Epicycloid

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 %2B r) \cos \theta - r \cos \left( \frac{R %2B r}{r} \theta \right)
y (\theta) = (R %2B r) \sin \theta - r \sin \left( \frac{R %2B r}{r} \theta \right),

or:

x (\theta) = r (k %2B 1) \cos \theta - r \cos \left( (k %2B 1) \theta \right) \,
y (\theta) = r (k %2B 1) \sin \theta - r \sin \left( (k %2B 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]

Contents

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 the two condition, 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%2Br \right)\cos \theta -r\cos\left( \theta%2B\alpha \right) =\left( R%2Br \right)\cos \theta -r\cos\left( \frac{R%2Br}{r}\theta \right)
y=\left( R%2Br \right)\sin \theta -r\sin\left( \theta%2B\alpha \right) =\left( R%2Br \right)\sin \theta -r\sin\left( \frac{R%2Br}{r}\theta \right)

See also

References

External links