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/
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