xtiler 0.1.0

Coordinate system for tiling with hexagons.

Hexa-tiling

Coordinate system for tiling with hexagons.

What does the grid look like?

How does it work?

Rings and angles

The grid can be viewed as accumulation of concentric rings. On the image, the origin is labelled 0, surrounded by the ring #1 composed of tiles 1-6, surrounded by the ring #2 composed of tiles 7-18.

Numbering

Tiles are numbered from 0, in increasing order, starting at the origin (center of the grid), spiraling around the origin, with the first tile of each ring at 0°.

Coordinates

Polar coordinates can be used to find a tile. Each tile has a radius and an angle.

• The radius is the distance to the origin, corresponding to the rank of the ring the tile belongs to.
• The angle, in degrees, is measured between the direction "noon" (or "up", "north", etc) and the position of the tile.

The maths behind it

Ring length (Perimeter)

Let the ring length be $P$ and its rank $n$.

$$P(0) = 1$$

$$P(1) = 6$$

$$\forall n > 1, \quad P(n) = 2P(n-1)$$

$$\forall n > 0, \quad P(n) = 6 * 2^{n-1}$$

Grid size (Area)

Let the number of tiles for a grid with $n$ rings be $A$.

$$A(n) = \sum^{n}_{i = 0}{P(i)}$$

$$= 1 + \sum^{n}_{i = 1}{6 * 2^{i-1}}$$

$$= 1 + 6 * \sum^{n-1}_{i = 0}{2^{i}}$$

$$= 1 + 6 * (2^n - 1)$$

$$= 6 * 2^n - 5$$

Number from coordinates

Let the number of a tile be $X$, its radius $n$ and its angle $\alpha$.

$$X(0, \alpha) = 0$$

$$\forall n > 0, \quad X(n, \alpha) = A(n - 1) + \frac{\alpha}{360}P(n)$$

$$= 6 * 2^{n-1} - 5 + \frac{\alpha}{360}*6 * 2^{n-1}$$

$$= 6 * 2^{n-1}(1 + \frac{\alpha}{360}) -5$$

Radius

To find the radius $n$ given the number $X$:

$$X = 0 \Leftrightarrow n = 0$$

$$0 \le \alpha < 360$$

$$\implies \forall n > 0, \quad X(n, 0) \quad \le \quad X(n,\alpha) \quad < \quad X(n, 360)$$

$$\implies \forall n > 0, \quad A(n - 1) + \frac{0}{360}P(n) \quad \le \quad X \quad < \quad A(n - 1) + \frac{360}{360}P(n)$$

$$\implies \forall n > 0, \quad A(n - 1) \quad \le \quad X \quad < \quad A(n - 1) + P(n) = A(n)$$

$$\implies \forall n > 0, \quad 6 * 2^{n-1} - 5 \quad \le \quad X \quad < \quad 6 * 2^n - 5$$

$$\implies \forall n > 0, \quad 2^{n-1} \quad \le \quad \frac{X + 5}{6} \quad < \quad 2^n$$

$$\implies \forall n > 0, \quad n-1 \quad \le \quad \log _2(\frac{X + 5}{6}) \quad < \quad n$$

$$\implies \forall n > 0, \quad n = \lfloor \log _2(\frac{X + 5}{6}) \rfloor + 1$$

Angle

Tile 0 has no angle.

$$\forall n > 0, \quad X(n, \alpha) = 6 * 2^{n-1}(1 + \frac{\alpha}{360}) -5$$

$$X + 5 = 6 * 2^{n-1}(1 + \frac{\alpha}{360})$$

$$\frac{X + 5}{6 * 2^{n-1}} = 1 + \frac{\alpha}{360}$$

$$360(\frac{X + 5}{6 * 2^{n-1}} - 1) = \alpha$$

$$360(\frac{X + 5}{6 * 2^{\lfloor \log _2(\frac{X + 5}{6}) \rfloor + 1-1}} - 1) = \alpha$$

$$360(\frac{X + 5}{6 * 2^{\lfloor \log _2(\frac{X + 5}{6}) \rfloor }} - 1) = \alpha$$