... 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... [Jane Q. Public, 2014-09-15]

But if the chamber walls are also at 150F, they're not cooler than the source and the input required to heat the source to 150F is zero.

... 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. [Jane Q. Public, 2014-09-15]

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.

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

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?

Ask the Europeans that constantly tell us Americans we are too enslaved to the notion that we all need our own car.

You just made that up. I don't know if you've ever driven around a European city, but car ownership is pretty widespread, at least judging by driving through Rome/London/Paris/etc.

It's funny what some Americans think about Europe. They've got this AM talk radio version of Europe knocking around in their heads. "Yeah, they're all dying in the streets because of socialized medicine and everybody's gay and you can't get a decent hamburger anywhere. And they're a bunch of carpoolers who don't realize that we fought and died so that people could drive their own 4500lb vehicle like God intended." "You betcha, Mack. Next up is Fred from Midland. So, what grinds your gears about Europeans, Fred?"

I guess that would be an answer if nobody ever took the time to show you how to defend yourself with your hands.

I think you have been listening to people, that have gleefully lied to you.

I think you are projecting.

A far cry from "proven to make up data and conceals data that doesn't fit his ideology".

No. Not being able to produce the data that your most important work is based on is not a far cry from making up data and concealing data that doesn't fit his ideology.

If a researcher can't produce his data, his work is not taken seriously. The scientific method includes making your data available so other people can review your work.

You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F. The answer is YES, and here is why: You are proposing to bring the whole system up to a level of higher thermodynamic energy, rather than just the heat source. And you are somehow proposing that it doesn't take more energy to do that. But of course it does. The power required to bring the heat source up to 150F remains the same, because the Stefan-Boltzmann law says it has to be. But NOW, you are ALSO bringing the walls up to that higher temperature, and THAT would require even more power (because of the slightly larger surface area). [Jane Q. Public, 2014-09-15]

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.

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.

Either way, as long as the chamber walls are held at 150F, the heat source would need absolutely no electrical heating power to remain at 150F. Zero. Period.

You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F. The answer is YES... [Jane Q. Public, 2014-09-15]

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

They sort of look and feel like apple stores.

I do not want to buy my expensive Tesla from a smelly "genius" walking around with a corporate-logo polo shirt snug around the belly that hangs over his belt, which sports an iPhone holster. I'd rather just order the damn thing on-line and have USPS deliver it to my front door.

You're either disputing conservation of energy, or you're not calculating the actual electrical heating power. If you're calculating the actual electrical heating power, your calculation has to account for radiation from the chamber walls because it passes in through that boundary. That's why the electrical heating power would be zero if the chamber walls were also at 150F!

Nonsense. This is textbook heat transfer physics. We have a fixed emissivity. Therefore, according to the Stefan-Botlzmann radiation law, the ONLY remaining variable which determines radiative power out is temperature. NOTHING else. That's what the law says: (emissivity) * (S-B constant) * T^4. That's all. Nothing more. This makes it stupidly easy to calculate the radiative power out, and therefore the necessary power in. [Jane Q. Public, 2014-09-15]

It's "stupidly easy" to calculate radiative power out and power in through what boundary? The boundary you're describing has to include the source's radiative power passing out through it, without including radiative power from the chamber walls passing in. I think that's impossible, but feel free to explain exactly where such a boundary would be drawn.

One question only: do you agree with the Stefan-Boltzmann relation: power out P = (emissivity) * (S-B constant) * T^4 ?? No more bullshit. "Yes" if you agree that equation is valid, or "No" if you deny that it is valid. Just that and no more. I'm not asking your permission. I'm just trying to find out whether you're actually crazy or just bullshitting. [Jane Q. Public, 2014-09-15]

Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law. But I've obviously failed to communicate that the power from the chamber walls has to pass in through that boundary, so you're only using half the equation to calculate the electrical heating power.

The REASON there would not be as great a power DIFFERENCE if the chamber walls were also at 150F, is that the walls would themselves be radiating more power out, so there would be less heat transfer (in that case 0). It is NOT, as you assert, because the heat source would be using less power. That's false, by the S-B equation. Its power output remains the same because (Spencer's stipulation) the power input remains the same. The reason my solution does not violate conservation of energy, is that the power consumption of the chamber wall is allowed to vary. THAT is where the change takes place, not at the heat source. Again, this is a stipulation of Spencer's challenge. Once again: power out of heat source remains constant, because P = (emissivity) * (S-B constant) * T^4. There is nothing in these conditions that changes this at all. Therefore, BECAUSE the power out and power in at the heat source remain constant, so does the temperature. It's all in that one little equation. [Jane Q. Public, 2014-09-15]

Once again, no. Draw 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

"Power in" has to include the radiative power passing in through the boundary. Otherwise energy isn't conserved, because power in = power out through any boundary where nothing inside that boundary is changing with time.

