... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... I did NOT claim the net radiative power through the boundary was zero. What I wrote was that the radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. ... The actual radiant power through that boundary isn't zero, because the radiant power output of the heat source still goes through it of course. Leaving a NET positive transfer of energy OUTWARD through the boundary. ... [Jane Q. Public, 2014-10-11]

This is just more dishonest out-of-context nonsense again. I clearly told you that the context of my statement was radiation from the walls through a boundary. I did not claim the net radiation across that boundary was zero. I claimed the net radiation across the boundary from the wall is zero, because it just goes right back out. This is a wonderful example of how you distort context, in order to make it appear someone else is saying something they actually did not. That is a form of lying. [Jane Q. Public, 2014-10-11]

Since Jane keeps bolding "from the wall" and claiming that "radiative power from the walls through that boundary cancels itself out," Jane seems to be saying:

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

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

It certainly seems like that's what Jane's saying. If "radiative power from the walls through that boundary cancels itself out" then those terms cancel to produce Jane's final energy conservation equation:

Jane's power in = electrical heating power
Jane's power out = radiative power out from source

Jane, is that what you're saying by "net radiation across the boundary from the wall is zero"?

... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... I did NOT claim the net radiative power through the boundary was zero. What I wrote was that the radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. [Jane Q. Public, 2014-10-11]

Again, Jane must using some kind of Sky Dragon Slayer definition of the word "net". In physics, "net radiative power out" means "radiative power out" minus "radiative power in". This is only zero when the source and chamber walls are at the same temperature.

Similarly, "net radiative power in" means "radiative power in" minus "radiative power out". Again, this is only zero when the source and chamber walls are at the same temperature.

The actual radiant power through that boundary isn't zero, because the radiant power output of the heat source still goes through it of course. Leaving a NET positive transfer of energy OUTWARD through the boundary. [Jane Q. Public, 2014-10-11]

That's ridiculous, Jane. Notice that "net radiative power out" equals negative "net radiative power in". Since Jane seems to agree that "net radiative power out" is positive, "net radiative power in" can't be zero. It has to be negative, which just means more radiative power is flowing out than flowing in.

So Jane must not be using the physics definition of "net". What's the Sky Dragon Slayer definition of "net"? And how is it possible for the "net power in" to be zero when the source is hotter than the chamber walls?

... 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. ... Energy can be absorbed and re-emitted... [Jane Q. Public, 2014-09-28]

... I do not deny that some radiation is absorbed; but then it's just re-emitted. ... [Jane Q. Public, 2014-10-03]

... Here is a fundamental principle of thermodynamics, as related to radiant energy: net incoming radiation from cooler bodies is ALL either reflected, transmitted, or scattered. Any absorption and re-transmission is part of the "transmitted" term. ... [Jane Q. Public, 2014-10-10]

Jane can't quote a textbook stating this "fundamental principle" because it's nonsense. For instance, the "transmitted" term describes a body's transparency, not its absorption and re-emission. Here's an introduction:

"A body's behavior with regard to thermal radiation is characterized by its transmission t, absorption a, and reflection p. ...

An opaque body is one that transmits none of the radiation that reaches it, although some may be reflected. That is, t = 0 and a + p = 1

A transparent body is one that transmits all the radiation that reaches it. That is, t = 1 and a = p = 0."

Jane, absorption and re-emission isn't part of the "transmitted" term. They're totally different. The "transmitted" term is zero for opaque bodies like aluminum or blackbodies. If absorption and re-emission were part of the "transmitted" term, blackbodies would be transparent because they absorb all radiation that hits them. If absorption and re-emission were part of the "transmitted" term then both terms would equal 1. But once again, blackbodies can't be transparent.

Also, it's bizarre that Jane insists he's accounting for absorbed and re-emitted radiation in a "transmitted" term that isn't even in his energy conservation equation.

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. ... that ZERO of the radiative power output ... THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

... First, there is no NET radiative power absorbed by a body at one thermodynamic temperature from another body at a lower temperature. ... Those are the statements I made. Anything else is a logical extension of those two principles. ... [Jane Q. Public, 2014-10-10]

That's a serious problem, because Jane's first principle is wrong. Net radiative power would only be zero if the source and the chamber walls were at the same temperature.

Jane might consider replacing his incorrect first principle with "conservation of energy" which means power in = power out through a boundary where nothing inside is changing.

