4 ms·

We tried to figure out how to cut a planet into hexes, turns out you can't (no surprise). Footballs use pentagons. Well, I guess you can just make equal area s

We tried to figure out how to cut a planet into hexes, turns out you can't (no surprise). Footballs use pentagons. Well, I guess you can just make equal area square map and cut that. http://en.wikipedia.org/wiki/Gall%E2%80%93Peters_projection http://en.wikipedia.org/wiki/Gall%E2%80%93Peters_projection Also, it's curious nobody mentioned Battle of Wesnoth. Battle of Wesnoth (together with Fantasy General and Battle Isle) is a shining example of hex-based strategic wargame.

e12e 13y ago
Couldn't you map hexagons onto a Icosahedron, leaving some hexagons "flat" and some with a "rise" towards the top (by subdividing each triangle into one "centered" hexagon, and three triangles of one-third (well, one ninth, I guess) size, forming parts of other hexagons ? , / \ /___\ / \ the dots (.,') mark centers of "new" / . \ hexagons, each of the ,' would be /\ /\ "peaks", the one centered on "." would /__\_____/__\ be "flat". ' ' You'd probably need to subdivide the hexagons further, and decide what those "peaks" should "mean" -- or ignore them... Not sure if that would be easier than using pentagons, though... Another way too look at it, is to say the hexes would be distorted wrt. area -- Not sure if that would matter though...
- e12e 13y ago
  Replying to myself: all those sides would be of equal length, defying Eculid -- or rather I'm missing the fact that five triangles meet in the original icosahedron. So those "corners" would be pentagons, not hexagons.
  - lmm 13y ago
    Indeed that is one way to derive the same "football projection".
- guard-of-terra 13y ago
  I can't say I understand what you mean, but hexagons should be more-or-less equal in size and possible for players to understand.
  - e12e 13y ago
    Yes, but as has been mentioned by others, that is impossible. So either one can use pentagons and hexagons (as I inadvertently did above ;-) or just pentagons. My goal was trying to get a hex grid, with roughly "ignorable" distortion. I think the most straight forward way would be to use four "pseudo" (irregular) hexagons, and wrap them into a cube (ie: four "hexagons" made from a square and two triangles with 45 and 90 degree angels). You would then have a "round" map that could be subdivided. It would be far from perfect, but very straightforward -- and if mapped onto a planet "north-south" (aka pointy-side up) -- I think it would yield a similar distortion to what we are used to from various map projections: /\ /\ /\ /\ <- distorted sides /__\ /__\ /__\ /__\ ________ _______ | A | B | C | D | / /\ |\ c /| | __ | __ | __ | __ | / N /bb\ |d \ / | \ / \ / \ / \ / /_______/bbbb\ | /'\b | \/ \/ \/ \/ \aaaaaaa\bbbb/ |/__a__\| \aaaaaaa\b / top down view \_______\/ ("North pole") When wrapped up into a cube, the south and north poles would lie at the intersection of the four hexagons. Unless I'm tricking myself, this has the interesting property that the distance from A to B and D is 1, while from A to C is 2 -- whichever way you "circumnavigate". Being hexagons, they could be further subdivided into hexagons (as for most purposes this grid would be too coarse). I'm not convinced it would be very useful as a "proper" map projection -- but might work for a game.
    - nappatak 13y ago
      With triangles alone you could produce a Dymaxion map.
      - e12e 13y ago
        Yep, but not ones that would (directly) map "back" to hexagons. One of the things that went through my mind as I read the article, was that triangles might make for useful maps as well.
xyzzyz 13y ago
For anyone wondering why you cannot tile a sphere with hexes: this is a matter of Euler characteristic. Suppose that there is a tiling of sphere with N hexes. Each hex has 6 edges, but every edge is shared by two hexes, so there are 6N/2 = 3N edges. Similarly, there are 6N/3 = 2N vertices.Thus, the Euler characteristic would be 2N - 3N + N = 0. Now, the sphere has Euler characteristic 2, so we get a contradiction. Incidentally, a torus has Euler characteristic 0, and indeed it turns out that you can tile torus with hexes.
- marshray 13y ago
  Carbon nanotubes are mostly hexagons. They certainly could make toruses, if they wanted to.
- omaranto 13y ago
  There's no need to assume exactly 3 hexagons meet at each vertex: if you only assume at least 3 hexagons meet at each vertex, then 6N >=3V, and we get that the Euler characteristic is <=0, which is still a contradiction. Of course if you allow vertices with only two hexagons meeting at them you can make a sphere: for example, take two hexagons and glue them around the edges, then stuff the middle to get a hexagonal cushion.

