Easy experiment shows there is no heat gain by backradiation.

polarbear

I eat morons
Jan 1, 2011
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The sun warms the ground and the ground warms the air. That we can all agree on.
Heat can only be transferred from a warmer mass to a colder one, but the AGW advocates argue that heat can also be transferred from the colder mass to the warmer one. Climate "scientists" must insist on it else they don't have a case.
That means that the air has to be able to warm the warmer ground with "back radiation", which they say has increased with the increase in CO2.
They don't worry about mass or specific heat and simply apply black body physics for the radiative heat transfer while eyeballing heat conduction and convection yet claim that their results are accurate within a fraction of a degree.
Any scientist and/or engineer who has to deal with heat transfer in the real world knows how ridiculous this claim is.
Since 7/10 th of the earth surface is water and air at 400 ppm is available and free it is very easy to prove that
All it takes is 2 soda cans filled with water, one hot an another one cold, take note of the room temperature and observe the heat transfer.
Wait till the can with the warm water has cooled to 10 deg C above room temperature and note the time.
After 1 hour note the water temperature. Next measure the exact dimensions of the soda can and the amount of water in it. That allows you to get the cals per second or the # of watts.
The pop can surface area was 292.3 cm^2 the water in it was 370 ccm and it cooled from 34.5 C to 30 C in 1 hour. So it transferred 1665 calories (6996 watt sec) in 1 hour to the air that was 10.5 deg C cooler.
With the STB equation for this temperature difference you get 65 Watt/m^2
I observed 66.5 Watt/m^2 which of course includes heat conduction from the water through the thin Alu skin and into still air (convection=negligeble)....which are the extra 1.5 watt/m^2 (heat conduction) over the theoretical 65 W radiative transfer
If the climate warmers were right I should have only had 65(StB) - 1.8 (back radiation)+ my 1.5 conduction = 64.7 Watts/m^2
Which would be less cooling ergo more warming....but that did not happen, neither is global warming by CO2 back radiation.
 
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The sun warms the ground and the ground warms the air. That we can all agree on.
Heat can only be transferred from a warmer mass to a colder one, but the AGW advocates argue that heat can also be transferred from the colder mass to the warmer one. Climate "scientists" must insist on it else they don't have a case.
That means that the air has to be able to warm the warmer ground with "back radiation", which they say has increased with the increase in CO2.
They don't worry about mass or specific heat and simply apply black body physics for the radiative heat transfer while eyeballing heat conduction and convection yet claim that their results are accurate within a fraction of a degree.
Any scientist and/or engineer who has to deal with heat transfer in the real world knows how ridiculous this claim is.
Since 7/10 th of the earth surface is water and air at 400 ppm is available and free it is very easy to prove that
All it takes is 2 soda cans filled with water, one hot an another one cold, take note of the room temperature and observe the heat transfer.
Wait till the can with the warm water has cooled to 10 deg C above room temperature and note the time.
After 1 hour note the water temperature. Next measure the exact dimensions of the soda can and the amount of water in it. That allows you to get the cals per second or the # of watts.
The pop can surface area was 292.3 cm^2 the water in it was 370 ccm and it cooled from 34.5 C to 30 C in 1 hour. So it transferred 1665 calories (6996 watt sec) in 1 hour to the air that was 10.5 deg C cooler.
With the STB equation for this temperature difference you get 65 Watt/m^2
I observed 66.5 Watt/m^2 which of course includes heat conduction from the water through the thin Alu skin and into still air (convection=negligeble)....which are the extra 1.5 watt/m^2 (heat conduction) over the theoretical 65 W radiative transfer
If the climate warmers were right I should have only had 65(StB) - 1.8 (back radiation)+ my 1.5 conduction = 64.7 Watts/m^2
Which would be less cooling ergo more warming....but that did not happen, neither is global warming by CO2 back radiation.
Anybody who thinks energy can move from colder to warmer media did not take any chemistry ever, neither in high school nor in college.

Funny !!
 
The Earth's climate is warming.

Most people attribute this to the Sun.

We don't know why.

We just know it is happening.

We don't know if the insects, animals, and humans on the Earth can do anything about it. I am betting that we cannot.
 
