# Epicycloid

(Redirected from Epicycloids)
The red curve is an epicycloid traced as the small circle (radius r = 1) rolls around the outside of the large circle (radius R = 3).

In geometry, an epicycloid is a plane curve produced by tracing the path of a chosen point of a circle — called an epicycle — which rolls without slipping around a fixed circle. It is a particular kind of roulette.

If the smaller circle has radius r, and the larger circle has radius R = kr, then the parametric equations for the curve can be given by either:

$x (\theta) = (R + r) \cos \theta - r \cos \left( \frac{R + r}{r} \theta \right)$
$y (\theta) = (R + r) \sin \theta - r \sin \left( \frac{R + r}{r} \theta \right),$

or:

$x (\theta) = r (k + 1) \cos \theta - r \cos \left( (k + 1) \theta \right) \,$
$y (\theta) = r (k + 1) \sin \theta - r \sin \left( (k + 1) \theta \right). \,$

If k is an integer, then the curve is closed, and has k cusps (i.e., sharp corners, where the curve is not differentiable).

If k is a rational number, say k=p/q expressed in simplest terms, then the curve has p cusps.

If k is an irrational number, then the curve never closes, and forms a dense subset of the space between the larger circle and a circle of radius R + 2r.

The epicycloid is a special kind of epitrochoid.

An epicycle with one cusp is a cardioid.

An epicycloid and its evolute are similar.[1]

## Proof

We assume that the position of $p$ is what we want to solve, $\alpha$ is the radian from the tangential point to the moving point $p$, and $\theta$ is the radian from the starting point to the tangential point.

Since there is no sliding between the two cycles, then we have that

$\ell_R=\ell_r$

By the definition of radian (which is the rate arc over radius), then we have that

$\ell_R= \theta R, \ell_r=\alpha r$

From these two conditions, we get the identity

$\theta R=\alpha r$

By calculating, we get the relation between $\alpha$ and $\theta$, which is

$\alpha =\frac{R}{r} \theta$

From the figure, we see the position of the point $p$ clearly.

$x=\left( R+r \right)\cos \theta -r\cos\left( \theta+\alpha \right) =\left( R+r \right)\cos \theta -r\cos\left( \frac{R+r}{r}\theta \right)$
$y=\left( R+r \right)\sin \theta -r\sin\left( \theta+\alpha \right) =\left( R+r \right)\sin \theta -r\sin\left( \frac{R+r}{r}\theta \right)$