Ari-Matti Leppanen wrote:
> 
> "Rune" <run### [at] mobilixnetdk> wrote in message
> news:3cb81031$1@news.povray.org...
> 
> : I don't know how big to make the bounding objects.
> 
> For a Dodecahedron and Icosahedron I'd take a sphere { 0 1.2577 } and
> for
> a Octahedron a sphere { 0 1.73 }. The Tetrahedron is tricky.. a box?
> Someone quick in math could calculate the difference of the volumes of
> the shapes and the smallest box/sphere/whatever they fit in and see
> thats the best result.
> 
> Ari-Matti

A quick look at shapes2.inc learns that those polyhedra are
circumscribed to a unit sphere.
Knowing the radius of the inscribed sphere it's rather easy to obtain
the radius of the circumscribed sphere for each of the polyhedra (not
the same sphere of course).
look at
http://mathforum.org/dr.math/faq/formulas/faq.polyhedron.html
calculate the ratio of the radius of the circumscribed and the inscribed
sphere.
You can bound the polyhedron by a sphere with a radius a little bit
larger than this ratio.

-- 
Herman Serras
Gent (Belgium)
http://cage.rug.ac.be/~hs/

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