From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Cycloid fun
Date: 1 Apr 1999 16:04:53 -0500
Newsgroups: sci.math
Keywords: Kepler's equation and cycloids

In article <37038642.D65556F9@online.no>,
Michael Andersen  <micander@online.no> wrote:
>Hello. I wonder if it's possible to express a cycloid as a function
>y(x). I can easily set up a parameterprojection of one (with radius and
>'wheel' speed = 1):
>
>x = t - sin t
>y = 1 - cos t
>
>But if I try to express y with x without t, I get an equation I can't
>solve. For example:
>
>x = arccos(1 - y) - sin [arccos(1 - y)]
>
>This is not a homework problem, I'm just curious.


The problem reduces to inverting the relation x = t - sin(t) , which leads
to an extreme case of Kepler's equation. The solution involves a periodic
function with period 2*pi which I will call g (not knowing if there is an
official name for it); 

this g(x) is defined by  g(x) = sin(x + g(x))

and the solution to the first equation would be

 t = x + g(x)

Then the rest is easy:  y = 1 - cos(x + g(x))

Predictably, this g has a fractional power singularity at integer
multiples of 2*pi; at 0, its expansion around x=0 starts with

 g(x) = (6*x)^(1/3) * (1 + x^(2/3) * power series in x^(2/3))

That prevents g from being analytic, but perhaps g^3 (with a change of
variable) will be analytic.

A "formula in finite terms" for g is very unlikely to exist. There is some
literature about the periodic functions that help solve Kepler's equation.
Consult handbooks on special functions and references therein.

Incidentally, the parametric expression is excellent for plotting; if you
use dots rather than solid line, it reveals the speed of the point we
follow on the wheel (clustered dots indicate slower movement).

If you want the dots evenly placed, you need (in the first arch) to
parametrize by arclength. It is a good exercise to verify that if s is the
arclength,  0 <= s <= 8,  then  t = 4 * arcsin(sqrt(s/8)).

Also, a numerically reliable equation of x in terms of y, during the first
half-arch only (!), is

  x = 4 * arctan(sqrt(y) / (sqrt(2) + sqrt(2-y))) - sqrt(y*(2-y))

Graphically, though, the highest points will appear as "moving very
fast".

Have fun, ZVK(Slavek).