Gear Reducer Sizing: 5 Steps With a Worked Conveyor Example

Gear reducer sizing is arithmetic, not judgement — but it is arithmetic done in a specific order, and skipping a step is what produces drives that stall, run hot, or fail inside a year. This article sets out a five-step method for sizing a reducer, then runs a complete worked example on an inclined belt conveyor so you can see every number appear. If you already know the theory and only want the sequence, start with the step table and then check your own figures against the example at the end.

The Five Steps, in Order

Step 1 and Step 2 are pure geometry and mechanics. Step 3 is where experience enters, because it encodes what the machine does that the drawing does not show. Step 4 and Step 5 are catalogue work plus checks — and Step 5 is the one most selections skip.

Step 1: Convert the Machine’s Speed Requirement into Shaft Speed

Machines are usually specified by a process speed — belt speed in metres per second, mixer tip speed, door travel time — while reducers are specified by output r/min. The conversion for a pulley-driven machine is:

n = 60 × v ÷ (π × D)

with v in m/s and D, the effective pulley or drum diameter including lagging, in metres. Use the effective diameter: a 200 mm drum with 5 mm of lagging is 210 mm, which changes the answer by 5% and pushes the belt speed out of tolerance on a machine with a tight process window. For screw conveyors, use the screw’s actual handling requirement rather than a belt-speed analogy; for anything driven through a chain or timing belt, include the sprocket ratio at this step so that the reducer sees the true output speed.

Step 2: Calculate Running Torque, Not Running Power

Two routes reach the same number. If you can estimate the forces, work from force and radius:

T = F × r

where F is the effective tangential force at the pulley in newtons and r is the pulley radius in metres. If you know the power the machine absorbs and the speed, work from power:

T = 9550 × P ÷ n

The force route is better for conveyors, lifts and anything with a gravity component, because it forces you to account for incline and friction separately. For a horizontal or inclined belt conveyor, the steady-state resisting force is approximately:

F = m × g × (sin θ + μ × cos θ)

where m is the total mass being moved, including belt, and μ is the friction coefficient of the belt against its support — in the region of 0.05–0.10 for a slider bed, lower for roller-supported belts. Add the force needed to accelerate the belt and product from rest, and add any scraper, plough or transition drag. Treat the result as a screening figure and confirm with the machine designer where safety or product damage is at stake.

Step 3: Apply a Service Factor That Matches the Duty

The running torque is the load on a good day with a warm machine running steadily. The service factor covers everything else: frequent starts, jams, reversing, dusty conditions and long daily hours. For electric-motor drives the practical bands are:

• 1.0–1.25 — uniform duty, centrifugal pumps, light fans, up to about 10 hours a day.
• 1.25–1.5 — light shock, loaded conveyors, machine tools, most packaging machinery.
• 1.5–2.0 — moderate shock, screw conveyors, mixers, indexers, 24-hour operation.
• 1.75–2.5+ — heavy shock, crushers, reversing drives, kilns.

Adjust upward for direct-on-line starting under load, and upward again where the machine can jam — a jam stalls the motor at locked-rotor current, and the reducer sees the full motor torque multiplied by the ratio.

Step 4: Choose the Ratio and the Frame

The nominal ratio is motor speed divided by output speed, using the motor’s rated speed — 1400–1450 r/min for a four-pole motor at 50 Hz, 1700–1750 at 60 Hz — not its synchronous speed. Then pick the nearest standard ratio and recalculate the true output speed, because catalogues rarely offer the exact number you wanted:

n_actual = n_motor ÷ i_selected

Always resolve this before checking torque, since the available torque depends on the speed you actually get. Then look up the frame whose rated output torque at that speed meets or exceeds the selection torque from Step 3. If the calculation suggests a power rating, remember that the motor power follows from torque and speed, not the other way round: the same reducer output can be reached by a small fast motor with a high ratio or a larger motor with a lower ratio, and the two options differ in cost, efficiency and starting behaviour.

