Guest Post by Willis Eschenbach
Pushed by a commenter on another thread, I thought I’d discuss the R. W. Wood experiment, done in 1909. Many people hold that this experiment shows that CO2 absorption and/or back-radiation doesn’t exist, or at least that the poorly named “greenhouse effect” is trivially small. I say it doesn’t show anything at all. Let me show you the manifold problems with the experiment.
To start with, let me give a curious example of the greenhouse effect, that of the Steel Greenhouse. Imagine a planet in the vacuum of space. A residue of nuclear material reacting in the core warms it to where it is radiating at say 235 watts per square metre (W/m2). Figure 1 shows the situation.
Figure 1. Planet in outer space, heated from the interior. Drawing show equilibrium situation
This planet is at equilibrium. The natural reactor in the core of the planet is generating power that at the planet’s surface amounts to 235 W/m2. It is radiating the same amount, so it is neither warming nor cooling.
Now, imagine that without changing anything else, we put a steel shell around the planet. Figure 2 shows that situation, with one side of the shell temporarily removed so we can look inside.
Figure 2. As in Figure 1, but with a solid steel shell surrounding the planet. Near side of the shell temporarily removed to view interior. Vertical distance of the shell from the surface is greatly exaggerated for clarity—in reality the shell and the shell have nearly the same surface area. (A shell 6 miles (10 km) above the Earth has an exterior area only 0.3% larger than the Earth’s surface area.)
[UPDATE: Misunderstandings revealed in the comments demonstrated that I lacked clarity. To expand, let me note that because the difference in exterior surface area of the shell and the surface is only 0.3%, I am making the simplifying assumption that they are equal. This clarifies the situation greatly. Yes, it introduces a whopping error of 0.3% in the calculations, which people have jumped all over in the comments as if it meant something … really, folks, 0.3%? If you like, you can do the calculations in total watts, which comes to the same answer. I am also making the simplifying assumption that both the planet and shell are “blackbodies”, meaning they absorb all of the infrared that hits them.]
Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium. The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of power is emitted to space as in Figure 1.
And that is all that there is to the poorly named greenhouse effect. It does not require CO2 or an atmosphere, it can be built out of steel. It depends entirely on the fact that a shell has two sides and a solid body only has one side.
Now, this magical system works because there is a vacuum between the planet and the shell. As a result, the planet and the shell can take up very different temperatures. If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.
Another way to lower the efficiency of the system is to introduce an atmosphere. Each watt of power lost by atmospheric convection of heat from the surface to the shell reduces the radiation temperature of the surface by the same amount. If the atmosphere can conduct the surface temperature effectively enough to the shell, the surface ends up only slightly warmer than the shell.
Let me summarize. In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.
Now, remember that I started out to discuss the R. W. Wood experiment. Here is the report of that experiment, from the author. I have highlighted the experimental setup.
Note on the Theory of the Greenhouse
By Professor R. W. Wood (Communicated by the Author)
THERE appears to be a widespread belief that the comparatively high temperature produced within a closed space covered with glass, and exposed to solar radiation, results from a transformation of wave-length, that is, that the heat waves from the sun, which are able to penetrate the glass, fall upon the walls of the enclosure and raise its temperature: the heat energy is re-emitted by the walls in the form of much longer waves, which are unable to penetrate the glass, the greenhouse acting as a radiation trap.
I have always felt some doubt as to whether this action played any very large part in the elevation of temperature. It appeared much more probable that the part played by the glass was the prevention of the escape of the warm air heated by the ground within the enclosure. If we open the doors of a greenhouse on a cold and windy day, the trapping of radiation appears to lose much of its efficacy. As a matter of fact I am of the opinion that a greenhouse made of a glass transparent to waves of every possible length would show a temperature nearly, if not quite, as high as that observed in a glass house. The transparent screen allows the solar radiation to warm the ground, and the ground in turn warms the air, but only the limited amount within the enclosure. In the “open,” the ground is continually brought into contact with cold air by convection currents.
