... 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?

The pessimism and dystopia in sci-fi doesn't come from a lack of research resources on engineering and science. It mainly comes from literary fashion.

If the fashion with editors is bleak, pessimistic, dystopian stories, then that's what readers will see on the bookshelves and in the magazines, and authors who want to see their work in print will color their stories accordingly. If you want to see more stories with a can-do, optimistic spirit, then you need to start a magazine or publisher with a policy of favoring such manuscripts. If there's an audience for such stories it's bound to be feasible. There a thousand serious sci-fi writers for every published one; most of them dreadful it is true, but there are sure to be a handful who write the good old stuff, and write it reasonably well.

A secondary problem is that misery provides many things that a writer needs in a story. Tolstoy once famously wrote, "Happy families are all alike; every unhappy family is unhappy in its own way." I actually Tolstoy had it backwards; there are many kinds of happy families. Dysfunctions on the other hand tends to fall into a small number of depressingly recognizable patterns. The problem with functional families from an author's standpoint is that they don't automatically provide something that he needs for his stories: conflict. Similarly a dystopian society is a rich source of conflicts, obstacles and color, as the author of Snow Crash must surely realize. Miserable people in a miserable setting are simply easier to write about.

I recently went on a reading jag of sci-fi from the 30s and 40s, and when I happened to watch a screwball comedy movie ("His Girl Friday") from the same era, I had an epiphany: the worlds of the sci-fi story and the 1940s comedy were more like each other than they were like our present world. The role of women and men; the prevalence of religious belief, the kinds of jobs people did, what they did in their spare time, the future of 1940 looked an awful lot like 1940.

When we write about the future, we don't write about a *plausible* future. We write about a future world which is like the present or some familiar historical epoch (e.g. Roman Empire), with conscious additions and deletions. I think a third reason may be our pessimism about our present and cynicism about the past. Which brings us right back to literary fashion.

That actually makes sense.

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.

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?

(store-comments comment /dev/null)

"We might stop worshiping veterans and start questioning if all the wars we're in are necessary "
1. It is respecting veterans. They do not decide which wars are just and which are not the voters and elected officials do.

Maybe but isn't a great thing that we have a peace loving president in the Whitehouse.....

I can shoot you in the head and kill you but I can not just intentionally blind you?

Actually it seems like a simple enough technical problem. When you go to fire the first burst is a range finder burst and then you set the power for the range. Of course this would all be done by the weapon and not the user.

"ad hominem" is not always a logical fallacy if the person expressing the opinion is some how has value added to because of position or expertise and is applicable when issues involving morality or ethics.
Musician that posts material that is offensive to a large segment of the population complains about the , tastefulness, morality and or ethics of getting a free album from popular band. That does seem to fit a valid use for an ad hominem based reply.
 

What negative social actions has U2 done?
It would be more like what if George Strait album showed up on my phone. A musical act that I do not really like but has done nothing extremely annoying or offensive.
I would just delete it and by on my way.

Put them on YouTube, DropBox and so on.

Maybe they are still hip and cool because.
1. They still have lots of fans.
2. They still get lots of airplay.
3. All the social causes that Bono and the other members of the band are involved in like Amnesty International.

Sure they are not cool and hip like Tyler The Creator with content and actions that push for social change like Homophobia and Violence towards women but they still have some fans.