I need an inexpensive, low backlash, compact, gear reduction which can be produced with imprecise 3D printed parts and can achieve very high ratios in a small space. As far as I know, this doesn’t exist, so I’ll design it myself. This is a difficult set of design constraints.

The combination of low backlash and imprecise 3D printed parts rules of spur gears. Timing belts, however, are inexpensive, relatively precise, inherently low backlash, and the pulleys have geometry which lends itself well to 3D printing. So we’ll start our design around widely available Gates GT2 belting.

The design I’ll eventually end up on can be adapted to pretty much whatever ratio you’d like, but in particular I settle on roughly 12000:1. The usual way to achieve such high ratios is with a compound drive train: one reduction stage feeds another. With a two stage reduction, to achieve 12000:1, each stage needs to provide roughly 110:1 reduction. Using 20 tooth minor pulleys (about the minimum practicable), this gives 2200 tooth major pulleys. These major pulleys then have a diameter of 1400mm! We would need to go to a four stage reduction to keep the pulley sizes reasonable. In that case we would have four stages of 20:209, where the 209 tooth pulley has a diameter of 133mm. We can do better by abandoning this sequential compound design.

If we build a differential^{1} (a mechanical device which can output the difference of two inputs) then we can achieve very high reduction ratios by driving the two inputs to the differential at slightly different speeds.

The housing required for an automotive style differential is poorly suited to 3D printing, and they require printed (bevel) spur gears, which we have already ruled out for similar reasons.

We can however build an epicyclic differential^{2} to negate the need for a housing, and use belting in place of gearing. We will use 20 tooth GT2 idlers as planets, since they are readily available for not much more than the cost of the ball bearings that go into them. We will use glued in place closed GT2 belts to add teeth to the sun and ring gears. In a planetary gear system, the ring gear needs to have a radius equal to the sum of the radius of the sun gear and the **diameter** of a planet. This means that the belt we use for the ring gear will need to have \(2*20\) teeth more than the sun. With GT2 having a 2mm pitch, that means the ring gear belt will be 80mm longer than the sun gear belt. So we need to find off-the-shelf belt sizes which fit that constraint. Luckily 200mm and 280mm belts are readily available.

We’ll fix on those 200mm and 280mm belts, with 20 tooth idlers. We can start nailing down some dimensions:

\[ Ring_{diam} = \frac{280mm}{\pi} - 1.01mm = 88.13mm \]

\[ Sun_{diam} = \frac{200mm}{\pi} - 1.01mm = 62.67mm \]

\[ PlanetCircle_{diam} = \frac{200mm}{\pi} + \frac{20t * 2\frac{mm}{t}}{\pi} = 76.39mm \]

where \(1.01mm\) is the distance from the back of GT2 belt to half the tooth height.

Due the the particulars of the application, we’ll use the ring and planet carriers as input and take our output from the sun. Note that the drive ratios of these two inputs differ. With a fixed ring gear the planet carrier drives the sun at a (+ 1 (/ 140 100.0)) = 2.4 ratio. While with a fixed planet carrier, the ring drives the sun at a (* -1 (/ 140 100.0)) = -1.4 ratio (note the minus sign, this will be useful).

Since the planet carrier drives the sun faster than the ring does, to null out for high resolution differential drive, the planet carrier needs to be driven by a slower pulley ratio than the ring does. The ring gear needs to be of some minimum size in order to accommodate the planets. While the planet carrier must extend beyond the ring gear in order to be geared down sufficiently.

We’ll set the target size for the driven surface of the ring gear by taking the diameter of inner surface (88.13mm) and adding an allowance, say 10mm radially, for structural reasons. This gives us (/ (* (+ 20 88.13) pi) 2) ~= 170 teeth. I already have some 167 tooth rings from early experiments.

In the end the goal is sub arc-second resolution when driven by a standard 1.8° stepper motor with no microstepping.

We’ll use a bit of Haskell to choose suitable values for all the pulleys (ghci is great for small finite optimization problems).

```
import Data.List (sortOn)
arcSecondsPerStep step ringDrive ringPulley planetDrive planetPulley =
(1/step) * -- microstepping factor
((2.4 * planetDrive/planetPulley) - (1.4 * ringDrive/ringPulley)) * -- differential ratio
(360 * 60 * 60) * -- Convert turns to arc-seconds
(1/200) -- 200 steps per turn standard stepper motor
main = print $
take 10
. sortOn (\(a,_,_,_,e) -> abs e) -- sort by smallest planetPulley
. filter (\(a,_,_,_,_) -> abs a < 1 && abs a > 0.1) -- looking for sub arc-second
$ [ ( arcSecondsPerStep step ringDrive ringPulley planetDrive planetPulley
, ringDrive
, ringPulley
, planetDrive
, planetPulley
)
| ringDrive <- [50..100]
, ringPulley <- [167] -- Use the ring pulleys I already have
, planetDrive <- [50..100]
, planetPulley <- [50..250]
, ringPulley < planetPulley
, step <- [1]]
```

This produces:

```
[(-0.5477985206056246,96.0,167.0,57.0,170.0),(0.8523069468000433,91.0,167.0,55.0,173.0),(-0.38588017335448477,87.0,167.
0,55.0,181.0),(-0.3816629037012298,97.0,167.0,62.0,183.0),(0.9228677779576967,82.0,167.0,53.0,185.0),(-0.99069945215932
26,99.0,167.0,65.0,188.0),(0.4850299401200431,82.0,167.0,55.0,192.0),(-0.4850299401200431,85.0,167.0,57.0,192.0),(0.960
0592629176674,90.0,167.0,61.0,194.0),(0.3545396516613586,93.0,167.0,64.0,197.0)]
```

The first entry here seems fine. It will require a 167 tooth ring sprocket driven by a 96 tooth sprocket, and a 170 tooth planet carrier sprocket driven by a 57 tooth sprocket; and will produce a ~0.55 arc-second resolution.

It can be useful to know the exact transmission ratio. We can compute that with another snippet of Haskell.

```
import Data.Ratio
ratio ringDrive ringPulley planetDrive planetPulley = recip (((24 % 10) * (planetDrive % planetPulley)) - ((14 % 10) * (ringDrive % ringPulley)))
main = print $ ratio 96 167 57 170
```

This produces: (-70975) % 6. 70975:6 is a reduction of 11829.1667.

Using the belt length calculations from a previous post, we find that we need 170.43mm center to center spacing for the 96:167 system, and 184.34mm for the 57:170 system in order to use readily available 610mm belts.

Now all that’s left is to build, test and program!