Maybe they should just offer a link to a free YYZ download with every Netflix membership. Problem solved.

I have said nothing that contradicts this. Not only do I freely admit this, my calculations relied on that fact. I kept the power (and hence energy over time) input into the plate from the electric heater completely constant. Which we may freely do, since it was a stipulation of Spencer's experiment.

No, I didn't "stumble" over that point, YOU are stumbling over it. Everything changes at thermal equilibrium. The "heated" body is no longer warmer than its surroundings and can begin taking on energy from its surroundings. And it is not a "gradual" change: the Stefan-Boltmann law says a warmer body DOES NOT absorb net radiant energy from its surroundings. That only begins to happen at thermal equilibrium. BUT thermal equilibrium does not apply to this experiment, anywhere, at any time. This is just another straw-man argument. Which you are very good at, by the way. Not good enough to sucker me in, though.

Of course I realize that, and have all along. The error lies in your implication that this is a gradual change.

It isn't a gradual change. It's a result of Ta^4 - Tb^4 = 0. A transition from non-zero to 0.

That's the only reason. The transition between non-zero and zero is a profound change which affects everything, and there is nothing gradual about it. But it doesn't apply in this context. The surfaces are never at thermal equilibrium. And 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. The sign of the result is everything here.

If body (a) is warmer than body (b), Ta^4 - Tb^4 > 0, and net heat transfer is ONLY from (a) to (b).

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.

If body (a) is at a lower temperature than body (b), Ta^4 - Tb^4 < 0, which means there is net transfer of heat from (b) to (a).

The third condition is the ONLY one in which there is any input to (a) from its surroundings. But that condition never occurs in Spencer's experiment because the heat source is always hotter than its surroundings.

Knock off the BS. Time to admit you were wrong. I repeat: anything else is a violation of the Stefan-Boltzmann radiation law.

For $8 a month they offer a pretty good selection.

I agree, assuming that selection includes movies I actually want to see. Now it mostly doesn't, so its value to me is approaching $0, sadly.

Yes. Unfortunately, as my trust in them goes up, their useful library continues to shrink.

No, that is not correct. You made assumptions that are, to be blunt, bullshit nonsense.

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.

By the Stefan-Boltzmann law, since the power in remains constant, then UNLESS power is taken up from some other source, the temperature will remain constant. This follows directly from the S-B radiation law, which you seem to be disputing.

Another requirement of the S-B law, and also of thermodynamics: since EVERY other object in the system is at a lower temperature than the heat source, NET heat transfer is in ONLY one direction: from hotter to colder.

Therefore, no energy is flowing "backward" to boost the output of the heat source.

Yet another fact that follows directly from the S-B law, is that nearby cooler bodies have zero effect on the output of the heat source. They don't "suck" power from it, nor (see above) do they "lend" power to it.

The only logical conclusion -- the only physically possible conclusion, unless you dispute the Stefan-Boltzmann radiation law, is that the heat source does not change temperature. Power out = power in, and is constant. Everything else is cooler, so it remains a constant. There is no further energy or power flowing "backward" the heat source.

The Stefan-Boltzmann law clearly shows that no NET radiation from cooler objects is absorbed; it is either transmitted, reflected, or scattered. Since these are diffuse gray bodies, they do not transmit. That leaves reflection and scattering. For our purposes, the net effect is that it is all reflected.

You are imagining some kind of power input to the heat source that doesn't exist. Further, if the heat source became even hotter, as you assert, it would require even MORE power, because as you say, power in = power out. That was YOUR assertion. Draw your boundary around the heat source itself. There is no net radiation absorbed from outside, and the supplied power remains constant.

It this whole "proof" of yours, I have shown where you have contradicted yourself at least 3 different ways.

I don't know where you get this idea, because I did. I used the S-B equation to find my solution. I used the textbook equations for heat transfer. Yes, I ignored area because the areas were so similar. But it was still a reasonably accurate approximation. I checked my work, and it wasn't off by more than a fraction of a percent.

Because there isn't any. Your own "boundary" principle says so. This isn't a matter of differential equations at this point. 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.

Which is not only false (the S-B relation again, which says it only relies on its radiant temperature, not the temperature of cooler bodies nearby), but another straw man, because the chamber walls aren't warmed. They are held at a constant 255.37K.

No, they don't. Gray body radiant power vs. temperature is expressed by S-B equation, and we already know that gray body absorptivity = emissivity. I was using the proper equation, and you were using it too (if improperly). Are you trying to tell me that the equation YOU have been using is invalid?

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. We were assuming diffuse gray bodies. Further:

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.

But in this context it is already "dirt simple", as I pointed out before. These are diffuse gray bodies. (1 - emissivity) is assumed to be the "reflection", which in this context also includes scattering but no transmission. This is already accounted for in the equations, such as the heat transfer equation you borrowed from Wikipedia.

