Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.
I didn't say they were the same. They don't need to be the same.
This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature.
If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy.
What "I propose" is the textbook answer to this question. It's not even "my" idea, as I clearly showed you just yesterday. YOU are the one going against "established" physics here. So I daresay it's up to you to prove your point, rather than arguing with me about it.
Which you will never do, because you're wrong. If you could actually show how the physics textbook idea of heat transfer was wrong, you would be world famous by now. Instead, you're arguing ineffectively with some person on Slashdot, about something every textbook on the subject, as well as other sources, say your are wrong about.
Instead of calling me a blathering religious zealot liar who wasn't ever actually a physicist, could you calmly explain why you disagree with this energy conservation equation?
I already did so, several times. What, do you honestly think that If I fail to refute this idea just one more time, it will somehow magically become correct?
The heat transfer scenario I presented, and my calculations of temperatures, were correct within a reasonable degree of precision. Yours, on the other hand, were not.
By what stretch of your imagination am I obligated to KEEP refuting your same, lame arguments? This is all old news now. You can read about it all again later, when I write this all up and publish it. In the meantime, if you want answers to these questions AGAIN, you can go back and read our prior discussion of the matter.
Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.
You're just re-hashing old arguments that I've already shot down.
Why are you doing that, if your purpose is not dishonest?
I've explained that net heat transfer = radiative power out - radiative power in, so of course they're not the same.
Your BS "explanations" are not informative to readers who actually want to be educated.
Once again, I'm just saying that "radiative power out" is different than "electrical heating power".
No, you aren't, because then your "explanation" re-introduces the dependency. Which is what I have been saying all along (and repeatedly): your methodology contradicts itself.
I'm not even going to bother answering the rest of your blather. Because your whole argument was PUT to rest weeks ago and your failure to understand that (or at least admit it) is rather like a zombie which hasn't quite realized it is dead yet.
I repeat: I have documented this all. I have the reputable and credible (and MAINSTREAM, "ACCEPTED") references which show you to be wrong.
For a while I thought explaining this in different ways would show you that you were wrong. But over time, I have come to accept that you simply won't admit it, no matter what. That's too bad, because I had really hoped you would listen to the actual accepted SCIENCE behind this, and further accept that it was right and you were wrong.
I no longer hold any such hope. I have myself come to accept that you are either a religious zealot, or a self-interested liar.
And I very seriously doubt that you were ever actually a physicist.
It's all irrelevant. Even if 2 centuries before Columbus, he's still even more centuries after other explorers who have already been shown to have been here at the North American continent.
For other readers (not for you, because despite your claims you've seen this already several times), from Wikipedia (edited here for clarity given Slashdot's character handling):
A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity, epsilon
j* = epsilon * sigma * T^4
In the above equation, using the "dot" notation which YOU pointed out to ME, j is energy and j* is power. This isn't just Wikipedia. It is very easy to find this relation in other sources as well:
Since all bodies are continuously receiving and radiating thermal energy, energy radiating from unit area (all this energy is absorbed by the black surroundings) = sigma * emissivity * T^4
The example goes on to express heat transfer between long co-axial cylinders using heat transfer equations similar to those we discussed before. But heat transfer is NOT the same as the radiative power of a SINGLE gray body at steady-state. Power out is a function of emissivity and temperature ONLY. Heat transfer from one surface to another requires 2 bodies, or 2 surfaces of the same body. But note that the equation for power out clearly implies it is independent of transfer to cooler bodies.
You can also find it here. In this case, note that it gives the equation for power output as distinct from radiation "loss" (heat transfer). BECAUSE THEY ARE TWO DIFFERENT THINGS. One is the power output of a SINGLE gray body at a given temperature. The other is radiative transfer to another body. One requires ONLY emissivity and temperature to calculate. The other involves 2 bodies.
Is this clear yet? Or are you going to continue to erroneously claim that radiative POWER output is dependent on the presence of cooler bodies? Do you really need more examples, or are you finally willing to admit you have been proved wrong? If you need more examples of this, you can find them with a quick search of the 'net. I just did, since I don't have a good way to link to my textbooks.
Anti-competitive practices do not require "conspiracy". They only require mutual complicity.
