But again: if you need to "draw a boundary", it needs to be drawn around the passive plate itself. We have already firmly established that your "boundary" around the heat source and the "enclosing shell" is even thermodynamically incorrect. It leads to an erroneous result of very close to (within a few thousandths) DOUBLE the radiative power from that surface that actually exists.
I have explained this to you 3 times now. If you can't get it through your head, that's your problem and nobody else's.
As I said, I am going to write this up more thoroughly, elsewhere. But I have presented enough here for anybody who is really interested to figure it out without too much difficulty. Present company apparently excepted.
We have already shown that your particular application of "drawing boundaries" here was a MISAPPLICATION of the principle you are trying to use.
The "enclosing shell" (if by that you mean the passive plate that was inserted) is acted upon only by radiation. You should have drawn your shell around THAT, and that alone. And you should at least have tried drawing your boundary around your own goddamned heat source, both for initial conditions and your final result, to check your work. But you didn't. What you got was a universe-busting violation of conservation of energy.
But of course you are still trying to defend something which YOU claimed earlier is not valid to do. There is a word for that.
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.
One more correction and again it's minor, but again I plea that I have been very busy and had to dash this off in a hurry:
In my list of "clues" toward the bottom, Clue #1:
The bit about "Barring any issues like 'view factor'" was an irrelevant comment because that relates to interactions between bodies and my whole point was that the S-B equation for radiative power from a body does not include other bodies. The part about "view factor" can be left out of that passage without affecting its correctness.
Further, if you have a problem with my equations (including my minor corrections) you are welcome to do your own and prove me wrong.
But you aren't going to, because I'm not wrong, in any basic way. I might have gotten a hundredth or a thousandth off here or there, but unlike you I did double-check your work. All while (according to you) I was spouting something off on Facebook or something.
I spent about an hour and a half on this, give or take, between my regular work, and working on the car. That includes identifying your errors, and determining what the right way to do it was.
As I said, in the unlikely event that you wrote down equations, they'd violate conservation of energy. Thermodynamic thinking like this leads one back to reality, not astray. Draw a boundary inside the inner surface of the enclosing shell at your steady-state values. Since nothing inside that boundary is changing, power in = power out. But that's completely impossible. Your solution violates conservation of energy, as predicted.
Violating conservation of energy is something you do quite well, you complete bozo.
You ASSUMED that power at the surface of the heat source = power at the outer surface of the hollow sphere. You did NOT take into account that the sphere radiates from BOTH surfaces. You can even fucking add 2 + 2.
THEN, you neglected to do the most basic check of your work, such as: applying the S-B relation to your new, hotter heat source. You just hoist yourself by your own petard, dude, because our input is a FIXED amount of power, but you upped your output by almost 100 degrees F. Draw your line JUST around your heat source. Where is that extra power coming from? S-B law says that power is ONLY related to temperature and emissivity, and the emissivity hasn't changed.
So where is that extra power coming from? Thin goddamned air?
I made a couple of very minor errors, which I have corrected. My basic proof still stands, and you are still wrong.
In fact, you're a complete loon.
I either included or referenced all the equations that were necessary to solve the problem. It's not my fault if you still can't do it.
Holy crap, I just discovered another error. Pardon me, folks, but despite khayman80's nasty remarks elsewhere, I really have been busy. Between work, doing work on the car that had to be done today, and schooling a physicist on why his physics is awful, I've been very, very busy.
This is what I wrote:
Think of it this non-tecnical way: radiant heat transfer is a function of difference in temperatures. If you have two differences, A = X - Y and B = Y - Z, then A - Z = (X - Y) plus (Y - Z).
What I meant was this:
Think of it this non-tecnical way: radiant heat transfer is a function of difference in temperatures. If you have two differences, A = X - Y and B = Y - Z, then (A + B) = (X - Z).
Another minor correction: I stated that "the difference in area" between 2 and 3 was only about 0.001, or some such. That was not quite correct. What I meant was in the process of doing my calculations, I discovered that the difference made by the difference in area was only about 0.001. So I determined that it could safely be ignored, given the fairly loose precision of our other calculations.
Jane's obligations include continuing to spread misinformation about ocean acidification even after I've repeatedly debunked him.
So I predict that Jane's answer won't include any equations that could be used to calculate the enclosed source temperature. Instead, he'll probably grace us with another lengthy, incoherent rant about "problems" in my analysis which are (as usual) too vague to be expressed in equations. In the extremely unlikely event that Jane musters up the courage and competence to actually write down an equation that could be used to calculate the enclosed source temperature, it will almost certainly violate conservation of energy.
