Dumbest Math Theory Ever

Mainstream climate scientists believe in the dumbest math theory ever devised to try and explain physical reality. It is called the Greenhouse Effect. It’s so silly and unbelievable that I don’t even want to give it the honor of calling it a scientific theory, because it is nothing but ideological mathematics that has never been empirically validated. In fact it is nothing but a post hoc fallacy: the surface is hotter than what the sun alone can do, therefore greenhouse gases did it!

Today we will play with this silly math theory called the greenhouse effect. Here are two examples of its typical canonical depiction:

Let’s get started. Please create a new file called gheffect, and paste the following into it:

``````# bash gheffect
# Zoe Phin, 2020/03/03

[ -z \$TSI ] && TSI=1361
[ -z \$ALB ] && ALB=0.306

echo \$1 | awk -vALB=\$ALB -vTSI=\$TSI 'BEGIN {
SIG = 5.67E-8 ; CURR = LAST = SUN = TSI*(1-ALB)/4
printf "Sec | Upwelling |   Temp    | GH Effect |  Trapped  | To Space\n"
} {
for (i=1 ;; i++) {
printf "%3d | %7.3f W | %7.3f C ", i, CURR, (CURR/SIG)^0.25-273.16

CURR = SUN + \$1*LAST/2 ; GHE = SUN - (LAST*(1-\$1))

printf "| %7.3f W | %7.3f W | %07.3f W\n", GHE, CURR-LAST, CURR-GHE

if ( sprintf("%.3f", CURR) == sprintf("%.3f", LAST) ) break

#if ( CURR==LAST ) break

LAST = CURR
}
}'``````

Now run it with atmospheric emissivity = 0.792:

``````\$ bash gheffect 0.792

Sec | Upwelling |   Temp    | GH Effect |  Trapped  | To Space
1 | 236.133 W | -19.125 C | 187.018 W |  93.509 W | 142.625 W
2 | 329.642 W |   2.971 C | 167.568 W |  37.030 W | 199.104 W
3 | 366.672 W |  10.419 C | 159.866 W |  14.664 W | 221.470 W
4 | 381.336 W |  13.212 C | 156.816 W |   5.807 W | 230.327 W
5 | 387.142 W |  14.296 C | 155.608 W |   2.300 W | 233.834 W
6 | 389.442 W |  14.722 C | 155.130 W |   0.911 W | 235.223 W
7 | 390.352 W |  14.890 C | 154.940 W |   0.361 W | 235.773 W
8 | 390.713 W |  14.957 C | 154.865 W |   0.143 W | 235.991 W
9 | 390.856 W |  14.983 C | 154.835 W |   0.057 W | 236.077 W
10 | 390.912 W |  14.994 C | 154.824 W |   0.022 W | 236.111 W
11 | 390.935 W |  14.998 C | 154.819 W |   0.009 W | 236.125 W
12 | 390.944 W |  14.999 C | 154.817 W |   0.004 W | 236.130 W
13 | 390.947 W |  15.000 C | 154.816 W |   0.001 W | 236.132 W
14 | 390.949 W |  15.000 C | 154.816 W |   0.001 W | 236.133 W``````

W is shorthand for W/m². Parameters are taken from NASA Earth Fact Sheet.

As you can see, by delaying outgoing radiation for 14 [¹] seconds [²], we have boosted surface up-welling radiation by an additional ~66% (154.8/236.1 W/m²). Amazing, right? That’s what my program shows, and that’s what is claimed:

This is zero in the absence of any long‐wave absorbers, and around 155 W/m² in the present‐day atmosphere [Kiehl and Trenberth, 1997]. This reduction in outgoing LW flux drives the 33°C greenhouse effect …

Attribution of the present‐day total greenhouse effect

The main prediction of the theory is that as the atmosphere absorbs more infrared radiation, the surface will get warmer. Let’s rerun the program with a higher atmospheric emissivity = 0.8

``````\$ bash gheffect 0.8

Sec | Upwelling |   Temp    | GH Effect |  Trapped  | To Space
1 | 236.133 W | -19.125 C | 188.907 W |  94.453 W | 141.680 W
2 | 330.587 W |   3.168 C | 170.016 W |  37.781 W | 198.352 W
3 | 368.368 W |  10.746 C | 162.460 W |  15.113 W | 221.021 W
4 | 383.481 W |  13.614 C | 159.437 W |   6.045 W | 230.088 W
5 | 389.526 W |  14.738 C | 158.228 W |   2.418 W | 233.715 W
6 | 391.944 W |  15.184 C | 157.745 W |   0.967 W | 235.166 W
7 | 392.911 W |  15.361 C | 157.551 W |   0.387 W | 235.747 W
8 | 393.298 W |  15.432 C | 157.474 W |   0.155 W | 235.979 W
9 | 393.453 W |  15.461 C | 157.443 W |   0.062 W | 236.072 W
10 | 393.515 W |  15.472 C | 157.431 W |   0.025 W | 236.109 W
11 | 393.539 W |  15.477 C | 157.426 W |   0.010 W | 236.124 W
12 | 393.549 W |  15.478 C | 157.424 W |   0.004 W | 236.130 W
13 | 393.553 W |  15.479 C | 157.423 W |   0.002 W | 236.132 W
14 | 393.555 W |  15.479 C | 157.423 W |   0.001 W | 236.133 W``````

A 1% rise in atmospheric emissivity (0.8/0.792) predicts a 0.479 °C rise in surface temperature.

You would think such intelligent and “correct” mathematics would be based on actual experiments, but you would be wrong; it is not based on anything other than its presuppositions, and has been so for more than a century by name, and two centuries by concept.

Let’s outline a very simple experiment to test whether the greenhouse effect is true:

``````          Solid Surface
v

1) Person   => |     IR Camera

2) Person   <- | ->  IR Camera

And repeats until "equilibrium"``````

Radiation leaves the body and strikes a screen. After absorption some radiation will go out to the IR camera, and the rest will go back to the person, thereby warming them up further, according to greenhouse effect theory. Note that we don’t even need absorption, merely reflecting back a person’s radiation should warm them up.

Let’s assume the human body emits 522.7 W/m² (37 °C) (Emissivity: 0.9961, [Sanchez-Marin 2009]). For compatibility with my program, we multiply this figure by 4, and call it TSI. Let’s assume the screen and air in between together has a total emissivity of 0.9. Now run:

``````\$ TSI=2090.8 bash gheffect 0.9
Sec | Upwelling |   Temp    | GH Effect |  Trapped  | To Space
1 | 362.754 W |   9.658 C | 326.478 W | 163.239 W | 199.515 W
2 | 525.993 W |  37.188 C | 310.154 W |  73.458 W | 289.296 W
3 | 599.451 W |  47.498 C | 302.809 W |  33.056 W | 329.698 W
4 | 632.507 W |  51.830 C | 299.503 W |  14.875 W | 347.879 W
5 | 647.382 W |  53.725 C | 298.016 W |   6.694 W | 356.060 W
6 | 654.076 W |  54.566 C | 297.346 W |   3.012 W | 359.742 W
7 | 657.088 W |  54.943 C | 297.045 W |   1.356 W | 361.398 W
8 | 658.443 W |  55.112 C | 296.909 W |   0.610 W | 362.144 W
9 | 659.053 W |  55.188 C | 296.848 W |   0.274 W | 362.479 W
10 | 659.328 W |  55.222 C | 296.821 W |   0.124 W | 362.630 W
11 | 659.451 W |  55.238 C | 296.809 W |   0.056 W | 362.698 W
12 | 659.507 W |  55.244 C | 296.803 W |   0.025 W | 362.729 W
13 | 659.532 W |  55.248 C | 296.801 W |   0.011 W | 362.743 W
14 | 659.543 W |  55.249 C | 296.799 W |   0.005 W | 362.749 W
15 | 659.548 W |  55.250 C | 296.799 W |   0.002 W | 362.752 W
16 | 659.550 W |  55.250 C | 296.799 W |   0.001 W | 362.753 W
17 | 659.552 W |  55.250 C | 296.799 W |   0.000 W | 362.753 W``````

We see that the screen is “trapping” a lot of human radiation from reaching the IR camera, and we expect an extra 296.8 W/m² greenhouse effect, bringing us up to 55°C. Merely placing a screen in front of us should make us feel as if we’re stepping inside a sauna.

These people must be really feeling the heat. But they don’t, and for good reason: preventing radiation from reaching a colder place does not cause heating back at the source. Had these people had thermometers strapped to them, they would note the virtually zero temperature rise (due to blocked convection). Look very closely at the videos. Note the seconds the screens are placed in front of their faces and notice the lack of any thermal reading changes. None!

All empirical evidence shows the opposite of the claims of the greenhouse effect.

So the question remains, why is the surface hotter than the sun can make it alone?

If we look at the energy budget, we can see a dependency loop between surface and atmosphere: Surface -> Atmo = 350 and Atmo -> Surface = 324. So which came first, the chicken or the egg? This is nonsense. You can’t have a dependency loop for heat flow. Let’s try a theory that does not cause mental anguish and lacks empirical evidence. For this, we ignore the climate “scientists”, and go to the geophysicists:

Here we see that Earth’s geothermal energy is capable of delivering 0 °C to the surface; This is equivalent to 315.7 W/m². We add the sun and subtract latent+sensible heat:

315.7 + 168 – 24 – 78 = 381.7 = Upwelling Radiadtion

Now we get a figure that that’s 390 – 381.7 = 8.3 W/m² off, but that’s OK because latent and sensible heat are not directly measured but estimated with certain physical assumptions, and/or the 0 °C geothermal is an approximation too.

Now we finally realize that the greenhouse effect is a hoax, and nothing but geothermal flipped up-side down. There is no Downwelling Radiation, there is only Upwelling-from-measurement-instrument Radiation (See here). Those who read Why is Venus so hot?, probably already saw where I was going. Now doesn’t it make more sense than backradiation temperature raising? Reality shows abolutely normal geothermal and solar combining to produce what we observe. We see all normal heating, and no ugly backwards zig-zag heating.

Let’s summarize:

``````     Upwelling
^
|      |       ^        ^
v      |       |        |
===============================
|    Latent  Sensible
Solar ---+     Heat      Heat
|       ^        ^
|       |        |
+------ Geothermal``````

Now which explanation does Occam’s Razor favor?

I hope you have enjoyed the return to sanity.

Sincerely, -Zoe

Notes

[¹] We only care about matching 3 decimal places. If we want to extend it to IEEE754 64-bit precision, it takes 40 seconds. Not that this matters much; Most work is accomplished in the first 5 seconds.

[²] I debated with myself whether to use the term seconds or iterations. Real physical calculations would take mass and heat capacity into account, but since greenhouse theorists don’t use these, I won’t either. Their simple model is in seconds.

https://phzoe.com

421 thoughts on “Dumbest Math Theory Ever”

1. Max Polo says:

I thought I had responded to Rocky’s two latest posts…but I don’t find my responses. This is a test.

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1. Rocky says:

I see this response Max. I set up a calculation for your plate measurements and my calculations show that the irradiated plate will only increase in temperature by a fraction of a degree C. Given that you have puffs of air that cause bigger fluctuations than that, I don’t think that your set up can measure the effect you are looking for.

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1. Max Polo says:

With my choice of panes size/distance, back-radiation heating should be 2.5-2.7°C (against 4.7°C in a vacuum) so very well measurable with a thermometer that detects 0.1°C changes.
Here-below is my reply to your previous message.

“Agreed, but it is also far more difficult to interpret the results because you are unable to isolate the different effects. For example, let’s say you had an electric circuit with two resistors in parallel, red and blue. You are trying to measure the resistance of the red resistor, which has a large resistance in comparison to the blue one. The most accurate way to measure the red resistance is to take the blue resistor out of the circuit entirely. Furthermore, if the blue resistance is very small, then it may be impossible to measure the red resistance if your instruments are not sensitive enough. Radiative and convective heat transfer sort of act like the red and blue resistors in your experiment.”

For my set-up, the mutual panes view factor was ranging from 62% to 68% (depending on panes distance: 4 to 5 cm), i.e. 62-68% of the heat flowing from one side of the warmer pane was getting absorbed by the cooler pane. Therefore 32% to 38% of the heat flow from the same side of the pane was going straight to the background. On the other side of the pane, obviously, 100% of the emerging energy was flowing straight to the background. As far as convection is concerned – my calculations show it to be approximately 70% of the total heat exchanged for the case of the single pane or 58-60% for the case of the front (irradiated) double pane arrangement. Therefore the radiative heat is not that negligible, and its effect should be well measurable in the amount I have said (2.5 – 2.7°C theoretical back-radiation heating).

“That’s really no better than using the sun if you are interested in doing a controlled experiment.”

I agree.

“Perhaps more reliable than a heat lamp, but not more reliable than a controlled, directed, and measured heat source. With the heat lamp, you don’t know how much of its energy is actually being absorbed by the panel. The same goes with the sun.”

My procedure consists of reverse-engineering the sun radiation input based on the measured temperature of the front pane using heat balance equations. In this way, I can get rid of the sun’s uncertainty. Since the single and double panes are set side-to-side, one can be pretty sure that the same amount of irradiation is reaching both panes. This is true even if the panes are not perfectly square to the sun rays because the incidence angle would be the same for both of them.

“I’m skeptical of that claim. Duly accounting for convection would require highly detailed numerical calculations.”

Simple and reliable textbook still air convection correlations are widely used with good results in HVAC engineering. You do not need to go through complicated Nusselt/Grashof/Reynolds calculations to get a good ballpark value for a vertical plate in calm air. With no calculations, typically suggested numbers for vertical plates are around 5 W/m2/°C. With my correlation, I get 4 to 5 W/m2/°C depending on actual Delta-T (pane-air) after convergence is reached (obviously the equations are non-linear).

“Can you please show me the details of that calculation? I’d like to see if I agree with your methods.”

I’m prepared to disclose the entire report once I’ve finished. I’m happy to be challenged because my only wish is to grab as much truth as possible. Some equations (although not fully updated) are visible via the old link that I have shared last year.

“That’s not surprising to me with the geometry that you have. It is dominated by convective heat transfer. Thus I highly doubt that your calculations are correct.”

See above.

“I suggest you try a vacuum. I’m willing to bet a significant amount of money that it will show up.”

That would be my possible next plan. We’ll see.

“You also have radiation to the rear pane from the surroundings. Did you account for that? Do you know what the temperature of the surroundings was in your experiment?”

Obviously, I have duly taken into account the effect of the surroundings, dividing it into two “portions”: “ground” (measured at about 25-38°C depending on the day/time) and “sky” (measured at about 0°C-8°C depending on day/time), whose “weight” is 50%-50% based on textbook view factors.

“Again, I’d love to see your calculations. I’m highly skeptical that you have accounted for the radiative and convective heat transfer properly. It’s clear that you are not testing the simple plate configuration that has been analyzed by others.”

I’ll be back with my revised report maybe in a couple of weeks.

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1. Rocky says:

“With my choice of panes size/distance, back-radiation heating should be 2.5-2.7°C”

Thus statement makes zero sense. There are two steady state temperatures, one with the second plate and one without. Introducing the second plate alters more than just the “backradiation”. You cannot assign a certain amount of back radiation to a temperature change. It doesn’t work that way. Convection is going to keep the second plate close to the temperature of the surroundings and so the radiation from the surroundings is simply replaced by the radiation from the plate and thus will have little affect on the temperature of the irradiated plate.

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2. Max Polo says:

For some reason, I am unable to post my pointwise reply to your earlier message. Maybe it is too long. I’ll try posting a link to a file with my answer in another post. However, with my set up back-radiation heating should be 2.5-2.7°C (4.7°C in a vacuum) so without any doubt, it’s measurable with a thermometer that reacts to 0.1°C variations.

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1. Rocky says:

I disagree with your assessment. My calculations show that the effect will be a fraction of a degree.

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2. Rocky says:

Have you done measurements on the fluctuations of the temperature on the single plate? It makes no difference if the thermometer can measure 0.1C if the fluctuations are 2C. The same goes for fluctuations in the surroundings. Did you measure the temperature of the surroundings?

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3. Max Polo says:

@Rocky – my answer to your earlier post on the “screen”.

If your body is at 36°C and the room walls are at 25°C, the screen will settle at an intermediate temperature depending on distance/view factors. Let’s assume for instance that the screen temperature is 28°C. Since 28°C > 25°C, the GHE supporter will claim that your body will be burning fewer calories to keep you at 36°C because the screen works as insulation and you are losing less heat to it than you would to the walls (due to the lower temperature difference). This seems correct at a first glance, but it is not, because it’s your own body that is “sustaining” with its energy the higher temperature of the screen by virtue of which you can give off less heat to the screen itself. The “consensus” approach forgets a part of this energy. So in reality your body will burn the exact same calories (assuming the screen mass is negligible).

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4. Max Polo says:

Thanks, Zoe. I’m trying to post my reply to Rocky again.

