--- "R. William Gosper" <rwg@spnet.com> wrote: There are 90 vertices, all tetravalent,

so maybe there is a C_90 fullerane (as opposed to fullerene).

I also made a pentacontahedral candidate for C_48 fullerane with 30 squares, 8 equilateral triangles, and twelve rhombi. --rwg

These would be very strained molecules; all the normally tetrahedrally pointing bonds would be forced to lie on one side of a plane. However a C60H60 fullerane should exist as a fairly unstrained molecule. Likewise, perazadodecahedrane N20 and perazafullerane N60 would be less strained than octaazacubane N8. I'm not aware that any of these molecules have been systhesized (and they might not tell us about the nitrogen ones). Another point is that the more symmetries a molecule possesses, the lower is its energy, and the higher are the energy barriers to rearrangement. __________________________________________________ Do you Yahoo!? Yahoo! Mail Plus - Powerful. Affordable. Sign up now. http://mailplus.yahoo.com

"R. William Gosper" wrote:

with its vertices on a sphere and its edge lengths all equal have more than 92 faces? I'm looking at one (a doenneacontahedron?) with 12 regular pentagons (pentaga?), 60 equilateral triangles, and 20 non-regular (but equilateral) hexagons. There are 90 vertices, all tetravalent, so maybe there is a C_90 fullerane (as opposed to fullerene).

A polyhedron has planar faces. If it is inscribable, the vertices of a given face lie on the circle where the plane intersects the sphere. So how can you have a non-regular equilateral hexagon inscribed in a circle? George http://www.georgehart.com/