# [pstricks] pst-3dplot

camus.philippe at free.fr camus.philippe at free.fr
Fri Jul 4 18:18:24 CEST 2008

Selon camus.philippe at free.fr:

> Selon Christoph Bersch <usenet at bersch.net>:
>
> > camus.philippe at free.fr schrieb:
> > >
> > > \pspicture(-5,-5)(5,5)
> > > \psset{Alpha=150,Beta=40}
> > > \pstThreeDCoor[xMin=0,xMax=4,yMin=0,yMax=4,zMin=0,zMax=4]
> > > \def\al{60}\def\be{130} %just because I can't use \Alpha and \Beta !
> > > \FPeval{a}{cos(\be)*sin(\al)}
> > > \FPeval{b}{cos(\be)*cos(\al)}
> > > \FPeval{c}{sin(\be)}
> > >
> > > \pstThreeDLine(\a,\b,\c)(0,0,0)
> > >
> > > \endpspicture
> > >
> > > I expect that the 3DLine is reduced to a point (the origin).
> >
> > I don't known why you expect this. \a \b and \c each contain a number
> > (which are all != 0 for your angles), so you plot a line from the point
> > specified by theses coordinates to the origin.
> >
> > Christoph
>
> Because the transformation frome 3D to 2D is made through the matrix :
>      (-cos(al)         sin(al)          0      )
>    M=(                                         )
>      (-sin(al)sin(be)  -cos(al)sin(be)  cos(be))
> and (if my computations are OK),
>      (\a)  (0)
>    M* (\b)= ( ).
>      (\c)  (0)
>
> (this is what I mean by "kernel" of the matrix.
> More generally, every projection is made following a direction, and I'm
> looking
> for this direction...
>
> Philippe
>
Sorry
the answer is sometimes too obvious !
fp seems to work with radians when postscript uses degrees !
that's ll !
but I'd still be glad to know how to use Alpha and Beta directly in the
computation.

Philippe