All those lead thickness options are too thick by at least a factor of 5. Even if that NASA study is "retarded", RockDoctor just mentioned that Earth's atmosphere protects us with only ~10 tons/m^2. Since lead's density is 11 tons/m^3, a lead shield wouldn't have to be thicker than ~0.9 meter.
"Passive shielding is known to work. The Earth's atmosphere supplies about 10 t/m^2 of mass shielding and is very effective. Only half this much is needed to bring the dosage level of cosmic rays down to 0.5 rem/yr. In fact when calculations are made in the context of particular geometries, it is found that because many of the incident particles pass through walls at slanting angles a thickness of shield of 4.5 t/m^2 is sufficient."
Water could be an effective shield, and would be especially easy to apply and repair. Just melt it and let it freeze in place. That's how most of the lighthuggers in Revelation Space were shielded, as well as the starship in Songs of Distant Earth.
The only downsides I can think of would be the low tensile strength, so a water shield couldn't spin with a rotating ship, and the fact that if the ship overheats then its radiation shield sublimates away...
Centrifuges need to rotate no faster than 1 rpm to avoid inducing motion sickness. That's for long-term colonies, so maybe 2 or 3 rpm would be acceptable for astronauts selected for their resistance to motion sickness. Maybe even faster if they're hibernating the whole way. But regardless, the centrifuge would still have to be quite large.
If the centrifuge is inside the shielding, that makes the shield enormously bigger and heavier. Alternatively, only the hibernation/living chamber at the end of the centrifuge could be shielded. But that requires that the shielding mass be attached to the centrifuge, which vastly increases its required tensile strength. That's why the NASA study placed the colony's centrifuge inside a separate shield: if the shield rotates with the centrifuge then the centrifuge would have to be built out of carbon nanotubes. If the shield is separate then the centrifuge can be built out of aluminum.
I repeat: I have already answered these questions several times. You have no legitimate purpose in asking them again somewhere else. And yes, repeating your questions here after they have already been answered is ill behavior on your part. [Jane Q. Public, 2014-10-04]
Jane, you're still wrongly insisting that electrical heating power per square meter = (e*s)*T1^4. Once again, Jane's equation violates conservation of energy. That's why I'm trying to understand why you keep insisting it's correct. At first I thought you agreed that power in = power out, but that we only disagreed about which terms to include:
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".
Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]
Jane's statement originally made me think that Jane is reasoning like this:
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
But now it seems like our disagreement is even more fundamental:
I am not going to get drawn into an argument that you have already lost. I repeat that the equation you show is for HEAT TRANSFER, not "radiative power out". You are just plain wrong about that and any heat transfer textbook will you so.... [Jane Q. Public, 2014-10-03]
This objection is completely different than Jane's "A = A" objection above, which at least seemed to acknowledge that we should start with the principle of conservation of energy, where power in = power out. But now Jane even seems to dispute that starting point.
I'm starting to suspect that Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a fancy way of saying "electrical heating power". Is that how Jane "derived" his incorrect equation that electrical heating power per square meter = (e*s)*T1^4?
If so, that's kind of a boring mistake because "radiative power out" isn't just a fancy way of saying "electrical heating power". They're completely different. To find electrical heating power, Jane needs to use conservation of energy, where power in = power out. That results in a heat transfer equation, not just an equation for "radiative power out".
Jane, if you don't agree with the "power in" and "power out" that I've tried to glean from your rants, just fill in the following blanks like I did. It'll be much faster than accusing me of ill behavior.
Jane's power in = ?
Jane's power out = ?
Or, explain why we shouldn't start with the principle of conservation of energy which results in a heat transfer equation. Or, (more
We've been over this before, and you already know the answers I've given you. Stop being a grandstanding asshole. I don't have to keep repeating my answers every time you demand them. That's called ASSHOLE behavior, asshole. You have already seen my calculations and my answers to all these questions. By bringing them up and demanding them AGAIN in a different forum, you are advertising your own dishonesty. It didn't work. Don't worry, as I promised this will all be published when I find the time. [Jane Q. Public, 2014-10-03]
Jane, the answers you've given don't make any sense. That's why I'm asking you for a very simple equation describing the required electrical heating power. Again, filling in the following blanks would be be much faster than repeatedly calling me an asshole.
I never said we should ignore extrasolar particles. I was just showing that even using angel'o'sphere's assumption that the sun is the main hazard, the shielding mass decreases for a hibernating crew. In other words, I was defending you, Jane. Even though I can't be trusted to build a bridge over a creek.
But since you brought up those other arguments...
There is no reason to "guess" at my reasoning. I spelled it out quite clearly when we had our "argument" (which you lost). You do realize this is all going to be published, right? I warned you not just once or twice, but many times now. Every time you pull this kind of BS will be just another instance of widespread public knowledge of your dishonesty. [Jane Q. Public, 2014-10-03]
I have to guess at your reasoning because what you've said doesn't make any sense.
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".
Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]
I have to guess at what Jane meant by this, because it's not in equation form. In physics, statements in equation form are easier to analyze.
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
I am not going to get drawn into an argument that you have already lost. I repeat that the equation you show is for HEAT TRANSFER, not "radiative power out". You are just plain wrong about that and any heat transfer textbook will you so.... [Jane Q. Public, 2014-10-03]
Once again, to calculate "electrical heating power" you need to use a heat transfer equation which accounts for power in and power out. That's because power in = power out through any boundary where nothing inside is changing. Once again, the equation Jane's using is only valid for "radiative power out" which is completely different than "electrical heating power". That's why I'm starting with the principle of "conservation of energy" and trying to understand what Jane's saying, in equation form.
