Sizing a Unit-Load Friction-Drive Conveyor

Not every conveyor carries a bed of bulk material. A huge class of machine-building conveyors moves a single discrete load — a circuit board, a tray, a machined part, a pallet — along a short belt or roller run driven by friction on a pulley. Sizing one of these is a different, simpler problem than the ISO 5048 troughed-belt calculation: there is no material cross-section, trough angle or idler-mass build-up. Instead you work from the load pressing on the belt, the friction that resists its motion, and the pulley the motor drives through a gearbox.

This guide walks that sizing end to end: from the load and friction to the resistance force, from the resistance to the torque at the drive pulley, then to the gearhead output torque that has to cover it with margin, and finally to the belt transfer speed the motor and gear ratio deliver. A worked example ties the steps together, and the free calculator runs the same numbers so you can drop in your own.

Resistance force — what the belt has to overcome

On a level run, the only thing resisting the load's motion is friction between the load (or its carrier) and the belt: the resistance force is the friction coefficient times the load pressing on the belt. Both are things you know — the load is the workpiece plus any carriage weight, and the friction coefficient comes from the belt/part material pair. On an incline the load's own weight adds a component: raising the load adds the full gravity term, so an inclined transfer needs more force than a level one at the same load.

A decline is where care is needed. Going downhill, gravity helps drive the load — but past a certain angle it more than cancels friction and the load starts to drive the belt rather than the other way round. That is an overhauling condition, and a drive motor sized only to push the load will not hold it back: the descent needs a brake. Size the resistance at the real incline of your run, and treat a decline steep enough to overhaul as a braking problem, not a driving one.

Resistance force at the belt

F = μ·W·cos(θ) + W·sin(θ)

where F = tangential force to move the load [N]; μ = friction coefficient (belt–load); W = load pressing on the belt (workpiece + carriage) [N]; θ = incline angle (+ raising, − declining). On the level, F = μ·W. If F goes negative the decline overhauls — a brake is required.

From resistance to drive-pulley torque

The motor drives the belt through a pulley, so the resistance force becomes a torque at the pulley shaft: the force times the pulley radius. Divide by the drive efficiency — the small losses in the pulley bearings and belt — and you have the torque the gearhead must actually deliver at its output. Dividing by efficiency raises the required torque, which is the conservative direction: it never flatters the drive.

This torque is a lower bound in one specific sense: it assumes the drive pulley grips the belt without slipping. A friction drive with a small wrap angle or a low pulley friction can slip, in which case the real drive needs more torque or higher belt tension than this figure. The calculator flags this assumption; if slip is a concern, increase the wrap (a snub pulley) or the pretension so the pulley can transmit the force.

Required torque at the drive pulley

T = |F|·(D/2) / η_drive

where T = torque the gearhead output must supply [N·m]; F = resistance force [N]; D = drive pulley diameter [m]; η_drive = drive efficiency (pulley bearings, belt). Efficiency divides — it raises the required torque (conservative). Assumes no belt/pulley slip.

Gearhead margin and transfer speed

A geared motor delivers, at its output, the motor's rated torque multiplied by the gear ratio and the gearhead efficiency. Compare that available output torque against the required load torque: the ratio is your safety factor, and you want it comfortably above 1 — a target of 1.5 or more, because a friction drive loses reserve to slip and to the extra torque a cold, static belt needs to break away at startup. A margin of exactly 1.0 is a drive that only just moves on a good day.

The same gear ratio that multiplies torque divides speed. The belt transfer speed is the motor speed reduced by the gear ratio, converted to a surface speed through the pulley circumference. That lets you check both things a conveyor has to get right at once: enough torque to move the load, and the right line speed to meet the cycle time. The MechanixCalc conveyor calculator's Unit-Load Drive tab computes the resistance, the pulley torque, the gearhead margin and the transfer speed together, so you can trade pulley size, gear ratio and motor against all of them.

Gearhead output torque, margin and belt speed

T_G = T_motor·G·η_gear SF = T_G / T V = (N / G)·π·D / 60

where T_G = gearhead output torque [N·m]; T_motor = rated motor torque [N·m]; G = gear ratio; η_gear = gearhead efficiency; SF = torque margin (target ≥ 1.5); V = belt transfer speed [mm/s]; N = motor speed [rpm]; D = pulley diameter [mm].

