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Re: 6 CIM Drivetrains... What's Your Experience?
Quote:
Originally Posted by Nate Bloom
What's the math for a traction limited drive?
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If:
- the wheels are all the same type and diameter and are chained (or belted) together on each side, and
- you are powering each side with a single gearbox with N CIMs, and
- you assume that traction force = normal force * coefficient of static friction (where coefficient is a constant), and
- the center of mass lies along the vehicle longitudinal axis, then:
A = [(4*D/G*Istall*mu*W)/(eff*Tstall)]/N
where:
A = amps per CIM required to break traction and spin the wheels
W = weight (lbs) of vehicle
mu = coefficient of static friction
D = diameter of wheel in inches
G = the total speed reduction gear ratio from motor to wheel (including sprockets/pulleys)
eff = drivetrain torque efficiency fraction
Tstall = CIM spec stall torque (oz_in)
Istall = CIM spec stall amps
Last edited by Ether : 07-12-2013 at 16:59.
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