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Thanks, but what I meant is this:
Take perfect cube. set each edge on a grinding wheel such that the normal of
the two faces on this edge make a 45 degree angle with the normal of the
grinding wheel. Do it for just a second or two.
David Fontaine wrote:
> "Greg M. Johnson" wrote:
>
> > I want the coordinates for a truncated cube, in terms of triangles.
>
> If you have a cube going from <-1,-1,-1> to <1,1,1> and you want to truncate
> it so each of the original six faces becomes a regular octagon, then the
> vertices would be at
>
>
> (2^3+2^3+2^3=24 vertices)
>
> --
> David Fontaine <dav### [at] faricy net> ICQ 55354965
> My raytracing gallery: http://davidf.faricy.net/
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