Forgot your password?
typodupeerror

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 56

by khayman80 (#48015949) Attached to: Wanxiang May Give 2012's Fisker Karma a Relaunch

Maybe the Slayers could explain how uncooled IR detectors see cooler objects?

Straw-man. Our argument involved gray bodies, not detectors of specific wavelengths or electronics that take advantage of specific quantum effects. But I have an answer anyway: they measure DIFFERENCES, not absolute radiation. ... [Jane Q. Public, 2014-09-28]

This isn't a quantum effect. The reason IR detectors measure DIFFERENCES, not absolute radiation, is because electrical heating power = (e * s) * (Ta^4 - Tb^4). If that weren't true, there would be no way to detect this difference, so uncooled IR detectors wouldn't be able to see cooler objects. And we couldn't have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How?

... 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. What I claimed, I repeat, is that no NET radiative energy transfer occurs from cooler bodies to warmer. That concept does not conflict with the ability of infrared cameras or pyrometers to detect "cooler" radiation. Energy can be absorbed and re-emitted... and often (for non-gray-bodies) it is re-emitted in different wavelengths. But the fact remains that there is still no NET energy transfer from cooler to warmer. If there were, it would violate the second law of thermodynamics. My argument has always been about NET heat transfer. I have explained to you many times that I do NOT claim no radiation from cooler bodies is ever absorbed. My argument is, and has been, about NET. ... [Jane Q. Public, 2014-09-28]

Once again, Jane insists electrical heating power = (e * s) * (Ta^4). Once again, Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.

If Jane would reconsider conservation of energy and include a term for "radiative power in", then Jane could honestly say he was only claiming that no net radiative power is absorbed by the source. Until then, Jane's equation claims that no radiation is absorbed by the source at all. And since Jane seems to think he's only saying no "NET" radiative power is absorbed, Jane will probably never be able to recognize his error, let alone correct it.

... And further, contrary to your own assertions, since the NET energy transfer from cooler bodies is ZERO, it is not included in the "radiative power out" term of heat transfer equations. Which is a concept that (apparently, if we assume you're being honest, which I doubt) you have had supreme difficulty getting through your head. ... [Jane Q. Public, 2014-09-28]

Once again, it's not included in the "radiative power out" term of heat transfer equations because it's included in the "radiative power IN" term.

I'm having supreme difficulty getting your concept through my head because it's Sky Dragon Slayer nonsense. The fact that more heat flows from warm to cold than vice-versa doesn't mean we can ignore the smaller amount of heat flowing from cold to warm. In fact, as I've repeatedly stressed, ignoring that heat violates conservation of energy.

Here's one way to see that: draw a boundary around a heated blackbody source. It's heated by constant electrical power flowing in. Blackbody cold walls at 0F (T4 = 255.4K) also radiate power in. The source at 150F (T1 = 338.7K) radiates power out. At steady-state, power in = power out:

electricity + (s)*T4^4 = (s)*T1^4 (Eq. 1J.2)

But Jane's equation is:

electricity = (s)*T1^4 (Jane's equation)

Because Jane's equation completely ignores radiative power flowing in, Jane's equation violates conservation of energy. How does Jane justify this violation?

... 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]

... because T(p) < T(s), no matter now much of the radiation from P strikes S, no net amount is absorbed; it is all reflected, transmitted, or scattered according to S-B. ... [Jane Q. Public, 2014-09-04]

Once again, no. This Slayer talking point can't possibly apply to blackbodies. How does Jane rationalize ignoring "radiative power in" when the blackbody source can only absorb that radiation, not reflect or scatter it? Once again, remember that the gray body equation has to reduce to the blackbody equation when emissivity = 1.

... In incorrectly "solving" Spencer's challenge, YOU ignored basic textbook methods and math to get your answer. You used an imaginary "khayman80" method of arriving at your answer, which not only contradicts everything engineering textbooks say about heat transfer, your methodology directly contradicts the Stefan-Boltzmann radiation law, even though you used it yourself in calculations. Talk about hypocrisy. I repeat: I checked your final "answer" for temperature of the heat source and it violates both the Stefan-Boltzmann law and the second law of thermodynamics. [Jane Q. Public, 2014-09-28]

Good grief, Jane. Once again, my solution doesn't violate the Stefan-Boltzmann law or the second law of thermodynamics. But it's fascinating that Jane/Lonny Eachus keeps regurgitating this baseless Slayer talking point. It seems like Prof. Brown was right to say that arguing with Slayers is a pointless waste of time.

... Venus proves nothing about CO2-based warming on Earth. If you ASSUME it's causing warming here, then you can ASSUME it causes warming there, in proportion. Such assumptions prove nothing. ... [Jane Q. Public, 2014-09-26]

Again, if CO2 isn't the reason, then why is Venus hotter than Mercury? This isn't an assumption, it's a real-world example which any true skeptic should ponder before dismissing mainstream physics in favor of Sky Dragon Slayer brainwashing. Is Venus hotter than Mercury because of CO2, gray Oreos, or basketball player gloves?

... Funny, because he's contradicting just about every argument behind the whole idea of AGW. I like how he makes these claims but isn't able to show how it actually works. He claims you can show warming via back-radiation WITHOUT the S-B equation? When it is absolutely fundamental to the very "energy transfer" he is asserting? What garbage. ... [Jane Q. Public, 2014-09-28]

That's odd. Just yesterday Jane had no argument with Prof. Brown. Now Jane claims that Prof. Brown is spreading "garbage" that contradicts just about every argument behind the whole idea of AGW. But Jane certainly isn't arguing with Prof. Brown or Dr. Shore or even me. Perish the thought.

Jane, don't you see how ironic it is to accuse three physicists (and the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society, etc.) of "FUCKING UP" their physics, while also claiming that:

.. I'll take the physicists' word for it, thank you very much. .. [Jane Q. Public, 2012-07-05]

.. I consult "the experts". When it's a question of physics, for example, I look to references from physicists, not climatologists. After all, physicists are "the experts" when it comes to physics. [Jane Q. Public, 2013-11-15]

If Jane wanted to be consistent, he'd have to retract one of these claims. Jane is either a Sky Dragon Slayer accusing physicists of "FUCKING UP" their physics, or he believes physicists are "the experts" when it comes to physics and Jane takes the physicists' word for it. But not both!

Also, STOP sock-puppet modding down my comments. THAT'S AGAINST SLASHDOT'S RULES and it's just plain an asshole thing to do. [Jane Q. Public, 2014-09-28]

I haven't used moderator points in over a year. But the fact that Jane is so convinced I am that he's cussing and screaming in ALL CAPS is emblematic of Jane's reasoning problems, just like when Jane was absolutely convinced that I'm a six-headed hydra.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 56

by khayman80 (#48000785) Attached to: Wanxiang May Give 2012's Fisker Karma a Relaunch

... As usual, you distort reality. Prof. Brown had nothing in the way of refutation or rebuttal or even retort to my second comment? Don't you find that interesting? I do. ... [Jane Q. Public, 2014-09-26]

It's not that interesting that Prof. Brown decided to ignore Jane/Lonny Eachus, given that he later said:

"Wow, Joel, I gotta say (after reading some of the replies on this thread) that this really is pointless. These folks have no conception of the FIRST law of thermodynamics, let alone the second. The argument for warming doesn't even require mentioning the SBE, it only requires the first law, the second law, and a monotonic relation between temperature difference in ANY channel and the rate of energy transfer in that channel, subject to very broad constraints.

