... 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.

You're right: if you change what it says by deleting some of the words, then it says something different.

In the next sentence, it says in particular what you're not allowed to use or run, including proxy services.

Use or run: It's not merely that you're not allowed to run proxy services: you're not allowed to use them, either.

If that's stupid-- well, how about that.

As I said: the interpretation of this text could be ambiguous. You could do the lawyer thing and claim to interpret it the way you say. But the clear straightforward text is: proxies are listed on the list of things you are specifically not allowed to use or run.

... 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?

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.

PS: Bad form to reply to own post but I'd also like to say I agree with your post, those consideration are a matter of due-diligence in my mind.

Of course you will also want to apply the same standards to the claims made by those upholding the status quo. After all, the FF industry is one of the most powerful economic entities on the planet, it has a lot more power and wealth than Gore, they are at least an order of magnitude higher in assets than the clean energy industry as a whole.

We have already seen the US senate abused in a failed attempt to discredit a single scientific paper. What I would like to see now is a repeat of the "tobacco trials" for the coal industry and their pet politicians (yes senator Inhofe, I'm looking at you)..

Another way of looking at it is that Gore puts his money where his mouth is, and considering the profits are fed back into his educational charity it's hard to see how it has given him more power and wealth than (say) a post-political career as a corporate lobbyist. For the most part I find American's in particular are generally for/against his charitable work based not on the contents of his documentaries, but on the perceived colour of his politics. The rest of the world don't really know him as as a politician, and are therefore less inclined to instinctively "shoot the messenger".

Sigh, the scientific community almost unanimously came out of the lab to praise the documentary because they felt it was a "bloody accurate" representation of their work.

Agree, now if you do some fact checking you will find the vast majority of climate scientists have already come out of their labs to loudly defend Gore's work, I'm not sure what your reading/viewing habits are, but you obviously missed the last 10yrs of debate, so the question is now - what will you tell your kids? - Can you set a good example by demonstrating a true skeptic changes his mind when confronted with inconvenient facts, or will misplaced pride take you down the creationist road?

Increasing seasonal sea ice in Antarctica is not "contra-evidence", it's a prediction that most models have been making for over 20yrs now, the mechanism that causes the counter intuitive result is well understood. So called "skeptics" are flogging a dead horse in their attempts to cite it as some sort of "smoking gun" that climate scientists are attempting to hide. The often intentionally misleading claim is ranked at #10 on skeptical sciences list of most popular climate myths.

As for Al Gore, any internet idiot can play "gottcha science" by taking words out of context and deliberately misinterpreting them. However the scientists who were lead authors of the IPCC reports that Gore's documentary was based on gave it a good review for it's representation of the report. Of course there were minor errors, and yes, the scientists pointed them out. The reason Gore shared the Nobel prize with the IPCC is that he put the IPCC's monumental lit-review effort squarely at the center of public policy debate.

Useful idiots? - As someone who has followed climate science with interest since the late 70's, Gore's documentary was an excellent (but imperfect) explanation of the science and it's real world consequences. It's a shame so many slashdotters mindlessly join in when the Gore bashing starts, he's the only well educated geek that has come close to sitting in the whitehouse for a very long time. History will admire his charitable public education efforts, even if most american's currently do not.

Disclaimer: I've been well known on slashdot for commenting on climate related stories for around 15yrs now, I'm not and have never been an "AGW advocate", I'm a science advocate.

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?

Companies have been paying the post office to shove stuff in my mailbox for years. That actually causes physical annoyance, as I have to shovel it into the recycle bin and then toss it. Then there are those crazy people who hand out free samples on the street. I don't have to take it, but I still have to see them.

Whoever tagged this "first world problems" was dead on.

...that the Earth is flat.

"It stands to reason", "it just makes sense", "it's common sense"... these are not just not arguments, they are anti-arguments: anyone using them is saying loudly and clearly "I have nothing to contribute to this discussion but here's some noise to dilute the signal."

Any time you find yourself offering an opinion based only on your imagination, please don't. Get some data, learn some modelling, do some statistics before you speak.

Philosophers attempted to understand the world for thousands of years based on what "just makes sense" and failed completely and utterly. After three hundred years of scientists showing us a better way--and showing that what "stands to reason" has absolutely nothing at all to do with the way the world actually is--there is really very little excuse for continuing to promulgate this erroneous and basically useless way of knowing.

Here is the text of what's forbidden, from TFA. Note the bold face on the word use (bold is from the original):

use or run dedicated, stand-alone equipment or servers from the Premises that provide network content or any other services to anyone outside of your Premises local area network (“PremisesLAN”), also commonly referred to as public services or servers. Examples of prohibited equipment and servers include, but are not limited to, email, web hosting, file sharing, and proxy services and servers;

Agreed, the interpretation of this text could be ambiguous. The straightforward reading, however, is that it is forbidden to use proxy services. You're also not allowed to run them, but that's specified separately.

This doesn't surprise me.

Personally, I prefer to use the bus and catch up my reading when I go into town. Otherwise, I walk to the shops and back.

You are right about the rain that falls on the ocean but I don't see how you're right when the rain falls on the land.

There is a lot more (About 3 times) area of ocean as there is land. And, as pointed out elsewhere, bunker oil is normally not burned until you're well out to sea, for precisely this reason. It's a perfectly good reason. Which is already covered.