If you like, you can use the preferred method (according to Wikipedia) for calculating the respective radiant output of the surfaces: the Radiosity Method. That method explicitly accounts for reflection (1 - emissivity). And I already know that it confirms my solution. So go ahead. I simply didn't show it in my brief write-up because I intended it to be a brief write-up. I do intend to show it in the fuller version.

More nonsense. The S-B relation says that the radiative power out of a body is P = (epsilon)*(sigma)*T^4. It is not wrong. It is a simple equation that is well-known to physicists. You claim to be a physicist, so why don't you know it?

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?

I repeat: I checked my solution using Wikipedia's equation, including the areas AND the view factors AND the reflections. It checked out just fine, thank you very much. Why don't you try it yourself and see?

But after using Jane's equation in pointless attempts to illustrate more fundamental problems in Jane's analysis, I wanted to stress once again that MIT's equation is more appropriate for enclosing chamber walls because it accounts for reflections.

It doesn't matter. It still checks out. Although I'd say that Wikipedia's equation is more correct because it includes area and view factor, which MIT's equation does not.

Other than your mention of the equations in the latter part of your comment, it is easy to show that EVERYTHING ELSE is just plain nonsense. You are trying to dispute the Stefan-Boltzmann radiation law and its corollaries. Excuse me, but that didn't work in the beginning, and it still isn't working. You've added nothing worthwhile to the conversation since.

You've been owned, man. BE enough of a man to admit it. Because everybody's going to know it anyway.

There's a huge problem with Android device encryption.

Unlike Apple's usual forms of encryption, once an Android device is encrypted, it is not reversible. There is no way to UN-encrypt it, except to back up all your programs, flash your original unencrypted OS back to the phone, then restore the programs. And that requires unlocking and rooting the phone.

There are LOTS of problems caused by that.

Don't forget the "War on Terrorism".

What they should do is use the ocean version of "pumped storage": build a giant vertical cylinder in the ocean, and when you have surplus electricity you pump water OUT of the chamber. Then when usage peaks and you need more electricity, you let water run back in and turn turbines to generate it.

It's probably a hell of a lot cheaper than batteries. Pumped storage has been an up-and-coming technology for 20 years now. I worked on one project in which they hollowed out an entire stone mountain, creating huge chambers to store water for a pumped-storage system.

When I was a child, we had a nice wood boat. A ChrisCraft. The finish was getting pretty weather-worn so my father took it to a guy who refinished boats to get it done. He specified brass screws, just like the original. The refinisher said, "Everybody uses stainless steel these days. They're just as good." My father reluctantly let him use the stainless steel screws.

The boat was moored by strong chains to a dock in the ocean. (You had to leave lots of play in the chains so the boat could ride up and down with the tide.) A few weeks later, by family got a call from the SeaBees. They had found the boat, dangling underwater by the chains holding it to the dock pilings.

The seawater had eaten the stainless steel screws right up. It only took a few weeks.

Small waves and ripples carry a lot of energy. Even the small waves that are not, on average, enough to budge a houseboat at all, can charge batteries and power lights, pumps, TVs, etc.

Is FDD here to stay?

It seems like you're extrapolating from that experience, to thinking "FDD" is a current trend. AFAIK it's not. A small number of dysfunctional shops like that has virtually always existed. I'm going to go out on a limb and guess that you've only been doing software development for a few years, so you're working from a limited sample size.

I have been in a few jobs where the managers were verbally and/or emotionally abusive. In both cases I left ASAP.

This has always puzzled me why some developers list this as a negative. What is wrong with wearing a suit?

They're expensive. They generally need dry-cleaning. Spilling stuff on them is expensive. They're typically less comfortable than some alternatives. They tend to be hotter in the summer than what I'd normally wear.

Every professional workplace has an expectation of a formal atire.

Either you have an unusually narrow definition of "workplace", or your statement is just factually incorrect.

No, no, no, no!

COBOL is still in use because because mid-to-large corporations spend many millions of dollars on systems that WORKED, and now it's far cheaper to keep them working, the same old way, than it is to do it all over again with modern equipment and languages.

This is called "installed base" and it's a particular problem for COBOL because that was one of the first business languages, and has one of the largest, large-corporation "installed bases".

COBOL has nothing to offer that newer languages don't do better. Not. One. Thing.

No, they don't. This is a statement made by someone about ready to REtire.

Most high-paying tech jobs today do not require a suit and many not even an office to go into. Often you can work at home in your pajamas, if you like.

Yes, really.

The lifetime has been exaggerated from Day 1. Further, multiplying this problem manyfold, is that when an SSD fails, it tends to fail totally. In contrast, when a hard drive i failing, you tend to get a few bad sectors which flag an impending problem, and you main lose a file or two. Bad SSD usually means "everything gone with no warning".

If you use SSD you should have a good HDD backup.