> At 05:26 PM 7/6/03 -0400, Erik Reuter wrote:
> >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
Good assumptions. By the way, the original post was also
asking about weather. One way to produce a temperature gradient
is to have the air near the axis cooled by being near the "window",
while the air at the rim is heated by the sunlight. But the degree
of cooling depends on details of the habitat which are not
specified...
> >
> >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,
The formula above seems to be assuming that pressure at
the axis is zero, which is unrealistic.
At a guess, I would say that the air pressure throughout
the habitat will be pretty much the same. And it will be at whatever
value the habitat is PRESSURIZED to. The extra pressure at the
rim of the habitat is caused by the "weight" of the air above it.
The radius of the habitat is 5 km, so a small region at the rim
has a 5 km column of air above it. But note that weight goes as
distance from the center, so higher parts of the column don't
contribute as much to the pressure as they would on Earth. Also
note that the "column" over the region is really a wedge, bringing
another factor of r/5 (or 1 - h/R) into play.
---David
Integral omitted, I don't work on Sundays. : )
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