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## Summary

In geometry, an epicycloid 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. 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).

## 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:

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

in a more concise and complex form

$z(\theta )=r(e^{i(k+1)\theta }-(k+1)e^{i\theta })$

where

• angle $\theta$  is in turns: $\theta \in [0,2\pi ].$
• smaller circle has radius r
• the larger circle has radius kr

## Area

(Assuming the initial point lies on the larger circle.) When k is a positive integer, the area of this epicycloid is

$A=(k+1)(k+2)\pi r^{2}.$

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

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

sketch for proof

We assume that the position of $p$  is what we want to solve, $\alpha$  is the angle from the tangential point to the moving point $p$ , and $\theta$  is the angle 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 angle (which is the rate arc over radius), then we have that

$\ell _{R}=\theta R$

and

$\ell _{r}=\alpha r$ .

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 $p$  on the small circle clearly.

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