Jane, if you don't agree with the "power in" and "power out" that I've tried to glean from your rants, just fill in the following blanks like I did. It'll be much faster than accusing me of dishonesty, fraud, and libel.
There is no reason to "guess" at my reasoning. I spelled it out quite clearly when we had our "argument" (which you lost). You do realize this is all going to be published, right? I warned you not just once or twice, but many times now. Every time you pull this kind of BS will be just another instance of widespread public knowledge of your dishonesty. [Jane Q. Public, 2014-10-03]
I have to guess at your reasoning because what you've said doesn't make any sense.
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".
Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]
I have to guess at what Jane meant by this, because it's not in equation form. In physics, statements in equation form are easier to analyze.
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
I am not going to get drawn into an argument that you have already lost. I repeat that the equation you show is for HEAT TRANSFER, not "radiative power out". You are just plain wrong about that and any heat transfer textbook will you so.... [Jane Q. Public, 2014-10-03]
Once again, to calculate "electrical heating power" you need to use a heat transfer equation which accounts for power in and power out. That's because power in = power out through any boundary where nothing inside is changing. Once again, the equation Jane's using is only valid for "radiative power out" which is completely different than "electrical heating power". That's why I'm starting with the principle of "conservation of energy" and trying to understand what Jane's saying, in equation form.
Jane, if you don't agree with the "power in" and "power out" that I've tried to glean from your rants, just fill in the following blanks like I did. It'll be much faster than accusing me of dishonesty, fraud, and libel.
If the main hazard is the sun, that requires thicker shielding on the sunward side. Minimum shielding mass would then be obtained by putting 4.41 tons/m^2 on the sunward side, which given moon dust density equals a ~2.4 meter thick shield on the sunward side. If the people are perpendicular to the sun, that shield is heavier. The people are awake and moving around, that shield is much heavier.
A hibernating crew could be closely packed and aligned with their feet towards the sun, reducing the required shielding area and mass at constant thickness. That's because only the hibernation chamber would need to be shielded, not the entire ship.
NASA found that 441 grams/cm^2 of silicon dioxide (Moon dust) would be sufficient shielding, which equals 4.41 tons/m^2. Hibernation dangers and personal preference regarding books may vary, of course.
Reliable shielding isn't impossible. Shielding of 4.41 tons/m^2 is sufficient. Putting the crew in hibernation does reduce shielding because otherwise the entire back side of the spacecraft (at least) has to be covered with 4.41 tons/m^2 of shielding. In hibernation, the crew could be closely packed and aligned with their feet towards the sun, reducing the required shielding area and mass.
It's not a "black body" source, it's a "gray body" source, as per our agreement when this discussion first started. And I showed you my equations not just once but many times. You're just lying again. [Jane Q. Public, 2014-10-03]
Again, Jane's gray body equation has to reduce to the black body equation when emissivity = 1, so this is a way to check Jane's work. But since Jane seems convinced that checking his work is "lying" let's write down both equations simultaneously.
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Now use the Stefan-Boltzmann law to describe the radiative terms, one at a time. First for Jane's gray body:
Because "radiative power in from chamber walls" is emitted by graybody walls at temperature T4, the Stefan-Boltzmann law says:
gray electrical heating power + (e*s)*T4^4 = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Now for Jane's black body check:
Because "radiative power in from chamber walls" is emitted by blackbody walls at temperature T4, the Stefan-Boltzmann law says:
black electrical heating power + (s)*T4^4 = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane probably won't write down an equation describing electrical heating power for a blackbody source, so I'll try to guess at Jane's reasoning.
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".
Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]
Draw a boundary around the blackbody heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Charming. As I just explained, IR detectors don't have to depend on quantum effects. Classical mainstream physics allows a temperature-controlled source to detect IR from the cooler chamber walls as follows:
electricity = (e*s)*(T1^4 - T4^4)
If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).
If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.
If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.
If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.
If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.
... Further, I repeat for about the 100th time that I do not deny that some radiation is absorbed; but then it's just re-emitted. Sometimes, in a non-gray body, in a slightly different form. And ALL of that is straw-man irrelevancy, since no NET radiation absorption occurs from colder bodies to warm, which was the subject under discussion.... [Jane Q. Public, 2014-10-03]
If you don't deny that some radiation is absorbed, then it should be very easy to write down a simple equation describing the required electrical heating power (not the radiative power out) of a blackbody source.
I don't need to "agree" with you about anything. I've already demonstrated how TEXTBOOK PHYSICS proved you wrong. That doesn't require any kind of "agreement". I'm just wondering when you're going to stop the dishonesty and admit you were wrong. The whole world is going to see it soon anyway, so you might as well "come clean", as they say. [Jane Q. Public, 2014-10-03]
Jane, if we can't agree on the meaning of the term "NET", why are you still capitalizing the word "NET"? Screaming the word louder and louder is unlikely to be productive.
1. Can we agree that net heat transfer through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net heat transfer always contains terms in both directions?
3. Can we agree that just because "radiative power out" > "radiative power in", that doesn't mean "radiative power in" = 0?
If Jane answers "no" to any of those three yes/no questions... why?