If by "most people attribute this to the sun" then you're running with a crowd of hermits. Most people attribute this to carbon dioxide produced by humans burning coal and petroleum. We CAN do something about it, but it is going to take a big effort and it is going to take time. If we do NOT do anything about it, its going to be very bad for our children and theirs.
 
Wait till the can with the warm water has cooled to 10 deg C above room temperature and note the time.
After 1 hour note the water temperature. Next measure the exact dimensions of the soda can and the amount of water in it. That allows you to get the cals per second or the # of watts.
The pop can surface area was 292.3 cm^2 the water in it was 370 ccm and it cooled from 34.5 C to 30 C in 1 hour. So it transferred 1665 calories (6996 watt sec) in 1 hour to the air that was 10.5 deg C cooler.

Okay so far. 7000 joules. Don't overdo the precision.

7000J/3600S/0.0292m^2 = 67 W/m^2. So far, so good.

With the STB equation for this temperature difference you get 65 Watt/m^2

That's going to vary considerably as the temperature cools..

0.9 * 304.5^4 *5.67E-08 = 438 W/m^2 (warmer can) (emissivity of aluminum = 0.9)

0.9 * 300^4 * 5.67E-08 = 413 W/m^2 (cooler can)

0.9 * 294.5^4 * 5.67E-08 = 383 W/m^2 (room)

So, 55W/m^2 net radiation loss at the start, 30W/m^2 at the end.

If conduction losses are added to radiation losses, that would match the 67 W/m^2 average loss, so the experiment is consistent with backradiation. Without the 383 W/m^2 backradiation contribution, your can would have been losing 400+W/m^2, and would have been starting to freeze by the end of the hour.

I observed 66.5 Watt/m^2 which of course includes heat conduction from the water through the thin Alu skin and into still air (convection=negligeble)....which are the extra 1.5 watt/m^2 (heat conduction) over the theoretical 65 W radiative transfer
If the climate warmers were right I should have only had 65(StB) - 1.8 (back radiation)+ my 1.5 conduction = 64.7 Watts/m^2

Where did you get 1.8 W/m^2 for the backradiation contribution? That's totally wrong, so the conclusion is totally wrong.

Which would be less cooling ergo more warming....but that did not happen, neither is global warming by CO2 back radiation.

Quite the contrary. Your can did not start to freeze, hence your experiment proves backradiation was present.
 
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The Earth's climate is warming.

Most people attribute this to the Sun.

We don't know why.

We just know it is happening.

We don't know if the insects, animals, and humans on the Earth can do anything about it. I am betting that we cannot.
Which is why we need to move the smarter people to a different planet. ;)
 
The sun warms the ground and the ground warms the air. That we can all agree on.
Heat can only be transferred from a warmer mass to a colder one, but the AGW advocates argue that heat can also be transferred from the colder mass to the warmer one. Climate "scientists" must insist on it else they don't have a case.
That means that the air has to be able to warm the warmer ground with "back radiation", which they say has increased with the increase in CO2.
Any scientist and/or engineer who has to deal with heat transfer in the real world knows how ridiculous this claim is.

Actually, poop4brains, your fantasies about what scientists "know" is based entirely on your ignorance and stupidity.

In the real world, from, straight from the darling of the deniers, Dr. Roy Spencer.....

Help! Back Radiation has Invaded my Backyard!
Measuring The (Nonexistent) Greenhouse Effect in My Backyard with a Handheld IR Thermometer and The Box

by Roy W. Spencer, Ph. D.
August 6th, 2010


Laypersons are no doubt confused by all of our recent esoteric discussions regarding radiative transfer, and whether global warming is even possible from a theoretical standpoint.

So, let’s take a break and return to the real world, and the experiments you can do yourself to see evidence of the “greenhouse effect”.


One of the claims of greenhouse and global warming theory that many people find hard to grasp is that there is a large flow of infrared radiation downward from the sky which keeps the surface warmer than it would otherwise be.

Particularly difficult to grasp is the concept of adding a greenhouse gas to a COLD atmosphere, and that causing a temperature increase at the surface of the Earth, which is already WARM. This, of course, is what is expected to happen from adding more carbon dioixde to the atmosphere: “global warming”.

