Jane's "interest" in that NAS report evaporated after I showed that Jane had been fooled by "Steven Goddard" once again. So let's return to Jane's confusion about basic thermodynamics.

But net radiative power out of a boundary around the source = "radiative power out" minus "radiative power in", so the equation Jane just described also says:

NO!!!!! As I have explained to you innumerable times now, you can also consider your heat source, by itself, that "sphere". The only NET radiative power out comes from the electrical power in. Further, the cooler walls do not contribute any of that NET power out. That's what net means. [Jane Q. Public, 2014-12-16]

I've already pointed out that Jane's hopelessly confused about the word "net", but that's just one of the mistakes Jane packed into these few sentences.

Jane's also wrong to imply that energy conservation across one choice of boundary could somehow contradict energy conservation across another boundary choice. That's impossible. Many boundary choices are inconvenient but they all have to be consistent. Otherwise, how could we possibly tell which boundary choice was correct?

So Jane can't object to the simple energy conservation equation I derived by claiming that some other boundary choice would somehow contradict my equation. That's completely impossible, and if Jane doesn't understand that point then he should learn about conservation of energy: example (backup), example (backup), example (backup).

As you can tell after reading those introductions, here's how to apply conservation of energy. 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 through any boundary where nothing inside is changing:

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

I put the boundary around the heat source so the boundary is in vacuum. That's because radiation can't travel through opaque solids like the heat source. So the only way to obtain an energy conservation equation with radiative terms is to place the boundary around the heat source.

For example, I calculated the enclosing shell's inner temperature by drawing the boundary within the enclosing shell. This boundary was inside aluminum, so heat transfer through it was by thermal conduction, not radiation. Notice that even this boundary choice leads to a conduction equation where electrical heating power depends on the cooler chamber wall temperature. That's because all boundary choices have to be consistent. They can't contradict each other unless one of them is wrong.

After I asked Jane to explain exactly where his boundary would be drawn, Jane replied:

... You can draw the boundary right around the heat source. Electric power comes in, radiative power goes out. There is no contradiction, and no inconsistency. ... [Jane Q. Public, 2014-09-15]

Nonsense. I've repeatedly explained that my boundary is drawn around the heat source, so it's in vacuum and therefore contains radiative terms both for radiation going out and radiation going in.

Choosing to put the boundary somewhere else, like inside the heat source, leads to an energy conservation equation with conduction rather than radiative terms. But even those conduction equations agree that electrical heating power depends on the cooler chamber wall temperature. They can't contradict each other. Putting the boundary somewhere else might be inconvenient, but it couldn't possibly contradict the fact that electrical heating power depends on the cooler chamber wall temperature.

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

Once again, Jane's wrong. There is literally no choice of boundary which will lead to his absurd equation. Once again, it really sounds like Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a fancy way of saying "electrical heating power".

At least, that's the most charitable explanation. Once again, I'm trying to rule out less charitable explanations like the disturbing possibility that Jane isn't honestly confused about basic thermodynamics. Maybe Jane/Lonny Eachus has simply betrayed humanity by deliberately spreading civilization-paralyzing misinformation.

Jane/Lonny Eachus could help convince posterity that he was just honestly confused by thinking carefully about conservation of energy, explaining exactly where his boundary lies, and carefully listing all the power going in and out of that boundary.

Or Jane/Lonny Eachus could help convince posterity that he's betrayed humanity by continuing to spread civilization-paralyzing misinformation.

Jane's "interest" in that NAS report evaporated after I showed that Jane had been fooled by "Steven Goddard" once again. So let's return to Jane's confusion about basic thermodynamics.

But net radiative power out of a boundary around the source = "radiative power out" minus "radiative power in", so the equation Jane just described also says:

NO!!!!! As I have explained to you innumerable times now, you can also consider your heat source, by itself, that "sphere". The only NET radiative power out comes from the electrical power in. Further, the cooler walls do not contribute any of that NET power out. That's what net means. [Jane Q. Public, 2014-12-16]

I've already pointed out that Jane's hopelessly confused about the word "net", but that's just one of the mistakes Jane packed into these few sentences.

Jane's also wrong to imply that energy conservation across one choice of boundary could somehow contradict energy conservation across another boundary choice. That's impossible. Many boundary choices are inconvenient but they all have to be consistent. Otherwise, how could we possibly tell which boundary choice was correct?

So Jane can't object to the simple energy conservation equation I derived by claiming that some other boundary choice would somehow contradict my equation. That's completely impossible, and if Jane doesn't understand that point then he should learn about conservation of energy: example (backup), example (backup), example (backup).

As you can tell after reading those introductions, here's how to apply conservation of energy. 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 through any boundary where nothing inside is changing:

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

I put the boundary around the heat source so the boundary is in vacuum. That's because radiation can't travel through opaque solids like the heat source. So the only way to obtain an energy conservation equation with radiative terms is to place the boundary around the heat source.

