On Sun, Jul 06, 2003 at 07:22:39PM +0000, Robert J. Chassell wrote: > What is the air pressure near the center of a spinning O'Neil type > space habitat when the pressure at the rim is 1 atmosphere?
> What is the equation that tells you the pressure? Tricky. The only way I know of to get a simple equation is to make a couple approximations: (1) The temperature of all the air is the same, 300K (2) The air all rotates with the endcaps, speed proportional to radial distance from center In that case, the formula I come up with is P = P0 ( 1 - h/R ) ^ ( m v^2 / k / T ) where h is the distance above the "floor" of the habitat. If we make another approximation that the air is just N2 molecules, then m=4.8e-26 kg, and v^2 = 10R, and k=1.38e-23, and T=300K, then P = P0 ( 1 - h/R ) ^ 0.58 In reality, I'm not sure how good this formula will do. There is such gradient in air velocity with height. I think it may be necessary to take into account wind currents, and depending on the heating, thermal currents may be important. You may need to do a full simulation to get useful results. > Am I right about the spin? My membory is that A = v^2/r, where A is > the acceleration, equal to circumference/time-of-a-rotation, and v is > the tangential velocity of the rim. > ------------ > | 4 pi^2 r > Or, put another way, T = period-of-rotation = \ | ---------- > \| A > That is correct. -- "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/ _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l