... The experiment we were discussing was Spencer's radiation experiment. Not "global warming". You keep trying to apply my arguments about Spencer's challenge to the broader issue of global warming, aka "climate change", and it's not valid to do so.... [Jane Q. Public, 2014-10-25]
Once again, how bizarre. The whole reason Slayers deny that an enclosed source warms is because that implies greenhouse gases can't warm the surface:
.. the CO2-warming model rely on the concept of "back radiation", which physicists (not climate scientists) have proved to be impossible. I'm happy to leave actual climate science to climate scientists. But when THEIR models rely on a fundamental misunderstanding of physics, I'll take the physicists' word for it, thank you very much... [Jane Q. Public, 2012-07-05]
... The only reason I agreed to work through the Spencer experiment with you was because I already knew you were wrong, and wanted the chance to show that to everybody, unequivocally. Well, I got that chance. And as soon as I get it written up (which as I have stated before will take a while), I fully intend to show everybody. You asked me if I really was willing to publish the results, no matter the outcome. Well, now that in fact it didn't go well for you, sour grapes isn't going to get you anywhere.... [Jane Q. Public, 2014-10-25]
If Jane is so sure that his Sky Dragon Slayer nonsense is correct, why can't he write down a simple energy conservation equation around the heated source without wrongly "cancelling" terms? Ironically, this is the very first equation needed to understand Spencer's experiment. And Jane can't even get the first equation right. Prof. Cox is right: this isn't even degree-level physics.
Jane, if you tried just once to write down an energy conservation equation for a boundary around the source without wrongly "cancelling" terms, you'd realize all this Slayer nonsense is wrong.
... 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?
... If you want to ask him about what amounts to a pretty straightforward textbook radiation problem, go right ahead. But I already know the answer -- which, in fact, I got from textbooks on the subject -- so I don't have to bet. You go ahead, if you want to.... [Jane Q. Public, 2014-10-]
In other words: bok bok bok BOKKKKK. That's what I thought. Jane/Lonny Eachus is chicken.
If Jane/Lonny Eachus were a real skeptic, he'd at least consider the possibility that Jane's "radiant power output" equation doesn't describe "electrical heating power". Jane's textbooks don't say to use a "radiant power output" equation to describe "electrical heating power".
That's why Jane is too chicken to ask Prof. Cox if electrical heating power depends on the cooler vacuum chamber wall temperature. Because Jane's afraid that Prof. Cox will say yes. If not, why did Prof. Cox say all these things?
... 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?
... 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?
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
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]
... 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.
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".
... 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. Fascina
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?
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.
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.
... 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.
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.
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.
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]
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
... 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 theor
... 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"?
... 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?
... 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
... there is no NET radiative power absorbed by a body at one thermodynamic temperature from another body at a lower temperature. Doing so would violate the Second Law of Thermodynamics. I'm using the standard definition of "net", which is to say "all inputs minus all outputs".... [Jane Q. Public, 2014-10-10]
If net radiative power is "all inputs minus all outputs" then net radiative power is only zero if all the inputs equal all the outputs. That only happens if the source is at the same temperature as the chamber walls.
But Jane claims that net radiative power is zero when the source is hotter than the chamber walls.
NOWHERE did I state "zero net radiative output". I don't believe I used that phrase at all, but if I did, you present it here out of context. [Jane Q. Public, 2014-10-10]
Hmm...
... 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]
Again, net radiative power is only zero if the source is at the same temperature as the chamber walls.
NOWHERE did I state "zero net radiative output". I don't believe I used that phrase at all, but if I did, you present it here out of context. [Jane Q. Public, 2014-10-10]
Oh, so you're saying net radiative power isn't zero? For some odd reason I thought you were saying it was zero.
... 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]
By the way, just in case it wasn't obvious from the fact that I was responding to Jane's claims of zero net radiation absorbed and zero net radiative power output, I was talking about net radiative power because that's what Jane seemed to be talking about. That's why my equation only has radiative terms. Here's a less ambiguous version:
So you're not going to retract your claim that net radiative power is zero when the source is warmer than the chamber walls?