I don't need to "agree" with you about anything. I've already demonstrated how TEXTBOOK PHYSICS proved you wrong. That doesn't require any kind of "agreement". I'm just wondering when you're going to stop the dishonesty and admit you were wrong. The whole world is going to see it soon anyway, so you might as well "come clean", as they say. [Jane Q. Public, 2014-10-03]
If you're so confident that you're right, why not prove it by taking a few seconds to write dow
... Almost Latour's entire thesis is that S-B law says net heat transfer is either 0 or in one direction, from the hotter area to the colder. If the roles are reversed, and the colder item becomes the hotter, then the sign changes and the net heat transfer is still only in one direction... from hotter to colder.... [Jane Q. Public, 2014-07-29]
... At no time in this experiment are the temperatures equal, so net heat transfer is always in one direction and only one direction.... [Jane Q. Public, 2014-09-04]
... There is heat transfer which is energy, which represents NET flow in one direction.... there IS a net, non-zero flow of energy (heat transfer) THROUGH that boundary in one direction from the hollow enclosing plate to the chamber wall. This is a net, non-zero quantity. [Jane Q. Public, 2014-09-08]
... According to the S-B equation itself, net heat transfer is either 0, or only in one direction. Yes, we are talking NET here.... [Jane Q. Public, 2014-09-10]
... 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.... [Jane Q. Public, 2014-09-15]
... 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.... [Jane Q. Public, 2014-09-19]
... When A is warmer than B, (Ta^4 - Tb^4) yields a positive number. Which means all NET radiative energy transfer goes from A to B. That is clearly indicated by the minus sign, and is further dictated by the Second Law of Thermodynamics. There is no NET energy going from B to A. Only when B is hotter than A does any NET energy transfer in the other direction.... [Jane Q. Public, 2014-10-01]
... You could not NOT understand it, unless you are 100% clueless about what the term NET means.... [Jane Q. Public, 2014-10-01]
It's beginning to seem like we disagree about the meaning of the term "NET".
1. Can we agree that net heat transfer through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net heat transfer always contains terms in both directions?
3. Can we agree that just because "radiative power out" > "radiative power in", that doesn't mean "radiative power in" = 0?
If we can agree on all those points, that's great. Maybe this will help Jane write down a simple equation describing the electrical heating power required t
That reference shows the object (i.e. chamber wall) temperature has an effect on the temperature controlled cavity (i.e. source). Which Jane denies:
... Radiation from the cooler walls has no effect on the heat source whatsoever. This is a basic requirement of thermodynamics!... [Jane Q. Public, 2014-09-21]
No, that's Sky Dragon Slayer nonsense. If radiation from the cooler walls really had no effect on the heat source whatsoever, the IR thermometer wouldn't work because the cooler object temperature would have no effect on the temperature controlled cavity whatsoever.
When the source temperature is held constant, its required electrical heating power is an IR thermometer.
Here's one way to see that: draw a boundary around a heated aluminum source. It's heated by constant electrical power flowing in. Aluminum cold walls at some unknown temperature T4 also radiate power in. The source at 150F (T1 = 338.7K) radiates power out. At steady-state, power in = power out. Using the equation which neglects reflections:
electricity = (e*s)*(T1^4 - T4^4)
If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).
If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.
If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.
If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.
If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.
That's why when the source temperature is held constant, its required electrical heating power is an IR thermometer. At least, it's a thermometer when using mainstream physics. But Jane's equation is:
electricity = (e*s)*T1^4 (Jane's equation)
Since Jane's equation doesn't depend on the chamber wall temperature, uncooled IR detectors can't see cooler objects in Janeland. And we couldn't possibly have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How? This must be inexplicable to Slayers who are brainwashed into believing that:
... 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.... [Jane Q. Public, 2014-09-20]
No. Again, mainstream physics shows that electrical heating power gradually decreases to zero as the chamber wall temperature increases. That's how uncooled IR detectors can see cooler objects.
Once again, Jane insists electrical heating power = (e * s) * (Ta^4). Once again, Jane's ridiculous e
Maybe the Slayers could explain how uncooled IR detectors see cooler objects?
Straw-man. Our argument involved gray bodies, not detectors of specific wavelengths or electronics that take advantage of specific quantum effects. But I have an answer anyway: they measure DIFFERENCES, not absolute radiation.... [Jane Q. Public, 2014-09-28]
This isn't a quantum effect. The reason IR detectors measure DIFFERENCES, not absolute radiation, is because electrical heating power = (e * s) * (Ta^4 - Tb^4). If that weren't true, there would be no way to detect this difference, so uncooled IR detectors wouldn't be able to see cooler objects. And we couldn't have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How?
... An object that is radiating at a certain black-body temperature WILL NOT absorb a less-energetic photon from an outside source. This is am extremely well-known corollary of the Second Law.... [Jane Q. Public, 2013-05-30]
... I have NOT been claiming that no radiation from a cooler body is absorbed by a warmer body. What I claimed, I repeat, is that no NET radiative energy transfer occurs from cooler bodies to warmer. That concept does not conflict with the ability of infrared cameras or pyrometers to detect "cooler" radiation. Energy can be absorbed and re-emitted... and often (for non-gray-bodies) it is re-emitted in different wavelengths. But the fact remains that there is still no NET energy transfer from cooler to warmer. If there were, it would violate the second law of thermodynamics. My argument has always been about NET heat transfer. I have explained to you many times that I do NOT claim no radiation from cooler bodies is ever absorbed. My argument is, and has been, about NET.... [Jane Q. Public, 2014-09-28]
Once again, Jane insists electrical heating power = (e * s) * (Ta^4). Once again, Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.
If Jane would reconsider conservation of energy and include a term for "radiative power in", then Jane could honestly say he was only claiming that no net radiative power is absorbed by the source. Until then, Jane's equation claims that no radiation is absorbed by the source at all. And since Jane seems to think he's only saying no "NET" radiative power is absorbed, Jane will probably never be able to recognize his error, let alone correct it.
