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GearMotorsUK

Guide · 7 min read

How to size a gear motor: speed, torque, service factor and mounting

· GearMotorsUK technical desk

Size a gear motor from the load, not from the motor: decide the output speed the machine needs, work out the torque at the output shaft, multiply by a service factor for the duty (1.0 for a smooth eight-hour load, up to 1.75 or more for shock loads or 24-hour running), pick the mounting and shaft, and only then read off the motor power. A unit chosen by kW alone is the most common reason gear motors fail early.

Step 1: output speed

Start at the machine. A conveyor belt at 0.5 m/s on a 200 mm drum turns the drum at about 48 rpm (speed divided by drum circumference). A mixer paddle might want 60 rpm, a screw feeder 20 rpm. That output speed, divided into the motor speed (about 1,400 rpm for a four-pole motor on 50 Hz), gives the ratio: 1,400 divided by 48 is roughly 29:1. Gearboxes come in fixed ratios, so you will pick the nearest, for example 30:1, and accept 47 rpm.

Step 2: output torque

Torque is what actually moves the load. For a conveyor: the pull needed on the belt (friction, incline and acceleration) times the drum radius. For a rotating load: the force at the radius it acts on. If you only know the power of the old motor, torque in Nm is 9,550 times kW divided by output rpm: a 1.5 kW unit at 47 rpm delivers about 305 Nm. Write down the running torque and, separately, the starting torque if the load starts under full weight.

Step 3: service factor, where most mistakes live

Every gearbox is rated for a nominal torque at a nominal duty. Real machines are harsher: they start and stop, they run 24 hours, they take shock loads. The service factor multiplies the required torque to bring the rating up to the real conditions. Manufacturers publish their own tables; the shape is always the same.

Load typeUp to 8 hours a day8 to 16 hoursOver 16 hours
Uniform (fans, light conveyors, agitators)1.01.251.5
Moderate shock (belt conveyors with lumps, mixers, packaging)1.251.51.75
Heavy shock (crushers, presses, reversing drives)1.51.752.0 or more

Starts per hour push the factor up again above about ten. If a 305 Nm conveyor runs two shifts with lumpy material, the factor is 1.5 and the gearbox must be rated for about 460 Nm. Skip this step and the unit runs hot, the oil degrades and the gears pit within a year.

Step 4: mounting, shaft and position

  • Solid output shaft with a key, when you fit a coupling, sprocket or pulley.
  • Hollow output shaft, when the unit hangs on the driven shaft with a torque arm (parallel shaft and worm units). Give the bore diameter.
  • Foot mounted (B3) on a base, or flange mounted (B5) straight onto the machine.
  • Mounting position (M1 to M6, or the maker's code): it sets the oil level and the breather, so a unit fitted output-shaft-up with the wrong plug can run dry.
  • Type of gearing: worm for compact right-angle and intermittent duty, inline helical for straight-line continuous duty, bevel helical for right-angle continuous duty, parallel shaft for conveyors.

Step 5: motor power, last

With torque, speed and gearbox efficiency known, the motor power follows: kW equals torque times rpm divided by 9,550, divided by efficiency. A helical unit at 96 percent barely changes the number; a worm unit at 70 percent needs a motor almost half as big again. Then choose poles (four-pole is the default), voltage (400 V three phase, or 230 V single phase for small units), IE3 efficiency, IP55 as a minimum, and whether an inverter will drive it (which affects cooling at low speed).

What to send us for a quote that is right first time: output speed or ratio, output torque or the load description, hours per day and starts per hour, shaft type and size, mounting position, supply voltage, and whether it is inverter driven. A photo of the existing nameplate covers most of it.

Two worked examples

Belt conveyor, 12 m long, 600 mm wide, 0.4 m/s, 200 mm drum, moderate lumps, two shifts. Drum speed 38 rpm, ratio about 37:1, running torque about 180 Nm, service factor 1.5, gearbox rating 270 Nm, motor 0.75 kW four-pole, parallel shaft unit with 40 mm hollow bore and torque arm, IP55, IE3.

Paddle mixer for dry powder, 60 rpm, 24-hour duty, ten starts an hour. Ratio about 23:1, running torque 350 Nm, factor 1.5 for the duty plus an allowance for the starts, rating 560 Nm, motor 3 kW (2.2 kW of shaft work plus gearbox losses leaves no margin on a 2.2 kW motor), inline helical unit with a 35 mm solid shaft and flange mounting. A worm unit would be smaller and cheaper on the day and cost more in electricity every month after.

Questions we get asked

Can I size a gear motor by the old one's kW?

Only if the old one was right and the duty has not changed. Nameplate matching is fine for a like-for-like swap; for a new machine, size from torque and duty.

What service factor should I use if I am not sure?

1.5 covers most industrial duties without oversizing badly. If the load has real shock or reverses, go to 1.75. If it is a fan running eight hours, 1.0 is fine.

Does an inverter change the sizing?

Two ways: at low speed a self-cooled motor loses airflow and may need forced ventilation or derating, and at high speed the gearbox input speed limit must not be exceeded. Tell us the speed range and we account for both.

What is the ratio of a gearbox?

The input speed divided by the output speed. A 1,400 rpm motor through a 20:1 gearbox gives 70 rpm at the output, with torque twenty times higher minus losses.

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