Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.

Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.

I suspect that glazing a ship in ice would always be cheaper than building a network of chambers to act as a radiation shield. That's because it seems like improvements in robotic construction (prefabrication, etc.) could also be applied to the glazing process.

It seems even more likely that the repair costs of a "glazed" shield would be lower than a shield made out of water chambers. If a "large" micrometeorite blasts a chunk of a glazed shield away, you just send a robot with a water tank out to the hole and let it spray more water on the shield. If that micrometeorite hits a shield made out of water chambers, you have to repair or replace whatever chambers and pumps were damaged in the explosion.

Again, metric tons per square meter = thickness * density. That means the half-value layer should be inversely proportional to the shield's density. So if metric tons per square meter are relevant to the half-value layer, the half-value layer should be inversely proportional to the shield's density.

Yes, on a first glance it should. But in fact it does not. It highly depends on the material you use.

Again, to a good approximation, those half-value layers are inversely proportional to the shield's density.

I plotted those half-value layers against the inverse densities of concrete, steel, lead, tungsten and uranium. The blue squares are for the iridium source, and the red circles are for the cobalt source. Since those points lie close to a straight line, radiation absorption is determined primarily by density. So metric tons per square meter is a good first order approximation, at least for those materials.

What do you mean by that? What are "people like me"? "Laymen"?

Obviously...

Why are you making assumptions about who I am? If you clicked on my homepage, it would only take a few seconds to realize that you're wrong. But more importantly, it's not necessary or productive to accuse someone of being a layman. There's no reason to be nasty. Just discuss the science, and leave your assumptions about who the other person is out of it.

I linked you the 'half value layer' articles ... metric tons per square meter are irrelevant.

No. Metric tons per square meter = thickness * density, so if density is relevant then metric tons per square meter is also relevant.

Relevant is how dense the material is and what its actual properties are to 'break' or capture cosmic rays. A ton of water simply does not equal a ton of lead, even if you believe so after you got missleaded by that NASA article :)

The NASA article I showed you explicitly calculated the required shielding using silicon dioxide (Moon dust) as I've failed to explain. They're not saying a ton of water exactly equals a ton of lead, and neither am I.

no one uses tons per m^2 to describe radiation absobtion. A measure like that would imply the material used is irrelevant, which it is not. The correct material is the prime shielding factor.

No, density is the prime shielding factor. That means metric tons per square meter is a good first order approximation.

That is a laymen explanation for people like you. I linked you the 'half value layer' articles ... metric tons per square meter are irrelevant. Relevant is how dense the material is and what its actual properties are to 'break' or capture cosmic rays. A ton of water simply does not equal a ton of lead, even if you believe so after you got missleaded by that NASA article :)

Again, metric tons per square meter = thickness * density. That means the half-value layer should be inversely proportional to the shield's density. So if metric tons per square meter are relevant to the half-value layer, the half-value layer should be inversely proportional to the shield's density.

Did you try plotting those half-value layers against the inverse densities for concrete, steel, lead, tungsten and uranium? If you did, you'd notice that they're all close to a straight line. So metric tons of shielding per square meter is a good first order approximation.

Also, you claimed I mixed up the travel time, but you still haven't shown that my 3.5 day travel time to Mars at 0.25g is somehow wrong. What travel time did you get?

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. [Jane Q. Public, 2014-10-06]

Once again, your textbooks don't say I'm wrong. They just say that "radiative power out per square meter = (e*s)*T^4". Once again, I agree with that statement. So how am I going against "established" physics or arguing with "every textbook on the subject"?

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. [Jane Q. Public, 2014-10-06]

Jane, you're saying:

electrical heating power out per square meter = (e*s)*T1^4.

But the Stefan-Boltzmann law in your textbooks actually says:

radiative power out per square meter = (e*s)*T1^4.

Jane, don't you see how your equation for electrical heating power would only be true if "radiative power out = electrical heating power"? If you "didn't say they were the same" then why does your equation depend on them being the same?

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. [Jane Q. Public, 2014-10-06]

Jane seemed to try to explain why he disagrees here by saying "A = A" and helpfully asking if I knew what a zero was. But as usual Jane refused to actually write down what Jane considered to be the "correct" energy conservation equation. When I asked what equation Jane meant, Jane said that wasn't it. So Jane's never written down an energy conservation equation around the heated source, which is the first step to calculating the required electrical heating power.

Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out through any boundary where nothing inside is changing:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Instead of calling me dishonest, could you calmly explain why you disagree with this energy conservation equation? If you disagree with that equation, it would be very easy and very fast to write down the energy conservation equation you think is correct. Just fill in these blanks:

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

You're just re-hashing old arguments that I've already shot down. Why are you doing that, if your purpose is not dishonest? [Jane Q. Public, 2014-10-06]

Dishonest? Shot down? Have you even considered the possibility that radiative power out might actually be 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. Is saying that dishonest?

This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature. Is saying that dishonest?

If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy. Is saying that dishonest?

Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out through any boundary where nothing inside is changing:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Instead of calling me dishonest, could you calmly explain why you disagree with this energy conservation equation?

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

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.

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.

Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out through any boundary where nothing inside is changing:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

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?

