Hello,
I have a small DC motor. i want to measure the torque. NO datasheet of any kind. I used 15lbs (6.8Kg) weight attached to 1.1 cm of the shaft (motor geared at 1/720 so it turns about 8 rpm). Can we say that the motor has a torque of at least 7.48Kg/cm ? The torque is as high as the motor can rotate at the specific voltage? in my case 12v.
Ken
Torque would be kg*cm. It would be 7.48 kg*cm at the shaft where you have that weight applied, or 0.733 N-m. If the motor is hiding behind a gear ratio of 720, it would see only about .001 N-m, plus an additional amount due to friction in the gears. This is assuming the weight is and remains applied at an exact tangent to the 1.1 cm circle, like lifting a weight hanging on a fishline attached to a pulley of that size.
-- John
** You mean the maximum torque ??? ** Gobbledegook.
What exactly has the dimension of 1.1 cm ??
** Who knows - you have not defined anything. ** Torque = force / radiusMax torque is found when any increase in the load causes the DC motor to stall.
........ Phil
I just wanted to make sure about the calculation, When i mentionned 1.1 cm it is the distance from the center of the shaft to the exterior where the rope was. I was a able to lift a max of 20Lbs or 9.1Kg. So I guess I have a 11Kg-cm max torque motor
k
Re: The radial distance from the shaft center or moment arm is 1.1 cm or
0.011m. The force acting perpendicular to the moment arm and downward is (6.8kg)(9.8m/s^2)=66.7N. Therefor, the torque developed is about (66.7N)(0.011m)=0.733 N*m at the reduction gear output. If the reduction gear ratio is indeed 720:1, then the motor proper output torque is approx. 0.733/720=0.001N*m (neglecting gearing losses).Dan Akers
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