Talk:Prismatoid

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Polyhedra

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pentagrammic crossed-antiprism edit

Latest comment: 15 years ago3 comments3 people in discussion

Is the 4th object pictured, the pentagrammic crossed-antiprism, really a prismatoid? If so, why? It seems to have vertices in places other than the two parallel planes. mg 17:00, 30 November 2007 (UTC)Reply

See Prismatic uniform polyhedron for more examples. There's just two planes of vertices, but the edges intersect so it may look like there's more vertices - just like the pentagram has 5 vertices but could be 10 if you thought there was vertices at the intersections. Tom Ruen 18:06, 30 November 2007 (UTC)Reply

I altered the definition of prismoid to insist that the side faces are quadrilaterals. See, e.g., http://mathworld.wolfram.com/Prismoid.html. There is some ambiguity out there on this topic, but I believe simply requiring the same number of vertices is too weak a condition. Joseph O'Rourke (talk) 23:24, 16 September 2008 (UTC)Reply

Volume edit

Latest comment: 14 years ago2 comments2 people in discussion

This was added anonymously to the intro, corrected by someone else, and there's no evidence where it comes from or its correctness. Tom Ruen (talk) 01:33, 17 January 2010 (UTC)Reply

If the areas of the two parallel faces are A₁ and A₃, the cross-sectional area of the intersection of the prismatoid with a plane midway between the two parallel faces is A₂, and the height (the distance between the two parallel faces) is h, then the volume of the prismatoid is given by $V=h{\frac {(A_{1}+A_{3})}{6}}+h{\frac {2A_{2}}{3}}$ .

Tom Ruen

The formula follows by very elementary means. I have restored it with an explanation. JamesBWatson (talk) 14:50, 20 January 2010 (UTC)Reply

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