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.
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:
or:
in a more concise and complex form[1]
where
(Assuming the initial point lies on the larger circle.) When k is a positive integer, the area of this epicycloid is
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 on the small circle) varies up and down as
R <= OP <= (R + 2r)
R = radius of large circle and
2r = diameter of small circle
k = 1 a cardioid
k = 2 a nephroid
k = 3 a trefoiloid
k = 4 a quatrefoiloid
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.
We assume that the position of is what we want to solve, is the angle from the tangential point to the moving point , and is the angle from the starting point to the tangential point.
Since there is no sliding between the two cycles, then we have that
By the definition of angle (which is the rate arc over radius), then we have that
and
From these two conditions, we get the identity
By calculating, we get the relation between and , which is
From the figure, we see the position of the point on the small circle clearly.