But seriously, just a waste of time. When people just make stuff up and reject the contents of ELEMENTARY textbooks on the subject because they just don't like the conclusion those contents lead to, how can you argue with them? If somebody tries to solve the light bulb problem while pretending that it doesn't primarily cool via radiation and completely ignoring radiation, what can you do?

Get them to say "oops"?

Never happen. It's a religious issue, not a scientific one."

In other words, Prof. Brown gave up trying to educate Slayers like Jane/Lonny Eachus because it's a "waste of time."

... As for Joel Shore, again he was mis-applying an equation for heat transfer when he should have been using the equation for radiant power out. Both you and Shore insist on mis-applying this equation in a way that violates the Second Law of Thermodynamics. It's rather amusing that you brought him up, because you both FUCKED UP YOUR PHYSICS in a similar way. ... [Jane Q. Public, 2014-09-26]

That's odd. Just yesterday Jane had no argument with Dr. Shore. Now Jane claims that Dr. Shore "FUCKED UP" his physics.

... As for Joel Shore, again he was mis-applying an equation for heat transfer when he should have been using the equation for radiant power out. Both you and Shore insist on mis-applying this equation in a way that violates the Second Law of Thermodynamics. It's rather amusing that you brought him up, because you both FUCKED UP YOUR PHYSICS in a similar way. ... Engineers the world over do the math the way I did. So far that hasn't resulted in you either freezing or burning to death in your home. If they're all crazy, you might want to ask yourself why. [Jane Q. Public, 2014-09-26]

Physicists have "FUCKED UP" their physics, and only the Slayers can save the day! Or maybe the Slayers are crackpots. How could anyone tell, unless maybe Dr. Shore explained that:

"Actually, the idea that radiation goes only from the warmer to colder objects is an invention of the Slayers. It appears nowhere in the physics literature. I don't know about the exact history of our understanding, but my physics textbook from 1983 (Serway, "Physics for Scientists and Engineers", after introducing the law P = sigma*A*e*T^4 says

"A body radiates and also absorbs electromagnetic radiation at rates given by Eq. 17.11. If this were not the case, a body would eventually radiate all of its internal energy and its temperature would reach absolute zero. The energy that the body absorbs comes from the surroundings, which also emit radiant energy. If the body is at a temperature T and its surroundings are at a temperature T_0, the net power gained (or lost) as a result of radiation is given by

P_net = sigma*A_*e*(T^4 - T_0^4) (17.12)

When a body is in equilibrium with its surroundings, it radiates and absorbs energy at the same rate and so its temperature remains constant. When a body is hotter than its surroundings, it radiates more energy than it absorbs, and so it cools..."

Maybe the Slayers could explain how uncooled IR detectors see cooler objects? Using the equations Dr. Shore and I are using, the source's required electrical heating power depends on the chamber wall temperature. That's how uncooled IR detectors can see cooler objects. But Jane's "Slayer physics" insists that the chamber wall temperature doesn't affect the source's electrical heating power. So how do uncooled IR detectors see cooler objects? In particular, how did we detect the 2.7K cosmic microwave background radiation with warmer detectors?

... Venus proves nothing about CO2-based warming on Earth. If you ASSUME it's causing warming here, then you can ASSUME it causes warming there, in proportion. Such assumptions prove nothing. ... [Jane Q. Public, 2014-09-26]

If CO2 isn't the reason, then why is Venus hotter than Mercury? This isn't an assumption, it's a real-world example which any true skeptic should ponder before dismissing mainstream physics in favor of Sky Dragon Slayer brainwashing. Is Venus hotter than Mercury because of CO2, gray Oreos or basketball player gloves?

The Slayers have their own incoherent answer, while mainstream physics is presented by the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.

.. Be a man for a change and admit it. .. [Jane Q. Public, 2014-09-15]

.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it. .. [Jane Q. Public, 2014-09-19]

... Time to act like a man and admit that you were wrong. ... [Jane Q. Public, 2014-09-24]

Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 56

by khayman80 (#47996613) Attached to: Wanxiang May Give 2012's Fisker Karma a Relaunch

... These are just straw-man arguments, as usual. I have no argument with these other physicists. It was about Spencer's challenge and how YOU got it wrong, nothing more. Have you asked them, personally, about Spencer's experiment? (No, you haven't, or you would know you were wrong.) ... [Jane Q. Public, 2014-09-25]

Does Jane have the memory of a goldfish? Of course Jane has argued with these other physicists. Jane personally asked Prof. Brown about Sky Dragon Slayerism, but wasn't able to "educate" him. Lonny Eachus personally asked Dr. Joel Shore about Sky Dragon Slayerism, but wasn't able to "educate" him. And now Jane/Lonny Eachus fantasizes that these physicists agree with his Sky Dragon Slayerism? Maybe Jane/Lonny Eachus should read those exchanges again, and notice that Prof. Brown and Dr. Shore told Jane/Lonny Eachus the same things I am. That's because Prof. Brown, Dr. Shore and I are simply reiterating elementary mainstream physics.

... Bringing up OTHER arguments like greenhouse gases won't win THAT argument for you. You have already lost it. ... [Jane Q. Public, 2014-09-25]

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]

That's why Jane, Dr. Latour and the rest of the Slayers disagree with the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.

Again, how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?

If Sky Dragon Slayers could answer these questions without resorting to gray Oreos or basketball player gloves, physicists might take the Slayers more seriously.

.. Be a man for a change and admit it. .. [Jane Q. Public, 2014-09-15]

.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it. .. [Jane Q. Public, 2014-09-19]

... Time to act like a man and admit that you were wrong. ... [Jane Q. Public, 2014-09-24]

Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 56

by khayman80 (#47989827) Attached to: Wanxiang May Give 2012's Fisker Karma a Relaunch

... I mean, didn't it send up a red flag when you took your answer and fed it back into standard heat transfer equations and it didn't balance? Oh, that's right... you didn't. But I did. ... [Jane Q. Public, 2014-09-24]

Completely backwards, as usual. I've already shown that my solution keeps electrical heating power constant. 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%.

... because ALL of the incoming cooler radiation is reflected or scattered, and no NET amount is absorbed... [Jane Q. Public, 2014-09-24]

Good grief, Jane. How did the Sky Dragon Slayers brainwash you into endlessly regurgitating this nonsense? Once again, radiation is absorbed by any surface with absorptivity > 0. Jane's either hopelessly confused about the very term "NET" which he keeps capitalizing, or Jane/Lonny Eachus has betrayed humanity by deliberately spreading civilization-paralyzing misinformation.

Again, how do Slayers think we detected the 2.7K cosmic microwave background radiation with warmer detectors? How do Slayers think uncooled IR detectors see cooler objects? Again, why do Slayers think Venus is hotter than Mercury?

... I'm not arguing with you now and I'm not going to again. You're either a fool or a liar, and I do not care which. I have already proved it and I intend to publish that for the world to see. Along with textbook explanations and diagrams showing exactly where and how you went wrong. [Jane Q. Public, 2014-09-24]

Again, Jane/Lonny Eachus actually means that he intends to show where mainstream physics "went wrong" according to the Sky Dragon Slayers. There are many ignorant, stupid physicists that Jane/Lonny Eachus needs to educate: Prof. Brown, Dr. Joel Shore, the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society, etc.