I almost forgot to mention another aspect of khayman80's folly.
He could have discovered he was wrong almost 2 years ago, almost immediately, by checking his work. He didn't. And this was the result:
He assumed that the total power output of the heat source was available on the OUTSIDE of the enclosing hollow sphere. We now know this was incorrect. He then used that to calculate a hotter temperature for the heat source itself. That may have made sense, giving his (incorrect) assumption about how "power in = power out" worked. BUT... he got a value for temperature almost 100 degrees F hotter than before.
Remember the S-B law? That means the POWER output of that heat source was much greater. It doesn't matter where it was coming from... according to his own reasoning (which would be correct for the heat source itself), it can only be as hot as the power that is input lets it. Power out p = (epsilon)(sigma)T^4. Period. No two ways about it. So if there is a greater T, there is higher power output, and it has to come from somewhere. His heat source is going to draw more power from whatever is powering it.
But wait! There's more!
All he had to do was continue this magical thinking. Now transfer this power out back to the outside of the enclosing sphere, the way he did it the first time. Then he can back-calculate, the same way he did the first time, and get an even HOTTER figure for the heat source. Then he can transfer this power back out to the outside of the enclosing sphere again, and continue ad infinitum!
It's hilarious, because khayman80 accused me of spreading "civilization-paralyzing misinfomation". But if the universe really worked according to the information HE has been spreading, it would result in a thermal runaway and destroy itself in an extremely brief period of time. For the price of a few kilowatts.
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.
Correction: "reflecting on CLUE #2, and the fact that the "view factor" from 1 to 2 (F12) = 1, PLUS our assertion that area of 2 is close enough to the area of 3 to ignore, we get the rate of heat transfer from surface 1 to surface 4"
Should have just been: "reflecting on CLUE #2, and the fact that the "view factor" from 1 to 2 (F12) = 1, we get the rate of heat transfer from surface 1 to surface 4"
The relationship between areas 2 and 3 are not relevant until later.
The first thing I want to do here is ask a question of khayman80. Did he take over 2 years to fully -- and at least somewhat clearly -- explain the methodology for his "solution" to this problem because he honestly thought he had the correct answer, or because he's just a trolling, malicious, lying son of a bitch?
I do not know the answer for certain, but for a number of reasons I am strongly convinced of the latter.
I ask because for 2 years now he has berated me, publicly derided and taunted me, and (in my strong opinion) libeled me, based on my position regarding Spencer's challenge, even though I knew he was wrong all along, but in order to prove it to everybody else in an understandable way, he had to explain his methods clearly. It's hard to disprove something when it's not clearly defined.
But now he has. And now 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. Because, you see, as I know from experience, it isn't enough to show people the right way. At the same time it is necessary and desirable to show beyond doubt that "global warming alarmist" bullshit is just that: bullshit.
What's funny, khayman80, is that you may have thought I was being funny or incompetent with my interjections, but I was actually feeding you hints all along the way about the right way to do this and the correct answer, but you didn't take any of those hints. Not one. Did you really think I was refusing to "agree" with your assumptions because I was stupid? Again: was that due to 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.
Let's summarize briefly: I suggested to you earlier that the problem was simpler than how you were treating it. I also gave you some good hints that some of your assumptions were incorrect. And I even gave you a QUOTE from an engineering textbook explaining that if you aren't careful, "thermodynamic" reasoning could get you into a lot of trouble when it comes to heat transfer.
I'm going to start by showing the correct answer. I am going to show my basic work but not all the intermediate steps. You are capable of doing those on your own. Later I'm going to do a full writeup and post this on the internet, some time later. After my brief explanation of the correct answer, I'm going to explain more thoroughly where you went wrong.
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.
I am also going to make one assumption myself (which is not really an assumption, but a calculation): given the dimensions stipulated by khayman80, the difference between the outside and inside areas of the enclosing hollow sphere are not enough to matter given the precision of our other calculations. (About 0.0001 m^2, give or take.) This simplifies our equations quite a bit.
Here are the initial conditions we agreed upon: a central sphere of dimensions listed below, with enough power input to heat it to 150 deg. F (338.71K) in radiative steady-state with chamber walls that are actively cooled to 0 deg. F (255.37K).
As for variable names, I do not intend to fully use "standard" notation because this explanation isn't for physicists, it's for others wh
No, Jane. I linked to Wikipedia's equation [wikipedia.org] for radiative heat transfer, which is in Watts, not Joules.
