Let's add another rule: quadrilaterals that meet at a zero angle at their edges become a larger planar polygon; edges that meet at a zero angle become a longer edge. So a tetrahedron can't be quadrangulated. Can a smooth surface -- e.g., a hemisphere -- be approximated by a convex quadrangulated surface? The answer is trivially yes -- the approximation is a cube -- but is there a way to quadrangulate it with smaller & smaller quadrilaterals to get a better & better approximation? At 09:00 AM 1/6/2013, Henry Baker wrote:
I was thinking more along the lines of approximating smooth surfaces.
I guess one could break up the triangles on the surface of the tetrahedron into quadrilaterals, but that method would enable any triangulated surface to be broken up in the same way.
I was hoping for something less trivial.
At 08:53 AM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com> A usual trick in modeling 3D surfaces is to "triangulate" them with triangles that could in principle be hooked together with hinges.
Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) quadrilaterals "hinged" together ?
(A quadrilateral has to have 4 non-trivial sides, so no triangles as limiting cases.)
How would we quadrangulate a tetrahedron?