Estimating Load/No Load RPM Difference for Gearing

mikegrundvig

10 mW
Joined
Sep 6, 2009
Messages
24
Hi all; I was wondering if there is a rule of thumb to convert the no-load RPM to a real-world loaded RPM for the sake of creating proper gearing. I have several motor options, one brushless and one brushed and I know the theoretical RPM of each but that's always no-load. I want to design a gearing solution to get me to the right speed but that would take knowing the more real-world load RPM. Is there a rule of thumb for this somewhere? I've heard 25% loss passed around at one point but don't know if that's valid or not. Thanks!

-Mike
 
Like everything else, there is no simple answer because it depends on the load. In a thread the other day someone was surprised by the increase in top speed when using two motors, with the same voltage as the single motor had used. The answer was, each motor had less load to haul, so they were able to get closer to their no load speed.

It also depends on the winding and gearing. On my trike with the motor driving the IGH, low gear allows the motor to get very close to unloaded speed, whereas in high gear there is a more nominal difference.
 
Miles said:
Simplified:
The peak efficiency point is where the copper losses (useable torque) are at parity with the parasitic losses. At greater power levels than that, copper losses start to dominate. Copper losses go up as the square of the torque. For a PMDC motor, the efficiency at maximum power out (not maximum continuous) is close to 50%. Below 50% of no load rpm (maximum power out) power drops off but increasing torque mean increasing inefficiency.

The above assumes no current limit, fixed voltage, of course.

Parasitic losses:
Bearing losses are proportional to RPM
Hysteresis losses are proportional to RPM
Eddy current losses are proportional to RPM²
Aerodynamic losses are proportional to RPM³
 
Back
Top