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Quote:
12:45 Ratio, so 19.65 * (45 / 12) = 73.7 in-lbs. This 13 tooth gear has a radius of (13/20) / 2, or 0.325 inches. Hence, the applied force is 73.7 in lbs / 0.325 lbs or 226.8 lbs. A quick Lewis Bending Equation I'm showing says this is not fairing too well. This admittedly conservative equation is showing 0.68 factor of safety, assuming 40 kpsi yield stress on your 'common cold rolled steel'. (Estimated Stress: 58 kpsi for 20 degree Pressure Angle. 14.5 will be roughly 20-30% weaker) A few things to ask: I used the stall torque of the CIMs. Is it possible for this geartrain to actually 'Lock Up'? Are you doing anything fancy that could cause locking or braking down the line in your drivetrain? If not, and your gear the drivetrain properly to ensure you don't trip breakers while pushing up against a wall, then the true peak torque on these gears will be less than what I used. (Perhaps even only 40% of the torque I used.) A few things you can do to reduce stress on the gear:
Calcs attached (EDIT: for 20 degrees, its what I had handy) for reference. Bottom Line: I won't make any promises, but assuming a standard drive train setup with slipping tires right around stall current, I'd say this gear, as is, is on on the edge. I'd sleep better with 1/2" face width. Matt Last edited by Matt Adams : 22-01-2006 at 03:27. |
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