“Agreed, but it is also far more difficult to interpret the results because you are unable to isolate the different effects. For example, let’s say you had an electric circuit with two resistors in parallel, red and blue. You are trying to measure the resistance of the red resistor, which has a large resistance in comparison to the blue one. The most accurate way to measure the red resistance is to take the blue resistor out of the circuit entirely. Furthermore, if the blue resistance is very small, then it may be impossible to measure the red resistance if your instruments are not sensitive enough. Radiative and convective heat transfer sort of act like the red and blue resistors in your experiment.”

For my set-up, the mutual panes view factor was ranging from 62% to 68% (depending on panes distance: 4 to 5 cm), i.e. 62-68% of the heat flowing from one side of the warmer pane was getting absorbed by the cooler pane. Therefore 32% to 38% of the heat flow from the same side of the pane was going straight to the background. On the other side of the pane, obviously, 100% of the emerging energy was flowing straight to the background. As far as convection is concerned – my calculations show it to be approximately 70% of the total heat exchanged for the case of the single pane or 58-60% for the case of the front (irradiated) double pane arrangement. Therefore the radiative heat is not that negligible, and its effect should be well measurable in the amount I have said (2.5 – 2.7°C theoretical back-radiation heating).

“That’s really no better than using the sun if you are interested in doing a controlled experiment.”

I agree.

“Perhaps more reliable than a heat lamp, but not more reliable than a controlled, directed, and measured heat source. With the heat lamp, you don’t know how much of its energy is actually being absorbed by the panel. The same goes with the sun.”

My procedure consists of reverse-engineering the sun radiation input based on the measured temperature of the front pane using heat balance equations. In this way, I can get rid of the sun’s uncertainty. Since the single and double panes are set side-to-side, one can be pretty sure that the same amount of irradiation is reaching both panes. This is true even if the panes are not perfectly square to the sun rays because the incidence angle would be the same for both of them.

“I’m skeptical of that claim. Duly accounting for convection would require highly detailed numerical calculations.”

Simple and reliable textbook still air convection correlations are widely used with good results in HVAC engineering. You do not need to go through complicated Nusselt/Grashof/Reynolds calculations to get a good ballpark value for a vertical plate in calm air. With no calculations, typically suggested numbers for vertical plates are around 5 W/m2/°C. With my correlation, I get 4 to 5 W/m2/°C depending on actual Delta-T (pane-air) after convergence is reached (obviously the equations are non-linear).

“Can you please show me the details of that calculation? I’d like to see if I agree with your methods.”

I’m prepared to disclose the entire report once I’ve finished. I’m happy to be challenged because my only wish is to grab as much truth as possible. Some equations (although not fully updated) are visible via the old link that I have shared last year.

“That’s not surprising to me with the geometry that you have. It is dominated by convective heat transfer. Thus I highly doubt that your calculations are correct.”

See above.

“I suggest you try a vacuum. I’m willing to bet a significant amount of money that it will show up.”

That would be my possible next plan. We’ll see.

“You also have radiation to the rear pane from the surroundings. Did you account for that? Do you know what the temperature of the surroundings was in your experiment?”

Obviously, I have duly taken into account the effect of the surroundings, dividing it into two “portions”: “ground” (measured at about 25-38°C depending on the day/time) and “sky” (measured at about 0°C-8°C depending on day/time), whose “weight” is 50%-50% based on textbook view factors.

“Again, I’d love to see your calculations. I’m highly skeptical that you have accounted for the radiative and convective heat transfer properly. It’s clear that you are not testing the simple plate configuration that has been analyzed by others.”

I’ll be back with my revised report maybe in a couple of weeks.

Like

5. Max Polo says:

@Rocky

“This statement makes zero sense. There are two steady state temperatures, one with the second plate and one without.”

What I did is to calculate the two steady-state temperatures, assuming that the equations work as you claim they should. “Theoretical” back-radiation heating is defined as Twsp – Top
Twsp : front pane temperature WITH second plate
Top : pane temperature WITHOUT second plate
Generally, I found Twsp to be 2.5-2.7°C higher than Top. That was with the convection effect working. Simulating the vacuum conditions without changing all the other input data including surroundings temperature (ground/sky), I found that, obviously, both temperatures were quite higher, with their difference settling around 4.7°C, almost double the other case. This confirms your earlier suggestion to do the test in a vacuum. But even in presence of air, the result is not dramatically different, since 2.5°C are without any doubt immediately detectable by the thermometer (if they were real).

“Introducing the second plate alters more than just the “backradiation”. You cannot assign a certain amount of back radiation to a temperature change. It doesn’t work that way.”

I did not assign a certain back-radiation. I calculated the equilibrium temperature using the non-linear system of equations applied in the way “consensus” science applies them.

“Convection is going to keep the second plate close to the temperature of the surroundings and so the radiation from the surroundings is simply replaced by the radiation from the plate and thus will have little affect on the temperature of the irradiated plate”

The temperature of the surroundings (radiative) was composed of two main portions (sky and ground) quite different than air temperature (convective), as I explained in my earlier post. Convections tend to keep all plates closer to the temperature of the air, but obviously surroundings play a role in adjusting the panes equilibrium temperature to an intermediate value depending on all of them (ground, sky, air).

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1. Rocky says:

“I calculated the equilibrium temperature using the non-linear system of equations applied in the way “consensus” science applies them.”

So show what equations you solved. Why is it taking you so long to just do that?

What do your equations say the temperature of a single plate should be if it is shaded from the radiation from the sun? That is an easy temperature for you to measure. Do your equations match the measurements for that case?

Non of your old links work. Can you repost them?

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1. Max Polo says:

Here is a link to the equations:

Here below is a link to the calculation of the single plate temperature shaded from the radiation of the sun. Since there is still the diffused radiation you’ll find 116 W/m2 as an input (estimated from: https://instesre.org/Solar/BirdModelNew.htm). In the test that I did today: calculated plate temp 27.9°C, measured 28°C. To be honest, the calculated temperature, in this case, is quite affected by the f factor (fraction of the background that has the temperature of the sky Ts, as opposed to the 1-f part of the background that has the temperature of the ground). I have chosen f = 0.3 (and not the standard 0.5 which I used in my earlier tests) because of the large building walls close by that are covering some part of the sky.

I’m going to post a link to my revised report (an excerpt of which can be seen at the above link of the equations) maybe early next week.

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1. Rocky says:

“Here below is a link to the calculation of the single plate temperature shaded from the radiation of the sun. … In the test that I did today: calculated plate temp 27.9°C, measured 28°C.”

So this calculation includes “backradiation” from the atmosphere. Yet you still question if backradiation exists.

Do you see how strange that is?

Thanks for the details of your calculation. I will go over it soon.

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2. Rocky says:

“Since there is still the diffused radiation you’ll find 116 W/m2 as an input”

So you mean diffuse radiation that is being reflected from the sun. So that should be an entry into the energy balances for both panes in the 2 pane case.

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6. Max Polo says:

@Rocky
“yet you still question if backradiation exixs”: not really: in my view, back-radiation heating of a warmer object exists when it comes from a colder source whose temperature is not dependent on the energy received from the warmer object. A couple of cases where back-radiation heating exists. Case 1: a cold source that is energized to be maintained at constant temperature provides back-radiation heating to a warmer object (at the expense of the energy supplied to the cold source). Case 2: a cold source that has a heat capacity so much larger than the warmer object so that its heat cannot affect the cold source’s temperature.
Rear pane (atmosphere) temperature is totally dependent/determined by front pane (earth) temperature and is certainly not fixed like in Case 1 and 2.

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1. Tom Dayton says:

The reason for the word “back” in the standard term “back radiation” for radiation sent by Object 2 to Object 1, is to refer specifically to radiation that previously was sent from Object 1 to Object 2. That terminology is used regardless of whether Object 1 or Object 2 is warmer.

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1. Max Polo says:

Noted. Thus the fact that such a term is non-existent in physics textbooks (at least those of my time) maybe gives a clue…

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1. Rocky says:

That’s pretty much irrelevant to the governing equations. Between two graybody objects the radiant heat transfer is:

q = F12 sigma epsilon_eff (T1^4 – T2^4)

The “backradiation” is right there staring you in the face. Colder objects can and do emit radiation that is absorbed by hotter objects. Why is this concept so controversial? I’m guessing that the physics textbooks of your time teach this very concept.

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2. Tom Dayton says:

Max, “back radiation” as I defined it indeed is present in physics textbooks, insofar as it is an inherent, implicit part of any system of multiple objects that are exchanging energy, as Rocky pointed out. It is called out as the explicit phrase “back radiation” only when useful to describe it, which is the case for example when discussing planetary greenhouse gas effect. The term and even concept is irrelevant when describing a singe object that is not exchanging energy with any other object.

Your personal, idiosyncratic definition of “back radiation” including a constraint based on the temperatures of multiple objects that are exchanging energy, is not present in any physics textbooks, because that constraint does not exist.

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3. Tom Dayton says:

Max, what *does* exist is a constraint on the *relative temperatures* of two objects that are exchanging energy. You seem to be confusing relative temperatures with energy that is radiated and absorbed.

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2. Rocky says:

“in my view, back-radiation heating of a warmer object exists when it comes from a colder source whose temperature is not dependent on the energy received from the warmer object”

Why? What does it matter how the colder object arrived at its temperature? The colder object will emit radiation according to its temperature. Period.

You are trying to question the existence of “backradiation” from the atmosphere. Yet, your calculation that apparently matched your measurements has backradiation from the sky in your energy balance equations.

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1. Max Polo says:

“why? …” because no one can lift himself up by his hair. Or because there is such a thing called “irreversibility”.

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1. Rocky says:

How is this “lifting himself up by his hair”? How does it violate irreversibility? The heat transfer is from hot to cold, which is in agreement with irreversibility.

q = F12 sigma epsilon_eff (T1^4 – T2^4)

Do you not understand that this formula has “backradiation” built right into it? Do you not understand that this formula is consistent with EVERYTHING that was in your physics textbooks?

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2. Rocky says:

@Max – Do you understand that when you place two things at equal temperature next to one another that microscopic energy transfers do not cease? It’s simply that the two objects pass energy back and forth equally to one another.

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3. Rocky says:

@Max Polo – Try the following with your equations. Turn the radiation heat transfer between panes 1 and 2 off completely and see what temperatures you get. Your equations for the convective heat transfer and how that changes when the second pane is put into place are far more significant than the radiative transfer aspects of your equations.

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1. Max Polo says:

Will do it tomorrow.

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2. Max Polo says:

There is no need to turn Q12 off to see %convection vs % radiation, as the program already provides this info for the single pane. I have run two cases (link below): the first has the convection coefficient for calm air (around 4 W/m2/°C) while the second is 25 W/m2/°C for 3 knots wind speed.

1° — Calm air : 32% convection, 68% radiation.
2° — 3 knots wind : 80% convection, 20% radiation.

If you look at the calculation, what rules out the second case is the abnormal sun flux (1600 W/m2) that would be needed to reach the measured temperature, vs the normal value (a bit less than 700 W/m2, approximately in line with Bird’s model prediction, see the previous link I shared) of the first case.

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1. Rocky says:

The simple fact is that this experimental set-up has far too many factors that are out of your control and that you do not know how to measure.

How much of the diffuse solar radiation is blocked when the second plate is put into place?

How much does convection change from the first plate when the second is put into place?

What is the temperature of the air in the intervening gap?

All of these are questions that you cannot answer. And finally, ALL of your calculations have BACKRADIATION from the sky and the ground.

Your entire premise is filled with contradictions.

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2. Rocky says:

“There is no need to turn Q12 off to see %convection vs % radiation”

Turn it off for the two pane case to see what your calculations tell you.

You could make all of this so much simpler if you would simply use a controlled heat source, use the cylindrical geometry, and do it all in a vacuum. I will take all bets against those claiming that the inside cylinder will not warm when the outside cylinder is put into place.

If you really wanted to refute decades of knowledge on heat transfer, then you are going to have to do a clean experiment. Not one where you have no idea what effect is dominating.

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7. Max Polo says:

@Rocky – I agree with you. Diffused solar irradiation should be added to all panes. I’m going to update the formulas to include it.

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8. Max Polo says:

@Rocky “Do you not understand that this formula has “backradiation” built right into it?” I think I understand this formula. If you write the heat balance equation for body 2, you will see that “q” does not cover all the energy needed to sustain the temperature of body 2 (assuming T2 is dependent on T1). Per Stefan-Boltzmann law, body 2 at T2 will emit the radiant power sigma*T2^4. The transferred heat q does not cover the entire Stefan-Boltzmann emission of body 2, but only the part sigma*(1-F12)*T2^4 i.e. the part viewing “directly” the background. The heat q is an “excess of energy” flowing from 1 to 2, but – as any excess – it requires a “starting level” that in our case is F12*sigma*T2^4. This “starting level” of energy is still provided by body 1, it wouldn’t exist without it. If you forget it, you’ll be forced to invent the famous T1 bump-up called back radiation heating, otherwise, your balance equations won’t work.

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1. Tom Dayton says:

Max, back radiation is real. If you pretend it does not exist, you must invent something else to make up for its absence.

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1. Max Polo says:

I agree it is. That also has the capacity to increase the temperature of the same body that got rid of it….. well, that is a different story. That’s why I’m making the experiment 🙂

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1. Jarle says:

I cheer for you Max. Remember, every great scientist became great scientists exactly because they deviated from established science. This becomes difficult if you let someone lead you by the nose, and trying to imprint you about what you want to investigate. There is ALWAYAS a chance to find something that everybody else missed. And sometimes what is found is not even the subject of investigation. You have the correct scientific attitude. Few people are like you.

Liked by 1 person

2. gbaikie says:

I think it best to think of it as entropy:
“A consequence of entropy is that certain processes are irreversible or impossible, aside from the requirement of not violating the conservation of energy, the latter being expressed in the first law of thermodynamics. Entropy is central to the second law of thermodynamics, which states that the entropy of isolated systems left to spontaneous evolution cannot decrease with time, as they always arrive at a state of thermodynamic equilibrium, where the entropy is highest.”
https://en.wikipedia.org/wiki/Entropy

Or it’s expressed as huge amount of energy [more than sunlight] but it doesn’t do work, or it does heat nor can one make it heat anything.
It’s like ocean water, but one could find ways of making ocean water provide heat energy.
I would use ocean heat to power a rocket launch, for example. But I need liquid air to use this energy.
I call it a pipe launcher, which provide more lifting power then most powerful rocket engines.
Or could get a lot energy from tropical ocean surface water, though also for water at say 5 to 10 C
but warm water like say 60 C water is better [and quite easy to get].
So in theory one could get energy from entropy, but no one figured out to get useful energy from back radiation. It more useful to think of as a blanket or insulation rather than warming anything.

Liked by 1 person

1. gbaikie says:

Also kind of like solar wind [it doesn’t heat anything] and might able to use this energy to generate
a vast amount of energy, but at moment it’s roughly considered be like entropy which isn’t even enough to even think of it as a blanket. But solar wind seems to me to have far more potential of using it, compared to Earth’s back radiation.
I guess mainly because in space one could have things of massive scale- assuming we were space faring and didn’t lift stuff off earth in relatively small boxes with a lot gee forces involved {and very expensive to do].

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2. Tom Dayton says:

Max, I’ll leave it to Rocky to evaluate your calculations, because he’s way more qualified than I am. But from just this paragraph you wrote, it *appears* to me that *maybe* you decided that back radiation does not exist, despite the lack of any physical basis for that decision. And then maybe you decided to compensate for that by inventing a “starting level” of energy that also has no physical basis.

You have agreed that back radiation does exist, per the definition I gave earlier. It appears that you introduced an artificial and arbitrary limitation on back radiation, based on nothing but your personal preference that such a limitation should exist. That forced you to invent a fictional additional factor to compensate for the energy gap left by the limitation on back radiation.

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1. Max Polo says:

Actually, it’is the other way around. I started one year ago making the experiment, basically …I don’t trust anyone so I decided to experiment with something simple to see what really happens. After one year the results are telling always the same story, which is not the one I hear from “consensus” science. Based on the results, I’m trying to figure out why they are like they are.

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1. Rocky says:

“so I decided to experiment with something simple to see what really happens”

The problem is that you have not tried something simple. Your experiment has so many effects that are out of your control that you do not know what effect is responsible for your observations.

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3. Rocky says:

@Max
“I think I understand this formula.”

Clearly you don’t.

“body 2 at T2 will emit the radiant power sigma*T2^4” – This is true if body 2 is a blackbody.

“The transferred heat q does not cover the entire Stefan-Boltzmann emission of body 2”

Very good, that is correct. The heat transfer between two blackbody objects is the radiation emitted by body 2 and absorbed by body 1 minus the radiation emitted by body 1 and absorbed by body 2. It’s this second part that is the “backradiation” that you are claiming should not exist.

“This “starting level” of energy is still provided by body 1, it wouldn’t exist without it.”

This is an irrelevant comment. You need to figure out why. Objects emit radiation according to their temperature. It makes no difference how an object attained its temperature, it emits all the same.

“If you forget it” – You are wrong. It makes ZERO difference how an object arrived at its temperature.