... And further, contrary to your own assertions, since the NET energy transfer from cooler bodies is ZERO, it is not included in the "radiative power out" term of heat transfer equations. Which is a concept that (apparently, if we assume you're being honest, which I doubt) you have had supreme difficulty getting through your head.... [Jane Q. Public, 2014-09-28]
Once again, it's not included in the "radiative power out" term of heat transfer equations because it's included in the "radiative power IN" term.
I'm having supreme difficulty getting your concept through my head because it's Sky Dragon Slayer nonsense. The fact that more heat
... As usual, you distort reality. Prof. Brown had nothing in the way of refutation or rebuttal or even retort to my second comment? Don't you find that interesting? I do.... [Jane Q. Public, 2014-09-26]
It's not that interesting that Prof. Brown decided to ignore Jane/Lonny Eachus, given that he later said:
"Wow, Joel, I gotta say (after reading some of the replies on this thread) that this really is pointless. These folks have no conception of the FIRST law of thermodynamics, let alone the second. The argument for warming doesn't even require mentioning the SBE, it only requires the first law, the second law, and a monotonic relation between temperature difference in ANY channel and the rate of energy transfer in that channel, subject to very broad constraints.
But seriously, just a waste of time. When people just make stuff up and reject the contents of ELEMENTARY textbooks on the subject because they just don't like the conclusion those contents lead to, how can you argue with them? If somebody tries to solve the light bulb problem while pretending that it doesn't primarily cool via radiation and completely ignoring radiation, what can you do?
Get them to say "oops"?
Never happen. It's a religious issue, not a scientific one."
In other words, Prof. Brown gave up trying to educate Slayers like Jane/Lonny Eachus because it's a "waste of time."
... As for Joel Shore, again he was mis-applying an equation for heat transfer when he should have been using the equation for radiant power out. Both you and Shore insist on mis-applying this equation in a way that violates the Second Law of Thermodynamics. It's rather amusing that you brought him up, because you both FUCKED UP YOUR PHYSICS in a similar way.... [Jane Q. Public, 2014-09-26]
That's odd. Just yesterday Jane had no argument with Dr. Shore. Now Jane claims that Dr. Shore "FUCKED UP" his physics.
... As for Joel Shore, again he was mis-applying an equation for heat transfer when he should have been using the equation for radiant power out. Both you and Shore insist on mis-applying this equation in a way that violates the Second Law of Thermodynamics. It's rather amusing that you brought him up, because you both FUCKED UP YOUR PHYSICS in a similar way.... Engineers the world over do the math the way I did. So far that hasn't resulted in you either freezing or burning to death in your home. If they're all crazy, you might want to ask yourself why. [Jane Q. Public, 2014-09-26]
Physicists have "FUCKED UP" their physics, and only the Slayers can save the day! Or maybe the Slayers are crackpots. How could anyone tell, unless maybe Dr. Shore explained that:
"Actually, the idea that radiation goes only from the warmer to colder objects is an invention of the Slayers. It appears nowhere in the physics literature. I don't know about the exact history of our understanding, but my physics textbook from 1983 (Serway, "Physics for Scientists and Engineers", after introducing the law P = sigma*A*e*T^4 says
"A body radiates and also absorbs electromagnetic radiation at rates given by Eq. 17.11. If this were not the case, a body would eventually radiate all of its internal energy and its temperature would reach absolute zero. The energy that the body absorbs comes from the surroundings, which also em
... These are just straw-man arguments, as usual. I have no argument with these other physicists. It was about Spencer's challenge and how YOU got it wrong, nothing more. Have you asked them, personally, about Spencer's experiment? (No, you haven't, or you would know you were wrong.)... [Jane Q. Public, 2014-09-25]
Does Jane have the memory of a goldfish? Of course Jane has argued with these other physicists. Jane personally asked Prof. Brown about Sky Dragon Slayerism, but wasn't able to "educate" him. Lonny Eachus personally asked Dr. Joel Shore about Sky Dragon Slayerism, but wasn't able to "educate" him. And now Jane/Lonny Eachus fantasizes that these physicists agree with his Sky Dragon Slayerism? Maybe Jane/Lonny Eachus should read those exchanges again, and notice that Prof. Brown and Dr. Shore told Jane/Lonny Eachus the same things I am. That's because Prof. Brown, Dr. Shore and I are simply reiterating elementary mainstream physics.
... Bringing up OTHER arguments like greenhouse gases won't win THAT argument for you. You have already lost it.... [Jane Q. Public, 2014-09-25]
How bizarre. The whole reason Slayers deny that an enclosed source warms is because that implies greenhouse gases can't warm the surface:
.. the CO2-warming model rely on the concept of "back radiation", which physicists (not climate scientists) have proved to be impossible. I'm happy to leave actual climate science to climate scientists. But when THEIR models rely on a fundamental misunderstanding of physics, I'll take the physicists' word for it, thank you very much... [Jane Q. Public, 2012-07-05]
If Sky Dragon Slayers could answer these questions without resorting to gray Oreos or basketball player gloves, physicists might take the Slayers more seriously.
... I mean, didn't it send up a red flag when you took your answer and fed it back into standard heat transfer equations and it didn't balance? Oh, that's right... you didn't. But I did.... [Jane Q. Public, 2014-09-24]
Completely backwards, as usual. I've already shown that my solution keeps electrical heating power constant. Once again, Jane's solution halved the electrical heating power. Jane didn't notice this because he calculated net transfer incorrectly, which led him to the absurd conclusion that Jane was only off by about 0.1% when Jane was actually off by ~100%.
