I am trying to find the parametric equations of a cycloidal curve, which, instead of using a circle, uses an ellipse to oscillate around a base circle. Below are equations of the standard, circular epicycloid and hypocycloid parametric equations generally used.

Does anybody have any ideas?


Hypocycloid: http://mathworld.wolfram.com/Hypocycloid.html

$$\begin{align} x &= (a−b)\cos\phi + b \cos\left(\frac{a−b}{b}\;\phi\right) \\[6pt] y &= (a−b)\sin\phi − b \sin\left(\frac{a−b}{b}\;\phi\right) \end{align}$$

Epicycloid: http://mathworld.wolfram.com/Epicycloid.html

$$\begin{align} x &= (a+b) \cos\phi−b \cos\left(\frac{a+b}{b}\;\phi\right) \\[6pt] y &= (a+b) \sin\phi−b \sin\left(\frac{a+b}{b}\;\phi\right) \end{align}$$

  • $\begingroup$ When an ellipse rolls on a straight line it's focus locus is given by $\kappa_1+\kappa2=const. $ But it is a particular point only. Please show your derivation work at first. . The equation should be derived. One more constant is involved..you already gave name to it! $\endgroup$
    – Narasimham
    May 28, 2016 at 5:20
  • 1
    $\begingroup$ As there is no closed formula for the perimeter of the ellipse it would be very hard to calculate how much the ellipse is rotated as it is rounding the circle. You can find some kind of solution admitting that the ellipse slides on the circle. $\endgroup$
    – N74
    May 28, 2016 at 9:29
  • $\begingroup$ @N74, there's a closed form for the arclength function itself (see e.g. this inverse problem); the problem is that the inverse of the arclength function is the one that doesn't have a closed form. $\endgroup$ May 26, 2017 at 19:24


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.