I haven't "baselessly" accused anyone of anything. I make sure I have very good bases when I make actual accusations. If anything, your comment was a "baseless accusation". And those emails were almost certainly not "hacked". Evidence strongly suggests the leak was an "inside job", as the saying goes See what I mean, folks?. [Jane Q. Public, 2015-03-23]

Good grief, Jane. Of course Sky Dragon Slayers like you mistakenly believe the emails were almost certainly not "hacked". And since you're almost completely incapable of admitting when you're wrong, you'll never be reasoned out of that position.

But surely even Jane can see that unless the scientists Jane was quoting and attacking "leaked" the emails, then Jane was using private emails to attack scientists. And apparently that's cool, but responding to public comments is illegal, unethical, despicable lowlife sociopathic behavior.

Janeland is a funny place. For instance:

... I mean look: you want evidence that this guy is short of a full load? Some of those comments are from someone he thinks is me, and not even from Slashdot. ... [Jane Q. Public, 2015-03-23]

It's so adorable how Jane keeps hiding behind a pole.

... call it revenge for not getting to name Charon "Goofy". [Lonny Eachus, 2014-11-07]

... If left up to me, I would have really, honestly, named new 9th planet "Goofy". [Lonny Eachus, 2013-07-28]

... You're talking to somebody who thought that other rock should have been named "Goofy". [Lonny Eachus, 2013-07-28]

I want the definition to go back the way it was. That way, maybe we will finally get to name its companion "Goofy", rather than that dumbass Charon moniker. [Jane Q. Public, 2015-02-28]

Dumbass Charon moniker? James Christy named Charon because his wife's nickname is "Char".

In fact, "Charon is informally pronounced "SHAR-on," similar to the name of the discoverer's wife, Charlene."

The man who discovered Pluto wanted to name its moon after his wife, and astronomers pronounce it "SHAR-on" to honor his choice. Why does Jane/Lonny Eachus call this a dumbass moniker?

... I have explained to you many times that I was not disputing the definition of "net". So STOP LYING. Because that's what you are doing. [Jane Q. Public, 2015-03-23]

Once again, you disputed my simple substitution of the standard physics definition of the term "net" into your equation, and simultaneously insisted that you don't dispute the standard physics definition.

Who do you think that's going to fool, Jane? I don't think it's likely to fool anyone who understands what a "definition" is. Not to mention the fact that you repeatedly pretended not to understand how a crayon mark representing cooler power doesn't pass through a boundary inside the cooler wall.

If you're actually this confused about basic physics, why are you lecturing physicists about physics?

If you're not actually confused, why has Jane/Lonny Eachus betrayed humanity?

I mean look: you want evidence that this guy is short of a full load?

Charming. Jane, keep in mind that you're saying this right after adamantly rejecting the standard physics definition of the word "net", and pretending not to understand how a crayon mark representing cooler power doesn't pass through a boundary inside the cooler wall.

Looks like you were lying yet again, when you claimed that you'd be happy to declare to everyone that you were wrong about your Latour Sky Dragon Slayer nonsense. Or maybe you'd like to keep disputing basic physics definitions and the esoteric art of using crayons in a coloring book?

I'm sorry for disrupting the discussion. But I'm a little more concerned about the fact that Jane's spreading civilization-paralyzing misinformation about one of the defining issues of our time, and libeling scientists by repeatedly and baselessly accusing them of fraud.

Every minute Jane spends cussing and screaming at me is one minute he can't spend trying to confuse someone else, or attacking yet another scientist.

He also doesn't seem to realize I do not have a similar obsession about him. Quite the opposite: I would appreciate it very much if he went away and left me alone. It's kind of ironic that he posts here where the topic is harassment.

Once again, leave you in peace so you can keep baselessly accusing scientists of fraud? If you want to keep baselessly accusing scientists of fraud, you're going to have to go through me first.

