Darren New <dne### [at] sanrrcom> wrote:
> A few questions: ARe you talking about gravity in the surronding space 
> (i.e., flying a space ship around this planet) or are you talking about 
> gravity on the surface?

  There's no difference between those two. The gravity at any point in
space is calculated in the same way regardless of where that point is.

> Just off the top of my head, it doesn't seem like it would be terribly 
> difficult to solve the gravity equation analytically (integrating either 
> over cylinders or disks-with-holes)

  If I understand correctly, the only way to solve it analytically is to
solve a rather complicated triple integral (ie. a volume integral). Unlike
with a sphere, the volume integral most probably cannot be simplified into
a simpler one (or if it can, it would have to be proved, and proving it
most probably requires solving the triple integral and seeing that it
reduces to a simpler integral, so you are not going to save anything).

  When talking about a spherical body the situation is much simpler.
If you solve the volume integral for the gravity field (outside) of the
sphere, you'll find out that the sphere radius plays no part on the
result (as long as the radius is smaller or equal to the distance between
the point being tested and the center of the sphere). Thus the radius can
be reduced to zero (while retaining the mass), significantly simplifying
calculations. However, that's probably not the case with a torus because
it does not have the required symmetries. (If it is the case, it would
have to be demonstrated, again, probably solving the triple integral.
One could perhaps assume that the minor radius of the torus could be
reduced to zero without affecting the result, but that would have to be
proved before you can make the assumption.)

> The way you're talking about slicing things up, you *are* talking about 
> integrating the volume to find the gravity in a particular direction. You're 
> just trying to figure out how to do an integration with easier shapes.

  No. What I'm talking about is distributing sample points evenly inside
the torus (or, more precisely, making the masses of these sample points
such that the overall mass density is even throughout the torus) for the
purpose of performing a numerical approximation.

  It has nothing to do with integration.

> A quick google turns up http://www.mathpages.com/home/kmath402/kmath402.htm
> But you probably already did that.

  Yes, but I'm not completely convinced that the solution presented is
valid, both because I don't fully understand all the math, and because
it may be that the author is making simplifications without proving that
they can be made without affecting the result.

  Anyways, I'm not so interested in the end result as in the *process* of
getting there by numerical approximation, which is the whole point. :)

> It also seems if you made *enough* test point masses, worrying about how to 
> subdivide it wouldn't be worthwhile. If your torus has radius 1, and you 
> slice it into 10,000 test masses, isn't that going to give you enough 
> accuracy? Use the digitalness of your compuations to advantage. Put mass in 
> a sphere if the center of the sphere is inside the torus, and calculate from 
> there.

  The amount of points is not the problem. It's their distribution that is
(either spatial or mass).

  You can't just put points of equal mass in whichever way you want inside
the torus because you easily end up with uneven density. You have to either
distribute the points evenly, or scale their masses according to the local
point density. That's the problem I'm trying to figure out.

-- 
                                                          - Warp

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