Predicting Spirograph patterns with complex numbers

Clearing out my grandparents' loft I found an old Spirograph-style set: a plastic ring with 96 teeth on the inside and twelve toothed wheels. You pin the ring down, put a pen through a hole in a wheel and roll the wheel round inside the ring. Some wheels give a flower with a few fat petals, others a dense lace of thirty or more, and some take so long to close that I gave up on my first attempt before the pattern had joined up. I wanted to know if I could predict the pattern before drawing it.

Aim. To model the path of the pen with complex numbers and use the model to predict (i) how many lobes (“petals”) a pattern has, (ii) how many times the wheel must go round the ring before the pattern closes, and (iii) how pointed the lobes are, and then to test the predictions against my own drawings.

I chose complex numbers because in class the exponential form \(re^{i\theta}\) made rotations easy, and rolling is a rotation on top of a rotation.

I put the centre of the ring at the origin of the complex plane. The teeth are equally spaced; I measured 10 teeth as 31.4 mm, so the spacing is about \(\pi\) mm and a gear with \(n\) teeth has (pitch) radius \(\frac{n\pi}{2\pi} = \frac n2\) mm. So the ring has radius \(R = 48\) mm and a wheel with \(n\) teeth has radius \(r = \frac{n}{2}\) mm. The pen is in a hole at distance \(\rho\) from the wheel's centre, with \(\rho < r\).

When the point of contact has moved round the ring by an angle \(\theta\), the wheel's centre is at \((R - r)e^{i\theta}\). Because the teeth mesh, the wheel does not slip, so the arc it has rolled along the ring, \(R\theta\), equals the arc of its own rim that has touched the ring. The wheel has therefore turned by \(\frac{R\theta}{r}\) clockwise relative to the line from its centre to the contact point, and that line itself has turned \(\theta\) anticlockwise. The net rotation of the wheel is \(\theta - \frac{R\theta}{r} = -\frac{(R - r)\theta}{r}\). Starting with the pen at its furthest point from the origin, its position is

\[ z(\theta) = (R - r)e^{i\theta} + \rho\, e^{-i\frac{(R - r)}{r}\theta}. \qquad (1) \]

The model's pictures look exactly like my drawings, which gave me confidence to use (1) for predictions.

A lobe has its tip where the pen is furthest from the centre. Using \(|z|^2 = z\bar z\) with (1),

\[ |z|^2 = (R-r)^2 + \rho^2 + (R - r)\rho\left(e^{i\frac{R}{r}\theta} + e^{-i\frac{R}{r}\theta}\right) = (R-r)^2 + \rho^2 + 2(R-r)\rho\cos\left(\frac{R\theta}{r}\right). \qquad (3) \]

This is largest when \(\cos\frac{R\theta}{r} = 1\), i.e. \(\theta = \frac{2\pi r m}{R}\) for integers \(m\), and smallest when \(\cos\frac{R\theta}{r} = -1\). So the outer diameter of a pattern is \(2(R - r + \rho)\) and the inner diameter is \(2|R - r - \rho|\). During the \(k = \frac ng\) turns before the pattern closes, \(\theta\) runs from 0 to \(\frac{2\pi n}{g}\), and the tips occur every \(\frac{2\pi r}{R}\), so the number of lobes is

\[ \frac{2\pi n/g}{2\pi r/R} = \frac{n}{g}\cdot\frac{96}{n} = \frac{96}{\gcd(96, n)}. \qquad (4) \]

I also noticed that the patterns have rotational symmetry. Replacing \(\theta\) by \(\theta + \frac{2\pi r}{R}\) in (1) multiplies both terms by \(e^{i\frac{2\pi r}{R}}\) (the second because \(e^{-i\frac{2\pi(R-r)}{R}} = e^{-2\pi i}e^{i\frac{2\pi r}{R}}\)), so \(z\left(\theta + \frac{2\pi r}{R}\right) = e^{i\frac{2\pi r}{R}}z(\theta)\): moving on by one lobe rotates the whole picture by the angle \(\frac{2\pi n}{96}\). This rotation is a power of \(\omega = e^{2\pi i g/96}\), a \(\frac{96}{g}\)th root of unity, and I checked in a spreadsheet for all my wheels that its powers give every \(\frac{96}{g}\)th root of unity, so each pattern has rotational symmetry of order \(\frac{96}{g}\) — the same as the number of lobes.

The complex-number model \(z(\theta) = (R - r)e^{i\theta} + \rho e^{-i(R-r)\theta/r}\) predicts Spirograph patterns well. For a wheel with \(n\) teeth in the 96-tooth ring, the pattern has \(\frac{96}{\gcd(96,n)}\) lobes and closes after \(\frac{n}{\gcd(96,n)}\) turns, with rotational symmetry of the same order as the number of lobes. The lobes are most pointed when the pen hole is near the rim, because the pen's speed at the tips is \((R - r)(1 - \frac{\rho}{r})\). All six of my drawings had the predicted number of lobes, and the diameters were within 1.5 mm of the predictions.

The main limitation is that my measurements were done with a ruler and the differences were all in the same direction, which I could not fully explain. The model also assumes the wheel never slips and the pen is exactly at the hole's centre. In the future I would like to look at wheels rolling on the outside of the ring (epitrochoids), and at patterns made with two wheels.