CIM quiz

[theprgramerdude]

Well, lets see.... Assuming the motor is mostly linear in performance, The No-Load Current is 2.7 amps, the constant (lets use this) D determining the ratio of current to torque is (64-0)/(27-2.7)= 2.633 oz-in per amp, the no-load 12v speed is 5310 rpm, with it losing (5310-4320)/(64) rpm per oz-in, or 5310-4320/27-2.7 per amp.
Oh, and 5310 RPM per 12 volts equates to 442.5 RPM's per volt.
Based on this:
5310 RPM no load
2.7 amps no load
2.633 oz-in per amp of torque
-15.46 rpm per oz-in torque, or 40.74 rpm per amp
442.5 RPM per volt, no load.

So, lets answer number two first. In order for there to be 100 oz-in of torque, the current applied (which in turn causes the force) must be the same, and I'm hoping this math was done correct before, 40.64 amps.


Which in turn brings us back to #1. The voltage loss across a .1 ohm resistor at 40.64 amps would be equal to V=IR, or .1*40.64= 4.064 volts (can't believe I used my TI-89 on that.)
12-4.064 volts = 7.936 volts left for the CIM motor.
A no load speed would be 3511.68 rpm. Subtracting the torque losses would result in an actual speed of -15.46 * 100 + 3511.68 = 1965.68 RPM.


Number 3: Power input to the CIM = VI= 7.936 V * 40.64 amps = 322.5 watts.

Numbah 4: Power output, mechanical: Speed * Force. 1966 RPM * 100 oz-in ~ using a random converter = 100 oz *.278N per oz* (32.76 revs * 2 pi inches * .0254m/in) per sec = OMG = 145.3 ? Watts.

Number 5: I believe I had to answer the second question first. You don't have to, but it eliminates a few extra steps of variable making.