.. Be a man for a change and admit it. .. [Jane Q. Public, 2014-09-15]

.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it. .. [Jane Q. Public, 2014-09-19]

... Time to act like a man and admit that you were wrong. ... [Jane Q. Public, 2014-09-24]

Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 56

by khayman80 (#47987163) Attached to: Wanxiang May Give 2012's Fisker Karma a Relaunch

... I used the proper equation for radiative power, which at steady-state doesn't depend on other bodies. So there is no "difference" term. Just temperature. That's simple physics. You are trying to use a heat transfer equation to calculate power out of a single body at known temperature. That's just plain WRONG. ... [Jane Q. Public, 2014-09-24]

No, Jane tried to use an equation that only calculates radiative "power out" when Jane needs to use an equation for heat transfer that calculates radiative "power out minus power in".

If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".

Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]

Jane's accounting for "power out" without including a term for "power in". That's not A = A, it's A = 0 because one of the terms has been ignored. It's led Jane to the absurd conclusion that electrical heating power doesn't depend on the cooler chamber wall temperature. If that's the case, then how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?

... All the radiation going IN from the cooler body just goes right back OUT again, making the NET radiation crossing your boundary from the cooler body zero. If that were not so, then you'd have net energy being transferred from a cooler body to a hotter one, which is a violation of the second law of thermodynamics. As I've explained to you many times now. You're just plain wrong. ... [Jane Q. Public, 2014-09-24]

This is complete gibberish, Jane. Power radiated in from the chamber walls needs to be accounted for using one term. Power radiated out from the source needs to be accounted using another. Once again, accounting for power flowing in doesn't violate the second law of thermodynamics or somehow imply net energy transfer from cool to hot, no matter how many times Jane wants to assert that nonsense. However, failing to account for power flowing in does violate conservation of energy, because power in = power out through any boundary where nothing inside is changing.

So Jane refuses to retract his absurd claim that view factors vary as the radius ratio, which violates conservation of energy. A cynic might have expected as much, given how Jane flagrantly violates conservation of energy by incoherently ignoring radiative power passing in through a boundary around the heat source.

I made no such claim, you liar. As you well know, the view factor from the surface of the inner sphere to the inner surface of the outer sphere is 1. The calculated view factor from the outer sphere to the inner was 0.9998... [Jane Q. Public, 2014-09-24]

Jane made no such claim? Jane keeps making that absurd claim! Again, the link I've repeatedly given Jane shows that for smaller radius R1, F21 = (R1/R2)^2 = 0.9978.

If the view factor varied as the radius ratio like Jane claims, energy really wouldn't be conserved. The view factor has to vary as the area ratio, which is the square of the radius ratio.

Jane's campaign of educating ignorant, stupid physicists about physics has only just begun. Jane still needs to educate Prof. Brown and Lonny Eachus still needs to educate Dr. Joel Shore.

No, I don't need to educate either one. They can both pick up a textbook on heat transfer and see that I am correct. I'm not arguing with them. [Jane Q. Public, 2014-09-24]

Of course Jane argued with Prof. Brown and wasn't able to "educate" him. Of course Lonny Eachus argued with Dr. Joel Shore and wasn't able to "educate" him. Why not, Jane? Do those physicists not have heat transfer textbooks, or are they just ignorant and stupid?

... I did NOT make broad claims in this recent exchange about "greenhouse gas" or any such thing. So I'm not arguing with those other people. I simply showed YOU to be wrong. ... [Jane Q. Public, 2014-09-24]

But Jane does make broad claims:

.. 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]

Jane/Lonny Eachus insists that an enclosed source doesn't warm, which means CO2 emissions couldn't cause warming. That's why Jane/Lonny Eachus needs to educate the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.

.. Be a man for a change and admit it. .. [Jane Q. Public, 2014-09-15]

.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it. .. [Jane Q. Public, 2014-09-19]

... Time to act like a man and admit that you were wrong. ... [Jane Q. Public, 2014-09-24]

Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

by khayman80 (#47981183) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

.. By the Stefan-Boltzmann radiation law, the chamber walls add no net power in. It just goes right back out through your boundary again. How many times must I explain this to you? .. [Jane Q. Public, 2014-09-23]

If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out". For instance:

There is no net "radiative power in" from cooler to hotter. It's against the second law of thermodynamics, and it violates the S-B radiation law: (e * s) * (Ta^4 - Tb^4). [Jane Q. Public, 2014-09-23]

That's exactly the equation Jane should be using to calculate electrical heating power! It has separate terms for "power in" and "power out" so it can describe power entering and exiting a boundary. If Jane would use that equation, he'd honestly be only saying there is no net "radiative power in" from cooler to hotter.

Instead, Jane insists that electrical heating power = (e * s) * (Ta^4). Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.

There is nothing more to say. You have been proved wrong. You can write books about your nonsense "physics", and it won't make your bullshit theory any more correct. .. The textbooks all say you're wrong. Goodbye. [Jane Q. Public, 2014-09-23]

So Jane refuses to retract his absurd claim that view factors vary as the radius ratio, which violates conservation of energy. A cynic might have expected as much, given how Jane flagrantly violates conservation of energy by incoherently ignoring radiative power passing in through a boundary around the heat source.

.. I honestly -- and I mean that: honestly -- don't believe you could be this stupid and possess a degree in physics. .. [Jane Q. Public, 2014-09-15]

.. I only replied on the off-chance that you really were ignorant and could be educated. .. [Jane Q. Public, 2014-09-20]

Jane's campaign of educating ignorant, stupid physicists about physics has only just begun. Jane still needs to educate Prof. Brown and Lonny Eachus still needs to educate Dr. Joel Shore.

Then, Jane/Lonny Eachus needs to educate the "ignorant" and "stupid" American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.

.. 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]

.. I consult "the experts". When it's a question of physics, for example, I look to references from physicists, not climatologists. After all, physicists are "the experts" when it comes to physics. [Jane Q. Public, 2013-11-15]

All those professional physics societies agree that our CO2 emissions are causing warming, which Slayers like Jane/Lonny Eachus deny. Jane's claimed that physicists are "the experts" when it comes to physics, and that Jane "takes the physicists' word for it." I'm skeptical.

.. Be a man for a change and admit it. .. [Jane Q. Public, 2014-09-15]

.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it. .. [Jane Q. Public, 2014-09-19]

Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.

