Talk:Rhombic dodecahedral honeycomb

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Polyhedra

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It's edge-uniform, isn't it, as well as cell-uniform? --Tamfang 06:02, 16 January 2006 (UTC)[reply]

I'm sure you're right, face-uniform too! Feel free to add content as you can. I'm still stretching for sources and deciding what information to include in the tables. Tom Ruen 06:09, 16 January 2006 (UTC)[reply]

How do you get eight square pyramids around one vertex? — Start with a 3D checkerboard; remove the cubes of one color and replace each with six pyramids, joined to the remaining cubes. Is the result not a tiling of rh.dod.? —Tamfang 09:24, 16 January 2006 (UTC)[reply]

Sorry, you're right. 4 and 6 cells per vertex. Glad I have your help! :)

Shortly after shutting down last night I saw where you went astray. The facet of the dual is the dual of the vertex figure, not the vf itself. If you're accustomed to dealing with regular 3-polytopes, whose vertex figures are self-dual, that's easy to forget! --Tamfang 22:17, 16 January 2006 (UTC)[reply]

dual of this honeycomb?[edit]

What is the dual of this honeycomb?18.187.1.80 10:07, 25 April 2007 (UTC)[reply]

I assume you mean "what does the dual mean?" since the dual is linked. It is same as the dual of a tiling - examples list of uniform tilings. The Tetrahedral-octahedral honeycomb is a vertex-transitive honeycomb, so all vertices are identical. In the dual, vertices and cells are interchanged, and faces and edges are interchanged, so this honeycomb is cell-transitive. Does that help? Tom Ruen 17:59, 25 April 2007 (UTC)[reply]