“you’ll be forced to invent the famous T1 bump-up called back radiation heating, otherwise, your balance equations won’t work”

I haven’t had to invent anything. My balance equations work perfectly. In fact, for the steel greenhouse problem they are IDENTICAL to the ones that Zoe has provided.

YOU have been using “backradiation” from the atmosphere to the panes in ALL of your calculations.

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1. Max Polo says:

“Very good, that is correct”

OK, so you agree that body 2 (needless to say I am always talking of blackbodies) emits more energy than the heat q it receives from body 1. But aren’t we supposed to be in a steady state status (equilibrium) where energy-in = energy-out also for body 2? then, who provides body 2 with such missing radiant power? I do not think this energy should magically appear out of nothing.

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1. Rocky says:

“OK, so you agree that body 2 (needless to say I am always talking of blackbodies) emits more energy than the heat q it receives from body 1.”

Body 2 can only do this if it is also receiving heat from other entities like the sky, ground, diffuse solar radiation, etc.

“But aren’t we supposed to be in a steady state status (equilibrium) where energy-in = energy-out”

Indeed. That is what all of the calculations are assuming. Whether or not your experimental conditions have achieved steady state is another question. You have no control over this.

“then, who provides body 2 with such missing radiant power?”

I have answered that.

Have you ever solved the steel greenhouse problem? Does your solution procedure agree with Zoe’s and mine?

“I do not think this energy should magically appear out of nothing.”

It never does in AGW. If you think otherwise, then you don’t understand AGW. Why don’t you tell me where there is energy magically appearing in the steel greenhouse solution?

No shell : q_in=240 W/m^2, Tsphere = 255K, Sphere emits 240 to space. Heat into sphere is balanced by energy out.

With shell: q_in=240 W/m^2, Tsphere = 303K, Tshell = 255K
Energy in to sphere from source = 240
Energy emitted by sphere to shell = 480
Energy absorbed by sphere from shell = 240
Net for sphere = 0

Energy absorbed by shell from sphere = 480
Energy emitted by shell to sphere = 240
Energy emitted by shell to space = 240
Net for shell = 0

So where exactly do you think that energy has magically appeared?

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2. Rocky says:

I should have said, “Body 2 can only do this if it is also receiving ENERGY from other entities …”

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9. Max Polo says:

@gbaikie : thanks, much appreciated ! Indeed, entropy should play a role, like irreversibility as you mention. I will think more about that.

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10. Max Polo says:

@Rocky – this is the last one for today – I’m leaving :
“you have been using backradiation from the atmosphere to the panes in ALL of your calculations”

Correct. My pane weighs approximately 70 grams. The atmosphere weighs approximately 5.15×10^18 kilograms. There are enough orders of magnitude in between, to assume that we are in Case 2 of one of my earlier posts, i.e. the atmosphere is a cold source at a fixed temperature FOR THE PANE.

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1. Rocky says:

Why on Earth do you think that the atmosphere is at fixed temperature?

You really are confused. Why don’t you try to solve the steel greenhouse problem?

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1. Max Polo says:

this is really my last one for today: “why on earth do you think the atmosphere is at a fixed temperature” ? I said “for the pane” it acts like it had a fixed temperature, i.e. the tiny bit of heat it exchanges with the pane doesn’t alter the atmosphere’s temperature, but the pane’s temperature. Try modeling the heat transient behavior of two bodies whose mass (heat capacity) is orders of magnitudes different, and that are mutually exchanging heat, and you will see it.

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11. Rocky says:

“I said “for the pane” it acts like it had a fixed temperature”

I can also say that for the pane it acts like it provides a fixed source of radiation.

“the tiny bit of heat it exchanges with the pane doesn’t alter the atmosphere’s temperature”

Actually the tiny bit of heat it exchanges alters the temperature a tiny bit. But who cares? The first law of thermodynamics only cares about how much energy is provided by an object. The amount of radiation provided by an object depends on its temperature. PERIOD. You still don’t seem to understand this very simple concept.

“Try modeling the heat transient behavior of two bodies whose mass (heat capacity) is orders of magnitudes different, and that are mutually exchanging heat, and you will see it.”

Feel free to provide a clear problem statement and I will solve it. There is nothing for me to see.

The solution to the steel greenhouse problem is EXACTLY the same whether the additional shell has a heat capacity that is a billion times that of the sphere or if it is one billionth that of the sphere. Heat capacity does not affect the steady state temperatures. Perhaps they didn’t teach that concept when you were taking physics.

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1. Max Polo says:

“The simple fact is that this experimental set-up has far too many factors that are out of your control and that you do not know how to measure.
How much of the diffuse solar radiation is blocked when the second plate is put into place?
How much does convection change from the first plate when the second is put into place?
What is the temperature of the air in the intervening gap?”

Some clarification is needed here: global solar input (direct + diffuse) does not represent an uncertainty, because I’m reverse-calculating it based on the actual measured single pane temperature. This is easy to measure with great accuracy, as opposed to the solar radiation that depends on so many weather factors. But I was surprised to find that even with this method, the resulting (inferred) solar input is close to Bird’s model prediction. The same goes for the calculation of the pane temperature in a shade that you requested. This should tell that the model works with reasonable accuracy, despite the many factors.
With regard to the effect of the gap temperature/convection: initially I had introduced a correction factor here because I wasn’t able to reproduce the measured temperature of the rear pane (measured value was 7-8°C greater than predicted). Then I realized that – due to the sun elevation – some direct solar irradiation was “leaking” and going straight to the rear pane. I then introduced this energy input in the second pane, gauging it by the percentage of the rear pane area that was hit by sun rays. That percentage was from 25 to 30%, depending on the hour. After introducing such energy input, I was able to reproduce back pane temperature with quite a good accuracy. So, I decided to get rid of such “gap correction”, because the mere introduction of this measurable and controllable energy input in the model was able to provide a good forecast of the rear pane temperature.
The funny thing is, that this “back-radiation heating” didn’t show up even if there was this additional energy boost to the rear pane.

About the effect of diffused radiation (good point, thanks), when I have time, I’m going to make a few tests to evaluate how much this potential inaccuracy can affect the predicted back-radiation heating. I have already made a sensitivity analysis to some inaccuracies (measured temperature, film coefficient, distance of the panes…) to determine a mean probable global error finding that it is around 1/7 of the predicted backradiation heating….but I’ll need to add this one…

“I have answered that – body 2 can only do this (emit more energy than the heat q it receives from body1 ) if it is also receiving heat from other entities, like sky, ground…”

Noted. So let’s assume that the two bodies are in a vacuum (deep space), hence no atmosphere, no ground, no sky, no diffused radiation…do you believe that in such conditions body 2 does not emit the entire radiant power sigma*T2^4? I’m asking this because the heat Q12 it receives from body 1 is only a portion of this emission, even in deep space.
Maybe the diagram at the below link clarifies what I am trying to explain :

“Whether or not your experimental conditions have achieved steady state is another question. You have no control over this.”

I think I had control over this aspect. For a pane of 70 grams, and a convection coefficient around 5 W/m2/°C, the temperature should settle to its equilibrium value (with max error < 1 tenth of a degree) in about 2.5 seconds. I waited for no less than a couple of minutes before gauging the temperature.

Enough for today.

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1. Rocky says:

“Some clarification is needed here: global solar input (direct + diffuse) does not represent an uncertainty, because I’m reverse-calculating it based on the actual measured single pane temperature.”

This is nonsense. You can’t claim to be experimentally confirming or denying anything if your conclusions are based on calculations that are fraught with assumptions.

You HAVE to design an experiment that ISOLATES the effect you are attempting to prove/disprove.

This is BASIC experimental design.

But on another note, I have an honest question. If you are using experimentally determined correlations for the convective heat transfer, don’t those correlations already have the effects of radiative heat transfer with the environment that they were established in?

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1. Max Polo says:

“this is nonsense”
I would (almost) agree it is if hadn’t added “the resulting (inferred) solar input is close to Bird’s model prediction”

“I have an honest question…”: good question. Indeed, several times practical correlations try to include also radiation effects. This is not the case of this one, that is proportional to DT^0.5 which is typical for convection (like the more complicated one from which it derives, based on Nusselt and Rayleigh numbers). For comparison, this paper gives a ballpark value (still air) of 4 W/m2/°C (we are close but we have a dependence on DT) :
https://www.sciencedirect.com/topics/engineering/convection-coefficient
You can double check that on other online resources like https://quickfield.com/natural_convection.htm

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2. Rocky says:

@Max

Can you please provide your solution procedure and rationale for the steel greenhouse problem?

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1. Max Polo says:

I haven’t done an experiment on this one (yet) so I am in trouble 😉
but I can tell you what’s wrong (in my view) with your solution: first of all, you are summing up fluxes with nonchalance, but I seem to recall that if you do that you won’t get a correct Planck curve for the higher temperature, and you’ll violate Wien’s law. This can be seen with a simple spreadsheet that plots Planck’s curve for the two fluxes and for the resulting one. This seems to be an algebraic manipulation with no real physical significance. But aside from that, a practical experience I have is about heat treatment of large plates. I used to work for a company that made solution annealing heat treatment on large/thick stainless steel plates at 1038°C. The plates would come out of the furnace and put stacked on the floor with spacers, parallel like our little panes. So intermediate plates were seeing the two other plates close by with a view factor of almost 1, capable of transferring almost the entire 167600 W/m2 to the plate in the middle. Summing the two fluxes from the adjacent plates we would obtain a warm-up of 248°C in the middle plates. Inertia and losses will dampen that somewhat, but anyways you should get a massive warm-up, that should have been well detected by our QC people that were monitoring cooling rates which are critical for plate mechanical properties and corrosion resistance. But they never highlighted what could have been a big issue for the quality of the plates. If instead, your method of summation is right …..kudos to you! then it’s time to file a patent for some device that based on this principle can melt steel using much lower temperatures than used in the industry.

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2. Rocky says:

“first of all, you are summing up fluxes with nonchalance”

What the hell does that mean? Point to the object where the energy in is not balanced by the energy out. Simple addition and subtraction are just that, SIMPLE. There is nothing “nonchalant” about basic arithmetic.

“but I seem to recall that if you do that you won’t get a correct Planck curve for the higher temperature, and you’ll violate Wien’s law”

WTF are you talking about? The Planck curve for a blackbody gives the emitted radiation as sigma T^4. There is no violation of Wien’s law in that at all.

“This seems to be an algebraic manipulation with no real physical significance.”

Again, WTF are you talking about? You have done similar algebraic manipulations in all of your calculations. The physical significance is that the FIRST LAW OF THERMODYNAMICS requires you to ADD all energies into a system and SUBTRACT all energies out from the system. The net result at STEADY STATE must be zero. THAT is the physical significance of adding up all of the energies.

Why don’t you put down some equations and EXPLAIN your procedure for solving this simple problem?

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3. Max Polo says:

Rocky, I decided to overcome my troubles and provide my solution procedure and rationale for the steel greenhouse problem. Here it is.

1) No Shell

Fixed input flux to Sphere = W

Sphere energy in = Sphere energy out :
W = sigma*T10^4 [1]
T10 = [W/(sigma)]^0.25
The heat released into the background is then Q120 = W

2) With Shell (Shell Radius almost equal to Sphere radius)

Fixed input flux to sphere = W

Shell energy in = Shell energy out :
sigma*T1^4 = 2*sigma*T2^4 [2a] ==> T2 = T1/(2)^0.25 [2b]

The heat Q12 transferred from Sphere to Shell is given by :
Q12 = sigma*(T1^4-T2^4) [3]

Substituting the expression [2a] in [3] we get :

Q12 = sigma*T2^4 [4]

from [4] we see that the heat that the Sphere is transferring to the Shell, makes up for just one-half of Shell’s total emission; the other half, emitted towards the inside, must be still provided by the Sphere since there are no other energy sources. So, this is the total radiant power W’ that the Sphere must provide:
W’ = Q12 + sigma*T2^4 = sigma*(T1^4-T2^4)+sigma*T2^4 = sigma*T1^4 [5]

The statement of the problem says that the power provided to the sphere must remain constant, so : W = W’

Equating the expressions of W [1] and W’ [5] :

T1 = T0 [6]

Comment: the heat flow Q12 is less than the heat flow Q120 that would occur if the Shell wasn’t present, but this does not yield an increase in the temperature of the Sphere, because the Shell requires more energy than just Q12 to keep its balance. In other words, Q12 is not sufficient to sustain T2. A different case would be that in which T2 is fixed (heat sink) : T2 would be sustained by an external energy source and would not “suck out” anything more than just Q12 from the Shell.

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4. Rocky says:

“1) No Shell -> T10 = [W/(sigma)]^0.25
The heat released into the background is then Q120 = W”

Good. We agree on this.

“2) With Shell (Shell Radius almost equal to Sphere radius) -> sigma*T1^4 = 2*sigma*T2^4 [2a] ==> T2 = T1/(2)^0.25”

So far, so good.

“The heat Q12 transferred from Sphere to Shell is given by :
Q12 = sigma*(T1^4-T2^4)”

Again, we agree.

“from [4] we see that the heat that the Sphere is transferring to the Shell, makes up for just one-half of Shell’s total emission”

WRONG. The HEAT that the Sphere is transferring to the Shell makes up for ALL of the Shell’s emission to space. The ENERGY that the Sphere is transferring to the Shell accounts for both the ENERGY that the Shell transmits to space AND the ENERGY that the Shell radiates BACK to the Sphere. The HEAT transfer is the ENERGY that the Sphere emits to the Shell minus the ENERGY that the Shell emits BACK to the sphere.

You need to perform a conservation of energy analysis of the Sphere. It’s really not that difficult.

Energy in = Energy out

W + sigma*T2^4 = sigma*T1^4

or you can rearrange this to give

W = sigma*(T1^4-T2^4)

Energy in from source = Heat out

“the heat flow Q12 is less than the heat flow Q120 that would occur if the Shell wasn’t present”

This is again WRONG. The heat flow through ALL surfaces enclosing the source must be W at steady state. Otherwise that means that energy must be accumulating within the system.

Finally, I will add that your approach is in disagreement with Zoe’s.

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3. Rocky says:

“So let’s assume that the two bodies are in a vacuum (deep space), hence no atmosphere, no ground, no sky, no diffused radiation…do you believe that in such conditions body 2 does not emit the entire radiant power sigma*T2^4? I’m asking this because the heat Q12 it receives from body 1 is only a portion of this emission, even in deep space.”

Body 2 emits sigma*T2^4 times its surface area. So does body 1.

Is that something that we agree on?

“I’m asking this because the heat Q12 it receives from body 1 is only a portion of this emission, even in deep space.”

Indeed. Apparently this is also something that we agree on.

Both objects will cool. However they will cool MORE SLOWLY than if these objects were not in close proximity to one another. Is that something that we also agree on?

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1. Max Polo says:

Full agreement! Except that I was intending to move the experiment to space, which means that you should add irradiation on the front plate, and wait for the temperatures to settle. That was the premise of my question.

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2. Rocky says:

“Full agreement!” – That’s good Max.

“Except that I was intending to move the experiment to space, which means that you should add irradiation on the front plate, and wait for the temperatures to settle. That was the premise of my question.”

Fine. Without the second plate (which I assume is in the shade of the first) the temperature of the first plate will be lower that with the second plate.

This should not be controversial. If you disagree then provide your equations and solution procedure that shows otherwise.

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4. Rocky says:

“the temperature should settle to its equilibrium value (with max error < 1 tenth of a degree) in about 2.5 seconds"

Only if ALL of your energy inputs and outputs are steady in time. You have NO CONTROL over that.

Seriously, why don't you at least do this inside in a controlled environment with a controlled heat source? You will still not be able to get around the complexities of how convection changes and if the correlations you are using already have radiation built into them, but at least you can remove the complications of performing an experiment in an entirely uncontrolled outside environment.

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1. Max Polo says:

Dear Rocky, that is a great hint, but I also have a job to look after. Anyhow I will take this into due consideration. I think I will take a break from this discussion…I am busy, and arguing in a language that is not my own is … tiring and time-consuming. Thanks a lot for all the comments and suggestions.

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2. Max Polo says:

“Feel free to provide a clear problem statement and I will solve it. There is nothing for me to see.

The solution to the steel greenhouse problem is EXACTLY the same whether the additional shell has a heat capacity that is a billion times that of the sphere or if it is one billionth that of the sphere. Heat capacity does not affect the steady state temperatures. Perhaps they didn’t teach that concept when you were taking physics.”

Here is the problem statement.
You have two panes in deep space, geometrically equal to the ones of my experiment (height = 0.25 m, width = 0.18 m, distance 0.05 m). The only difference is that the rear pane is made out of a material having the heat capacity of the atmosphere (Cfront = 112 J/Kg ; Crear = 5×10^21 J/Kg) but that has still the emissivity of a blackbody. The front pane is irradiated with 1236 W/m2 on one side.

The equilibrium solution, calculated obviously without caring about the thermal inertia, would be (per the “consensus” version) :
Tfront = 58.2°C
Trear = -26.2°C

According to the “rationale inferred by my experiment” (bear with me…) it would be :
Tfront = 50°C
Trear = -32°C

I don’t care (for the point I’m making) whether and which solution is correct or which is wrong.