Is Jane using some kind of special Sky Dragon Slayer definition of the word "net"? In physics, net radiative power through a boundary around the source = "radiative power out" minus "radiative power in".
So you're not going to retract your claim that net power is zero when the source is warmer than the chamber walls?
... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. This is a requirement of the Second Law of Thermodynamics. Now, this is NOT the same as saying "no radiation absorbed at all". But when you put the two points above together, what it does mean is that ZERO of the radiative power output from the above equation is coming from other bodies. THE AMOUNT OF POWER OUTPUT IN THIS EQUATION DOES NOT NEED TO ACCOUNT FOR POWER FROM THE WALLS, BECAUSE THE NET IS ZERO.... [Jane Q. Public, 2014-10-10]
Is Jane using some kind of special Sky Dragon Slayer definition of the word "net"? In physics, net power through a boundary around the source = "radiative power out" minus "radiative power in".
My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]
... at steady-state, the relation given above already accounts for any radiative power being absorbed from other bodies.... [Jane Q. Public, 2014-10-10]
Only if "already accounts for" means "completely ignores" in Janeland.
... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. This is a requirement of the Second Law of Thermodynamics. Now, this is NOT the same as saying "no radiation absorbed at all". But when you put the two points above together, what it does mean is that ZERO of the radiative power output from the above equation is coming from other bodies. THE AMOUNT OF POWER OUTPUT IN THIS EQUATION DOES NOT NEED TO ACCOUNT FOR POWER FROM THE WALLS, BECAUSE THE NET IS ZERO.... [Jane Q. Public, 2014-10-10]
Once again, it seems like we disagree about the meaning of the term "NET".
1. Can we agree that net power through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net power through a boundary is only zero if "radiative power out" equals "radiative power in"?
3. Can we agree that "radiative power out" only equals "radiative power in" if the source and the chamber walls are at the same temperature?
If we can agree on those three points... how can the net power be zero when the source is warmer than the chamber walls?
I spent more time than I'd care to admit trying to think of a way to arrange the solar panels that doesn't require a special magnetic rotary joint.
Maybe the solar panels could be physically attached to the midpoint, and arranged in a circle with the same surface normal as the plane of rotation.
Ordinarily this would result in no solar power, regardless of whether the spheres' plane of rotation has the same surface normal as the ecliptic plane (the "parked" configuration) or if its surface normal points along the orbital velocity vector (the "Hohmann transfer" configuration).
But a large cheap mirror could reflect sunlight onto the circular solar panel, eliminating the need for a special magnetic rotary joint, and the inefficiency of microwave or inductive power transfer.
The mirror could be held in place against solar pressure using VASIMR drives. When the spheres are under thrust, the total fuel needed to move the mirror should be negligible compared to the fuel needed to move the heavily shielded spheres. Moving the mirror independently would allow the spheres' plane of rotation to change without reconfiguring its solar panels each time.
I've been considering Hohmann transfer orbits because they only require thrust that's completely tangential to the Sun. In that case, the spheres' plane of rotation would be perpendicular to the ecliptic plane.
Sorry, this is ambiguous. Here's a better explanation. I've been considering Hohmann transfer orbits because they only require thrust that's completely tangential to the Sun. In that case, the spheres' plane of rotation would be perpendicular to the ecliptic plane, with its surface normal pointing along the (circular) orbital velocity vector.
... Claiming that I "probably won't" write down a black body equation is a form of lying by implication, because we weren't discussing black bodies! By your own insistence. It was just another straw-man argument, AND blatant dishonesty at the same time. I have a copy of our AGREEMENT to treat all the materials as gray bodies, in black and white. So your claim that I "probably won't" include a black-body equation, when I HAD shown you the gray-body equations I used, is just another dishonest way to distort the argument.... [Jane Q. Public, 2014-10-09]
Good grief, Jane. As I've repeatedly explained, the gray body equation has to reduce to the black body equation when emissivity = 1. I wasn't lying or being blatantly dishonest. I was trying to show you how to check your work.
