EngineerBenchRelating motor power, torque, and RPM
Shaft power is torque multiplied by angular speed. At a given power, a shaft turning more slowly carries more torque. The calculation converts RPM to angular speed so you can find the missing value from power, torque, and speed without mixing revolutions per minute with radians per second.
Shaft power and electrical input are different
Mechanical power is the useful power at the shaft. Electrical input must also cover losses. Enter efficiency as a percentage when estimating input demand; do not assume a motor's electrical rating is the power delivered to the driven machine.
Check the operating point
A calculated torque is not a motor's torque-speed curve. Starting torque, intermittent overload, cooling, and drive limits can govern the selection. The service factor is an entered allowance, not evidence that a particular motor can sustain the duty.
Working between watts and newton-meters
At 1,000 RPM, a shaft transmitting 10 N·m delivers about 1,047 W. The conversion uses two pi radians per revolution and 60 seconds per minute. If the drive efficiency is 80 percent, the electrical input needed for that mechanical operating point is about 1,309 W. Dividing by efficiency increases input demand; multiplying by efficiency would instead calculate useful output from a known input.
Use continuous shaft power for a continuous-duty comparison. A short peak power figure, a drive's electrical input rating, and a motor's mechanical nameplate output describe different things. If the required speed changes while load torque stays constant, required shaft power changes in direct proportion to speed. If required power stays constant, torque changes inversely with speed. Neither relationship proves that a particular motor can operate throughout that range. Match the calculated operating point to the motor and drive data, including the supply, duty cycle, and cooling conditions.
Formula
P = T × 2π × RPM / 60. Electrical power divides mechanical power by efficiency.