Well, it is one of the marvels of our electronic age that you can buy a very sensitive handheld IR thermometer for only $50 and observe the effect for yourself.

These devices use a thermopile, which is an electronic component that measures a voltage which is proportional to the temperature difference across the thermopile.

If you point the device at something hot, the higher-intensity IR radiation heats up the hot-viewing side of the thermopile, and the IR thermometer displays the temperature it is radiating at (assuming some emissivity…my inexpensive unit is fixed at e=0.95).

If you instead point it at the cold sky, the sky-viewing side of the thermopile loses IR radiation, cooling it to a lower temperature than the inside of the thermopile.


For instance, last night I drove around pointing this thing straight up though my sunroof at a cloud-free sky. I live in hilly territory, the ambient air temperature was about 81 F, and at my house (an elevation of 1,000 feet), I was reading about 34 deg. F for an effective sky temperature.

If the device was perfectly calibrated, and there was NO greenhouse effect, it would measure an effective sky temperature near absolute zero (-460 deg. F) rather than +34 deg. F, and nighttime cooling of the surface would have been so strong that everything would be frozen by morning. Not very likely in Alabama in August.

What was amazing was that driving down in elevation from my house caused the sky temperature reading to increase by about 3 deg. F for a 300 foot drop in elevation. My car thermometer was showing virtually no change. This pattern was repeated as I went up and down hills.

The IR thermometer was measuring different strengths of the greenhouse effect, by definition the warming of a surface by downward IR emission by greenhouse gases in the sky. This reduces the rate of cooling of the Earth’s surface (and lower atmosphere) to space, and makes the surface warmer than it otherwise would be.

If you have a day where there are patches of blue and clouds, you can point the thermometer at the clouds and pick up a warmer reading than the surrounding blue sky.

I did it this morning (see photo, above). When I moved from a view of the blue sky to the patch of clouds, the sky-viewing side of the thermopile became warmer…even though the thermopile is already at a higher temperature than the sky. The display would read a few degrees warmer than the reading looking at blue sky.

If you perform this experiment yourself, you need to be careful about the elevation angle above the horizon you are pointing being about the same. Even in a clear sky, as you move from the zenith (overhead), down toward the horizon the path length of sky the IR thermometer sees increases, and so you measure radiation from lower altitudes, which are warmer. This makes the effective sky temperature goes up. (This is ALSO evidence of the greenhouse effect, since looking at the sky above the horizon is like adding greenhouse gases to the atmosphere overhead. The (apparent) concentration of greenhouse gases in the lower atmosphere goes up, and so does the intensity of the back radiation.)

Even earlier in the morning, about 5:30, the middle-level clouds were thicker, and I measured a sky temperature in the 50’s F. We will see more evidence of that using air temperatures, below.

This shows that the addition of an IR absorber/emitter, even at a cold temperature (the middle level clouds were probably somewhere around 30 deg. F), causes a warm object (the thermopile) to warm even more! This is the effect that some people claim is impossible.

Remember, the IR thermometer calibrated temperature output is based upon real temperatures, the temperatures on either side of the thermopile.

And if you think this is just an effect of some sunlight reflecting off the cloud….read on.

Evidence from The Box

I have been seeing the same effect in “The Box”, which is my attempt to use the greenhouse effect to warm and cool a thin aluminum plate coated with high-emissivity paint, that is heavily insulated from its surroundings in order to isolate just the radiative transfers of energy between the sky and the plate. This can be considered a clumsy, inefficient version of the IR thermometer. But now, *I* am making actual temperature measurements.

The following plot (click on it for the full-size version) shows data from the last 2 days, up through this morning’s events. The plate gets colder at night than the ambient temperature because it “sees” the cold sky, and is insulated from heat flow from the surrounding air and ground.


In the lower right, I have also circled where thin middle-level clouds came over, emitting more IR radiation downward than the clear sky, and causing a warming of the plate. Since the plate is mostly isolated from heat exchanges with the surrounding air and warm ground, it responds faster than the ambient air temperature to the intensity of “back radiation” downwelling from the sky.