... If you're publishing an equation for calculating P, and you have an additive term on one side of the equation, which is exactly cancleled out by an additive term on the other side of an equation, you don't include them, you cancel them.... it's already known that X cancels out!!! There is no NET absorption of radiative power from cooler bodies. WE KNOW THIS FROM THERMODYNAMICS. So any radiative power that comes in, goes right back out. That's YOUR power in = power out!... [Jane Q. Public, 2014-10-08]
Careful physicists write down all the possible terms in their equations first.
Sure. When it's relevant. But I don't have to write down extra terms when the equation is already a known physical law. See my comment above about volume of a sphere. The formula is already known and other "terms" are not relevant.
You appear to be trying to imply that I left something out. So if YOU think that, then why don't you write what you think it is here? Hint: we already know what it is, because you've already made that erroneous claim, several times. So you're just re-hashing fallacious old news again. But if you want to "explain" how you think it works again, go right ahead. If it's the same as last time, I reserve the right to laugh at you again. [Jane Q. Public, 2014-10-09]
Jane still refuses to write down his energy conservation equation before he "canceled" terms, so I still have to guess at his original equation. This still seems like the only energy conservation equation consistent with what Jane's saying above:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Again, Jane appears to be saying that "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out". If that's the case, then those terms would cancel as Jane claims. That's the only way to get from "power in = power out" to Jane's final equation:
My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]
Really? Let's use the Stefan-Boltzmann law to describe the radiative terms, one at a time. Let's start with a term we can probably agree on. Because "radiative power out from source" is emitted by a graybody source at temperature T1, the Stefan-Boltzmann law says:
electrical heating power per square meter + radiative power in from chamber
I think we got into this discussion talking about rotating ships, to provide midi-gravity. We know that microgravity requires a lot of effort to counteract, so... you're going to need some major engineering reasons to not go down the spin-for-pseudo-gravity route.
Yes, centrifugal gravity seems like the only way to stay healthy in space. I pointed out that long-term colonies shouldn't rotate faster than 1 rpm in order to avoid inducing motion sickness. That imposes such serious tensile strength requirements that it seems like the shield can't spin with the ship unless the ship is made of carbon nanotubes.
Take a close look at the design of the ISS (because I've seen those designs online ; other spacecraft will have the same issues) : the radiators protrude in one direction radial to the Sun, but the solar panels are perpendicular to the Sun. If you rotate the system by 90 degrees, then the solar panels are useless and the radiators become heat absorbers. That's probably a large part of the reason for not rotating the ISS, but... comments above about the effort needed to avoid the health problems of microgravity.
Yes, the ISS is a useful example. I'm proposing a modular design, where a sphere with interior radius of 10.7 meters has enough living and garden space to support 4 people. One sphere alone couldn't provide centrifugal gravity, but in that configuration the solar panels would be unfolded perpendicular to the Sun, and the radiators would be unfolded behind the sphere, radially away from the Sun.
But two spheres could dock and attach tethers at the top of each sphere. Then if they separate to a distance of 1800 meters, they could rotate at 1 rpm around their shared center of mass to produce 1g of centrifugal gravity.
If they're not going anywhere, their plane of rotation should probably be the ecliptic plane. Otherwise the Sun's orientation would change as they orbit the Sun. Each sphere's radiators could be attached to the tethers, parallel to the ecliptic plane so they never face the Sun.
During the docking procedure, each sphere's solar panel would be detached and remain at the midpoint between the spheres. They'd have to be able to move along the tether in case one of the spheres becomes heavier and moves the center of mass. The solar panels would be kept perpendicular to the Sun as the spheres rotate, so they'd have to be kept in place magnetically and transfer power to the spheres using induction or microwaves.
I still don't like relying on rotary joints, particularly coaxial ones. I'd use them where unavoidable, but I'd avoid them where possible. And in life-support, they'd scare me.
Yeah, me too. That's why I spent more time than I'd care to admit trying to think of a way to arrange the solar panels that doesn't require a special magnetic rotary joint. At first I thought the sphere's plane of rotation should have a surface normal that points directly at the Sun. That way the solar panels could be attached directly to the tethers on the side that always faces the Sun, and the radiators could also be attached directly to the tethers, but at 90 degrees so they never face the Sun. They could also be attached to the side of the sphere which never faces the Sun.
