It would also be good to have a picture of the pseudo-octahedron. I've put Scott Kim's picture of it at http://mathenchant.org/pseudo-octahedron.png; there should be a realization of it as a motley dissection of the surface of the sphere that has all the rotational symmetries of the octahedron (but none of the reflection symmetries). Can Christian (or anyone else) construct a virtual model? In some ways it might be harder to visually parse the pseudo-octahedral motley dissection than the pseudo-cuboctahedral motley dissection, even though the former is simpler, because the visible hemisphere offers less information. But an interactive rotatable version (or just a GIF that shows the object rotating) would solve that problem. Jim On Wed, Jul 3, 2019 at 2:49 PM James Propp <jamespropp@gmail.com> wrote:
Wow; this is great!
Is there a way to rotate the view?
Jim
On Wed, Jul 3, 2019 at 3:38 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
I've made a 3d model in OpenSCAD and uploaded it to thingiverse: https://www.thingiverse.com/thing:3726912. I've put a simple rendering of the shape on that page; I'm going to both 3d print it and do a nicer CGI rendering now.
On Tue, 2 Jul 2019 at 19:24, James Propp <jamespropp@gmail.com> wrote:
The link to the video apparently didn't survive the conversion from HTML to PDF, so here's the URL for Scott Kim's video:
Jim
On Tue, Jul 2, 2019 at 12:03 PM James Propp <jamespropp@gmail.com> wrote:
At http://faculty.uml.edu/jpropp/mathenchant/motley-draft3.pdf you'll find a draft of an essay that will be published on July 11th. Comments are welcome, but even more welcome would be a good picture of the "spherical pseudo-cuboctahedron"! In principle I know how to make such a picture, sort of: take the motley dissection of the surface of the cube (shown in the essay), push all the vertices out onto a sphere, and draw spherical arcs joining up those vertices. This seems like a job for Mathematica, but my mastery of 3D graphics isn't up to the job. I'll gratefully acknowledge anyone who can create a picture I can use!
Thanks,
Jim Propp
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