When I woke this morning before sunrise, around 5:30, I saw these mid-level clouds (I used to be a certified aviation weather observer), I measured about 50 deg. F from the handheld IR thermometer.

This supports what people already experience…cloudy nights are, on average, warmer than clear nights. The main reason is that clouds emit more IR downward, change the (im)balance between upwelling and downwelling IR, and if you change the balance between energy flows in and out of an object, its temperature will change. Conservation of Energy, they call it.

(WARNING: a technical detail about the above measurements and their importance to greenhouse theory follows.)

What this Means for the Miskolczi “Aa=Ed” Controversy

Except for relatively rare special cases, the total amount of IR energy downwelling from the sky (Ed) will ALWAYS remain less than the amount upwelling from below and absorbed by the sky (Aa). As long as (1) the atmosphere has some transparency to IR radiation (which it does), and (2) the atmosphere is colder than the surface (which it is), then Ed will be less than Aa…even though they are usually close to one another, since temperatures are always adjusting to minimize IR flux divergences and convergences.

But it is those small differences that continuously “drive” the greenhouse effect.
 
Wait till the can with the warm water has cooled to 10 deg C above room temperature and note the time.
After 1 hour note the water temperature. Next measure the exact dimensions of the soda can and the amount of water in it. That allows you to get the cals per second or the # of watts.
The pop can surface area was 292.3 cm^2 the water in it was 370 ccm and it cooled from 34.5 C to 30 C in 1 hour. So it transferred 1665 calories (6996 watt sec) in 1 hour to the air that was 10.5 deg C cooler.

Okay so far. 7000 joules. Don't overdo the precision.
7000J/3600S/0.0292m^2 = 67 W/m^2. So far, so good.

With the STB equation for this temperature difference you get 65 Watt/m^2

That's going to vary considerably as the temperature cools..

0.9 * 304.5^4 *5.67E-08 = 438 W/m^2 (warmer can) (emissivity of aluminum = 0.9)

0.9 * 300^4 * 5.67E-08 = 413 W/m^2 (cooler can)

0.9 * 294.5^4 * 5.67E-08 = 383 W/m^2 (room)

So, 55W/m^2 net radiation loss at the start, 30W/m^2 at the end.

If conduction losses are added to radiation losses, that would match the 67 W/m^2 average loss, so the experiment is consistent with backradiation. Without the 383 W/m^2 backradiation contribution, your can would have been losing 400+W/m^2, and would have been starting to freeze by the end of the hour.

I observed 66.5 Watt/m^2 which of course includes heat conduction from the water through the thin Alu skin and into still air (convection=negligeble)....which are the extra 1.5 watt/m^2 (heat conduction) over the theoretical 65 W radiative transfer
If the climate warmers were right I should have only had 65(StB) - 1.8 (back radiation)+ my 1.5 conduction = 64.7 Watts/m^2

Where did you get 1.8 W/m^2 for the backradiation contribution? That's totally wrong, so the conclusion is totally wrong.

Which would be less cooling ergo more warming....but that did not happen, neither is global warming by CO2 back radiation.

Quite the contrary. Your can did not start to freeze, hence your experiment proves backradiation was present.
Yes the 1.8 W/m^2 is actually less than what they claim:
Carbon_Dioxide_radiative_forcing.png

It's you who has the math all wrong.
(emissivity of aluminum = 0.9)
0.9 * 300^4 * 5.67E-08 = 413 W/m^2 (cooler can)