.. I set out to have a scientific discussion, not to argue about your religion. [Jane Q. Public, 2014-09-21]

As Jane expands his endless campaign to educate ignorant, stupid physicists about physics, Jane might set out to have a scientific discussion without writing things like this:

".. non-person.. disingenuous and intended to mislead .. he is either lying .. dishonest .. intellectually dishonest .. intellectually dishonest .. Khayman80's intellectual dishonesty .. Pathetic. .. you've come out the loser in every case.. you can't win a fucking argument. You don't know how. You don't understand logic. You've proved this many times. Get stuffed, and go away. The ONLY thing you are to me is an annoyance. I have NO respect for you either as a scientist or a person. .. cowardice .. odious person .. you look like a fool .. utterly and disgustingly transparent .. Now get lost. Your totally unjustified arrogance is irritating as hell. .. You are simply proving you don't know what you're talking about. .. Jesus, get a clue. This is just more bullshit. .. spewing bullshit .. You're making yourself look like a fool. .. Hahahahahaha!!! Jesus, you're a fool. .. a free lesson in humility.. you either misunderstand, or you're lying. After 2 years of this shit, I strongly suspect it is the latter. .. Now I KNOW you're just spouting bullshit. .. if we assume you're being honest (which I do not in fact assume) .. I wouldn't mind a bit if the whole world saw your foolishness as clearly as I do. .. stream of BS.. idiot .. Your assumptions are pure shit. .. I'm done babysitting you.." [Jane Q. Public]

"Jesus, you're a dumbshit. .. your adolescent, antisocial behavior .. keep making a fool of yourself. .. you're being such a dumbass .. your analysis of it is a total clusterfuck. .. you're so damned arrogant you think I'm the one being stupid. .. you were too goddamned stupid .." [Jane Q. Public]

".. what a despicable human being you are .. after you are gone, I will quite happily reveal those things and your "legacy" won't be quite what you thought it was. .. get stuffed. I am far beyond tired of your incessant BULLSHIT. If you want to contemplate something before you die, I would suggest starting with meditating on why you have been such an incorrigibly rude, insufferable human being .. You'd at least expect a "physicist" to get that much right. .. Now I have given you your bone, doggie. GO AWAY. .. a clusterfuck pretending to be physics .. simply bad math .. you haven't even managed to ride your tricycle without falling off .. either you're not competent to analyze this, or (probably more likely), you are attempting yet again to misdirect from the real science .. weasel out of it .. you had to obfuscate it and throw n all this other bullshit. Every goddamned time. .. you can go knowing that you abdicated on a chance to prove to the world that you can solve "civilization-paralyzing misinformation". And I will know that you went exactly as you (from what you have shown me, anyway) deserve: unknown and deservedly so. .. you refuse to lose like a man .. you're STILL full of shit, you pretender. .. you're STILL full of shit, you pretender. This is the most ludicrous thing I've heard coming from someone who claims to be a real scientist in years. .. It is A WASTE OF MY TIME to argue with you. You don't learn. I won't do it any more. And I'm going to give a copy of this to my grandchildren. .. bullshit .. weaseling .. all your misdirection .. I am willing to concede that you really are a Kool-Aid drinker, and can't accept that the dogma isn't what you thought it was. That's preferable to believing that you're simply a malicious lying sonofabitch. I am fucking well done here. .. Same shit different day. .. you won't do it because you know you're wrong. .. you're wrong by default .. Why don't you just shut up and do it? Why have you been so mightily struggling, like a fish on a hook, to avoid it? .. BS excuse .. Same shit different day. .. I consider that to be an admission of defeat. .. bullshit excuse .. I guess you do admit defeat. .. your analysis is completely full of shit. .. absolute fantasy .. I'm really not sorry to say this after your past behavior, but showing you're wrong is just plain dirt simple. And not JUST wrong, but so ridiculously wrong that I can (and will, believe me!) use it as entertainment for certain of my friends. .. a pretty major concession that I don't think you deserve. .. Bullshit. .. you're still falling off your tricycle .. simple damned algebra .. You're just clownishly hand-waving again.. START OVER AND DO IT RIGHT .. you're full of bull, and you have been all along. Either you are incapable of doing this properly, or you're just bullshitting everybody for reasons of your own. .. Hahahahaha! .. just more bullshit .. no more bullshit .. of course you still won't, because you're not capable. .. if you don't want me to keep calling you (and showing you to others to be) nothing more than a clown. .. I want to show other people just how much a clown you actually are. .. shut up .. you want to try to mischaracterize everything I say.. you were just messing with me. .. fantasy .. It feels as though I'm explaining to a high-school student who has never seen a physics problem before. .. supposed to have been a physics major. .. Stop being obtuse. .. SIMPLE MULTIPLICATION .. No matter how you try to bullshit your way around this, it is still WRONG. .. provably bullshit .. I'm just plain tired of your bull. .. Jesus, I'm glad you weren't one of my physics profs. .. That's your goddamned problem, and you don't get to complain about it. I'm really looking forward to showing this latest exchange to my friends. .. There is no way to weasel out of this, man. You're trying to output more power than you're putting in. This isn't even 11th-grade physics. Let's try it at something more like your level: You have 200 beans equally distributed among 10 squares. If you now take those beans, and divide them equally among 25 squares of the same size, how many beans do you now have per square? Show your work. .. THERE'S NOTHING "CUTE" ABOUT IT! IT'S AN ACCURATE ASSESSMENT OF YOUR ERROR! This is not "approximation", it's fucking logical error! JESUS CHRIST, man, you can't talk your way around this. .. You can violate thermodynamics all you want, and it doesn't prove a damned thing. .. STOP THE BULLSHIT. .. If you continue to just bullshit your way around, as I have stated I will declare you in default and damned few reasonable people would disagree. .. NO. See my comment above. One more bullshit comment like this, and as I said, I will just call you a clown and few reasonable people will disagree. .. you are deliberately trying to make things difficult. .. It is dirt simple to show you are wrong. .. you're throwing a fit .. Are you drunk? .. Get the hell on with it.. I am very, very close to calling you full of shit and posting this where everyone can see it. .. YOU are the one who is trolling.. You simply wanted to waste more of my time. .. You're finally proving that you were full of bull all along. .. You're just plain wrong. .. you are quite clearly throwing a fit.. How could I possibly be "wrong"? .. I called bullshit.. prepare to be publicly declared a charlatan. .. plenty of reason to call you both wrong and a liar. .. I am going to declare you a fraud and a failure. .. I'm still going to declare you a failure.. he's just a trolling, malicious, lying son of a bitch.. he has berated me, publicly derided and taunted me, and (in my strong opinion) libeled me.. I can show clearly, to someone with high school level math skills, that he was utterly, abjectly, and rather pathetically wrong, and the "Slayers", as he calls them, were right all along. .. "global warming alarmist" bullshit is just that: bullshit. .. mere incompetence and arrogant belief in your own abilities and contempt for others? Or was it because you were protecting your political ideology, or global warming religion, or maybe JPL grant money? I really don't know, and I really don't care, but now I can show the world very clearly, using your own words, that you were wrong the whole time. I would thank you for that but you don't deserve thanks. .. I am not going to judge here whether he was honestly mistaken or he was just a malicious bullshitter, but in all honesty it's hard to imagine someone who calls himself a physicist unintentionally getting it so badly wrong so many ways. Unless his "global warming" religion would simply not allow him mentally to accept the right answer. .. I could go on, but this was my BRIEF analysis of khayman80's folly. As I sincerely promised him, I will be writing up a more complete discussion of his errors later on "the interwebz". Spencer and khayman80 were wrong. Latour was right, and I was correct to stick to my guns and say so, despite all of khayman80's public bullying and insults and braying like an ass. .. another aspect of khayman80's folly. .. khayman80, otherwise known as Bryan Killett, you're either a liar of a fool. As I said before, I don't know which, but I've proved that it MUST be one of the two. .. khayman80's nasty remarks .. schooling a physicist on why his physics is awful.. You can't even fucking add 2 + 2. .. you complete bozo. .. you're a complete loon. .. I'm not wrong, in any basic way. .. Face it. You've been spouting the wrong answer for 2 years, and using it to justify calling OTHER PEOPLE names, and bullying them online, and other nasty antisocial behavior. But even if I made a small mistake somewhere (I did NOT make a large one), you're still busted. .." [Jane Q. Public]