To test the matter I constructed two enclosures of dead black cardboard, one covered with a glass plate, the other with a plate of rock-salt of equal thickness. The bulb of a thermometer was inserted in each enclosure and the whole packed in cotton, with the exception of the transparent plates which were exposed. When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass. In order to eliminate this action the sunlight was first passed through a glass plate.
There was now scarcely a difference of one degree between the temperatures of the two enclosures. The maximum temperature reached was about 55 oC. From what we know about the distribution of energy in the spectrum of the radiation emitted by a body at 55 o, it is clear that the rock-salt plate is capable of transmitting practically all of it, while the glass plate stops it entirely. This shows us that the loss of temperature of the ground by radiation is very small in comparison to the loss by convection, in other words that we gain very little from the circumstance that the radiation is trapped.
Is it therefore necessary to pay attention to trapped radiation in deducing the temperature of a planet as affected by its atmosphere? The solar rays penetrate the atmosphere, warm the ground which in turn warms the atmosphere by contact and by convection currents. The heat received is thus stored up in the atmosphere, remaining there on account of the very low radiating power of a gas. It seems to me very doubtful if the atmosphere is warmed to any great extent by absorbing the radiation from the ground, even under the most favourable conditions.
I do not pretend to have gone very deeply into the matter, and publish this note merely to draw attention to the fact that trapped radiation appears to play but a very small part in the actual cases with which we are familiar.
Here would be my interpretation of his experimental setup:
Figure 3. Cross section of the R. W. Wood experiment. The two cardboard boxes are painted black. One is covered with glass, which absorbs and re-emits infrared. The other is covered with rock salt, which is transparent to infrared. They are packed in cotton wool. Thermometers not shown.
Bearing in mind the discussion of the steel greenhouse above, I leave it as an exercise for the interested reader to work out why this is not a valid test of infrared back-radiation on a planetary scale … please consider the presence of the air in the boxes, the efficiency of the convective heat transfer through that air from the box to the cover plates, the vertical temperature profile of that air, the transfer of power from the “surface” to the “shell” through the walls of the box, and the relative temperatures of the air, the box, and the transparent cover.
Seems to me like with a few small changes it could indeed be a valid test, however.
Best regards,
w.
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Ryan says (February 8, 2013 at 8:16 am): “Go back to Willis’ shell system, but instead of a thermo-nuclear reaction consider a massive ball of water at a smidgen under 100Celsius. In the first diagram the water is cooling into space quite freely. But what about the second diagram. You are now reflecting a lot of IR back to the water, i.e. it seems to be receiving a lot more energy. Does that mean the water will now boil, or will it just stay a smidgen under 100Celsius?”
Neither. The water will gradually cool to the “temperature” of space, because the system as a whole (inner sphere plus shell) is still cooling to space via radiation from the shell.
Your question suggests that you don’t fully understand Willis’s example.
Ryan says:
“Thus the NET emissions from the planets surface are exactly as they were before. Thus the temperature of the planet’s surface is exactly the same – all that is happening is that the back-radiation that is reflected towards the planet from the shell is simply reflected back towards the shell again (maybe the energy is absorbed and then re-emitted again but from an energy balance point of view we don’t need to worry about those details). In effect, Willis simple energy balance model demonstrates that back radiation of IR cannot heat the surface of the planet, even when it is 100% – it simply increases the local radiative losses from the surface of the planet.”
Yes.
It seems with shell you could trap a lot low energy photons. I think one effect of doing this would
it would interfere with thermal goggles ability to see very far. And tend warm things which were fairly cool. Question is how bad could this IR pollution get and could get to point where there some mass to all this photons.
Greg House says:
February 7, 2013 at 10:01 pm
Willis Eschenbach says, February 7, 2013 at 9:11 pm: ” Show up with science in your hand, citations saying cold objects CAN’T radiate to hot objects, or don’t show up at all.”
=============================================================
It is not about whether colder objects can or can not radiate to warmer objects.
