Forgot your password?
typodupeerror

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 81

by khayman80 (#48218191) Attached to: Fiber Optics In Antarctica Will Monitor Ice Sheet Melting

... I repeat: your use of a heat transfer equation, rather than a radiant power equation, to calculate the radiant power output of the hottest object in an isolated vacuum environment is just laughable. Your own "power in = power out" claim shows it to be wrong. It contradicts your own calculations, which I showed to be wrong 3 different ways. Hell, you even got some simple math wrong. ... [Jane Q. Public, 2014-10-23]

Once again, Jane confuses "radiant power output" with "electrical heating power". Since "electrical heating power" is zero if the chamber walls are at the same temperature as the source, Jane is simply wrong to use a "radiant power output" equation to describe "electrical heating power". As I just explained, mainstream physicists and even most climate contrarians agree that "electrical heating power" has to account for the chamber wall temperature.

If Jane tried just once to write down an energy conservation equation for a boundary around the source without wrongly "cancelling" terms, he'd realize that this Slayer nonsense is wrong.

Or maybe Jane could listen to Prof. Brian Cox. Jane/Lonny Eachus likes Prof. Brian Cox and is very bothered by the fact that Prof. Cox agrees with mainstream physics. Jane/Lonny urges Prof. Cox to take time from his obviously busy schedule to review the actual state of the science on this extremely important subject.

Jane/Lonny seems to think that physicists just need to be told the glorious Sky Dragon Slayer "truth" and then they'll happily abandon conservation of energy. Maybe Jane/Lonny Eachus could convince physicists like Prof. Cox by finally writing down an energy conservation equation for a boundary around the source without wrongly "cancelling" terms? Or maybe Jane/Lonny could just ask Prof. Cox if the required electrical heating power depends on the cooler vacuum chamber wall temperature?

I bet Jane/Lonny Eachus $100 that Prof. Cox answers "yes" to the previous question. Is Jane/Lonny Eachus chicken?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 81

by khayman80 (#48217751) Attached to: Fiber Optics In Antarctica Will Monitor Ice Sheet Melting

... Would you all like to see his dumbass failure at trying to school me in thermodynamics? All you have to do is follow his comments back a ways. A long ways... because he kept making the same nonsense arguments, over, and over, and over again, even after he had been shown how wrong they were. ... [Jane Q. Public, 2014-10-22]

Jane keeps insisting that this Sky Dragon Slayer equation describes electrical heating power:

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

Once again, that violates conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Jane's equation wrongly cancels "radiative power in" with a nonexistent term.

The BASIS of “greenhouse warming” -- back radiation -- has been SCIENTIFICALLY shown to be a load of hogwash. [Lonny Eachus, 2014-10-14]

No, Jane/Lonny Eachus's Slayer nonsense has been scientifically shown to violate conservation of energy. Unless, of course, Jane/Lonny can finally write down an energy conservation equation before wrongly "cancelling" terms?

It's fascinating that Jane/Lonny Eachus keeps insisting that mainstream physics is a hogwash dumbass failure. Jane/Lonny just needs to inform "dumbasses" like Prof. Brown, Dr. Joel Shore, physicists in the National Academies of Science, the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, the European Physical Society, etc.

Jane/Lonny's Sky Dragon Slayer nonsense is so ridiculous that even prominent climate contrarians are rational enough to back away from the Slayers:

  • Dr. Fred Singer finds it "surprising that this simplistic argument is used by physicists, and even by professors who teach thermodynamics. One can show them data of downwelling infrared radiation from CO2, water vapor, and clouds, which clearly impinge on the surface. But their minds are closed to any such evidence." The comments prove his point.
  • Dr. Roy Spencer "clearly demonstrates that IR absorbing gases (greenhouse gases) reduce the Earth's ability to cool to outer space. No amount of obfuscation or strawman arguments in the comments section, below, will be able to get around this fact."
  • Anthony Watts banned one of the original authors because of his nutty comments and later called the argument "rubbish ".
  • In a thread that was deleted after a lawsuit threat, Prof. Judith Curry observed that the Sky Dragon group "damaged the credibility of skepticism about climate change and provides a convenient target when people want to refer to 'deniers' and crackpots."
  • Lucia Liljegren calls the book muddled and confusing after being overwhelmed by the errors in a single chapter.
  • Jeff Condon "will no longer discuss or even acknowledge his fake radiation nonsense here ... the second law backradiation bullshit needs to go away until a proper mathematical foundation supports it."
  • Even Lord Monckton finds the Sky Dragon arguments exasperating, because they "fly in the face of experiments that even children can perform". As Joel Shore notes: "You know how far out in space your scientific reasoning is when Lord Monckton looks like a paragon of scientific thought by comparison!"