".. you were "hanging yourself", as the saying goes. Hoist by your own petard. .. You are busted. .. I'll be here watching and laughing all the way. .. It doesn't matter how you try to squirm and twist this. You have been owned. End of story. .. I repeat that you can twist and squirm all you want, but unless you can come up with a "khayman80 law" to replace the Stefan-Boltzmann law, this IS the answer, it is known, and it is unequivocal. .. Introduce all the complications, and prevarications and half-assed reasoning you want. I have already shown you the correct answer according to established physics. Give it up lest you make yourself look more of a fool than you already are. Because as I promised you, all of this is being recorded and will be made public, with your name displayed prominently. I promised that I would do that regardless of how it turned out. You have no reason to complain just because you lost. Further, I'm going to INVITE people who teach heat transfer to examine my write-up, and evaluate it. I already know what they will say about your half-assed thermodynamic reasoning. To be honest, I still don't see why YOU don't see, where I showed that you were clearly wrong. But again, I suspect that your CO2-based greenhouse gas religion will not let you accept the clearly established facts. I have said all I need to say here. Nothing you say will change it, and no, I do not agree with your fallacious "reasoning". I'll stick with the engineering textbooks, thanks very much. .. Have I reminded you lately that your grasp of logic seems a bit off? .. It's just bullshit. You're squirming like a fish on a hook. You just don't seem to realize you have already been flayed, filleted, and fried in batter. You're owned, man. .. PROOF that you're bullshitting everybody.. You keep making the same bullshit assertions, after I have proved them false. Why do you do this? You're just going to look that much more foolish later. .. YOU are disputing the Stefan-Boltzmann law. But it is a known physical law, and this is a textbook demonstration of it. You lose. .. Your calculations contradict themselves, and your methodology contradicts itself. .. no matter how you cut it, your answer is wrong, by your own rules. .. I find it highly amusing that you derive your own calculations from the Stefan-Boltzmann law, then deny that it is valid. Every time you try to squirm out of this you just contradict yourself again. I am further amused that you find it "adorable" that you've been proven wrong. Be a man for a change and admit it. .. No more bullshit. .. I'm just trying to find out whether you're actually crazy or just bullshitting. .. Are you REALLY the moron you make yourself out to be? .. You are giving physicists a bad name, and I repeat that I am going to show this to all the world to see. .. This is so utterly obvious that I honestly don't believe you don't get it. .. I have finally concluded that you are just a very good troll. I honestly -- and I mean that: honestly -- don't believe you could be this stupid and possess a degree in physics. .. You're just wrong about how this works. And not just a little bit wrong, but completely out there in lala-land wrong. And you have made it perfectly obvious that I am wasting my time talking to you. You are either crazy, or stupid, or a very talented troll. Based on my experience, I vote for that last one, but I think that necessarily implies a little bit of the first, too. So we're done. I'm going to write this up as it stands here. I don't need anything else, and you've made it very clear that anything else would be further waste of my time. You refuse to change your tune, so fine. I'll just write it up that way. Don't worry: I am going to include your exact words. .. You DO know what a minus sign is, yes? .. You made assumptions that are, to be blunt, bullshit nonsense. .. Do you think we're all idiots? .. knock off the bullshit, because I see right through it, and so will the others I show this to. .. Yet again, you have contradicted yourself. You're a great bullshitter but I've caught you out and you've already been proved wrong. All this trying to twist out from under the obvious any way you can only confirms that you were bullshitting all along. Be a man and admit the truth, because people ARE going to see this. Why do you want to look more foolish than you do already? .. Complete bullshit again. .. It is a simple equation that is well-known to physicists. You claim to be a physicist, so why don't you know it? .. You've been owned, man. BE enough of a man to admit it. Because everybody's going to know it anyway. .. This is just another straw-man argument. Which you are very good at, by the way. Not good enough to sucker me in, though. .. your assertion is only "obvious" if you're not a heat transfer engineer or a physicist, you pretender. Heat transfer is not a science of the obvious. Intuition (and, as pointed out before, "thermodynamic thinking") can easily lead you astray. .. Knock off the BS. Time to admit you were wrong. .. I've already proved you wrong, mathematically, logically, and thermodynamically. The fact that your "global warming" religion will not let you accept the reality of the Stefan-Boltzmann radiation law is not my problem. But you have sure as hell tried hard to make it everyone else's problem. .. You're either incompetent or a liar. As I said before: I don't know for sure which, but I strongly suspect the latter. It's a done deal. You have been proved wrong. You have been owned. Your ranting means nothing. I only replied on the off-chance that you really were ignorant and could be educated. But it seems that you are determined to promote your ignorance (or more likely: ignorant act and propaganda) to everyone else. So be it. No more replies. You haven't earned any; you don't deserve any. .. NOW what kind of bullshit are you trying to pull? Do you understand what NET means, or do you not? I assure you that a lot of people do. You claimed before that you did. Why are you doing this? Are you really trying to make yourself look more ridiculous than before? .. I'm going to ask you again: WHY do you continue to spout this violation-of-physics bullshit? What do you think you're accomplishing other than wasting my time? I have concluded that is all you are trying to do. .. If you are sincere (you certainly haven't been acting like you are), then you must be postulating some kind of "tractor beam" effect that allows the chamber wall to "suck" power out of the heat source from a distance. I assure you that at least at out current level of technology, we have not managed to build such a sucking device. The heat source radiates out what it radiates out, and nothing around it is "sucking" any power from it. Although you seem to be doing your very best at "sucking" my time away over stupid bullshit. .. NONSENSE. .. What's ridiculous is your constant repetition of this bullshit idea. .. If you're being honest, then it's really too bad that you still don't understand the clear implications of the Stefan-Boltzmann radiation law. But at the same time, it makes me wonder how you got your degree. I'm done. If all you're going to do is keep repeating these incorrect assertions, after why they are incorrect has been clearly explained to you many times, this is indeed just a waste of my time. I set out to have a scientific discussion, not to argue about your religion. .. NO!!! This is just plain bullshit. .. You are VERY good at trying to make it appear I have been saying things I actually haven't. But it isn't going to fly. It's just bullshit. .. Why do you keep disputing textbook physics laws? Stop lying. Because that's all you're doing now. .. What I object to is your insane insistence that the electrical power to the heat source requires a term for the chamber walls. This is sheer nonsense. .. YOU are the one who is getting them confused, not me. .. Look it the hell up. .. Apparently you don't understand the concept of NET, even though you have derided me for supposedly "ignoring" it. .. This is textbook stuff, and you just aren't getting it straight. Are you sure you're a physicist? .. Why don't you look it up in a textbook and discover that for yourself? .. I repeat: if you truly don't understand this, due to your "greenhouse gas religion" or something, that's just too bad. I'm using textbook physics for situations like this. You are not. You are espousing magical net power transfer from cold to hot, rather than actual physics. .. That's complete bullshit. Doesn't happen. Knock off the fantasy physics and pick up a textbook. .. There is nothing more to say. You have been proved wrong. You can write books about your nonsense "physics", and it won't make your bullshit theory any more correct. I have 3 heat transfer textbooks here, and they all say you're wrong. I'll stick with the well-known and established physics, thanks very much, and dismiss the nonsense from the cheap seats. Funny, but for years you talked about "consensus" and "established science", but whenever the established physics disagrees with you, you will write pages and pages about why they're wrong and you're right. There's a word for that. The word is "hypocrisy". There are other words for what you do, too, but I'll let other readers decide on those. Well, it didn't work and it won't work. The textbooks all say you're wrong. Goodbye. .." [Jane Q. Public]

