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?

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

**Moderators:** gmalivuk, Moderators General, Prelates

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

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.

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

Ah right, I hadn't thought to consider examples like that to sanity-check whether a general formula would even be possible.cyanyoshi wrote:Consider two parallel 2-gons

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

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

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)*(πr

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)*(πr

^{2}+ s^{2}+ 2rs).- gmalivuk
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### Re: Volume of the convex hull of parallel (convex) shapes?

That confirms the geometric intuition I had of the square frustum - pyramid + cone pieces that make up the shape.cyanyoshi wrote:Integrating the area of the slices from z=0 to h gave me a volume of (h/3)*(πr^{2}+ s^{2}+ 2rs).

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