Ahah. Very well, you caught me on that one. I was not aware of the dot-notation. I never pretended to be a physicist. I don't see that notation in the engineering references I have handy.
Your comment about my radiative power equation though was referencing that that same page, but that wasn't what *I* was actually referencing. It might be relevant to black bodies in some way but my own sources (and Wikipedia, too, at the link I showed you) say it is the radiative power out of a gray body at temperature "T" and emissivity "epsilon", and I used it to compute power out of the heat source initially, remember? Our radiated power figures for the heat source in initial conditions agreed, even if you calculated it a different way.
The final answer for the enclosed source at steady-state is 385.4 K (234.1 F). Anyone with a calculator could have verified this based on my comment yesterday.
Certainly, I could have found it with my calculator easily enough. But I had reasons for wanting YOU to post it yourself. Among those reasons, but not the only one, was that it was YOUR claimed refutation, not mine. You get to take either credit or blame, not me.
So! Great news! You have finally completed your claimed refutation of Latour. I shall examine it in detail and get back to you. Probably tomorrow some time.
Just so we are absolutely clear on what your claim is: starting at the agreed-upon initial conditions, heat source at 150F, when a hollow sphere is suddenly inserted into the chamber, completely surrounding the heat source, of the specified dimensions, then when allowed to reach steady-state the actual temperature of the heat source is 234.1 degees F.
Did I summarize that accurately enough? I don't want to re-hash the initial conditions we agreed upon. I still agree with them.
If so, will you really be happy to declare to everyone that you were wrong? Or was that a lie?
Of course. I don't mind admitting it when I'm wrong. But I have to check your work first. I already strongly suspect that there is a large hole in your reasoning, but I will not have time to check it until tomorrow. Stay tuned.
I am not convinced. Even Android required interaction for an NFC payment, and the researchers were able to extract confidential credentials without user interaction.
I'm not claiming you're wrong. It could be so. But I'd have to see more information before I was convinced that it is internally much if any different.
Let me put it another way: if it was significantly different from other NFC transactions, it would require new terminals. Are stores going to buy different terminals just for iPhones?
Sigh. I am short of time so I am reduced to repeating what I wrote to the other responder above, and the others will have to be satisfied with this answer as well:
That is incorrect. Credentials were hacked from NFC-enabled devices even before NFC was very common. There were several different methods used, but apparently it was not difficult at all. If I recall, it only required a device to spoof a legitimate terminal.
That is only one reference. The original researcher was named Christopher something. He was the same one who mounted an antenna in his car for reading RFIDs at a distance.
Haha. Got me on CFAA. I was trying to remember the name of the law and the wrong one came up.
I admit that I haven't looked it up lately, but to the best of my understanding, the server does not have to be in the United States. Just the person who accessed it.
It's an interesting question, though, and I may have to look it up again after all. If a person standing in the United States shoots someone on the other side of the Canadian (or Mexican) border dead, has he or she committed a crime in the United States, or in that other country?
I honestly don't know. It seems to me I did know the answer to that, but I have long since forgotten.
You're just confirming what I said earlier. You're finally proving that you were full of bull all along.
We'll obviously have to agree to disagree that I explicitly used the equation for net radiative power, and linked to an equation described as: "The radiative heat transfer from one surface to another is equal to the radiation entering the first surface from the other, minus the radiation leaving the first surface."
We're not "agreeing to disagree" on anything. You're just plain wrong. You just confirmed exactly what *I* said above: that this is a heat transfer equation, but you called it a "net radiative power" equation.
We've agreed that net radiative power is power out minus power in through a boundary, but we'll obviously have to agree to disagree that Wikipedia's radiative heat transfer is "equal" to net radiative power.
We will most certainly have to disagree on that, because it's wrong. That equation is for finding Q, the net heat transfer, which is not "equal" to power at all. It is energy in Joules.
If you are using it to calculate "net radiative power", then clearly show here how you have manipulated the equation to solve for power instead, so that other people can check your work.
If I was "throwing a fit" by saying we'll obviously have to agree to disagree, then what's this?
Yes, you are quite clearly throwing a fit, and comparing it to something else doesn't change that, or make anything else you have said correct. I had plenty of good reason for stating what I did in the quotes you post here, and those reasons are ALSO soon to be part of a widely-read (very possibly and we can hope) public record. AND... that is all completely beside the real point. You're stalling, and obfuscating again.
No, you haven't even taken a single solitary step towards solving for the enclosed source temperature. But I've repeatedly tackled the full problem in real detail.
