Erik Reuter wrote:
>
> On Sun, Jul 06, 2003 at 09:28:52PM -0400, David Hobby wrote:
>
> > Sure there is something to provide the pressure! The habitat
> > won't let air out, and you've pumped it up to the point where there is
> > 1 atm at the rim. Or?
>
> No, that is counteracted by the rotational motion.
Sorry, I have no idea what you mean here.
>
> > But look, here's a quote from the original question:
> >
> > At 5.5 km (18,000 feet), atmospheric pressure is about 0.5 bar.
> >
> > This is for Earth, and sounds about right. So even for
> > Earth, you get a pressure of .5 atm at around 5 km high for a
> > "surface" pressure of 1 atm. But the gradient will be weaker in
> > the habitat, for the two reasons I mentioned in my previous post:
> > 1) Weight falls off closer to the axis.
> > 2) The region of space "above" a given square on the "surface" is
> > wedge-shaped instead of box-shaped, tapering as it goes "up".
>
> Bad inference. The earth is much larger than the habitat described. I
> calculated the pressure based on my assumptions. If you have different
> assumptions, then state them. If you think my formula is wrong based on
> the assumptions, then state a better one.
I'm going straight from Robert Chassell's post. The habitat
is a cylinder with r = 5 km. It rotates so that there is 1 g at the
rim. It is full of just enough air so that the pressure at the rim
is 1 atmosphere. I am assuming room temperature, not that this is
very critical. His first question was about the pressure distribution.
No, an analogy with the Earth is fine. I'm not using the
entire Earth, just a box shape of air over a small square on the
surface. If you want, I'll surround it by curtains to make sure the
air is undisturbed. But I can calculate the pressure distribution
inside the box, and the result will be a good approximation of the
actual pressure distribution since all I'm neglecting are winds.
As I said, I'm not calculating the distribution just now.
It's not really a fair tactic to say that I must.
> Mine probably isn't wrong, however, given the assumptions. It is a
> fairly simple calculation. The chemical potential of an ideal gas is
> proportional to k T ln[ n[h] / nq ], where n is the concentration of
> particles per unit volume. The potential energy in the reference frame
> due to the rotation is m v^2 ln[1 - h/R ]. Add that on to the internal
> potential of the ideal gas and solve for n assuming the chemical
> potential is constant with height (which it is in equilibrium, by
> definition of chemical potential). By the ideal gas law at constant
> temperature, pressure is proportional to n. Thus the result I quoted.
Oh, so this is what you were doing. This sounds like a fine
approach, although I'm shaky on your potential energy term. If I
differentiate the potential energy, I should get the centrifical force.
To simplify, let's use the distance from the axis as the variable:
r = (R-h). Then your potential energy is m v^2 ln[r/R], and its
derivative with respect to r is C/r for some constant C. This is
wrong, force should go linearly with r. Right?
---David
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