These are Schlegel Nets; that is, one face (usually the one with the most edges) has been selected as a base and the polyhedron flattened into a plane within the enclosing polygon. To help with identifying faces, they are color-coded as follows:
Also, we are only concerned with topologically distinct polyhedra, that is, differing in number or type of faces and vertices. Thus, a triangular prism and a tetrahedron with one vertex truncated are topologically equivalent 5-hedra, a cube and rhombohedron are topologically equivalent 6-hedra, and so on.
|With one quadrilateral face, there are four free vertices|
|Three quadrilaterals forming a strip, which may or may not close to form a prism.|
we need auxiliary diagrams. These
polyhedra have a vertex where three quadrilaterals meet.
In the lower diagrams, the four meeting quadrilaterals are moved to the back of the polyhedron (hidden edges shown in green). Since those three quadrilaterals have 6 free edges, the remaining faces can be shown drawn on an hexagonal net. Thus each polyhedron is shown twice, once in a square net and again in the hexagonal net immediately below it.
|Here again, each polyhedron is shown twice, once in a square net and again in the hexagonal net immediately below it.|
|Here we have three quadrilaterals and a triangle meeting at a vertex. These faces have 7 free edges, so the remaining faces can be drawn in a heptagonal net.|
|These solids have three quadrilaterals forming a triangular prism except that it closes only at one vertex. The topology of each solid is unique enough that no auxiliary diagram is needed.|
|Here we have two quadrilaterals meeting at an edge. The auxiliary diagram below looks into the open wedge formed by the two quadrilaterals (the common edge, which runs from end to end, is not shown) and shows the remaining faces filling the opening.|
|With one pentagonal face, there are obviously three remaining free vertices. If the quadrilateral shares an edge with the pentagon there is one free vertex left.|
|If the quadrilateral shares only a vertex with the pentagon there is no free vertex left. Obviously it is impossible to have the pentagon and quadrilateral completely separate and still have only 8 vertices.|
|With one hexagonal face, there are obviously two remaining free vertices.|
Created 10 June 1998, Last Update 7 June 1999
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