Torque vs Weight: Calculate Force for 6.156 kgf cm, 20 rpm, 5 cm Rad

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cloudsword654
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Hey, I was wondering
If I have a motor that can supply 6.156 kgf cm of torque operating at 20 rpm, how much weight can that engine move forward given that the radius of the tires will be 5 cm? Please provide equations. Thanks
 
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Hi Cloudswords-
First, let's convert the motor parameters to power. the torque 6.186 kgf cm is 0.604 Newton-meters, so the power is 0.604 Nm x 20 x 2 pi/60 = 1.26 watts. The power to move the vehicle on a level surface is the velocity times the force to push it, which is 0.01 times the weight W (in Newtons) times the velocity (m/sec), where 0.01 is the expected rolling resistance coefficient of the rubber tires. Using direct drive to the tire, the velocity = [STRIKE]0.104[/STRIKE] 0.658 meters/sec, so the required power is 0.01 W x [STRIKE]0.104[/STRIKE] 0.658 = [STRIKE]0.00104[/STRIKE] 0.00658 W watts.

So 1.26 watts = [STRIKE]0.00104[/STRIKE] 0.00658 W watts
So W = [STRIKE]1211[/STRIKE] 191 Newtons ([STRIKE]122[/STRIKE] 19.5 Kg)

[Edit] See table of rolling resistance coefficients in Table near bottom of
http://en.wikipedia.org/wiki/Rolling_resistance

Bob S
 
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