... because ALL of the incoming cooler radiation is reflected or scattered, and no NET amount is absorbed... [Jane Q. Public, 2014-09-24]
Good grief, Jane. How did the Sky Dragon Slayers brainwash you into endlessly regurgitating this nonsense? Once again, radiation is absorbed by any surface with absorptivity > 0. Jane's either hopelessly confused about the very term "NET" which he keeps capitalizing, or Jane/Lonny Eachus has betrayed humanity by deliberately spreading civilization-paralyzing misinformation.
Again, how do Slayers think we detected the 2.7K cosmic microwave background radiation with warmer detectors? How do Slayers think uncooled IR detectors see cooler objects? Again, why do Slayers think Venus is hotter than Mercury?
... I'm not arguing with you now and I'm not going to again. You're either a fool or a liar, and I do not care which. I have already proved it and I intend to publish that for the world to see. Along with textbook explanations and diagrams showing exactly where and how you went wrong. [Jane Q. Public, 2014-09-24]
... I used the proper equation for radiative power, which at steady-state doesn't depend on other bodies. So there is no "difference" term. Just temperature. That's simple physics. You are trying to use a heat transfer equation to calculate power out of a single body at known temperature. That's just plain WRONG.... [Jane Q. Public, 2014-09-24]
No, Jane tried to use an equation that only calculates radiative "power out" when Jane needs to use an equation for heat transfer that calculates radiative "power out minus power in".
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".
Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]
Jane's accounting for "power out" without including a term for "power in". That's not A = A, it's A = 0 because one of the terms has been ignored. It's led Jane to the absurd conclusion that electrical heating power doesn't depend on the cooler chamber wall temperature. If that's the case, then how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?
... All the radiation going IN from the cooler body just goes right back OUT again, making the NET radiation crossing your boundary from the cooler body zero. If that were not so, then you'd have net energy being transferred from a cooler body to a hotter one, which is a violation of the second law of thermodynamics. As I've explained to you many times now. You're just plain wrong.... [Jane Q. Public, 2014-09-24]
This is complete gibberish, Jane. Power radiated in from the chamber walls needs to be accounted for using one term. Power radiated out from the source needs to be accounted using another. Once again, accounting for power flowing in doesn't violate the second law of thermodynamics or somehow imply net energy transfer from cool to hot, no matter how many times Jane wants to assert that nonsense. However, failing to account for power flowing in does violate conservation of energy, because power in = power out through any boundary where nothing inside is changing.
So Jane refuses to retract his absurd claim that view factors vary as the radius ratio, which violates conservation of energy. A cynic might have expected as much, given how Jane flagrantly violates conservation of energy by incoherently ignoring radiative power passing in through a boundary around the heat source.
I made no such claim, you liar. As you well know, the view factor from the surface of the inner sphere to the inner surface of the outer sphere is 1. The calculated view factor from the outer sphere to the inner was 0.9998... [Jane Q. Public, 2014-09-24]
.. By the Stefan-Boltzmann radiation law, the chamber walls add no net power in. It just goes right back out through your boundary again. How many times must I explain this to you?.. [Jane Q. Public, 2014-09-23]
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out". For instance:
There is no net "radiative power in" from cooler to hotter. It's against the second law of thermodynamics, and it violates the S-B radiation law: (e * s) * (Ta^4 - Tb^4). [Jane Q. Public, 2014-09-23]
That's exactly the equation Jane should be using to calculate electrical heating power! It has separate terms for "power in" and "power out" so it can describe power entering and exiting a boundary. If Jane would use that equation, he'd honestly be only saying there is no net "radiative power in" from cooler to hotter.
Instead, Jane insists that electrical heating power = (e * s) * (Ta^4). Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.
There is nothing more to say. You have been proved wrong. You can write books about your nonsense "physics", and it won't make your bullshit theory any more correct... The textbooks all say you're wrong. Goodbye. [Jane Q. Public, 2014-09-23]
So Jane refuses to retract his absurd claim that view factors vary as the radius ratio, which violates conservation of energy. A cynic might have expected as much, given how Jane flagrantly violates conservation of energy by incoherently ignoring radiative power passing in through a boundary around the heat source.
.. I honestly -- and I mean that: honestly -- don't believe you could be this stupid and possess a degree in physics... [Jane Q. Public, 2014-09-15]
.. I only replied on the off-chance that you really were ignorant and could be educated... [Jane Q. Public, 2014-09-20]
Jane's campaign of educating ignorant, stupid physicists about physics has only just begun. Jane still needs to educate Prof. Brown and Lonny Eachus still needs to educate Dr. Joel Shore.
.. the CO2-warming model rely on the concept of "back radiation", which physicists (not climate scientists) have proved to be impossible. I'm happy to leave actual climate science to climate scientists. But when THEIR models rely on a fundamental misunderstanding of physics, I'll take the physicists' word for it, thank you very much...
Slashdot Mirror
All those lead thickness options are too thick by at least a factor of 5. Even if that NASA study is "retarded", RockDoctor just mentioned that Earth's atmosphere protects us with only ~10 tons/m^2. Since lead's density is 11 tons/m^3, a lead shield wouldn't have to be thicker than ~0.9 meter.
Yep. That's the same sanity check used by that NASA study:
"Passive shielding is known to work. The Earth's atmosphere supplies about 10 t/m^2 of mass shielding and is very effective. Only half this much is needed to bring the dosage level of cosmic rays down to 0.5 rem/yr. In fact when calculations are made in the context of particular geometries, it is found that because many of the incident particles pass through walls at slanting angles a thickness of shield of 4.5 t/m^2 is sufficient."