The main thing that's actually "suppressing" competition is regulations, which make it costly and risky to create alternative Internet providers in an area.
If that were true, then the municipalities that created public infrastructure would not be as tremendously successful as they actually are. These "regulations" you refer to are often the result of lobbying (read: "bribery") by the cable operators to keep other players out. Not only has it been in the news, I've watched it happen, pretty damned transparently, with my own City Council.
That's not "conspiracy". It's just all-too-common, unethical, anti-competitive business practice. If you want to call lobbying and expensive presents "conspiracy", then you're saying that most of Congress are conspirators.
Jane, mainstream physics is based on conservation of energy. That means power in = power out through any boundary where nothing inside is changing. If your textbook doesn't agree with that principle, it's either wrong or you're misinterpreting what it says. For instance:
I know how this works. Stop trying to be insulting. I'm not the one who got it wrong. YOUR answer (checked 3 different ways) violated conservation of energy. Not mine. Again: I checked both my work and yours. Your "answer" didn't even check out using your own heat transfer equations.
No Jane, you've misinterpreted your textbook. Energy is always conserved, so power in = power out through any boundary where nothing inside is changing. This isn't a "very rare exception". It's a fundamental law called "conservation of energy". Does Jane seriously think his textbook says that using a fundamental law like "conservation of energy" is "doomed to fail"?
I know energy is always conserved, you insufferable ass. I have already proved that my answer conserved energy and yours did not. Your constant blathering about it elsewhere (like here) does not change that.
Is that how Jane "derived" his incorrect equation that electrical heating power per square meter = (e*s)*T1^4?
Your insistence on "electrical heating power" is a red herring. Energy is energy. Your misguided attempt to include the power used to cool the chamber walls does not change that. Spencer stipulated "electrical power to the heat source". It is neither necessary nor called for to calculate the power used to cool the chamber walls in order to find the temperatures of the other bodies.
(e*s)*T1^4 is often called the "Stefan-Boltzmann relation", which is derived from the Stefan-Botlzmann radiation law, and which describes the relationship between thermodynamic temperature and radiative power output of a single gray body. I repeat: you can find this equation in heat transfer textbooks and I also showed you where it is in Wikipedia. Stop pretending ignorance about things I already explained to you clearly several times. I can only conclude that you're doing this in order to harass.
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".
NO, it does NOT result in a heat transfer equation. There is no need to account for other, cooler bodies when calculating radiative power out. What, do you imagine that these cooler bodies are somehow "sucking" power away from the heat source? And that a warmer body (but still cooler than the source) "sucks" less power than colder ones do? That seems to be what you're saying here.
Just no. That's not the way it works, man. At steady-state, radiative power out can be calculated from temperature and emissivity alone. Other factors (such as heat transfer) are affected by nearby bodies, but radiative power out of a gray body at steady-state is related ONLY to temperature and emissivity. It's that simple, and claiming otherwise is just wrong.
I repeat: look it the hell up. I have not just one but 4 textbooks here, plus Wikipedia, plus the testimony of experts in the field of heat transfer. They ALL disagree with you. It's that simple.
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
Just no. That is not even remotely what I meant, and I explained this to you clearly at least several times already. I have no reason to continue to re-explain it just because you keep asking.
Instead I'm going to repeat something else I have stated several times: pick up a textbook on heat transfer, and see what the accepted, textbook, "consensus" science says about it. Hint: they don't agree with you.
I don't appreciate this constant harassment over something that has been explained to you clearly many times over. If you truly still don't understand it, that is sad but it is also not my problem. A textbook might help.
You basically have to weight how often a craft will be hit by really high energy particles.
But this is the whole point.
"Really high energy particles" do not come exclusively from the sun. While we can agree that MOST of them do.
So -- again I think we agree -- shielding in non-sun-facing parts does not have to be anywhere near as heavy. HOWEVER... regardless of whether you are referring to sun shielding or shielding from the rest of space, reducing the volume (and cross-sectional area) of the crew area that needs to be shielded can make a big difference in the mass. And, as you probably know, reduce the volume and mass of the crew area and the entire ship can be a lot smaller from start to finish.
However, the companies said that they are in different regions and are not going to suppress the competition.