Where to be careful

This is a unit-load model — it sizes a discrete part on a friction-driven pulley, not a bulk material bed. For a trough of sand, ore or grain use the ISO 5048 bulk calculation (the conveyor tool's Belt & Load tab) instead, which builds the load from a material cross-section. Keep the torque margin honest against startup and slip, not just steady running. Confirm the drive pulley actually grips — wrap angle and pretension decide whether the belt slips before the motor stalls. And on any decline, check for overhauling and size a brake or holding device separately; a drive motor is not a brake.

Worked example

A transfer conveyor moves an 80 N tray (part + carriage) along a level belt, friction coefficient 0.25, on a Ø60 mm drive pulley. A geared motor delivers 0.5 N·m at 3000 rpm through a 30:1 gearhead (efficiency 0.7). Check the drive and the belt speed.

Given

Load on belt W80 N
Friction coefficient μ0.25
Drive pulley diameter D60 mm
Drive efficiency η_drive0.9
Motor torque / speed0.5 N·m @ 3000 rpm
Gearhead30:1, η_gear 0.7

Result

Resistance force20 N
Required pulley torque0.667 N·m
Gearhead output torque10.5 N·m
Torque margin SF15.75 (target ≥ 1.5)
Transfer speed314 mm/s

Resistance force (level): F = μ·W = 0.25 × 80 = 20 N.
Required pulley torque: T = F·(D/2)/η_drive = 20 × 0.030 / 0.9 = 0.667 N·m.
Gearhead output torque: T_G = T_motor·G·η_gear = 0.5 × 30 × 0.7 = 10.5 N·m.
Torque margin: SF = T_G / T = 10.5 / 0.667 = 15.75 — far above the 1.5 target, so the drive is amply sized for torque.
Transfer speed: V = (N/G)·π·D/60 = (3000/30) × π × 60 / 60 = 314 mm/s. Confirm that meets the required line speed; if it is too fast, raise the gear ratio (which also raises the torque margin further).

These numbers are produced by the live MechanixCalc unit-load conveyor engine. It is an engineering estimate — no governing standard covers unit-load friction-drive sizing — and the torque assumes the drive pulley grips without slipping. On an incline the raising gravity term is added; a decline that overhauls is failed for a brake.

Do it on your own numbers

Enter the load, friction, pulley and geared motor to get the resistance, pulley torque, gearhead margin and belt speed together. Free 30-minute preview, no sign-up.

Open the Conveyor Calculator — Unit-Load Drive tab

Frequently asked questions

How is this different from a bulk-material conveyor calculation?

A bulk conveyor (ISO 5048 / CEMA) sizes a bed of loose material from its cross-section, density, trough angle and idler masses. A unit-load friction drive moves a single discrete part, so the load is a stated force (part + carriage weight) and the resistance is simply friction times that load. The torque, gearhead margin and belt speed then follow from the pulley and gear ratio. Use the unit-load model for trays, boards and pallets; use the bulk model for sand, ore or grain.

What torque does the motor need?

The gearhead output torque must exceed the load torque at the pulley, which is the resistance force times the pulley radius divided by the drive efficiency. The available output torque is the motor's rated torque times the gear ratio times the gearhead efficiency. Aim for a margin of at least 1.5× so the drive covers startup break-away and any belt slip, not just steady running.

How do I get the belt transfer speed?

Reduce the motor speed by the gear ratio to get the pulley speed, then convert to a surface speed through the pulley circumference: V = (N/G)·π·D/60 in mm/s with N in rpm and D in mm. The same gear ratio that raises torque lowers speed, so torque margin and line speed trade against each other — the calculator shows both at once.

What about a downhill (declining) conveyor?

On a decline gravity helps drive the load, but past a certain angle it overhauls — the load drives the belt and a drive motor alone cannot hold it back. The calculator flags this and fails the drive-only case, reporting the braking torque required. Size a brake or holding device separately; do not rely on motor torque to control an overhauling descent.