I looked more into tesselating triangles with vertices of degree 6 on a sphere and found that you would need minimum of 12 extraordinary points of degree 5, which is the same as the number of pentagons on a soccer ball. On further reading I found https://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Hopf_theorem which shows that the sum of the indices "how much a the vector field turns around near the point", which in this case is (6-degree)/2 should equal to the euler characteristic i.e. 2 for a sphere. Thus if we have degree 5 extraordinary points and a euler characteristic of 2, we would need 12 of those to model a sphere
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This theorem is a direct consequence of Euler's polyhedron formula. We could tile a plane with vertices of degree 6. Most parts of our mesh are very close to a plane, so most vertices are not extroardinary. But out mesh is topologically equivalent to a sphere, rather than a plane.
I looked more into tesselating triangles with vertices of degree 6 on a sphere and found that you would need minimum of 12 extraordinary points of degree 5, which is the same as the number of pentagons on a soccer ball. On further reading I found https://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Hopf_theorem which shows that the sum of the indices "how much a the vector field turns around near the point", which in this case is (6-degree)/2 should equal to the euler characteristic i.e. 2 for a sphere. Thus if we have degree 5 extraordinary points and a euler characteristic of 2, we would need 12 of those to model a sphere
This theorem is a direct consequence of Euler's polyhedron formula. We could tile a plane with vertices of degree 6. Most parts of our mesh are very close to a plane, so most vertices are not extroardinary. But out mesh is topologically equivalent to a sphere, rather than a plane.