That might be an emergency configuration if the magnetic rotary joint fails, but the sphere's plane of rotation stays fixed as they orbit the Sun. That means that in 4 months the configuration will have shifted by 90 degrees, making the solar panels useless.
It would be too expensive to continually use fuel to keep the sphere's plane of rotation in place relative to the Sun. Maybe an electrodynamic tether could work, but I haven't looked at that possibility in detail.
Slashdot Mirror
Once again, how bizarre. The whole reason Slayers deny that an enclosed source warms is because that implies greenhouse gases can't warm the surface:
If Jane is so sure that his Sky Dragon Slayer nonsense is correct, why can't he write down a simple energy conservation equation around the heated source without wrongly "cancelling" terms? Ironically, this is the very first equation needed to understand Spencer's experiment. And Jane can't even get the first equation right. Prof. Cox is right: this isn't even degree-level physics.
Jane, if you tried just once to write down an energy conservation equation for a boundary around the source without wrongly "cancelling" terms, you'd realize all this Slayer nonsense is wrong.
In other words: bok bok bok BOKKKKK. That's what I thought. Jane/Lonny Eachus is chicken.
If Jane/Lonny Eachus were a real skeptic, he'd at least consider the possibility that Jane's "radiant power output" equation doesn't describe "electrical heating power". Jane's textbooks don't say to use a "radiant power output" equation to describe "electrical heating power".
That's why Jane is too chicken to ask Prof. Cox if electrical heating power depends on the cooler vacuum chamber wall temperature. Because Jane's afraid that Prof. Cox will say yes. If not, why did Prof. Cox say all these things?
Remember,
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?
Jane keeps insisting that this Sky Dragon Slayer equation describes electrical heating power:
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.
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:
Do you really want to ask for that link? Watch what happened the last time someone asked Jane/Lonny Eachus for that link:
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.
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.
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.
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. Fascina
Jane objected:
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?
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.
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:
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.
Jane already actually said:
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.
No. Once again, gray body equations have to reduce to black body equations where there are no reflections.
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
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.
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
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.
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.
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.
Electrical power isn't radiative power, so it wouldn't be included in net radiative power.
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.
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 theor
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
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"?
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.
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?
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.
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.
If net radiative power is "all inputs minus all outputs" then net radiative power is only zero if all the inputs equal all the outputs. That only happens if the source is at the same temperature as the chamber walls.
But Jane claims that net radiative power is zero when the source is hotter than the chamber walls.
Hmm...
Again, net radiative power is only zero if the source is at the same temperature as the chamber walls.
Oh, so you're saying net radiative power isn't zero? For some odd reason I thought you were saying it was zero.
By the way, just in case it wasn't obvious from the fact that I was responding to Jane's claims of zero net radiation absorbed and zero net radiative power output, I was talking about net radiative power because that's what Jane seemed to be talking about. That's why my equation only has radiative terms. Here's a less ambiguous version:
So you're not going to retract your claim that net radiative power is zero when the source is warmer than the chamber walls?
Is Jane using some kind of special Sky Dragon Slayer definition of the word "net"? In physics, net radiative power through a boundary around the source = "radiative power out" minus "radiative power in".
So you're not going to retract your claim that net power is zero when the source is warmer than the chamber walls?
Is Jane using some kind of special Sky Dragon Slayer definition of the word "net"? In physics, net power through a boundary around the source = "radiative power out" minus "radiative power in".
Only if "already accounts for" means "completely ignores" in Janeland.
Once again, it seems like we disagree about the meaning of the term "NET".
1. Can we agree that net power through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net power through a boundary is only zero if "radiative power out" equals "radiative power in"?
3. Can we agree that "radiative power out" only equals "radiative power in" if the source and the chamber walls are at the same temperature?
If we can agree on those three points... how can the net power be zero when the source is warmer than the chamber walls?
Maybe the solar panels could be physically attached to the midpoint, and arranged in a circle with the same surface normal as the plane of rotation.