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

by khayman80 (#47976775) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

But Jane's equation is different:

electrical power per square meter = (s)*(e)*Ta^4

YES!!! This is a different equation! It's not an equation for heat transfer! It's the Stefan-Botlzmann RELATION between radiative power out and temperature for gray bodies. It is used for calculating RADIATIVE POWER OUT versus TEMPERATURE and vice versa. It is not for heat transfer and I'm not using it for heat transfer. YOU are the one who is getting them confused, not me. This other equation shows that radiative power is dependent ONLY on emissivity and temperature. It does not depend on other bodies. For the third time (today): it's a temperature vs. power equation, not a heat transfer equation. Further, "electrical" is your own addition. The equation is for power. It doesn't specify "electrical". [Jane Q. Public, 2014-09-22]

My equation for electrical power is different than the equation for radiative power out, which is why it's bizarre that Jane keeps using the equation for radiative power out to determine electrical power. That's what I've been trying to tell Jane: we don't disagree about the equation for radiative power out. The equation for radiative power out is simply a part of the equation for conservation of energy: power in = power out through a boundary where nothing inside is changing. That's why we need to use a heat transfer equation to determine electrical heating power, not just an equation for radiative power out.

... it is not necessary to account for cooler bodies in the temperature versus power out equation. ... The second equation you cited above is the STANDARD equation for calculating radiative power out of a gray body. I showed you where it was in Wikipedia. It also just happens to be in my heat transfer textbooks. The answer is 82.12 W/m^2. It is the textbook answer. It isn't going to change. Why don't you look it up in a textbook and discover that for yourself? ... Radiative power out of the warmer body is dependent ONLY on emissivity and thermodynamic temperature. Anything else violates the second law of thermodynamics. It isn't controlled or mitigated by nearby cooler bodies. ... [Jane Q. Public, 2014-09-22]

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

Since power in = power out:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

This doesn't violate the equation for radiative power out. It simply uses that equation to account for the power flowing out of the boundary, and uses that same equation for radiative power to describe radiative power flowing into the boundary.

... I will repeat: I did not and do not claim that no radiation is absorbed. Just no net radiative power. Any that does get absorbed is just re-transmitted... [Jane Q. Public, 2014-09-22]

Jane's been calculating the required electrical heating power, which requires using a net heat transfer equation to describe power in = power out through a boundary around the source. Because Jane's equation doesn't even include a term for "radiative power in", Jane's equation does claim that no radiation is absorbed at all.

If Jane would reconsider conservation of energy and include a term for "radiative power in", then Jane could honestly say he was only claiming that no net radiative power is absorbed by the source. Until then, Jane's equation claims that no radiation is absorbed by the source at all.

But Jane's equation is nonsense, because absorption is controlled by absorptivity. So we could only ignore the power radiated from the chamber walls if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source.

Look at your S-B equation above. What does it say? No net radiative power is absorbed by warmer bodies from cooler bodies. You said so yourself. But NOW, you're claiming that it is. You contradict yourself. ... [Jane Q. Public, 2014-09-22]

No, I said the source has to absorb some radiation as long as it has absorptivity > 0. I never said the source would absorb more radiation than it emitted. In fact I said the opposite "happens automatically".

... The chamber walls neither transfer any of their net radiative power to the heat source, nor do they cause the net radiative power of the heat source to be any less. They have NO EFFECT. Net energy flows only FROM the heat source to the walls, and the temperature of the walls effects heat transfer only, not radiative power of the heat source. ... [Jane Q. Public, 2014-09-22]

If the temperature of the walls affects heat transfer, they also affect how much electrical heating power is required to keep the source at 150F. Note that I said "electrical heating power" and not "radiative power out" because these are two very different things. Calculating "radiative power out" just requires writing down the Stefan-Boltzmann law. Calculating "electrical heating power" requires drawing a boundary around the heat source at steady-state, and setting power in = power out.

... Do you think we're all idiots? Power in = power out. Your Newmann and Dirichlet boundary conditions are just more straw men. We don't need them to find the answer to this. Plain old algebra works just fine, because everything is at steady-state. So knock off the bullshit, because I see right through it, and so will the others I show this to. ... [Jane Q. Public, 2014-09-19]

Of course boundary conditions are needed to find the answer, because they determine what "plain old algebra" is used, even at steady-state. We're all applying Dirichlet boundary conditions to the chamber walls, but Jane mistakenly applied them to the source as well, instead of the correct Neumann boundary conditions. Jane also continues to wrongly insist that Jane held electrical heating power constant as well as holding source temperature constant. So apparently in Janeland there's no difference between Neumann and Dirichlet boundary conditions. If that's true, why do physicists and engineers use different names for Neumann and Dirichlet boundary conditions?

... The areas in his equation were unnecessary ... Therefore the areas were irrelevant and about all he accomplished with his large equation was to further confuse the issue. ... [Jane Q. Public, 2014-09-10]

After I originally solved a simple equation without areas, Jane objected that neglecting areas was a "fucking logical error!". That's why I had to solve the more accurate large equation with areas, even though I warned Jane that it wouldn't substantially change the answer.

The large equation with areas was also necessary because:

... 788.01 W != 721.44 W (!!!) Power is not conserved. ... the inner surface of the cavity has twice as much area, so the total power radiated is twice as much. Power is not conserved. ... [Jane Q. Public, 2014-09-07]

Jane confused himself about areas so badly that he claimed "power is not conserved". So I explained that Wikipedia’s equation correctly takes into account areas and view factor.

Wikipedia's equation conserves energy because the view factor from chamber walls to enclosed source equals the area ratio. If the view factor didn't vary exactly like that, energy really wouldn't be conserved.

But the chamber wall completely encloses the source, so its view factor is 1.

No. If the surfaces are numbered 1, 2, 3, 4 as I did in my solution, F12 = F34 = 1. In the other direction (as you already know, and so do I) it is R1/R2, where R1 is the smaller diameter. F21 = F43 = 0.9989. [Jane Q. Public, 2014-09-19]

As I said, the view factor from enclosed source to chamber walls is 1. If Jane wants to calculate the view factor in the other direction, the link I've repeatedly given Jane shows that for smaller radius R1, F21 = (R1/R2)^2 = 0.9978.

If the view factor varied as the radius ratio like Jane claims, energy really wouldn't be conserved. The view factor has to vary as the area ratio, which is the square of the radius ratio.

... I'd say that Wikipedia's equation is more correct because it includes area and view factor, which MIT's equation does not. ... [Jane Q. Public, 2014-09-19]

If only I'd mentioned that repeatedly.

