The pseudo science of man-made global warming...

[IanC]

Yanking your own chain like that is called mental masturbation....you let me know when you get an observed, measured instance of back radiation gathered with an instrument at ambient temperature.

As we've pointed out before, anyone can buy a longwave IR camera, one that works at ambient temperature, which does exactly that. Point it at the sky, it displays in image showing the differing temperatures of clouds and sky. It is clearly measuring the IR radiation coming down from the very cold sky. That is, the backradiation.

We've also pointed out that sensors never needed to be chilled to measure backradiation. Chilling just lowered the thermal noise, making the image clearer. Modern electronics found around around that need for chilling.

Now, wait until you see how SSDD handwaves that away. It's both hilarious and pathetic.

That brings an interesting quiz to mind.
Suppose you point an IR sensor at a target which is an ice cube right next to another same sized object which is at +20C. How many watts/m^2 should you see on the instrument`s display?
The ice cube @ 0 C radiates just under 314 w/m^2 and the 20 degC object radiates almost 418 w/m^
I take it you would say that the instrument would display the sum of both, a total of 732 w/m^2 ?
And if I can get only 400 watts/m^2 you would say that there is something wrong with my IR gun?
Buy one and try it out. Maybe you can come up with an explanation why it`s never higher than that. Off the shelf these are calibrated for Temperature but from that you can calculate the watts/m^ 2. Mine has a laser pointer so it`s quite easy to center it and you can find out the proper distance just as easily if you move back till the indicated temperature deviates to below the known temperature of the target.
I also have a thermal imaging camera, but it does not have a USB to connect to another storage device else I would upload some pictures.
If I put a cold object next to a warm one, the side of the warm object facing the cold one shows up cooler than the rest of that object...conforming to "a cold object can make a warm object even warmer" my camera should not see what it is seeing...explain that

Been drinking Bernie?

Accusing crick of believing that a gallon of 10C water added to a gallon of 40C water will result in 50C water is pretty low even for you.

If it is minus 20C outside your warm object will be warmer if you put a block of ice at only minus 5C next to it. So what was your point?

 

[SSDD]

why are you insulting Wuwei while at the same time ignoring SSDD's nonsense?

you did the same thing years ago when you fought me tooth and nail while ignoring wirebender's equally absurd nonsense. actually I think SSDD and wirebender are one and the same but that is a different topic.

What is this obsession you have with me and a poster who is no longer here?...We have been through this before...the moderators checked and confirmed that me and the other guy are not the same person. At your suggestion, I read quite a few of his posts and his arguments and ideas are different from mine although he regularly pointed out that there was no such thing as back radiation also. In addition, he used mathematics far more heavily than I ever could. Get over your obsession and admit that there is no such thing as back radiation.

 

[polarbear]

Yanking your own chain like that is called mental masturbation....you let me know when you get an observed, measured instance of back radiation gathered with an instrument at ambient temperature.

As we've pointed out before, anyone can buy a longwave IR camera, one that works at ambient temperature, which does exactly that. Point it at the sky, it displays in image showing the differing temperatures of clouds and sky. It is clearly measuring the IR radiation coming down from the very cold sky. That is, the backradiation.

We've also pointed out that sensors never needed to be chilled to measure backradiation. Chilling just lowered the thermal noise, making the image clearer. Modern electronics found around around that need for chilling.

Now, wait until you see how SSDD handwaves that away. It's both hilarious and pathetic.

That brings an interesting quiz to mind.
Suppose you point an IR sensor at a target which is an ice cube right next to another same sized object which is at +20C. How many watts/m^2 should you see on the instrument`s display?
The ice cube @ 0 C radiates just under 314 w/m^2 and the 20 degC object radiates almost 418 w/m^
I take it you would say that the instrument would display the sum of both, a total of 732 w/m^2 ?
And if I can get only 400 watts/m^2 you would say that there is something wrong with my IR gun?
Buy one and try it out. Maybe you can come up with an explanation why it`s never higher than that. Off the shelf these are calibrated for Temperature but from that you can calculate the watts/m^ 2. Mine has a laser pointer so it`s quite easy to center it and you can find out the proper distance just as easily if you move back till the indicated temperature deviates to below the known temperature of the target.
I also have a thermal imaging camera, but it does not have a USB to connect to another storage device else I would upload some pictures.
If I put a cold object next to a warm one, the side of the warm object facing the cold one shows up cooler than the rest of that object...conforming to "a cold object can make a warm object even warmer" my camera should not see what it is seeing...explain that

Been drinking Bernie?