Except for your final answer. And whether *I* have done it is completely immaterial. You claimed Latour was wrong, and that you had successfully refuted him. So where is your final answer for the temperature of the heat source at stead-state? THAT was what you said you were calculating, so where is it?
This particular Slashdot thread is not MY show, it is yours. You demanded it. I have given you all the opportunity you asked for. So where's the punchline? Where's the finale?
It would only take you a few minutes to write down the equation and values that could be used to solve for the enclosed source temperature. Is the only reasonable conclusion to be drawn here that you won't do it because you know you're wrong?
How could I possibly be "wrong"? I'm not doing anything. This is YOUR claim, not mine. This is what YOU said YOU could do. It has nothing to do with me, except that I (very reluctantly, for reasons we're seeing now) agreed to be your audience.
In the time it's taken you to write all these incoherent rants and talk about washing people's balls, you could've written down the equation and values describing the enclosed source temperature. In fact, you could've done that many times over. Instead, you cuss in ALL CAPS. Are you really going to give a copy to your grandchildren?
Why should I do that? YOU said you were going to refute Latour. It wasn't my claim. You got partway through, now you refuse to finish, and you're trying to blame ME somehow? How do you figure?
You know what Latour's claims and math were. (They're not mine, they're his.) You proclaimed loudly that he was wrong (actually your words were much stronger than that), and that you had refuted him. I called bullshit, and now here we are. You wanted an opportunity to show that you were honest. Here it is.
My honest opinion is that YOU are the one who is trolling, and never intended to actually refute anybody at all. You simply wanted to waste more of my time.
Once again, we'll obviously have to agree to disagree about the net heat transfer between two gray surfaces.
What the HELL are you talking about? I understand the equation from Wikipedia. I just happened to mention that you called it a power equation rather than a heat transfer equation. THAT IS ALL. Then later, you called my emittance equation a heat transfer equation. That's not anything I did, that's something you did.
Here are your words, complete with the link I was referring to:
As before, that net radiative power is described by Wikipedia's equation which accounts for areas and view factors.
Anybody who follows that link can see that it is a heat transfer equation, not a "net radiative power" equation.
THEN I gave you an equation for radiant emittance: (epsilon)(sigma)T^4, and you called it a "heat transfer" equation having something to do with 0K black bodies, which is simply false. That equation, for gray bodies, can be found here.
Now, when I simply pointed out these apparent MISTAKES in terminology to you, in order to try to keep things straight, you're throwing a fit. Well, don't try to blame this on me. I was just explaining why the things YOU have been saying lead to confusion. I will not apologize for simply trying to sort out basic misunderstandings.
Again, you seem to be asserting that Jane's equation should be used instead of Wikipedia's equation. Is that the case? If so, all you need to do to catch up is to list the values you'll plug into that equation, like I did. This would only take a few minutes. If you're confused and need help, just ask.
Why would you think that? Have I said anything like this? Answer: no, I have not, as just about any reader should have little trouble understanding. In fact I told you twice now that equation was incorrect. I've stated it right here in black and white.
I did not assert any other equation is "correct". I'm letting YOU show me YOUR methods. That's what you said you were going to do, right?
So, I don't know what the hell is going on. Are you drunk?
I will repeat what I have already stated several times: the only things I "insisted" upon were that we agree on the initial conditions of the problem. I do not insist you use any particular equations. This is YOUR show, which I am participating in only as a courtesy. I'm just following along.
Either get on with it, or not. But if you refuse to do what you told me I was coming here to see you do (refute Latour), then you refuse. That has nothing to do with me, and you don't get blame it on me.
Get the hell on with it, or not. Whichever you do, it's YOUR choice. I am very, very close to calling you full of shit and posting this where everyone can see it.
Ironically, you actually are judging my method based on its physics, which is actually a step forward:
NO!!! I was not "judging your method". I was describing your nasty habit of confusing the issues. Two very different things.
Once again, your equation is only for net radiative power (or net heat transfer) to a 0K blackbody.
NO, it is not. It is the equation for THE RADIATIVE EMITTANCE (often called radiative power) of a surface. Its instantaneous value does not depend on surrounding conditions or nearby bodies. The only variables are emissivity and temperature.
I repeat: that equation has nothing directly to do with heat transfer, though heat transfer equations may rely upon it.
This is just another example of how you have tended to obfuscate things. What I stated was a very straightforward, textbook equation that has nothing at all to do at all with "nearby 0K blackbodies".
