Octagon geometry

This derives the dimensions of a regular octagon that the 4.8.8 layout is built on.

Side, width and diagonal

Diagram courtesy of Penny Nom [1].

In the diagram $s$ is the side length, $w$ is the width across the flats, and $d$ is the longest diagonal, from one corner to the opposite one.

Segment x

Extending the sides of the octagon cuts a right isosceles triangle off each corner, such as $EBD$. Its two equal legs are the segments $x$ and its hypotenuse is a side of the octagon. By the Pythagorean theorem:

\[s^2 = 2x^2 \qquad x = \frac{s}{\sqrt{2}}\]

Width w

The width runs from one flat to the opposite one. It is a side plus a segment $x$ at each end:

\[w = s + 2x = s + \frac{2s}{\sqrt{2}} = s(1 + \sqrt{2})\]

Diagonal d

$ABC$ is a right triangle with the right angle at $A$. Its legs are the width, $AB = w$, and a side, $AC = s$. Its hypotenuse is the diagonal:

\[d^2 = s^2 + w^2 = s^2 + s^2(1 + \sqrt{2})^2\] \[d = s\sqrt{4 + 2\sqrt{2}}\]

Summary

Quantity	In terms of the side
Segment	$x = s / \sqrt{2}$
Width	$w = s(1 + \sqrt{2})$
Diagonal	$d = s\sqrt{4 + 2\sqrt{2}}$

How the library uses them

The library sizes an octagon by radius, the distance from the centre to a corner. The diagonal is twice that, which gives the side length that appears in src/semi_regular/tess_4_8_8.scad:

\[s = \frac{2 \cdot \text{radius}}{\sqrt{4 + 2\sqrt{2}}}\]

Neighbouring octagons in the 4.8.8 tiling touch along a flat, so their centres are one width apart. That is the grid pitch, total_width in the source:

\[w = s(1 + \sqrt{2})\]

The squares have the same side length $s$ as the octagons. Each sits where four octagons meet, half a pitch along both axes from the octagon centres, and stands on a corner.

Chord across two sides

A rotated octagon layout would need two more lengths. The library does not use them yet.

The chord $AC$ joins two corners that are two sides apart. In triangle $ABC$ the sides $AB$ and $BC$ are both $s$, and the angle between them is the octagon’s interior angle, 135°. By the law of cosines:

\[AC^2 = 2s^2 - 2s^2\cos(135^\circ)\]

Since $\cos(135^\circ) = -\frac{\sqrt{2}}{2}$,

\[AC = s\sqrt{2 + \sqrt{2}}\]

Let $M$ be the midpoint of $AC$. $BM$ is perpendicular to $AC$, and in the right triangle $ABM$:

\[BM^2 + \left(\frac{AC}{2}\right)^2 = s^2\] \[BM = \sqrt{s^2 - \frac{s^2(2 + \sqrt{2})}{4}} = \frac{s}{2}\sqrt{2 - \sqrt{2}}\]

So the chord is $y = s\sqrt{2 + \sqrt{2}}$ and the height of the corner above it is $h = \frac{s}{2}\sqrt{2 - \sqrt{2}}$.

References

1. Quandaries and queries, Math Central, University of Regina.
2. Octagon, Cuemath.