Accusing crick of believing that a gallon of 10C water added to a gallon of 40C water will result in 50C water is pretty low even for you.

If it is minus 20C outside your warm object will be warmer if you put a block of ice at only minus 5C next to it. So what was your point?

No I have not been accusing anybody of anything and in addition to that (if you read my post) that was addressed to mamooth not Crick
It`s a very simple question to answer if you got an IR sensor. Since you decided to play referee here you might as well try it out and answer that question.
It`s a simple experiment. Put 2 soda cans in the fridge and let them cool off let 2 more warm up to room temperature. Then point the IR detector at center between the 2 cans which are say at +20C and note the reading + the distance from the cans
Then use the cold cans also side by side and repeat the procedure. After that replace one of the cans with a cold one and tell me what you get for a reading.
If you don`t want to get an instrument like that then just do the math.
Would you do average the 2 temperatures [in K of course] first and then multiply the 4 th power of that average by σ or not ?
If you choose to do it like that then explain why you would not do it the same way if you know the temperature of one body and the watts/m^2 of the other one...as is the case if you wanted to calculate how a X-amount of radiation from a colder source would heat another body which is warmer.
And also do explain why in your mind a simple physics quiz could be an insult. Mamooth whom I addressed did not complain about it. Anyway forgive me I`m currently not involved with any of the so called universities and "progressive" campus policy, so did not get the memo how to avoid intruding into the "safe space" of sissies

 

[polarbear]

And while I was at it I also came across this:

is the effective emissivity of earth, about 0.612
If you disagree then take it up with the authors of this page:

You are wrong about sigma regarding absorption.
Emissivity is wavelength dependent. The area of interest is IR from 3 to 20 microns. It varies around 0.95.

You should heed common sense which tells you "if you are in a hole quit digging"
But you persist digging deeper and doubling down.
For the last time σ has nothing to do with absorption and an emission spectrum of a substance has nothing to do with absorption !!!
And then there is that picture you posted with no link to the source,
I found it anyways, it was:
Far-infrared surface emissivity and climate
No wonder you concealed it because they clearly state that this graph is not based on actual data but is a model:

Far-IR emissivity can be measured from spectrally resolved observations, but such measurements have not yet been made. ...
there is no comprehensive knowledge of angularly averaged terrestrial surface emissivity outside of the laboratory at far-IR wavelengths because there are no comprehensive measurements at those wavelengths. What is known about far-IR surface emissivity is derived from laboratory measurements and research on planetary environment

Then you pulled this out of your hat: "The area of interest is IR from 3 to 20 microns."
The author of that graph clearly stated this:

Terrestrial emission plays a critical role in the climate system (1), and over 99% of this radiation occurs in the wavelength range from 5 to 100 μm (2,000 cm−1 to 100 cm−1). However, there have been very few spectrally resolved measurements of terrestrial emission at wavelengths between 15.4 μm and 100 μm (650 cm−1 to 100 cm−1), often referred to as the far infrared,

That graph shows 5μm and only up to 50μm but no matter that`s where they say 99% of the IR energy is radiated...their "area of interest" extends to 15.4 μm because of the CO2 absorption band.
So let`s re-hash why and what exactly I did not agree with in your original statement where you said:
"The average emissivity of earth water,ice, dirt and plants is 0.95" and now you finally admit that it is wavelength dependent because it dawned on you that it does depend on the material which emits the radiation. I also showed you with a link that the earth average emissivity is ~ 0.612
Right here:
Climate model - Wikipedia

is the effective emissivity of earth, about 0.612

But you still persist that it is 0.95
And I don`t know why you can`t wrap your head around the difference between absorption and emission.
Both are depending on which substance you are dealing with...the entire field of spectroscopy hinges on that and a black body does not care about the wavelength. It absorbs at every wavelength and emits a continuous frequency spectrum that depends only on the body's temperature.
If you can`t understand that it`s your problem and I don`t really care what you believe.
I wish there were an english word for "rechthaberisch" it would describe you and others here to the "T" but since there isn`t I`ll quote Albert Einstein who said "Manche glauben dass Einbildung eine Ausbildung ist"
Auf wiedersehen und merry Christmas

 

[Crick]

Try "know-it-all"

And I think the problem may be that 0.612 is the EFFECTIVE emissivity. The actual emissivity may be something completely different and may be precisely what is called for depending on the application.

