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## Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1)159

by Jane Q. Public (#47947019) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

No. Once again, in this experiment [archive.today] 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. ..."

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.

Jane's even stumbled across this point:

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! 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?

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.

## Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1)159

by Jane Q. Public (#47943375) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Instead of holding electrical heating power constant, Jane held the source's radiative power output constant. That held source temperature constant and forced electrical heating power to change.

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.

Jane might wonder why he can't derive a single equation which works for all these cases.

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.

But Jane can't even admit there's a difference between holding electrical heating power constant and holding the source's radiative power output constant.

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.

Again, warming the chamber walls is like partially closing the drain on a bathtub where water is flowing in at a constant rate

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.

Hopefully these are just more badly-worded sentences because they all require absorptivity = 0.

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?

But for gray bodies it's just an approximation because it ignores reflections. After obviously failing to explain that we need to account for reflections, I decided to agree to disagree. For two gray bodies interacting with small view factors (e.g. Earth's tiny view factor of the Sun) reflections can be safely neglected. But the chamber wall completely encloses the source, so its view factor is 1. That's why MIT's equation is more accurate here: it accounts for reflections.

Complete bullshit again. We were assuming diffuse gray bodies. Further:

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.

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.

Since Jane's proposed equation is missing the "(s)*T4^4" term, it doesn't reduce to this simpler Eq. 1J.2 for blackbodies where (e) = 1. So it's wrong.

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.

## Comment: Re:Really? (Score 1)125

by Jane Q. Public (#47941143) Attached to: Next Android To Enable Local Encryption By Default Too, Says Google
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.

## Comment: Re:Oregon... (Score 2)190

by Jane Q. Public (#47930027) Attached to: Wave Power Fails To Live Up To Promise
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.

## Comment: Re:When doing anything involving the ocean (Score 4, Interesting)190

by Jane Q. Public (#47929969) Attached to: Wave Power Fails To Live Up To Promise
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.

## Comment: Re:Lots of problems with it (Score 1)190

by Jane Q. Public (#47929879) Attached to: Wave Power Fails To Live Up To Promise
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.

## Comment: Re:COBOL: Why the hate? (Score 2)268

by Jane Q. Public (#47924493) Attached to: College Students: Want To Earn More? Take a COBOL Class

However, since if it is still being used, then it still has some capability that is not available in other solutions.

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.

## Comment: Re:The UK Cobol Climate Is Very Different (Score 4, Insightful)268

by Jane Q. Public (#47924475) Attached to: College Students: Want To Earn More? Take a COBOL Class

Every professional workplace has an expectation of a formal atire.

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.

## Comment: Re:Lifetime at 16nm? (Score 1)64

by Jane Q. Public (#47924301) Attached to: Micron Releases 16nm-Process SSDs With Dynamic Flash Programming

seems like the average life expectancy of SSDs are well beyond the needs of most people at the moment, unless you're doing some serious content creation with massive amounts of read/writes.

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.

## Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1)159

by Jane Q. Public (#47914835) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting
And, last comment here: you have confirmed that you have not abandoned your incorrect (and actually quite ludicrous) version of heat transfer, which violates the Stefan-Boltzmann radiation law on its very face.

That was all I needed. I am now done. Have a nice day. You can have the last word all you like; it won't make you any more correct.

## Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1)159

by Jane Q. Public (#47914825) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting
And no, I don't have to ask myself that, because it doesn't happen.

I have already found the solution to a reasonable degree of precision. Your solution, as stated (approximately 241 degrees F for the central heat source) does not check out, even using your own equations.

## Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1)159

by Jane Q. Public (#47914807) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Once again, if the electrical heating power is held constant, the heat source has to warm. Once agin, Jane's heat source keeps the source temperature constant by halving its electrical heating power. Jane/Lonny Eachus might ask himself why his required electrical heating power goes down by a factor of two after the enclosing shell is added.

That is neither correct, or an answer to my question.

## Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1)159

by Jane Q. Public (#47914803) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

In other words, the electrical heating power is determined by drawing a boundary around the heat source: power in = electrical heating power + radiative power in from the chamber walls power out = radiative power out from the heat source

Since power in = power out:

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

Right?

No. Not right. Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. Because the only power transfer taking place here is heat transfer, which is a function of (emissivity) * (S-B constant) * (Ta^4 - Tb^4).

You DO know what a minus sign is, yes?

Since emissivity doesn't change the input required to heat source to achieve 150F is constant, regardless of where it comes from. But as long as the walls of the chamber are cooler than the source, NONE of the power comes from the chamber walls, because of that minus sign in the equation above. Nothing has changed in that respect, and that's what the Stefan-Boltzmann law requires.

The only time that changes is if the walls are at an equal temperature, in which case heat transfer is 0 and you can begin to use "ambient" temperature as input. You are still supplying the same input power, you are just supplying it a different way.

If the chamber walls were hotter than the central source, then heat transfer would be in the other direction (because the sign of the solution to the equation above changes), and only THEN are you getting net heat transfer TO the central sphere.

And BOTH of those situations are a violation of Spencer's conditions.

## Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1)159

by Jane Q. Public (#47914693) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting
But wait. I take that back. Before I declare that I am done and go away, I just want to ask you: do you still maintain that after the enclosing passive sphere is inserted, the central heat source raises in temperature to approximately 241 degrees F? You haven't said anything about that in a while, so I'm just checking.

## Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1)159

by Jane Q. Public (#47914669) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Again, that's completely ridiculous. I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant.

Nonsense. It would take power to bring the chamber walls up to 150F (338.71K). How else do you expect them to get to that temperature? Where are you getting that power from? This is so utterly obvious that I honestly don't believe you don't get it.

For example, you could simply place the vacuum chamber somewhere with an ambient temperature of 150F. That would require zero power, but once again it doesn't matter even if the vacuum chamber were on Pluto. Because that power never crosses the boundary.

You could, but we haven't. Regardless, it still remains the same. Power output at that temperature remains constant because P = (emissivity) * (S-B constant) * T^4 says it has to.

The only thing you are doing is ADDING energy to the system by putting it in an ambient environment of 150F. That's not irrelevant at all, because if you're at thermal equilibrium, there is no heat transfer. Since this is all about heat transfer, how could it be irrelevant?

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.

The ONLY time the power output changes is if you change the temperature. You can do that by making the walls HOTTER than the "heat source", thereby causing a net heat transfer TO it from the walls, OR you can input more electrical power to the heat source, thereby making it hotter, but that would be a violation of the conditions Spencer stipulated.

Here's our disagreement. Conservation of energy demands that a heat source at 150F requires no electrical heating power inside 150F vacuum chamber walls.

That's not our disagreement at all. Not even frigging close. Of course it wouldn't need a separate heat source if its environment were maintained at 150 degrees. I just got done saying that. But it still does have power input. It' just that it comes from the environment in this case rather than an electrical element.

Because its radiant output power remains constant according to the Stefan-Boltzmann law. All you have done is raise the environment's output power to match, and raised the input to that environment enough to achieve that temperature. Big deal. That takes energy of its own, and proves exactly nothing. You haven't proved that it needs no power, you just changed the source of that power. And used up even more power in the process, because the environment is larger than the central sphere.

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.

"It's curtains for you, Mighty Mouse! This gun is so futuristic that even *I* don't know how it works!" -- from Ralph Bakshi's Mighty Mouse

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