It is about whether radiation of colder objects affects temperature of warmer objects. That is the point, please, note that.
I have never seen any valid link to a real scientific experiment proving it does. Nobody I talked to on various blogs was able to present such a link. No link, no description of a real scientific experiment proving that point directly, by actual temperature measurements was presented. People did present fictional stories or general references to the same unproven claims, but nothing real. The fictional stories about plates, shells etc have been around for years, but again, nothing real has come up until now.
There’re plenty of such experiments and they have been documented ad nauseam, your refusal to read texts etc. doesn’t make them fiction it makes you a liar!
One I have described here involves the use of a thermocouple. If you immerse a thermocouple in flame gases it will measure a temperature which is lower than the adiabatic flame temperature because it is losing heat via radiation to its surroundings. If you mount a radiation shield around the ThC which still allows conduction and convection between the flame and ThC. The radiation shield gets hotter than the surroundings and so heats the ThC up and a higher temperature is recorded than in the absence of the shield, double shields increase it further. Well researched by NACA back in the 50’s for example. Below is a link with some examples of heat transfer problems from a text book, 12-89 shows the calculations for such a shield. Now shut up!
http://tinyurl.com/asffsbx
Genius; All horses are purple.
Horse Breeder; Uh…. no they aren’t.
Genius; Yes they are. I read it in a book from 1906.
Horse Breeder; Uhm… I’m a horse breeder, I’ve studied horses my whole life, and they aren’t purple.
Genius; The book from 1906 proves they are.
Horse Breeder; Look, here are 16 text books on horses, everything from the various breeds and strains to special cases like zebra and even half breed strains like mules, and not one of them mentions purple horses.
Genius; Those were all proven wrong by the book from 1906 that I read.
Horse Breeder; Uhm….tell you what. I have a horse in my trailer. Let’s go have a look.
Genius: No, I don’t want to look.
Horse Breeder; Why not? You said all horses are purple, I have a horse right here that you can look at and verify that it isn’t purple….
Genius; You’ve just proven my point.
Horse Breeder; Uhm…. and you figure that…how?
Genius; You said it isn’t purple. So it isn’t a horse. How stupid do you think I am?
Horse Breeder; Well, let’s just say I have to revise my initial opinion….
mkelly says:
February 8, 2013 at 11:04 am
Willis Eschenbach says:
February 7, 2013 at 9:11 pm
q = A C s T1^4 – A C s T2^4 (5.10)
Thanks for proving my point. When T1 and T2 are equal q goes to zero and heat transfer stops. So the shell cannot warm the globe.
You really don’t understand the physics at all! When q equals zero the globe will heat up like crazy and q will no longer be zero, q is the rate of heat transfer from globe to shell, when the globe is being heated as in Willis’s example Th will only equal Tc when Th is 0K.
mkelly says: ” Thanks for proving my point. When T1 and T2 are equal q goes to zero and heat transfer stops. So the shell cannot warm the globe.”
Let’s think about this for a moment. The inner sphere was at T1 = 254 K with a 235 W/m^2 heater when there was no shell around it. Now lets add that shell at T2. If T2 = T1 = 254 K, then there would indeed be no heat going from the shell to the surface, nor from the surface to the shell — heat transfer between the two stops just as you claim. BUT THERE IS STILL 235 W of heat (from the heater inside the sphere) going to sphere. Since the sphere is losing no energy to the shell, but gaining 235 W/m^2, it will necessarily warm up. It will warm up until it is radiating 2×235 W/m^2 @ur momisugly 302 K.
Apropos to many internet discussions is the “Dunning–Kruger Effect”.
Shawnhet says:
February 8, 2013 at 10:21 am
“Oh, come on. You ask for specific pages of textbooks to give you an idea of where others are coming from and then once you are given them, you respond by arguing against a straw man position that has nothing to do with what you have been pointed to. I guess you can point someone to dozens of textbooks but you can’t make them understand any of them.
BTW, where does the extra heat come from that warms someone after they put on a parka when it is cold outside. Does that extra heat become any less real if you call it “backradiation”?