 

[ding]

Why do you believe that our present temperature is NOT 1.4C to 2.4C below the peak temperature of three of the past four interglacials?

Because, as has been pointed out over and over, your graph doesn't show the present.

And it points at a single location on earth, so no conclusions from global temperature can be drawn from it.

And the time scale is too compressed for any rate-of-change data to be drawn from it.

This is issue has been studied in detail, by very smart people, and good summary is here. Read it. Try to learn about the many, many ways that you've faceplanted here.

The Last Interglacial Part Two - Why was it so warm?

Look... it's catching on, lol

Late fall and -10 F in Chicago...global warming?

 

[Wuwei]

I also showed you with a link that the earth average emissivity is ~ 0.612

Crick is right. The source was looking at a simple model that gave an average that included clouds. Your source says,

This is because the above equation represents the effective radiative temperature of the Earth (including the clouds and atmosphere) ... This very simple model is quite instructive, ...​

Space Images | NASA Spacecraft Maps Earth's Global Emissivity

Narrowband emissivities less than 0.85 are typical for most desert and semi-arid areas due to the strong quartz absorption feature (reststrahlen band) between 8-9.5 μm range, whereas the emissivity of vegetation, water and ice cover are generally greater than 0.95 and spectrally flat in the 8-12 μm range.​

http://www.star.nesdis.noaa.gov/smcd/emb/lst/Ogawa2004_emissivity.pdf

The range of the broadband emissivity was found to be between 0.85 and 0.96 for the desert area. [Sahara]​

https://www-cave.larc.nasa.gov/pdfs/Wilber.NASATchNote99.pdf
Look at page 13-14. For emissivities of all kinds of earth coverage. All of them are way above 0.612

Please try to understand what your source is actually saying.

I am not interested in continuing your discourse on emissivity. You really don't understand it. I don't know why you are so obsessed with it. It has no bearing on the physical principles of the S-B equation since it is already explicitly included in the equation.

For the last time σ has nothing to do with absorption and an emission spectrum of a substance has nothing to do with absorption !!!

Sigma is a constant. It has nothing to do with spectra. The emissivity is a function of wavelength. However your original question was.

If you have a physics book in which you have been reading that the Stefan-Boltzmann constant has anything to do with absorption then you better chuck it out.

You are wrong. Sigma is in the equation derived from the SB equation. Sigma and emissivity are the same for both absorption and emission of energy. The equation has two terms. One involves emission and the other involves absorption.

 

[SSDD]

You are wrong. Sigma is in the equation derived from the SB equation. Sigma and emissivity are the same for both absorption and emission of energy. The equation has two terms. One involves emission and the other involves absorption.

You really don't have a clue do you? That equation has nothing whatsoever to do with absorption....That equation EXPLICITLY states: The amount of energy that a black body radiates is equal to its emissivity times the S-B constant times its area times the difference between its own temperature and the temperature of its surroundings to the 4th power....there is no mention in that equation about absorption...it is all about the amount of energy the black body is radiating....hang it up guy...you keep proving beyond question that you don't have even a loose grip on the basics.

Just for entertainment's sake though, how about you state in straight forward terms what you think that equation is stating...in the same format as I just told you what it is actually saying.

 

[Wuwei]

Just for entertainment's sake though, how about you state in straight forward terms what you think that equation is stating...in the same format as I just told you what it is actually saying.

There you go:

This is by far the clearest way of thinking about it

 

[SSDD]

Just for entertainment's sake though, how about you state in straight forward terms what you think that equation is stating...in the same format as I just told you what it is actually saying.

There you go:

This is by far the clearest way of thinking about it

I was pretty sure that you didn't have a clue....thanks for confirming...I asked you to state in your own words what you thought the equation YOU POSTED said...You cut and paste a different equation...as an explanation for what you believe the equation YOU POSTED actually said....you understand equations like crick understands graphs....that would be not at all.