Water could be an effective shield, and would be especially easy to apply and repair. Just melt it and let it freeze in place. That's how most of the lighthuggers in Revelation Space were shielded, as well as the starship in Songs of Distant Earth.
The only downsides I can think of would be the low tensile strength, so a water shield couldn't spin with a rotating ship, and the fact that if the ship overheats then its radiation shield sublimates away...
Centrifuges need to rotate no faster than 1 rpm to avoid inducing motion sickness. That's for long-term colonies, so maybe 2 or 3 rpm would be acceptable for astronauts selected for their resistance to motion sickness. Maybe even faster if they're hibernating the whole way. But regardless, the centrifuge would still have to be quite large.
If the centrifuge is inside the shielding, that makes the shield enormously bigger and heavier. Alternatively, only the hibernation/living chamber at the end of the centrifuge could be shielded. But that requires that the shielding mass be attached to the centrifuge, which vastly increases its required tensile strength. That's why the NASA study placed the colony's centrifuge inside a separate shield: if the shield rotates with the centrifuge then the centrifuge would have to be built out of carbon nanotubes. If the shield is separate then the centrifuge can be built out of aluminum.
Jane, you're still wrongly insisting that electrical heating power per square meter = (e*s)*T1^4. Once again, Jane's equation violates conservation of energy. That's why I'm trying to understand why you keep insisting it's correct. At first I thought you agreed that power in = power out, but that we only disagreed about which terms to include:
Jane's statement originally made me think that Jane is reasoning like this:
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
But now it seems like our disagreement is even more fundamental:
This objection is completely different than Jane's "A = A" objection above, which at least seemed to acknowledge that we should start with the principle of conservation of energy, where power in = power out. But now Jane even seems to dispute that starting point.
I'm starting to suspect that Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a fancy way of saying "electrical heating power". Is that how Jane "derived" his incorrect equation that electrical heating power per square meter = (e*s)*T1^4?
If so, that's kind of a boring mistake because "radiative power out" isn't just a fancy way of saying "electrical heating power". They're completely different. To find electrical heating power, Jane needs to use conservation of energy, where power in = power out. That results in a heat transfer equation, not just an equation for "radiative power out".
Jane, if you don't agree with the "power in" and "power out" that I've tried to glean from your rants, just fill in the following blanks like I did. It'll be much faster than accusing me of ill behavior.
Jane's power in = ?
Jane's power out = ?
Or, explain why we shouldn't start with the principle of conservation of energy which results in a heat transfer equation. Or, (more
Jane responds.
Jane, the answers you've given don't make any sense. That's why I'm asking you for a very simple equation describing the required electrical heating power. Again, filling in the following blanks would be be much faster than repeatedly calling me an asshole.
Jane's power in = ?
Jane's power out = ?
I never said we should ignore extrasolar particles. I was just showing that even using angel'o'sphere's assumption that the sun is the main hazard, the shielding mass decreases for a hibernating crew. In other words, I was defending you, Jane. Even though I can't be trusted to build a bridge over a creek.
But since you brought up those other arguments...
I have to guess at your reasoning because what you've said doesn't make any sense.
I have to guess at what Jane meant by this, because it's not in equation form. In physics, statements in equation form are easier to analyze.
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Once again, to calculate "electrical heating power" you need to use a heat transfer equation which accounts for power in and power out. That's because power in = power out through any boundary where nothing inside is changing. Once again, the equation Jane's using is only valid for "radiative power out" which is completely different than "electrical heating power". That's why I'm starting with the principle of "conservation of energy" and trying to understand what Jane's saying, in equation form.
Jane, if you don't agree with the "power in" and "power out" that I've tried to glean from your rants, just fill in the following blanks like I did. It'll be much faster than accusing me of dishonesty, fraud, and libel.
Jane's power in = ?
Jane's power out = ?
I have to guess at your reasoning because what you've said doesn't make any sense.
I have to guess at what Jane meant by this, because it's not in equation form. In physics, statements in equation form are easier to analyze.
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Once again, to calculate "electrical heating power" you need to use a heat transfer equation which accounts for power in and power out. That's because power in = power out through any boundary where nothing inside is changing. Once again, the equation Jane's using is only valid for "radiative power out" which is completely different than "electrical heating power". That's why I'm starting with the principle of "conservation of energy" and trying to understand what Jane's saying, in equation form.
Jane, if you don't agree with the "power in" and "power out" that I've tried to glean from your rants, just fill in the following blanks like I did. It'll be much faster than accusing me of dishonesty, fraud, and libel.
Jane's power in = ?
Jane's power out = ?
If the main hazard is the sun, that requires thicker shielding on the sunward side. Minimum shielding mass would then be obtained by putting 4.41 tons/m^2 on the sunward side, which given moon dust density equals a ~2.4 meter thick shield on the sunward side. If the people are perpendicular to the sun, that shield is heavier. The people are awake and moving around, that shield is much heavier.
A hibernating crew could be closely packed and aligned with their feet towards the sun, reducing the required shielding area and mass at constant thickness. That's because only the hibernation chamber would need to be shielded, not the entire ship.
NASA found that 441 grams/cm^2 of silicon dioxide (Moon dust) would be sufficient shielding, which equals 4.41 tons/m^2. Hibernation dangers and personal preference regarding books may vary, of course.
