Finding the vertices of a rhombic dodecahedron I'm trying to figure out a straightforward way to find the vertex (x,y,z) coords for a rhombic dodecahedron. Besides starting with a rhombus and rotating it around at the proper angles, I have no idea.
I need it for a little graphics project I'm working on, so I'd prefer the center to be at the origin to make it easier to place accurately in 3-space. I'd rather not have to import a .obj file from blender or something like that. Projects not really big enough to warrant the extra code for file parsing.
Is there an easy way to do this?
 A: Assuming we are talking about the rhombic dodecahedron described in this wikipedia article, then it would appear that the coordinates for the vertices are given (later in the article) as the points $(\pm 1, \pm 1, \pm1)$ and the permutations of $(0, 0, \pm 2)$.
This would then give the length of each edge of the rhombic dodecahedron as $\sqrt{3}$, and it is straightforward to see which vertices are joined to which, by using this fact together with the 3-dimensional version of Pythagoras.
Edit:
In a bit more detail, these are the vertices:


*

*$(1, 1, 1), (1, 1, -1), (1, -1, 1), (1, -1, -1), (-1, 1, 1), (-1, 1, -1), (-1, -1, 1), (-1, -1, -1)$ 

*$(0,0,2), (0,0,-2), (0,2,0), (0,-2,0), (2,0,0), (-2,0,0)$


The vertices in list (1) are the 8 vertices where 3 faces meet, and the vertices in list (2) are the 6 vertices where 4 faces meet.
Also, for example, the vertices connected to the vertex $(0, 0, 2)$ by an edge are: $(1,1,1), (1,-1,1), (-1,1,1), (-1,-1,1)$, and the vertices connected to $(1, 1, 1)$ by an edge are $(2, 0, 0), (0, 2, 0), (0, 0, 2)$.