But again, for entertainment's sake, tell me how you believe

is making the same mathematical statement as

 

[Wuwei]

I asked you to state in your own words what you thought the equation YOU POSTED said...You cut and paste a different equation...as an explanation for what you believe the equation YOU POSTED actually said....you understand equations like crick understands graphs....that would be not at all.

My own words would be very similar to the way science says it: equation 3. The final equation you referred to is at the far right.

But again, for entertainment's sake, tell me how you believe

is making the same mathematical statement as

You make a statement that is incorrect and expect me to respond to it for entertainment's sake.
It's not making the same mathematical statement It's only making half. Your first equation references Re, which is emission only. A similar equation Ra represents absorption. The net energy transfer is the difference, as in your second equation, and derived in equation 3.

 

[mamooth]

The ice cube @ 0 C radiates just under 314 w/m^2 and the 20 degC object radiates almost 418 w/m^
I take it you would say that the instrument would display the sum of both, a total of 732 w/m^2 ?

No, I wouldn't.

If each surface filled part of the aperture view, you should get a number somewhere between the two individual numbers.

Just like you did.

 

[ding]

The ice cube @ 0 C radiates just under 314 w/m^2 and the 20 degC object radiates almost 418 w/m^
I take it you would say that the instrument would display the sum of both, a total of 732 w/m^2 ?

No, I wouldn't.

If each surface filled part of the aperture view, you should get a number somewhere between the two individual numbers.

Just like you did.

You ever figure out what the present AGT is, lol?

 

[westwall]

I also showed you with a link that the earth average emissivity is ~ 0.612

Crick is right. The source was looking at a simple model that gave an average that included clouds. Your source says,

This is because the above equation represents the effective radiative temperature of the Earth (including the clouds and atmosphere) ... This very simple model is quite instructive, ...​

Space Images | NASA Spacecraft Maps Earth's Global Emissivity

Narrowband emissivities less than 0.85 are typical for most desert and semi-arid areas due to the strong quartz absorption feature (reststrahlen band) between 8-9.5 μm range, whereas the emissivity of vegetation, water and ice cover are generally greater than 0.95 and spectrally flat in the 8-12 μm range.​

http://www.star.nesdis.noaa.gov/smcd/emb/lst/Ogawa2004_emissivity.pdf

The range of the broadband emissivity was found to be between 0.85 and 0.96 for the desert area. [Sahara]​

https://www-cave.larc.nasa.gov/pdfs/Wilber.NASATchNote99.pdf
Look at page 13-14. For emissivities of all kinds of earth coverage. All of them are way above 0.612

Please try to understand what your source is actually saying.

I am not interested in continuing your discourse on emissivity. You really don't understand it. I don't know why you are so obsessed with it. It has no bearing on the physical principles of the S-B equation since it is already explicitly included in the equation.

For the last time σ has nothing to do with absorption and an emission spectrum of a substance has nothing to do with absorption !!!

Sigma is a constant. It has nothing to do with spectra. The emissivity is a function of wavelength. However your original question was.

If you have a physics book in which you have been reading that the Stefan-Boltzmann constant has anything to do with absorption then you better chuck it out.

You are wrong. Sigma is in the equation derived from the SB equation. Sigma and emissivity are the same for both absorption and emission of energy. The equation has two terms. One involves emission and the other involves absorption.

I notice that you all rely on "simple" models. Get back to me when you have a model at least as complex as what they use to model aero pieces for an F1 race car. THEIR models actually model reality. Unlike yours.

 

[polarbear]

The ice cube @ 0 C radiates just under 314 w/m^2 and the 20 degC object radiates almost 418 w/m^
I take it you would say that the instrument would display the sum of both, a total of 732 w/m^2 ?

No, I wouldn't.

If each surface filled part of the aperture view, you should get a number somewhere between the two individual numbers.

Just like you did.

Good answer and I never REALLY thought you would add them. That was meant just as a joke anyways.
Matter of fact you do get a number somewhere between the 2 individual numbers.
But here is the problem and why I posted and addressed it to you:
Look at Spencer`s experiment
Experiment Results Show a Cool Object Can Make a Warm Object Warmer Still « Roy Spencer, PhD
Experiment Results Show a Cool Object Can Make a Warm Object Warmer Still
August 28th, 2016 by Roy W. Spencer, Ph. D.

