views:

1758

answers:

8

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.

+7  A: 

Hexagonal tiles are too complicated for regular geometry as applied to geospatial uses. Check out HTM for a similar thing with triangles or google for "Hierarchical Triangular Mesh" for other sources.

Erich Mirabal
+3  A: 

The first website that comes to mind is Amit's Game Programming Information and its collection of links on hexagonal grids.

Kristo
+5  A: 

It is impossible to cover a sphere with regular tiles (except for long and thin "orange slices". This is actually a pretty difficult research problem.

One sort of tiling used very often (in astrophysics) is the HEALPIX pixelisation: http://healpix.jpl.nasa.gov/

This pixelization satisfies the equal-area requirement; it's impossible to make everything equidistant, however.

Another pixelization is "GLESP", which has some different properties (and isn't as polished a software package): http://www.glesp.nbi.dk/

Andrew Jaffe
I had never heard of that one. Pretty awesome.
Erich Mirabal
very nice, and never heard about it before
David Lehavi
dumb question time: Is this why we see Geodesics being popular on college campi ( nominative plural of campus ) - that it is thought to be unsolvable or known to be unsolvable leads to known solutions in Geodesics ... ( ? ) I always wondered about those architect students who could work the issue and wish to identify "pretty difficult research problem"(s) so as to avoid deep excursions in time wasters that have no solution.
Nicholas Jordan
+4  A: 

You can't cover a sphere with equal hexagons, but you could cover it with a geodesic, which is mostly hexagons, with 12 pentagons at the vertices of an icosohedron, and the hexagons slightly distorted to make it bulge into a sphere.

GoatRider
i wish i could picture it
carrier
GoatRider
Or think of a soccer ball: it's made up of pentagons each of which is surrounded by hexagons. Or, look here for the truncated icosahedron: http://en.wikipedia.org/wiki/File:Uniform_polyhedron-53-t12.png
Frank Shearar
+4  A: 

Getting a sphere to divide into equal parts is a tough nut. Because of this, you end up with Geodesic shapes, which are not composed of shapes that can be in turn composed of triangles of equal size. Breaking down all of the hexagons and pentagons into triangles, you end up with triangles that have different interior angles, leading to a loss of symmetry.

The one consolation that I can give you is that all of the shapes will have a limited number of triangles that can be catagorised, which means for a small geodesic, that 5 or 6 triangles can be used repeatedly to describe all of the hexagons and pentagons required for the geodesic. While distances will not be equal from the "center" of each triangle/shape, you can at least divide the handling of each triangle into a discrete case, lending to a potential work-around in code.

Avery Payne
+2  A: 

Well lots of people have made the point that you can't tile the sphere with hexagonal tiles - maybe you are wondering why.

Well - Euler statet (and there are lots of interesting and diffenernt proofs and even a hole book) that given a tile of the sphere in x Polygons with y Edges total and z vertices total (for examble a cube has 6 polygons with 12 egdes and 8 vertices) the formula "x - y + z = 2" allways hold (mind the minus).

(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 but one edge is shared by two hexagons - so we only want to count 3*x of them, and 6*x vertieces but again each of them is shared by 3 hexagons so you end up with 2*x edges.

No 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 he does - of course modern ones are more fancy but the basic fact remains.

CKoenig
interesting. any links for that proof?
Erich Mirabal
+2  A: 

Read "Geodesic Discrete Global Grid Systems" by Kevin Sahr, Denis White, and A. Jon Kimerling

You can find it here...

Sphaerica
+1  A: 

The old Traveller roleplaying game used to map planet surfaces as ikosahedra (cut open for printing in a book). This produced a big distortion at the corner hexes (they have to become pentagons). You might find some such material when searching for GURPS Traveller.

Svante