Slashdot Mirror
But again: if you need to "draw a boundary", it needs to be drawn around the passive plate itself. We have already firmly established that your "boundary" around the heat source and the "enclosing shell" is even thermodynamically incorrect. It leads to an erroneous result of very close to (within a few thousandths) DOUBLE the radiative power from that surface that actually exists.
I have explained this to you 3 times now. If you can't get it through your head, that's your problem and nobody else's.
As I said, I am going to write this up more thoroughly, elsewhere. But I have presented enough here for anybody who is really interested to figure it out without too much difficulty. Present company apparently excepted.
s/should have drawn your shell/should have drawn your boundary
We have already shown that your particular application of "drawing boundaries" here was a MISAPPLICATION of the principle you are trying to use.
The "enclosing shell" (if by that you mean the passive plate that was inserted) is acted upon only by radiation. You should have drawn your shell around THAT, and that alone. And you should at least have tried drawing your boundary around your own goddamned heat source, both for initial conditions and your final result, to check your work. But you didn't. What you got was a universe-busting violation of conservation of energy.
But of course you are still trying to defend something which YOU claimed earlier is not valid to do. There is a word for that.
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.
One more correction and again it's minor, but again I plea that I have been very busy and had to dash this off in a hurry:
In my list of "clues" toward the bottom, Clue #1:
The bit about "Barring any issues like 'view factor'" was an irrelevant comment because that relates to interactions between bodies and my whole point was that the S-B equation for radiative power from a body does not include other bodies. The part about "view factor" can be left out of that passage without affecting its correctness.
Further, if you have a problem with my equations (including my minor corrections) you are welcome to do your own and prove me wrong.
But you aren't going to, because I'm not wrong, in any basic way. I might have gotten a hundredth or a thousandth off here or there, but unlike you I did double-check your work. All while (according to you) I was spouting something off on Facebook or something.
I spent about an hour and a half on this, give or take, between my regular work, and working on the car. That includes identifying your errors, and determining what the right way to do it was.
s/you can even add/you can't even add
My keyboard needs cleaning.
As I said, in the unlikely event that you wrote down equations, they'd violate conservation of energy. Thermodynamic thinking like this leads one back to reality, not astray. Draw a boundary inside the inner surface of the enclosing shell at your steady-state values. Since nothing inside that boundary is changing, power in = power out. But that's completely impossible. Your solution violates conservation of energy, as predicted.
Violating conservation of energy is something you do quite well, you complete bozo.
You ASSUMED that power at the surface of the heat source = power at the outer surface of the hollow sphere. You did NOT take into account that the sphere radiates from BOTH surfaces. You can even fucking add 2 + 2.
THEN, you neglected to do the most basic check of your work, such as: applying the S-B relation to your new, hotter heat source. You just hoist yourself by your own petard, dude, because our input is a FIXED amount of power, but you upped your output by almost 100 degrees F. Draw your line JUST around your heat source. Where is that extra power coming from? S-B law says that power is ONLY related to temperature and emissivity, and the emissivity hasn't changed.
So where is that extra power coming from? Thin goddamned air?
I made a couple of very minor errors, which I have corrected. My basic proof still stands, and you are still wrong.
In fact, you're a complete loon.
I either included or referenced all the equations that were necessary to solve the problem. It's not my fault if you still can't do it.
This is what I wrote:
Think of it this non-tecnical way: radiant heat transfer is a function of difference in temperatures. If you have two differences, A = X - Y and B = Y - Z, then A - Z = (X - Y) plus (Y - Z).
What I meant was this:
Think of it this non-tecnical way: radiant heat transfer is a function of difference in temperatures. If you have two differences, A = X - Y and B = Y - Z, then (A + B) = (X - Z).
Another minor correction: I stated that "the difference in area" between 2 and 3 was only about 0.001, or some such. That was not quite correct. What I meant was in the process of doing my calculations, I discovered that the difference made by the difference in area was only about 0.001. So I determined that it could safely be ignored, given the fairly loose precision of our other calculations.
Jane's obligations include continuing to spread misinformation about ocean acidification even after I've repeatedly debunked him.
So I predict that Jane's answer won't include any equations that could be used to calculate the enclosed source temperature. Instead, he'll probably grace us with another lengthy, incoherent rant about "problems" in my analysis which are (as usual) too vague to be expressed in equations. In the extremely unlikely event that Jane musters up the courage and competence to actually write down an equation that could be used to calculate the enclosed source temperature, it will almost certainly violate conservation of energy.
Wrong again. Or perhaps I should say STILL.
I almost forgot to mention another aspect of khayman80's folly.
