Talk:Hemipolyhedron
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Are U75 and Skilling hemipolyhedra?
[edit]Are U75 and Skilling hemipolyhedra, since their duals extend to infinity? Professor M. Fiendish, Esq. 23:45, 10 September 2009 (UTC)
- I'd say no, since not in the name. It's worth reading from mathworld, content from Johnson's unpublished book there, [1]. Tom Ruen (talk) 23:59, 10 September 2009 (UTC)
- The "hemi" names are given to indicate that the diametral faces are half as many as some convex polyhdron whose faces lie parallel. For example the tetrahemihexahedron has three diametral faces at right angles, half of the six-sided cube or hexahedron. The diametral faces of Skilling's beast do not double up in the same way to make a known convex figure. Hence, it does not have "hemi" in its name. For Heaven's sake, do not take Mathwrld as authoritative on the subject of polyhedra - it is a terrible purveyor of old folkore. For example neither Wenninger (who first described them) nor Johnson (whose book remains unpublished anyway) thinks of the hemi duals as proper "polyhedra" - something which quite slips the mind of whoever wrote that Mathworld article (I notice it continues to appear all over Wikipedia, too - I do hope you guys can go back and edit out some of your more misplaced enthusiasms). -- Cheers, Steelpillow (Talk) 09:29, 13 September 2009 (UTC)
- I think the hemispherical faces there make a nonconvex octahedron. (An alternate way to represent the duals is the one given by Polyhedra 2001 - I like that better!) 4 T C 04:07, 28 February 2010 (UTC)
- Where is the "there" that has 8 hemi faces? The kind of representation of the hemi duals one "likes" must take second place to the questions, what exactly is a polyhedron? and what exactly is this particular process of dualising? Only then can we figure out whether the hemis have duals, what form these take, and whether they too are polyhedra. The representations on the Polyhedra 2001 site have no faces, which is not consistent with the idea that vertices dualise to faces, and so rules them out as proper duals (by most theories). -- Cheers, Steelpillow (Talk) 18:21, 28 February 2010 (UTC)
- I agree that they would probably have to be ruled out as proper duals. At that time, I thought that that representation looked rather elegant, but I don't think so now. (I'm not so sure about the nonconvex octahedron; I wrote that because I had misread some website and hence thought that the only faces passing through the origin were eight squares.) Double sharp (talk) 12:12, 22 February 2012 (UTC)
- Where is the "there" that has 8 hemi faces? The kind of representation of the hemi duals one "likes" must take second place to the questions, what exactly is a polyhedron? and what exactly is this particular process of dualising? Only then can we figure out whether the hemis have duals, what form these take, and whether they too are polyhedra. The representations on the Polyhedra 2001 site have no faces, which is not consistent with the idea that vertices dualise to faces, and so rules them out as proper duals (by most theories). -- Cheers, Steelpillow (Talk) 18:21, 28 February 2010 (UTC)
- I think the hemispherical faces there make a nonconvex octahedron. (An alternate way to represent the duals is the one given by Polyhedra 2001 - I like that better!) 4 T C 04:07, 28 February 2010 (UTC)