Shaft power from torque and speed, or from force and velocity — and the motor power once drive losses are added.
Belts, chains or gearboxes between motor and load.
Example Result
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Power is the rate of doing work. On a rotating shaft it is torque multiplied by angular speed; on anything moving in a straight line it is force multiplied by velocity. Choose rotational or linear mode, enter torque and rpm or force and speed in SI or US units, and the calculator returns the power in watts, kilowatts, mechanical horsepower and metric horsepower (PS). Add the efficiency of each belt, chain or gearbox between the motor and the load to see how much power the motor itself has to deliver and how much is lost on the way. The examples below size a motor for a pump drive, convert a US torque figure, and work a conveyor from belt pull and speed.
Choose rotational (torque and rpm) or linear (force and speed), and SI or Imperial.
Enter the torque or force needed at the load, and its speed.
Optionally add the efficiency of each belt, chain or gearbox stage to get the power the motor must deliver.
Press Calculate. Power is shown in W, kW, mechanical hp and metric hp (PS).
By O.E. — Partner, chairman of the board and CEO of a metal products manufacturer for 21 years; the company exports to 45 countries.
Drafted with AI assistance; calculations checked against primary sources and tested with the site's own code. About FlowUtil
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Rotational: P = T × ω with ω = 2π × n ÷ 60 — in practical units P [kW] = T [N·m] × n [rpm] ÷ 9,549, or P [hp] = T [lb·ft] × n [rpm] ÷ 5,252. Linear: P = F × v. Through a drive, the motor must deliver the load power divided by the efficiency of every stage. 9,549 is 60,000 ÷ 2π; 5,252 is 33,000 ÷ 2π (one horsepower is 33,000 ft·lbf per minute). ω must be in radians per second — using rpm directly overstates the power by a factor of 9.55.
1A pump needs 50 N·m at 1,450 rpm, driven through a V-belt (95%) and a gearbox (97%).
ω = 151.84 rad/s; 50 × 151.84 = 7.59 kW; ÷ (0.95 × 0.97)
Load 7.59 kW (10.2 hp, 10.3 PS) · motor shaft 8.24 kW (11.0 hp) · 0.65 kW lost in the drive
Choose a motor whose rated output is at least 8.24 kW, with a margin for starting torque and overloads; the electrical supply must then cover the motor's own losses.
236 lb·ft at 1,750 rpm.
36 × 1,750 ÷ 5,252
12.0 hp (8.94 kW)
In reverse, a 10 hp motor at 1,750 rpm delivers 10 × 5,252 ÷ 1,750 = 30 lb·ft.
3A belt conveyor needs 2,000 N of pull at 1.5 m/s.
2,000 × 1.5
3.0 kW at the drive pulley
The pull is the sum of friction, lift on an incline and the force to accelerate the load at start-up.
| Unit | Watts |
|---|---|
| Mechanical horsepower (hp, 550 ft·lbf/s) | 745.7 |
| Metric horsepower (PS, CV, ch) | about 735.5 |
| Kilowatt | 1,000 |
7.59 kW is 10.18 hp but 10.32 PS. US and UK motor plates use hp; European engine data often use PS.
At constant power, torque falls as speed rises. A 7.5 kW motor delivers about 49 N·m at 1,450 rpm but about 25 N·m at 2,900 rpm; a gearbox that drops 1,450 rpm to 145 rpm raises the torque tenfold, minus its losses. That is why a high-speed motor with a gearbox is usually smaller and cheaper than a low-speed motor of the same power.
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