Again, even those contrarians are able to recognize that the correct equation (neglecting reflections) is:

electrical heating power per square meter + (e*s)*T4^4 = (e*s)*T1^4

Note that this equation obeys conservation of energy because it accounts for the temperature of the chamber walls.

Even the analysis Jane reviewed himself and found well-designed uses that equation on page 16 to describe the radiative component of heating power. Note that they don't use Jane's Slayer equation which stubbornly refuses to account for the temperature of the ambient chamber walls, which violates conservation of energy.

So not only do the vast majority of physicists disagree with Jane's Slayer nonsense, most climate contrarians are also rational enough to back away from the Slayers. Even the convicted scam artist who conned Jane and Lonny Eachus knows better than to ask Sky Dragon Slayers to prop up his latest scam.

So why is Jane/Lonny Eachus still regurgitating Sky Dragon Slayer nonsense, when all Jane/Lonny has to do to recognize they're wrong is try to write down an energy conservation equation without wrongly "cancelling" terms?

Comment: Re:WTF, the antarctic gets FO before me? (Score 1) 81

by khayman80 (#48207951) Attached to: Fiber Optics In Antarctica Will Monitor Ice Sheet Melting

Granted, Goddard got some things wrong in the beginning, but lately he's been getting a lot more right, as even GISS has admitted. [Jane Q. Public]

Have they. Do you have a link to this admission? [Truth_Quark]

Do you really want to ask for that link? Watch what happened the last time someone asked Jane/Lonny Eachus for that link:

GISS ADMITTED GODDARD WAS RIGHT. YOU DIDN’T KNOW. YOU’RE IGNORANT OF THE FACTS. LEARN THEM. MEANTIME, GO AWAY. [Lonny Eachus, 2014-08-30]

Try Google, dumbshit. Unless you don’t know how. It took me all of 20 seconds. ... Why? Why should I do this for you? Would you like me to wash your balls too? Answer: no. ... The fact I WON’T wash your balls for you is not evidence that they don’t exist. The fact that YOU won’t, IS. ... Correct. To all outside observers, so far, your balls don’t exist. Why don’t you prove that they do? show us. ... Should we just ASSUME it? Or, like you, should we require that you SHOW US? ... To make an even better analogy: there is a picture of them that has been posted online by your girlfriend. ... BUT we don’t believe you really have any. Should we ask you to prove they’re yours? Every time we discuss it? [Lonny Eachus, 2014-08-30]

Sorry, dude. You aren’t going to get me to wash your balls. The rest of us are looking at pictures of your girlfriend. wondering when you’re going to say “I won’t hang them out again just for you. Look it up.” [Lonny Eachus, 2014-08-30]

Are you REALLY that fucking stupid? [Lonny Eachus, 2014-08-30]

I’ve insulted you because you deserve it. Arguments were made. Your inability to absorb them is not evidence. [Lonny Eachus, 2014-08-30]

I guarantee something: that doesn’t make ME an asshole. I’ll leave it up to others what it does mean. [Lonny Eachus, 2014-08-30]

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48135547) Attached to: NASA Study: Ocean Abyss Has Not Warmed

... I have repeatedly demonstrated that this person who calls himself "kayman80" has been blatantly dishonest about past conversations that have occurred here on Slashdot and elsewhere. And that he has a habit of deliberately distorting what other people say, for reasons of his own. I have ceased feeding the troll. I recommend that you do so as well. [Jane Q. Public, 2014-10-13]

Instead of endlessly accusing me of blatantly dishonest deliberate distortions, a real skeptic would write down an energy conservation equation before wrongly "cancelling" terms. Actually, a real skeptic would've done that months ago, but better late than never.

Jane, I believe in you. I believe you can learn how conservation of energy works, but first you have to take a baby step of your own by writing down an energy conservation equation before wrongly "cancelling" terms. Just try it. You might learn something. On the other hand, endlessly accusing me of dishonesty probably isn't very educational.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48135193) Attached to: NASA Study: Ocean Abyss Has Not Warmed

I haven't expended ANY energy to avoid writing anything down. I've written down the proper and necessary equations not just once but many times now. [Jane Q. Public, 2014-10-13]

Ironically, Jane's still trying hard to avoid writing down his energy conservation equation before wrongly "cancelling" terms. If he'd try to write down that equation just once, he might realize that the nonsensical equation he's written down many times isn't proper or necessary.

