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You can't make a sphere entirely out of hexagons; the math doesn't work [1]. In the article they note that they include 12 pentagons too, to solve this problem.

by jpab 7y ago

You can't make a sphere entirely out of hexagons; the math doesn't work [1]. In the article they note that they include 12 pentagons too, to solve this problem. Neatly, they arrange to have all the pentagons in areas of water where they presumably won't have to analyse traffic patterns for a while. (I mention it mostly because I think it's an interesting little mathematical factoid.) [1] https://math.stackexchange.com/q/2121175 https://math.stackexchange.com/q/2121175

na85 7y ago
That link is a great example of why I stopped using stack exchange. The OP asked a question about tiling spheres with pentagons and the third response is basically "you can tile a sphere if it's not a sphere." I had analogous experiences every time I asked a question there. One would ask a very clear question like, say, "how do I print to stdout in C?" And the first or second answer you get is inevitably about taking input from stdin. Or polymorphism.
- ant6n 7y ago
  Maybe people just like to hijack threads and change the topic to what they like to talk about.
  - alluro2 7y ago
    Ouch :) Thanks for that, nice one :)
jacobolus 7y ago
One thing you can do though is tile an octahedron by just hexagons. At each of the 6 corners you end up with 2 hexagons which border each-other along 2 edges (instead of the usual 1). If you blow this octahedron up into a sphere those hexagons appear to be pentagons, because two of their edges are colinear (i.e. the same great-circle arc). This can be nicer in some cases: the edge case your hexagon-grid algorithms have to deal with is having a hexagon with one of the same neighbors twice, instead of needing to worry about pentagons per se.
- alanbernstein 7y ago
  Would you care to explain this? When you mention tiling an octahedron with hexagons, the first thing I imagine is a truncated octahedron, which is composed of hexagons and squares.
  - jacobolus 7y ago
    You can start with 4 hexagons, each one covering an octant and a third of the neighboring octants. Or another way to say this: if you start with 4 hexagons, with each glued together with each other along two adjacent edges, and you add the appropriate folds, you can make an octahedron. Then you can subdivide each of those starting hexagons into n hexagons for any of these numbers, https://oeis.org/A003136 https://oeis.org/A003136 (power-of-4 sizes may be the most convenient among these, so that the overall grid has 2^n by 2^n size)

by jpab 7y ago

na85 7y ago

That link is a great example of why I stopped using stack exchange. The OP asked a question about tiling spheres with pentagons and the third response is basically "you can tile a sphere if it's not a sphere." I had analogous experiences every time I asked a question there. One would ask a very clear question like, say, "how do I print to stdout in C?" And the first or second answer you get is inevitably about taking input from stdin. Or polymorphism.

ant6n 7y ago
Maybe people just like to hijack threads and change the topic to what they like to talk about.
- alluro2 7y ago
  Ouch :) Thanks for that, nice one :)

jacobolus 7y ago

One thing you can do though is tile an octahedron by just hexagons. At each of the 6 corners you end up with 2 hexagons which border each-other along 2 edges (instead of the usual 1). If you blow this octahedron up into a sphere those hexagons appear to be pentagons, because two of their edges are colinear (i.e. the same great-circle arc). This can be nicer in some cases: the edge case your hexagon-grid algorithms have to deal with is having a hexagon with one of the same neighbors twice, instead of needing to worry about pentagons per se.

alanbernstein 7y ago
Would you care to explain this? When you mention tiling an octahedron with hexagons, the first thing I imagine is a truncated octahedron, which is composed of hexagons and squares.
- jacobolus 7y ago
  You can start with 4 hexagons, each one covering an octant and a third of the neighboring octants. Or another way to say this: if you start with 4 hexagons, with each glued together with each other along two adjacent edges, and you add the appropriate folds, you can make an octahedron. Then you can subdivide each of those starting hexagons into n hexagons for any of these numbers, https://oeis.org/A003136 https://oeis.org/A003136 (power-of-4 sizes may be the most convenient among these, so that the overall grid has 2^n by 2^n size)