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. > 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. 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. -- "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/ _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l