I don't need to write down a "conservation of energy equation" in regard to Spencer's experiment. I don't refuse to do it because I can't, as you have clearly implied. I refuse to do it because this is a dead issue. You were proved wrong weeks ago, and your demands for additional proof from me are just laughable. [Jane Q. Public, 2014-10-13]

If Jane actually could write down an energy conservation equation before wrongly "cancelling" terms, Jane would see that "radiative power from the walls" can't cancel out.

Once again, the only way Jane's final term could cancel with the radiative power in term "(e*s)*T4^4" to obtain Jane's final equation would be if "radiative power from chamber walls, re-emitted back out" equals "(e*s)*T4^4". But it's being emitted by the source, which is at temperature T1. If reflections confuse you, just remember that the gray body equation has to reduce to the black body equation where there aren't any reflections at all. In that case, all that power is being absorbed and re-emitted, not reflected.

If Jane would write down an energy conservation equation and think about it, he might realize that he's been endlessly crowing about "proving me wrong" using Sky Dragon Slayer nonsense that violates conservation of energy and/or the Stefan-Boltzmann law.

But since Jane's Slayer brainwashing is so thorough that he can't bring himself to write down that equation, Jane will probably keep endlessly crowing about "proving me wrong".

Ironically, if Jane's Slayer nonsense was right, Jane would also have "proven wrong" Prof. Brown, Dr. Joel Shore, the National Academies of Science, the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, the European Physical Society, etc.

... YOU are the one going against "established" physics here. ... If you could actually show how the physics textbook idea of heat transfer was wrong, you would be world famous by now. ... [Jane Q. Public, 2014-10-06]

No, I'd have to get in line behind all those other physicists who agree that adding CO2 warms Earth's surface, which is equivalent to saying that enclosing a heat source warms it. This is probably the most fascinating part of Jane's delusion. Not only does Jane completely misunderstand fundamental physics, Jane seems to earnestly believe that his crackpot Slayer conspiracy theory represents "established" physics. Fascinating.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48128761) Attached to: NASA Study: Ocean Abyss Has Not Warmed

Jane's power in = electrical heating power + radiative power in from chamber walls

NO, it doesn't, and fucking well STOP claiming that it is. If YOU want to assert that, go ahead, but stop putting my name on it. I did not say that, and I do not say that, so stop putting my name on it. DO YOU UNDERSTAND??? Holy fuck, you're a dimwit. [Jane Q. Public, 2014-10-13]

Oh, okay. So Jane completely denies that the chamber walls emit radiation in through a boundary around the source. ...

Jane objected:

NO, I VERY CLEARLY AND REPEATEDLY EXPLAINED THAT I DENY NO SUCH THING. I don't have any patience for your lying anymore. Goodbye. I will record any responses, at least for a while, but I won't reply. Jesus, you're an ass. I mean the most incredible ass I've ever had the misfortune to meet online. I mean that very, very sincerely. [Jane Q. Public, 2014-10-13]

Only if "VERY CLEARLY AND REPEATEDLY EXPLAINED" means that Jane/Lonny Eachus clearly and repeatedly explained that he would never write down an energy conservation equation, and that only dimwits would claim that Jane thinks that "radiative power in from chamber walls" should be included in "Jane's power in".

It's fascinating how much effort Jane/Lonny Eachus has expended just to avoid writing down the power flowing into and out of a boundary around the heat source. If Jane/Lonny Eachus is so sure that he's right, why not just write down that obviously correct energy conservation equation?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48128681) Attached to: NASA Study: Ocean Abyss Has Not Warmed

No. Once again, gray body equations have to reduce to black body equations where there are no reflections.

No, they don't, because you still have an emissivity (which is the same as absorptivity, in a gray body). You're trying to have it both ways again. There is also a "scattering" term which you're ignoring, which is not the same as reflection. [Jane Q. Public, 2014-10-13]

Gray bodies have emissivities between 0 and 1. So black bodies are one limit of gray bodies. Black bodies don't scatter or reflect radiation, they only absorb it.

