37kW 1800rpm motor torque — 196 N·m
A 37kW 1800rpm motor delivers 196 N·m of rated torque. That is 37000 W divided by an angular velocity of 188.5 rad/s, the same torque written as 20 kgf·m or 145 lb·ft. In horsepower, 37 kW is 50.3 PS on the metric definition and 49.6 hp on the mechanical one.
Poles and mains frequency fix the speed
The synchronous speed of an induction motor is not a choice: it is 120 × frequency ÷ poles. The same four-pole motor therefore turns at 1800 rpm on 60 Hz and 1500 rpm on 50 Hz — Korea, the United States and eastern Japan run 60 Hz, while Europe, China and India run 50 Hz. Same power, speeds in a 6 : 5 ratio, so the torques land in a 5 : 6 ratio. In service the rotor lags by the slip, which is why the nameplate reads something like 1746 rpm.
- 60 Hz · 1800 rpm196 N·m
- 50 Hz · 1500 rpm236 N·m
- Rated output
- 37 kW · 37000 W
- Synchronous speed
- 1800 rpm
- Mains frequency
- 60 Hz
- Poles
- 4
- Angular velocity
- 188.5 rad/s
- Torque
- 196 N·m
- Torque in kgf·m
- 20 kgf·m
- Torque in lb·ft
- 145 lb·ft
- Metric horsepower (PS)
- 50.3 PS
- Mechanical horsepower (hp)
- 49.6 hp
- Full-load speed
- 1746 rpm · 202 N·m
Through a gearbox
Put a reduction ratio i in the path and the output shaft turns at 1/i of the speed while the torque grows i-fold. Power does not grow — it shrinks by the efficiency η, so the torque is i × η times the input. The figures here assume η = 0.95, typical of a helical gearbox. A worm drive falls to 0.5–0.8 and delivers far less. More torque never means more energy: the same power is simply traded for slower, heavier turning.
- Gear ratio 3 · 600 rpm559 N·m
- Gear ratio 5 · 360 rpm932 N·m
- Gear ratio 10 · 180 rpm1860 N·m
- Gear ratio 20 · 90 rpm3730 N·m
Three-phase current by voltage
For a three-phase induction motor the current is I = P ÷ (√3 × V × cosφ × η). Rated output is what leaves the shaft, so the electrical input is larger by the efficiency, and only the working share of that counts through the power factor. Supply voltage differs by country, so the same motor draws 1.7 times as much at 220 V as at 380 V. The power factor and efficiency used here are representative values by frame size; the nameplate carries the real ones and should be read first.
- 220 V · cosφ 0.87 · η 0.92121 A
- 380 V · cosφ 0.87 · η 0.9270.2 A
- 460 V · cosφ 0.87 · η 0.9258 A
Nearby cells
Same output, other speeds
Same speed, other outputs
Worth knowing
- Torque (N·m) is power (W) divided by angular velocity (rad/s), and that velocity is 2π × rpm ÷ 60.
- The 9550 in T = 9550 × P(kW) ÷ n(rpm) is 60000 ÷ 2π rounded up.
- At constant power, doubling the speed halves the torque — they are inversely proportional.
- Metric horsepower (PS) is 735.49875 W and mechanical horsepower (hp) is 745.699872 W: different numbers.
Related tables
Common questions
Q. How much torque does a 37kW 1800rpm motor make?
196 N·m — 37000 W divided by an angular velocity of 188.5 rad/s. The same figure is 20 kgf·m or 145 lb·ft.
Q. What happens through a 10:1 gearbox?
The speed drops to 180 rpm and the torque rises to 1860 N·m. It is not a clean 10-fold gain because the gearbox efficiency of 0.95 is applied as well.
Q. How much three-phase current does it draw?
On 60 Hz: 121 A at 220 V, 70.2 A at 380 V, 58 A at 460 V. This assumes a representative power factor of 0.87 and efficiency of 0.92, so use the nameplate figures when you have them.