Am I trying to do the impossible? (3d guru's apply)

[Pazil]

I don't think that doing the hexagons ALL around the sphere is possible...

I'm not too sure at the moment about exactly how the geometry fits together, but you can make an IcoSphere with triangles, or maybe pentagons. Anything with a higher edge count is out of the question.

Or then again, you can texture map that onto a sphere, after (probably should check out that link render posted last) you run some calculations on the texture image to stretch out the 2 ends of it, so it texture maps perfectly, like they have those world maps stretched out at the top and bottom!

P.

[rachil0]

I think euler's formula for polyhedra indicates it can't be done, at least not with a texture that's a regular tiling of hexagons (ie each vertex having valence 3).

For any genus G polyhedron you must have V-E+F=2-2G, where V is the number of vertices, E is the number of edges and F is the number of facets. The sphere (and any polyhedron homeomorphic to a sphere) has genus 0.

But if you consider the regular tiling of hexagons in 2D, if you have F hexagons then you have 6F/2 edges (each hex has 6 edges but we double counted them because each edge is shared among two hexes). Also, the number of vertices is 6*F/3, each hex has 6 vertices and each vertex is shared 3 ways. Plug these numbers into eulers formula:

V-E+F = 2-2G
6F/3-6F/2+F = 2-2G
2F-3F+F = 2-2G
0 = 2-2G
G= 1

So when you tile regular hexagons like that you can only make genus 1 shapes - a torus. You can envision this tiling by taking the texture and wrapping it up into a cylinder, then rolling that cylinder back onto itself. You won't be able to texture or facetize a sphere seamlessly with hexagons unless you have some nodes with valence!=3. (Vertices like that are called extraordinary points in subdivision surfaces - in the infinitesimal limit they have weaker curvature guarantees that the other spots on the surface).

// brain.explode();

[deezbucks]

actually the euler characteristic of a sphere (or any surface homeomorphic to S^2) is 2. you are mistaking the euler characteristic of a torus of genus 1 (i.e. X(T) = 2 - 2g = 2-2(1) = 0).

for regular hexagons your counterexample argument still applies but you will arrive at the step 0 = 2. if the hexagons are not regular, the problem is slightly more complicated (find a relationship e and f, then use the given condition to put an upper bound on v in terms of f. This will
give you an upper bound on the euler characteristic, which will gie a contradiction), but it is still impossible to subdivide the surface of a sphere (or any surface homeomorphic to S^2) with 6-gons regular or not.

there are only 5 regular polyhedra, and the proof arises from a basic combinatorial argument. this can be found on google.

I think euler's formula for polyhedra indicates it can't be done, at least not with a texture that's a regular tiling of hexagons (ie each vertex having valence 3).

For any genus G polyhedron you must have V-E+F=2-2G, where V is the number of vertices, E is the number of edges and F is the number of facets. The sphere (and any polyhedron homeomorphic to a sphere) has genus 0.

But if you consider the regular tiling of hexagons in 2D, if you have F hexagons then you have 6F/2 edges (each hex has 6 edges but we double counted them because each edge is shared among two hexes). Also, the number of vertices is 6*F/3, each hex has 6 vertices and each vertex is shared 3 ways. Plug these numbers into eulers formula:

V-E+F = 2-2G
6F/3-6F/2+F = 2-2G
2F-3F+F = 2-2G
0 = 2-2G
G= 1

So when you tile regular hexagons like that you can only make genus 1 shapes - a torus. You can envision this tiling by taking the texture and wrapping it up into a cylinder, then rolling that cylinder back onto itself. You won't be able to texture or facetize a sphere seamlessly with hexagons unless you have some nodes with valence!=3. (Vertices like that are called extraordinary points in subdivision surfaces - in the infinitesimal limit they have weaker curvature guarantees that the other spots on the surface).

// brain.explode();