... No NET radiative input from chamber walls means anything crossing your precious boundary inward goes right back out. As I have explained to you many times now. ... [Jane Q. Public, 2014-10-10]

Jane, it's not my "precious" boundary. It's a general principle called "conservation of energy". Drawing a boundary is needed to apply conservation of energy. Here are some introductions: example (backup), example (backup), example (backup).

Note that drawing a boundary is needed to apply conservation of energy in all those introductions, just like I've repeatedly explained.

If that's what you think, could you take a few seconds to write down the energy conservation equation (before cancelling terms) that you think is correct?

I could, but I will not. ... I'm not going to take valuable time out of my day ... I'm not going to waste my time. ... [Jane Q. Public, 2014-10-09]

If Jane really could write down an energy conservation equation before wrongly "cancelling" terms, that would only take a few seconds. Endlessly screaming in ALL CAPS and accusing me of lying wastes more of Jane's valuable time than writing down an energy conservation equation for a boundary around the source:

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

... there is no NET radiative power absorbed by a body at one thermodynamic temperature from another body at a lower temperature. Doing so would violate the Second Law of Thermodynamics. I'm using the standard definition of "net", which is to say "all inputs minus all outputs". ... [Jane Q. Public, 2014-10-10]

If net radiative power is "all inputs minus all outputs" then net radiative power is only zero if all the inputs equal all the outputs. That only happens if the source is at the same temperature as the chamber walls.

But Jane claims that net radiative power is zero when the source is hotter than the chamber walls.

NOWHERE did I state "zero net radiative output". I don't believe I used that phrase at all, but if I did, you present it here out of context. [Jane Q. Public, 2014-10-10]

Hmm...

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. ... that ZERO of the radiative power output ... THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

... no NET radiative power ... [Jane Q. Public, 2014-10-10]

Again, net radiative power is only zero if the source is at the same temperature as the chamber walls.

NOWHERE did I state "zero net radiative output". I don't believe I used that phrase at all, but if I did, you present it here out of context. [Jane Q. Public, 2014-10-10]

Oh, so you're saying net radiative power isn't zero? For some odd reason I thought you were saying it was zero.

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. ... that ZERO of the radiative power output ... THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

By the way, just in case it wasn't obvious from the fact that I was responding to Jane's claims of zero net radiation absorbed and zero net radiative power output, I was talking about net radiative power because that's what Jane seemed to be talking about. That's why my equation only has radiative terms. Here's a less ambiguous version:

So you're not going to retract your claim that net radiative power is zero when the source is warmer than the chamber walls?

Is Jane using some kind of special Sky Dragon Slayer definition of the word "net"? In physics, net radiative power through a boundary around the source = "radiative power out" minus "radiative power in".

So you're not going to retract your claim that net power is zero when the source is warmer than the chamber walls?

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. This is a requirement of the Second Law of Thermodynamics. Now, this is NOT the same as saying "no radiation absorbed at all". But when you put the two points above together, what it does mean is that ZERO of the radiative power output from the above equation is coming from other bodies. THE AMOUNT OF POWER OUTPUT IN THIS EQUATION DOES NOT NEED TO ACCOUNT FOR POWER FROM THE WALLS, BECAUSE THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

Is Jane using some kind of special Sky Dragon Slayer definition of the word "net"? In physics, net power through a boundary around the source = "radiative power out" minus "radiative power in".

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

... at steady-state, the relation given above already accounts for any radiative power being absorbed from other bodies. ... [Jane Q. Public, 2014-10-10]

Only if "already accounts for" means "completely ignores" in Janeland.

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. This is a requirement of the Second Law of Thermodynamics. Now, this is NOT the same as saying "no radiation absorbed at all". But when you put the two points above together, what it does mean is that ZERO of the radiative power output from the above equation is coming from other bodies. THE AMOUNT OF POWER OUTPUT IN THIS EQUATION DOES NOT NEED TO ACCOUNT FOR POWER FROM THE WALLS, BECAUSE THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

Once again, it seems like we disagree about the meaning of the term "NET".

1. Can we agree that net power through a boundary around the source = "radiative power out" minus "radiative power in"?

2. Can we agree that net power through a boundary is only zero if "radiative power out" equals "radiative power in"?

3. Can we agree that "radiative power out" only equals "radiative power in" if the source and the chamber walls are at the same temperature?

If we can agree on those three points... how can the net power be zero when the source is warmer than the chamber walls?

I spent more time than I'd care to admit trying to think of a way to arrange the solar panels that doesn't require a special magnetic rotary joint.

