Many strategy games use hexagonal tiles. One of the main advantages is that the distance between the center of any tile and all its neighboring tiles is the same.
I was wondering if anyone has any thoughts on marrying a hexagonal tile system with the traditional geographic system (longitude/latitude). I think it would be interesting to cover a globe with hexagonal tiles and be able to map a geographic coordinate to a tile.
Has anyone seen anything remotely close to this before?
UPDATE
I'm looking for a way to subdivide the surface of a sphere so that each division has the same surface area. Ideally, the centers of adjacent sub-divisions would be equidistant.
You can't tile a sphere with uniform regular square or hexagon tiles (Euler said so!). If you try to tile with hexagons (by subdividing an icosahedron), you end up with twelve pentagons left over.
A hexagon tiling is a tiling of the plane by identical hexagons. The regular hexagon forms a regular tessellation, also called a hexagonal grid, illustrated above. There are at least three tilings of irregular hexagons, illustrated above.
Question from Angela, a student: A sphere is made of 112 hexagons. Each side measures 10 cm. The width of the border between each hexagon is 5 cm.
Take a look at vraid/earthgen; it uses hexagons (plus a few pentagons) and includes source code (see planet/grid/create_grid.cpp).
As of 2018 a new version is available based on racket.
Well, lots of people have made the point that you can't tile the sphere with hexagonal tiles - maybe you are wondering why.
Euler stated (and there are lots of interesting and different proofs, and even a whole book) that given a tile of the sphere in x Polygons with y Edges total and z vertices total (for example, a cube has 6 polygons with 12 edges and 8 vertices) the formula
x - y + z = 2
always holds (mind the minus sign).
(BTW: it's a topological statement so a cube and a sphere - or, to be precise, only their border - is really the same here)
If you want to use only hexagons to tile a sphere, you end up with x hexagons, having 6*x edges. However, one edge is shared by each pair of hexagons. So, we only want to count 3*x of them, and 6*x vertices but, again, each of them is shared by 3 hexagons so you end up with 2*x edges.
Now, using the formula:
x - 3*x + 2*x = 2
you end up with the false statement 0 = 2
- so you really can't use only hexagons.
That's why the classical soccer ball looks like it does - of course modern ones are more fancy but the basic fact remains.
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