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.

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.

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.

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.

No, they don't. This is a statement made by someone about ready to REtire.

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.

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.

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.

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.

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

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.

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.

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.

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.

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.

Are you REALLY the moron you make yourself out to be? NET radiation from a cooler surface that passes the boundary is reflected, transmitted, or scattered and passes right back out through the boundary. This is a corollary of the Stefan-Boltzmann radiation law, which states that NET heat transfer is always from hotter to cooler.

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.

And again: by that same law, it just passes right back out again because the same NET amount of radiative power that crosses the boundary and intercepts the smaller sphere is either reflected, transmitted, or scattered. (Since we are discussing diffuse gray bodies here, we can consider it all reflected or scattered because there is no transmissivity.) The radiation that crosses the boundary that does not strike the smaller sphere due to view factor also just passes right back out. You are ignoring (e*s) * (Ta^4 - Tb^4). Anything other than what I described does not add up.

Just NO. Net heat transfer is ALL from hotter to colder, by (e*s) * (Ta^4 - Tb^4).

Let me put it another way: we can easily show how you have gotten your thermodynamics backward by referring to a question you asked earlier. 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).

This clearly illustrates your ass-backward thermodynamic thinking. The radiative power output of the heat source does not change due to the temperature of the walls. At all. The only thing that changes as the wall temperature changes is the heat transfer, which would lessen as you brought up the temperature of the walls. But that isn't because the heat source is using less power, it is because you are putting more power into raising the wall temperature. You are creating a more thermodynamically energetic environment, and that requires power.

Just like your other arguments: you invent power in out of thin air, and claim you can do that because it's "moving" in the opposite direction in which heat transfer is actually taking place.

You are giving physicists a bad name, and I repeat that I am going to show this to all the world to see.

GISS is precisely the dataset that has been accused of the the most egregious "adjustments".

Further, it was recently found that GISS was improperly averaging in "missing" data over a period of years, which they admitted to about 2 months ago.

It is interesting that the historical HCN data disagree quite a bit with the modern versions of the data sets.

Actually NASA (or was it NOAA?) changed their tune again and are saying it was 1937.

Gotta keep up with this stuff, man.

The raw, unadjusted temperature records always have said 1937. It's the adjustments that are questionable, not the historical record.