According to you the cooling rate of the water in the Alu can is supposed to be "adjusted" for the emissivity.
As if the emissivity of the pop can that the water was in did not slow the cooling process down by a factor of 0.9 already. Where would I get a pop can that has an emissivity of 1 ???
The other blunder you committed was picking the temperature after the can had cooled for an entire hour to get your numbers.
Worst of all is your statement:
That's going to vary considerably as the temperature cools..
The can transferred 1665 calories (6996 watt sec) in 1 hour to the air.
At the start the caloric content was 1665 calories more than after 1 hour and while it did loose these 1665 calories (6996 Watt seconds) during this hour it happened in this universe obeying all physics laws in it.
So it is entirely legitimate to say that a body that emits 66.5 W/m^2 will have emitted exactly the same amount of energy as a body with a 0.0292 m^2 surface that lost 6996 watt seconds in 1 hour.
If you got a problem with that then you got a problem with everything else. Like how much energy you can get out of a battery or a tank of fuel. It won't matter if the rate of consumption varies while you consume the energy the total available energy you used was how many K or M watt hours you had in the reservoir.
According to you a battery or a tank of fuel contains more energy the slower the rate of consumption and less energy at a higher rate of consumption.
And then there is the most ridiculous thing you said:
Okay so far. 7000 joules. Don't overdo the precision.
7000J/3600S/0.0292m^2 = 67 W/m^2. So far, so good.

No, that is not good it was equal to 66.5 W/m^2 not 67!
And the precision was way better than any of the fractional degree crap you guys quote after it's been washed several times by "averaging"
I have used my measuring measuring calipers to machine metal parts to within 1/1000's of an inch.
My scale is accurate to within 10 milligrams and my thermometer is accurate within 1/10th of a degree.
On top of that I have done a whole pile of Calorimetry and differential Calorimetry determinations during my employ. Only somebody who has no clue how these analysis are performed would deny that level of precision and claim they "overdo the precision"
When I looked at this thread before I signed in and the idiot I blocked was visible I noticed that idiotic picture where Spencer pointed an IR thermometer at the sky and got some ridiculously high temperature...but on the picture you can't read the display. When I did that at a ground temperature of + 37 C I got a sky temperature of -18.5 C and I did post that a couple of years ago...and you commented on it, saying that was expected
 
Wait till the can with the warm water has cooled to 10 deg C above room temperature and note the time.
After 1 hour note the water temperature. Next measure the exact dimensions of the soda can and the amount of water in it. That allows you to get the cals per second or the # of watts.
The pop can surface area was 292.3 cm^2 the water in it was 370 ccm and it cooled from 34.5 C to 30 C in 1 hour. So it transferred 1665 calories (6996 watt sec) in 1 hour to the air that was 10.5 deg C cooler.

Okay so far. 7000 joules. Don't overdo the precision.
7000J/3600S/0.0292m^2 = 67 W/m^2. So far, so good.

With the STB equation for this temperature difference you get 65 Watt/m^2

That's going to vary considerably as the temperature cools..

0.9 * 304.5^4 *5.67E-08 = 438 W/m^2 (warmer can) (emissivity of aluminum = 0.9)

0.9 * 300^4 * 5.67E-08 = 413 W/m^2 (cooler can)

0.9 * 294.5^4 * 5.67E-08 = 383 W/m^2 (room)

So, 55W/m^2 net radiation loss at the start, 30W/m^2 at the end.

If conduction losses are added to radiation losses, that would match the 67 W/m^2 average loss, so the experiment is consistent with backradiation. Without the 383 W/m^2 backradiation contribution, your can would have been losing 400+W/m^2, and would have been starting to freeze by the end of the hour.

I observed 66.5 Watt/m^2 which of course includes heat conduction from the water through the thin Alu skin and into still air (convection=negligeble)....which are the extra 1.5 watt/m^2 (heat conduction) over the theoretical 65 W radiative transfer
If the climate warmers were right I should have only had 65(StB) - 1.8 (back radiation)+ my 1.5 conduction = 64.7 Watts/m^2

Where did you get 1.8 W/m^2 for the backradiation contribution? That's totally wrong, so the conclusion is totally wrong.

Which would be less cooling ergo more warming....but that did not happen, neither is global warming by CO2 back radiation.