I care about what happens in transient conditions.
Let’s suppose the sun is suddenly turned off at time 0 and let’s see what happens to front and rear panes temperatures over time (grant me this simplification: panes are totally emitting to space…otherwise things get a bit too messy…. but you are smart and you’ll understand the point or you can build a better model).

Time 0 :
Tfront/Trear = 58.2°C/-26.2°C

Time 0.5 sec:
Tfront/Trear = 56.8°C/-26.2°C

Time 5 sec:
Tfront/Trear = 45.5°C/-26.2°C

Time 50 sec:
Tfront/Trear = -20°C/-26.2°C

Time 8 min :
Tfront/Trear = -134°C/-26.2°C

Time 20 min :
Tfront/Trear = -165°C/-26.2°C

etc etc etc
…..
…..

you can go ahead for hours (or days ?…didn’t try that, I used a 0.1 s time interval to integrate numerically…..) before you start seeing 1/100 °C variation for rear pane temperature.

But our experiments on planet earth don’t last for days, they last minutes or hours: so what do we measure? We measure an atmosphere that looks like a heat sink, i.e. it can be considered to be at a constant temperature, or in other words its temperature is not affected by the front pane for the timeframe of the experiment.

This is why the atmosphere can affect the pane temperature (as I assumed in my calculations) : it is like a portion of the back-ground that has a fixed temperature (heat sink) for the pane under examination. The front pane is unable to make even the slightest change to the rear pane (atmosphere) temperature in the short practical timeframes.

So, heat capacity makes a ton of difference even for measurements in steady state conditions, especially if the masses of the two objects exchanging heat are orders of magnitude different. Practical timeframes for us poor humans are unfortunately so short.

My understanding from the experiment (bear with me again…) is that back-radiation heating only works if the back-radiating body meets the above definition of “heat sink”. Otherwise, if it has a temperature that is sustained by (depending on) the radiating body, the warming-up won’t happen.

Think of it as if you were trying to lean against a rickety wall…as opposed to a solid and stable one.

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1. Rocky says:

“I care about what happens in transient conditions.”

Why? Everything that you have done up to now has been for steady state. If you want to look at the transient conditions then you are going to have to solve a set of coupled partial differential equations. I don’t think that you know how to do that. As for the convection problem, it would require numerical solutions.

“So, heat capacity makes a ton of difference even for measurements in steady state conditions”

COMPLETELY WRONG. If you have steady state conditions then heat capacity make ZERO difference to measurements of temperature in steady state. ZERO.

If your argument is that you are measuring TRANSIENT conditions in your experiment then you have to compare your results to a TRANSIENT analysis. You have not performed any transient analyses.

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1. Max Polo says:

Let me try to explain my point in a different way: before reaching equilibrium, you will have to pass thru a continuous series of unstable states (changes of temperatures of the two bodies). But if the temperature of Body 2 is “pinned” due to its huge inertia, it is Body 1 the one that will have to adjust its temperature towards the equilibrium. The “pinned” Body 2 will work as a “background” or “heat sink” for the emitting Body 1. This was just to explain why I have introduced the backradiation from the sky (Body 2) in my model, in apparent contradiction with my statement that backradiation cannot increase the temperature of the emitting Body 1 (Pane). There is no contradiction because in my earlier statement I added “unless Body 2 is a heat sink”, and for the timeframe of my experiment, there is no doubt that the atmosphere works as a heat sink = constant temperature (see my thermal transient calculation).
An explanation of my statement “backradiation does not produce heating unless Body 2 is a heat sink (constant temperature)” is provided in my solution of the Steel Greenhouse Ruse (other post).

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2. Rocky says:

@Max
“before reaching equilibrium, you will have to pass thru a continuous series of unstable states (changes of temperatures of the two bodies).”

You do not reach equilibrium, you approach steady state. You go through a transient as you approach equilibrium.

“There is no contradiction because in my earlier statement I added “unless Body 2 is a heat sink””

Call it what you want. You are arbitrarily changing the “rule” in order to deny that “backradiation” exists. “Bacjradiation” is nothing more than the radiation that it emitted by ALL objects that are above absolute zero temperature.

Your claims are physically nonsensical.

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12. gbaikie says:

–Mainstream climate scientists believe in the dumbest math theory ever devised to try and explain physical reality. It is called the Greenhouse Effect.–
Who was the author of this dumbest math theory?

–It’s so silly and unbelievable that I don’t even want to give it the honor of calling it a scientific theory, because it is nothing but ideological mathematics that has never been empirically validated. In fact it is nothing but a post hoc fallacy: the surface is hotter than what the sun alone can do, therefore greenhouse gases did it!–
I think any scientific theory requires an author.
As far as know the greenhouse effect silliness doesn’t say sunlight can’t heat the ground surface of Earth to 70 C. But in regard to Venus they will say the sunlight can’t heat the rocky surface as hot as
it is with sunlight,
It seems to me greenhouse effect silliness went off the rails, when some idiots failed to understand why Venus surface was so hot.
One thing about Venus rocky surface is that it has very dim sunlight reaching it.
Or Venus having very hot surface is not because sunlight is trapped at it rocky surface.
It’s sort of have a hot oven, and there is light bulb shining into it, and imagining the light bulb is warming the hot oven.
Though “mainstream climate scientists” seem to be quite aware that Venus is heated in it’s upper atmosphere where there is intense sunlight. And these “mainstream climate scientists” also claim
some of Earth’s heating is done in the ozone layer [also upper atmosphere of earth}.

Anyhow this thing called Greenhouse Effect, does not have an author, but it has “fathers” and perhaps a mother or two- which seems quite idiotic.
Considering how important the greenhouse effect is, if had an author, the author should be thought
highly of- or least someone would know their name.
I think it’s fair to say that there is not greenhouse effect theory.
But seems we talk about greenhouse effect without saying it’s a theory.
For instance I could say there could more than one greenhouse effect.
Or more than one mechanism that traps heat. Or somehow retains heat better as compare to something else.
Of course one put a stove in a house of glass and the burning stove is a part of greenhouse effect,
or source of heat which kept in house, as long as you don’t leave the door open.
In terms of Venus being hot due geothermal heat, the first problem is we not idea of the amount
geothermal heat, Venus has. But it appears Venus has far less volcanic activity as compare to Earth,
but it could hidden somehow.
Though I don’t think it’s likely, it possible Venus being hot is due to impactors hitting it.
When idiots were wondering about why Venus was so surprising hot, it wasn’t “known” that Earth and
all planets were regularly hit with space rocks. And even at this late date, it seems few have much idea of how much Venus is being impacted by space rocks.
But I think Venus is hot mostly because it warmed by sunlight in it upper atmosphere or don;’t think
it’s due to geothermal or impact energy of impactors [though some of it is].
And also think it would be quite simple to make Venus colder than Earth. Because you just need
to block a portion of sunlight reaching it.
I also agree roughly the “mainstream climate scientists” that Earth is in a Ice Age.
Why Earth is so cold, has been something I wonder about.
And like “fathers” of greenhouse effect I think 5 C warmer would a good thing,
Though it seems to me, that 5 C of warming would be close to impossible.
Or is impossible if you want it, within a century or two.

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1. Jarle says:

CERES claims a 0.8W/m2 radiation imbalance. Compare that to 340W/m2 from the sun. Of course it is proof of absolutely nothing. It drowns in inaccuracies. GHE cannot be backed up by even the most sophisticated instrument.

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1. Rocky says:

“GHE cannot be backed up by even the most sophisticated instrument.”

Of course it can. Go look up the outgoing radiation spectrum at the top of the atmosphere. When you see that big dip around the wavelengths that CO2 absorbs, THAT is experimental verification of the GHE.

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1. Jarle says:

I disagree. GHE is the theory that our planet becomes uninhabitable unless we stop burnings fossil fuels. (This is what GHE has become, and this is consequently what we have to treat).
This is a hypohesis, and those who have extraordinary claim, have to provide extraordinary evidence. There is no evidence, yet. What we need is this spectrum of yours, (representative of the WHOLE earth) in successive measurements to see the development of the narrowing of the dip. And even then we would need to know that thera are no counteracting effects at play. And even if it could be established that there are no counteracting effects, we would need to prove that burning the rest of the available resources of oil,gass,coal would push us far enought to really cause what they say it is goimg to cause. If we burned all of the remaining resources ovrenight, then we would get only around 670PPM CO2. (50% absorption of the ocean not accounted for – guess this takes some time). Uninhabitable planet? Irreversible climate disaster? Prove it.

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2. Rocky says:

“I disagree. GHE is the theory that our planet becomes uninhabitable unless we stop burnings fossil fuels.”

Wrong. The greenhouse effect is simply the fact that there are some gases that allow shortwave radiation to pass through and absorb longwave radiation, which causes the surface of the planet to be warmer than if those gases were transparent to longwave radiation.

Your ignorance of the term and what it means is noted.

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3. Rocky says:

“What we need is this spectrum of yours, (representative of the WHOLE earth) in successive measurements to see the development of the narrowing of the dip.”

No we don’t. The existence of the dip alone is proof of the GHE.

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2. gbaikie says:

So in terms more than one greenhouse effect.
What is the simple definition of greenhouse effect?
“The greenhouse effect is the way in which heat is trapped close to Earth’s surface by “greenhouse gases.” These heat-trapping gases can be thought of as a blanket wrapped around Earth, keeping the planet toastier than it would be without them.”
And:
“the trapping of the sun’s warmth in a planet’s lower atmosphere, due to the greater transparency of the atmosphere to visible radiation from the sun than to infrared radiation emitted from the planet’s surface.”
So ozone layer is not trapping heat “near surface” and it’s doesn’t have greater transparency to sunlight, rather it absorbs sunlight and doesn’t block infrared light. But O3 is called a “greenhouse gas”
Anyhow, a greenhouse effect is transparent to sunlight and not transparent to infrared light created
sunlight energy being absorbed.
Earth has ocean which is very transparent to sunlight. Most energy striking the Earth surface, goes thru the transparent ocean surface, and heat generated by sunlight is trapped under the transparent
ocean.
So the greatest amount of sunlight is not heating land surface, only a small portion is heating the land, so ocean is second greenhouse effect which traps more 80% of all sunlight pass thru the mostly transparent atmosphere, and the very transparent ocean surface.
Or Earth has 3 greenhouse effect [at least] the ocean, the lower atmosphere, and Ozone layer.
And Venus has one, it’s upper cloudy atmosphere.
As said 3 at least, one also have Earth’s clouds being a greenhouse effect. And clouds can trap heat in lower atmosphere. And clouds can do lot’s of things. And effects of clouds a huge variable in terms of surface air temperature- which are said to overwhelm the weak effect of the CO2 greenhouse gas.
Or unpredictable nature of clouds, is a factor in all errors in global climate future projections.
Though one could say the various oceanic effect are even worst than clouds.
Though what some of what European might thought as warming effect and thought had of do
with a greenhouse effect, was actually the warming effect of Europe by the Gulf Stream.

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13. Max Polo says:

@Rocky

As I said earlier, in case my theory is wrong, I have a plan B ready: filing a patent for an innovative, energy-efficient, environment-friendly metal melting process based on the AGW rationale.
Let’s consider for instance Aluminum melting industry (this example is inspired by my customer, an Alu furnace manufacturer).

We can think of the following set up.
Take two stainless steel plates heated by resistors. Put an Aluminum sheet in between, in physical contact with both steel plates. Turn the electric resistance on and heat the steel plates up to 512°C. Wait for the Aluminum temperature to settle to 512°C (thermal equilibrium). Move the two steel plates away a few centimeters from the Alu plate (of course, some supports are required under the Alu plate). This is the moment in which the massive power of the AGW rationale comes into place: the Aluminum plate temperature will bump-up to 660°C, thanks to the 21500+21500 = 43000 W/m2 hitting the Aluminum plate. At this point, you’ll see the Aluminum plate starting to melt like ice cream in the sun (660°C = Alu melting point).

Advantages : no need for refractory because stainless steel can manage temperatures around 500°C, significant energy saving because no additional power is consumed to increase the Alu sheet temperature from 512°C to 660°C, thanks to the amazing “teamwork” of the two fluxes.

No doubt it will be a breakthrough.

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1. Rocky says:

“I have a plan B ready: filing a patent for an innovative, energy-efficient, environment-friendly metal melting process based on the AGW rationale.”

The problem for you is that you don’t understand AGW rationale. If you did, then you would know that it does not claim to do what you are implying. Your problem is you don’t understand the simple concept of conservation of energy.

“This is the moment in which the massive power of the AGW rationale comes into place: the Aluminum plate temperature will bump-up to 660°C”

No it won’t. Let’s neglect end effects and view factors to make our lives simpler and assume this can be approximated as a set of infinite plates. Prior to separating the plates the electrical energy that you must supply per unit area of a single side at steady state is:

W = 2 sigma T0^4 (here T0 = 512C) -> T0 = (W/2/sigma)^(1/4)

Let’s assume that this means that each plate has electrical energy in of W/2 = sigma T0^4. Now let’s separate the plates and maintain the SAME electrical energy input. Due to the symmetry of the problem we will get a new temperature for the steel plates, Tsteel, and a new temperature for the aluminum sheet, Talum.

Left (or right) steel plate energy balance:

W/2 = sigma (Tsteel^4 – Talum^4) + sigma Tsteel^4

The first term on the right is the heat transferred from the steel to the Aluminum, and the second term is the energy emitted from the other side of the steel to space.

Aluminum sheet energy balance:

2 sigma (Tsteel^4 – Talum^4) = 0

The factor of 2 come from the Aluminum receiving heat from the steel on both sides.

Well look at that! Our energy balance equations tell us that AT STEADY STATE there is zero HEAT transfer from the steel to the aluminum, that the aluminum sheet will be the SAME temperature as the steel plates, and finally that the temperature of the steel plates REMAINS AT 512°C.

So THAT is what AGW rationale says will happen, which YOU obviously do not understand.

“No doubt it will be a breakthrough.”

Indeed. Because what you have described is not the correct solution.

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1. Rocky says:

“No doubt it will be a breakthrough.”

Sorry, I misread that. Clearly I have proven that what you have stated is physically incorrect, and so no breakthrough is forthcoming.

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2. Max Polo says:

Fine, I agree with your solution.
But in the statement of the problem, I was never worried about the electrical input required. I should have said, “let’s take two plates at a fixed temperature, whatever the power requirement is”

Let’s assume that we are in this case: the temperatures are fixed.

Do you agree that they both emit a power flux sigma*T^4 towards each other?
If you agree, why are you not summing the two fluxes, since you claimed this to be the right thing to do?

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1. Rocky says:

“Let’s assume that we are in this case: the temperatures are fixed.”

You really are slow. The solution shows that when you separate the plates the temperatures remain the same and the power requirement remains the same.

Do try to keep up.

“Do you agree that they both emit a power flux sigma*T^4 towards each other?”

Once the plates are separated each plate emits this flux out from each of their surfaces.

“If you agree, why are you not summing the two fluxes, since you claimed this to be the right thing to do?”

The heat flux is the VECTOR sum of the two fluxes. I AM summing all of the fluxes to get the heat flux. Are vector sums something that you learned when you studied physics?

The HEAT transfer from one side of the aluminum to the facing side of the steel is the flux emitted by the aluminum to the steel MINUS the flux emitted by the steel to the aluminum. Since the aluminum and the steel are all the same temperature these fluxes all cancel one another out to give zero heat flux.

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14. Max Polo says:

@Rocky

you keep claiming that I said that “back radiation” does not exist. I never intended to say this (bear with me, I am not a native english speaker…).

What I claim is the following: backradiation (in the sense of the algebraic term sigma*T2^2) does exist, but whether or not it is capable of raising the temperature of the emitting body T1, depends on whether or not this T2 is kept fixed. If T2 is fixed, then backradiation can push the temperature of the emitting body up, otherwise, it cannot. The reason is that in the former case, new energy is introduced into the system (via the heat sink), while this does not hold true in the latter case.

I am OK if the above is nonsensical: I have that exciting plan B in my pocket (my other post on the AGW Aluminum melting furnace).

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1. Rocky says:

“What I claim is the following: backradiation (in the sense of the algebraic term sigma*T2^2) does exist, but whether or not it is capable of raising the temperature of the emitting body T1, depends on whether or not this T2 is kept fixed.”

This is absolute nonsense. ALL energy input is capable of raising the temperature of a body. Whether or not the temperature of a body actually doe increase depends on whether the energy input exceeds the energy output. That’s it. There are no other considerations required. In fact that is ALL that you can consider as DICTATED by the first law of thermodynamics. PERIOD. FULL STOP.

If you agree that “the algebraic term sigma*T2^2 does exist” then you MUST account for it in the fist law of thermodynamics and it DOES have an effect on temperature.