By surreptitiously "agreeing" to operate and different areas, they ARE suppressing competition. In most areas your "one" cable operator is the only game in town for broadband.
DirecTV might be great for TV, but not so much for internet.
Jane, the answers you've given don't make any sense.
They don't make any sense to you. This much is obvious.
That's why I'm asking you for a very simple equation describing the required electrical heating power.
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.
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.
As I implied elsewhere, when you minimize solar radiation you are eliminating most of the energetic radiation/particles, but by no means all. We already know this from Spacelab and ISS experience. If you ignore extrasolar energetic particles you're just being stupid. Unless you plan a 1-way trip. Which has been suggested.
Certainly most of the shielding should be between the sun and the crew. But it's not all the shielding necessary. And though the "other" shielding need not be as heavy, its area is much larger so it still contributes a lot to the overall mass of the vehicle.
Based on other arguments with khayman80, to be honest I would not trust him to build a bridge over a creek, much less a spaceship. That's just the truth.
Unlike you, I didn't "fail" to read. The direction most of the dangerous radiation comes from is not irrelevant, BUT you seem to think that JUST because most comes from the sun, that's the only significant shielding needed.
Bullshit.
We already know better from experience. Why didn't YOU know that?
America is quickly reaching a point, if we are not there already, where their are two systems of law; one for the wealty, the privledged and the corporations and one for the rest of us. Guess who is getting fucked hard ?
I'm not going to claim I disagree with you, but whether we agree is not the real point.
The only answer to being fucked hard is to fuck back. You don't have to enjoy it, you just have to make sure THEY enjoy it a lot less.
Then, when the worst offenders are too sore to continue, you can work on building societal justice.
Republicans claim they want to "set the private sector loose" and it causes more income inequality.
Nope. Sorry. Income inequality has snowballed under Obama and his "Progressive" crony-capitalist friends. The actual historical record says you're simply wrong here. And understand that I don't like saying that, because I like you. There aren't many people I can honestly say that to on Slashdot.
The main difference I see between current Democrats and Republicans is the excuses they give for the policies they push
This we can agree on. More on that in a moment.
I don't think it's liberal hypocrisy. I think it's pretty accurate. What passes for conservativism in the US is reactionary and they do cling to a notion of some Ozzie and Harriet past when everything was good and fair and the greatest generation blah blah
Yes, but so what? The "other side" has been pushing alternatives that have been even worse. We can see it in our economy, and our prices, and the incessant wars. I repeat: Republicans are bad, but the Democrats have shown themselves to be even worse.
Obaman "inherited" one "war". Including that one, he has been involved in at least 6 since. He can't blame those on Republicans. Hell, he didn't even ask them.
Yes, it's hypocrisy. But yes, it's on both "sides". The problem is that they aren't really 2 "sides" anymore. It's oppressive government (whether Democrat or Republican) vs. everybody else. And today, "everybody else" is 60% of the voting public.
Windows is designed to be very secure and have its security policies easily managed but most corporate IT departments don't implement the best security measures because of the huge headache that creates.
That's the most hilarious thing I've read in days.
ALL major operating systems offer the very same protections. This is a matter of IT and how they set it up, not the software at all.
You're trying to school someone who has been doing this for 30 years. You're not just wrong, but laughably wrong.
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?)
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.
Every reply you have given the past couple of weeks has demonstrably been a lie, in one form or another: presenting principles which you know to be not representative of the real situation (e.g., heat transfer in place of the proper "radiated power" equation), or claims that I stated something that I provably did not.
One might be characterized as fraud, and the other as libel. And you expect anyone to take you seriously?
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:
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 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.
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.
What is wrong with you? I ask this question very seriously. You were very clearly shown to be wrong, using textbook physics methodology, yet you continue this bullshit. Why? I'd really like to know. (And it was indeed textbook physics. I have 3 different textbooks here... wait, make that 4... which all disagree with you.)
You replied not by admitting you were wrong, but by lying about what I wrote and refusing to accept the clear demonstration that your own brand of "physics" as you applied it to this problem is a blatant violation of the Second Law of Thermodynamics.
You leave me no choice but to conclude that either you are one of the "True Believers", and no facts will sway you, or that you're simply being dishonest. I quite literally have no other options.
