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The epicycloid is constructed from a circle rolling around a stationary circle

Let $R$ be the stationary circle's radius

Let $r$ be the rolling circle's radius

Let $a$ be the ratio $\frac{R}{r}$

Let $d$ be how far to the left of the rolling circle's center we place the point whose location is being traced by the epicycloid (negative $d$ means it is to the right instead of to the left)

Parametrize the epicycloid by

$$\left(\left(a+1\right)r\cos\left(t\right)-d\cos\left(\left(1+a\right)t+\theta_{0}\right),\left(a+1\right)r\sin t-d\sin\left(\left(1+a\right)t+\theta_{0}\right)\right)$$

Then the quantity I want to derive is the curvature at $t=-\frac{\theta_{0}}{a+1}$

WLOG, consider the case where $\theta_0=0$

$$\left(\left(a+1\right)r\cos\left(t\right)-d\cos\left(\left(1+a\right)t\right),\left(a+1\right)r\sin t-d\sin\left(\left(1+a\right)t\right)\right)$$

Then to obtain my desired quantity I can instead derive the curvature at $t=0$

Once I have my desired expression for the curvature in terms of $a$, $d$ and $r$, I can then ask the following: for some specified value of $r$, what must be the relationship between $a$ and $d$ in order for the curvature at $t=0$ to be zero? The answer I'm looking for will then be in the form of an expression that defines the relationship between $a$, $d$ and $r$

In other words, the question is: if the curvature equals zero, then what is the expression relating $a$, $d$ and $r$

Have I correctly set myself up with a valid process for arriving at the answer? I don't think I have it in me to provide the actual expression, because the algebra just seems to turn out to be way too unwieldy.

Here I have a graph of the epicycloid as well as (now that I know the answer) the osculating circle at $t=0$

Simon M
  • 657

2 Answers2

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The following show standard results for epitrochoid and hypotrochoid.

  • Curvature of parametric curve

$$\kappa = \frac{x' y''-y' x''}{(x'^2+y'^2)^{3/2}}$$

  • Epitrochoid

\begin{align} x &= (a+b) \cos t - h\cos \frac{(a+b)t}{b} \\[3pt] y &= (a+b) \sin t - h\sin \frac{(a+b)t}{b} \\[3pt] \kappa &= \frac {b^3-bh(a+2b) \cos \dfrac{at}{b}+(a+b)h^2} {(a+b)\left( b^2-2bh\cos \dfrac{at}{b}+h^2 \right)^{3/2}} \end{align}

  • Hypotrochoid

\begin{align} x &= (a-b) \cos t + h\cos \frac{(a-b)t}{b} \\[3pt] y &= (a-b) \sin t - h\sin \frac{(a-b)t}{b} \\[3pt] \kappa &= \frac {b^3+bh(a-2b) \cos \dfrac{at}{b}-(a-b)h^2} {(a-b)\left( b^2-2bh\cos \dfrac{at}{b}+h^2 \right)^{3/2}} \end{align}

The rest is left as an exercise.

Ng Chung Tak
  • 18,990
  • My parametrization is perfectly intuitive, nothing atypical going on. Take for instance what you provided as the x-coordinate $$\begin{align} x &= (R+r) \cos t - d\cos \frac{(R+r)t}{r} \[3pt] \end{align}$$

    Now substitute $a=R/r$ $$\begin{align} x &= (a+1)r \cos t - d\cos ((a+1)t) \[3pt] \end{align}$$

     – Simon M Sep 25 '23 at 14:15
  • additionally, the wikipedia for the epicycloid lists my parametrization as an alternate parametrization, where they use $k$ instead of $a$. So it turns out my parametrization not atypical at all; wikipedia acknowledges it – Simon M Sep 25 '23 at 14:26
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Using the expression for curvature provided by Ng Chung Tak, then substituting $a=R/r$, the curvature at $t=0$ is $$k=\frac{r^{3}+\left(a+1\right)rd^{2}-r^{2}d\left(a+2\right)}{r\left(a+1\right)\left(r^{2}+d^{2}-2rd\right)^{3/2}}$$

Thus, if the curvature at $t=0$ happens to be zero, it must be the case that $$0=\frac{r^{3}+\left(a+1\right)rd^{2}-r^{2}d\left(a+2\right)}{r\left(a+1\right)\left(r^{2}+d^{2}-2rd\right)^{3/2}}$$

Which means $$\boxed{0=r^{2}+\left(a+1\right)d^{2}-rd\left(a+2\right)}$$

This is the answer I was looking for

Simon M
  • 657