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Unread 24-04-2016, 23:57
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asid61 asid61 is offline
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AKA: Anand Rajamani
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Re: Calculating optimal weight efficiency ratios

Quote:
Originally Posted by GeeTwo View Post
While weight of the bull gear would eventually be an issue, the decision to go from single-stage to double (or triple) stage is more commonly driven by geometric considerations (the SIZE of the gearbox, esp. relative to the wheels). Pinions for most of the FRC motors typically have 10-15 teeth. Smaller numbers of (spur gear) teeth would result in poor pressure angles. Gears with more than about 50 or 60 teeth (again, in FRC-typical sizes) are often larger than the wheels they drive, leading to untenable gearboxes. That is, at somewhere around 3.5:1, a two-stage gearbox is preferable to drive wheels of 4" or larger. At somewhere around 14:1, a three-stage gearbox is preferable to drive 4" or larger wheels.

With planetary gearboxes, these ratios are somewhat different, of course. Planetary gearboxes can reduce up to 10:1 (theoretically a bit higher, but no more than 14.9:1) per stage.
This is a purely academic exercise, not really one that's practical.
I totally agree with you though. I'm working a geared flipped-cim gearbox right now, and my main constraint is the combined area of the gears and the mounting on the 2x1, not the weight.
I usually run 2 stages for 14:1, something like a 12:50 x 14:50, but it varies based on the application.

Why do you say a planetary can only go up to 14.9:1? The formula should just be R/S + 1, and as R (teeth on ring) gets larger it better approximates a gear rack, so S (teeth on sun) can get smaller while avoiding interference.
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