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Selecting a motor for a roller conveyor rests on three questions: at what speed the line will flow, what load the rollers will carry with what friction, and by which route the drive will be transmitted to the rollers. Line speed is calculated from motor speed together with the gearbox ratio, the sprocket ratio and the roller diameter; the power required depends less on the weight of the load than on roller rolling resistance and inertia at start-up. This article explains how to set up this calculation for driven roller conveyors and how to choose the gearbox.
How is power distributed to the rollers on a driven roller conveyor?
On a belt conveyor a single drum pulls the whole belt. On a roller conveyor, however, each roller carries the load above it, and the drive is shared among the rollers by a chain or belt arrangement. The arrangements most often seen on site are:
- Roller-to-roller chain: Each roller has a double sprocket at its end, and neighbouring rollers are linked by short chains. Common for heavy loads, pallet and crate lines. Chain losses accumulate as the number of rollers increases.
- Continuous chain: A single long chain runs beneath the roller sprockets. There are fewer parts, but chain tension needs attention along the line.
- Line shaft and O-ring belts: Each roller is driven by flexible belts from a shaft rotating along the line. Seen on light carton and crate lines; the belts allow a certain amount of slip.
Whatever the arrangement, the motor is usually at one end or in the middle of the line, turning the main sprocket through a gearbox. Long lines are divided into several sections, each with its own motor.
Motorised rollers fitted inside the roller (drum motors or low-voltage motorised rollers) are a separate product group. This task requires a special motor; it may fall outside the scope of the standard series in HEM Motor's 2026 catalogue. A standard three-phase motor is intended for arrangements that drive the rollers from outside by chain or belt.
Calculating line speed
On a roller line the product speed equals the peripheral speed of the rollers; in a belt arrangement, slight slip reduces this speed a little. The calculation chain is: motor rated speed ÷ gearbox ratio = gearbox output speed; this speed is converted to roller speed through the sprocket ratio; roller speed × roller circumference (π × diameter) = line speed.
The values below are a hypothetical example. Let the target line speed be 20 m/min and the roller diameter 60 mm. Since the roller circumference is about 0.188 m, the roller speed must be 20 / 0.188 ≈ 106 rpm. If the gearbox output and roller sprocket have the same number of teeth (1:1), the gearbox ratio required with a 4-pole motor of 1,400 rpm rated speed is 1,400 / 106 ≈ 13.2. The nearest standard ratio on the market is chosen and the actual line speed is recalculated for that ratio.
| Changing element | Effect on line speed |
|---|---|
| Roller diameter increases | At the same speed, line speed increases in proportion to diameter. |
| Gearbox ratio increases | The line slows down and torque at the roller increases. |
| Sprocket on the roller side gets larger | The line slows down; this is an additional reduction after the gearbox. |
| 6-pole motor instead of 4-pole | The line slows by about one third; the ratio must be reselected. |
Load, friction and starting torque
When a roller conveyor carries a load horizontally, the force the motor overcomes is a small fraction of the weight of the products: roller bearing friction, the rolling resistance of the product on the rollers and chain losses. This fraction varies greatly with the base structure of the product. A hard-based plastic crate rolls easily; a soft-based cardboard box or wooden pallet creates more resistance. That is why two lines with the same weight may need different power.
Take the friction coefficient to be used in the power calculation from the roller and chain manufacturer's data; do not invent a value by guesswork. The following information is needed to set up the calculation:
- The total load on the line in the worst case (with the line completely full)
- Number of rollers, roller weight and roller diameter
- Chain arrangement and number of chains (each chain loop adds losses)
- Line speed and the target acceleration time at start-up
At start-up the motor overcomes not only friction but also the inertia of all the rollers and products on the line. On a full, long roller line, roller mass can be close to product mass. For this reason, whether the starting torque is sufficient is checked separately from continuous running power. You can see the motor's rated torque on the product page or in the torque table.