... EVEN IF we accepted your idea that the "electrical" power required to be input to the heat source is dependent on the temperature difference between the heat source and chamber wall (a violation of the S-B law), you still contradict yourself because your answer of a hotter heat source would still then require MORE power, because the difference is greater. But that is not allowed by the stated conditions of the experiment, and you keep glossing over that simple check of your own work which proves it wrong. So no matter how you cut it, your answer is wrong, by your own rules. ... [Jane Q. Public, 2014-09-15]

Once again, no. I've already shown that the electrical power in my solution remains constant.

Once again, that's because I'm correctly applying the principle of conservation of energy to determine the electrical heating power.

It seems like we can't agree that "power in" includes the radiative power passing in through a boundary around the heat source. Is that because you disagree that power in = power out through any boundary where nothing inside that boundary is changing with time? Or is it because you disagree that the radiative power from the chamber walls passes in through a boundary around the heat source?

The REASON there would not be as great a power DIFFERENCE if the chamber walls were also at 150F, is that the walls would themselves be radiating more power out, so there would be less heat transfer (in that case 0). It is NOT, as you assert, because the heat source would be using less power. ... [Jane Q. Public, 2014-09-15]

That's absurd. A 150F plate surrounded by 150F chamber walls wouldn't need an electrical heater at all. Period. The electrical heating power would be exactly zero. Maybe you're mistaking "electrical heating power" with "radiative power out"? Or maybe you're missing half the equation necessary to calculate the required electrical heating power, and it's leading you to bizarre conclusions?

You failed to demonstrate that more than one of these many annual nonfatal injuries involves a penis being shot off.

Most US gun owners don't have good enough aim to shoot off their own penises. That's why they need semi-automatic weapons. It raises the odds of being able to actually hit that tiny thing.

Pathology? There is nothing pathological about a person wanting to employ all self-defensive measures to secure life or liberty

Gun ownership in the US has very little to do with "life or liberty". Be honest with yourself. If it was really about protecting your "life or liberty", you wouldn't have clown shows like this.

http://www.rawstory.com/rs/blo...

There is nothing invalid about using defensive uses of guns by police and against animals, since if there were no gun available

Yes, there is. No gun control proposal in the US has suggested taking guns away from law enforcement, the military or people who live where there are wildlife attacks. It's completely invalid. The technical term is "red herring". Look it up. Lott's book was about civilian ownership of guns. Kleck's work was designed to support civilian ownership of guns and has been used to attack all gun control laws. What's worse, his sloppy work and broad assumptions were used by the Supreme Court in the Heller decision, which began this entire notion of the Second Amendment being about civilian ownership of guns. Remember, until the '80s, there were no legal scholars who believed in this absolutist notion of the Second Amendment. Even Robert Bork, the sainted patron of the modern conservative, believed the Second Amendment did not apply to a right of every civilian to own (not to mention carry) a gun.

This entire argument is an artifact of Edwin Meese, the NRA and the Reagan Administration. There was a time when the NRA's literature quoted the entire Second Amendment, including the militia clause. Now, the quote above their headquarters door leaves that entire clause out. People who act like this so-called "right" goes back to the founding fathers are dizzy.

Let's assume Lott's figure is in the ball park just for argument sake.

No. Why should we do that when we know it's an imaginary number.

Your assertion that the U.S. may be the most lawless country in the world is ludicrous.

That's not my assertion, it's Lott's. His results imply that many hundreds of thousands of murders should have been occurring when a private gun was not available for protection. Yet guns are rarely carried, less than a third of adult Americans personally own guns, and only 27,000 homicides occurred in 1992. He assumes that there were 2.5 million attempted crimes that were thwarted by gun ownership. If that's true, and without those guns those crimes would have occurred, it would make the United States the most lawless country in the world. Do the math yourself. Assume for a moment that gun ownership is banned. Add 2.5 million to the crime statistics. That would just about triple the crime rate in the US.

Secondly, my neighbor travels to South America regularly and used to live in Argentina.

Do you know what "anecdotal" means? I lived in Sao Paolo when my wife was doing a math fellowship at a university there. The crime statistics in Brasil are about 30% higher than the US. Not double, not triple.

So as far as i'm concerned I'd believe him before believing your generalization.

What generalization? I cited a list of researchers and their studies that have refuted Lott. Are you going to believe your neighbor over published studies, too?

After all, you could turn the cops laser right back at him...