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1. Max Polo says:

“ *from [4] we see that the heat that the Sphere is transferring to the Shell, makes up for just one-half of Shell’s total emission* — WRONG. The HEAT that the Sphere is transferring to the Shell makes up for ALL of the Shell’s emission to space…….this is again WRONG”

Sorry, I don’t understand this.
Didn’t you say that you agreed on my equations (2a), and (3) ?
From (2a) : sigma*T2^4 = (sigma*T1^4)/2
To get equation (4), I’ve just substituted the value of sigma*T2^4 from (2a) into equation (3), that you both said you agreed on.
This gets you the following expression for Q12 :
Q12 = sigma*T2^4.

Is Q12=sigma*T2^4 not equal to the half of 2*sigma*T2^4^4?

Is one not equal to the half of two?

“Finally, I will add that your approach is in disagreement with Zoe’s”

She is way above me.
This time I humbly beg to disagree.

Let’s say I agree with her …to the 97% 🙂

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1. Max Polo says:

Obviously I meant to say :
To get equation (4), I’ve just substituted the value of sigma*T1^4 from (2a) into equation (3).

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2. Rocky says:

Here is your equation 3: “Q12 = sigma*(T1^4-T2^4) [3]”

This is the HEAT transferred from the sphere to the shell. The first law at steady state dictates that the energy supplied by the source to the sphere must be equal to the HEAT transferred out from the sphere to the shell.

That means that Q12 must be equal to W.

So, your equation 3 becomes W = sigma*(T1^4-T2^4). Now plug in your sigma*T2^4 = (sigma*T1^4)/2 to arrive at W = sigma T1^4 – (sigma*T1^4)/2 = (sigma*T1^4)/2.

So according to your equations T1 = (2W/sigma)^(1/4).

“Is Q12=sigma*T2^4”

Yes T2 = (W/sigma)^(1/4)

“…not equal to the half of 2*sigma*T2^4”

Yes, W/sigma is 1/2 of 2W/sigma. I’m hoping that eventually you will get it. You’re pretty slow though.

“She is way above me.”

Indeed. You can’t even understand this simple problem and solve it correctly.

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3. Rocky says:

Have you still not figured out that Q12 = W?

The energy supplied to the sphere, W, must be balanced by the HEAT, Q12, out from the sphere. This is dictated by the first law of thermodynamics.

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15. Jarle says:

Rocky,

Only by applying the scientific method can we determine whether CO2 will cause what it is said to cause. And climate science has a veeeery long way to go in that sense. I agree that the GHE is what you say it is. It is on the borderline of strawman technique to imply that I don’t. You are missing my point, and you love to miss my point. BTW. In my previous post I meant “reserves” and not “resources”

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1. Rocky says:

“It is on the borderline of strawman technique to imply that I don’t.”

That’s funny. You accuse me of a strawman when in fact it was you who set up a strawman definition of what the GHE is.

“Only by applying the scientific method can we determine whether CO2 will cause what it is said to cause.”

Been there, done that.

“You are missing my point”

No. I am refuting your inaccurate and nonsensical claims.

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16. Jarle says:

Maybe you should read Donna Labframboise’s book the Delinquent Teenager to get a sense of how climate science operates.

Liked by 1 person

1. Rocky says:

Maybe you should take physics 101 to get a sense of how science operates. Your arguments are pathetic.

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1. Jarle says:

They are not pathetic as clearly demonstrated by the climate gate emails. I was referring to CLIMATE science, and not science in general. Climate science must be the least successful science of all sciences, judging by the absence of all its doomsaying and failing of models. It is shameful. A disgrace to all of science how they operate.

Liked by 1 person

1. Rocky says:

They certainly are pathetic. You have a juvenile understanding of science and most likely base you opinions on the topic at hand from your political views.

“I was referring to CLIMATE science, and not science in general.”

Yes, I know. And if you had any understanding of science in general then your would understand that climate science follows the same tenets.

“Climate science must be the least successful science of all sciences”

How do you explain that James Hansen provided predictions for the amount of warming that we should see for an amount of added GHG forcing back in 1988, and 30 years later his predictions turned out to be correct?

The disgrace here is you. You either purposefully or ignorantly misrepresent the subject.

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2. Do not harass my regular commentors. I will boot you. I have allowed you to voice your opinion, but I will not tolerate you if you persist like this. Tone it down. If you’re confident in your opinion, you don’t need to deride people you feel are wrong. Thanks

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3. Tom Dayton says:

Zoe, you routinely call people “stupid,” “retard,” and “liar.”

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4. Yes. My house, my rules. I can’t let my guests harass other guests. Do you own any property? Have you ever invited people over?

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17. Jarle says:

Why do you think that someone goes to the extent of leaking hundreds of emails in the first place? It is because they possess knowledge that something is terribly wrong. We have seen that Zoe presents net solar@surface as a very good candidate for the observed recent warming. The climate gate emails reveal how they discuss that Henrik Svensmark should not be allowed to be an author in the IPCC reports. What a surprise.

Liked by 1 person

1. Rocky says:

What was wrong?

What SCIENCE was uncovered to be WRONG?

Your arguments are absolutely pathetic. You avoid discussing the actual science at all costs. Oh wait, I forgot, you actually tried to make a scientific statement but you got that horrendously wrong. Then you go on a Gish Gallup just like deniers of science across all field do. Seriously, it’s pathetic.

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2. Rocky says:

“We have seen that Zoe presents net solar@surface as a very good candidate for the observed recent warming.”

The net solar at surface is affected by cloud cover. The reduction in albedo is PREDICTED by climate models. So yet again you prove that you have no idea what you are talking about or how the scientific method works. Changes in ASR are a known feature of AGW.

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1. Jarle says:

And the same goes for the theory of Henrik Svensmark. It PREDICTS less clouds. So why would they conspire to prevent him from being an author? When did it become scientific to operate like this?

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1. Rocky says:

Damon and Laut (2004) “Pattern of strange errors plagues solar activity and terrestrial climate data” EOS vol. 85 pp. 370-374 discusses the problems with Svensmark’s work.

Here’s a plot of cosmic ray activity. Are temperatures supposed to go up or down with cosmic rays?

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2. Really? So CO2 now reduces clouds, is that correct? That’s the NEW “greenhouse effect”?

I’ve seen many papers discuss CO2 and ASR in parallel, but they have yet shown how the causation works. The idea is by talking about both things simultaneously, extremely stupid people will just tie them together and nod their heads. It’s just a rhetorical trick: a logic fallacy.

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1. Rocky says:

“Really? So CO2 now reduces clouds, is that correct? That’s the NEW “greenhouse effect”?”

No Zoe. Warming is associated with feedbacks. The warming that is initiated by CO2 has feedbacks that increase ASR.

If you want a paper the go find, Donohoe et al. (2014) “Shortwave and logwave contributions to global warming under increasing CO2”, PNAS vol. 111 pp.16700-16705.

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2. You just confirmed what I said and didn’t notice it.

If CO2 actually caused increased ASR (which you obfuscate with other words like “feedback” and “initated” while saying “No Zoe”) then we would see LONGWAVE exceed SHORTWAVE, not longwave perfectly match shortwave.

The textbook theory is that CO2 can increase surface longwave WITHOUT any changes in shortwave.

Yes, that paper you linked is exactly the kind of rhetorical trickery I’m talking about.

Jarle can easily see it. Why can’t you?

If longwave exceeded shortwave, then I would honestly have to accept AGW. But it didn’t and so I can’t.

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3. Rocky says:

I’m curious if you and I are now in agreement on the steel greenhouse problem for the case where the internal energy source maintains a fixed energy flux.

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4. Rocky says:

“You just confirmed what I said and didn’t notice it.”

How so? It’s not CO2 that reduces clouds, it’s warming. I suggest you read the paper that I provided the citation to in order to understand the distinction. CO2 initiates warming and warming induces cloud cover and water vapor changes. It’s a causality chain.

“The textbook theory is that CO2 can increase surface longwave WITHOUT any changes in shortwave.”

I suggest that you read the paper so that you can expand beyond such simplified descriptions. Again, there is a causality chain. CO2 increases first. This causes a reduction in the net radiation out at TOA. This causes warming inside the TOA. Warming causes water vapor and cloud cover changes which increase ASR.

“Yes, that paper you linked is exactly the kind of rhetorical trickery I’m talking about.”

Rhetorical trickery? It shows the theory and explains the logical chain of causality. It’s really not that difficult to follow.

“Jarle can easily see it.”

Jarle doesn’t even understand what the greenhouse effect is. I don’t put much stock in what he “sees”.

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5. You really lack critical thinking skills. You confuse models with observations. Look at the data for heaven’s sake.

“This causes a reduction in the net radiation out at TOA.”

lol. never happened

It blows my mind how gullible you are. You will support any rhetoric that amounts to “whatever we observed … CO2 did it”.

Seriously, anyone with integrity can see what these charlatans are doing.

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6. Rocky says:

“You really lack critical thinking skills. You confuse models with observations. Look at the data for heaven’s sake.”

It’s not me that is confused. The data shows increased ASR and the models show that is to be expected during AGW. What exactly is confusing about that?

““This causes a reduction in the net radiation out at TOA.”
lol. never happened”

Why do you make that baseless claim?

The rest of your comment is spewing bile. I provided you with the paper that explains all of this clearly. You are not refuting it.

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7. “that is to be expected during AGW.”

Rhetoric. Whatever happened … AGW “explains it”.

“Why do you make that baseless claim?”

Because I looked at the data, dummy.

https://phzoe.com/2022/06/10/20-years-of-climate-change/

“I provided you with the paper that explains all of this clearly.”

I pre-empted your paper.

Dumb dumb, can you explain why they avoid showing a model of the surface but focus on TOA?

I’ll repeat this again so you can properly focus your energy:

If CO2 really did initiate ASR, then surface longwave changes should EXCEED shortwave changes. It doesn’t. PERIOD.

Read my “20 years” link carefully. Look at the data.

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8. Rocky says:

“Rhetoric. Whatever happened … AGW “explains it”.”

I don’t think you understand what “rhetoric” means. The physical models, i.e. models based on physics, predict that AGW results in increased ASR. Shouting “rhetoric” does not refute the physical models. In fact, all you are doing can be considered “rhetoric”. Try some science.

“Because I looked at the data, dummy.”

Calm down Zoe, there’s no need for name calling. CO2 started increasing far longer than 20 years ago. I thought that would be obvious to you. Again, the models predict what is seen in the observations.

“I pre-empted your paper.”

Actually you didn’t. I have to wonder if you have even read the paper, let alone understand it.

“Dumb dumb, can you explain why they avoid showing a model of the surface but focus on TOA?”

So you’ve confirmed that you do not understand the paper. See Equation 1.

“If CO2 really did initiate ASR, then surface longwave changes should EXCEED shortwave changes. It doesn’t. PERIOD.”

I suggest you read the paper for what should be observed and when. You 20 years of observations does not suffice.

If you’d like I can teach you how to create a simple modified time-dependent steel greenhouse model that shows all of these effects that the GCMs show.

Are you game?

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9. It’s like arguing with a communist. No thanks. You are not convincing anyone.

ASR alone can explain it all. Talking about CO2 simultaneously doesn’t show causation.

My data shows outgoing TOA increasing by 0.15 W/m^2. Your paper says CO2 should decrease TOA and then slowly crawl up to zero. Then they show ASR, and your gullible brain is supposed to fill the gap with AGW. lol

I can’t argue with such gullible people. It’s best to avoid idiots.

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10. I’ll give you another shot.

Let’s say CO2 did induce an increase of longwave from surface by 0.8 W/m^2. This then melted some ice (changed albedo) … somehow reduced cloud cover … and these 2 effects allowed an extra 1.2 W/m^2 solar to reach and be absorbed by the surface.

What is the total available extra flux at the surface?

It should be 2 W/m^2. Do you agree? If not, explain.

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11. Rocky says:

“You are not convincing anyone.”

The scientific community has already been convinced. Non-scientists have their own non-scientific reasons for not accepting the scientific findings that they cannot refute.

“ASR alone can explain it all.”

Sorry Zoe, but perhaps this piece of logic has eluded you. ASR alone cannot explain why ASR increased. AGW does explain why ASR has increased and you have not refuted any of that science.

“My data shows outgoing TOA increasing by 0.15 W/m^2.”

Indeed, and the paper that I provided to you shows a similar increase in outgoing LWR after sufficient time.

“Your paper says CO2 should decrease TOA and then slowly crawl up to zero.”

I assume that you mean it shows the outgoing LWR should decrease and then slowly crawl up to zero. That’s right. You don’t have the data that goes back far enough to show that this did not happen. Hence your claims that it never happened are baseless.

“I can’t argue with such gullible people. It’s best to avoid idiots.”

That’s quite a lot of venom and bile Zoe. I’m sorry to have to break the news to you, but no one of consequence is convinced by your rhetoric.

Are we in agreement for our solutions to the steel greenhouse problem with a constant energy source? Max Polo does not appear to agree with the procedure that you posted.

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12. “AGW does explain why ASR has increased ”

No, it actually doesn’t. Where is that in the paper? An ASSERTION is not evidence. Modeling two things simultaneously is not causation.

I can see ASR increasing 0.15 W/m^2 on its own (details omitted), YOU need proof that CO2 reduced OLR by 0.5 and ASR increased it by 0.65 (made up numbers, but illustrates the point). You have proof for NEITHER and yet you strut around as if you do because your model says so. Well guess what? I ran 40mph yesterday and set a world record. Unforunately there was a lot of wind so it looked like 18mph. When am I going to get the recognition?

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13. Rocky says:

“I’ll give you another shot.
Let’s say CO2 did induce an increase of longwave from surface by 0.8 W/m^2.”

You got step 1 wrong. That’s not what happens at the beginning.

Would you like to take another shot?

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14. CO2 already changed emissivity is my first step. Quit stalling and answer the simple question.

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15. Rocky says:

@Zoe – “Explain”

I’m happy to walk you through a simple steel greenhouse model that shows all of these effects and can answer all of your questions.

This type of thing seems to be something that you like working on and coding. It should be fun.

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16. Quit stalling and answer the question.

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17. Rocky says:

You need to understand the basic physics in order to understand the answer to your question.

Are you ready to start?

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18. Rocky says:

“Where is that in the paper?”

Are you kidding?

“An ASSERTION is not evidence.”

Indeed. All you have been doing here is making assertions. The paper that I provided you provided a quantitatively supported explanation of the observed behavior. You have done nothing to refute it.

“Modeling two things simultaneously is not causation.”

This is a meaningless statement. The model shows exactly how the fundamental physics associated with AGW predicts the observations. You have done nothing to refute it. All you have done is make baseless assertions.

Go through the model yourself. It’s not that difficult.

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19. The model says (Fig 1A) “let’s assume OLR is instantaneously reduced by 4 W/m^2” … “it will take ~140 years to get back to ZERO”.

It is popularly known that CO2 has increased from ~280 ppm to ~420 ppm in the last ~250 years.

Do you disagree with this general assessment?

The forcing should be:

5.35 * ln( 420 / 280) = 2.17 W/m^2

Disagree with this basic formula? Remember, they are not modeling any water vapor feedbacks, so 2.17 W/m^2 is all they have here.

If it takes 140 years to return to zero for a sudden 4 W/m^2 drop …

What would happen in 250 years of gradual combined drops totaling 2.17 W/m^2 ?

“The model shows exactly how the fundamental physics associated with AGW predicts the observations.”

LMFAO

How you could think this model is informative of “exact” reality is so hilarious.

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20. Rocky says:

You’re the one stalling here Zoe. The first step is to confirm that you are in agreement with the principles associated with solving the steel greenhouse model. Then we can add complications. We can give the shell emissivity that increases over time. We can give the sphere a heat capacity. We can have the energy input to the sphere depend on the temperature of the shell. Finally, we can generate the transient solution and see how all of the energies move through the system.

Are you ready, or are you going to continue stalling?

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21. You don’t agree that an increase in CO2 will result in higher surface temperature and higher longwave flux? OK then

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18. Tom Dayton says:

Jarle, Svensmark has not been prevented from being an author. He has published. His publications have been criticized by scientists is subsequent publications. His claims unequivocally have been shown empirically to be incorrect. An easy entry to that literature is here. After you read the Basic tabbed pane, read the Intermediate one and then the Advanced one. Links to scientific publications are in those tabs.

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1. Ah, so cosmic rays must match temperature perfectly, but CO2 doesn’t. Interesting standard.

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2. The CHART shows GCRs declined while Temps rose, exactly as predicted. While the text says:

“On the contrary, while GCRs are up, global temperatures are also way up, and temperatures in recent years reached record highs.”

The text doesn’t match the chart. Lying straight to your face – a favorite technique of skepticalscience.

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1. Tom Dayton says:

Zoe, you are interpreting the GCR vertical axis upside down. Read the caption.

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2. So nice of them to flip it upside down!