I recorded temperatures every 5 secs with the plate alternately exposed to a view of the ice for 5 minutes, then with the ice covered for 5 minutes. This cycling was repeated five times. The results are shown in Fig. 3. What we see is just what I would expect, that the temperature of the hot plate increases with time when its view of the ice is blocked by the room-temperature sheet.

And he leads off by saying this:
The experiment shown below does not prove that greenhouse gases in the atmosphere perform such a function, only that it is not a violation of the 2nd Law of Thermodynamics for a cooler object emitting infrared radiation to keep a warm object warmer that it would otherwise be if the cooler object was not present.
His conclusion is:

Conclusion
There is no violation of the 2nd Law of Thermodynamics in the experiment; a cool object can make a warm object even warmer still through infrared radiative effects. The phenomenon can only happen, though, if the cool object replaces something that is even colder, and thereby reduces the rate at which the warm object loses infrared energy to its surroundings. In this experiment, the room temperature plate takes the place of the ice which still emits at around 300 Watts per sq. meter; in the climate system, the atmosphere takes the place of deep space, which emits energy at close to 0 Watts per sq. meter.

So what do you think Spencer proved with his experiment?
Certainly not that a cold object can make a warm object even warmer. All it did prove is, that when the warm object was exposed to the uncovered ice box is that the colder object cooled off the warmer one and that a warm object next to a warmer one reduces the cooling of the warmer object...but not that this second object heated up the warmer one even more. That would imply that the second object is a heat source
And this is a meteorologist who is or was the Principal Research Scientist at the U of Alabama and a "former NASA Scientist"
No way would or should make a real physicist commit such a blunder. Amazing how low the bar is set at NASA for climate "scientists". That`s why some people call it pseudo science.
And you say?

 

[Crick]

Why do you people have such a problem with net transfer? Until I came here, I never met ANYONE who didn't understand the process instantly.

 

[SSDD]

Why do you people have such a problem with net transfer? Until I came here, I never met ANYONE who didn't understand the process instantly.

Still waiting for some observed, measured, quantified evidence of net energy transfer taken with instruments at ambient temperature...till that happens, net transfer only exists within unobservable, unmeasurable, untestable mathematical models...and models aren't reality.

 

[Wuwei]

Why do you people have such a problem with net transfer? Until I came here, I never met ANYONE who didn't understand the process instantly.

Still waiting for some observed, measured, quantified evidence of net energy transfer taken with instruments at ambient temperature...till that happens, net transfer only exists within unobservable, unmeasurable, untestable mathematical models...and models aren't reality.

If you don't believe in net transfer of radiation and you disagree that entropy allows net transfer. Do you disbelieve the properties of entropy too? Just what do you believe that stops photons from being emitted from colder objects to warmer objects.

Since entropy is a mathematical model. What substitute to a mathematical model do you have that makes you disbelieve net thermal energy flow?

 

[SSDD]

If you don't believe in net transfer of radiation and you disagree that entropy allows net transfer. Do you disbelieve the properties of entropy too? Just what do you believe that stops photons from being emitted from colder objects to warmer objects.

Since entropy is a mathematical model. What substitute to a mathematical model do you have that makes you disbelieve net thermal energy flow?

The second law of thermodynamics....
It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.

Let me know when they rewrite the law to support your view.

 

[IanC]

The ice cube @ 0 C radiates just under 314 w/m^2 and the 20 degC object radiates almost 418 w/m^
I take it you would say that the instrument would display the sum of both, a total of 732 w/m^2 ?

No, I wouldn't.

If each surface filled part of the aperture view, you should get a number somewhere between the two individual numbers.

Just like you did.

Good answer and I never REALLY thought you would add them. That was meant just as a joke anyways.
Matter of fact you do get a number somewhere between the 2 individual numbers.
But here is the problem and why I posted and addressed it to you:
Look at Spencer`s experiment
Experiment Results Show a Cool Object Can Make a Warm Object Warmer Still « Roy Spencer, PhD
Experiment Results Show a Cool Object Can Make a Warm Object Warmer Still
August 28th, 2016 by Roy W. Spencer, Ph. D.

