Hexagon geometry

This derives the spacing used by the hexagon layouts and the number of tiles in the radial one.

Width and side length

In the diagram, $w$ is the width of the hexagon across its flats and $s$ is its side length.

A regular hexagon is made of six equilateral triangles, such as $ABC$. $D$ is the midpoint of $AB$, so in the right triangle $CDB$ the angle at $B$ is 60° and

\[\tan(60^\circ) = \frac{CD}{DB}\]

$CD$ is half the width and $DB$ is half a side. Since $\tan(60^\circ) = \sqrt{3}$,

\[\sqrt{3} = \frac{w/2}{s/2} = \frac{w}{s}\]

which gives the width in terms of the side, and the reverse:

\[w = s\sqrt{3} \qquad s = \frac{w}{\sqrt{3}}\]

A hexagon 8 units across the flats, for example, has sides of $8/\sqrt{3} \approx 4.62$ units.

The triangles are equilateral, so the distance from the centre to a corner, which the library calls radius, equals the side length $s$.

Spacing of the centres

The layouts place the hexagons with a corner pointing up, so the flats face left and right.

Neighbours in a row touch along a flat, so their centres are one width apart:

\[\text{column pitch} = w = \text{radius} \cdot \sqrt{3} = 2 \cdot \text{radius} \cdot \cos(30^\circ)\]

The next row sits in the notches of the first. Its centres are half a column to the side and one and a half radii up:

\[\text{row pitch} = \text{radius} + \text{radius} \cdot \sin(30^\circ) = 1.5 \cdot \text{radius}\]

These are offset_step and offset_y in src/regular/hexagons.scad. offset_x is half the column pitch, $\text{radius} \cdot \cos(30^\circ)$.

hexagons_centers_rect() applies the two pitches directly, shifting odd rows by offset_x.

hexagons_centers_radial() builds the middle row first. It has $2L - 1$ hexagons for $L$ levels and starts at

\[dx = -(L - 1) \cdot 2 \cdot \text{offset}_x\]

so that it is centred on the origin. Each row above it has one hexagon fewer and starts offset_x further right, and the rows below mirror the rows above.

Growth rate

Let $H(L)$ be the number of hexagons in a radial layout of $L$ levels. One level is the central hexagon:

\[H(1) = 1\]

Each further level is a ring around the previous pattern. Ring $L$ has $L - 1$ hexagons along each of its six sides:

\[H_{\text{added}}(L) = 6(L - 1)\]

The total is the sum of the rings:

\[H(L) = 1 + \sum_{i=2}^{L} 6(i - 1)\]

Using the sum of the first $n$ natural numbers,

\[\sum_{i=1}^{n} i = \frac{n(n + 1)}{2}\]

the total becomes

\[H(L) = 1 + 6 \cdot \frac{(L - 1)L}{2} = 1 + 3L(L - 1)\]

and so

\[H(L) = 3L^2 - 3L + 1\]

The pattern grows quadratically: 1, 7, 19, 37, 61 hexagons for one to five levels.

References

1. Hexagon tangent problem, Math Central, University of Regina. The diagram is from this page.