I'm not saying they're not, but at the same time I've read that the Pentagon has rejected Windows for certain "critical" operations.
It must be kept in mind that for the U.S. government, Windows is not "closed". But for other governments -- even most other governments -- it is a different story.
They don't know the internals and don't trust them. In the same way that the U.S. has (finally... it took a hell of a long time) started looking at Chinese ICs as a threat.
Personally (not having the direct inside view our government has), I wouldn't use Windows for ANYTHING critical. And I feel very sorry for any troops of ours who have to.
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 to keep a blackbody source at 150F inside 0F chamber walls. Remember that "electrical heating power" is different than "radiative power out". Also remember that blackbodies can only absorb radiation, not reflect or scatter it. Finally, remember that the graybody equation has to reduce to the blackbody equation when emissivity = 1.
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.
That reference [omega.com] shows the object (i.e. chamber wall) temperature has an effect on the temperature controlled cavity (i.e. source). Which Jane denies:
Via a QUANTUM EFFECT, you fucking moron. 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.
It's a combination of your historical tendency to straw-man argue, and outright lies about what I wrote.
This is the only kind of reply you're going to get from me, as long as you keep up your dishonest bullshit.
Maybe we should try for the Moon? It's a lot closer, and it would give us time to work out these types of issues?
I'm with you on that.
Seems to me, the "cold sleep" option mainly solves the problems of crew space, resources, and radiation. Those are not small things.
A long-term space expedition must have room to move and exercise. That's a lot of size and mass. Then it needs food to promote exercise and waking function, and waste disposal to match. And THEN all that has to be wrapped in effective radiation shielding, which adds a lot more mass.
Eliminate the exercise, confine the crew to a small space, feed intravenously, and shield only that small part is FAR more efficient.
On the other hand, as many have pointed out, it comes with some serious cost as well.
I think landing on Mars would be a great accomplishment, but establishing a permanent moon base would be a vastly greater accomplishment.
Slashdot Mirror
Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.
I didn't say they were the same. They don't need to be the same.
This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature.
If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy.
What "I propose" is the textbook answer to this question. It's not even "my" idea, as I clearly showed you just yesterday. YOU are the one going against "established" physics here. So I daresay it's up to you to prove your point, rather than arguing with me about it.
Which you will never do, because you're wrong. If you could actually show how the physics textbook idea of heat transfer was wrong, you would be world famous by now. Instead, you're arguing ineffectively with some person on Slashdot, about something every textbook on the subject, as well as other sources, say your are wrong about.
Instead of calling me a blathering religious zealot liar who wasn't ever actually a physicist, could you calmly explain why you disagree with this energy conservation equation?
I already did so, several times. What, do you honestly think that If I fail to refute this idea just one more time, it will somehow magically become correct?
The heat transfer scenario I presented, and my calculations of temperatures, were correct within a reasonable degree of precision. Yours, on the other hand, were not.
By what stretch of your imagination am I obligated to KEEP refuting your same, lame arguments? This is all old news now. You can read about it all again later, when I write this all up and publish it. In the meantime, if you want answers to these questions AGAIN, you can go back and read our prior discussion of the matter.
Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.
You're just re-hashing old arguments that I've already shot down.
Why are you doing that, if your purpose is not dishonest?
I've explained that net heat transfer = radiative power out - radiative power in, so of course they're not the same.
Your BS "explanations" are not informative to readers who actually want to be educated.
Once again, I'm just saying that "radiative power out" is different than "electrical heating power".
No, you aren't, because then your "explanation" re-introduces the dependency. Which is what I have been saying all along (and repeatedly): your methodology contradicts itself.
I'm not even going to bother answering the rest of your blather. Because your whole argument was PUT to rest weeks ago and your failure to understand that (or at least admit it) is rather like a zombie which hasn't quite realized it is dead yet.
I repeat: I have documented this all. I have the reputable and credible (and MAINSTREAM, "ACCEPTED") references which show you to be wrong.
For a while I thought explaining this in different ways would show you that you were wrong. But over time, I have come to accept that you simply won't admit it, no matter what. That's too bad, because I had really hoped you would listen to the actual accepted SCIENCE behind this, and further accept that it was right and you were wrong.