Shawnhet: I am NOT arguing that cooler objects can contribute energy to warmer ones. I’m arguing with the shell diagram. The “pages” w provided do nothing to support the shell diagram.
And your parka works primarily by INSULATING you. There is no “extra heat” there, and I doubt that the backradiation has much effect.
Phil. says:
February 8, 2013 at 10:56 am
Thanks, if the photon is thermalized, will the molecule’s kinetic temperature increase proportional to the energy of the photon?
Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2.
Two points:
1:
How does the shell acquire more W/m2 than is being generated by the planet?
At T=0, every thing is at 0K, then the reactor is turned on.
The planet warms up and eventually emits 235 W/m2, the shell follows this warming due to it eventually receiving 235 W/m2.
Then you say this shell radiates twice the power :- 235 W/m2 down and up!
How can the shell now deliver twice the power than that which it receives?
Valid equations can show you doubling the power but I would suggest that in this case it is the misapplication of valid equations is sending people down the wrong path.
2:
In the two bodies proposed experiment, it is suggested that a cold body can increase the temperature of a nearby hot body.
This is true if the colder body is treated as an insulator.
We can all agree that when fed with power, the hot body will continue to radiate and it will reach an equilibrium temperature with its heat sink.
If an insulator/reflector/path disruptor is placed between the hot plate and it heat sink, then the hot plate will increase in temperature until it can dissipate the same amount of energy as before via a more difficult path (via the insulator/cold body).
Its just insulation.
I would modify the two body experiment:
Hot plate as before (but made of aluminium), the cold plate was a sheet of super thin aluminium.
Cool the cool plate, insert into experiment,
I suggest that the radiation from/to the cold plate will have no effect except to get the cold plate up to the hot plate temperature, and have no measurable effect on the hot plate temperature.
Why?
Because the thin cold plate will have negligible insulation effect on the hot plate, so causing no temperature rise in the hot plate.
When I have run this experiment, I will let you know.
Steve Richards
“Willis Eschenbach says:
February 6, 2013 at 2:33 pm
Michael Moon says:
February 6, 2013 at 1:11 pm
Silver Ralph,
You just flunked your first hourly in Thermo. The cooler radiator would be warmed by the warmer radiator, and begin radiating more. If you think this would warm the warmer radiator, then you will fail all your hourlies and never get through school.
Wilis, Joe Public has it exactly right. If you want to know what happens to the flux from a cooler source when it hits a warmer source, the answer is exactly nothing. It is not absorbed, but immediately re-emitted, transferring NO heat.
All these analogies are amusing but ignore Second Law.
So you agree that it is re-emitted. To be re-emitted, it had to be absorbed. You just say it happens really fast.
Actually, once energy is absorbed, there’s no way to distinguish it from any other energy, so there’s no way to tell when that particular energy was re-emitted.”
I think point is are heating something?
Heating something with light requires transformation of energy.
So with solids and liquids they have a molecular structure- unlike gases.
Heat is vibrations of the molecular structures. Whereas heat for gas is the kinetic
motion of gas molecule. Gas molecules can freely move, the molecules of
liquids and solids are bound by their molecular structure.
So the conversion of electromagnetic energy into heat is the electromagnetic
energy increasing the vibrational energy of liquids and solids.
And this vibrational energy of liquids or solids is conducted to entire molecular
structure [thereby being heated]. This conduction of vibration not is not immediate
or at speed of light.
Putting cool frying pan on heating element does not convey the heat of element to
top surface of frying pan immediately- it takes a moment to heat up.
Better the conductor the faster this occurs.
Photon re-emitted can lack the time to heat something and also could not add to vibrational
energy of a solid or liquid- not convert the electromagnetic energy into heat.
Say you some surface which is 15 C. And heat it up by say 5 C, so it’s 20 C. The 20 C surface will emit more of wavelengths and more higher wavelength as compared to 15 C. The higher wavelength are a characteristic of the higher temperature, whereas more of lower wavelengths is not a characteristics of higher temperature- different material may emit more of certain wavelength as compared to other material with same temperature.