I recorded temperatures every 5 secs with the plate alternately exposed to a view of the ice for 5 minutes, then with the ice covered for 5 minutes. This cycling was repeated five times. The results are shown in Fig. 3. What we see is just what I would expect, that the temperature of the hot plate increases with time when its view of the ice is blocked by the room-temperature sheet.

And he leads off by saying this:
The experiment shown below does not prove that greenhouse gases in the atmosphere perform such a function, only that it is not a violation of the 2nd Law of Thermodynamics for a cooler object emitting infrared radiation to keep a warm object warmer that it would otherwise be if the cooler object was not present.
His conclusion is:

Conclusion
There is no violation of the 2nd Law of Thermodynamics in the experiment; a cool object can make a warm object even warmer still through infrared radiative effects. The phenomenon can only happen, though, if the cool object replaces something that is even colder, and thereby reduces the rate at which the warm object loses infrared energy to its surroundings. In this experiment, the room temperature plate takes the place of the ice which still emits at around 300 Watts per sq. meter; in the climate system, the atmosphere takes the place of deep space, which emits energy at close to 0 Watts per sq. meter.

So what do you think Spencer proved with his experiment?
Certainly not that a cold object can make a warm object even warmer. All it did prove is, that when the warm object was exposed to the uncovered ice box is that the colder object cooled off the warmer one and that a warm object next to a warmer one reduces the cooling of the warmer object...but not that this second object heated up the warmer one even more. That would imply that the second object is a heat source
And this is a meteorologist who is or was the Principal Research Scientist at the U of Alabama and a "former NASA Scientist"
No way would or should make a real physicist commit such a blunder. Amazing how low the bar is set at NASA for climate "scientists". That`s why some people call it pseudo science.
And you say?

Conclusion
There is no violation of the 2nd Law of Thermodynamics in the experiment; a cool object can make a warm object even warmer still through infrared radiative effects. The phenomenon can only happen, though, if the cool object replaces something that is even colder, and thereby reduces the rate at which the warm object loses infrared energy to its surroundings. In this experiment, the room temperature plate takes the place of the ice which still emits at around 300 Watts per sq. meter; in the climate system, the atmosphere takes the place of deep space, which emits energy at close to 0 Watts per sq. meter.

Spencer proved exactly what he set out to prove. Your so-called criticism agrees with what he stated.


Personally I think Spencer should have emphasized that the warm object has a heat source, and that the temperature of the warm object is a combination of both energy input and energy output. Without an energy source everything just cools.

Look at the bottom graph. with the ice shielded the temperature jumps two degrees, when the ice is exposed it drops down two degrees. the local conditions changed and the temperature equation changed and the temperature moved to reflect that change. It is easier to visualize how the heated plate quickly cools and stabilizes when exposed to the ice than it is to understand what is happening when the ice is shielded and the plate starts to warm up. Where does the energy needed to warm up the plate come from?

It comes from the energy NOT lost to the environment. That energy is stored in the plate and is expressed as an increase in temperature. The energy stored is exactly the same as the extra energy released when the plate is exposed to the ice and cools down.

Now switch over to the Earth and its atmosphere. There is a tremendous amount of energy stored in the atmosphere as kinetic and potential energy. Energy that would be directly lost to deep space if no solar input was present to keep it aloft. Everything above zero degrees Kelvin radiates according to its temperature. Everything can be either warmer or cooler than its surroundings but it is always radiating. The atmosphere is cooler than the surface but warmer than space but it sends radiation to both the surface and to space.

Just like Spencer's experiment, the atmosphere is cooler than the surface (plate) but not as cool as space (ice). The presence of the atmosphere increases the surface temperature by lowering heat loss to space, the energy not lost to space is the source of temperature change at the surface.

Still dont believe me? Imagine what would happen if solar input just stopped. The Earth would continue to radiate and cool. Until all the stored energy was lost to space as the atmosphere collapsed into a frozen crust on the surface.

Matter at any temperature can be a net absorber or emitter of radiation. A glass of ice water melts above 0C or freezes solid below 0C but it is giving off the same amount of radiation until it does one or the other.