# Irregular Gears

Posted: February 26, 2013 / Updated: May 20, 2016

## The Mathematics of Irregular Gears

I'm basing my approach to this problem on the humble epitrochoid. Unlike the point-based approach, here we are keeping one gear stationary and rolling the other around its surface.

For a standard epitrochoid, the paramaterized coordinates are found using a parameter $$t_1 \in \left[ -\pi, \pi \right]$$. Don't be fooled by the limits of $$t_1$$, it is NOT the polar angle of the resulting epitrochoid. Anyway: $$\begin{array}{l} \displaystyle\phantom{y}x(t_1) = (R+r)\cos{(t_1)}-d\cos{\left(\frac{R+r}{r}t_1\right)}\\ \displaystyle\phantom{x}y(t_1) = (R+r)\sin{(t_1)}-d\sin{\left(\frac{R+r}{r}t_1\right)} \end{array}$$ where $$x$$ and $$y$$ are the cartesian coordinates with respect to $$t_1$$; $$R$$ and $$r$$ are the radii of the fixed (central) circle and the rolling circle, respectively; and $$d$$ is the distance from the center of the rolling circle to the point being traced.

Applying this to irregular gear design, we must consider that the distance $$d$$ varies along the profile of the input gear. Unfortunately, just calculating the epitrochoid radius at each $$t_1$$ will probably not give us a gear that meshes. We must also consider the profile of the input gear in nearby the contact point. To do this, we need to introduce a second parameter, $$t_2$$, which will be some kind of "sweep angle" that allows us to get the $$x$$ and $$y$$ positions of any point on the input gear, instead of a single point per $$t_1$$ value. This is shown below. $$\begin{array}{l} \displaystyle\phantom{y}x(t_1,\,t_2) = (R+r)\cos{(t_1)}-d(t_2)\cos{\left(\frac{R+r}{r}t_1+t_2\right)}\\ \displaystyle\phantom{x}y(t_1,\,t_2) = (R+r)\sin{(t_1)}-d(t_2)\sin{\left(\frac{R+r}{r}t_1+t_2\right)}\\ \end{array}$$

I can smell that this is going to turn into a parameter optimization problem, so we're going to need at least a new function and a new variable. Let $$\theta$$ be the polar angle of the output gear, and let $$G(\theta)$$ be the radius of said output gear at that polar angle.

We want $$G(\theta)$$ to be equal to the smallest radius returned by any combination of $$t_1$$ and $$t_2$$, provided the radius is reachable at angle $$\theta$$. As an equation, this becomes $$\begin{array}{l} \displaystyle G(\theta) = \min_{\substack{t_1 \in \left[-\pi,\,\pi\right]\\ t_2 \in \left[-\pi,\, \pi\right]}}{\left\{ x(t_1,\,t_2)^2 + y(t_1,\, t_2)^2 \right\}}\\ \displaystyle\operatorname{subject\,to:}\,\tan^{-1}{\left(\frac{y(t_1,\,t_2)}{x(t_1,\,t_2)}\right)}=\theta. \end{array}$$

After some trigonometric identities, the equations above give the following. $$\begin{array}{l} \displaystyle G(\theta) = \min_{\substack{t_1 \in \left[-\pi,\,\pi\right]\\ t_2 \in \left[-\pi,\, \pi\right]}} \left\{ \left(d(t_2)\right)^2 -2d(t_2)(R+r)\cos{\left( \frac{R}{r}t_1+t_2 \right)} \right\}\\ \displaystyle\operatorname{subject\,to:}\,\tan^{-1}\left(\frac {(R+r)\sin{(t_1)}-d(t_2)\sin{\left( \frac{R+r}{r}t_1+t_2 \right)}} {(R+r)\cos{(t_1)}-d(t_2)\cos{\left( \frac{R+r}{r}t_1+t_2 \right)}} \right)-\theta = 0. \end{array}$$

It's hard to go further without knowing the actual function $$d$$ to be used. But at least now we've got something to plop into MATLAB's optimization function fmincon. You know. If you were so inclined. But that's another post for another time.