Or to heat up a gas, one needs higher vibration of the molecular structure- and less molecular vibration will cool the gas [a lot more of lower vibration state will not make the gas warmer].
jae says:
February 8, 2013 at 8:40 am
Since I gave citations to three different references above, including page numbers, I have no clue what you are complaining about.
In that comment, I also said, and I say again:
As a result, you showing up to tell me what you “suspect” is useless. Come back with references or I’m not interested in the slightest.
w.
Ryan says:
February 8, 2013 at 2:38 am
Thanks, Ryan. I’ve cut off your post above at the site where you made your mistake in logic.
The surface temperature is related, not to the NET radiation, but to the total radiation. How could it be otherwise, since the surface has no idea where the radiation is coming from or going to. So if a surface is radiating at e.g. 470 W/m2 the surface can’t say “well, half of these are from the shell, so they don’t count.”
470 W/m2 is 470 W/m2, and the temperature of the surface is related to only that, not the net radiation.
Hope this clarifies things,
w.
Will says:
February 8, 2013 at 2:49 am
A colder sky cannot possibly heat a warmer surface, any more than a trillion 1 ton blocks of frozen CO2 radiating at 194.56 K could melt a 1 inch water ice cube radiating at 273.15 K.
Thanks, Will. Consider this question. Which one will result in a warmer earth, either looking up from the surface to see:
1. An sky which is radiating something like 300 W/m2 back to the surface from the presence of GHGs, or
2. Outer space, which is radiating … well … nothing back to the surface.
When we replace the cold of outer space with the warmth of a GHG-containing atmosphere, the planetary temperature rises. In common parlance this is described as “the atmosphere warming the planet”. While this is not literally correct, it is the usual usage … but if you prefer, you can think of it as being equivalent to “the earth is warmer because of the presence of a cold atmosphere than it would be without it”. They mean the same thing.
w.
Tjfolkerts,
Thanks for taking the time to provide the shell energy budget. I can see where I went wrong. At the lowest level I had missed one of the back flows.
Regards, vc
Steve Richards says:
February 8, 2013 at 12:28 pm
Steve, it doesn’t. As the shell warms up it radiates back to the planet. As stated in the article,
So the shell receives 470 W/m2. I know this is difficult to grasp if you haven’t come across it before. But take a cold shower then check it out, this is what happens and everything is in balance.
I cannot express it clearer than explained by Willis:
No energy is being created and no energy is been destroyed.
Steve Richards says :
“At T=0, every thing is at 0K, then the reactor is turned on.
OK
“The planet warms up and eventually emits 235 W/m2, the shell follows this warming due to it eventually receiving 235 W/m2.”
Good so far. The sphere has warmed to 254 K.
“Then you say this shell radiates twice the power :- 235 W/m2 down and up!”
No, that is not what we are saying. The shell has TWICE the area of the sphere — it has an inside and an outside surface. So the shell radiates 235/2 W/m^2 out to space and 235/2 W/m^2 down to the surface = the power it receives. The shell is much cooler than the sphere: 213 K.
The presence of the “warm” shell (I call it “warm” because it is much warmer that outer space) adds 235/2 W/m^2 in addition to the 235 W/m^2 from the heater. This means the sphere is receiving 1.5 x 235 W/m^2 = 352 W/m^2.
So the sphere will warm up until it radiates 352 W/m^2 (from 254 K to 281 K). The shell will be radiating 352/2 W/m2 up and 352/2 down (warming from 213 K to 236 K).
But now the sphere is receiving 235 + 352/2 = 411 W/m^2, so it will warm up until it radiates 411 W/m^2 (warming from 281 K to 292 K) … but then the shell … but then the sphere …
We have an infinite series where the temperature of the sphere approaches 302 K ( 2X235 W/m^2) and the shell approaches 254 K (235 W/m^2). Energy is conserved at every location at every instant. Heat always flows from warmer to cooler.
