If you have a complicated object, it's common to build a conservative collision cage around the object using discrete oriented polytopes (k-DOPs) and the common $18$-DOP will contain hexagons and squares. I can't think of any plausible reason you would want a hexagonal (or mostly-hexagonal) tesselation of the sphere for simulation or collision detection purposes. These tesselation are well-known (consider the soccer ball, or geodesic dome). Note that you could instead use a mix of hexagons and pentagons.
For example, you could take two hexagons, glue them on top of each other, and call that a "sphere." I would consider any such tiling to be degenerate, but if that's what you really want, it is at least possible to construct. To get Euler characteristic two you must therefore have some vertices where only two hexagons meet. What if you allow three or more hexagons to meet at a vertex, instead of exactly three? You get that $V \leq 2F$ and you're in the same boat, since now $V-E+F$ is still $\leq 0$. Plugging into the Euler characteristic formula you get
If you have $F$ hexagons, this means you must have $3F$ edges (since each hexagon has six edges, shared by two hexagons) and $2F$ vertices (since each hexagon has six vertices, shared by three hexagons). It is impossible to tessellate (regularly or otherwise) a sphere with hexagons, at least when three hexagons meet at each vertex.
0 kommentar(er)