For example, I calculated the enclosing shell's inner temperature by drawing the boundary within the enclosing shell. This boundary was inside aluminum, so heat transfer through it was by thermal conduction, not radiation. Notice that even this boundary choice leads to a conduction equation where electrical heating power depends on the cooler chamber wall temperature. That's because all boundary choices have to be consistent. They can't contradict each other unless one of them is wrong.

After I asked Jane to explain exactly where his boundary would be drawn, Jane replied:

... You can draw the boundary right around the heat source. Electric power comes in, radiative power goes out. There is no contradiction, and no inconsistency. ... [Jane Q. Public, 2014-09-15]

Nonsense. I've repeatedly explained that my boundary is drawn around the heat source, so it's in vacuum and therefore contains radiative terms both for radiation going out and radiation going in.

Choosing to put the boundary somewhere else, like inside the heat source, leads to an energy conservation equation with conduction rather than radiative terms. But even those conduction equations agree that electrical heating power depends on the cooler chamber wall temperature. They can't contradict each other. Putting the boundary somewhere else might be inconvenient, but it couldn't possibly contradict the fact that electrical heating power depends on the cooler chamber wall temperature.

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

Once again, Jane's wrong. There is literally no choice of boundary which will lead to his absurd equation. Once again, it really sounds like Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a fancy way of saying "electrical heating power".

At least, that's the most charitable explanation. Once again, I'm trying to rule out less charitable explanations like the disturbing possibility that Jane isn't honestly confused about basic thermodynamics. Maybe Jane/Lonny Eachus has simply betrayed humanity by deliberately spreading civilization-paralyzing misinformation.

Jane/Lonny Eachus could help convince posterity that he was just honestly confused by thinking carefully about conservation of energy, explaining exactly where his boundary lies, and carefully listing all the power going in and out of that boundary.

Or Jane/Lonny Eachus could help convince posterity that he's betrayed humanity by continuing to spread civilization-paralyzing misinformation.

It has been well known that all Corporate IT security is a complete joke. CIO refuses to spend the money on it, COO refuses to make users actually follow real security procedures, and the CFO loves the "it wont happen to us" line that means they will not have to actually spend money on real IT security.

This is not new, I'm just glad that it's happening in a very public way so that maybe the worthless executives out there will actually listen to their IT experts about the fact that we NEED to spend the money to try and keep the bad guys out.

The old guy there was trying to get internet to the island and they threw him in jail. Let's start with forcing the Cuban Government to take bi-polar meds first?

I find it stupid. My problem is, I can't quite figure out who is being stupid here, except that it doesn't seem to have much to do with N Korea, if I'm any judge. They may be idiots, but they have been clever enough to hold on to power for decades in the most astonishingly ridiculous circumstances - a bit like the goings-on in the declining Roman empire. But it clearly doesn't add up, the idea that North Korea are somehow able to threaten the US into submission. I would be very interested in knowing the actual truth of the matter.

Not saying there is evidence it has happened. Saying it is a great idea and should have happened.

Have you ever retired a riddle by mistake?

The Federal numbers are an average for cars that cost $500,000 to $25,000 my 2007 civic will lose less than $3.00 for the 3000 miles added to it, it's already at the bottom of the curve and even adding 10,000 miles will not change it's "resale value" that has no real meaning as I dont intend to sell it.

And "major repairs" don't come from miles, they come from abuse and lack of proper maintenance.

Now my Ferrari F40, that would have a much higher depreciation for those miles.

Lol, in Gary, you're at much higher risk from industrial pollution than you are from random hooligan violence on the street. That shit gets into your body every day, all day long. You can't shoot it back. And it's completely legal.

Exactly.

And if NK wanted to REALLY threaten Sony; they'd just assassinate Seth Rogan. Seems like the easiest and most sensible thing to do. Seriously. The rest of this just makes no fucking sense at all.

Not really, just keep a low profile long enough to make it to Mexico.u.

Yup, and they survived so they said "plausible" but if you read the after notes they both firmly believe they absolutely survived.

I think, if you actually read what the GP wrote, you'll see that he is expressing the same opinion. But let me tell you about how it is in my home country, Denmark:

- Nobody carries fire arms, except some criminals. In fact, most police officers aren't armed either.
- Gun shots are being fired so rarely that it makes the headlines when it happens. I don't actually recall last time that happened.
- School massacres? What is that?

In fact, one can argue that since nobody carries firearms, even the criminals don't feel they have to; they are not likely to be shot when they are 'at work'. You know, it isn't because Danes are particularly good-natured, or because we are a homogenous society; it's just that no firearms means less risk of gun related violence. It may be that you prefer to pay the price for everybody having high-powered guns, but if you argue that it somehow makes your country safer, you'll just end up looking silly. Again, you may prefer looking silly to the truth, but hey, that's your call.