Willis,
The problem of the steel shell is the steel and its 2 surfaces. Surface(or Side) A faces the planet, Surface (or Side) B faces cold outer space. You say at steady state both surfaces radiate 235 W/m^2.
Imagine the shell is perfectly transparent glass, then A radiates 0 and B radiates 235 that was passed through from A.
Imagine the shell is perfectly reflective silver, than A radiates 235 and B radiates 0.
In both cases, the shell is at 0 Kelvin because it can absorb no energy as a function of the sphere material.
But steel is a real material, so it reflects enery off sides A and B, transmits energy from side A to B, and over time absorbs energy and heats up. Also there is a non-zero distance from A to B and so a temperature gradient will develop between the surfaces.
The temperature of Side A will asymptotically approach the temperature of the planet. The temperature of Side B will asymptotically approach the temperature of outer space. Therefore, the sphere will radiate energy at a rate that will keep it about 1/2 the temperature of the surface of the planet. The exact temperature of the steel sphere is indeterminate from the information presented in the diagram since it partly depends on the reflective properties of the Side A and transmittal efficiency of Side B. At steady state, the radiation of Surface A + Surface B = 235 W/m^2 to satisfy the conservation of energy.
The other problem of the model is the surface of the planet at steady state has 2 temperatures corrsponding to 235 W/m^2 and 470 W/m^2. So, does the planet reflect all energy that it receives from the steel sphere or does it absorb it and then regulate its atomic reactions in such a way as to keep its surface temperature constant? Again, the model is indeterminate.
It does seem the planet surface will be warmer with the sphere but impossible to say by how much.
(A third material to consider is black hole stuff. Side A radiates 0 and Side B radiates 0 and the temperature of the sphere increases until the planet’s reactor is exhausted. So now we have a very hot black body radiating nothing.)
The problem with Willis’ steel BB is he’s never clear what it is he’s trying to relate. He has done this every time I’ve seen the BB posited.
It seems he takes a well known fact: that radiative conduction through solids is slower than free space
and tries to claim this, is the Greenhouse Gas Effect pertinent to infrared-resonant gas.
Yet he mistakenly – yet correctly – refers to the principles he discusses relevant to the BB as reflection on the “Greenhouse Effect” – which IS a real thing, it’s the “Atmospheric Greenhouse Effect.”
So by the time he’s discussed Woods’ work – the seminal expose on Arrhenius’ proven-wrong hypothesis that was tweaked by other failures until it’s claimed that solely infrared-resonant gases lend any additional temperature to the atmosphere at all – preposterous on it’s face yet touted far and wide among eco-wackos –
by the time he even invokes’ Woods’ work alone, he’s called one thing: Greenhouse Gas Effect –
another thing, the Atmospheric Greenhouse Effect.
This alone makes the article impossible to decipher and lends the idea Willis doesn’t know the difference. It wouldn’t be aggravating but I’ve seen him refer to these two utterly separate and distinct concepts by the same name, many times.
I’ve given him years as he’s dragged out the BB and that’s not ever been clear.
So that I’M clear, Willis
(1) refers to the Greenhouse Gas Effect, which Woods was checking, as the Greenhouse Effect, which Woods was NOT checking.
(2)refers to the model: which he then proceeds to describe under false tenets he ascribes very clearly as though he believes he’s talking about the Atmospheric Greenhouse Effect: come to think of it he later SAYS, “you don’t have to have CO2” – so I see now.
Willis thinks Woods was checking the Atmospheric Greenhouse Effect.
Willis also thinks the Atmospheric Greenhouse Effect is the Greenhouse Gas Effect.
Ok, so that’s now clearer.
—————
On to the next problem.
In Willis’ description of the BB he ascribes, arbitrarily, a 50-50 guaranteed absorption rate for radiation at the inner surface of his imaginary sphere. WHaT???
In Willis’ description of the BB he ascribes to the BB that it is steel: which has propagation delays due to interstitial resonance geometry irregularities, everybody knows why energy radiates through solids, slower.