Jane's power in = electrical heating power + radiative power in from chamber walls

NO, it doesn't, and fucking well STOP claiming that it is. If YOU want to assert that, go ahead, but stop putting my name on it. I did not say that, and I do not say that, so stop putting my name on it. DO YOU UNDERSTAND??? Holy fuck, you're a dimwit. [Jane Q. Public, 2014-10-13]

Oh, okay. So Jane completely denies that the chamber walls emit radiation in through a boundary around the source. That's what I suspected from the beginning, but Jane kept coyly saying things like this:

... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. ... [Jane Q. Public, 2014-10-11]

... net radiation across the boundary from the wall is zero, because it just goes right back out. ... [Jane Q. Public, 2014-10-11]

These statements made me think Jane was rational enough to see that "power in" through a boundary around the source would have to include radiative power from the chamber walls. Jane just seemed to wrongly think it cancelled, because Jane kept refusing to write down the energy conservation equation. But now Jane completely denies that radiation from the chamber walls passes in through a boundary around the heat source.

Fascinating.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48128619) Attached to: NASA Study: Ocean Abyss Has Not Warmed

Jane is saying what Jane already actually said, not this distorted nonsense of yours. [Jane Q. Public, 2014-10-13]

Jane already actually said:

... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. ... [Jane Q. Public, 2014-10-11]

... net radiation across the boundary from the wall is zero, because it just goes right back out. ... [Jane Q. Public, 2014-10-11]

Jane's already actually described the following energy conservation equation:

Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

If Jane didn't mean to describe that equation, Jane would've written down his actual equation. But Jane hasn't, because Jane can't.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48128523) Attached to: NASA Study: Ocean Abyss Has Not Warmed

... none (NET) is absorbed in the first place. Again, this is what I have been saying all along. It is reflected or scattered. As I have stated before, this is a requirement of the Second Law of Thermodynamics. ... [Jane Q. Public, 2014-10-13]

No. Once again, gray body equations have to reduce to black body equations where there are no reflections.

... Are you trying to claim that radiation from the chamber walls is absorbed, and NOT re-emitted? ... [Jane Q. Public, 2014-10-13]

No, I'm claiming that any real physicist would write down an energy conservation before wrongly "cancelling" terms. Jane refuses to do that, so:

Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

... radiation inward from the chamber walls does cancel, because it is reflected or scattered and goes right back out. (A small part of it misses the inner sphere completely.) So no matter how you look at it, it is still a zero sum. ... [Jane Q. Public, 2014-10-13]

No, Jane. Once again, gray body equations have to reduce to black body equations where there are no reflections.

Once again, if Jane would simply write down his energy conservation equation before wrongly "cancelling" terms, he would see that they can't cancel.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48128423) Attached to: NASA Study: Ocean Abyss Has Not Warmed

You've tried to claim that POWER IN to the heat source is somehow magically dependent on the chamber walls. And the justification you gave for this was a heat transfer equation, as I described above. [Jane Q. Public, 2014-10-13]

It's physics, not magic. Radiation from the chamber walls passes in through a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Jane, however, seems to say this:

Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

Jane's saying that "radiative power from the walls through that boundary cancels itself out" so Jane claims those terms cancel to produce Jane's final energy conservation equation:

Jane's power in = electrical heating power
Jane's power out = radiative power out from source

The only way Jane's final term could cancel with the radiative power in term "(e*s)*T4^4" to obtain Jane's final equation would be if "radiative power from chamber walls, re-emitted back out" equals "(e*s)*T4^4". But it's being emitted by the source, which is at temperature T1. If reflections confuse you, just remember that the gray body equation has to reduce to the black body equation where there aren't any reflections at all. In that case, all that power is being absorbed and re-emitted, not reflected.

The acknowledged formula for finding radiative power from temperature is just (sigma epsilon)T^4. There are no other factors involved... [Jane Q. Public, 2014-09-05]

That's why radiation re-emitted by the source at temperature T1 is (e*s)*T1^4. There are no other factors involved. The source can't re-emit radiation at (e*s)*T4^4, so those terms in Jane's equation can't cancel. And the last term double-counts radiation emitted by the source, so it's zero.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48128351) Attached to: NASA Study: Ocean Abyss Has Not Warmed

You insist that the radiant power output calculation of the heat source has to take into account the cooler temperature of the chamber walls. [Jane Q. Public, 2014-10-13]

No, I've repeatedly agreed that radiative power out only depends on emissivity and temperature.

Once again, I'm just saying that "radiative power out" is different than "electrical heating power".