Once again, don't flatter yourself. Debunking misinformation and defending scientists against baseless attacks are my unhealthy obsessions. It's hardly my fault that you're one of the most prolific misinformers I've ever seen. If you didn't want people responding to your claims, you probably should've written them in a notebook instead of on a public website. It's also strange that you call my responses to your public comments "stalking and harassment" while quoting hacked private emails from years ago to baselessly attack scientists.

I think you've been on the Internet long enough to know that just because someone uses a female handle/nick, doesn't mean they are actively trying to deceive others into believing they are female.

Right, I've explicitly said that pseudonyms don't constitute lying.

The reasons I say Jane's actively trying to deceive people are too numerous to list here. For starters, Jane has made it clear that he's either a woman or a "flamer" (his words). Jane insists that most people who bothered to look have referred to him as a gal, etc. Jane/Lonny Eachus didn't need to pepper his comments with all these lies, which is why I call them pathological lies.

Apparently you feel that you have a right to be offended, but not I. [Lonny Eachus, 2010-12-06]

There is no "right to not be offended". The very notion is a mockery of American values. ... [Lonny Eachus, 2013-12-22]

Some folks got this weird idea they have a right to not be offended. [Lonny Eachus, 2013-12-28]

THERE IS NO “RIGHT” TO NOT BE OFFENDED. THERE IS AN OBLIGATION TO BE HUMBLE WHEN YOU OFFEND EGREGIOUSLY, THOUGHTLESSLY, OR GRATUITOUSLY. [Lonny Eachus, 2014-05-03]

I’ve met lot of people who seem to think “Right To Not Be Offended” is in the 1st Amendment. ... [Lonny Eachus, 2014-06-24]

This is very sad. It implies that people have some kind of “right” to not be offended, which is ludicrous. [Lonny Eachus, 2014-10-03]

Some people need to get over the ridiculous notion that they have some kind of "right" to not be offended. ... [Jane Q. Public, 2015-03-22]

No one is complaining that they should have the "right" not to be offended.

Wrong. LOTS of people do it. I see that kind of crap from one person or another on social media almost every day. And I have gotten it at work, too. Not for a long time, but it did happen. ... [Jane Q. Public, 2015-03-22]

No, Jane. You might think you're seeing "that kind of crap" almost every day, but as usual you're just putting words in peoples' mouths.

I worked in an office that the women basically ran, and believe me, they were sexist tyrants! They could tell jokes, men could not. I was told that ANYTHING they found offensive (in any way) would be considered sexual harassment. [Lonny Eachus, 2009-04-30]

... While discussing harassment at the workplace, the bookkeeper (who I found to be a pretty offensive person herself) said "ANYTHING I consider to be offensive is sexual harassment." Thinking she didn't mean that quite the way she said it, I said "You mean anything sexual you find offensive is sexual harassment." She gave me a rather nasty look and said: "No. ANYTHING I find offensive is sexual harassment." ... [Jane Q. Public, 2015-03-22]

Poor Jane/Lonny Eachus. Was this why you started posing as a woman on the internet? To get back at the sexist tyrant women by making all women look bad?

I've repeatedly failed to explain that the power to the cooled walls you keep talking about is completely irrelevant because it doesn't pass through that boundary.

No, you haven't "failed to explain" this. What you did -- typically in your fashion, in my experience -- was change your story when you realized that it was not a viable avenue of attack. ... [Jane Q. Public, 2015-03-20]

Don't be ridiculous, Jane. Anyone who clicks that link will see that I've consistently told you that only power which passes through a boundary is included in its energy conservation equation. Again, the source heating power passes through that boundary, but the exterior wall cooler power doesn't pass through that boundary.

It's just like crayons in a coloring book, Jane.

I repeat: I have all this already on record. [Jane Q. Public, 2015-03-20]

Gosh, really? Before you give a copy of your cussing and screaming to your grandchildren, you might want to consider giving it to them before they've mastered coloring books. Otherwise "Grandma Jane" will have to answer a lot of awkward questions.