In Willis’ description of the BB he then goes on to describe the steel BB as being a perfect black body object, with dramatically lower propagation speeds through it’s interstices.
So : there are at least three different descriptions of the functional radiative transfer qualities of the sphere:
(a) it’s steel
(b) it’s a blackbody
(c) it’s reflectance is 50% no matter what temperature it’s at, no matter what temperature the light coming in is at, and no matter what temperature the field on the outside is at, in spite of the fact it’s a blackbody which is a perfect absorber of infrared.
That’s PURE, SIMPLE, gobble-dee-guke
Willis then goes on to try to amateur-freestyle discuss energy migration,
to, eventually,
lay claim to the STUNNING reality he and apparently david hoffer believe fervently:
“The mathematics I do as an electronic engineer, regarding equilibria in everything I do in my job – which is calculating and checking the radiant transfer of electromagnetic energy through the atmosphere, through vacuum of space, and through industrial compounds needed to make that happen –
are LYING. There has been ALL ALONG, a TWO-WAY FLOW that neither mathematics, nor empirical demonstration ever showed before,”
and that only the people who believe in – whatever it is, they believe in – are smart enough to know.
That entropic migration during charge equalization
is actually not traveling at the speed of math.
It’s traveling at the speed of math one way, and too fast for anybody to have ever calculated,
the other way.
===============
That
is not science.
Not the part where Willis misunderstood what Woods checked: the question whether infrared-resonant gases are primary heat holders in atmospheric thermal cycling –
not the part where Willis described his Magic BB as a steel absorber,
then a 50/50 arbitrarily-always half & half absorber, whatever that is,
then a perfect blackbody absorber
which must by definition absorb all, conduct through and radiate perfectly out the other side, at the speed with which it would move through a vacuum adding only the additional time to make the change in direction which would allow the sphere to have that geometry –
that being said, what kind of amateur,
claims to have a perfect absorber & radiator, then talks about it heating up, because it can’t radiate?
Then claims others’ arguments don’t persuade ?
One who hasn’t ever had to pay the dues, to get a formalized education in energy transfer through radiative emissions law, guessing like a blind man in a crowd where his credibility went.
It wasn’t science, when Willis tried to tell me that when I became an electronic engineer and calculated radiative transfer as per entropy hundreds of times for a grade,
the math I learned was only part of the story.
No Willis the math you learned from those “introductory” textbooks you and david hoffer love so much,
was only part of the story.
The full story is that charge density dissipation travels one way because where there is no charge, there is nothing to counter an existing one. Therefore there really is a difference between five energy quanta, and three.
And when I have 5 in one object, and 3 in another, and calculate that 5 charges, giving up one charge, makes those two objects equal in energy contained,
I didn’t actually have a whole bunch of little unmeasurable-hence unwritable charges, secretly flying between the two, while I calculated the 1, migrating over to the three.
Now Willis you can pretend you and david hoffer didn’t spend an afternoon telling me that when I have
5 joules of energy in one globe, and 3 in another,
and that while 1 joule was migrating over to establish equality,
that 5 joules and 3 joules were furiously, but undetectably, swapping sides, between bars,
until just when that magically NOT furiously swapping sides 1, migrated OVER
but it’s not going to stop me pointing out that you tried to claim you think that happened.
None of that crap is science,
none of it is radiative transfer.
It’s gobble-dee-guke.
That starts out wrong and then goes to the point of telling me I don’t know how heat dissipates from an entity, and that while one value migrates, magically indiscernible and incalculable anti-quanta are hopping sides in my equations and in all objects in the universe.
jae:” I am NOT arguing that cooler objects can contribute energy to warmer ones. I’m arguing with the shell diagram. The “pages” w provided do nothing to support the shell diagram.
And your parka works primarily by INSULATING you. There is no “extra heat” there, and I doubt that the backradiation has much effect.”