Step 5: Verify the Three Limits Torque Does Not Cover

• Thermal rating. The heat the housing can shed continuously is often lower than the mechanical rating, especially at low output speed and high ambient. If the unit runs near its thermal limit, expect oil degradation and seal leaks rather than broken teeth.
• Starting and breakaway torque. An AC induction motor delivers roughly two to two and a half times rated torque at standstill. If the machine needs more than that to break away — a loaded incline, a settled product, a cold lubricant — the drive will not start even though it runs fine once moving.
• Overhung load. Any pulley or sprocket on the output shaft adds a radial load with its own rating. A generous torque margin does not compensate for a shaft that is being bent by an over-tensioned belt.

Worked Example: Inclined Belt Conveyor

Given: belt speed 0.5 m/s, head pulley 200 mm effective diameter, total moving mass 250 kg including belt and product, incline 10°, slider-bed friction coefficient 0.08, four-pole motor at 1400 r/min on a 50 Hz supply, two-stage helical reducer, 16 hours a day with loaded starts.

Reading the result. The power calculation alone says 310 W at the pulley, which rounds to a 0.4 kW motor once gearbox efficiency of about 0.92 is included. But the 0.4 kW frame delivers only about 82 N·m at 46.7 r/min, an effective service factor of 1.32 against the 1.4 required. The honest options are to accept the 0.4 kW frame where the conveyor cannot jam and a torque limiter is fitted, or to move to the 0.75 kW frame and carry a real margin. Starting torque is adequate in both cases: 620 N translates to 62 N·m at the pulley, and an AC motor delivers well over that at standstill.

This is the most common outcome of careful sizing — the torque check, not the power check, decides the frame. It also explains why two suppliers can quote very different prices for the “same” conveyor: one sized on running power, the other on selection torque.

Sizing Mistakes That Repeat

• Using motor synchronous speed instead of rated speed, which throws the ratio out by 3–4%.
• Forgetting that a jam stalls the motor: locked-rotor torque multiplied by the ratio is the worst case the reducer sees.
• Sizing on power and buying on price, without ever comparing selection torque to rated torque at the actual speed.
• Ignoring the incline’s gravity component because the conveyor “looks flat”.
• Skipping the thermal check on slow, high-ratio, high-ambient drives.

When the numbers are in hand, the remaining work is matching them to a catalogue: how to calculate gear ratio covers the ratio arithmetic in more depth, and the gear motor size chart gives frame dimensions and torque ratings for standard sizes so the selected frame can be checked against the available space.

Conclusion

Gear reducer sizing comes down to one sequence: speed, torque, service factor, ratio, verification. Do it in that order and the frame size falls out of the arithmetic instead of being guessed and justified afterwards. If you have the belt speed, pulley diameter, moving mass and incline, the AC gear motor and reducer range can be checked against your selection torque at the calculated output speed.

FAQ

What is the difference between nominal torque and selection torque?

Nominal torque is what the machine needs in steady running. Selection torque is that figure multiplied by a service factor for starts, shock and duty hours, and it is the number the reducer’s rated torque must exceed.

Do I size the reducer on motor power or on load torque?

On load torque. Motor power is an output of the calculation, not an input: the drive must deliver the required torque at the required speed, and power follows from those two numbers.

How much margin should I leave for acceleration?

Add the torque needed to accelerate belt and product from rest to full speed within the desired ramp time, then apply the service factor to the total. For most conveyors the acceleration term is modest; for high-cycle indexers it dominates and should be modelled properly.

Why does a higher output speed reduce torque?

Because torque equals power divided by angular speed. At fixed motor power, doubling output speed halves the available torque, which is why speeding a machine up by changing a pulley so often ends in a stalled drive.

Can a reducer be too large?

It can. An oversized unit costs more, takes more space, runs at lower efficiency and, for worm gearboxes, may lose the self-locking behaviour that was relied on. Size to the selection torque rather than to instinct.