... The equation you are trying to use there is a partial equation for heat transfer, not radiant power output. They're not the same things. The proper equation for power out given radiant temperature is right there in the above paragraph. It can be found in any heat transfer textbook and many physics books. Didn't you notice that MIT's equation is essentially the SAME equation as Wikipedia's heat transfer equation, except for areas? I sure did. Why didn't you notice that? ... [Jane Q. Public, 2014-09-19]

Of course I noticed that they're both net heat transfer equations, which is why I used them both in the same way to get nearly identical answers. I'm using MIT's and Wikipedia's equations because they yield radiative "power out minus power in". These net heat transfer equations are the proper equations for applying conservation of energy to a boundary around the source.

In contrast, Jane's clinging to an equation for "power out" and incoherently trying to justify ignoring "power in" through that boundary.

... I will make use of only ONE of your assumptions: that the enclosing plate (hollow sphere) is, due to thermal conductivity, approximately the same temperature on both sides. It's only 1mm thick after all, and the thermal conductivity of aluminum was a stipulation of yours so it will be the same to a couple of decimal places, give or take. So the answer won't be exact, but it will be reasonably accurate. Certainly close enough to demonstrate the concept. ... [Jane Q. Public, 2014-09-10]

When I approximated the enclosing shell as a thermal superconductor, Jane insisted that there's no way to demonstrate anything with it without leading to a contradiction, and that it was nothing but misdirection and a fantasy ultimate straw-man argument.

When Jane approximates the enclosing shell as a thermal superconductor, it's reasonably accurate and certainly close enough to demonstrate the concept.

A cynic might suspect double standards.

... I have already explained how your "boundary" assumed that all the power was output from the outside of the enclosing sphere. ... you neglected to account for the fact that the hollow sphere has TWO surfaces it is radiating from. You left out half the m^2 in A, so your figure for W/m^2 was off by very nearly 100%. Q.E.D. [Jane Q. Public, 2014-09-11]

Once again, Jane's completely wrong. When I held the source temperature constant, I reproduced Jane's result. So we're actually disagreeing about what to hold constant. If Jane's hilarious "Q.E.D." were correct, I wouldn't have been able to reproduce Jane's result simply by changing what variable I held constant.

... Add them together for the total heat transfer: 27.7832 + 27.7813 = 55.5645 total heat transfer. This checks against our initial calculation which was 55.5913. The difference is only 0.0268, or about 0.1%. Close enough for what we're doing. ... [Jane Q. Public, 2014-09-10]

Ironically, Jane's off by ~100%. Again, Jane's total heat transfer dropped to 27.8 W/m^2 after the shell was added, so Jane's meaningless 55.6 W/m^2 value is ~100% higher than the actual value.

... Factor out (e*s) from both sides. (Despite khayman80's assertion that we cannot do this, yes we can. It is the same scalar and the same constant on both sides.) ... [Jane Q. Public, 2014-09-10]

Once again, I never asserted that. In fact, I repeatedly showed Jane an equation derived by factoring out the sigmas and epsilons from both sides. Only Jane/Lonny Eachus could repeatedly quote that equation and even agree with it, then accuse me of asserting the opposite.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

by khayman80 (#47971031) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't repeatedly object to including a term for radiation from the chamber walls in his calculation of required electrical power. Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.

NO!!! This is just plain bullshit. I do NOT object to a term for electrical power. ... I don't object to a term for "electrical power" and never have. ... [Jane Q. Public, 2014-09-22]

I never said Jane objected to a term for "electrical power". I said Jane repeatedly objects to including a term for radiation from the chamber walls in his calculation of required electrical power. And Jane continues to do this:

... I simply asserted a physical truth: in our isolated system, the electrical power to the heat source, called for by Spencer, has zero dependency on the chamber walls. It is this nonsense dependency on the chamber walls that I have disputed, nothing else. That is a violation of the Stefan-Boltzmann law. ... My only objection is your insistence that the power input to the heat source is somehow related to radiation from the chamber walls. If these are treated as gray bodies: just no. That's a violation of Stefan-Boltzmann. [Jane Q. Public, 2014-09-22]

Ranting about imaginary violations of the Stefan-Boltzmann law won't help Jane understand physics. It might help Jane to draw a boundary around the heat source and think carefully about exactly why Jane keeps ignoring the heat radiated in from the chamber wells. Accounting for that radiation doesn't "violate the Stefan-Boltzmann law" but ignoring it violates conservation of energy.

... The power output is not dependent on the chamber walls, therefore the power input is not dependent on the chamber walls. ... [Jane Q. Public, 2014-09-21]

Why does Jane think the second part follows from the first? It doesn't. For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls. Maybe Jane could explain why he wrote "therefore" when his reasoning fails to describe even a simple black body problem? (Keep in mind that the gray body equation has to reduce to the black body equation when emissivities = 1.)

Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.

NO!!! Repeat, for about the 100th time now: no NET radiative power input from cooler objects. That is ALL I have claimed, and it's a direct result of the Stefan-Botlzmann radiation law. Why do you keep disputing textbook physics laws? Stop lying. Because that's all you're doing now. [Jane Q. Public, 2014-09-22]

Jane/Lonny Eachus can capitalize "NET" all he wants, but it doesn't change the fact that Jane's equation assumes warmer objects absorb no radiation from colder objects. Here's an equation which only says there's no NET radiative power input from cooler objects:

electrical power per square meter = (s)*(e)*(Ta^4 - Tb^4)

The above equation satisfies conservation of energy and says there's no NET radiative power input from cooler objects.

But Jane's equation is different:

electrical power per square meter = (s)*(e)*Ta^4

Jane's equation doesn't just say there's no NET radiative power input from cooler objects. That happens automatically. Jane's equation violates conservation of energy by completely ignoring the term describing radiative "power in" from the chamber walls. So Jane's equation says warmer objects absorb no radiation from colder objects.

But Jane's equation is nonsense, because absorption is controlled by absorptivity. So we could only ignore the power radiated from the chamber walls if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source.

Jane/Lonny Eachus can capitalize "NET" all he wants, but it doesn't change this fact. Unless Jane/Lonny Eachus would like to correct his equation for required electrical heating power and derive an answer other than 82 W/m^2?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

by khayman80 (#47960791) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... What's ridiculous is your constant repetition of this bullshit idea. Yes, the cooler walls radiate inward but they have no effect whatsoever on the heat source. ALL of that radiation is reflected or scattered by the heat source. (It is not transmitted because we're dealing with diffuse gray bodies of significant mass.) ... [Jane Q. Public, 2014-09-21]

It's truly surreal to watch Jane repeatedly double-down on nonsense which Jane claims is too ridiculous even for Sky Dragon Slayers (as if that were possible!).

... You took a badly-worded sentence or two and jumped on them as though Latour made a mistake. But his only mistake was wording a couple of sentences badly. He does in fact NOT suggest that warmer objects absorb no radiation, and he has written as much many times. ... You have refuted NOTHING but a couple of unfortunately-worded sentences, which Latour himself publicly corrected shortly after that post appeared. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't repeatedly object to including a term for radiation from the chamber walls in his calculation of required electrical power. Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.

