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.

Stereographic projection of a snub cube




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