On Wed, Jul 09, 2003 at 09:11:22PM -0400, David Hobby wrote:
> "Robert J. Chassell" wrote:
>
> > We may have interpreted the configuration differently. I interpreted
> > C as meaning a torus, or donut, or `like the inner tube of a tire'.
>
> Agreed.
>
> > .... The short columns must have the same pressure distribution as
> > the long columns in the spokes, since they are in equilibrium with
> > each other at any given height. Now C is nothing but short
> > columns--again nothing changes.
> >
> > Except that this `inner tube' or torus arrangement has no long columns
> > of air within spokes.
>
> Yes, but how do the short columns "know" that the long columns
> aren't there? It doesn't matter what the other columns are!
>
> > Let me put this another way:
> >
> > Given (by the specification) that the pressure at the rim is 1 bar
> > and the surface acceleration is 10 m/s^2,
> >
> > Case 1: the spinning tuna can
> >
> > The air column above a point on the rim is 10 km, going to
> > the other side, and it is 5 km to the central spin axis.
> >
> > Case 2: the spinning donut
> >
> > The air column above a point on the rim is 1 km, although the
> > diameter of the torus is 10 km.
> >
> > In each case, what is the air pressure at an altitude of 1 km from
> > the rim?
> >
> > For case 1, based on what Erik wrote, the pressure is 0.988 of the
> > rim pressure. What is the air pressure for case 2?
>
> The same as in case 1.
Yes, I agree.
P/P0 = exp[ - ( h / R )^2 / 3.45 ]
Since h/R = 1/5 = 0.2, P/P0 = 0.988
> (Although a pressure of .988 bar seems a bit high--a kilometer of
> height makes a much larger pressure difference on Earth.)
As I said before, it does not make sense to make direct numerical
comparisons with Earth. Earth has a different potential gradient and is
much larger than a 5km habitat. You have a better physical intuition
than I do, David, but I think your refusal to work with actual equations
and numbers is hampering you here.
The potential energy at a height h above the Earth is
U = m g h / ( 1 + h / R_e )
and the resulting equation for pressure
P/P0 = exp[ -( h / R_e )( R_e m g / k / T ) / ( 1 + h / R_e ) ]
but since R_e = 6370km, and h = 1km, (1 + h / R_e) = 1 is an excellent
approximation so the formula becomes
P/P0 = exp[ -( h / R_e )( R_e m g / k / T ) ]
= exp[ -739 ( h / R_e )]
At 1km on Earth, P/P0 = 0.89, but it is worth repeating again that the
formula is different, exp[-h] dependence instead of exp[-h^2], and the
radius used in each formula is vastly different. So it is a bad idea to
make direct numerical comparisons of pressure gradients between Earth
and small, spinning habitats.
> I'll try one last time. You are free to add all the partitions
> between parts of the habitat you want, and it won't affect the
> pressure. So go from Case 1 to Case 2 by adding a "ceiling" partition
> at 1 km height. It makes no difference!
Let me expand on this a little. At any point inside the habitat, there
is a gas pressure (from many atoms randomly bouncing around and hitting
whatever is measuring or feeling the pressure). But in equilibrium, the
pressure exerted in one direction must be equal to the pressure in the
opposite direction (if it were not, there would not be equilibrium and
in the absence of any transient driving forces the system would change
until it did reach such an equilibrium).
The point I am making is that at any height the gas pressure pushing up
must equal the gas pressure pushing down at equilibrium.
One more point. If you push your hand against the wall, the wall pushes
back with an equal and opposite force -- i.e., the force exerted by the
wall is not constant, it depends on how hard you press. Some people find
this counter-intuitive, but the wall is really no different than a very,
very stiff spring. The more you squeeze (or pull) a spring the harder it
becomes to squeeze (or pull) it further. The same for the wall, but it
is so stiff you cannot see it compress (unless you are very strong or
the wall is very cheap!).
So, when the gas column above 1km is replaced by a wall, nothing is
different about the pressures, since the gas molecules exert a pressure
on the wall and the wall pushes back with the same pressure (which is
also the same pressure as the 4km gas column above used to press on the
1km gas column below).
--
"Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/
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