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3. gbaikie says:

https://spaceweatherarchive.com/2019/10/03/cosmic-rays-are-nearing-a-space-age-maximum/

We are in beginning Solar Max, and amount GRC is less, but not low for past Solar Max [before solar cycle 24]. Zero is average over last 50 year or so. Now, it’s at +1% in last couple week dip down as low at -2%. With Solar Max, getting well below -5% was typical during Solar Max, and getting well above +5% was not typical [until recently] with Solar Min.
For highest amounts GCR it would be 2007 to 2011.
But we have seen low levels since 2005, so for long term trend it could 2005 to now or could look between 2007 to 2011 most year of higher GRC.
And shouldn’t look at temperature, should look if more clouds were formed or occurring where don’t normally happen as often. Or something. We not having much hurricane forming, I wouldn’t imagine
this due to GCR levels, but I know very little about it, but yet to see strong evidence supporting it, but could a lot GCR starting after about 4 to 5 year when in solar min and possible we don’t see low levels of GRC within next few decades.
BUT solar weather is still as unpredictable as Earth weather.

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1. Jarle says:

GCR at TOA down => less clouds=> more GCR comes down to the surface.

Could be the explanation.

In most of the previous century we had what is referred to as a grand solar maximum. Meaning a very active sun that must have shielded us from cosmic rays if that meant also a stronger magnetic field of the sun.

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1. gbaikie says:

A hydrogen molecule or compound of H, as in H20 is shielding to GCR. Whereas lead, steel or something without Hydrogen will create Secondary Radiation or gives neutron counts which is how GRC is measured.
Or it’s only when GCR hit something and violently explode, that it can be detected.
It’s like when we accelerate a proton and smash it into stuff creating smaller particles.
A very very small but very very violent explosion.
The sun also accelerates protons but never get as high velocity as GCR, but I am not sure how the moderately high velocity solar protons would effect neutron counts.

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19. gbaikie says:

But if you want to look at temperature, wiki, Global warming hiatus:
“Publicity has surrounded claims of a global warming hiatus during the period 1998–2013. The exceptionally warm El Niño year of 1998 was an outlier from the continuing temperature trend, and so subsequent annual temperatures gave the appearance of a hiatus: by January 2006, it appeared to some that global warming had stopped or paused. A 2009 study showed that decades without warming were not exceptional,[6] and in 2011 a study showed that if allowances were made for known variability, the rising temperature trend continued unabated.[6] There was increased public interest in 2013 in the run-up to publication of the IPCC Fifth Assessment Report, and despite concerns that a 15-year period was too short to determine a meaningful trend, the IPCC included a section on a hiatus, which it defined as a much smaller increasing linear trend over the 15 years from 1998 to 2012, than over the 60 years from 1951 to 2012.”
https://en.wikipedia.org/wiki/Global_warming_hiatus

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1. gbaikie says:

And we in an even shorter “hiatus” right now.
In terms of decades the trend it is .13 C per decade, it think it lower to .12 C within 1/2 year or a year of time. Which some might call “massive cooling” but I would not say it is.
Instead, it would confirm to me, that we have stopped recovering from the Little Ice Age.
Or presently I think we have stopped recovering for LIA AND future measurement of average temperature will confirm my wild and uneducated guess.

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20. Max Polo says:

“ **first of all, you are summing fluxes with nonchalance…** What the hell does that mean? Point to the object where the energy in is not balanced by the energy out. Simple addition and subtraction are just that, SIMPLE. There is nothing nonchalant about basic arithmetic…….. WTF are you talking about etc etc etc”

In heat transfer textbooks, there are these things called “re-radiating surfaces”. These are defined as surfaces facing each other that are kept at the same temperature T and do not exchange heat. Example: two energized panes like those in my (fake) “Aluminum melter”. Textbooks, and Stefan-Boltzmann, say that both panes emit the entire sigma*T^4 towards each other. No textbook (as far as I know) says that each surface emits (sigma*T^4)/2 to maintain the temperature by summing the two half-fluxes. They couldn’t say this because they would violate Stefan-Boltzmann law.

Why summing fluxes doesn’t work? I have found this paper from Ross Mc Leod at the link below (excerpt) that gives some hints: we know that sigma*T^4 comes from the integration of the Planck curve for the temperature T. If you sum wavelength-by-wavelength the two Planck curves for the two fluxes that you are adding to obtain T, then what you get is not a Planck curve as you would expect, but something else. This non-Planck curve has a peak at a wavelength that does not even respect Wien law. Then what you get with the sum is not the Planck emission of a body at temperature T.

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1. gbaikie says:

We are living in interglacial period, which is called Holocene.
Prior to this warming period, about 20,000 years, Earth average temperature was much cooler.
But perhaps more important is was a lot more drier world.
Or a cool earth is dry earth, and warm Earth is wet earth.
We live in an Ice Age which means Earth is cooler and drier.
So, 20,000 years ago, Earth is colder and drier than it is now.
And now is a dry world with 1/3 land area as deserts.
The highest temperature ever officially recorded:
“Currently, the highest officially registered temperature is 56.7C (134F), recorded in California’s Death Valley back in 1913.”
Which doesn’t mean there was higher global temperature back in 1913
And instead is suggests 1913 was drier [and lower global temperature].
Or warmer global temperature doesn’t mean highest daytime temperatures
recorded, but rather quite the opposite.
Or India has average temperature of about 25 C, and India doesn’t break world highest temperature
records.
So, if you back in time 20,000 years ago on Earth when is was coldest Earth is known to ever been,
and properly measured air temperature, you might get air temperature in some day, higher
than “California’s Death Valley back in 1913”
Or drier gives more temperature extremes- both cold and hot.
And about 10,000 years ago, the sahara deserts, was mostly grassland, with lakes and rivers and trees that disappeared 5000 years ago, when the Earth cooled.
Or if we could get enough global warming, at some point the Sahara desert would green and Earth would have much warmer global average temperature.
But that seems unlikely. But we could make the Sahara green {there are people are trying], and artificially making the Sahara desert, green, would likewise increase global temperatures by some amount. And such greened areas [if covering large enough area] would have less temperature extremes- and a higher average temperature than the Sahara desert, now.

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2. Max Polo says:

Thanks, Jarle, much appreciated. I’ll surely go through it. Have a nice day.

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1. Rocky says:

“Textbooks, and Stefan-Boltzmann, say that both panes emit the entire sigma*T^4 towards each other”

Indeed. That is EXACTLY what my solutions have. Look at the equations carefully and read the explanations carefully.

“Why summing fluxes doesn’t work?”

Summing fluxes DOES work. You are confused if you believe otherwise.

“If you sum wavelength-by-wavelength the two Planck curves”

That’s not summing the total fluxes Max. That is not what I have done in my calculation. The total flux is the AREA UNDER THE PLANCK CURVE. You have to add the total areas under the curve to get the total flux.

Face it. You are wrong.

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21. Rocky says:

“You don’t agree that an increase in CO2 will result in higher surface temperature and higher longwave flux? OK then”

So you’ve now been reduced to arguing like a 12 year-old.

Look at Figure 1. CO2 does eventually lead to higher temperature and higher longwave flux. INITIALLY the longwave flux out of the TOA is reduced.

Seriously, read and UNDERSTAND the paper.

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1. I’m talking about the SURFACE. Stop referring to the paper and TOA, and answer the simple question. Guess you can’t.

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1. Rocky says:

The paper talks about the surface to Zoe.

You can’t even formulate a proper question because you don’t understand the causality chain.

Again, I can walk you through a simple sphere-shell model the explains all of this.

INITIALLY, increasing CO2 causes a DECREASE in outgoing LWR. THAT causes an INCREASE in the temperature of the surface. An INCREASE in the temperature of the surface causes an increase in the temperature of the atmosphere, an increase in the outgoing LWR, and a reduction in albedo which causes an INCREASE in ASR. All of these changes are TIME DEPENDENT.

Are you starting to get it yet? If not then I suggest that you SOLVE A MODEL for yourself so you can UNDERSTAND it.

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1. Retard, my example is after time has already passed. Do you understand this? Now answer my simple question, and stop referring back in time.

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2. Rocky says:

More childish name-calling Zoe? What a shame. You are missing an opportunity to learn.

“my example is after time has already passed”

How much time has passed Zoe?

Again, the paper, and the model that I can teach you how to solve show how all of the quantities evolve in time.

Do you understand this?

“Now answer my simple question, and stop referring back in time.”

You will first have to ask an intelligible question.

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3. The amount of time that passed doesn’t matter. I am showing you the results. Results that can be measured by satellite. You as a scientist now need to make sense of these values. What is your response? “Zoe stop looking at numbers, muh time dependency!”

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4. Rocky says:

“The amount of time that passed doesn’t matter.”

Of course it does Zoe. Look at Figure 1d. LWR and ASR have very interesting variations in time. Interestingly the simple sphere-shell model that I developed shows the same qualitative behavior.

“I am showing you the results.”

Yes, you are showing results after a certain point in time.

“Results that can be measured by satellite.”

Indeed. Unfortunately we did not have satellites measuring these results back when we started pumping CO2 into the atmosphere.

“You as a scientist now need to make sense of these values.”

Indeed. I have. As has Donohoe. I can’t really help it if you don’t have the ability to understand how these values make sense.

“What is your response? “Zoe stop looking at numbers, muh time dependency!””

Correction, my response is to challenge you to solve a model so that you too can gain the ability to understand the numbers you are looking at.

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5. You first need to convert Donohoe’s model to the SURFACE. When you do that then you can begin to understand what I’m asking. But you can’t do that, because you’re nothing but a troll.

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6. Rocky says:

“You first need to convert Donohoe’s model to the SURFACE.”

Are you kidding? Ts in Donohoe’s paper is the SURFACE temperature.

Again, if you will step up to the challenge I can show you a similar sphere-shell model that analogizes the surface and atmosphere DIRECTLY. It give the SAME QUALITATIVE PICTURE as Donohoe’s paper.

“When you do that then you can begin to understand what I’m asking.”

It’s been DONE Zoe. If you don’t understand that then you don’t even understand why your questions are unintelligible.

Seriously, why are you avoiding learning this simple sphere-shell model?

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7. The model in Fig 1 is not the surface, and I’m talking about surface flux changes, not any Ts he may compute. But Ts changes are obviously flux changes, and so why don’t YOU put your attention into that to answer my question.

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8. Wait, are you for real? The Sphere-Shell model is now a good way to model ASR? OMG you are bonkers.

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9. Rocky says:

“The model in Fig 1 is not the surface, and I’m talking about surface flux changes, not any Ts he may compute.”

Oh my Zoe. Are you unable to relate surface temperature to surface flux?

You require far too much hand-holding for something so simple.

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10. I am certainly able to do that. That’s why I know the flux should increase. You keep referring me to TOA, where flux drops and then returns to zero.

Now answer my question to show that YOU can understand surface fluxes!

Let’s say CO2 did induce an increase of longwave from surface by 0.8 W/m^2. While that happened … some ice melted (changed surface albedo) … somehow cloud cover was reduced (changed atmo albedo)… and these 2 effects allowed an extra 1.2 W/m^2 solar to reach and be absorbed by the surface.

What is the total available extra flux at the surface?

It should be 2 W/m^2. Do you agree? If not, explain.

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11. Rocky says:

“Wait, are you for real? The Sphere-Shell model is now a good way to model ASR? OMG you are bonkers.”

It’s a good way for you to learn about the causality change.

Do try to stay away from straw man arguments Zoe.

Why don’t you step up to the challenge and learn how to solve the sphere-shell model?

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2. Rocky says:

“That’s why I know the flux should increase.”

Yes Zoe, we all know that when the surface warms it emits more radiation.

So what?

“Let’s say CO2 did induce an increase of longwave from surface by 0.8 W/m^2. ”

So one example of this, if we approximate the surface as having an emissivity of ~0.95, would be that the surface warmed from 288K to 288.16K.

“While that happened … some ice melted (changed surface albedo) … somehow cloud cover was reduced (changed atmo albedo)… and these 2 effects allowed an extra 1.2 W/m^2 solar to reach and be absorbed by the surface.”

OK, so ASR increases by 1.2W/m^2, and the outgoing radiation increases by 0.8W/m^2.

“What is the total available extra flux at the surface?”

The total extra flux into the surface would be the extra amount absorbed from the sun, 1.2 W/m^2, minus the extra amount emitted by the surface, 0.8 W/m^2, plus any extra amount of radiation absorbed from the atmosphere, which you have not specified, minus any extra latent and sensible heat transferred from the surface to the atmosphere, which you also have not specified.

It should be 2 W/m^2. Do you agree? If not, explain.”

See above. I don’t think you have a sufficient understanding of the energy balance for the surface.

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1. That’s an interesting solution. I said CO2 will cause extra 0.8 W/m^2 outgoing. You don’t think the extra 1.2 W/m^2 from ASR will also cause its own 1.2 W/m^2 to be emitted?

Would you like to try again, or are you sticking with your answer.

“extra latent and sensible heat transferred from the surface to the atmosphere, which you also have not specified.”

That’s included. Assume CERES standard. We have full accounting.

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22. Rocky says:

“The model says (Fig 1A) “let’s assume OLR is instantaneously reduced by 4 W/m^2” … “it will take ~140 years to get back to ZERO”.

What is your problem with this? It is using the model to probe the behavior.

“It is popularly known that CO2 has increased from ~280 ppm to ~420 ppm in the last ~250 years.
Do you disagree with this general assessment?”

Yes, this is a known fact..

“The forcing should be:
5.35 * ln( 420 / 280) = 2.17 W/m^2”

So what? The model is being used to illustrate behavior. The GCMs have already been used to study the historical evolution.

“Disagree with this basic formula?”

No. That formula is fine.

“Remember, they are not modeling any water vapor feedbacks, so 2.17 W/m^2 is all they have here.”

Again, they are not studying the historical evolution in this paper. That has already been done. They are introducing a simplified model to illustrate the causality chain that occurs in the physical system.

“If it takes 140 years to return to zero for a sudden 4 W/m^2 drop …”

That is only for the case if there is no SW feedback. We live in a system where there is SW feedback. You need to read the paper more carefully. I suggest that you actually solve a model. That way you will get a better understanding of what is going on.

“What would happen in 250 years of gradual combined drops totaling 2.17 W/m^2 ?”

This question makes no sense.

““The model shows exactly how the fundamental physics associated with AGW predicts the observations.”
LMFAO”

Again, you are acting like a 12 year old.

“How you could think this model is informative of “exact” reality is so hilarious.”

How can you think this model is not informative or reality. It provides you with the causality chain that you were not knowledgable of. It explains what is seen in more detailed climate models which predict the observation that AGW results in increased ASR.

I’m sorry that you thought that increased ASR was a dagger to the heart of AGW. The reality is that you didn’t understand the details of the science well enough to know that AGW had already predicted this observation.

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1. Wow! You are complete garbage.

The paper does not prove a SW feedback. It merely models LW and SW side by side. Where is the causality?

SHOW IT!

There isn’t.

You don’t have the brains to eithet analyze or explain in your own words.

You are DONE!

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1. Rocky says:

As always you resort to childish name-calling when you can’t address the science.

“It merely models LW and SW side by side. Where is the causality?”

Indeed it does model it. I’ve explained the causality to you multiple times now. WARMING, no matter what the cause, causes changes in water vapor and cloud cover that increases ASR.

“SHOW IT!”

Go look at the physics that is included in GCMs.

“You don’t have the brains to eithet analyze or explain in your own words.”

Correction, I don’t have the ability to explain it at a level that someone with your lack of expertise can understand.

“You are DONE!”

Typical.

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1. “I’ve explained the causality to you multiple times now. WARMING, no matter what the cause, causes changes in water vapor and cloud cover that increases ASR.”

Comedy gold right there!

You explained to me multiple times that no matter the cause causes what you want it to.

The physical mechanism? “Doesn’t matter”

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2. Rocky says:

Do you understand that warming causes changes in water vapor and cloud cover Zoe?

That is the mechanism you are asking for. I’ve already explained this to you. It is also explained in Donohoe’s paper.

You can lead a horse to water, but you can’t make them drink.

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3. After globalists conquer CO2, it will be Argon that causes changes in water vapor and cloud cover. Don’t you know?

There is no explanation in Donohoe’s paper. It is nothing but a post-hoc and affirming-the-consequent logical fallacy. And you haven’t shown otherwise.

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4. Tom Dayton says:

Zoe wrote “There is no explanation in Donohoe’s paper. It is nothing but a post-hoc and affirming-the-consequent logical fallacy.”

Zoe, scientific papers do not repeat all the background for the particular points being explained. If they did that, every paper would be the length of several encyclopedias, and would bury the main points in the background. Instead, scientific papers refer readers to previous publications, for readers to learn the background. But papers do not even do that for background that is so fundamental that the intended audiences of the papers easily can be assumed to be familiar with it.

It appears you are unfamiliar with the centuries-long, gradual accumulation of knowledge that is background for the Donohue paper. A good source is “The Discovery of Global Warming” by physicist and science historian Spencer Weart: https://history.aip.org/climate/index.htm#contents

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5. I read Weart’s propaganda. The problem is you can’t think and explain things in your own words.

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6. Rocky says:

“After globalists conquer CO2, it will be Argon that causes changes in water vapor and cloud cover. Don’t you know?”

It’s basic climate science that warming causes changes in water vapor and cloud cover Zoe.