And again, inserting the standard physics definition of the word "net" into your equation reproduces the energy conservation equation you're still adamantly rejecting. That's another independent way to see that you should consider the possibility that only power passing through a boundary should be included in the energy conservation equation across that boundary.

No, the electrical power input is however many watts are sent in through the boundary around the heat source. That's why it's included in the energy conservation equation through that boundary.

You have just contradicted yourself AGAIN, because I have records of you clearly arguing that the input power was to maintain a temperature difference between the heat source and the walls, while I was arguing that the input to the heat source was constant but the power to the cooled walls was not stipulated and could be variable. [Jane Q. Public, 2015-03-20]

Again, the reason the electrical input power heating the source is included in the energy conservation equation through a boundary around the heat source is because it passes through that boundary. That's the important point.

You still don't seem to understand that power which doesn't pass through that boundary isn't included in that energy conservation equation. I've repeatedly failed to explain that the power to the cooled walls you keep talking about is completely irrelevant because it doesn't pass through that boundary.

Jane, this is on the level of "drawing within the lines." Does the power pass through the boundary or not? Just think about whether a crayon line crosses the lines in a coloring book. If it does, that power gets included in the energy equation through that boundary.

Seriously, take a few seconds to write down an energy conservation equation for a boundary around the source without wrongly "cancelling" terms. You'd quickly find that:

(1) The power to the cooled walls is irrelevant.
(2) Because only the power passing through the boundary is included, the electrical power heating the source maintains a temperature difference between the heat source and the walls.

And again, inserting the standard physics definition of the word "net" into your equation reproduces the energy conservation equation you're still adamantly rejecting. That's another independent way to see that you should consider the possibility that only power passing through a boundary should be included in the energy conservation equation across that boundary.

... Spencer's experiment stipulated that the outer wall be kept at a constant temperature. Given that it is being given input from interior heat sources, it would take energy (over time, power of course) to maintain that low temperature. This was obviously Spencer's attempt to model the radiation "escaping to space". ... [Jane Q. Public, 2015-03-20]

Again, any power used to maintain that low temperature is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant.

... However, YOU have repeatedly stated that your electrical power input was considered to be maintaining a temperature difference between the heat source and the outer wall. In fact that was the stated basis for many of your arguments about conservation of energy. ... [Jane Q. Public, 2015-03-20]

No, the electrical power input is however many watts are sent in through the boundary around the heat source. That's why it's included in the energy conservation equation through that boundary.

... But your input energy was supposed to be constant. So you're either violating the parameters of the experiment, or you are creating energy from nothing. You don't get to have it both ways, and again your "solution" contradicts itself. ... [Jane Q. Public, 2015-03-20]

The electrical power input which crosses the boundary around the heat source is constant. Any power which doesn't cross that boundary is irrelevant, because it isn't included in that energy conservation equation.

And again, inserting the standard physics definition of the word "net" into your equation reproduces the energy conservation equation you're still adamantly rejecting. Would it really be so hard to take a few seconds to write down an energy conservation equation for a boundary around the source without wrongly "cancelling" terms? That's another way to see that you should consider using the standard physics definition of the word "net".

This is really basic physics, Jane. If you're actually this hopelessly confused, maybe you shouldn't be lecturing physicists about physics.

And for your sake I hope you actually are just confused. It's difficult to understand why anyone would deliberately spread misinformation about what the National Academy of Sciences calls "one of the defining issues of our time."

... I will do you a favor here, and say: don't bother to go calculating the energy, either. The problem is that an analysis of this kind, based on the assumption that power-in = power-out, is doomed to fail except in coincidental cases. Even conservation of energy can give very misleading results. The black body example I gave shows why your "energy conservation just inside the surface" won't work. ... power-in = power-out is not necessarily true, and in fact that is probably a very rare exception. ... [Jane Q. Public, 2014-09-07]

Energy is always conserved, so power in = power out through any boundary where nothing inside is changing.

