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Comment: Re:COBOL: Why the hate? (Score 2) 220

by Jane Q. Public (#47924493) Attached to: College Students: Want To Earn More? Take a COBOL Class

However, since if it is still being used, then it still has some capability that is not available in other solutions.

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.

Comment: Re:The UK Cobol Climate Is Very Different (Score 4, Insightful) 220

by Jane Q. Public (#47924475) Attached to: College Students: Want To Earn More? Take a COBOL Class

Every professional workplace has an expectation of a formal atire.

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.

Comment: Re:Lifetime at 16nm? (Score 1) 56

by Jane Q. Public (#47924301) Attached to: Micron Releases 16nm-Process SSDs With Dynamic Flash Programming

seems like the average life expectancy of SSDs are well beyond the needs of most people at the moment, unless you're doing some serious content creation with massive amounts of read/writes.

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.

Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by Jane Q. Public (#47914835) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting
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.

Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by Jane Q. Public (#47914825) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting
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.

Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by Jane Q. Public (#47914807) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

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.

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

Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by Jane Q. Public (#47914803) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

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?

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.

Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by Jane Q. Public (#47914693) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting
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.

Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by Jane Q. Public (#47914669) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

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.

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.

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.

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.

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

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.

Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by Jane Q. Public (#47913945) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

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.

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.

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.

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.

Once again, no. Draw a boundary around the heat source: power in = electrical heating power + radiative power in from the chamber walls

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.

Comment: Re:Time for new terminology (Score 0) 590

by Jane Q. Public (#47913815) Attached to: Extent of Antarctic Sea Ice Reaches Record Levels
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.

Comment: Re:This may be the way to escape from Comcast (Score 2) 415

by Jane Q. Public (#47910443) Attached to: Comcast Allegedly Asking Customers to Stop Using Tor

In the end, you signed a contract and are legally bound to continue to pay for almost any type of service inturruption.

Except that I didn't. When my cable was installed I signed a small receipt acknowledging that the tech had been there. I signed no contract.

That might have been an oversight on their part, but that doesn't matter.

Further, the KIND of contract that Comcast has customers sign is known in the legal industry as a "contract of adhesion". What that means is that it was a non-negotiable, take-it-or-leave-it "contract". The problem being that contract law assumes that every party is free to negotiate before signing.

So in many genuine, legal senses of the term, it's not a "real" contract anyway, and honest judges are required in principle to view them "with a jaundiced eye", and lean toward the customer when a dispute arises.

I'm not saying all judges are honest enough to do that, but they're supposed to.

Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by Jane Q. Public (#47909703) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting
And one more thing I would like to make very clear:

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.

Comment: Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by Jane Q. Public (#47909379) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting
In fact let's just face this directly, with no mincing of words:

It's also adorable that Jane keeps ignoring the fact that his "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.

We are not AT thermal equilibrium, so that is a ridiculous straw-man argument.

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.

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