If we have to go all the way back to 1+1=2, it’s going to take you a while to learn all of this.

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7. Water vapor can absorb and reflect Shortwave.

Warming -> More water vapor -> More SW NOT reaching the surface.

The changes are not in your favor, you stupid troll.

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2. Rocky says:

“Water vapor can absorb and reflect Shortwave.”

Gases do not reflect radiation. The absorb or scatter radiation.

“Warming -> More water vapor -> More SW NOT reaching the surface.”

That’s not the physics Zoe. ASR includes solar radiation absorbed by the surface and the atmosphere. If ASR warms the atmosphere, that in turn also warms the surface. That is the benefit of looking at the TOA. Hopefully that helps clear up some of your confusion.

“The changes are not in your favor, you stupid troll.”

The science says they are in my favor. More name-calling Zoe? Grow up.

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1. “If ASR warms the atmosphere, that in turn also warms the surface.”

We know what ASR can do. The question is what CO2 can do.

CO2 is said to make it warmer. This is said to increase water vapor. Water vapor INCREASES atmo albedo, not reduces it. Therefore this should DECREASE ASR, not increase it.

Also, a warmer atmsophere will contain more gases. The thickness and pressure should go up. This allows for more layers of clouds to be present. This is why planets with high surface temperatures have thick atmospheres and lots of clouds that do not allow us to see through them. Again, this REDUCES ASR.

So pardon me for not seeing the physical mechanism by which CO2 increases ASR.

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23. Rocky says:

“I said CO2 will cause extra 0.8 W/m^2 outgoing.”

Right. If CO2 induces extra outgoing that means the the surface warms and emits more radiation. It is likely that the atmosphere also emits more radiation to the surface, but you did not specify how much.

“You don’t think the extra 1.2 W/m^2 from ASR will also cause its own 1.2 W/m^2 to be emitted?”

The surface emits according to its temperature. Any imbalance of energy in/out of the surface results in warming/cooling of the surface depending if there is more/less incoming energy.

“Would you like to try again, or are you sticking with your answer.”

My answer is correct. I suggest you show try learning more about what you are talking about. The best thing for you to do is to solve a model with a transient solution. You incorrectly think energy fluxes will balance during warming.

““extra latent and sensible heat transferred from the surface to the atmosphere, which you also have not specified.”
That’s included. Assume CERES standard. We have full accounting.”

It was not included in your statement. Whatever it is, plug it into my correct answer.

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1. So you don’t think that an extra 1.2 W/m^2 from ASR will raise temperature and cause more emission?

Would you like to try again?

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1. Rocky says:

“So you don’t think that an extra 1.2 W/m^2 from ASR will raise temperature and cause more emission?”

You’ve already specified how much more radiation the surface emits. Hence, in principle, we know how much the surface has warmed. As I explained to you, whether the surface warms or cools depends on the NET energy input to the surface, not only on one component.

“Would you like to try again?”

No. My answers are correct. You still don’t know what you are talking about.

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1. “You’ve already specified how much more radiation the surface emits.”

Here is exactly what I said:

“Let’s say CO2 did induce an increase of longwave from surface by 0.8 W/m^2”

This means this is what CO2, and CO2 alone theoretically did.

I did not suggest that is all.

“Hence, in principle, we know how much the surface has warmed.”

So you think 0.8 is all there is and the additional 1.2 goes nowhere?

Would you like to try again?

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2. Rocky says:

“This means this is what CO2, and CO2 alone theoretically did.”

OK. That is not how I read it.

Do you understand that it is impossible to measure what individual effect induces a change in emitted radiation if there are multiple effects at play?

So am I correct that what you meant to say is that CO2 causes an extra amount of radiation to be absorbed by the surface from the atmosphere of 0.8 W/m^2?

Am I to ignore any extra amounts absorbed from water vapor and other gases in the atmosphere?

“I did not suggest that is all.”

Perhaps there is a language barrier. Is English your first language?

“So you think 0.8 is all there is and the additional 1.2 goes nowhere?”

No. As stated the 1.2 is extra radiation that is ABSORBED by the surface. ABSORBED radiation does not cause the EMITTED radiation to be balanced when the system is warming or cooling.

“Would you like to try again?”

Nope. My responses continue to be correct. Would YOU like to try again? Try to be more precise with your wording. Here, this list will help.

Surface:
extra absorbed radiation from the sun =
extra absorbed radiation from CO2 =
extra absorbed radiation from other gases in the atmosphere =
extra latent and sensible heat transferred to the atmosphere =
extra radiation emitted by the surface =

Please fill in the blanks.

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3. No, English is my 3rd and best language.

At the surface:
extra absorbed radiation from the sun = 1.2
extra absorbed radiation from CO2 and all other gases = 0.8

SW induction = 1.2, LW induction = 0.8

Forget the latent and sensible heat. CERES incorporates that into general surface flux equivalents.

“ABSORBED radiation does not cause the EMITTED radiation to be balanced when the system is warming or cooling.”

Well that is interesting. You are saying the SKIN will not emit according to absorption, even though the SKIN has an exceedingly tiny heat capacity?

Satellites measure surface skin. They don’t really care what’s a meter below.

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4. Rocky says:

“No, English is my 3rd and best language.”

You still need to work on it.

“At the surface:
extra absorbed radiation from the sun = 1.2
extra absorbed radiation from CO2 and all other gases = 0.8
SW induction = 1.2, LW induction = 0.8”

OK. This is all fine.

“Forget the latent and sensible heat. CERES incorporates that into general surface flux equivalents.”

Huh? Where did you get that idea from? In any case, if you want to say the change in s+lh = 0 then that is fine.

““ABSORBED radiation does not cause the EMITTED radiation to be balanced when the system is warming or cooling.”
Well that is interesting.”

Is that news to you? It comes straight from the first law of thermodynamics. You might want to take thermo 101.

“You are saying the SKIN will not emit according to absorption, even though the SKIN has an exceedingly tiny heat capacity?”

Feel free to define the SKIN. Most solar radiation penetrates up to ~10m or so into the depths. That absorbed energy also gets conducted down to the depths. The heat capacity of the Earth is quite large. So your premise that the skin “has an exceedingly tiny heat capacity” is incorrect. Emission does not equal absorption when a system is warming or cooling.

“Satellites measure surface skin.”

Satellites measure the radiation exiting the top of the atmosphere. Some of that radiation is emitted by the surface. Some of the heat absorbed by the surface is transferred to lower depths.

“They don’t really care what’s a meter below.”

No they don’t. However, some of the heat absorbed by the surface is transferred a meter and more below. In the oceans solar radiation is absorbed more than a meter below.

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5. “In the oceans solar radiation is absorbed more than a meter below.”

That’s nice and all, but cold doesn’t warm hot. What is at the top is the hottest, and will emit to satellites.

What satellites observe is skin only. And that is where our numbers are from. And that is all we are commenting on.

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6. Rocky says:

That should have read “most solar radiation into the OCEANS …”

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7. Rocky says:

“That’s nice and all, but cold doesn’t warm hot. What is at the top is the hottest, and will emit to satellites.”

Irrelevant. The oceans have a massive heat capacity. Energy in is not balanced by energy out when the oceans are warming or cooling. This is not controversial. Your denial of this fundamental fact is telling.

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8. But we are not measuring emission from below the emitting surface (the “skin”).

The skin is maybe a millimeter thick if even that. A very small heat capacity.

Why are you interested in the entire depth of the ocean? Why not go further to the center of the Earth?

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2. Rocky says:

That should have been:

extra absorbed radiation from other matter in the atmosphere =

The surface also absorbs radiation from clouds and other particles in the atmosphere.

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1. Forget all that. We are only interested in the broad categories of SW and LW. Why must you over complicate for this example?

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2. Rocky says:

“Forget all that.”

The physics dictates that we are not allowed to forget all that.

“Why must you over complicate for this example?”

I’ve made this example as simple as it can possibly be. It cannot be simpler.

You need to fill in the blanks in order to proceed in a rational manner.

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3. “You need to fill in the blanks in order to proceed in a rational manner.”

That’s not how the audience will see it.

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4. Rocky says:

“That’s not how the audience will see it.”

Then the audience is scientifically incompetent because they don’t understand the first law of thermodynamics.

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5. They understand evasion. Trust me on that.

Look how much evasion you gave before you eventually started solving the problem.

Quit it and get on with it.

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6. Rocky says:

“They understand evasion. Trust me on that.”

Oh I do. You are evading. You refuse to fill in the blanks. You refuse to acknowledge the scientific fact that when a system is warming or cooling the energy in does not balance the energy out.

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7. You had enough info to give a very strange answer, and I gave you plenty of chances to correct it if you misunderstood. There is nothing holding you back.

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8. Rocky says:

“You had enough info to give a very strange answer, and I gave you plenty of chances to correct it if you misunderstood. There is nothing holding you back.”

I gave the correct answer given the information supplied.

I have given you plenty of chances to correct your problem statement by filling in ALL of the blanks. You continue to evade.

You also continue to evade admitting to the fact that when a system is warming or cooling the energy in does not balance the energy out.

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9. Remind me again what you said? Is it not …

For some reason you thought that because CO2 increased surface outgoing by 0.8, that means it didn’t receive it.

Then you though that an extra 1.2 from ASR didn’t do anything.

Then you though that an extra 1.2 from ASR is all there is. CO2 ignored.

Wrap it up for me. What’s your answer now?

And keep in mind your initial answer 50? comments ago could have been a simple “DISAGREE”.

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24. Rocky says:

“But we are not measuring emission from below the emitting surface (the “skin”).”

So what? There is still energy being transmitted and/or transferred below the skin. If you want to do an energy balance for the skin then you have to account for that energy.

“The skin is maybe a millimeter thick if even that. A very small heat capacity.”

So what? How much solar energy is transmitted through the skin to matter below the skin? How much heat is transferred from the warmer skin to the depths below?

Fill in those blanks Zoe.

“Why are you interested in the entire depth of the ocean? Why not go further to the center of the Earth?””

I’m interested in including any part of the system where a significant amount of energy change is occurring.

If you want to look at the skin then you will have to fill in these blanks:

Change in Solar radiation absorbed by skin =
Change in Solar radiation transmitted through skin =
Change in Atmospheric radiation absorbed by skin =
Change in Atmospheric radiation transmitted through skin =
Change in radiation emitted by the skin =
Change in non-radiative Heat transferred from the skin to the atmosphere =
Change in non-radiative Heat transferred from the skin to the depths =

Fill in ALL of the blanks Zoe.

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1. Somehow Donohoe didn’t need all that for his simple model in Fig 1. Why can’t you keep it simple and transfer to the surface?

I feel as though I’m talking to an imbecile.

If I shine a laser onto the ground such that the surface receives 1000 W/m^2, how long will it take for an thermometer to measure ~364K ? (albedo=0, emis=1).

According to you it will take a real loooooong time because the Earth is giant and its heat capacity is immense.

Even though in reality it will be quite fast.

Same thing if I shined that laser on top of water.

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1. Rocky says:

“Somehow Donohoe didn’t need all that for his simple model in Fig 1.”

That’s right, he didn’t. The energies into and out from the TOA dictate the behavior of the system inside the TOA and the change in surface temperature can be used to represent the change in internal energy of that system.

“Why can’t you keep it simple and transfer to the surface?”

Transfer what to the surface Zoe? The temperature of the surface is right there in the paper. If you think it is so simple to “transfer to the surface”, and by surface I assume you mean your negligible heat capacity skin, then go ahead and fill in the blanks.

“I feel as though I’m talking to an imbecile.”

I’m far more intelligent than you Zoe. You still don’t even understand why the questions that I am asking you are relevant and why yours are practically meaningless.

“According to you it will take a real loooooong time because the Earth is giant and its heat capacity is immense.”

Donohoe quantifies the appropriate heat capacity. He also provides the time scale. Is heat capacity difficult for you to understand?

“Even though in reality it will be quite fast.”

Quantify how fast Zoe. Put equations behind your claims. Climate scientists do that. You just spew empty rhetoric.

“Same thing if I shined that laser on top of water.”

If you shined a laser on the surface of water, how deep would the radiation from the laser penetrate? How much energy absorbed by the water would be conducted 1mm away after 1second? How much energy absorbed by the water would be conducted 10mm away after 10 seconds? Do you even know how to formulate the equations that can be solved to answer these questions Zoe?

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1. “change in surface temperature”

Which means changes in surface fluxes. So why can’t you discuss them?

“Donohoe quantifies the appropriate heat capacity.”

He uses 50m of ocean. Does the 50th meter not conduct to the 51st? Does not that conduct to the 52nd? … where is the limit? Where it’s hotter inside the Earth. So why not use that?

Perhaps you can explain why after billions of years there is such a difference between top and bottom ocean temperatures. Perhaps your conduction theory doesn’t work like you think it does!

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2. Rocky says:

“Which means changes in surface fluxes.”

Indeed it does.

“So why can’t you discuss them?”

I can discuss them. I have asked you to fill in the blanks. Why can’t you do that?

“He uses 50m of ocean. Does the 50th meter not conduct to the 51st?”

Indeed it does. How much energy do you estimate is transferred below 50m?

” … where is the limit?”

It depends how accurate you feel you need to be. Accounting for what is going below 50m is not going to change the quantitative answers beyond the number if significant figures we have been throwing around. But it’s good to see that you are starting to realize that energy is getting well below the skin. That’s progress for you.

“Perhaps you can explain why after billions of years there is such a difference between top and bottom ocean temperatures.”

Heat transfer explains it. This digression is simply more evasion on your part. But if this is the cross you want to die on then we can get into the mathematical details.

“Perhaps your conduction theory doesn’t work like you think it does!”

Conduction works pretty much exactly how I think it does. The governing equations have been well established for quite some time.

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3. What blanks are left to fill?

You seem to think 50m has some significance. But you do realize that the fact that the sun can penetrate through water means there is LESS differential between depths and hence less need for conduction?

Of course you didn’t, moron. The fact is that where there is no penetration – there is more need for conduction.

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25. Rocky says:

“For some reason you thought that because CO2 increased surface outgoing by 0.8, that means it didn’t receive it.”

You did not specify how much radiation was received from CO2. You stated that the surface increased emission by 0.8. Again, when a system is warming energy in does not equal energy out. That you continue to fail to understand this fundamental principle says quite a lot about you.

“Then you though that an extra 1.2 from ASR didn’t do anything.”

Nope. It was absorbed. Hence it contributed to the energy into the system. Why are you trying to set up straw men Zoe?

“Then you though that an extra 1.2 from ASR is all there is.”

You’re obviously still confused.

“CO2 ignored.”

By whom?

“Wrap it up for me. What’s your answer now?
And keep in mind your initial answer 50? comments ago could have been a simple “DISAGREE”.”

Indeed. Your implication that energy in balances energy out in a warming system is WRONG. You should grow up and admit your error.

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1. Retard,

In Donohoe’s paper LW (from CO2) and SW (from ASR) is simply added together. You can see that Fig 1C is just a merger of Fig 1A and Fig 1B.

Now Fig 1 is for TOA.

Can you do this for the surface without being a completely evasive slimeball?

“You stated that the surface increased emission by 0.8”

From CO2 alone. It sends out 0.8 because it receives 0.8 from CO2 in the atmosphere.

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1. Rocky says:

More childish name-calling Zoe. You should be better than that.

“In Donohoe’s paper LW (from CO2) and SW (from ASR) is simply added together.”

Those are TOA forcings. Learn to read scientific papers for comprehension.

“Can you do this for the surface without being a completely evasive slimeball?”

I can if you fill in all the blanks. You continue to evade. But you really should educate yourself on why the TOA numbers are the proper measures to use for this simple model.

We can also go to a sphere-shell model if you like. But you continue to evade doing that too.

Have to go. Will educate you some more tomorrow.

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1. So you can’t do surface fluxes because Donohoe does TOA fluxes and this is the proper way to do it because it’s not at the surface. Got it. Your logic is incredible.

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2. Rocky says:

“So you can’t do surface fluxes …”

Apparently it’s you who can’t do surface fluxes. You can’t fill in the blanks. If you think that surface fluxes can somehow overturn all of the points that I have made then go ahead and fill in the blanks.

TOA fluxes are sufficient to prove my point.

Satellites measure the fluxes at the TOA. You have NO measurements for how much solar radiation is absorbed by your negligible heat capacity skin.

Has it sunk in yet that your implication that energy in has to balance energy out in a warming system is wrong?

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3. Wow, now you think satellite surface measurements are all fraud. Can you get even more bonkers?

