VB6 was well-organized and coherent (but not entirely object-oriented).
But.NET, when it came around, was an attempt to do 2 things at the same time: [1] create a common underpinning (bytecode runtime) for all their IDE languages, and [2] insert object-orientation at the same time.
.NET was a mess. I liked VB6, as incomplete as it was, but.NET felt like a random conglomeration of just "stuff" thrown together to make it web-compatible. And I really hated that even when you created a pre-defined web object in.NET, you still had to manually define actions that should have been defaults for any such object. It is just plain weird.
In my personal opinion, having used both: VB6 was a great product for its time..NET was made to be a successor to it, but never quite made the grade.
Yes and no.
The whole work from home thing is becoming less common. Agile development is doing a good job killing it.
Utter nonsense. It was Agile Development that STARTED my work-at-home. Where did you get the idea that they are incompatible? In fact, it is Agile that made the whole thing possible.
When a hard drive fails, it is almost always the electronics or the bearings. The interface boards can be replaced, leaving the data on the drive intact. When bearings sieze, it is usually possible to free them up long enough to recover the data. As I mentioned before: I know because I've done it.
The only truly permanent, unrecoverable error on a hard drive is a catastrophic head crash, and those are extremely rare. But they do happen. I opened one up once to try to recover a guy's data and the surface of one platter was literally grooved by one head, which prevented the whole head assembly from moving properly. But because it tried, it was actually a series of concentric grooves.
It was completely toast. But as I say: this is extremely rare. In almost all cases it is possible to recover the vast majority of information from a hard disk that has failed.
Anandtech disagrees. Techreport. So, in fact, do huge numbers of user reports which suggest that SSDs really do last a long time.
This is not "disagreement". I didn't claim they don't last a long time. What I stated was that the claims of average lifetime have tended to be exaggerated. They can still last a long time.
I have seen this happen, but its not due to endurance of the flash cells but on the quality of the firmware / controller.
Absolute nonsense, and the manufacturers themselves will tell you. The issue *IS* endurance of the flash cells, and the tremendous improvement in firmware is a direct result of this limitation. The manufacturers have expended enormous effort to produce management schemes that mitigate the short lives of the cells, which in fact shortened even more from SLC to MLC.
what does it look like for an SSD or HDD or CPU or RAM to fail "not totally"? You get most of your bits back? All tech generally tends to fail catastrophically.
As someone who has been in the industry for decades, I can tell you this is COMPLETE bullshit. When a hard drive fails, it is almost always possible to recover most of the data. And almost all of that, for that matter. I have done it myself, more than once. Recovered everything on the drive but one or a few files.
But when an SSD really goes, it goes. Even sophisticated forensics are usually incapable of retrieving anything.
We've been over this before. I've already proved you wrong, mathematically, logically, and thermodynamically.
The fact that your "global warming" religion will not let you accept the reality of the Stefan-Boltzmann radiation law is not my problem. But you have sure as hell tried hard to make it everyone else's problem.
Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body.
No NET radiative energy. I did not claim "none at all", and I have repeatedly pointed this out to you. Just no NET transfer from cooler to warmer. This is a fundamental requirement of thermodynamics. It amazes me that you continue to deny this, no matter how you try to couch it in different terms.
You're either incompetent or a liar. As I said before: I don't know for sure which, but I strongly suspect the latter.
It's a done deal. You have been proved wrong. You have been owned. Your ranting means nothing.
I only replied on the off-chance that you really were ignorant and could be educated. But it seems that you are determined to promote your ignorance (or more likely: ignorant act and propaganda) to everyone else. So be it.
No more replies. You haven't earned any; you don't deserve any.
If Ta = Tb, no electrical heating power is required. But radiant power output of (a) doesn't change. So radiant power output can't be equal to electrical heating power. Using conservation of energy, can you write down an equation which yields the required electrical heating power given Ta and Tb?
If Ta=Tb, you're doing a different experiment. I've already stated that at that point, it requires no electrical heating power. But it's a straw-man for at least 2 reasons:
[1] it still requires the same amount of power, but once Ta=Tb, it can draw that power from the environment. Before that it can't, because Ta^4 - Tb^4 is a positive number so no net radiant energy is absorbed by (a) from (b). That means all the way up to the exact point thermal equilibrium is achieved, all radiant power is a result of electrical power, therefore the power input and power output are constant. It is not a "gradual" process.
