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Re: Omni vs Mecanum CoF ?
Quote:
Originally Posted by Ether
For one-CIM-per wheel:
amps=(5760*Istall*mu*V*W)/(pi*eff*Sfree*Tstall)
or
V=(pi*amps*eff*Sfree*Tstall)/(5760*Istall*mu*W)
where:
amps = current required to slip the wheel
Istall = CIM spec stall amps
Tstall = CIM spec stall oz_in
Sfree = CIM spec free RPM
mu = static coefficient
V = robot ft/sec speed at CIM free RPM
W = weight in lbs on the wheel
eff = drivetrain torque efficiency fraction
So, plugging in Oblarg's numbers:
amps=(5760*133*1.0*25*40)/(pi*1.0*5310*343.4) = 133.7
You're not going to get 133.7 amps in each of 4 motors, so if you gear for 25 ft/sec you definitely will not slip.
Let's try some more reasonable numbers and try again:
mu=1.0 V=12 W=37.5 eff=0.8
amps=(5760*133*1.0*12*37.5)/(pi*0.8*5310*343.4) = 75.2
Will the wheels slip? Well, 75 amps per motor times 4 motors is 300 amps. Unless your battery and wiring are in top shape, you're not going to get 75 amps per motor.
Let's try V=10:
amps=(5760*133*1.0*10*37.5)/(pi*0.8*5310*343.4) = 62.7
OK, maybe that will slip.
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Thanks for this. The calculation I did (obviously) placed no constraints on current draw on the motor.
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