For the hundredth time: nobody is disputing this. ... [Jane Q. Public, 2015-03-19]

"Nobody" is disputing this, apparently in the same way that "nobody" is disputing my simple substitution of the standard physics definition of the term "net" into Jane's equation.

... The equation for radiative power output of a body at steady-state does not change in the presence of cooler bodies. It remains exactly the same. It is dependent ONLY on emissivity, thermodynamic temperature, and the Stefan-Boltzmann constant. [Jane Q. Public, 2015-03-17]

... The statement that "no net energy is transferred to the warmer body from the colder" is exactly WHY the equation for radiant heat output does NOT change in the presence of colder bodies. BUT... you neglect the fact that it would have to, if it were absorbing and re-radiating radiation from those colder bodies. ... [Jane Q. Public, 2015-03-19]

No, Jane. The equation for radiant heat output is still the Stefan-Boltzmann equation. As I've repeatedly told you, we agree that it is dependent ONLY on emissivity, thermodynamic temperature, and the Stefan-Boltzmann constant.

However, temperature is determined by internal energy. That's determined by a fundamental law called "conservation of energy" which is necessarily true and certainly isn't a "very rare exception" or "doomed to fail".

The fact that conservation of energy determines temperature doesn't change the equation for radiant heat output, even in the presence of colder bodies. The Stefan-Boltzmann equation remains exactly the same, as I've repeatedly explained.

... even if you tried to maintain that it wasn't a feedback loop but just a new equilibrium, with the hotter temperature you would still have to increase the power to maintain the exterior walls at a constant temperature. Which means you would be extracting more thermal energy from the system... but not adding any more. Contradiction. ... [Jane Q. Public, 2015-03-19]

Jane, I've repeatedly failed to explain how "conservation of energy" works. Once again: any power used by the exterior wall cooler (or heater) is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant.

The word you're looking for isn't "contradiction". It's "irrelevant." As in, "Jane's objection is irrelevant because that power never crosses the boundary."

I've repeatedly explained that in every single equation I've derived, more power is radiated from hot to cold than vice versa. So my solution doesn't violate the second law of thermodynamics (or the first).

And this is where YOU are mis-stating what is meant by "net". You say "more" goes from hot to cold, but you have also been claiming that SOME goes from cold to hot. ... [Jane Q. Public, 2015-03-19]

Exactly! Because that doesn't violate the second law of thermodynamics. But ignoring the power passing through a boundary around the source violates the first law of thermodynamics.

... But what you're not getting is: given the nature of the experiment, that would be creating energy from nothing. You are increasing the thermodynamic energy of a hotter body by transferring heat to it from a colder body or bodies. And I'm not sorry to tell you: nature doesn't work that way. ... [Jane Q. Public, 2015-03-19]

Jane, do you agree that the way to tell if an equation is "creating energy from nothing" is to see if it satisfies the first law of thermodynamics, which is "conservation of energy"?

If your answer is no... how do you tell if an equation is creating energy from nothing?

If your answer is yes... that's why I've repeatedly asked you to write down an energy conservation equation for a boundary around the source without wrongly "cancelling" terms. Jane/Lonny Eachus adamantly refuses to take the very first step in applying the first law of thermodynamics to this problem, but as usual he's willing to endlessly insist that he's right.

Jane, if you'd just take a few seconds to apply conservation of energy to this problem, you'd see that your solution violates conservation of energy. That's the law which tells us energy isn't created from nothing, so it's extremely ironic that you keep refusing to take that very first step and yet continue to accuse me of creating energy from nothing, after I've repeatedly asked you to please write down an energy conservation equation for a boundary around the source without wrongly "cancelling" terms.

