Volume of the convex hull of parallel (convex) shapes?

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gmalivuk
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Volume of the convex hull of parallel (convex) shapes?

Postby gmalivuk » Tue Apr 05, 2016 6:01 pm UTC

In the square base cups thread, we're talking about a solid that's a circle on the top and (approximately) a square on the bottom. One of the likely constraints on alternative Solo cup designs is that the volume would be the same, but I don't know how one would calculate that for the square-base variety.

The formulas I've seen for the volume of a frustum all seem to be for the specific cone/pyramid case where the top and bottom are similar. Is there a "nice" formula for a (slightly) more general frustum, where the top and bottom are parallel and convex, but not necessarily similar, and the sides bound the convex hull of the two shapes in space?
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Re: Volume of the convex hull of parallel (convex) shapes?

Postby cyanyoshi » Wed Apr 06, 2016 1:52 am UTC

There might not be a nice formula in the general case, but they do exist for simple cases like antiprisms. Consider two parallel 2-gons, and the convex hull should be a trapezoid or something that has zero volume. Now rotate one of them by a little bit, and their convex hull is suddenly a tetrahedron, which has nonzero volume. It is not clear how to calculate the volume of arbitrary shapes with arbitrary orientations, but there may be some symmetry to exploit with the cups with square bases.

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Re: Volume of the convex hull of parallel (convex) shapes?

Postby gmalivuk » Wed Apr 06, 2016 2:13 am UTC

cyanyoshi wrote:Consider two parallel 2-gons
Ah right, I hadn't thought to consider examples like that to sanity-check whether a general formula would even be possible.
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Re: Volume of the convex hull of parallel (convex) shapes?

Postby cyanyoshi » Thu Apr 07, 2016 9:59 pm UTC

I'm fairly sure this is the cross section of the convex hull you are looking for.

Image

In this picture, r is the radius of the circle, s is the side length of the square, and h is the vertical distance between them. Integrating the area of the slices from z=0 to h gave me a volume of (h/3)*(πr2 + s2 + 2rs).

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Re: Volume of the convex hull of parallel (convex) shapes?

Postby gmalivuk » Fri Apr 08, 2016 1:15 am UTC

cyanyoshi wrote:Integrating the area of the slices from z=0 to h gave me a volume of (h/3)*(πr2 + s2 + 2rs).
That confirms the geometric intuition I had of the square frustum - pyramid + cone pieces that make up the shape.
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Re: Volume of the convex hull of parallel (convex) shapes?

Postby DiEvAl » Thu Apr 28, 2016 5:53 am UTC

It's (almost) a Prismatoid, so it's volume is h(A1+4A2+A3)/6.


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