Quite the contrary. Your can did not start to freeze, hence your experiment proves backradiation was present.
Yes the 1.8 W/m^2 is actually less than what they claim:
Carbon_Dioxide_radiative_forcing.png

It's you who has the math all wrong.
(emissivity of aluminum = 0.9)
0.9 * 300^4 * 5.67E-08 = 413 W/m^2 (cooler can)

According to you the cooling rate of the water in the Alu can is supposed to be "adjusted" for the emissivity.
As if the emissivity of the pop can that the water was in did not slow the cooling process down by a factor of 0.9 already. Where would I get a pop can that has an emissivity of 1 ???
The other blunder you committed was picking the temperature after the can had cooled for an entire hour to get your numbers.
Worst of all is your statement:
That's going to vary considerably as the temperature cools..
The can transferred 1665 calories (6996 watt sec) in 1 hour to the air.
At the start the caloric content was 1665 calories more than after 1 hour and while it did loose these 1665 calories (6996 Watt seconds) during this hour it happened in this universe obeying all physics laws in it.
So it is entirely legitimate to say that a body that emits 66.5 W/m^2 will have emitted exactly the same amount of energy as a body with a 0.0292 m^2 surface that lost 6996 watt seconds in 1 hour.
If you got a problem with that then you got a problem with everything else. Like how much energy you can get out of a battery or a tank of fuel. It won't matter if the rate of consumption varies while you consume the energy the total available energy you used was how many K or M watt hours you had in the reservoir.
According to you a battery or a tank of fuel contains more energy the slower the rate of consumption and less energy at a higher rate of consumption.
And then there is the most ridiculous thing you said:
Okay so far. 7000 joules. Don't overdo the precision.
7000J/3600S/0.0292m^2 = 67 W/m^2. So far, so good.

No, that is not good it was equal to 66.5 W/m^2 not 67!
And the precision was way better than any of the fractional degree crap you guys quote after it's been washed several times by "averaging"
I have used my measuring measuring calipers to machine metal parts to within 1/1000's of an inch.
My scale is accurate to within 10 milligrams and my thermometer is accurate within 1/10th of a degree.
On top of that I have done a whole pile of Calorimetry and differential Calorimetry determinations during my employ. Only somebody who has no clue how these analysis are performed would deny that level of precision and claim they "overdo the precision"
When I looked at this thread before I signed in and the idiot I blocked was visible I noticed that idiotic picture where Spencer pointed an IR thermometer at the sky and got some ridiculously high temperature...but on the picture you can't read the display. When I did that at a ground temperature of + 37 C I got a sky temperature of -18.5 C and I did post that a couple of years ago...and you commented on it, saying that was expected

Oh, poop4brains, your delusional denier cult pseudo-science OP and this last demented post all got completely debunked by actual real world experimental scientific evidence (two different experiments, actually) provided by a somewhat sceptical climate scientist, Dr. Roy Spencer, in post #10.

Now you are just making yourself look even more retarded than you always do.
 
Yes the 1.8 W/m^2 is actually less than what they claim:

No, totally wrong. 1.8 W/m^2 is not the backradiation. "Radiative forcing" is the net increase in energy absorbed by the earth. It's in the neighborhood of the change in the backradiation, not the total backradiation.

Since you botched that so totally, your results are a total botch.

Again, without backradiation, your can would have cooled enough over the hour by radiation enough to begin freezing (assuming no conduction either way). It didn't start freezing because backradiation was adding energy to it, and thus reducing the overall heat loss. Thus, your experiment was kind of a demonstration of backradiation.

As if the emissivity of the pop can that the water was in did not slow the cooling process down by a factor of 0.9 already.

0.9 emissivity slows down the emission by 10%, not 90%.

Where would I get a pop can that has an emissivity of 1 ???

Blacken it with soot. That will get you close. However, it doesn't matter. It has very little net effect, being it also effects incoming backradiation.

The other blunder you committed was picking the temperature after the can had cooled for an entire hour to get your numbers.

I calculated numbers at beginning and end. Did you even read my post?

So it is entirely legitimate to say that a body that emits 66.5 W/m^2 will have emitted exactly the same amount of energy as a body with a 0.0292 m^2 surface that lost 6996 watt seconds in 1 hour.

Yes, that average heat loss rate is correct, and I said so. It's a useful ballpark figure.

If you got a problem with that then you got a problem with everything else. Like how much energy you can get out of a battery or a tank of fuel. It won't matter if the rate of consumption varies while you consume the energy the total available energy you used was how many K or M watt hours you had in the reservoir.