... Your statement is a contradiction. Whether you claim it is a "raising" of energy of the warmer body, or "less loss" from the warmer body, the only other input power is the same. So you end up with a "hotter" hot body. But if your hot body is hotter, then its radiative output CHANGES, and so then does the temperature of the colder body, and you have created a feedback loop, not a new equilibrium. As I have already mentioned, you don't get to do that. You're adding energy from nowhere. ... [Jane Q. Public, 2015-03-19]

See above. If you want to prove that someone is "adding energy from nowhere" then you need to write down an energy conservation equation for a boundary around the source without wrongly "cancelling" terms.

If you ever find it in your heart to take those few seconds, you'll find that a new equilibrium (i.e. steady-state constant temperatures) is reached, exactly as I've repeatedly told you. Once again, the enclosed source temperature doesn't warm above 235F. Once it reaches that point, power in = power out, so the energy inside the boundary doesn't change. Thus the source reaches a new steady-state condition, where temperatures are constant.

Again, the universe is safe.

... A given input energy is only going to raise a body of given properties to a certain maximum temperature. ... [Jane Q. Public, 2015-03-19]

No, because "conservation of energy" says that power in = power out through any boundary where nothing inside is changing. Thus, both power flowing in the boundary over some time period (which Jane calls "input energy") and power flowing out of that boundary are required to determine the steady-state temperature of a body of given properties.

Again, warming the chamber walls is like partially closing the drain on a bathtub where water is flowing in at a constant rate. This raises the bathtub water level simply by reducing the water flow out. In exactly the same way, a source heated with constant electrical power warms when the chamber walls are warmed because that reduces the net power out.

Pretending that we only need to know the power flowing in and not the power flowing out is like pretending we only need to know a bathtub's faucet flow rate to determine the steady-state water level in the bathtub, and it doesn't matter if the drain is open or closed.

... It doesn't matter whether that energy is electricity or cow farts. Any other assertion is... well... hot air. ... [Jane Q. Public, 2015-03-19]

Jane, I just finished explaining that it doesn't matter if the energy is electricity or kerosene. I didn't see the point of addressing your "friction from a horse's ass" observation because that didn't seem productive, but I hoped it would be clear from my response that "horse ass friction" heating power would be treated exactly the same as electrical or kerosene heating power.

Sadly, once again I seem to have overestimated you. So I'll be more explicit.

I'm not saying the power source matters. Whether you're using an electrical source or a kerosene heater or "horse ass friction" or cow farts, the important point is that required heating power (electrical or kerosene) is zero if the chamber walls are at the same temperature as the heat source.

That's not true for "radiative power out", as I've explained ad nauseum. That's because "radiative power out" doesn't depend on the chamber wall temperature. But that's different than "heating power" (electrical or kerosene or "horse ass friction" or cow farts) which does depend on the chamber wall temperature. "Heating power" tells you how much electricity or kerosene you'd need to keep the source at a certain temperature, given the temperature of the chamber walls.

One way to check your solution is to make sure that your equation for heating power (electrical or kerosene or "horse ass friction" or cow farts) agrees that no heating power is necessary if the chamber walls are at the same temperature as the heat source. My equations have all passed that check. Jane's mistaken equation doesn't.

So you dispute my simple substitution of the standard physics definition of the term "net" into your equation, while simultaneously insisting that you don't dispute the standard physics definition?

I am disputing nothing at this time. I am NOT going to re-argue this with you. I have exactly zero reason or desire to do so. [Jane Q. Public, 2015-03-17]

Don't you see how the fact that you previously disputed my simple substitution of the standard physics definition of the term "net" into your equation looked like disputing that standard physics definition?

If you're really not disputing my simple substitution any longer, then you're now agreeing with my energy conservation equation. If so, that's great news!

Even Jane should be able to recognize that his 4 unnamed textbooks don't support him, because deep down even Jane should be able to tell that he's just endlessly blustering to cover up the fact that he can't produce any textbook quotes saying that electrical heating power = radiative power out.

