Epicycloid 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 or hypercycloid is a plane curve produced by tracing the path of a chosen point on the circumference of a circle—called an epicycle—which rolls without slipping around a fixed circle. It is a particular kind of roulette.

Equations

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:

$x(\theta )=r(k+1)\cos \theta -r\cos \left((k+1)\theta \right)\,$ $y(\theta )=r(k+1)\sin \theta -r\sin \left((k+1)\theta \right).\,$ (Assuming the initial point lies on the larger circle.)

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.

When measured in radian, $\theta$ takes value from $0$ to $2*\pi *{\frac {LCM(r,R)}{r}}$ where LCM is least common multiple.

The epicycloid is a special kind of epitrochoid.

An epicycle with one cusp is a cardioid, two cusps is a nephroid.

An epicycloid and its evolute are similar.

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.

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