e12e 13y ago

Couldn't you map hexagons onto a Icosahedron, leaving some hexagons "flat" and some with a "rise" towards the top (by subdividing each triangle into one "centered" hexagon, and three triangles of one-third (well, one ninth, I guess) size, forming parts of other hexagons ? , / \ /___\ / \ the dots (.,') mark centers of "new" / . \ hexagons, each of the ,' would be /\ /\ "peaks", the one centered on "." would /__\_____/__\ be "flat". ' ' You'd probably need to subdivide the hexagons further, and decide what those "peaks" should "mean" -- or ignore them... Not sure if that would be easier than using pentagons, though... Another way too look at it, is to say the hexes would be distorted wrt. area -- Not sure if that would matter though...

e12e 13y ago
Replying to myself: all those sides would be of equal length, defying Eculid -- or rather I'm missing the fact that five triangles meet in the original icosahedron. So those "corners" would be pentagons, not hexagons.
- lmm 13y ago
  Indeed that is one way to derive the same "football projection".
guard-of-terra 13y ago
I can't say I understand what you mean, but hexagons should be more-or-less equal in size and possible for players to understand.
- e12e 13y ago
  Yes, but as has been mentioned by others, that is impossible. So either one can use pentagons and hexagons (as I inadvertently did above ;-) or just pentagons. My goal was trying to get a hex grid, with roughly "ignorable" distortion. I think the most straight forward way would be to use four "pseudo" (irregular) hexagons, and wrap them into a cube (ie: four "hexagons" made from a square and two triangles with 45 and 90 degree angels). You would then have a "round" map that could be subdivided. It would be far from perfect, but very straightforward -- and if mapped onto a planet "north-south" (aka pointy-side up) -- I think it would yield a similar distortion to what we are used to from various map projections: /\ /\ /\ /\ <- distorted sides /__\ /__\ /__\ /__\ ________ _______ | A | B | C | D | / /\ |\ c /| | __ | __ | __ | __ | / N /bb\ |d \ / | \ / \ / \ / \ / /_______/bbbb\ | /'\b | \/ \/ \/ \/ \aaaaaaa\bbbb/ |/__a__\| \aaaaaaa\b / top down view \_______\/ ("North pole") When wrapped up into a cube, the south and north poles would lie at the intersection of the four hexagons. Unless I'm tricking myself, this has the interesting property that the distance from A to B and D is 1, while from A to C is 2 -- whichever way you "circumnavigate". Being hexagons, they could be further subdivided into hexagons (as for most purposes this grid would be too coarse). I'm not convinced it would be very useful as a "proper" map projection -- but might work for a game.
  - nappatak 13y ago
    With triangles alone you could produce a Dymaxion map.
    - e12e 13y ago
      Yep, but not ones that would (directly) map "back" to hexagons. One of the things that went through my mind as I read the article, was that triangles might make for useful maps as well.

xyzzyz 13y ago

For anyone wondering why you cannot tile a sphere with hexes: this is a matter of Euler characteristic. Suppose that there is a tiling of sphere with N hexes. Each hex has 6 edges, but every edge is shared by two hexes, so there are 6N/2 = 3N edges. Similarly, there are 6N/3 = 2N vertices.Thus, the Euler characteristic would be 2N - 3N + N = 0. Now, the sphere has Euler characteristic 2, so we get a contradiction. Incidentally, a torus has Euler characteristic 0, and indeed it turns out that you can tile torus with hexes.

marshray 13y ago
Carbon nanotubes are mostly hexagons. They certainly could make toruses, if they wanted to.
omaranto 13y ago
There's no need to assume exactly 3 hexagons meet at each vertex: if you only assume at least 3 hexagons meet at each vertex, then 6N >=3V, and we get that the Euler characteristic is <=0, which is still a contradiction. Of course if you allow vertices with only two hexagons meeting at them you can make a sphere: for example, take two hexagons and glue them around the edges, then stuff the middle to get a hexagonal cushion.