The Rhombohedron I am trying to model a rhombohedron (using Blender) as a first pass to building Dürer's solid so I am trying to calculate the (x,y,z) values for a given side length 'a' and angle 'theta' (starting with 72).  I want to express the (x,y,z) coordinates in terms of 'a' and 'Theta' so I can use Excel to calculate them and see the effect of changing 'theta'.  I think I have these relationships correct.
I am also trying to characterize the four interior diagonals and find the center of the Rhombohedron to determine where the cicumsphere would be for Durer's solid.
I can determine the diagonals of the flat sides, no problem, in terms of 'a' and 'theta', but I am completely stuck in trying to determine the lengths of the interior diagonals of the Rhombohedron and in finding the center, in terms of 'a' and 'theta'.  Any help would be greatly appreciated.
 A: If we want the solid to end up in the same orientation as on Dürer's engraving, then it will be convenient to stand up the rhombohedron with one end at $(0,0,0)$ such that its long (inner) diagonal is the $z$ axis. We can also decide that one of the lower sides must be in the $xz$-plane; you can always rotate it about a vertical axis later.
This is enough to decide that one of the lower corners of the rhombohedron must be at
$$ v_1 = (a\sin\phi, 0, a\cos\phi) $$
where $\phi$ is the angle between the sides and the long vertical diagonal. (The relation between $\phi$ and your $\theta$ will be determined later). The other two corners at that level will be
$$ v_2 = (-\frac12 a\sin\phi, \frac{\sqrt 3}{2} a\sin\phi, a\cos\phi) \\
v_3 = (-\frac12 a\sin\phi, -\frac{\sqrt 3}{2} a\sin\phi, a\cos\phi) $$
and the remaining corners of the rhombohedron will be at $v_1+v_2$ and $v_2+v_3$ and $v_3+v_1$ and finally the top corner at $v_1+v_2+v_3=(0,0,3a\cos\phi)$.
You can then translate everything downwards by $\frac32a\cos\phi$ if you want the center of the rhombohedron to be at $(0,0,0)$, but I'm not going to bother about that.
It remains to find the relation between $\phi$ and $\theta$. By the rules for dot products we can find
$$ a^2\cos\theta = v_1\cdot v_2 = a^2(\cos^2\phi-\tfrac12\sin^2\phi) = a^2(\tfrac32 \cos^2\phi - \tfrac12 ) = a^2(\tfrac34\cos2\phi + \tfrac14)$$
which we can solve to get
$$ \phi = \frac12 \arccos \frac{4 \cos\theta -1}{3}$$
or, equivalently,
$$ \cos\phi = \sqrt{\frac{1+ 2\cos\theta}{3}} \qquad \sin\phi = \sqrt{\frac{2-2\cos\theta}{3}} $$
