I just wanted confirmation. Does the motor 393 have a current limit at 1 amp? If so, would I just plug in the torque load formula to find out how many amps it can handle before it trips? Here is my math and the specs.
stall motor 1.67 n’m
free current 0.37 amps
stall current 4.8 amps
free speed 100 rpms
torque load= (current-free current)x stall torque/(stall current-free current)
torque load = (1 amp-0.37)x 1.67 n’m / (4.8 amps-0.37 amps)
torque load= .63 amps x 1.67 n’m / 4.43 amps
torque load = .237 n’m
Second, if that is my torque load, do I divide that by how much torque I need to hold up the object? For example, if my arm is .48 m and the force is 5.17 N, then I would multiple .48mx 5.17 N to get a torque of 2.48 n’m. Then, divide 2.48 n’m /.237 n’m = 10.4.
So would that mean I would need 11 motors or a gear ratio of 1:11 to hold up the object?
It appears that the VEX forum community has beaten us to the punch. In the linked thread there is a graph showing the Motor curve of the 393 Motor. Here is the link. To answer your questions:
- Does the motor 393 have a current limit at 1 amp?
A1. Nope.
- If so, would I just plug in the torque load formula to find out how many amps it can handle before it trips?
A2. The graph shown in the linked thread will provide what you are looking for.
- Second, if that is my torque load, do I divide that by how much torque I need to hold up the object? For example, if my arm is .48 m and the force is 5.17 N, then I would multiple .48mx 5.17 N to get a torque of 2.48 n’m. Then, divide 2.48 n’m /.237 n’m = 10.4.
A3. Newton Meters are as you have described, Newtons multiplied by Meters. If your robot arm is 0.48m long and their is a 5.17 N force perpendicularly to the arm at applied at the end, then your math is correct. If you would like to do advanced motor calculations fairly easily other people have already done most of the work for you. This link is to the JVN Mechanical Design Calculator which I recommend you download to keep handy.