Hex Grid Disk

Similar to HexGridSphere, GeographyExample, HexGridIcosahedron.

Goals:

Goals where coming close is good enough: Proposed solution:


Cell Neighbors

Unfortunately, the two hexagons can't be put together in such a way that the structure of the HexGrid? is preserved and homogeneous over the whole sphere. When tiling a sphere with cells that all have the same shape, it is impossible to make a mesh in which each node connects the same number of edges.


Rips in the Map


The size of the hex-grid can be chosen to let cities' hinterlands tile perfectly.

If each city has a hinterland with a radius of 2 cells, then the city can use a total of 19 cells (including its own cell).

Hex-grids can tile a plane: Each sub-grid has 6n neighbors (e.g., n/a, 6, 12, 18...). Each grid contains 1 + 6*(n+) sub-grids (e.g., 1, 7, 19, 37...)

The hex-grids do not tile a hexagon. Some of the grids extend slightly out-of-bounds; in other places, there are equal-sized gaps at the edges. These excesses and gaps are complementary, so that:

The hex-grids tile the disk.


Some convenient numbers of cells for the world:

 disks div-   hinter total  cell   hinterland  notes
      isions   land  cells  area   radius(es)
 -------------------------  ----   ----------  -----
 2 *            1 =      2                     (a trivial world)
 2 *            7 =     14                  1  (a 2-city world)
 2 *           19 =     38                  2  (a 2-city world)
 2 *       7 * 19 =    266                1,2  (a small world)
 2 *      19 * 19 =    722                  2
 2 *  7 *  7 * 19 =   1862  270,000 km^2  1,2
 2 * 37 * 37 * 19 =  52022    9,800 km^2  2,3  (each cell is about 1° across)

The Earth has an area of about 510,000,000 km^2


Formulas for a polar Lambert equal-area projection onto the hexagon:

 Each meridians is a straight line.
 Each parallel (of constant latitude) is a hexagon centered on the pole.
 When measured along a parallel, the scale is constant.

distance from center to point Let beta = -------------------------------- distance from center to equator, along the same meridian

sin(latitude) = 1 - (beta)^2


Error estimates:

Assume that the Earth is a sphere with a 40,000 km equator. (The error in treating the Earth as a sphere instead of an ellipsoid is negligible compared to the error in the HexGridDisk.)

This sphere has an area of 509,300,000 km^2 Each degree (along a great circle) has a length of 111.111... km

The HexGridDisk has an area of 509,300,000 km^2 (by the equal-area assumption). Each degree (along the equator) has a length of 165 km.

Assume that the shortest meridians are at 0, 60, 120... degrees longitude. Degrees measured along short meridians range in length from 75 km to 106 km.

The longest meridians are at 30, 90, 150... degrees longitude. Degrees measured along long meridians range in length from 86 km to 122.2 km.

Thus, the worst linear distortion is at the equator, where some distances are 48.5 % too long, and some distances are 32.7 % too short.


If the boundaries of the cells are suitable fractals, the grid can be iteratively (that is, fractally) subdivided.


A SquareGridDisk? can also be built. It will have greater distortion, as the equator will be twice as long as the great circle route over the poles.



Discussion:


See also: AustralianCulturalAssumption, HexGridSphere, HexGridIcosahedron, GeographyExample


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