He could have discovered he was wrong almost 2 years ago, almost immediately, by checking his work. He didn't. And this was the result:
He assumed that the total power output of the heat source was available on the OUTSIDE of the enclosing hollow sphere. We now know this was incorrect. He then used that to calculate a hotter temperature for the heat source itself. That may have made sense, giving his (incorrect) assumption about how "power in = power out" worked. BUT... he got a value for temperature almost 100 degrees F hotter than before.
Remember the S-B law? That means the POWER output of that heat source was much greater. It doesn't matter where it was coming from... according to his own reasoning (which would be correct for the heat source itself), it can only be as hot as the power that is input lets it. Power out p = (epsilon)(sigma)T^4. Period. No two ways about it. So if there is a greater T, there is higher power output, and it has to come from somewhere. His heat source is going to draw more power from whatever is powering it.
But wait! There's more!
All he had to do was continue this magical thinking. Now transfer this power out back to the outside of the enclosing sphere, the way he did it the first time. Then he can back-calculate, the same way he did the first time, and get an even HOTTER figure for the heat source. Then he can transfer this power back out to the outside of the enclosing sphere again, and continue ad infinitum!
It's hilarious, because khayman80 accused me of spreading "civilization-paralyzing misinfomation". But if the universe really worked according to the information HE has been spreading, it would result in a thermal runaway and destroy itself in an extremely brief period of time. For the price of a few kilowatts.
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.
Correction: "reflecting on CLUE #2, and the fact that the "view factor" from 1 to 2 (F12) = 1, PLUS our assertion that area of 2 is close enough to the area of 3 to ignore, we get the rate of heat transfer from surface 1 to surface 4"
Should have just been: "reflecting on CLUE #2, and the fact that the "view factor" from 1 to 2 (F12) = 1, we get the rate of heat transfer from surface 1 to surface 4"
The relationship between areas 2 and 3 are not relevant until later.
The first thing I want to do here is ask a question of khayman80. Did he take over 2 years to fully -- and at least somewhat clearly -- explain the methodology for his "solution" to this problem because he honestly thought he had the correct answer, or because he's just a trolling, malicious, lying son of a bitch?
I do not know the answer for certain, but for a number of reasons I am strongly convinced of the latter.
I ask because for 2 years now he has berated me, publicly derided and taunted me, and (in my strong opinion) libeled me, based on my position regarding Spencer's challenge, even though I knew he was wrong all along, but in order to prove it to everybody else in an understandable way, he had to explain his methods clearly. It's hard to disprove something when it's not clearly defined.
But now he has. And now 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. Because, you see, as I know from experience, it isn't enough to show people the right way. At the same time it is necessary and desirable to show beyond doubt that "global warming alarmist" bullshit is just that: bullshit.
What's funny, khayman80, is that you may have thought I was being funny or incompetent with my interjections, but I was actually feeding you hints all along the way about the right way to do this and the correct answer, but you didn't take any of those hints. Not one. Did you really think I was refusing to "agree" with your assumptions because I was stupid? Again: was that due to 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.
Let's summarize briefly: I suggested to you earlier that the problem was simpler than how you were treating it. I also gave you some good hints that some of your assumptions were incorrect. And I even gave you a QUOTE from an engineering textbook explaining that if you aren't careful, "thermodynamic" reasoning could get you into a lot of trouble when it comes to heat transfer.
I'm going to start by showing the correct answer. I am going to show my basic work but not all the intermediate steps. You are capable of doing those on your own. Later I'm going to do a full writeup and post this on the internet, some time later. After my brief explanation of the correct answer, I'm going to explain more thoroughly where you went wrong.
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.
I am also going to make one assumption myself (which is not really an assumption, but a calculation): given the dimensions stipulated by khayman80, the difference between the outside and inside areas of the enclosing hollow sphere are not enough to matter given the precision of our other calculations. (About 0.0001 m^2, give or take.) This simplifies our equations quite a bit.
Here are the initial conditions we agreed upon: a central sphere of dimensions listed below, with enough power input to heat it to 150 deg. F (338.71K) in radiative steady-state with chamber walls that are actively cooled to 0 deg. F (255.37K).
As for variable names, I do not intend to fully use "standard" notation because this explanation isn't for physicists, it's for others wh
If they don't archive this thread, I will have my answer for you a bit later today. I have other obligations.
No, Jane. I linked to Wikipedia's equation [wikipedia.org] for radiative heat transfer, which is in Watts, not Joules.
Ahah. Very well, you caught me on that one. I was not aware of the dot-notation. I never pretended to be a physicist. I don't see that notation in the engineering references I have handy.
