I finally had some time to play around with Knighty's new polychora scripts, and I am even beginning to understand a little of it.

These videos were a god help to understand stereographic projection of polychora:

http://www.dimensions-math.org/(I think one of the authors, Jos Leys, is a Fractal Forums member)

Now, for the original 3D polyhedron script, it is quite easy to find the settings for the five convex, regular polyhedrons (the platonic solids):

Type,U,V,W (<- Knighty's variable)

3,0,1,0 Tetrahedron

4,0,0,1 Cube

3,1,0,0 Octahedron

5,0,0,1 Dodecahedron

5,0,1,0 Icosahedron

But for the 4D polychora it is more difficult to find the 6 possible regular and convex polychora.

I tried using the 'polychora06.frag' script, and have found the following candidates:

Type,U,V,W,T

3,0,1,0 0 - 5-cell, 4-Simplex [Tetra cells]

4,0,1,0 0 - 8-cell, Hypercube [Cube cells]

4,0,0,1 0 - 24-cell * [Octa cells]

4,0,0,0 1 - 16-cell [Tetra cells]

5,0,1,0 0 - 120-cell [Dodeca cells]

5,0,0,0 1 - 600-cell [Tetra cells]

But I would like to have them confirmed, if other tried this - it is not easy to identify these for high vertex counts :-)

In the comments for the script it says the 24-cell symmetry group is not present, and Knighty posted a special script for this.

But in this "polychora-special.frag" fragment, the "24-cell" preset has 8 vertices and 24 edges, so it cannot be a 24-cell? (It could be the 16-cell, though). However, setting (U=1, and V,W,T=0) gives an object with 24 vertices, and 96 edges, which could be the 24-cell?

It does however look extremely similar to the one I got above using the 'polychora06.frag' script - see the attached picture, comparing the two scripts.