Amara Graps wrote:
>
> From: Spike Jones <spike66@ibm.net> Fri, 11 Feb 2000 :
>
> >From me:
> >> First of all, regarding the time it takes to fall from one
> >> end to the other (it takes 42 minutes each way) it doesn't
> >> matter _where_ the ends of the cylinder are located, i.e. where
> >> "yahoo_A" and "yahoo_B" are pitching their trash at each other.
> >> The time to fall from one end to the other (by gravity alone)
> >> is independent of the two places.
> >
> >Ja, assuming of course the tunnel passes thru the center
> >of the nonrotating sphere.
>
> No, no, that's one of the beautiful things about this problem.
> It _doesn't matter_ where the ends of the cylinder are located.
> It doesn't have to go through the center of the sphere. You
> simply have to assume that the density of the sphere is constant
> and that there's no rotation.
Only if you have a perfectly frictionless surface for the dropped object
to slide on.
"Imagine a spherical elephant on a frictionless plane..."
I thought Spike's original question wanted a closed-form solution for
the *exact* potential field for a sphere with a large diameter (Dhole =
0.1 * Dplanet) hole through it. This is quite a hairy problem, since
both spherical and cylindrical coordinates are needed. I declined to
work on the solution because I frankly have better things to do with my
time- at my current state of calculus rustiness, it would likely take me
several hours to find the solution... and I'm by no means certain it can
be solved analytically anyway.
I've already spent hours today working on a large worksheet to
interpolate from enthalpy tables the exact cooling capacity of nitrous
oxide. Indirect addressing, linear interpolation, and references to a
three dimensional table of scalar values reduces my enthusiasm for more
number crunching...
Forget AI, I want some IA.
-- Doug Jones Rocket Plumber, XCOR Aerospace http://www.xcor-aerospace.com
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