# Epicycloid

(Redirected from Epicycloidal)
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[citation needed] 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:

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

or:

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

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, two cusps is a nephroid.

An epicycloid and its evolute are similar.[1]

## Proof

sketch for proof

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

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

${\displaystyle \ell _{R}=\ell _{r}}$

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

${\displaystyle \ell _{R}=\theta R,\ell _{r}=\alpha r}$

From these two conditions, we get the identity

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

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

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

From the figure, we see the position of the point ${\displaystyle p}$ clearly.

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