Maybe the solar panels could be physically attached to the midpoint, and arranged in a circle with the same surface normal as the plane of rotation.

Ordinarily this would result in no solar power, regardless of whether the spheres' plane of rotation has the same surface normal as the ecliptic plane (the "parked" configuration) or if its surface normal points along the orbital velocity vector (the "Hohmann transfer" configuration).

But a large cheap mirror could reflect sunlight onto the circular solar panel, eliminating the need for a special magnetic rotary joint, and the inefficiency of microwave or inductive power transfer.

The mirror could be held in place against solar pressure using VASIMR drives. When the spheres are under thrust, the total fuel needed to move the mirror should be negligible compared to the fuel needed to move the heavily shielded spheres. Moving the mirror independently would allow the spheres' plane of rotation to change without reconfiguring its solar panels each time.

I've been considering Hohmann transfer orbits because they only require thrust that's completely tangential to the Sun. In that case, the spheres' plane of rotation would be perpendicular to the ecliptic plane.

Sorry, this is ambiguous. Here's a better explanation. I've been considering Hohmann transfer orbits because they only require thrust that's completely tangential to the Sun. In that case, the spheres' plane of rotation would be perpendicular to the ecliptic plane, with its surface normal pointing along the (circular) orbital velocity vector.

That means that in 4 months the configuration will have shifted by 90 degrees, making the solar panels useless.

Oops. In 3 months, or one quarter of an Earth year, the configuration will have shifted by 90 degrees.

... Claiming that I "probably won't" write down a black body equation is a form of lying by implication, because we weren't discussing black bodies! By your own insistence. It was just another straw-man argument, AND blatant dishonesty at the same time. I have a copy of our AGREEMENT to treat all the materials as gray bodies, in black and white. So your claim that I "probably won't" include a black-body equation, when I HAD shown you the gray-body equations I used, is just another dishonest way to distort the argument. ... [Jane Q. Public, 2014-10-09]

Good grief, Jane. As I've repeatedly explained, the gray body equation has to reduce to the black body equation when emissivity = 1. I wasn't lying or being blatantly dishonest. I was trying to show you how to check your work.

... If you're publishing an equation for calculating P, and you have an additive term on one side of the equation, which is exactly cancleled out by an additive term on the other side of an equation, you don't include them, you cancel them. ... it's already known that X cancels out!!! There is no NET absorption of radiative power from cooler bodies. WE KNOW THIS FROM THERMODYNAMICS. So any radiative power that comes in, goes right back out. That's YOUR power in = power out! ... [Jane Q. Public, 2014-10-08]

Careful physicists write down all the possible terms in their equations first.

Sure. When it's relevant. But I don't have to write down extra terms when the equation is already a known physical law. See my comment above about volume of a sphere. The formula is already known and other "terms" are not relevant.

You appear to be trying to imply that I left something out. So if YOU think that, then why don't you write what you think it is here? Hint: we already know what it is, because you've already made that erroneous claim, several times. So you're just re-hashing fallacious old news again. But if you want to "explain" how you think it works again, go right ahead. If it's the same as last time, I reserve the right to laugh at you again. [Jane Q. Public, 2014-10-09]

Jane still refuses to write down his energy conservation equation before he "canceled" terms, so I still have to guess at his original equation. This still seems like the only energy conservation equation consistent with what Jane's saying above:

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?)

Again, Jane appears to be saying that "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out". If that's the case, then those terms would cancel as Jane claims. That's the only way to get from "power in = power out" to Jane's final equation:

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

Really? Let's use the Stefan-Boltzmann law to describe the radiative terms, one at a time. Let's start with a term we can probably agree on. Because "radiative power out from source" is emitted by a graybody source at temperature T1, the Stefan-Boltzmann law says:

electrical heating power per square meter + radiative power in from chamber walls = (e*s)*T1^4 + radiative power from chamber walls, re-emitted back out (Jane's equation?)

Now for the next term. Because "radiative power in from chamber walls" is emitted by graybody walls at temperature T4, the Stefan-Boltzmann law says:

electrical heating power per square meter + (e*s)*T4^4 = (e*s)*T1^4 + radiative power from chamber walls, re-emitted back out (Jane's equation?)