Completely irrelevant. You found a temperature difference = power equation that applied to a completely different situation and you've been inappropriately applying it to this problem ever since. ... [Jane Q. Public, 2015-03-17]

Good grief, Jane. You've previously hallucinated conduction and convection terms in my equations describing conservation of energy through vacuum-filled spaces. If that's what you mean by "inappropriately applying" then you should look at my equations very carefully. Notice the complete lack of conduction and convection terms. Notice that my equations are based on a fundamental principle called "conservation of energy" that applies to all situations.

... Much like when you tried to call a heat transfer equation the equation for radiative power out. (Hint: it isn't.) [Jane Q. Public, 2015-03-17]

Good grief, Jane. This is your response to my comment explicitly and repeatedly telling you that radiative power out is different than electrical heating power? I've repeatedly told you that conservation of energy leads to heat transfer equations that describe electrical heating power, but the Stefan-Boltzmann equation can give you "radiative power out".

Once again: the Stefan-Boltzmann equation can give you "radiative power out" but only a completely different principle called "conservation of energy" can give you a totally different quantity known as "electrical heating power".

Once again: "radiative power out" isn't just a fancy way of saying "electrical heating power". They're completely different. To find electrical heating power, Jane needs to use conservation of energy, where power in = power out. That results in a heat transfer equation, not just an equation for "radiative power out".

Jane, I've been very clear that a heat transfer equation is used to find electrical heating power, not "radiative power out". And yet you keep claiming otherwise. Why, Jane?

... In order for YOUR argument to work, a sphere of one substance suspended in a vacuum cavity surrounded by the same substance at the same temperature, would spontaneously increase in temperature. If it did that, it would be at a higher temperature (i.e., radiate more power to the wall of the cavity), which would then itself become warmer, and you would have a universe-destroying positive feedback. ... [Jane Q. Public, 2015-03-17]

Nonsense. Here's the first energy conservation equation I derived:

electricity + sigma*T_c^4 = sigma*T_h^4 (Eq. 1)

Notice that if the two temperatures T_c and T_h are equal, the required electrical heating power per square meter (electricity) = 0. So my equation actually says that a sphere of one substance suspended in a vacuum cavity surrounded by the same substance at the same temperature, would stay at that temperature forever without any electrical heating power.

The universe is safe.

That wasn't so hard, was it? So why does Jane keep insisting otherwise? Notice that every equation I've derived says that electrical heating power goes to zero when the heat source and chamber walls are at the same temperature, including this more complicated solution.

But ironically, Jane's mistaken "energy conservation" equation doesn't explain why electrical heating power goes to zero when the heat source and chamber walls are at the same temperature:

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

Before Jane keeps insisting that my equations say something they obviously don't, Jane should note that his mistaken "energy conservation" equation can't explain why the heat source doesn't need electrical heating power if the chamber walls are at the same temperature.

Once again, note that conservation of energy through a boundary around the source leads directly to an equation describing the electrical power required to keep the source at temperature T1 inside chamber walls at temperature T4. This equation is valid for T1 > T4, T1 = T4, and T1 < T4. Jane might wonder why he can't derive a single equation which works for all these cases.

... the power source doesn't matter. Only power input matters. It doesn't matter whether that power is an electrical source, or a kerosene heater... [Jane Q. Public, 2015-03-17]

I'm not saying the power source matters. Whether you're using an electrical source or a kerosene heater, the important point is that required heating power (electrical or kerosene) is zero if the chamber walls are at the same temperature as the heat source.

That's not true for "radiative power out", as I've explained ad nauseum. That's because "radiative power out" doesn't depend on the chamber wall temperature. But that's different than "heating power" (electrical or kerosene) which does depend on the chamber wall temperature. "Heating power" tells you how much electricity or kerosene you'd need to keep the source at a certain temperature, given the temperature of the chamber walls.

