I need to draw spur gears to make a picture look cool... and I want to make them viable. I vaguely remembered that the teeth are involute curves to circles, but why and what exactly? I need formulae!

Why do I publish this online? I think it might be useful for others as well. Even my really really simple firefox add-ons which are meant for personal use have other users in the world.

For easier reading, compile the LaTeX.


I got myself a head start by reading that. Well, actually I made some attempts before reading it. It was well written and has illustrations, while mine doesn't. I try to use same notation where possible.


For spur gears, it is natural to require that the motion is transferred without sliding. Otherwise, the gears are prone to wear. Furthermore we require that the points of contact of two gears trace out a line and that the force between the two gears is along this line. In this way motion is transferred smoothly. The moment is not bumping here and there.

One Gear

With this simple requirement we derive the shape of the teeth. Let $L$ be a line which will be the line traced out by contact points. It is called the line of action. Assume that $L$ is tangent to a circle $C$ of radius $r$ and centre $O$. The centre of $C$ corresponds to the centre of the gear and $C$ is called the base circle.  There is a curve $S$, which will represent the shape of one side of a tooth, attached to the circle $C$ so that when $C$ rotates, $S$ and $L$ always intersect orthogonally. Now we let the tooth be our reference object. Thus $L$ rolls on the circle. The initial point of tangency of $L$ then traces out $S$. This construction shows that $S$ is the involute curve of the circle $C$. It is another matter to determine where we cut off the involute curve. Suppose the gear has $n$ teeth. Then we attach $n$ copies of $S$  to $C$ evenly.  We get a circle with $n$ spiralling spikes.

Two Gears Meshing

Now we bring in another gear and add subscripts $i=1$ or $2$ to the various symbols. Thus we have two circles $C_1$ and $C_2$ with dangling spikes. They are separated by some distance and share the same line of action. They are both tangent to $L$. Let $\alpha$ be the angle between the line connecting the centres of the two circles and the normal line to $L$. I heard that this angle is called the pressure angle. The intersection of $O_1O_2$ and $L$ is called the pitch point. We can consider the gear system to be two bigger circles at tangency at the pitch point and transferring motion without sliding. These bigger circles are called the pitch circles. Apparently the base circles are more fundamental and we don't really need the pitch circles in this discussion.

For the two gears to mesh, the spikes should be appropriately spaced. Given  two points of contact on gear $1$, gear $2$'s spikes should be in contact with the two points. By the property of  involute curves (`length of interval = length of arc'), this means that the spacings of spikes on both gears must be equal, i.e., $2\pi r_1 / n_1 = 2\pi r_2 / n_2 =:A$.  (We lack a schematic picture here. One should attempt to measure distances from the tangencies of $L$ and the base circles.)

It is natural to mirror the spikes and form teeth. It is not clear how wide each tooth should be. We don't want the gears to backlash. They should be engaged. Let $\ell_i$ be the width of a tooth on gear $i$ and $m_i$ bethe width of the empty space between two teeth where both quantities are measured along arc. Of course,
\ell_i + m_i = A.
Then some fiddling with triangles and involute curves shows that
\ell_1 - m_2 = 2 (r_1+r_2) (\tan (\alpha) - \alpha).

Many Gears Meshing Mutually

If we fix one gear with parameter $(r_0, n_0, \ell_0, m_0)$ and produce other gears that can mesh with it. Will any other two gears mesh? Suppose we produced $(r_1, n_1, \ell_1, m_1)$ and $(r_2, n_2, \ell_2, m_2)$. Let $c = 2 (\tan(\alpha) - \alpha)$. For these two gears to mesh we need
\ell_1 - m_2 = c(r_1+r_2).
We already have
\ell_0 - m_1 = c(r_0+r_1)\
\ell_0 - m_2 = c(r_0+r_2).
Thus we reach the constraint:
2\ell_0 - A = 2cr_0.
Thus given parameter $(r, n)$, we are forced to have $\ell = cr + \frac{1}{2}A$ in order to achieve mutual meshing. Recall that $c = 2 (\tan(\alpha) - \alpha)$ is determined by pressure angle while $A= 2\pi r / n$ is the circular base pitch (spacing along base circle of teeth). Note that meshing gears share the same $c$ and $A$.


In practice, we may also need to form dedendum for motion to be possible. We have not decided on the size of addendum either.

To visualise things, we need to plot the curves that bound the teeth. It remains to write down an equation to feed into mathematical software.

Make a mechanical watch!? (minus the part of getting hold of something that pulses like once per second, though I am OK with any rational number, well not with huge height.) I stared at my own watch for some time, but I can't make out what the shape of the teeth is. Are they involute or cycloidal? Let me keep this watch theoretic and focus on computing appropriate addendum and dedendum so that I can continue drawing.