Your comment about my radiative power equation though was referencing that that same page, but that wasn't what *I* was actually referencing. It might be relevant to black bodies in some way but my own sources (and Wikipedia, too, at the link I showed you) say it is the radiative power out of a gray body at temperature "T" and emissivity "epsilon", and I used it to compute power out of the heat source initially, remember? Our radiated power figures for the heat source in initial conditions agreed, even if you calculated it a different way.
The final answer for the enclosed source at steady-state is 385.4 K (234.1 F). Anyone with a calculator could have verified this based on my comment yesterday.
Certainly, I could have found it with my calculator easily enough. But I had reasons for wanting YOU to post it yourself. Among those reasons, but not the only one, was that it was YOUR claimed refutation, not mine. You get to take either credit or blame, not me.
So! Great news! You have finally completed your claimed refutation of Latour. I shall examine it in detail and get back to you. Probably tomorrow some time.
Just so we are absolutely clear on what your claim is: starting at the agreed-upon initial conditions, heat source at 150F, when a hollow sphere is suddenly inserted into the chamber, completely surrounding the heat source, of the specified dimensions, then when allowed to reach steady-state the actual temperature of the heat source is 234.1 degees F.
Did I summarize that accurately enough? I don't want to re-hash the initial conditions we agreed upon. I still agree with them.
If so, will you really be happy to declare to everyone that you were wrong? Or was that a lie?
Of course. I don't mind admitting it when I'm wrong. But I have to check your work first. I already strongly suspect that there is a large hole in your reasoning, but I will not have time to check it until tomorrow. Stay tuned.
You're either in a big office or you walk very slowly.
Do you know how far it is possible to walk in 5 minutes? Even if you're not particularly hurrying?
I don't necessarily disagree with you, but I still don't know too many bosses who would be cool with that.
I am not convinced. Even Android required interaction for an NFC payment, and the researchers were able to extract confidential credentials without user interaction.
I'm not claiming you're wrong. It could be so. But I'd have to see more information before I was convinced that it is internally much if any different.
Let me put it another way: if it was significantly different from other NFC transactions, it would require new terminals. Are stores going to buy different terminals just for iPhones?
Sigh. I am short of time so I am reduced to repeating what I wrote to the other responder above, and the others will have to be satisfied with this answer as well:
That is incorrect. Credentials were hacked from NFC-enabled devices even before NFC was very common. There were several different methods used, but apparently it was not difficult at all. If I recall, it only required a device to spoof a legitimate terminal.
That is only one reference. The original researcher was named Christopher something. He was the same one who mounted an antenna in his car for reading RFIDs at a distance.
Wrong. Look it up.
I don't have time right now, but HERE is one reference, though it might have been the same experiment.
Haha. Got me on CFAA. I was trying to remember the name of the law and the wrong one came up.
I admit that I haven't looked it up lately, but to the best of my understanding, the server does not have to be in the United States. Just the person who accessed it.
It's an interesting question, though, and I may have to look it up again after all. If a person standing in the United States shoots someone on the other side of the Canadian (or Mexican) border dead, has he or she committed a crime in the United States, or in that other country?
I honestly don't know. It seems to me I did know the answer to that, but I have long since forgotten.
We'll obviously have to agree to disagree that I explicitly used the equation for net radiative power, and linked to an equation described as: "The radiative heat transfer from one surface to another is equal to the radiation entering the first surface from the other, minus the radiation leaving the first surface."
We're not "agreeing to disagree" on anything. You're just plain wrong. You just confirmed exactly what *I* said above: that this is a heat transfer equation, but you called it a "net radiative power" equation.
We've agreed that net radiative power is power out minus power in through a boundary, but we'll obviously have to agree to disagree that Wikipedia's radiative heat transfer is "equal" to net radiative power.
We will most certainly have to disagree on that, because it's wrong. That equation is for finding Q, the net heat transfer, which is not "equal" to power at all. It is energy in Joules.
If you are using it to calculate "net radiative power", then clearly show here how you have manipulated the equation to solve for power instead, so that other people can check your work.
If I was "throwing a fit" by saying we'll obviously have to agree to disagree, then what's this?
Yes, you are quite clearly throwing a fit, and comparing it to something else doesn't change that, or make anything else you have said correct. I had plenty of good reason for stating what I did in the quotes you post here, and those reasons are ALSO soon to be part of a widely-read (very possibly and we can hope) public record. AND... that is all completely beside the real point. You're stalling, and obfuscating again.