I no longer hold any such hope. I have myself come to accept that you are either a religious zealot, or a self-interested liar.
And I very seriously doubt that you were ever actually a physicist.
It's all irrelevant. Even if 2 centuries before Columbus, he's still even more centuries after other explorers who have already been shown to have been here at the North American continent.
A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity, epsilon
j* = epsilon * sigma * T^4
In the above equation, using the "dot" notation which YOU pointed out to ME, j is energy and j* is power. This isn't just Wikipedia. It is very easy to find this relation in other sources as well:
Here is "A Textbook of Engineering Thermodynamics . The section on radiative power of a gray body:
Since all bodies are continuously receiving and radiating thermal energy, energy radiating from unit area (all this energy is absorbed by the black surroundings) = sigma * emissivity * T^4
The example goes on to express heat transfer between long co-axial cylinders using heat transfer equations similar to those we discussed before. But heat transfer is NOT the same as the radiative power of a SINGLE gray body at steady-state. Power out is a function of emissivity and temperature ONLY. Heat transfer from one surface to another requires 2 bodies, or 2 surfaces of the same body. But note that the equation for power out clearly implies it is independent of transfer to cooler bodies.
You can also find it here. In this case, note that it gives the equation for power output as distinct from radiation "loss" (heat transfer). BECAUSE THEY ARE TWO DIFFERENT THINGS. One is the power output of a SINGLE gray body at a given temperature. The other is radiative transfer to another body. One requires ONLY emissivity and temperature to calculate. The other involves 2 bodies.
Is this clear yet? Or are you going to continue to erroneously claim that radiative POWER output is dependent on the presence of cooler bodies? Do you really need more examples, or are you finally willing to admit you have been proved wrong? If you need more examples of this, you can find them with a quick search of the 'net. I just did, since I don't have a good way to link to my textbooks.
You make it sound like it's a conspiracy.
Only in your own mind.
Anti-competitive practices do not require "conspiracy". They only require mutual complicity.
The main thing that's actually "suppressing" competition is regulations, which make it costly and risky to create alternative Internet providers in an area.
If that were true, then the municipalities that created public infrastructure would not be as tremendously successful as they actually are. These "regulations" you refer to are often the result of lobbying (read: "bribery") by the cable operators to keep other players out. Not only has it been in the news, I've watched it happen, pretty damned transparently, with my own City Council.
That's not "conspiracy". It's just all-too-common, unethical, anti-competitive business practice. If you want to call lobbying and expensive presents "conspiracy", then you're saying that most of Congress are conspirators.
Jane, mainstream physics is based on conservation of energy. That means power in = power out through any boundary where nothing inside is changing. If your textbook doesn't agree with that principle, it's either wrong or you're misinterpreting what it says. For instance:
I know how this works. Stop trying to be insulting. I'm not the one who got it wrong. YOUR answer (checked 3 different ways) violated conservation of energy. Not mine. Again: I checked both my work and yours. Your "answer" didn't even check out using your own heat transfer equations.
No Jane, you've misinterpreted your textbook. Energy is always conserved, so power in = power out through any boundary where nothing inside is changing. This isn't a "very rare exception". It's a fundamental law called "conservation of energy". Does Jane seriously think his textbook says that using a fundamental law like "conservation of energy" is "doomed to fail"?
I know energy is always conserved, you insufferable ass. I have already proved that my answer conserved energy and yours did not. Your constant blathering about it elsewhere (like here) does not change that.
Is that how Jane "derived" his incorrect equation that electrical heating power per square meter = (e*s)*T1^4?
Your insistence on "electrical heating power" is a red herring. Energy is energy. Your misguided attempt to include the power used to cool the chamber walls does not change that. Spencer stipulated "electrical power to the heat source". It is neither necessary nor called for to calculate the power used to cool the chamber walls in order to find the temperatures of the other bodies.
(e*s)*T1^4 is often called the "Stefan-Boltzmann relation", which is derived from the Stefan-Botlzmann radiation law, and which describes the relationship between thermodynamic temperature and radiative power output of a single gray body. I repeat: you can find this equation in heat transfer textbooks and I also showed you where it is in Wikipedia. Stop pretending ignorance about things I already explained to you clearly several times. I can only conclude that you're doing this in order to harass.