That's why Jane will never write down his "energy conservation" equation before wrongly "cancelling" terms. If Jane ever did, he'd have to face the fact that Jane's terms don't cancel.

How ironic. Jane's terms would only cancel if radiant power output of the heat source took into account the cooler temperature of the chamber walls. But that's impossible because it would violate the Stefan-Boltzmann law.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48128029) Attached to: NASA Study: Ocean Abyss Has Not Warmed

That's ridiculous, Jane. Notice that "net radiative power out" equals negative "net radiative power in". Since Jane seems to agree that "net radiative power out" is positive, "net radiative power in" can't be zero. It has to be negative, which just means more radiative power is flowing out than flowing in.

Now you've just gone off the deep end. And by "deep end" I mean the deep end of the pit full of BS you've dug yourself. Just no. Any spherical boundary you draw within this system has additional input: your vaunted electrical power. I'm amazed that you finally got so caught up in your own bullshit that you made a mistake quite THAT fundamental. Get stuffed, troll. For that and actually quite a pile of other reasons that have built up over time, I still don't believe you're a real physicist. [Jane Q. Public, 2014-10-12]

Electrical power isn't radiative power, so it wouldn't be included in net radiative power.

... I have written down all I need to write down to answer Spencer's challenge. I solved for the correct temperature, and showed your own answer to be utterly wrong. ... [Jane Q. Public, 2014-10-11]

Once again, Jane's solution halved the electrical heating power. Jane didn't notice this because he calculated net transfer incorrectly, which led him to the absurd conclusion that Jane was only off by about 0.1% when Jane was actually off by ~100%.

So Jane hasn't written down all he needs to give the correct answer to Spencer's challenge. To give the correct answer, Jane has to draw a boundary around the heat source:
power in = electrical heating power + radiative power in from chamber walls
power out = radiative power out from source

This is the same answer that Prof. Brown and Dr. Joel Shore tried to explain to Jane. It's also the same answer that underlies the positions taken by the National Academies of Science, the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society, etc.

... YOU are the one going against "established" physics here. ... If you could actually show how the physics textbook idea of heat transfer was wrong, you would be world famous by now. ... [Jane Q. Public, 2014-10-06]

No, I'd have to get in line behind all those other physicists who agree that adding CO2 warms Earth's surface, which is equivalent to saying that enclosing a heat source warms it. This is probably the most fascinating part of Jane's delusion. Not only does Jane completely misunderstand fundamental physics, Jane seems to earnestly believe that his crackpot Slayer conspiracy theory represents "established" physics. Fascinating.

Maybe that's why Jane won't even take a few seconds to write down an energy conservation equation before he wrongly "cancels" terms. Deep down, maybe Jane suspects that he's wrong and mainstream physicists are right.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48121569) Attached to: NASA Study: Ocean Abyss Has Not Warmed

... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... I did NOT claim the net radiative power through the boundary was zero. What I wrote was that the radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. ... The actual radiant power through that boundary isn't zero, because the radiant power output of the heat source still goes through it of course. Leaving a NET positive transfer of energy OUTWARD through the boundary. ... [Jane Q. Public, 2014-10-11]

This is just more dishonest out-of-context nonsense again. I clearly told you that the context of my statement was radiation from the walls through a boundary. I did not claim the net radiation across that boundary was zero. I claimed the net radiation across the boundary from the wall is zero, because it just goes right back out. This is a wonderful example of how you distort context, in order to make it appear someone else is saying something they actually did not. That is a form of lying. [Jane Q. Public, 2014-10-11]

Since Jane keeps bolding "from the wall" and claiming that "radiative power from the walls through that boundary cancels itself out," Jane seems to be saying:

Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

It certainly seems like that's what Jane's saying. If "radiative power from the walls through that boundary cancels itself out" then those terms cancel to produce Jane's final energy conservation equation:

Jane's power in = electrical heating power
Jane's power out = radiative power out from source

Jane, is that what you're saying by "net radiation across the boundary from the wall is zero"?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48120703) Attached to: NASA Study: Ocean Abyss Has Not Warmed

... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... I did NOT claim the net radiative power through the boundary was zero. What I wrote was that the radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. [Jane Q. Public, 2014-10-11]

Again, Jane must using some kind of Sky Dragon Slayer definition of the word "net". In physics, "net radiative power out" means "radiative power out" minus "radiative power in". This is only zero when the source and chamber walls are at the same temperature.