And [2] because in Spencer's experiment, Ta=Tb doesn't happen.
No. Once again, in this experiment [archive.today] there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate...."
I have said nothing that contradicts this. Not only do I freely admit this, my calculations relied on that fact. I kept the power (and hence energy over time) input into the plate from the electric heater completely constant. Which we may freely do, since it was a stipulation of Spencer's experiment.
Jane's even stumbled across this point:
No, I didn't "stumble" over that point, YOU are stumbling over it. Everything changes at thermal equilibrium. The "heated" body is no longer warmer than its surroundings and can begin taking on energy from its surroundings. And it is not a "gradual" change: the Stefan-Boltmann law says a warmer body DOES NOT absorb net radiant energy from its surroundings. That only begins to happen at thermal equilibrium. BUT thermal equilibrium does not apply to this experiment, anywhere, at any time. This is just another straw-man argument. Which you are very good at, by the way. Not good enough to sucker me in, though.
Of course! That's why the variable Jane's holding constant isn't the electrical power supplied to the separate heat source. If Jane can realize that there's no need for a separate heat source if its environment were maintained at 150 degrees, why can't Jane see that his equation for required electrical power doesn't reflect this obvious fact?
Of course I realize that, and have all along. The error lies in your implication that this is a gradual change.
It isn't a gradual change. It's a result of Ta^4 - Tb^4 = 0. A transition from non-zero to 0.
That's the only reason. The transition between non-zero and zero is a profound change which affects everything, and there is nothing gradual about it. But it doesn't apply in this context. The surfaces are never at thermal equilibrium. And your assertion is only "obvious" if you're not a heat transfer engineer or a physicist, you pretender. Heat transfer is not a science of the obvious. Intuition (and, as pointed out before, "thermodynamic thinking") can easily lead you astray. The sign of the result is everything here.
If body (a) is warmer than body (b), Ta^4 - Tb^4 > 0, and net heat transfer is ONLY from (a) to (b).
If body (b) is brought up to the same temperature as (a), Ta^4 - Tb^4 = 0, and no net heat transfer takes place. Although radiant power output of (a) at that temperature doesn't change, as a corollary of that same law.
If body (a) is at a lower temperature than body (b), Ta^4 - Tb^4 < 0, which means there is net transfer of heat from (b) to (a).
The third condition is the ONLY one in which there is any input to (a) from its surroundings. But that condition never occurs in Spencer's experiment because the heat source is always hotter than its surroundings.
Knock off the BS. Time to admit you were wrong. I repeat: anything else is a violation of the Stefan-Boltzmann radiation law.
Instead of holding electrical heating power constant, Jane held the source's radiative power output constant. That held source temperature constant and forced electrical heating power to change.
No, that is not correct. You made assumptions that are, to be blunt, bullshit nonsense.
Since the emissivity for every object in our system is the same, power output is proportional to the T^4. Period. End of story.
Draw your boundary around the heat source. Power in = power out (your own principle). Therefore the power in is 41886.54 Watts, which is the power initially being radiated out.
SPENCER stipulated that this power is held constant. It wasn't my idea. It's a condition of the experiment.
By the Stefan-Boltzmann law, since the power in remains constant, then UNLESS power is taken up from some other source, the temperature will remain constant. This follows directly from the S-B radiation law, which you seem to be disputing.
Another requirement of the S-B law, and also of thermodynamics: since EVERY other object in the system is at a lower temperature than the heat source, NET heat transfer is in ONLY one direction: from hotter to colder.
Therefore, no energy is flowing "backward" to boost the output of the heat source.
Yet another fact that follows directly from the S-B law, is that nearby cooler bodies have zero effect on the output of the heat source. They don't "suck" power from it, nor (see above) do they "lend" power to it.
The only logical conclusion -- the only physically possible conclusion, unless you dispute the Stefan-Boltzmann radiation law, is that the heat source does not change temperature. Power out = power in, and is constant. Everything else is cooler, so it remains a constant. There is no further energy or power flowing "backward" the heat source.