That is a laymen explanation for people like you.

What do you mean by that? What are "people like me"? "Laymen"?

... heat transfer is NOT the same as the radiative power of a SINGLE gray body at steady-state. ... [Jane Q. Public, 2014-10-05]

I've explained that net heat transfer = radiative power out - radiative power in, so of course they're not the same.

... Power out is a function of emissivity and temperature ONLY. ... [Jane Q. Public, 2014-10-05]

I've repeatedly failed to communicate that I agree radiative power out is a function of emissivity and temperature only:

"Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary."

Once again, I'm just saying that "radiative power out" is different than "electrical heating power".

... But note that the equation for power out clearly implies it is independent of transfer to cooler bodies. ... [Jane Q. Public, 2014-10-05]

I've repeatedly failed to communicate that I agree radiative power out is independent of transfer to cooler bodies:

"Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law."

Once again, I'm just saying that "radiative power out" is different than "electrical heating power".

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

Exactly. Radiative power out is different than electrical heating power, because only electrical heating power goes to zero when the chamber walls are also at 150F. So electrical heating power involves 2 bodies, but radiative power out requires ONLY emissivity and temperature to calculate.

... are you going to continue to erroneously claim that radiative POWER output is dependent on the presence of cooler bodies? ... [Jane Q. Public, 2014-10-05]

I've never claimed that radiative power out is dependent on the presence of cooler bodies. Once again, I've repeatedly agreed that radiative power output doesn't depend on the presence of cooler bodies:

"I've been trying to tell Jane: we don't disagree about the equation for radiative power out."

Once again, I'm claiming that "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.

If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy.

Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out through any boundary where nothing inside is changing:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Instead of cussing and screaming, could you calmly explain why you disagree with this energy conservation equation?

... Your misguided attempt to include the power used to cool the chamber walls does not change that. ... 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. ... [Jane Q. Public, 2014-10-05]

Good grief, Jane. Once again, I never attempted to include the power used to cool the chamber walls! In fact, I've repeatedly told you it's irrelevant. Once again, that's not what "power out" means. Months ago, after I asked Jane if he agreed that power in = power out, Jane misunderstood my question and responded:

... As long as the power used by the source and the power used by the cooler are constant as required, any relationship between them has no bearing on the experiment. [Jane Q. Public, 2014-08-02]

So I explained that "I've never even mentioned the power used by the cooler of the chamber walls... none of these equations has anything to do with the power used by the cooler. ... Jane's also wrong to claim that the power used by the cooler is required to be constant. ..."

I tried again a month later: "I've repeatedly failed to explain that the power consumed by the refrigerator on the outside is irrelevant. So obviously we'll have to agree to disagree about that."

I tried once again: "... Jane might think I meant power in = electrical heating power, and power out = cooling power of the chamber walls. If so, that's not what I meant, and I'm sorry for not being more clear. I take full responsibility. Just to be clear, power in = power flowing into the boundary in question, and power in = power flowing out of that boundary. ... any power used by the cooler is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant."

I tried yet again: "I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant."

... Your misguided attempt to include the power used to cool the chamber walls does not change that. ... 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. ... [Jane Q. Public, 2014-10-05]

After I repeatedly explained that the power used to cool the chamber walls is irrelevant, it's bewildering that Jane accuses me of trying to include it.

... 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. ... 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]

... I know energy is always conserved, you insufferable ass. ... [Jane Q. Public, 2014-10-05]

Charming. So we can agree that an analysis based on the assumption that power in = power out isn't doomed to fail? We can agree that power in = power out is necessarily true for all boundaries where nothing inside is changing, not just for coincidental very rare exceptions? That's great. For some odd reason I thought you were disputing those points.

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?

... Your insistence on "electrical heating power" is a red herring. Energy is energy. ... (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. ... 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. [Jane Q. Public, 2014-10-05]

Once again, Jane, you have 4 textbooks that say "radiative power out per square meter = (e*s)*T^4". Since I've repeatedly agreed with that statement, those textbooks don't disagree with me.

Once again, I'm actually saying that "radiative power out" is different than "electrical heating power".

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

Once again, Jane, I never said we need to account for other, cooler bodies when calculating radiative power out.

Once again, I'm actually saying that "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.

If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy.

... 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. ... I have already proved that my answer conserved energy and yours did not. ... [Jane Q. Public, 2014-10-05]

Again, Jane got nonsensical answers and had to invent a new energy conservation law where power adds to the energy inside a boundary even if it never crosses that boundary. Correctly applying conservation of energy shows that Jane's electrical heating power drops in half after it's enclosed. But since Jane seems convinced that he held the electrical heating power constant, we clearly disagree about the principle of conservation of energy.

Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out through any boundary where nothing inside is changing:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Instead of cussing and screaming, could you calmly explain why you disagree with this energy conservation equation?

no one uses tons per m^2 to describe radiation absobtion.

Except NASA: "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."

You are mixing some stuff up :) actually lots of it.
Radiation absorption is not measured in tons per m^3.

If I'm mixing lots of stuff up, just explain how this NASA study was wrong to conclude that 4.41 tons/m^2 would be sufficient shielding.