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".
NO, it does NOT result in a heat transfer equation. There is no need to account for other, cooler bodies when calculating radiative power out. What, do you imagine that these cooler bodies are somehow "sucking" power away from the heat source? And that a warmer body (but still cooler than the source) "sucks" less power than colder ones do? That seems to be what you're saying here.
Just no. That's not the way it works, man. At steady-state, radiative power out can be calculated from temperature and emissivity alone. Other factors (such as heat transfer) are affected by nearby bodies, but radiative power out of a gray body at steady-state is related ONLY to temperature and emissivity. It's that simple, and claiming otherwise is just wrong.
I repeat: look it the hell up. I have not just one but 4 textbooks here, plus Wikipedia, plus the testimony of experts in the field of heat transfer. They ALL disagree with you. It's that simple.
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
Just no. That is not even remotely what I meant, and I explained this to you clearly at least several times already. I have no reason to continue to re-explain it just because you keep asking.
Instead I'm going to repeat something else I have stated several times: pick up a textbook on heat transfer, and see what the accepted, textbook, "consensus" science says about it. Hint: they don't agree with you.
I don't appreciate this constant harassment over something that has been explained to you clearly many times over. If you truly still don't understand it, that is sad but it is also not my problem. A textbook might help.
You basically have to weight how often a craft will be hit by really high energy particles.
But this is the whole point.
"Really high energy particles" do not come exclusively from the sun. While we can agree that MOST of them do.
So -- again I think we agree -- shielding in non-sun-facing parts does not have to be anywhere near as heavy. HOWEVER... regardless of whether you are referring to sun shielding or shielding from the rest of space, reducing the volume (and cross-sectional area) of the crew area that needs to be shielded can make a big difference in the mass. And, as you probably know, reduce the volume and mass of the crew area and the entire ship can be a lot smaller from start to finish.
That is all I was saying.
However, the companies said that they are in different regions and are not going to suppress the competition.
By surreptitiously "agreeing" to operate and different areas, they ARE suppressing competition. In most areas your "one" cable operator is the only game in town for broadband.
DirecTV might be great for TV, but not so much for internet.
Jane, the answers you've given don't make any sense.
They don't make any sense to you. This much is obvious.
That's why I'm asking you for a very simple equation describing the required electrical heating power.
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.
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.
As I implied elsewhere, when you minimize solar radiation you are eliminating most of the energetic radiation/particles, but by no means all. We already know this from Spacelab and ISS experience. If you ignore extrasolar energetic particles you're just being stupid. Unless you plan a 1-way trip. Which has been suggested.
Certainly most of the shielding should be between the sun and the crew. But it's not all the shielding necessary. And though the "other" shielding need not be as heavy, its area is much larger so it still contributes a lot to the overall mass of the vehicle.
Based on other arguments with khayman80, to be honest I would not trust him to build a bridge over a creek, much less a spaceship. That's just the truth.
You fail to read as well,
Unlike you, I didn't "fail" to read. The direction most of the dangerous radiation comes from is not irrelevant, BUT you seem to think that JUST because most comes from the sun, that's the only significant shielding needed.
Bullshit.
We already know better from experience. Why didn't YOU know that?
America is quickly reaching a point, if we are not there already, where their are two systems of law; one for the wealty, the privledged and the corporations and one for the rest of us. Guess who is getting fucked hard ?
I'm not going to claim I disagree with you, but whether we agree is not the real point.
The only answer to being fucked hard is to fuck back. You don't have to enjoy it, you just have to make sure THEY enjoy it a lot less.
Then, when the worst offenders are too sore to continue, you can work on building societal justice.
Republicans claim they want to "set the private sector loose" and it causes more income inequality.
Nope. Sorry. Income inequality has snowballed under Obama and his "Progressive" crony-capitalist friends. The actual historical record says you're simply wrong here. And understand that I don't like saying that, because I like you. There aren't many people I can honestly say that to on Slashdot.
The main difference I see between current Democrats and Republicans is the excuses they give for the policies they push
This we can agree on. More on that in a moment.