The Stefan-Boltzmann law clearly shows that no NET radiation from cooler objects is absorbed; it is either transmitted, reflected, or scattered. Since these are diffuse gray bodies, they do not transmit. That leaves reflection and scattering. For our purposes, the net effect is that it is all reflected.
You are imagining some kind of power input to the heat source that doesn't exist. Further, if the heat source became even hotter, as you assert, it would require even MORE power, because as you say, power in = power out. That was YOUR assertion. Draw your boundary around the heat source itself. There is no net radiation absorbed from outside, and the supplied power remains constant.
It this whole "proof" of yours, I have shown where you have contradicted yourself at least 3 different ways.
Jane might wonder why he can't derive a single equation which works for all these cases.
I don't know where you get this idea, because I did. I used the S-B equation to find my solution. I used the textbook equations for heat transfer. Yes, I ignored area because the areas were so similar. But it was still a reasonably accurate approximation. I checked my work, and it wasn't off by more than a fraction of a percent.
But Jane can't even admit there's a difference between holding electrical heating power constant and holding the source's radiative power output constant.
Because there isn't any. Your own "boundary" principle says so. This isn't a matter of differential equations at this point. Do you think we're all idiots? Power in = power out. Your Newmann and Dirichlet boundary conditions are just more straw men. We don't need them to find the answer to this. Plain old algebra works just fine, because everything is at steady-state. So knock off the bullshit, because I see right through it, and so will the others I show this to.
Again, warming the chamber walls is like partially closing the drain on a bathtub where water is flowing in at a constant rate
There's a huge problem with Android device encryption.
Unlike Apple's usual forms of encryption, once an Android device is encrypted, it is not reversible. There is no way to UN-encrypt it, except to back up all your programs, flash your original unencrypted OS back to the phone, then restore the programs. And that requires unlocking and rooting the phone.
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.
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.
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.
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".
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.
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.
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.
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.
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.
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.
Slashdot Mirror
I've done both.
.NET, when it came around, was an attempt to do 2 things at the same time: [1] create a common underpinning (bytecode runtime) for all their IDE languages, and [2] insert object-orientation at the same time.
.NET was a mess. I liked VB6, as incomplete as it was, but .NET felt like a random conglomeration of just "stuff" thrown together to make it web-compatible. And I really hated that even when you created a pre-defined web object in .NET, you still had to manually define actions that should have been defaults for any such object. It is just plain weird.
.NET was made to be a successor to it, but never quite made the grade.
VB6 was well-organized and coherent (but not entirely object-oriented).
But
In my personal opinion, having used both: VB6 was a great product for its time.
Yes and no. The whole work from home thing is becoming less common. Agile development is doing a good job killing it.
Utter nonsense. It was Agile Development that STARTED my work-at-home. Where did you get the idea that they are incompatible? In fact, it is Agile that made the whole thing possible.
I meant to add:
When a hard drive fails, it is almost always the electronics or the bearings. The interface boards can be replaced, leaving the data on the drive intact. When bearings sieze, it is usually possible to free them up long enough to recover the data. As I mentioned before: I know because I've done it.
The only truly permanent, unrecoverable error on a hard drive is a catastrophic head crash, and those are extremely rare. But they do happen. I opened one up once to try to recover a guy's data and the surface of one platter was literally grooved by one head, which prevented the whole head assembly from moving properly. But because it tried, it was actually a series of concentric grooves.
It was completely toast. But as I say: this is extremely rare. In almost all cases it is possible to recover the vast majority of information from a hard disk that has failed.
Anandtech disagrees. Techreport. So, in fact, do huge numbers of user reports which suggest that SSDs really do last a long time.
This is not "disagreement". I didn't claim they don't last a long time. What I stated was that the claims of average lifetime have tended to be exaggerated. They can still last a long time.
I have seen this happen, but its not due to endurance of the flash cells but on the quality of the firmware / controller.
Absolute nonsense, and the manufacturers themselves will tell you. The issue *IS* endurance of the flash cells, and the tremendous improvement in firmware is a direct result of this limitation. The manufacturers have expended enormous effort to produce management schemes that mitigate the short lives of the cells, which in fact shortened even more from SLC to MLC.
what does it look like for an SSD or HDD or CPU or RAM to fail "not totally"? You get most of your bits back? All tech generally tends to fail catastrophically.
