# Finding the right grid tessellation, for a chess board with elliptical geometry

I came up with the idea of non-euclidean chess(the chess board I'm working on will have 2D elliptical geometry) but I came across a problem. The question: What kind of grid do I put on it(what shapes will tesselate on it, how many are placed around a vertex, etc.) ? The grid on globes only cause parallel lines to intersect at only two points, the poles, which doesn't normally happen on this geometry. If I plot two points, from which I draw two parallel lines, they should intersect after a certain amount of distance, no matter where I place them on the sphere, aslong as the two points remain the same distance from each other, and the lines drawn from them are "parallel". I tried to do this, but I usually failed in one way or another. The goal is to find the right shape(or shapes) that will correctly tesselate on a sphere that will give me that property. I have a list of other desired properties, but it is most likely I won't get them all, and it's listed from most important to least:

• point A and B, are x distance apart, with two parallel lines drawn through each one, and the two lines will intersect after d distance from point C placed equidistant between point A and B. No matter where they are placed on the sphere.
• The tessellated shape used are quadrilaterals(with these shapes, making a pawn go forward is easy. One option. But with odd sided shapes, it gives them a choice, which breaks the idea of elliptical geometry, and could even be hyperbolic, in a way.)
• Rotationally symmetrical
• One type of shape used
• The angles of the type of shape used each have the same amount of degrees(Simply because I want to. The others have logical reasons. except for maybe the one before this one. It's technically not needed, it just has might some weird, unwanted effects on the board if this goal is not met)

Well, the cube seems to satisfy all your properties. I assume you want to have more than just 6 cells, though?

Unfortunately I do not understand your desired property of parallel lines, but I will try to help anyway.

You can define the curvature of a vertex as 360 degrees the sum of angles of its adjoining faces, for example:

• in a truncated icosahedron (aka soccerball), curvature of each (6,6,5) vertex = 360 - 120 - 120 - 108 = 12 degrees

• in a cube, curvature of each (4,4,4) vertex = 90 degrees

The idea here is that, while a "true" sphere has positive Gaussian curvature everywhere, a polyhedron has its curvature concentrated on vertices -- its faces and edges are flat. Just as the integral of the Gaussian curvature over the whole sphere is always $$4\pi$$, if you add up the curvatures of all vertices, you also get $$4\pi = 720$$ degrees (e.g. the cube has 8 vertices and the soccerball has 60 vertices). In case if faces are irregular, count their degrees them as if they were regular Euclidean polygons.

If you want to have more faces, you probably want to have no vertices where curvature is negative. Unfortunately you do not have many choices for the vertices with positive curvature. The best choices seem to be (6,6,4) (curvature 30 degrees), (6,6,5) (12 degrees), (4,4,4) (90 degrees), so you have only a limited number of vertices with positive curvature.

You can have arbitrarily many vertices with curvature 0. The easiest way is to subdivide each face of a cube into NxN squares. Then you have 8 corners (curvature 90) and all the other vertices have curvature 0.

Another solution is to apply the Goldberg-Coxeter construction to a cube or a dodecahedron. Some examples: GC(3,2) cube, GC(4,3) cube, GC(3,2) dodecahedron. (Press 'g' to mark vertices)

I think you are not satisfied by the subdivided cube, because two parallel lines on the same face will remain parallel. Unfortunately, I think this is unavoidable, because the areas where all the vertices have curvature 0 are "flat" -- the parallel lines here work as in the Euclidean plane. They will cross only when a vertex with positive curvature appears between them.

I have chosen co-prime parameters for my Goldberg-Coxeter variants, so that every straight line will eventually reach a vertex with positive curvature. However, this may happen after several rotations around the sphere, so this might not be sufficient. (Also, if you are doing the elliptic plane (not the sphere), the two parameters have to be of form $$(a,a)$$ or $$(a,0)$$, because $$(a,b)$$ generates chiral polyhedra, which does not agree with the non-orientability of the elliptic plane.)

Other than that, I think a random irregular Voronoi tiling on the sphere would be a good choice for some strategy games (because the irregular structure hides the corners of the underlying Platonic solid), but I think it would not work well for your purposes...

• So I guess there isn't any tessellation that will satisfy my goals. Also since chess pieces at most move in one of eight directions, using any other shape or putting more or less shapes around one vertex will complicate the game. I think I'll just have no grid, and each piece will be limited to how far they can move and in which direction. I wonder how that will affect the game....Anyways thanks. Oct 8 '18 at 1:21