I had no problem with it, which would be why I wrote "7000J/3600S/0.0292m^2 = 67 W/m^2. So far, so good." Please stop pretending that I said the exact opposite of what I actually said.

And then there is the most ridiculous thing you said:
Okay so far. 7000 joules. Don't overdo the precision.
7000J/3600S/0.0292m^2 = 67 W/m^2. So far, so good.

No, that is not good it was equal to 66.5 W/m^2 not 67!

You claimed a surface area of the can of 292.3 cm^2. That kind of precision is laughable, considering the other parameters. You didn't account for conduction. You didn't account for the bottom of the can being on the tabletop, which meant the table under it soon warmed to match the can, and the radiation exchange there was a wash.

And the precision was way better than any of the fractional degree crap you guys quote after it's been washed several times by "averaging"

So you don't understand basic statistics, check.

I have used my measuring measuring calipers to machine metal parts to within 1/1000's of an inch.
My scale is accurate to within 10 milligrams and my thermometer is accurate within 1/10th of a degree.
On top of that I have done a whole pile of Calorimetry and differential Calorimetry determinations during my employ. Only somebody who has no clue how these analysis are performed would deny that level of precision and claim they "overdo the precision"

4-digit precision, like you did with the surface area, is totally pointless when major parts of your experiment are a fudge factor. Overall, you didn't even manage 1-digit precision.

When I looked at this thread before I signed in and the idiot I blocked was visible I noticed that idiotic picture where Spencer pointed an IR thermometer at the sky and got some ridiculously high temperature...but on the picture you can't read the display. When I did that at a ground temperature of + 37 C I got a sky temperature of -18.5 C and I did post that a couple of years ago...and you commented on it, saying that was expected

Spencer was right when he measured about 0C. You measured something colder. it was backradiation both times. The sky isn't always the same temperature.
 
The sun warms the ground and the ground warms the air. That we can all agree on.
Heat can only be transferred from a warmer mass to a colder one, but the AGW advocates argue that heat can also be transferred from the colder mass to the warmer one. Climate "scientists" must insist on it else they don't have a case.
That means that the air has to be able to warm the warmer ground with "back radiation", which they say has increased with the increase in CO2.
They don't worry about mass or specific heat and simply apply black body physics for the radiative heat transfer while eyeballing heat conduction and convection yet claim that their results are accurate within a fraction of a degree.
Any scientist and/or engineer who has to deal with heat transfer in the real world knows how ridiculous this claim is.
Since 7/10 th of the earth surface is water and air at 400 ppm is available and free it is very easy to prove that
All it takes is 2 soda cans filled with water, one hot an another one cold, take note of the room temperature and observe the heat transfer.
Wait till the can with the warm water has cooled to 10 deg C above room temperature and note the time.
After 1 hour note the water temperature. Next measure the exact dimensions of the soda can and the amount of water in it. That allows you to get the cals per second or the # of watts.
The pop can surface area was 292.3 cm^2 the water in it was 370 ccm and it cooled from 34.5 C to 30 C in 1 hour. So it transferred 1665 calories (6996 watt sec) in 1 hour to the air that was 10.5 deg C cooler.
With the STB equation for this temperature difference you get 65 Watt/m^2
I observed 66.5 Watt/m^2 which of course includes heat conduction from the water through the thin Alu skin and into still air (convection=negligeble)....which are the extra 1.5 watt/m^2 (heat conduction) over the theoretical 65 W radiative transfer
If the climate warmers were right I should have only had 65(StB) - 1.8 (back radiation)+ my 1.5 conduction = 64.7 Watts/m^2
Which would be less cooling ergo more warming....but that did not happen, neither is global warming by CO2 back radiation.

You're either sandbagging here and you KNOW the truth or you have a bad case of purposely disregarding Radiative Physics. In the surface to sky RADIATIVE transfer of LW IR -- the colder sky ALWAYS WINS. The surface is losing thermal energy thru conduction, convection,and EM radiation 24 hours a day. So "back radiation" violates NOTHING. And the back radiation is only part of the measurement of the EM surface RADIATION loss.