As someone who has been in the industry for decades, I can tell you this is COMPLETE bullshit. When a hard drive fails, it is almost always possible to recover most of the data. And almost all of that, for that matter. I have done it myself, more than once. Recovered everything on the drive but one or a few files.
But when an SSD really goes, it goes. Even sophisticated forensics are usually incapable of retrieving anything.
The fact that your "global warming" religion will not let you accept the reality of the Stefan-Boltzmann radiation law is not my problem. But you have sure as hell tried hard to make it everyone else's problem.
Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body.
No NET radiative energy. I did not claim "none at all", and I have repeatedly pointed this out to you. Just no NET transfer from cooler to warmer. This is a fundamental requirement of thermodynamics. It amazes me that you continue to deny this, no matter how you try to couch it in different terms.
You're either incompetent or a liar. As I said before: I don't know for sure which, but I strongly suspect the latter.
It's a done deal. You have been proved wrong. You have been owned. Your ranting means nothing.
I only replied on the off-chance that you really were ignorant and could be educated. But it seems that you are determined to promote your ignorance (or more likely: ignorant act and propaganda) to everyone else. So be it.
No more replies. You haven't earned any; you don't deserve any.
state-blessed corporation with heavy state support.
Yep. Socialist.
And yes, as sad as it may be, the same is true, to a lesser extent, of some US corporations these days.
The "Communist" party isn't Communist. It's just bad Socialism, just like every other regime in history that called itself Communist.
Just calling yourself that doesn't make it so.
If Ta = Tb, no electrical heating power is required. But radiant power output of (a) doesn't change. So radiant power output can't be equal to electrical heating power. Using conservation of energy, can you write down an equation which yields the required electrical heating power given Ta and Tb?
If Ta=Tb, you're doing a different experiment. I've already stated that at that point, it requires no electrical heating power. But it's a straw-man for at least 2 reasons:
[1] it still requires the same amount of power, but once Ta=Tb, it can draw that power from the environment. Before that it can't, because Ta^4 - Tb^4 is a positive number so no net radiant energy is absorbed by (a) from (b). That means all the way up to the exact point thermal equilibrium is achieved, all radiant power is a result of electrical power, therefore the power input and power output are constant. It is not a "gradual" process.
And [2] because in Spencer's experiment, Ta=Tb doesn't happen.
"These aren't the drones you're looking for."
No. Once again, in this experiment [archive.today] there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."
I have said nothing that contradicts this. Not only do I freely admit this, my calculations relied on that fact. I kept the power (and hence energy over time) input into the plate from the electric heater completely constant. Which we may freely do, since it was a stipulation of Spencer's experiment.
Jane's even stumbled across this point:
No, I didn't "stumble" over that point, YOU are stumbling over it. Everything changes at thermal equilibrium. The "heated" body is no longer warmer than its surroundings and can begin taking on energy from its surroundings. And it is not a "gradual" change: the Stefan-Boltmann law says a warmer body DOES NOT absorb net radiant energy from its surroundings. That only begins to happen at thermal equilibrium. BUT thermal equilibrium does not apply to this experiment, anywhere, at any time. This is just another straw-man argument. Which you are very good at, by the way. Not good enough to sucker me in, though.
Of course! That's why the variable Jane's holding constant isn't the electrical power supplied to the separate heat source. If Jane can realize that there's no need for a separate heat source if its environment were maintained at 150 degrees, why can't Jane see that his equation for required electrical power doesn't reflect this obvious fact?
Of course I realize that, and have all along. The error lies in your implication that this is a gradual change.
It isn't a gradual change. It's a result of Ta^4 - Tb^4 = 0. A transition from non-zero to 0.
That's the only reason. The transition between non-zero and zero is a profound change which affects everything, and there is nothing gradual about it. But it doesn't apply in this context. The surfaces are never at thermal equilibrium. And your assertion is only "obvious" if you're not a heat transfer engineer or a physicist, you pretender. Heat transfer is not a science of the obvious. Intuition (and, as pointed out before, "thermodynamic thinking") can easily lead you astray. The sign of the result is everything here.