Since Jane probably won't even say yes or no, I'll keep trying to guess at Jane's reasoning. Now the next term for Jane's gray body:
Because "radiative power out from source" is emitted by the graybody source at temperature T1, the Stefan-Boltzmann law says:
gray electrical heating power + (e*s)*T4^4 = (e*s)*T1^4 + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Now the next term for Jane's black body check:
Because "radiative power out from source" is emitted by the blackbody source at temperature T1, the Stefan-Boltzmann law says:
black electrical heating power + (s)*T4^4 = (s)*T1^4 + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Reliable shielding isn't impossible. Shielding of 4.41 tons/m^2 is sufficient. Putting the crew in hibernation does reduce shielding because otherwise the entire back side of the spacecraft (at least) has to be covered with 4.41 tons/m^2 of shielding. In hibernation, the crew could be closely packed and aligned with their feet towards the sun, reducing the required shielding area and mass.
Again, Jane's gray body equation has to reduce to the black body equation when emissivity = 1, so this is a way to check Jane's work. But since Jane seems convinced that checking his work is "lying" let's write down both equations simultaneously.
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Now use the Stefan-Boltzmann law to describe the radiative terms, one at a time. First for Jane's gray body:
Because "radiative power in from chamber walls" is emitted by graybody walls at temperature T4, the Stefan-Boltzmann law says:
gray electrical heating power + (e*s)*T4^4 = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Now for Jane's black body check:
Because "radiative power in from chamber walls" is emitted by blackbody walls at temperature T4, the Stefan-Boltzmann law says:
black electrical heating power + (s)*T4^4 = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Jane probably won't write down an equation describing electrical heating power for a blackbody source, so I'll try to guess at Jane's reasoning.
Draw a boundary around the blackbody heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Charming. As I just explained, IR detectors don't have to depend on quantum effects. Classical mainstream physics allows a temperature-controlled source to detect IR from the cooler chamber walls as follows:
electricity = (e*s)*(T1^4 - T4^4)
If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).
If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.
If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.
If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.
If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.
If you don't deny that some radiation is absorbed, then it should be very easy to write down a simple equation describing the required electrical heating power (not the radiative power out) of a blackbody source.
Jane, if we can't agree on the meaning of the term "NET", why are you still capitalizing the word "NET"? Screaming the word louder and louder is unlikely to be productive.
1. Can we agree that net heat transfer through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net heat transfer always contains terms in both directions?
3. Can we agree that just because "radiative power out" > "radiative power in", that doesn't mean "radiative power in" = 0?
If Jane answers "no" to any of those three yes/no questions... why?
If you're so confident that you're right, why not prove it by taking a few seconds to write dow
It's beginning to seem like we disagree about the meaning of the term "NET".
1. Can we agree that net heat transfer through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net heat transfer always contains terms in both directions?
3. Can we agree that just because "radiative power out" > "radiative power in", that doesn't mean "radiative power in" = 0?
If we can agree on all those points, that's great. Maybe this will help Jane write down a simple equation describing the electrical heating power required t
That reference shows the object (i.e. chamber wall) temperature has an effect on the temperature controlled cavity (i.e. source). Which Jane denies:
No, that's Sky Dragon Slayer nonsense. If radiation from the cooler walls really had no effect on the heat source whatsoever, the IR thermometer wouldn't work because the cooler object temperature would have no effect on the temperature controlled cavity whatsoever.
When the source temperature is held constant, its required electrical heating power is an IR thermometer.
Here's one way to see that: draw a boundary around a heated aluminum source. It's heated by constant electrical power flowing in. Aluminum cold walls at some unknown temperature T4 also radiate power in. The source at 150F (T1 = 338.7K) radiates power out. At steady-state, power in = power out. Using the equation which neglects reflections:
electricity = (e*s)*(T1^4 - T4^4)
If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).
If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.
If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.
If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.
If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.
That's why when the source temperature is held constant, its required electrical heating power is an IR thermometer. At least, it's a thermometer when using mainstream physics. But Jane's equation is:
electricity = (e*s)*T1^4 (Jane's equation)
Since Jane's equation doesn't depend on the chamber wall temperature, uncooled IR detectors can't see cooler objects in Janeland. And we couldn't possibly have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How? This must be inexplicable to Slayers who are brainwashed into believing that:
No. Again, mainstream physics shows that electrical heating power gradually decreases to zero as the chamber wall temperature increases. That's how uncooled IR detectors can see cooler objects.
This isn't a quantum effect. The reason IR detectors measure DIFFERENCES, not absolute radiation, is because electrical heating power = (e * s) * (Ta^4 - Tb^4). If that weren't true, there would be no way to detect this difference, so uncooled IR detectors wouldn't be able to see cooler objects. And we couldn't have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How?
Once again, Jane insists electrical heating power = (e * s) * (Ta^4). Once again, Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.
If Jane would reconsider conservation of energy and include a term for "radiative power in", then Jane could honestly say he was only claiming that no net radiative power is absorbed by the source. Until then, Jane's equation claims that no radiation is absorbed by the source at all. And since Jane seems to think he's only saying no "NET" radiative power is absorbed, Jane will probably never be able to recognize his error, let alone correct it.
Once again, it's not included in the "radiative power out" term of heat transfer equations because it's included in the "radiative power IN" term.
