Designing pinions for an angled differential

Question

I'm designing a set of beveled gears where the input and output are oblique. It's simple to design a single pinion connecting them, two isn't hard, and I can model three as pitch cones, but getting a solution for three that works with real gears (re:integer tooth counts) is proving to be quite complex.

Below is my modelling approach. The gear module is 1 ($toothCount = diameter$) for the sake of calculations.

The diameter of the pinion and input/output gears are arbitrary integers. The angle between the input and output has some flexibility.

The plane's Z-axis rotation is arbitrary. The green pinion diameter is a driven value which must be an integer.

This smaller pinion is mirrored on the other side for a total of three pinions connecting the main gears.

These two arc lengths are the last driven values that need to be integers. Since the gear module is 1, the arc length can be evaluated by $smallPinionRadius * arcAngle / 180$.

In summary: the four chosen parameters are input/output angle, input/output diameter, large pinion diameter, and small pinion angle about the Z-axis. Critical driven values that need to be integers are the small pinion diameter, and the small pinion arc lengths between contact points on the input/output gears.

I was able to experimentally get a configuration that is only a few percent off from being true (used in the images), which is probably good enough for my application, but is there a formulaic way to find true solutions?

Answer 1

We consider two gears like your large gear and one of the small ones.

They both are cones with a side length of L and a base radius of R1, R2.

Let's say we make the large cone sharper by shortening the R but such that

$$Perimeter=2\pi R \ \in\Bbb Z $$

But the small gear perimeter will not necessarily be an integer If you maintain its Z axis. Unless you change the diameter of the small gear such that it is again an integer. But the axis will have changed.

Only a very limited number of positions will give the integer perimeter for both gears. In many cases never. unless the gears are very fine.

Answer 2

You are missing something. Not only must the pinions have an integer tooth count, the teeth of the three pinions have to mesh with both the driver and the driven gear. So the distance of the pitch circle loop running from the driver up one pinion, along the driven gear to the next pinion, back down the second pinion to the driver an back across the driver to the first pinion has to be an integer. Imagine loops of roller chain running around the pinions and attached to the rim of the driver and follower. Each loop has to have an integer pitch length.

Consider a driver, follower, and one pinion. All have an integer tooth count. So a path following the pitch circle of all the gears has an integer number of teeth. Now if you add a second pinion, and use two paths to cover the entire pitch circle, you have one pinion's worth more teeth, but the two loops, run the way to loops of chain would run, must each have an integer path length.

Define azimuth, $\theta $, of the pinion as the clocking angle of the pinion axis on the plane that bisects the driver and follower axes. This is zeroed at the intersection of the plane defined by the driver and follower axes.

Now some geometry - the angle on the pinion between the two contact points is $\beta = 180 +2\alpha sin(\theta)$ where $\alpha$ is the angle the driver and follower make to the pinion axis plane. (This is an approximation. The azimuth on the driver and follower is $\theta' = \theta + D pinion/D driver * \frac12(\beta-180)$, so you can iterate on Beta a couple times if you want.)

You now just have to advance $\theta$ from some random starting point until the teeth engage. Imagine pinion meshed on the driver and the pinion rolling until the follower meshes. This happens when the tooth count around the loop as described above total an integer. And we can now generate expressions for all four legs of that equation, so solve for $\theta$. There is obviously only a finite number of solutions, so you don't have perfect freedom with output angles of the follower, but it is pretty finely graded once the tooth counts are reasonable.