If body (a) is warmer than body (b), Ta^4 - Tb^4 > 0, and net heat transfer is ONLY from (a) to (b).
If body (b) is brought up to the same temperature as (a), Ta^4 - Tb^4 = 0, and no net heat transfer takes place. Although radiant power output of (a) at that temperature doesn't change, as a corollary of that same law.
If body (a) is at a lower temperature than body (b), Ta^4 - Tb^4 < 0, which means there is net transfer of heat from (b) to (a).
The third condition is the ONLY one in which there is any input to (a) from its surroundings. But that condition never occurs in Spencer's experiment because the heat source is always hotter than its surroundings.
Knock off the BS. Time to admit you were wrong. I repeat: anything else is a violation of the Stefan-Boltzmann radiation law.
Instead of holding electrical heating power constant, Jane held the source's radiative power output constant. That held source temperature constant and forced electrical heating power to change.
No, that is not correct. You made assumptions that are, to be blunt, bullshit nonsense.
Since the emissivity for every object in our system is the same, power output is proportional to the T^4. Period. End of story.
Draw your boundary around the heat source. Power in = power out (your own principle). Therefore the power in is 41886.54 Watts, which is the power initially being radiated out.
SPENCER stipulated that this power is held constant. It wasn't my idea. It's a condition of the experiment.
By the Stefan-Boltzmann law, since the power in remains constant, then UNLESS power is taken up from some other source, the temperature will remain constant. This follows directly from the S-B radiation law, which you seem to be disputing.
Another requirement of the S-B law, and also of thermodynamics: since EVERY other object in the system is at a lower temperature than the heat source, NET heat transfer is in ONLY one direction: from hotter to colder.
Therefore, no energy is flowing "backward" to boost the output of the heat source.
Yet another fact that follows directly from the S-B law, is that nearby cooler bodies have zero effect on the output of the heat source. They don't "suck" power from it, nor (see above) do they "lend" power to it.
The only logical conclusion -- the only physically possible conclusion, unless you dispute the Stefan-Boltzmann radiation law, is that the heat source does not change temperature. Power out = power in, and is constant. Everything else is cooler, so it remains a constant. There is no further energy or power flowing "backward" the heat source.
The Stefan-Boltzmann law clearly shows that no NET radiation from cooler objects is absorbed; it is either transmitted, reflected, or scattered. Since these are diffuse gray bodies, they do not transmit. That leaves reflection and scattering. For our purposes, the net effect is that it is all reflected.
You are imagining some kind of power input to the heat source that doesn't exist. Further, if the heat source became even hotter, as you assert, it would require even MORE power, because as you say, power in = power out. That was YOUR assertion. Draw your boundary around the heat source itself. There is no net radiation absorbed from outside, and the supplied power remains constant.
It this whole "proof" of yours, I have shown where you have contradicted yourself at least 3 different ways.
Jane might wonder why he can't derive a single equation which works for all these cases.
I don't know where you get this idea, because I did. I used the S-B equation to find my solution. I used the textbook equations for heat transfer. Yes, I ignored area because the areas were so similar. But it was still a reasonably accurate approximation. I checked my work, and it wasn't off by more than a fraction of a percent.
But Jane can't even admit there's a difference between holding electrical heating power constant and holding the source's radiative power output constant.
Because there isn't any. Your own "boundary" principle says so. This isn't a matter of differential equations at this point. Do you think we're all idiots? Power in = power out. Your Newmann and Dirichlet boundary conditions are just more straw men. We don't need them to find the answer to this. Plain old algebra works just fine, because everything is at steady-state. So knock off the bullshit, because I see right through it, and so will the others I show this to.
Again, warming the chamber walls is like partially closing the drain on a bathtub where water is flowing in at a constant rate
Which is not only false (the S-B relation
There's a huge problem with Android device encryption.
Unlike Apple's usual forms of encryption, once an Android device is encrypted, it is not reversible. There is no way to UN-encrypt it, except to back up all your programs, flash your original unencrypted OS back to the phone, then restore the programs. And that requires unlocking and rooting the phone.
There are LOTS of problems caused by that.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.