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Finding a minimum bounding sphere for a frustum
It's not square no, however I know that one of the sides will always be longer than the other. I think I've got it, actually. Simplifying it into 2D on the longest axis, and stipulating that the distance to the far vertex on the base and the far vertex on the truncated top should be equal (i.e. the bounding sphere touches the outside vertices), and the maths just kind of works out from there I think. I got that the distance from the base should be (Wn^s + Wf^2 +x^2)/2x, where Wn is the half-width of the narrow side, Wf is the half-width of the fat end, and x is distance from base to top. |
Feb
3 |
asked | Finding a minimum bounding sphere for a frustum |