Ordinarily this would result in no solar power, regardless of whether the spheres' plane of rotation has the same surface normal as the ecliptic plane (the "parked" configuration) or if its surface normal points along the orbital velocity vector (the "Hohmann transfer" configuration).
But a large cheap mirror could reflect sunlight onto the circular solar panel, eliminating the need for a special magnetic rotary joint, and the inefficiency of microwave or inductive power transfer.
The mirror could be held in place against solar pressure using VASIMR drives. When the spheres are under thrust, the total fuel needed to move the mirror should be negligible compared to the fuel needed to move the heavily shielded spheres. Moving the mirror independently would allow the spheres' plane of rotation to change without reconfiguring its solar panels each time.
Sorry, this is ambiguous. Here's a better explanation. I've been considering Hohmann transfer orbits because they only require thrust that's completely tangential to the Sun. In that case, the spheres' plane of rotation would be perpendicular to the ecliptic plane, with its surface normal pointing along the (circular) orbital velocity vector.
Oops. In 3 months, or one quarter of an Earth year, the configuration will have shifted by 90 degrees.
Good grief, Jane. As I've repeatedly explained, the gray body equation has to reduce to the black body equation when emissivity = 1. I wasn't lying or being blatantly dishonest. I was trying to show you how to check your work.
Jane still refuses to write down his energy conservation equation before he "canceled" terms, so I still have to guess at his original equation. This still seems like the only energy conservation equation consistent with what Jane's saying above:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Again, Jane appears to be saying that "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out". If that's the case, then those terms would cancel as Jane claims. That's the only way to get from "power in = power out" to Jane's final equation:
Really? Let's use the Stefan-Boltzmann law to describe the radiative terms, one at a time. Let's start with a term we can probably agree on. Because "radiative power out from source" is emitted by a graybody source at temperature T1, the Stefan-Boltzmann law says:
electrical heating power per square meter + radiative power in from chamber
Yes, centrifugal gravity seems like the only way to stay healthy in space. I pointed out that long-term colonies shouldn't rotate faster than 1 rpm in order to avoid inducing motion sickness. That imposes such serious tensile strength requirements that it seems like the shield can't spin with the ship unless the ship is made of carbon nanotubes.
Yes, the ISS is a useful example. I'm proposing a modular design, where a sphere with interior radius of 10.7 meters has enough living and garden space to support 4 people. One sphere alone couldn't provide centrifugal gravity, but in that configuration the solar panels would be unfolded perpendicular to the Sun, and the radiators would be unfolded behind the sphere, radially away from the Sun.
But two spheres could dock and attach tethers at the top of each sphere. Then if they separate to a distance of 1800 meters, they could rotate at 1 rpm around their shared center of mass to produce 1g of centrifugal gravity.
If they're not going anywhere, their plane of rotation should probably be the ecliptic plane. Otherwise the Sun's orientation would change as they orbit the Sun. Each sphere's radiators could be attached to the tethers, parallel to the ecliptic plane so they never face the Sun.
During the docking procedure, each sphere's solar panel would be detached and remain at the midpoint between the spheres. They'd have to be able to move along the tether in case one of the spheres becomes heavier and moves the center of mass. The solar panels would be kept perpendicular to the Sun as the spheres rotate, so they'd have to be kept in place magnetically and transfer power to the spheres using induction or microwaves.
Yeah, me too. That's why I spent more time than I'd care to admit trying to think of a way to arrange the solar panels that doesn't require a special magnetic rotary joint. At first I thought the sphere's plane of rotation should have a surface normal that points directly at the Sun. That way the solar panels could be attached directly to the tethers on the side that always faces the Sun, and the radiators could also be attached directly to the tethers, but at 90 degrees so they never face the Sun. They could also be attached to the side of the sphere which never faces the Sun.
That might be an emergency configuration if the magnetic rotary joint fails, but the sphere's plane of rotation stays fixed as they orbit the Sun. That means that in 4 months the configuration will have shifted by 90 degrees, making the solar panels useless.
It would be too expensive to continually use fuel to keep the sphere's plane of rotation in place relative to the Sun. Maybe an electrodynamic tether could work, but I haven't looked at that possibility in detail.