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. [Jane Q. Public, 2014-10-05]

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 will do you a favor here, and say: don't bother to go calculating the energy, either. The problem is that an analysis of this kind, based on the assumption that power-in = power-out, is doomed to fail except in coincidental cases. Even conservation of energy can give very misleading results. The black body example I gave shows why your "energy conservation just inside the surface" won't work. Aside from just "view factor" and a few other things, a certain amount of the power in (often a very significant amount) just ends up going right back out, but you often don't see that in the formulas. Quote from one of my references, "Fundamentals of Heat and Mass Transfer", by Inropera, et al., 6th edition, 2006, p13. I have to type this in by hand from the book so any typographical errors are mine. Emphasized words have been capitalized.

Relationship to Thermodynamics

At this point it is appropriate to note the fundamental differences between heat transfer and thermodynamics. Although thermodynamics is concerned with the heat interaction and the vital role it plays in the first and second laws, it considers neither the mechanisms that profide for heat exchange nor the methods that exist for computing the RATE of heat exchange. Thermodynamics is concerned with EQUILIBRIUM states of matter, where an equilibrium state necessarily precludes the existence of a temperature gradient. Although thermodynamics may be used to determine the amount of energy required in the form of heat to pass from one equilibrium state to another, it does not acknowledge that HEAT TRANSFER IS INHERENTLY A NONEQUILIBRIUM PROCESS. For heat transfer to occur, there must be a temperature gradient and, hence, thermodynamic nonequilibrium. The discipline of heat transfer therefore seeks to do what thermodynamics is inherently unable to do, namely, to quantify the RATE at which heat transfer occurs in terms of the degree of thermal nonequilibrium. This is done via the rate equations for the three modes ...

Heat transfer requires a temperature gradient, and therefore thermodynamic non-equilibrium (as we established early on). I was hoping you would catch on that this also implies that power-in = power-out is not necessarily true, and in fact that is probably a very rare exception. Therefore, you aren't going to prove anything with this approach. I wanted to stop you before you wasted more of your time. [Jane Q. Public, 2014-09-07]

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"?

Again, it really sounds like 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".

That makes sense. I was thinking in terms of the lighthuggers in Revelation Space which are literally "glazed" in water ice because that's cheaper than setting up a system of tanks and pumps. That shield would be very easy to repair even after a collision with a "large" micrometeorite because there would be no infrastructure. Just melt more water and apply.

The system you describe would also be useful as an alternate (albeit temporary) way to dump heat in case the external radiators were damaged. I've been thinking about a similar setup, but using loop heat pipes instead of steam pumps because heat pipes don't have moving parts.

If a VASIMR drive could sustain 0.25g acceleration, its fuel tanks would be enormous. It would also use a lot of power, requiring either a nuclear reactor or huge solar panels capable of supporting themselves at 0.25g.

But if it could be done, continuously accelerating at 0.25g to the midpoint then decelerating at 0.25g would result in an Earth-Mars travel time much shorter than 3 months. When Mars is closest to Earth, the travel time would only be 3.5 days. Even when Mars is on the other side of the Sun, the travel time would only be 9.4 days.

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.

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 likely) just keep calling me a fraudulent dishonest lying dumbshit fucking moron idiot asshole. But note that the last option says more about Jane than me.

Jane responds.

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.

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...

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.

Jane's power in = ?
Jane's power out = ?

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.

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?