One way to check your solution is to make sure that your equation for heating power (electrical or kerosene) agrees that no heating power is necessary if the chamber walls are at the same temperature as the heat source. My equations have all passed that check. Jane's mistaken equation doesn't.

... The textbooks say a great deal more than that. And you have been unwilling to admit that they're right about the rest of it, too. Your answer was wrong. I showed you where it was wrong. I used standard textbook radiant heat transfer equations to prove it. I explained to you WHY it was wrong. ... [Jane Q. Public, 2015-03-17]

Jane, the only time I've been "unwilling to admit" that your misinterpretation of the rest of your textbooks was right was when you claimed that your textbook "implies that power-in = power-out is not necessarily true, and in fact that is probably a very rare exception."

Once again: No Jane, you've misinterpreted your textbook. Energy is always conserved, so power in = power out through any boundary where nothing inside is changing. This isn't a "very rare exception". It's a fundamental law called "conservation of energy". Does Jane seriously think his textbook says that using a fundamental law like "conservation of energy" is "doomed to fail"?

Once again, no matter how many times Slayers are told that the second law of thermodynamics isn't violated because more power is radiated from hot to cold than vice-versa, that fact never seems to penetrate their skulls.

And once again, this is a mis-statement of the facts. Nobody I am aware of claims -- and I certainly did not claim -- that thermodynamics is violated because more power is radiated from hot to cold than vice-versa. Show me where somebody did say that. ... [Jane Q. Public, 2015-03-17]

As usual, that doesn't make any sense. Jane, you've been accusing me of violating the laws of thermodynamics. I've repeatedly explained that in every single equation I've derived, more power is radiated from hot to cold than vice versa. So my solution doesn't violate the second law of thermodynamics (or the first).

So your incessant accusations were baseless. If you're retracting those accusations, then that's great news!

... NET radiative heat transfer is always from warmer object to cooler. Anything else is a violation of the fundamental laws of thermodynamics. ... [Jane Q. Public, 2015-03-17]

... which just means that more power needs to be radiated from warmer to cooler than vice-versa. At least, that's the conclusion drawn using the standard physics definition of the term "net". (And once again, every equation I've derived satisfies that condition.)

... The equation for radiative power output of a body at steady-state does not change in the presence of cooler bodies. It remains exactly the same. It is dependent ONLY on emissivity, thermodynamic temperature, and the Stefan-Boltzmann constant. [Jane Q. Public, 2015-03-17]

Good grief.

Once again, I've already agreed that it's not necessary to account for cooler bodies in the temperature versus power out equation. Again, we're not disputing the equation for radiative power out. We're disputing the equation describing conservation of energy around a boundary drawn 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

There is no need to account for other, cooler bodies when calculating radiative power out. What, do you imagine that these cooler bodies are somehow "sucking" power away from the heat source? And that a warmer body (but still cooler than the source) "sucks" less power than colder ones do? That seems to be what you're saying here. [Jane Q. Public, 2014-10-05]

Once again, Jane, I never said we need to account for other, cooler bodies when calculating radiative power out.

Once again, I'm actually saying that "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.

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]

Once again, 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".

... If you want to prove yourself right, you're going to have to prove those textbooks wrong. ... [Jane Q. Public, 2015-03-17]

Once again, Jane has 4 textbooks that say "radiative power out per square meter = (e*s)*T^4". Since I've repeatedly agreed with that statement, those textbooks don't disagree with me.

Once again, Jane/Lonny Eachus just has 4 textbooks that say "radiative power out = (epsilon * sigma)*T^4*area". I bet Jane $100 that his textbooks don't claim that electrical heating power = radiative power out. That's Jane's incorrect Slayer assumption. Even Jane should be able to recognize that his 4 unnamed textbooks don't support him, because deep down even Jane should be able to tell that he's just endlessly blustering to cover up the fact that he can't produce any textbook quotes saying that electrical heating power = radiative power out.