I don't think it's liberal hypocrisy. I think it's pretty accurate. What passes for conservativism in the US is reactionary and they do cling to a notion of some Ozzie and Harriet past when everything was good and fair and the greatest generation blah blah
Yes, but so what? The "other side" has been pushing alternatives that have been even worse. We can see it in our economy, and our prices, and the incessant wars. I repeat: Republicans are bad, but the Democrats have shown themselves to be even worse.
Obaman "inherited" one "war". Including that one, he has been involved in at least 6 since. He can't blame those on Republicans. Hell, he didn't even ask them.
Yes, it's hypocrisy. But yes, it's on both "sides". The problem is that they aren't really 2 "sides" anymore. It's oppressive government (whether Democrat or Republican) vs. everybody else. And today, "everybody else" is 60% of the voting public.
Windows is designed to be very secure and have its security policies easily managed but most corporate IT departments don't implement the best security measures because of the huge headache that creates.
That's the most hilarious thing I've read in days.
ALL major operating systems offer the very same protections. This is a matter of IT and how they set it up, not the software at all.
You're trying to school someone who has been doing this for 30 years. You're not just wrong, but laughably wrong.
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?)
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.
Every reply you have given the past couple of weeks has demonstrably been a lie, in one form or another: presenting principles which you know to be not representative of the real situation (e.g., heat transfer in place of the proper "radiated power" equation), or claims that I stated something that I provably did not.
One might be characterized as fraud, and the other as libel. And you expect anyone to take you seriously?
Just asking.
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:
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.
Why should any 'sleep option' solve any radiation issue?
For reasons I explained but which you did not bother to read and understand.
I think it's hilarious that you blame me for your own failure to read.
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.
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.
What is wrong with you? I ask this question very seriously. You were very clearly shown to be wrong, using textbook physics methodology, yet you continue this bullshit. Why? I'd really like to know. (And it was indeed textbook physics. I have 3 different textbooks here... wait, make that 4... which all disagree with you.)
You replied not by admitting you were wrong, but by lying about what I wrote and refusing to accept the clear demonstration that your own brand of "physics" as you applied it to this problem is a blatant violation of the Second Law of Thermodynamics.
You leave me no choice but to conclude that either you are one of the "True Believers", and no facts will sway you, or that you're simply being dishonest. I quite literally have no other options.
I'm not saying they're not, but at the same time I've read that the Pentagon has rejected Windows for certain "critical" operations.
It must be kept in mind that for the U.S. government, Windows is not "closed". But for other governments -- even most other governments -- it is a different story.
They don't know the internals and don't trust them. In the same way that the U.S. has (finally... it took a hell of a long time) started looking at Chinese ICs as a threat.
Personally (not having the direct inside view our government has), I wouldn't use Windows for ANYTHING critical. And I feel very sorry for any troops of ours who have to.
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 to keep a blackbody source at 150F inside 0F chamber walls. Remember that "electrical heating power" is different than "radiative power out". Also remember that blackbodies can only absorb radiation, not reflect or scatter it. Finally, remember that the graybody equation has to reduce to the blackbody equation when emissivity = 1.
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.
That reference [omega.com] shows the object (i.e. chamber wall) temperature has an effect on the temperature controlled cavity (i.e. source). Which Jane denies:
Via a QUANTUM EFFECT, you fucking moron. 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.
It's a combination of your historical tendency to straw-man argue, and outright lies about what I wrote.
This is the only kind of reply you're going to get from me, as long as you keep up your dishonest bullshit.
Maybe we should try for the Moon? It's a lot closer, and it would give us time to work out these types of issues?
I'm with you on that.
Seems to me, the "cold sleep" option mainly solves the problems of crew space, resources, and radiation. Those are not small things.
A long-term space expedition must have room to move and exercise. That's a lot of size and mass. Then it needs food to promote exercise and waking function, and waste disposal to match. And THEN all that has to be wrapped in effective radiation shielding, which adds a lot more mass.
Eliminate the exercise, confine the crew to a small space, feed intravenously, and shield only that small part is FAR more efficient.
On the other hand, as many have pointed out, it comes with some serious cost as well.
I think landing on Mars would be a great accomplishment, but establishing a permanent moon base would be a vastly greater accomplishment.