Respectfully, jae, what Willis posted was exactly on point when it discussed how radiation always flows both ways between hot and cold bodies. His shell argument has nothing to do with having multiple low temperature torches heating something to higher than any one of them could get. Until you understand that, no amount of references will make sense to you.
The point about insulation and backradiation is that it is perfectly fine to think of GH gases as insulating (at least IMO) the Earth in pretty much the same way as your parka insulates your body in winter.
http://scienceofdoom.com/2012/07/23/how-the-greenhouse-effect-works-a-guest-post-and-discussion/
If, in fact, it is perfectly consistent to view GH gases as insulating the Earth, and you agree that insulation doesn’t produce extra heat, what is there left to disagree about at the end of the day?
Cheers, 🙂
I’m not here to be the most friendly nor most artistic skeptic. I’m here to be the sole person here, most likely, who’s calculated rate-of-dissipation of heat from one body to another THOUSANDS of times in my career.
I had to calculate radiant dissipation of energy through solids, and vacuum, and the atmosphere, by the end of 3-1/2 years’ combined school, on HUNDREDS of questions on which I was GRADED: give the right answer or FAIL.
These men on this forum, – Willis, david hoffer, all the usual
“entropy calculations don’t tell the truth, charge is actually flowing backwards while all ‘odd’ charges are migrating to equalization, forward”
crowd –
aren’t going to tell me there’s no mandate in charge based energy dynamics against unaided, reverse flow.
They’re just not.
No matter how many times they claim they have an “introductory text” that they think says undriven charge flow occurs.
My goodness… I forgot the /sarc at the end of that last comment. Thank you davidmhoffer for all of those kind words queueing me.
Allen B. Eltor:
I am writing this as a sincere attempt to be helpful.
I have read your long post at February 8, 2013 at 2:28 pm three times, and I still fail to understand what you are trying to say. I suspect that if I don’t ‘get it’ then others also won’t.
Perhaps you are trying to say several things at once and they are getting jumbled up?
I commend you to make a shorter post which attempts to make a single point in as clear and concise a manner as you can. You can make subsequent posts when that issue has been addressed.
I hope this helps.
Richard
A Willis Eschenbach starship.
Materials needed one asteroid- 1, 2, 5 or 20 Km diameter space rock.
Lots of steel sheeting.
Magnets. Lots of permanent magnets.
First make ceiling above asteroid.
Support ceiling with pillar which will needed to support the ceiling until
such time as it’s pressurized. After it’s pressured the pillars could be used
to hold the ceiling down. Make ceiling height: 30′.
The ceiling can made of steel and the thickness is related to how much you
want to pressurize the atmosphere. And it can be say less than 1/2″ thick
if you using enough pillars.
[One should keep in mind it’s very low gravity environment, less 1/100th of gee,
so making 1 ton car weigh less than 20 lb]
Above this ceiling we going use Willis Eschenbach’s insulation.
So 10 layers of very thin steel separated with a vacuum and supported
above the ceiling pillars with magnets {South South] which suspend
the weight of these spheres. Thus insuring there is no heat loss from
conduction of heat.
The question is once one has the ceiling pumped up with air and have 10 stacked
steel shells constructed, how much heating of 30′ room/atmosphere do we need?
And is there a certain temperature which dangerous- leading to dangerous
greenhouse effect?
Say we have floor heating which has a well insulated substrata, then 4″ concrete
with embedded thermal piping. With thermostat, which set to say 78 F [ 25.5 C 298.6 K]
which warmed by heated water.
So we know [and can adjust] temperature of the floor, the question what temperature of the ceiling and how much heating is needed.
We have low gravity so it seems ceiling fans which could provide well mixed air would
be good idea and it means air temperature at the ceiling will near temperature of the floor.
Now, a question arises by some [let’s call them deniers] of whether the ceiling
should have the conventional type insulation [say 6″ of fiberglass on the air
side of sphere ceiling] and others have faith it would inhibit Willis Eschenbach’s insulation,
thereby result in a increase of heating costs.
Which is the truth?
Willis: You must have missed a couple of my comments. I’m not challenging the citations you provided.