one measurable flow, from the warm object to the cool object.
When the energy is transferred by radiation, however, we can measure both individual flows as well as the net flow. The warm object cools, and the cool object warms (net flow). But there is a two way flow of radiation. This is an inescapable fact, because both objects are radiating. Radiation from the cool object strikes the warm object, and vice versa. Both radiation flows warm the object that they strike. The net flow goes from warm to cool, but there is an energy flow in both directions.
John Millett (15:57:15) :
There is no gradient. Temperature is a measure of the average of the speed of the molecules of what is being measured. But in a vacuum, there are no molecules …
Willis, you seem to have a knack of posting ideas which engender interesting and informative discussion. I’m looking forward to reading this one thoroughly.
Thank you for your great insights.
ok found something similar
Vijayan and Singh reported a convective heat transfer of 300-500 w/m2 to fry food in oil from frozen cited in “Advances in deep fat frying of food” 1997 CRC Press
I am just boggled by the number of commenters here who cannot follow Willis’ very simple “steel greenhouse” thought experiment. It really is nothing more than a typical simple problem given to students in an introductory thermodynamics course. (If I teach such a course again, I will use it — OK, Willis?) It looks like a lot of commenters would flunk such a course.
I guess those of us who have studied thermodynamics formally are used to the practice of defining different “control volumes” in a system, then applying the laws of thermodynamics to each. (Any such control volume you define must obey the laws — the skill comes in defining ones that are both useful and reasonably easy to analyze.)
People who work for me recently inadvertently created such a steel greenhouse, quite literally. Part of what I do is to design power electronic systems. One of the challenges always is to get the internally generated heat out of the power transistors so they don’t fry themselves. To that end they are typically mounted onto a metal heat sink that conducts the heat out of the transistor, then gets rid of the heat through some combination of radiation and convection.
Anyway, in this design, we did not have room or budget for fins on the heat sink or a fan, so the heat elimination would be more radiative than convective — just a flat plate. For this reason, I specified that the sheet metal cover for the product should not go over the side with the heat sink.
However, when the first prototypes came back, I found that the sheet metal cover did include the side with the heat sink. It was only about a millimeter thick and a millimeter away from the heat sink. The initial testing showed that the transistors ran hotter than we wanted. So I had my engineers remove this “steel greenhouse” from the heat sink side. With the identical heat generation level in the power transistors, they ran 20C cooler. The younger engineers were amazed. Of course, we ship the product with this side “open”.
(Yes, there is some convection involved here. But the radiative barrier effect is very important.)
The thought experiments Willis provides can easily be verified in a lab. (How much relevance they have to real AGW issues is a separate question.)
In physics one can only discuss measurable things. Radiation and its power flux can arbitrarily be decomposed into oppositely directed components, but this is not measurable. Any detector just above the earth will indicate that there is 235 W/m2 outward power flux in both situations of the first diagram. And so, provided the Earth and the steel behave as black bodies, the earth surface temperature will be exactly the same whether the steel is around or not.
Then this could be useful to explain that the greenhouse effect is not a simple blanket or mirror. One can only understand it by considering long- and shortwave radiation (which is perfectly measurable).
Anton Eagle (15:36:25) : “It COULD have twice the area if its very close to the surface (give or take a percent or two like you said)… but the model he proposes does not in any way require the shell to be close to the surface…… If you disagree, then show me where his model requires the shell to be at any particular distance from the planet.
Actually, he said it pretty clearly when he first describes the steel shell:
“Now imagine that the planet gets completely surrounded by a thin black steel shell, located a few thousand metres above the surface, as shown in a cutaway view in the picture above, and in Figure 1 below.”
——————————-
Note the “A FEW THOUSAND METRES ABOVE THE SURFACE”.
For a basic understanding of the greenhouse theory one can initially assume that the area of the earth’s surface and the area of ONE side of a steel shell just above the surface are equal.
The important point is that the total area of the shell is very close to TWICE that of the earth’s surface, with the inside surface of the shell radiating back towards the earth’s surface.
Anton,
Let’s do, and let’s also try to have a clear account of what the problem statement is.
Oops, you started out by failing the first question. If the outer shell is significantly larger than the planet, then the corresponding values at the planet surface will be significantly larger which will change all the equations. The only thing that will remain the same is the input/output relationship. Willis pointed out, several times now, that the reason we don’t have to adjust the values at the surface of the planet is precisely because the shell surface area is nearly (0.1%) the same as that of the planet and hence, the total surface is approximately 2:1 over the planet surface.
And thus, since you got the first point wrong, this one is wrong as well.
Correct, except that the values at the surface of the planet are different for a significantly larger shell, as well as the ratio of 2:1.
Given that the first assumption you made was incorrect, it follows that you cannot logically arrive at this conclusion, either.
We understood the basic premise, you did not.
No, actually, it doesn’t, at least not at the single shell level (the easiest to observe without some effort). John A is wrong, too, because he, like you, is missing one simple point (though different points).
Given that there is no contradiction, then you can’t defend your conclusions regarding the model. Period.
If you would like, I can post some pretty pictures, maybe even an equation or two, that explains how the feedback in such a system works. Heck, I can even give you some Excel commands.
Mark