I recently was reading about nautilus gears, and I was wondering what other irregular gears one can make. In particular, I'm interested in systems in which both gears are congruent (or at least similar).
What I know so far:
We've got a pair of gears with axles separated by some fixed distance $D$. If we forget the teeth, they're just two curves, and they have to roll without slipping. Gonna assume the contact point always lies on the line between the two axles.
If you know one gear has pitch curve $r_1(\phi_1)$, then you can derive the other one. (Details are here but I'm gonna abbreviate them).
Basically, let the first gear rotate through $\phi_1(\theta) = \theta$. The second gear rotates through (I'm using a different sign convention than the link): $$\alpha(\theta) = - \int_0^\theta \frac{r_1(\phi_1)}{D - r_1(\phi_1)}~d\phi_1$$
Since $r_1(\theta) + r_2(\alpha(\theta)) = D$ at all times, finding $r_2 \circ \alpha$ (i.e., $r_2$ as a function of $\theta$) is easy.
However, I want to find $r_2$ as a function of $\phi_2$, so I can actually compare the damn thing to $r_1$. It'd be nice if $\alpha$ were an easily-invertible function, but that's rarely the case... >:(
Example: circular gears
Say the first gear is circular, i.e., $r_1 \equiv A$. Then $$ \alpha(\theta) = - \int_0^\theta \frac{A}{D - A}~d\phi_1 = -\frac{A}{D - A} \theta $$
Since $B = D - A$ is the radius of the second gear, we get $\frac \alpha \theta = - \frac A B$, as we'd expect (the negative sign shows they rotate in opposite directions).
Okay, less trivial example: nautilus gears
The first gear has profile $r_1(\phi_1) = A \cdot B^{\phi_1 / 2 \pi}$. The distance between the gears is gonna be the sum of the biggest and smallest radius: $D = A + AB$.
This gives $$ \begin{align*} \alpha &= - \int_0^\theta \frac{A \cdot B^{\phi_1 / 2 \pi}}{A + AB - A \cdot B^{\phi_1 / 2 \pi}}~d\phi_1 \\ &= \int_0^\theta \frac{- B^{\phi_1 / 2 \pi}}{1 + B - B^{\phi_1 / 2 \pi}}~d\phi_1 \\ &= \left[ 2 \pi \frac{\ln(1 + B - B^{\phi_1 / 2 \pi})}{\ln B} \right]^\theta_0 \\ \alpha &= 2 \pi \frac{\ln(1 + B - B^{\theta / 2 \pi})}{\ln B} \\ \end{align*} $$
This one's actually invertible without too much fuss: $$ \begin{align*} B^{\alpha / 2 \pi} &= 1 + B - B^{\theta / 2 \pi} \\ AB^{\alpha / 2 \pi} &= A + AB - AB^{\theta / 2 \pi} \\ AB^{\alpha / 2 \pi} &= D - r_1(\theta) = r_2(\alpha(\theta)) \\ \end{align*} $$
So $r_2(\alpha) = AB^{\alpha / 2 \pi}$, which is congruent to the first gear.
Where I'm getting stuck is
- I couldn't seem to replicate this for elliptical gears. My integral is gross and nasty and requires elliptic integrals.
- Is there a way to set $r_1$ and $r_2$ equal (as functions of their respective $\phi$s) that doesn't involve inverting a nasty integral? I tried setting $r_1(\alpha(\theta)) = r_2(\alpha(\theta)) = r_1(\theta)$, but that didn't get me very far (guess putting an integral inside your function is a bit too bold?)