I'd like to understand the new SRI "Abacus" transmission, recently reported at http://spectrum.ieee.org/automaton/robotics/robotics-hardware/sri-demonstrates-abacus-rotary-transmission

I haven't been able to find any information on it that doesn't seem derivative of the IEEE story.

I'm curious about how one determines the reduction ratio, and what the "grooves" look like mathematically (for example, the function from angle to width).

Judging from the pictures, this one seems to have 6 setscrews on the inner race, 7 "beads", and 8 setscrews on the outer race.

It's hard to tell the number of cycles for the grooves, but the ones in the outer race seem to be aligned with the setscrews.

They say the beads can have any profile you like; it seems that must then determine the groove profiles, but how would you get from one to the other?

I've tried to reason out the relationship between number of groove cycles on each race, diameter profile of the beads, offset of the driveshaft's attachment to the inner cam, and the resulting reduction ratio. But my intuition is failing even after watching the video many times.

Can anyone explain the relationships involved here?



1 Answer 1


It's just a fancy rolling cycloidal drive (not that that being simple makes it any less cool). It really bothers me that all the available articles only compare it to harmonic drives when it only bears a passing resemblance to them. Probably because it's owned by Harmonic Drive LLC. The bearing races can be abstracted as wire ring races that have been bent sinusoidally. The reason the metal one has a conical bearing is so that the races are easier to manufacture, since they're just sinusoidal rather than cycloidal, like they'd almost certainly have to be for the round bearing, and good luck milling that. What also bothers me about the video posted about it is that the guy claims it is a 1:3 drive. Well, the formula for the gear ratio of a cycloidal drive is P-L/L where P is the number of outer ring pins, in this case the number of periods of the outer sinusoidal wave, and L is the number of lobes on the cycloidal disk, in this case the number of bearings. (I think you're right about the set screws) (8-7)/7 = 1/7, not 1/3.

  • $\begingroup$ @Auptimist, thanks much for giving me the search phrase "rolling cycloidal drive." Quite helpful for understanding this widget! $\endgroup$
    – Catalyst
    Commented Sep 23, 2017 at 11:53

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