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

[cloudsword654]

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

 

[Bob S]

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

 

[astrokreng]

I would be happy to assist with your question. To calculate the force generated by the motor, we can use the formula: Force = Torque / Radius. In this case, the torque is given in kilograms-force centimeters (kgf cm) and the radius is given in centimeters (cm). However, for consistency, it is important to convert the torque to a standard unit of torque, such as Newton-meters (N*m).

To convert kgf cm to N*m, we can use the conversion factor of 0.0980665 N*m per kgf cm. Therefore, the torque in N*m would be 6.156 kgf cm x 0.0980665 N*m/kgf cm = 0.603 N*m.

Plugging this value into our formula, we get: Force = 0.603 N*m / 5 cm = 0.1206 N.

So, with the given parameters, the motor can generate a force of 0.1206 Newtons. Keep in mind that this calculation assumes ideal conditions and does not take into account any external factors that may impact the force generated by the motor. I hope this helps answer your question.