"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. (Although a pressure of .988 bar seems
a bit high--a kilometer of height makes a much larger pressure
difference on Earth.)
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!
Or look at Erik's argument again. It makes no reference to
the height of air columns at all.
---David
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