The transfer of LWIR is a simple subtraction, backed by some more complex geometry to determine the radiation surfaces and which elements are actually in an optical path of that surface. So if the sky becomes slightly "warmer" by harboring gases capable of storing and releasing LWIR towards the ground -- it will only reduce the net loss to the sky by a small amount.

The "2 can" experiment you proposed cannot possibly recreate a realistic RADIATIVE "heat" exchange. Because of the small "geometry" of each other in total field of view where EM radiation can source from. And the materials involved. Most IR from the sky wouldn't even IMPINGE on the majority of the can surface.

Back rad is an instantaneous POWER calculation. NOT an energy calculation..
 
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The bogus OP and this whole bullshit thread got totally debunked in post #10. Only hard-core denier cult retards are still trying to push this anti-science nonsense.
 
The sun warms the ground and the ground warms the air. That we can all agree on.
Heat can only be transferred from a warmer mass to a colder one, but the AGW advocates argue that heat can also be transferred from the colder mass to the warmer one. Climate "scientists" must insist on it else they don't have a case.
That means that the air has to be able to warm the warmer ground with "back radiation", which they say has increased with the increase in CO2.
They don't worry about mass or specific heat and simply apply black body physics for the radiative heat transfer while eyeballing heat conduction and convection yet claim that their results are accurate within a fraction of a degree.
Any scientist and/or engineer who has to deal with heat transfer in the real world knows how ridiculous this claim is.
Since 7/10 th of the earth surface is water and air at 400 ppm is available and free it is very easy to prove that
All it takes is 2 soda cans filled with water, one hot an another one cold, take note of the room temperature and observe the heat transfer.
Wait till the can with the warm water has cooled to 10 deg C above room temperature and note the time.
After 1 hour note the water temperature. Next measure the exact dimensions of the soda can and the amount of water in it. That allows you to get the cals per second or the # of watts.
The pop can surface area was 292.3 cm^2 the water in it was 370 ccm and it cooled from 34.5 C to 30 C in 1 hour. So it transferred 1665 calories (6996 watt sec) in 1 hour to the air that was 10.5 deg C cooler.
With the STB equation for this temperature difference you get 65 Watt/m^2
I observed 66.5 Watt/m^2 which of course includes heat conduction from the water through the thin Alu skin and into still air (convection=negligeble)....which are the extra 1.5 watt/m^2 (heat conduction) over the theoretical 65 W radiative transfer
If the climate warmers were right I should have only had 65(StB) - 1.8 (back radiation)+ my 1.5 conduction = 64.7 Watts/m^2
Which would be less cooling ergo more warming....but that did not happen, neither is global warming by CO2 back radiation.

There is an easier experiment than that polar bear....take a solar oven ( I can provide plans to easily build one that works nicely...the cost of materials is about $20)....put a thermometer at the focal point and point the disk towards clear open sky...watch the temperature drop. If the claimed back radiation from the atmosphere were reality, the temperature would not be dropping. Now, if the ambient temperature is 45F or below, with the disk pointed at clear open sky, put a little container of water at the focal point...it will freeze. Again, if back radiation were a reality, that water would not be freezing when the ambient temperature is 13F above the freezing point.
 
Any scientist and/or engineer who has to deal with heat transfer in the real world knows how ridiculous this claim is.

They do? Can you show us one?

Practically all of them...the oregon petition is chock full of engineers who deal with actual physics all day every day...very few actual engineers are on the AGW crazy train but it seems that the train is well supplied with fake engineers.
 
The Oregon Petition makes no statement about heat gain by backradiation and makes absolutely no attempt to verify that signers are what they say they are. As far as a survey go, it's complete ratshit. And if you think even the bogus numbers on the petition are a significant sample of the world's engineers, you need a reality check.

Let's see a consensus among engineers and/or scientists who state that backradiation cannot or does not result in surface warming.
 
If by "most people attribute this to the sun" then you're running with a crowd of hermits. Most people attribute this to carbon dioxide produced by humans burning coal and petroleum. We CAN do something about it, but it is going to take a big effort and it is going to take time. If we do NOT do anything about it, its going to be very bad for our children and theirs.
That's typical doomsday cult drivel.
 

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