Hi!
By this, do you mean to find the power supply current, I should divide the motor’s rated power by the supply voltage of the power supply/battery?
Rated power, especially on cheap/hobby motors, is a bit optimistic. But yes, you can get a first-order approximation just by dividing the rated power by supply voltage. So for a 1.5kW motor with a 48V supply, you can budget 1.5kW/48 = 31A, maybe 40-50A to compensate for motor (in)efficiency.
Since my motor is rated for 660W, should I try to find a power supply rated for 1.25 to 1.5 times the rated motor power? If so, would this work?:
Yep, that’s fine. I personally have a lot of vevor kit and it’s generally fine quality (for the money). But why use a battery? Usually just going off a power supply will be easier for testing/setup/etc, e.g. 48V 600W High Peak Power Supply – ODrive Robotics US (600W nominal, can peak to 1.2kW).
Lastly, the spool diameter is 5 in. I’m working with another person on this project (he is a MechE major, I am an EE major), and he hasn’t mentioned any speed that we are aiming for, which makes sense to me, since a resistive machine shouldn’t be fast, it should be a struggle to pull.
Well, if someone does a really fast curl, then the spindle will have to have an appreciable speed to keep up, which changes the drawn power. So it’s good to figure out the max speed needed so you can calculate the power figures.
To elaborate a bit more:
That NEMA34 motor has a speed constant (KV) of 72 RPM/V. The torque constant will be 8.27/KV = 8.27/72 = 0.115 Nm/A. So for a 1Nm output, the motor will need 1/0.115 = 8.7A of current.
The motor has a phase resistance of 0.12 ohm. So given resistive losses = I^2 * R, the motor’s loss at 1Nm output will be 8.7A^2 * 0.12ohm = 9W (we’ll call this 9W figure our loss power).
So from a 48V supply, if you’re at zero speed, to output 1Nm of torque the motor will draw 9W → 9/48 = 0.19A from the power supply.
However, if the motor is spinning at e.g. 2000 RPM while outputting 1Nm, the math changes a bit.
Mechanical power = speed (RPM) * torque (Nm) * pi/30
So here the mechanical power is 2000 * 1 * pi/30 = 209W
Drawn supply power will be mechanical power plus loss power, so 209W + 9W = 218W. So at 2000RPM / 1Nm, it’ll draw 218/48 = 4.5A from a 48V supply.
If we run this motor at its limit of 3000RPM / 2.1Nm, we can do this math again:
mech power = 3000 RPM * 2.1Nm * pi/30 = 659W
motor current = 2.1/0.115 = 18.26A
loss power = 18.26A^2 * 0.12ohm = 40W
total power = 659W + 40W = 699W
supply current draw @ 48V = 699/48 = 14.56A
So something like the power supply I linked should be more than fine for typical use, given the fact it can peak to 1.2kW when needed.