Now for the final term. The only way the final term could cancel with the radiative power in term "(e*s)*T4^4" to obtain Jane's final equation would be if "radiative power from chamber walls, re-emitted back out" equals "(e*s)*T4^4". But it's being emitted by the source, which is at temperature T1. If reflections confuse you, just remember that the gray body equation has to reduce to the black body equation where there aren't any reflections at all. In that case, all that power is being absorbed and re-emitted, not reflected.

The acknowledged formula for finding radiative power from temperature is just (sigma epsilon)T^4. There are no other factors involved... [Jane Q. Public, 2014-09-05]

That's why radiation re-emitted by the source at temperature T1 is (e*s)*T1^4. There are no other factors involved. The source can't re-emit radiation at (e*s)*T4^4, so those terms in Jane's equation can't cancel. And the last term double-counts radiation emitted by the source, so it's zero.

So the correct equation (neglecting reflections) is:

electrical heating power per square meter + (e*s)*T4^4 = (e*s)*T1^4

I think we got into this discussion talking about rotating ships, to provide midi-gravity. We know that microgravity requires a lot of effort to counteract, so ... you're going to need some major engineering reasons to not go down the spin-for-pseudo-gravity route.

Yes, centrifugal gravity seems like the only way to stay healthy in space. I pointed out that long-term colonies shouldn't rotate faster than 1 rpm in order to avoid inducing motion sickness. That imposes such serious tensile strength requirements that it seems like the shield can't spin with the ship unless the ship is made of carbon nanotubes.

Take a close look at the design of the ISS (because I've seen those designs online ; other spacecraft will have the same issues) : the radiators protrude in one direction radial to the Sun, but the solar panels are perpendicular to the Sun. If you rotate the system by 90 degrees, then the solar panels are useless and the radiators become heat absorbers. That's probably a large part of the reason for not rotating the ISS, but ... comments above about the effort needed to avoid the health problems of microgravity.

Yes, the ISS is a useful example. I'm proposing a modular design, where a sphere with interior radius of 10.7 meters has enough living and garden space to support 4 people. One sphere alone couldn't provide centrifugal gravity, but in that configuration the solar panels would be unfolded perpendicular to the Sun, and the radiators would be unfolded behind the sphere, radially away from the Sun.

But two spheres could dock and attach tethers at the top of each sphere. Then if they separate to a distance of 1800 meters, they could rotate at 1 rpm around their shared center of mass to produce 1g of centrifugal gravity.

If they're not going anywhere, their plane of rotation should probably be the ecliptic plane. Otherwise the Sun's orientation would change as they orbit the Sun. Each sphere's radiators could be attached to the tethers, parallel to the ecliptic plane so they never face the Sun.

During the docking procedure, each sphere's solar panel would be detached and remain at the midpoint between the spheres. They'd have to be able to move along the tether in case one of the spheres becomes heavier and moves the center of mass. The solar panels would be kept perpendicular to the Sun as the spheres rotate, so they'd have to be kept in place magnetically and transfer power to the spheres using induction or microwaves.

I still don't like relying on rotary joints, particularly coaxial ones. I'd use them where unavoidable, but I'd avoid them where possible. And in life-support, they'd scare me.

Yeah, me too. That's why I spent more time than I'd care to admit trying to think of a way to arrange the solar panels that doesn't require a special magnetic rotary joint. At first I thought the sphere's plane of rotation should have a surface normal that points directly at the Sun. That way the solar panels could be attached directly to the tethers on the side that always faces the Sun, and the radiators could also be attached directly to the tethers, but at 90 degrees so they never face the Sun. They could also be attached to the side of the sphere which never faces the Sun.

That might be an emergency configuration if the magnetic rotary joint fails, but the sphere's plane of rotation stays fixed as they orbit the Sun. That means that in 4 months the configuration will have shifted by 90 degrees, making the solar panels useless.

It would be too expensive to continually use fuel to keep the sphere's plane of rotation in place relative to the Sun. Maybe an electrodynamic tether could work, but I haven't looked at that possibility in detail.

And on your general voyage (no, you don't design a vessel for only one voyage - craft design versus industrial production?) you are going to have a component of travel which is not radial to the Sun. Therefore, essentially all parts of the ship's surface are going to have alternating exposure to light and dark.

I've been considering Hohmann transfer orbits because they only require thrust that's completely tangential to the Sun. In that case, the spheres' plane of rotation would be perpendicular to the ecliptic plane. This way, engines on the "side" of each sphere could provide thrust for the Hohmann transfer. If that thrust is large enough to perceptibly affect the direction of "down" then the spheres could simply tilt to keep the total effective gravity vector "vertical" relative to the sphere's floors. To keep that total effective gravity at exactly 1g, the spheres' rotation rate would have to be slowed.