Similarly, "net radiative power in" means "radiative power in" minus "radiative power out". Again, this is only zero when the source and chamber walls are at the same temperature.

The actual radiant power through that boundary isn't zero, because the radiant power output of the heat source still goes through it of course. Leaving a NET positive transfer of energy OUTWARD through the boundary. [Jane Q. Public, 2014-10-11]

That's ridiculous, Jane. Notice that "net radiative power out" equals negative "net radiative power in". Since Jane seems to agree that "net radiative power out" is positive, "net radiative power in" can't be zero. It has to be negative, which just means more radiative power is flowing out than flowing in.

So Jane must not be using the physics definition of "net". What's the Sky Dragon Slayer definition of "net"? And how is it possible for the "net power in" to be zero when the source is hotter than the chamber walls?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 295

by khayman80 (#48118843) Attached to: NASA Study: Ocean Abyss Has Not Warmed

... An object that is radiating at a certain black-body temperature WILL NOT absorb a less-energetic photon from an outside source. This is am extremely well-known corollary of the Second Law. ... [Jane Q. Public, 2013-05-30]

... I have NOT been claiming that no radiation from a cooler body is absorbed by a warmer body. ... Energy can be absorbed and re-emitted... [Jane Q. Public, 2014-09-28]

... I do not deny that some radiation is absorbed; but then it's just re-emitted. ... [Jane Q. Public, 2014-10-03]

... Here is a fundamental principle of thermodynamics, as related to radiant energy: net incoming radiation from cooler bodies is ALL either reflected, transmitted, or scattered. Any absorption and re-transmission is part of the "transmitted" term. ... [Jane Q. Public, 2014-10-10]

Jane can't quote a textbook stating this "fundamental principle" because it's nonsense. For instance, the "transmitted" term describes a body's transparency, not its absorption and re-emission. Here's an introduction:

"A body's behavior with regard to thermal radiation is characterized by its transmission t, absorption a, and reflection p. ...

An opaque body is one that transmits none of the radiation that reaches it, although some may be reflected. That is, t = 0 and a + p = 1

A transparent body is one that transmits all the radiation that reaches it. That is, t = 1 and a = p = 0."

Jane, absorption and re-emission isn't part of the "transmitted" term. They're totally different. The "transmitted" term is zero for opaque bodies like aluminum or blackbodies. If absorption and re-emission were part of the "transmitted" term, blackbodies would be transparent because they absorb all radiation that hits them. If absorption and re-emission were part of the "transmitted" term then both terms would equal 1. But once again, blackbodies can't be transparent.

Also, it's bizarre that Jane insists he's accounting for absorbed and re-emitted radiation in a "transmitted" term that isn't even in his energy conservation equation.

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. ... that ZERO of the radiative power output ... THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

... First, there is no NET radiative power absorbed by a body at one thermodynamic temperature from another body at a lower temperature. ... Those are the statements I made. Anything else is a logical extension of those two principles. ... [Jane Q. Public, 2014-10-10]

That's a serious problem, because Jane's first principle is wrong. Net radiative power would only be zero if the source and the chamber walls were at the same temperature.

Jane might consider replacing his incorrect first principle with "conservation of energy" which means power in = power out through a boundary where nothing inside is changing.

... No NET radiative input from chamber walls means anything crossing your precious boundary inward goes right back out. As I have explained to you many times now. ... [Jane Q. Public, 2014-10-10]

Jane, it's not my "precious" boundary. It's a general principle called "conservation of energy". Drawing a boundary is needed to apply conservation of energy. Here are some introductions: example (backup), example (backup), example (backup).

Note that drawing a boundary is needed to apply conservation of energy in all those introductions, just like I've repeatedly explained.

If that's what you think, could you take a few seconds to write down the energy conservation equation (before cancelling terms) that you think is correct?

I could, but I will not. ... I'm not going to take valuable time out of my day ... I'm not going to waste my time. ... [Jane Q. Public, 2014-10-09]

If Jane really could write down an energy conservation equation before wrongly "cancelling" terms, that would only take a few seconds. Endlessly screaming in ALL CAPS and accusing me of lying wastes more of Jane's valuable time than writing down an energy conservation equation for a boundary around the source:

Jane's power in = ?
Jane's power out = ?

(1) Never draw what you can copy. (2) Never copy what you can trace. (3) Never trace what you can cut out and paste down.

Working...