I'm having supreme difficulty getting your concept through my head because it's Sky Dragon Slayer nonsense. The fact that more heat
It's not that interesting that Prof. Brown decided to ignore Jane/Lonny Eachus, given that he later said:
"Wow, Joel, I gotta say (after reading some of the replies on this thread) that this really is pointless. These folks have no conception of the FIRST law of thermodynamics, let alone the second. The argument for warming doesn't even require mentioning the SBE, it only requires the first law, the second law, and a monotonic relation between temperature difference in ANY channel and the rate of energy transfer in that channel, subject to very broad constraints.
But seriously, just a waste of time. When people just make stuff up and reject the contents of ELEMENTARY textbooks on the subject because they just don't like the conclusion those contents lead to, how can you argue with them? If somebody tries to solve the light bulb problem while pretending that it doesn't primarily cool via radiation and completely ignoring radiation, what can you do?
Get them to say "oops"?
Never happen. It's a religious issue, not a scientific one."
In other words, Prof. Brown gave up trying to educate Slayers like Jane/Lonny Eachus because it's a "waste of time."
That's odd. Just yesterday Jane had no argument with Dr. Shore. Now Jane claims that Dr. Shore "FUCKED UP" his physics.
Physicists have "FUCKED UP" their physics, and only the Slayers can save the day! Or maybe the Slayers are crackpots. How could anyone tell, unless maybe Dr. Shore explained that:
"Actually, the idea that radiation goes only from the warmer to colder objects is an invention of the Slayers. It appears nowhere in the physics literature. I don't know about the exact history of our understanding, but my physics textbook from 1983 (Serway, "Physics for Scientists and Engineers", after introducing the law P = sigma*A*e*T^4 says
"A body radiates and also absorbs electromagnetic radiation at rates given by Eq. 17.11. If this were not the case, a body would eventually radiate all of its internal energy and its temperature would reach absolute zero. The energy that the body absorbs comes from the surroundings, which also em
Does Jane have the memory of a goldfish? Of course Jane has argued with these other physicists. Jane personally asked Prof. Brown about Sky Dragon Slayerism, but wasn't able to "educate" him. Lonny Eachus personally asked Dr. Joel Shore about Sky Dragon Slayerism, but wasn't able to "educate" him. And now Jane/Lonny Eachus fantasizes that these physicists agree with his Sky Dragon Slayerism? Maybe Jane/Lonny Eachus should read those exchanges again, and notice that Prof. Brown and Dr. Shore told Jane/Lonny Eachus the same things I am. That's because Prof. Brown, Dr. Shore and I are simply reiterating elementary mainstream physics.
How bizarre. The whole reason Slayers deny that an enclosed source warms is because that implies greenhouse gases can't warm the surface:
That's why Jane, Dr. Latour and the rest of the Slayers disagree with the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.
Again, how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?
If Sky Dragon Slayers could answer these questions without resorting to gray Oreos or basketball player gloves, physicists might take the Slayers more seriously.
Completely backwards, as usual. I've already shown that my solution keeps electrical heating power constant. Once again, Jane's solution halved the electrical heating power. Jane didn't notice this because he calculated net transfer incorrectly, which led him to the absurd conclusion that Jane was only off by about 0.1% when Jane was actually off by ~100%.
Good grief, Jane. How did the Sky Dragon Slayers brainwash you into endlessly regurgitating this nonsense? Once again, radiation is absorbed by any surface with absorptivity > 0. Jane's either hopelessly confused about the very term "NET" which he keeps capitalizing, or Jane/Lonny Eachus has betrayed humanity by deliberately spreading civilization-paralyzing misinformation.
Again, how do Slayers think we detected the 2.7K cosmic microwave background radiation with warmer detectors? How do Slayers think uncooled IR detectors see cooler objects? Again, why do Slayers think Venus is hotter than Mercury?
Again, Jane/Lonny Eachus actually means that he intends to show where mainstream physics "went wrong" according to the Sky Dragon Slayers. There are many ignorant, stupid physicists that Jane/Lonny Eachus needs to educate: Prof. Brown, Dr. Joel Shore, the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society, etc.
No, Jane tried to use an equation that only calculates radiative "power out" when Jane needs to use an equation for heat transfer that calculates radiative "power out minus power in".
Jane's accounting for "power out" without including a term for "power in". That's not A = A, it's A = 0 because one of the terms has been ignored. It's led Jane to the absurd conclusion that electrical heating power doesn't depend on the cooler chamber wall temperature. If that's the case, then how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?
This is complete gibberish, Jane. Power radiated in from the chamber walls needs to be accounted for using one term. Power radiated out from the source needs to be accounted using another. Once again, accounting for power flowing in doesn't violate the second law of thermodynamics or somehow imply net energy transfer from cool to hot, no matter how many times Jane wants to assert that nonsense. However, failing to account for power flowing in does violate conservation of energy, because power in = power out through any boundary where nothing inside is changing.
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out". For instance:
That's exactly the equation Jane should be using to calculate electrical heating power! It has separate terms for "power in" and "power out" so it can describe power entering and exiting a boundary. If Jane would use that equation, he'd honestly be only saying there is no net "radiative power in" from cooler to hotter.
Instead, Jane insists that electrical heating power = (e * s) * (Ta^4). Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.
So Jane refuses to retract his absurd claim that view factors vary as the radius ratio, which violates conservation of energy. A cynic might have expected as much, given how Jane flagrantly violates conservation of energy by incoherently ignoring radiative power passing in through a boundary around the heat source.
Jane's campaign of educating ignorant, stupid physicists about physics has only just begun. Jane still needs to educate Prof. Brown and Lonny Eachus still needs to educate Dr. Joel Shore.
Then, Jane/Lonny Eachus needs to educate the "ignorant" and "stupid" American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.
After this thread is locked, this conversation can continue here.