One huge problem is that the shielding mass required to cover the sphere with 4.5 tons/m^2 is greater than 20,000 metric tons. Even if each sphere had a Saturn V rocket underneath it, it would only accelerate at 0.15 g for a few minutes. A fission or fusion rocket would probably be necessary to achieve any useful delta-v.

Again, I originally assumed black bodies because they're simpler: black bodies don't reflect any radiation. That means "power in" depends on the chamber walls and "power out" through that boundary only depends on the heat source.

I have it on record where you insisted that we assume gray bodies so that we could include a term for emissivity. Seriously. You insisted. I'm not going to look it up this late at night, because you are getting completely ridiculous. But I am sure as hell going to include it in my publication. [Jane Q. Public, 2014-10-08]

Remember to include that part where I originally assumed black bodies because they're simpler. And the part where Jane insisted that "we should use real materials with real emissivities and absorptivies. Just to keep everybody honest."

But black bodies aren't "dishonest". Also, Jane should make sure to include the part where Jane said I was "lying again" for considering a black body source.

But if Jane wants to work on the simpler black body problem that I originally proposed months ago, that's fine with me. It's simpler, and easier to learn from.

... the equation for radiant power emittance at steady state does NOT say "X + ( (epsilon * sigma) * T^4) - X". It simply says (epsilon * sigma) * T^4. Because it's already known that X cancels out!!! ... [Jane Q. Public, 2014-10-08]

No, that's because the equation for radiant power emittance doesn't have anything to do with conservation of energy, so those extra terms wouldn't be in the Stefan-Boltzmann equation in the first place.

That's what I've been trying to tell you, Jane. The Stefan-Boltzmann equation can give you "radiative power out" but only a completely different principle called "conservation of energy" can give you a totally different quantity known as "electrical heating power".

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

So you're saying electrical heating power is the same as radiative power out? What did that energy conservation equation look like before you cancelled terms? It's important.

Jane still hasn't written down an energy conservation equation for a boundary around the heated source which links "electrical heating power" to "radiative power out".

Yes, I have. I have done it at least several times before, and I just did it again. Not only did I give you the equations, I showed you my exact calculations. Why are you lying again, and trying to claim I did not do something that I very clearly and provably did do? In fact, since you seem to be so obsessed with archiving other people's comments, I am sure you have several instances of where I've showed this to you before. [Jane Q. Public, 2014-10-08]

Jane, just two days ago you claimed that you didn't say radiative power out was the same as electrical heating power, and that they don't need to be the same. Today you're saying they are the same.

... There is no NET absorption of radiative power from cooler bodies. WE KNOW THIS FROM THERMODYNAMICS. So any radiative power that comes in, goes right back out. That's YOUR power in = power out! ... [Jane Q. Public, 2014-10-08]

If any radiative power that comes in goes right back out, that means radiative power in = radiative power in that goes right back out. That means there should be a term for radiative power in, and a term for radiative power in that goes right back out. Maybe they cancel. But careful physicists would write them both down and think about them carefully before erasing them both because they "obviously" cancel.

You are trying to count power incoming from a cooler body as part of the radiative power out of the hotter body. That's counting it twice. I told you that... what was it now? 3 weeks ago? A month? And I told you not just once but several times. [Jane Q. Public, 2014-10-08]

If that's what you think, could you take a few seconds to write down the energy conservation equation (before cancelling terms) that you think is correct?

If there's no net radiative power coming in, that must mean all the "power in" from the chamber walls just goes back out. That would yield a net of zero. But as usual Jane didn't write down the power in = power out equation showing these terms before they supposedly cancel.

Why do that? Nobody does that. That's stupid. If you're publishing an equation for calculating P, and you have an additive term on one side of the equation, which is exactly cancleled out by an additive term on the other side of an equation, you don't include them, you cancel them. [Jane Q. Public, 2014-10-08]

Careful physicists write down all the possible terms in their equations first. Then the second step is to cancel out terms, if possible. Since we're disputing which terms cancel, the only equation we might be able to agree on is the equation before the terms are cancelled.

So let's see if we can agree on the energy conservation equation for a boundary around the heat source, before the contentious terms are cancelled:

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

This would be much faster than accusing me of lying again. It would only be two lines long. Just list Jane's "power in" and Jane's "power out" around the heat source. Just. Two. Lines.