Hello,

  Does any of you know how I could calculate the *exact* coordinates
of the tetrahedron points' coordinates? I mean, yes of course I looked
at include/shapes2.inc, but they seem to be approximations at 0.1 %, and I
need a greater precision (for a scientific job). So I'm looking for
the formulae that could give me these coordinates, or the equations from
which I could draw them.

  Any hint would be appreciated, even if it is just a geometry hint.

  E-mail answers will be summarized and posted here, news article answers
are welcome. Thanks in advance,

Roland.
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bob### [at] casimirrezelenstfr -- Linux, POV-Ray, LaTeX

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Roland Mas wrote:

>   Hello,
>
>   Does any of you know how I could calculate the *exact* coordinates
> of the tetrahedron points' coordinates?

<1   1   1>
<1 -1 -1>
<-1 1 -1>
<-1 -1 1>

Swiped from Jim Blinn (A Trip Down the Graphics Pipeline, p. 40).  It seems
that the a tetrahedron fits comfortably inside a cube such that two opposite
edges of the tetrahedron are diagonals of opposite faces of a cube (a unit
cube is convenient), and the vertices of the tetrahedron are the vertices of
the cube at those diagonals.

Rotate, translate, and/or scale to suit your needs, but be prepared to lose
precision in some cases.  Still, the above is exact, if you're willing to
accept that orientation.

-Mark Gordon

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Gosh, I knew it. It had just sneaked out of my mind.

  Thank you...
Roland.
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bob### [at] casimirrezelenstfr -- Linux, POV-Ray, LaTeX

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Mark T. Gordon wrote:

> Roland Mas wrote:
>
> >   Hello,
> >
> >   Does any of you know how I could calculate the *exact* coordinates
> > of the tetrahedron points' coordinates?
>
> <1   1   1>
> <1 -1 -1>
> <-1 1 -1>
> <-1 -1 1>

The simple demonstration we alsways try to find a complex answer when with hind
sight it is so simple ...
I'm impressed although I should not be - given the amount of maths I did,

Cheers,
Al.

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