No, you haven't even taken a single solitary step towards solving for the enclosed source temperature. But I've repeatedly tackled the full problem in real detail.
Except for your final answer. And whether *I* have done it is completely immaterial. You claimed Latour was wrong, and that you had successfully refuted him. So where is your final answer for the temperature of the heat source at stead-state? THAT was what you said you were calculating, so where is it?
This particular Slashdot thread is not MY show, it is yours. You demanded it. I have given you all the opportunity you asked for. So where's the punchline? Where's the finale?
It would only take you a few minutes to write down the equation and values that could be used to solve for the enclosed source temperature. Is the only reasonable conclusion to be drawn here that you won't do it because you know you're wrong?
How could I possibly be "wrong"? I'm not doing anything. This is YOUR claim, not mine. This is what YOU said YOU could do. It has nothing to do with me, except that I (very reluctantly, for reasons we're seeing now) agreed to be your audience.
In the time it's taken you to write all these incoherent rants and talk about washing people's balls, you could've written down the equation and values describing the enclosed source temperature. In fact, you could've done that many times over. Instead, you cuss in ALL CAPS. Are you really going to give a copy to your grandchildren?
Why should I do that? YOU said you were going to refute Latour. It wasn't my claim. You got partway through, now you refuse to finish, and you're trying to blame ME somehow? How do you figure?
You know what Latour's claims and math were. (They're not mine, they're his.) You proclaimed loudly that he was wrong (actually your words were much stronger than that), and that you had refuted him. I called bullshit, and now here we are. You wanted an opportunity to show that you were honest. Here it is.
I have no
My honest opinion is that YOU are the one who is trolling, and never intended to actually refute anybody at all. You simply wanted to waste more of my time.
Once again, we'll obviously have to agree to disagree about the net heat transfer between two gray surfaces.
What the HELL are you talking about? I understand the equation from Wikipedia. I just happened to mention that you called it a power equation rather than a heat transfer equation. THAT IS ALL. Then later, you called my emittance equation a heat transfer equation. That's not anything I did, that's something you did.
Here are your words, complete with the link I was referring to:
As before, that net radiative power is described by Wikipedia's equation which accounts for areas and view factors.
Anybody who follows that link can see that it is a heat transfer equation, not a "net radiative power" equation.
THEN I gave you an equation for radiant emittance: (epsilon)(sigma)T^4, and you called it a "heat transfer" equation having something to do with 0K black bodies, which is simply false. That equation, for gray bodies, can be found here.
Now, when I simply pointed out these apparent MISTAKES in terminology to you, in order to try to keep things straight, you're throwing a fit. Well, don't try to blame this on me. I was just explaining why the things YOU have been saying lead to confusion. I will not apologize for simply trying to sort out basic misunderstandings.
Again, you seem to be asserting that Jane's equation should be used instead of Wikipedia's equation. Is that the case? If so, all you need to do to catch up is to list the values you'll plug into that equation, like I did. This would only take a few minutes. If you're confused and need help, just ask.
Why would you think that? Have I said anything like this? Answer: no, I have not, as just about any reader should have little trouble understanding. In fact I told you twice now that equation was incorrect. I've stated it right here in black and white.
I did not assert any other equation is "correct". I'm letting YOU show me YOUR methods. That's what you said you were going to do, right?
So, I don't know what the hell is going on. Are you drunk?
I will repeat what I have already stated several times: the only things I "insisted" upon were that we agree on the initial conditions of the problem. I do not insist you use any particular equations. This is YOUR show, which I am participating in only as a courtesy. I'm just following along.
Either get on with it, or not. But if you refuse to do what you told me I was coming here to see you do (refute Latour), then you refuse. That has nothing to do with me, and you don't get blame it on me.
Get the hell on with it, or not. Whichever you do, it's YOUR choice. I am very, very close to calling you full of shit and posting this where everyone can see it.
Ironically, you actually are judging my method based on its physics, which is actually a step forward:
NO!!! I was not "judging your method". I was describing your nasty habit of confusing the issues. Two very different things.
Once again, your equation is only for net radiative power (or net heat transfer) to a 0K blackbody.
NO, it is not. It is the equation for THE RADIATIVE EMITTANCE (often called radiative power) of a surface. Its instantaneous value does not depend on surrounding conditions or nearby bodies. The only variables are emissivity and temperature.
I repeat: that equation has nothing directly to do with heat transfer, though heat transfer equations may rely upon it.
This is just another example of how you have tended to obfuscate things. What I stated was a very straightforward, textbook equation that has nothing at all to do at all with "nearby 0K blackbodies".