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 a positive integer, then the curve is closed, and has k cusps (i.e., sharp corners).

If k is a rational number, say k = p / q expressed as irreducible fraction, then the curve has p cusps.

To close the curve and
complete the 1st repeating pattern :
θ = 0 to q rotations
α = 0 to p rotations
total rotations of outer rolling circle = p + q rotations

Count the animation rotations to see p and q .

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 distance OP from (x=0,y=0) origin to (the point $p$ on the small circle) varies up and down as

R <= OP <= (R + 2r)

R = radius of large circle and

2r = diameter of small circle

Epicycloid examples
k = 1 a cardioid
k = 2 a nephroid
k = 3 - resembles a trefoil
k = 4 - resembles a quatrefoil
k = 2.1 = 21/10
k = 3.8 = 19/5
k = 5.5 = 11/2
k = 7.2 = 36/5

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 $p$ 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