... shortly after Latour published that blog post, it became clear that the language he used implied that no radiation at all was absorbed by the warmer body. So a reader could not reasonably be blamed for inferring that. But Latour quickly apologized for the unfortunate wording and corrected himself to make it very clear he was referring to net, not absolute, heat transfer. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body. Otherwise Jane's calculation of the required electrical power would include a term for radiation from the chamber walls. Since Jane adamantly insists that this term can't be included, Jane's calculation assumes that no radiation at all is absorbed by the source. None. Zero.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

by khayman80 (#47957813) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... Repeat: this ASSUMPTION of yours that the chamber walls must be accounted for in the power requirement of the heat source is a direct violation of the Stefan-Boltzmann law. There are no 2 ways around it. Established physics (the Stefan-Boltzmann law) says that the radiative power out (and therefore power in) of a gray body is dependent ONLY on emissivity and thermodynamic temperature. It is completely unrelated to any nearby cooler bodies. ... [Jane Q. Public, 2014-09-21]

Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary. But "power in" through a boundary around the heat source looks like this:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Jane refuses to account for the chamber wall radiative "power in" which would only be true if the source didn't absorb any of that radiation. Zero.

If you are sincere (you certainly haven't been acting like you are), then you must be postulating some kind of "tractor beam" effect that allows the chamber wall to "suck" power out of the heat source from a distance. I assure you that at least at out current level of technology, we have not managed to build such a sucking device. The heat source radiates out what it radiates out, and nothing around it is "sucking" any power from it. Although you seem to be doing your very best at "sucking" my time away over stupid bullshit. [Jane Q. Public, 2014-09-21]

That's ridiculous, Jane. I'm just noting that the chamber walls are hotter than 0K, so they emit radiation into a boundary around the heat source. Therefore Jane's wrong to ignore that radiation when applying the principle of conservation of energy:

... Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. ... [Jane Q. Public, 2014-09-15]

It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source.

So the only "heat source" where we could validly ignore the radiation from the chamber walls would be a perfectly reflective "bobble" from Vernor Vinge's Marooned in Realtime. I assure you that at our current level of technology, we haven't managed to build such a device. And even if we could, it wouldn't be a heat source.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

by khayman80 (#47957481) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... No NET radiative energy. I did not claim "none at all", and I have repeatedly pointed this out to you. Just no NET transfer from cooler to warmer. ... [Jane Q. Public, 2014-09-20]

Jane's equation claims "none at all":

electrical power per square meter = (s)*(e)*Ta^4

Since Jane's equation for required electrical power doesn't even include a term for radiation from the chamber walls, Jane's equation wrongly says that no radiation at all is absorbed by the source. None. Zero.

It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source. Slayer "physics" are incoherent nonsense.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

by khayman80 (#47955263) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... Before that it can't, because Ta^4 - Tb^4 is a positive number so no net radiant energy is absorbed by (a) from (b). That means all the way up to the exact point thermal equilibrium is achieved, all radiant power is a result of electrical power, therefore the power input and power output are constant. It is not a "gradual" process. ... [Jane Q. Public, 2014-09-20]

So Jane claims:

electrical power per square meter = (s)*(e)*Ta^4

The actual answer is:

electrical power per square meter = (s)*(e)*(Ta^4 - Tb^4)

Since Jane refuses to include a term accounting for radiation from the chamber walls, Jane's equation is saying that no radiation at all is absorbed by the warmer source. Why?

... Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. ... [Jane Q. Public, 2014-09-15]

Of course it is! Again, this is just Sky Dragon Slayer nonsense. Absorption doesn't work like Slayers imagine. It's controlled by the surface's absorptivity, which doesn't change if the source is slightly warmer or cooler than its surroundings. All that's required for the source to absorb radiation (from warmer or colder objects) is having absorptivity > 0. Since the source has absorptivity = 0.11, some radiative power from the chamber walls is absorbed by the heat source.

Jane's been regurgitating Slayer nonsense for years:

... Warmer objects cannot, and do not absorb lower-energy radiation from cooler objects. ... [Jane Q. Public, 2012-11-20]

Then how do uncooled IR detectors see cooler objects? How did we detect the 2.7K cosmic microwave background radiation with warmer detectors?

... explain how radiation that is of a LOWER "black-body temperature" will be absorbed by a body of a HIGHER black-body temperature. ... [Jane Q. Public, 2013-05-30]

... 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]

No, that's a Slayer fantasy. On the atomic scale, absorption of radiation doesn't depend on temperature because individual atoms don't have temperatures. Only very large groups of atoms have temperatures. Individual photons also don't have temperatures. Very large groups of photons from a 10C warm object have slightly different average wavelength curves than a -10C cold object, but they're very similar. This means that even if temperature somehow applied at the atomic scale of absorbing individual photons, an atom couldn't tell if a photon came from the 10C warm object or the -10C cold object.

... You took a badly-worded sentence or two and jumped on them as though Latour made a mistake. But his only mistake was wording a couple of sentences badly. He does in fact NOT suggest that warmer objects absorb no radiation, and he has written as much many times. ... You have refuted NOTHING but a couple of unfortunately-worded sentences, which Latour himself publicly corrected shortly after that post appeared. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't repeatedly object to including a term for radiation from the chamber walls in his calculation of required electrical power. Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.

... shortly after Latour published that blog post, it became clear that the language he used implied that no radiation at all was absorbed by the warmer body. So a reader could not reasonably be blamed for inferring that. But Latour quickly apologized for the unfortunate wording and corrected himself to make it very clear he was referring to net, not absolute, heat transfer. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body. Otherwise Jane's calculation of the required electrical power would include a term for radiation from the chamber walls. Since Jane adamantly insists that this term can't be included, Jane's calculation assumes that no radiation at all is absorbed by the source. None. Zero.

It's truly surreal to watch Jane repeatedly double-down on nonsense which Jane claims is too ridiculous even for Sky Dragon Slayers (as if that were possible!).

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

by khayman80 (#47947973) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... If body (b) is brought up to the same temperature as (a), Ta^4 - Tb^4 = 0, and no net heat transfer takes place. Although radiant power output of (a) at that temperature doesn't change, as a corollary of that same law. ... [Jane Q. Public, 2014-09-19]

If Ta = Tb, no electrical heating power is required. But radiant power output of (a) doesn't change. So radiant power output can't be equal to electrical heating power. Using conservation of energy, can you write down an equation which yields the required electrical heating power given Ta and Tb?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

by khayman80 (#47943463) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Since the emissivity for every object in our system is the same, power output is proportional to the T^4. Period. End of story. Draw your boundary around the heat source. Power in = power out (your own principle). Therefore the power in is 41886.54 Watts, which is the power initially being radiated out. SPENCER stipulated that this power is held constant. It wasn't my idea. It's a condition of the experiment. [Jane Q. Public, 2014-09-19]

No. Once again, in this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."

Jane's even stumbled across this point:

... Of course it wouldn't need a separate heat source if its environment were maintained at 150 degrees. ... [Jane Q. Public, 2014-09-15]

Of course! That's why the variable Jane's holding constant isn't the electrical power supplied to the separate heat source. If Jane can realize that there's no need for a separate heat source if its environment were maintained at 150 degrees, why can't Jane see that his equation for required electrical power doesn't reflect this obvious fact?

The Force is what holds everything together. It has its dark side, and it has its light side. It's sort of like cosmic duct tape.

Working...