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26. Rocky says:

Well, I have a dinner party to get to. We can resume tomorrow. – Best

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1. gbaikie says:

Zoe brought up time.
I think how quickly Earth surface cools is important question.
And I like to compared planets {or our Moon}
With Venus and radiate about 170 watts on average vs Earth’s 240 watts on average.
I think Venus cools quicker, despite it’s lower watts per square meter emission.
If that was true, one could say Earth has stronger “greenhouse effect” as compared to Venus.
Btw, related to this, I think Venus if at Earth distance {1 AU} would be colder than Earth is.
And Earth is in an Ice Age- or Earth has been in distance past much warmer than it is now.
So I think Venus at 1 AU distance from the sun, would be colder than Earth is now [in Ice Age]
And we just talk about Earth and what happens if the Sun disappears.
Zoe who doesn’t think the sun does much warming, would say Earth doesn’t cool very fast
and some think within mins of sun disappearing, Earth cools very quickly.
“In a way” I agree with both views.
It seems to me, if a fish is predator, it’s not going to have much problems is the Sun disappeared for
a month or even a year.
And if living in a deep cave, one might not notice much within a month or a year.
It seems all would tend agree about all that.
But climate things is largely concerned about global air temperature. And most people
live on land surface [and in houses of some kind]. And they sleep in their homes at night and wear
clothes to keep warm, even when in their warm homes. And their warm home generally have air temperature set at room temperature, which around 20 C [68 F] but room temperature is also a range of 20 to 30 C {68 F to 86 F} and winter people like it warmer and in summer they like colder- which not really healthy, but they tend to do this.
Some crazy people will go into sauna which is 100 C and then go jump into snow. They do polar bear swims, freezing temperature and jumping in cold water. It’s fun. But living in 15 C air is also way to torture people and 15 C water will eventually kill people. Humans hold a lot odd preferences. And most could not survive long without their house, Even if they imagine they have outdoor training skills the outside can be brutal. Fortunately, Musk is going to rid the world of dead zone for cell phones and save hundreds of lives. It take him a couple years.

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1. “Zoe who doesn’t think the sun does much warming”

That’s not a fair assessment of what I said.

In fact the trick is that most of geothermal is “stored solar”. I can get away with that because industry says the same thing. All shallow-Earth geothermal systems are still called “geothermal”, despite obviously being stored solar.

“Earth doesn’t cool very fast
and some think within mins of sun disappearing”

Such a cooling would cause a serious flux differential with the interior, and it’s more than likely cracks will form and magma spill out.

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1. gbaikie says:

Well I would say it cools quickly because if it’s noon and hot it would cool quickly, whereas if after midnight, you would not notice the sun disappearing. And night continues of sun disappeared for many hours. I also think there is big effect immediately in upper atmosphere and it is more a global
thing. So, if sun disappear when it’s midnight and you are somehow measuring upper atmosphere, you could measure the effect of sun disappearing almost immediately.
And since it immediately affects daytime high temperatures, it going have a large effect upon average temperature on land areas.
But most of Earth surface is ocean, and it have a smaller effect, so, globally a smaller effect.
And if it’s winter time and you living Canada, things going to get cold, quite quick.
But if in southern California in the summer, it would take days to freeze,
And if in tropics [regardless of seasons] weeks to freeze- particularly if surrounded by ocean.
But the tropical ocean heat engine, within days, stops working completely, and global heat engine isn’t warming the world. And in days, atmosphere will have skrunk quite a bit.
Or tropical troposphere was at 20 km, and and lowers by 5 km pretty quick. Or:
“The troposphere begins at the Earth’s surface and extends from 4 to 12 miles (6 to 20 km) high. The height of the troposphere varies from the equator to the poles. At the equator it is around 11-12 miles (18-20 km) high, at 50 °N and 50 °S , 5½ miles and at the poles just under four miles high.”
So tropical troposphere will still higher 6 km, as spinning effects it as well as it’s temperature. but 6 km polar tropical will drop as it gets colder and not have heat engine adding to it. Jet streams stop, and less mixing. But there would probably severe weather effect in parts of part, but tropics should remain without much those severe weather effects.
Or Sky will fall, but less end of world sky falling in tropics where will remain warm for quite long time.
And average ocean temperature of 3.5 C changes very little within a decade.

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2. gbaikie says:

http://nsidc.org/arcticseaicenews/

So how fast will that line climb in 24 hours after sun blinks out at this time of Aug 27, I would say about 4 times faster than ever climbed. So won’t rise out of average yearly variance in 24 hours.
But 6 month could one could greater polar sea ice extent than Earth has ever had.
And in 6 months how much will it effect the average ocean temperature of 3,5 C.
Very little.
How much will it affect Canada in 6 months. CO2 will be freezing out of the air. So, it’s quite cold.
Troposphere height less than 3 km in polar region
But we don’t have a Snowball Earth, yet
The supposive mythic state which Earth had- +300 or 700 million years ago. Or some other time long ago.
Or tropical ocean has not froze yet. Tropical land could well below freezing. Though not ice sheets added.
Globally it’s very dry.
Mars is very dry and it’s got average of 210 ppm of water vapor, so Earth’s probably got over 500 ppm of global water vapor. And Earth still has a huge liquid ocean [and Mars doesn’t- but mars does have CO2 freezing out the it’s atmosphere]

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3. gbaikie says:

So Earth as the star. and 6 months after sun is gone.
How much is Earth emitting into space?
I would wildly guess about as much as 170 watts on average.
But most due to ocean still being about 3.5 C.
Though geothermal heat is warming the ocean. But not warming the land, much.
Hot springs are merry places on a frozen wasteland, and life is roughly only alive due to heated vents. And creatures eating dead creatures.
Human could living, if not too stupid to live.
Though they would probably be worshiping nuclear power plants by this point in time.

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4. Maybe. We know that when sun activity is low, earthquake activity is high. No sun, and we may have crust-breaking convection?

This is never taken into account. Maybe it’s wrong? Don’t know.

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5. gbaikie says:

But what more interesting is what happens if sun blinks back on in 6 months.
This is opportunity climate manics to do their climate models.
Well, it should fully recover in about 3 months- though a lot animals could have gone extinct unless
humans save them. Plants with there seeds, could have survive it better. But a lot things would have died.
But a lot less would have died, if sun only blinked out for a month.

Another thing could be if Earth blinked out to Mars distance and happens in a month or
6 months.
And how fast would Venus cool at Earth distance in say 6 months or 5 years.
I think it might take more than 10 years to be colder than Earth, but I think in as little as
6 months Venus would change a lot.

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6. That is indeed interesting. Very hard to speculate though. Can’t argue with you.

Your mind goes everywhere. Sometimes hard to keep up. lol

Tell me, gbaikie, what do you think of this Rocky guy?

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27. gbaikie says:

What do I think of Rocky?
I haven’t given it much thought.
Explain to me what you think his point is.

Chances are, he disagree, that this is his point.

But try reading what he saying and get back to you.

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28. gbaikie says:

That too boring.
How about we go over what you said:
–Today we will play with this silly math theory called the greenhouse effect. Here are two examples of its typical canonical depiction:–
Well the problem is the focus is on global, whereas looking at tropics only, would be better,
People has done this, but I don’t where I could find it. See if I can find it.
This short but related:

Still looking energy budget for Tropics

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29. gbaikie says:

Well this “might” useful:
But more importantly it reminded me of Lindzen
I don’t find much to disgree with in regard to what Lindzen says.
Though Lindzen might very will disagree with whatever I say.
Cargo cult climate hates Lindzen.
Lindzen is Satan

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30. gbaikie says:

It could reasonable to think I would have a lot to disagree with about
this video:

I would interested in what is the biggest thing which seems the most incorrect.
But short list of top things, would also be good.

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31. Rocky says:

“Wow, now you think satellite surface measurements are all fraud. Can you get even more bonkers?”

Why do you insist on making straw man arguments Zoe? It’s childish.

Satellites are above the TOA, they are measuring what is coming out from or going in to the TOA.

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1. Rocky says:

They can measure radiation that passes through the atmospheric window.

Is that what you are asking? If you are trying to make a point then go ahead and make it.

Radiation at other wavelengths exiting the TOA is not being emitted from the surface.

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1. Yes, and not just the main atmospheric window, but 60 Ghz Oxygen is used primarily.

Mainstream scientists believe this is good enough to measure surface flux, so why don’t you? Why did you waste time going down this alley?

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2. Rocky says:

“Mainstream scientists believe this is good enough to measure surface flux”

Why don’t you quote EXACTLY what mainstream scientists state? Go ahead.

“Why did you waste time going down this alley?”

This is your alley Zoe. You are the one trying to deflect from your errors.

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3. They say: “Here’s our surface flux data. Please ignore it because Rocky knows satellites can’t measure it”.

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4. Rocky says:

Poor, poor Zoe. Do they say that they measure the flux from the surface? Or do they say that they measure the flux at the TOA and INFER the flux from the surface based on a radiative transfer MODEL?

Yet more math that is over your head.

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5. Some do that. Few. Most use atmo window and 60 GHz Oxygen line. The various methods are tested against each other.

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32. Rocky says:

“Some do that.”

Provide a citation to those that do not.

“Most use atmo window and 60 GHz Oxygen line.”

Any that use a 60 GHz Oxygen line are inferring temperature and/or fluxes based on a radiative transfer model.

“The various methods are tested against each other.”

Indeed. The accuracy and errors induced by the models needs to be quantified.

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1. So now we are questioning the validity of surface fluxes? And that’s why we must stick to TOA fluxes? To then infer surface fluxes, which you also have a problem with. And so you want to stick with TOA and are upset that I am discussing the surface. OK, weirdo.

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1. Rocky says:

“So now we are questioning the validity of surface fluxes?”

No. I am telling you what is measured and what is modeled.

“And that’s why we must stick to TOA fluxes? ”

It’s surprising that you are not smart enough to understand why the TOA fluxes are relevant. You need to define a system. The first law dictates that the change in internal energy of the system is equal to the net energy into that system. By accounting for the fluxes into and out from the TOA we can find the change in internal energy of the system inside the TOA.

It’s really not a difficult concept. Why is it taking you so long to grasp?

“And so you want to stick with TOA and are upset that I am discussing the surface.”

I’m not upset at all that you want to discuss the surface. However, you have been evading and throwing childish fits that I have asked you to fill in the blanks. You continue to evade.

I am off to mow the lawn now. Back later.

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1. I wanted to discuss the surface the whole time, and your problem is that I wanted to discuss the surface the whole time? So it’s my fault for wanting to discuss the surface the whole time?

You had no problem answering the question incorrectly without needing blanks filled. You could’ve done it correctly and we could’ve moved onto the next step. I filled in all your blanks, and you still couldn’t do it right.

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33. Rocky says:

“I wanted to discuss the surface the whole time, and your problem is that I wanted to discuss the surface the whole time?”

Not really. My problem is that you refuse to answer questions about the energies into, out of, and through the surface. So, in essence, you are now refusing to discuss that which you claim to want to discuss.

“You had no problem answering the question incorrectly without needing blanks filled.”

Correction, I answered the question as it was stated. You have since had to clarify what you actually meant by the amount of CO2 radiation and what exactly the “surface” is. Your implication that the radiation in must equal the radiation out remains wrong.

Your own-goal example of the laser heating the water proves that you do not even understand how to apply the first law of thermodynamics to a system.

“I filled in all your blanks, and you still couldn’t do it right.”

You have not filled in ANY of the blanks Zoe. Why are you making false statements?

Here are the blanks that you continue to refuse to fill in:

Change in Solar radiation absorbed by skin =
Change in Solar radiation transmitted through skin =
Change in Atmospheric radiation absorbed by skin =
Change in Atmospheric radiation transmitted through skin =
Change in radiation emitted by the skin =
Change in non-radiative Heat transferred from the skin to the atmosphere =
Change in non-radiative Heat transferred from the skin to the depths =

Fill in ALL of the blanks Zoe.

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34. Max Polo says:

@Rocky
I have followed your suggestion: I have done a series of tests indoor. Energy source was a 1 kW halogen lamp. I kept all the windows closed, no fans working, no AC turned on, and nobody was present (except myself) during the tests. I felt that the air / background conditions were much more in control this time. Air temperature ranged from 26 to 27°C, walls/ceiling/floor from 27 to 29°C. The only downside was the reduced power vs that of the sun. I could have put the panes closer to the source, but this wasn’t recommended by the lamp instructions (distance at least 1 m – fire hazard) and power flux uniformity was questionable. In addition to ensuring squareness and distances from the power source, I made an initial calibration by finely adjusting positions to get the same front pane temperature when the rear pane was removed. Once I would reach even temperatures (+/-0.2°C) I would start the test by adding the second pane.
At the link below, you’ll find some pictures of the arrangement and a typical record of the measured temperatures.
Again, no trend whatsoever was observed in the front pane temperature after the rear pane was added, provided a sufficient time would have waited for the temperatures to settle: take into consideration that just a simple touch or even breathing close to any pane would cause it to cool/change temp easily by half-degree °C or more. Nothing similar or even of lower magnitude was observed after the addition of the rear pane in steady state. No trace of any backradiation heating.
Correct application of the 1st Law of Thermodynamics gives a good explanation of this: see my other post.

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1. Rocky says:

Sorry Max, but this will unlikely get through because Zoe has blocked me.

You are still not doing what I suggested, and your experimental setup is still not expected to give considerably different results for the temperatures.

You have botched the first law terribly. Hopefully someone who Zoe has not blocked can help you.

Zoe: Yeah, I blocked him. He’s a troll.

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1. gbaikie says:

I would guess your 1 kW halogen lamp has reflector.
The reflector directs more energy in a direction.
If has something without a reflector which gave the same output, one probably
find a way increase out put, by roughly doing what reflector does.
Or having a reflector, diminishes the abilities to measure any effect which badly acts
as reflector does.

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35. Max Polo says:

“The first law at steady state dictates that the energy supplied by the source to the sphere must be equal to the HEAT transferred out from the sphere to the shell. That means that Q12 must be equal to W…….. Have you still not figured out that Q12 = W?…etc etc”

Rocky,
forcing Q12 = W won’t work for this system, I’m afraid.
This version of the 1st LOT is fine if the system does not increase its internal energy level during the process, which only happens when you have a single body and no DT = T1 – T2 develops in the system. Let me explain:
The full version of the Law is :
W-Q12 = DU
Or : W = DU + Q12
Where DU is the increase of energy level in the system.
When the temperature difference DT = T1-T2 settles in the two bodies, the relevant excess of energy Q12 gets transferred from 1 to 2, but this only happens in conjunction with the internal energy level sigma*T2^4 > 0 starting to develop. This “lower” level of energy must be overcome to create the heat flow. This sigma*T2^4 defines the DU, the new system energy level induced by the input W, so the correct version of the 1st LOT to be applied becomes W = Q12 + DU = Q12 + sigma*T2. This equation is just another way to see the entire sphere emission sigma*T1^4. This implies that Q12 is less than W, as you have seen from my solutions for the SGR that Q12 = ½*W, while for the double panes Q12 = (f-f^2/2)*W/2 where f is the view factor.

Your problem (said with all my respect) is that you treat the energy flux as if it was an extensive property: you apply the vector summation rule, and therefore you cancel out the two opposite emissions up to the T2 temperature for the mutually viewing surfaces. But this is wrong since the energy flux is an intensive property where 1 + (-1) = 1: the meaning of which is explained in the following hydraulic analogy.
Two communicating vessels (1 and 2) contain water at differing levels H1 > H2. At a certain depth, there is a pipe with a valve that connects the two vessels. There are two differing pressures P1 and P2 at each side of the valve, ready to push against each other. These two pressures start to “fight” as soon as you open the valve, and the water starts flowing from 1 to 2. The excess pressure DP = P1 – P2 drives the water speed as per Bernoulli’s law (1/2*density*velocity^2 = DP). DP is analogous to heat flow Q12, P1 and P2 are analogous to sigma*T1^4 and sigma*T2^4. If you put a pressure gauge in the valve when water is flowing, you’ll ready P = P2 > 0 in the display: you won’t read P=0, even if the two pressures are equilibrated and against each other. Vector summing won’t work for intensive properties like pressure. With your “consensus” method of solving the two panes problem (imposing W = Q12) you aren’t doing anything else than “annihilating” (via your opposing vectors that do not exist) the new internal energy level sigma*T2^4 (P2 in the analogy) that develops contextually while transferring energy Q12 (DP in the analogy). Dimensional analysis shows that pressure P (static energy) and flowing energy ½*density*velocity^2 have the units: (energy flux)*(mass flow rate) = (W/m2)*(Kg/s). Hence P has the same dimensions of an energy flux per unit of mass flow rate, which confirms the validity of the analogy.
To summarize: W = Q12 is a wrongly simplified version of the 1st LOT and does not represent the physics of two bodies exchanging heat via radiation. The full version must be used. When you apply the full version, it’s clear that the alleged heating of the emitting body has no reason to show up.

Conclusion :
Backradiation represents just an increase of internal energy in the system of the two bodies that are exchanging heat. This energy cannot be exchanged with the background nor with the emitting body and cannot induce any increase in the temperature of the emitting body. This energy stays somewhat hidden in the system or “trapped inside” if you like it, yet it is an effective entity that must be sustained by the external source, W.
So this “mighty” backradiation ultimately reveals its true nature: that of a threatening dragon, that ingloriously turns into a harmless little lizard trapped in a box.

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36. Jarle says:

According to Svensmark it is the muon-content of the cosmic rays that are relevant for lower level cloud formation.

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