This is a snub cube, radially projected onto its circumscribing sphere and then stereographically projected onto the plane. It's not quite a Lombardi drawing, because the angles of the quadrilateral faces are wider than the angles of the triangles. But I think it sets a good balance between optimizing the angular resolution around each vertex and preserving the property of the snub cube that, within each face, all the vertex angles are equal.

### Comments:

None:
2013-12-05T03:54:09Z

Do you know what are the three equations for the snub cube analogous to the icosahedral equation?

11011110:
2013-12-05T04:16:10Z

That's an equation for the coordinates of the points in the stereographic projection directly, right? But I used a different method for this drawing: the code I used to draw it constructs it in 3d, and then uses a general-purpose stereographic projection routine that would work for any set of points and great-circle arcs on the sphere.

The 3d vertex coordinates are permutations of $$(–1, \xi, 1/\xi)$$, where $\xi = \frac{(17 + 3\sqrt{33})^{1/3} - (-17 + 3\sqrt{33})^{1/3} – 1}{3}$ (a formula taken from the Wikipedia article).

None:
2013-12-05T04:39:56Z

Thank you for the response. Yes, that number is the reciprocal of the tribonacci constant. There is a nice continued fraction for it, by the way. Kindly see: http://sites.google.com/site/tpiezas/0012 P.S I find it amusing that the image for the guest post has a paper bag over the head. :) -- Tito