Sizing a Conveyor Drive
Short answer
- A conveyor drive is sized by totalling the moving mass of product, belt and hardware, converting it to force with friction and incline, then converting that force to torque and power.
- In the worked example, a 475 kg moving load at 1.5 m/s works out to about 82 N·m and 1.1 kW mechanical, which becomes about 1.7 kW of design power after efficiency and a service factor of 1.4.
- Starting torque, shock loading and accumulation must then be checked against locked-rotor torque, and a VFD or soft-starter is the safer route when the line sees frequent start-stop cycles or product buildup.
When a conveyor stalls or the motor runs hot, the first question is whether the drive was sized correctly for the actual load. The calculation starts with the moving mass, adds friction and any lift, then converts the resulting force to torque, power and finally to a suitable drive package.
How do you quantify a conveyor's total moving load?
The load that the drive must move is the sum of three components:
- Product weight – the mass of material on the belt at any instant. For a continuous belt, use the average linear load (kg/m) multiplied by the belt length that is occupied at steady state.
- Belt or chain weight – belt mass per unit length is supplied by the manufacturer (e.g., 2.5 kg/m for a 200 mm PVC belt). Multiply by the total span that is moving.
- Moving hardware – idlers, rollers, tensioning devices and the drive pulley itself. Their weight is usually small but must be included for long spans or heavy‑duty rollers.
Example: a 30 m conveyor carrying 250 kg of product, a 200 mm belt weighing 2.5 kg/m, and ten 15 kg rollers gives
Product: 250 kg Belt: 30 m × 2.5 kg/m = 75 kg Rollers: 10 × 15 kg = 150 kg Total moving mass = 475 kg
Convert mass to force using g = 9.81 m/s²:
Total force (W) = 475 kg × 9.81 m/s² ≈ 4 660 N
How do friction and incline add to conveyor load?
The drive must overcome both rolling/sliding resistance and any vertical component if the belt is inclined.
2.1 Friction from the support type
Typical coefficients of resistance (μ) are:
- Slider (flat) bed: μ ≈ 0.15–0.20
- Roller (or ball‑bearing) support: μ ≈ 0.02–0.05
Choose the value that matches the installed conveyor components. For a roller‑supported belt, use μ = 0.03 as a conservative figure.
2.2 Incline (lift) contribution
If the conveyor runs up an angle θ, the vertical component adds sin θ to the resistance term. The combined resistance factor becomes (μ + sin θ).
Example: a 5° incline gives sin θ ≈ 0.087. With μ = 0.03, the total factor = 0.117.
2.3 Tangential force required
F_t = W × (μ + sin θ)
F_t = 4 660 N × 0.117 ≈ 545 N
This is the steady‑state tangential force the drive pulley must deliver.
How do you convert conveyor force to torque and power?
The drive pulley radius (r) is a design choice; larger diameters reduce torque but increase belt speed for a given motor rpm. Assume a 0.15 m (6 in) drive pulley.
Torque (T) = F_t × r = 545 N × 0.15 m ≈ 82 N·m
Next, calculate the angular speed (ω) needed for the required belt speed (v). Belt speed is usually specified in m/min. Convert to rad/s:
v = 1.5 m/s (≈ 90 m/min) ω = v / r = 1.5 m/s / 0.15 m = 10 rad/s ≈ 95 rpm
Power (P) = T × ω
P = 82 N·m × 10 rad/s = 820 W ≈ 1.1 kW (mechanical)
Apply drive efficiency (η). A typical gearmotor efficiency is 0.90; a VFD‑fed motor can reach 0.95.
Electrical power = 1.1 kW / 0.90 ≈ 1.2 kW
Finally, include a service factor (SF) to cover start‑up shock, material accumulation, and minor overloads. Industry practice uses SF = 1.25–1.5 for bulk material conveyors.
Design power = 1.2 kW × 1.4 ≈ 1.7 kW
How do you select a conveyor drive package?
With the torque, speed and power numbers in hand, you can match a motor‑gearbox combination.
4.1 Gearmotor versus separate gearbox
A gearmotor integrates the motor and reducer in a single housing, simplifying alignment and reducing footprint. Choose a model whose rated output exceeds the design power (≈1.7 kW) and whose torque curve at 95 rpm meets or exceeds the 82 N·m requirement plus the service factor.
If the available gearmotor speeds do not match the desired belt speed, a dedicated gearbox can be added to step the motor speed down to the required 95 rpm while increasing torque proportionally.
4.2 Fixed‑speed motor or VFD?
Fixed‑speed gearmotors are robust and inexpensive, but they cannot accommodate load variations without excessive heating. A variable frequency drive (VFD) offers:
- Soft start – reduces mechanical shock and peak current.
- Speed control – useful when the process requires variable throughput.
- Energy savings – motor runs only as fast as needed for the load.
When the conveyor runs at a constant speed and the load is predictable, a gearmotor without VFD is acceptable. When the line experiences frequent start‑stop cycles, product buildup, or occasional speed changes, sizing a VFD‑fed motor is the safer route.
4.3 Backstops for inclined sections
On inclines, the belt may be prone to rollback if power is lost. A mechanical backstop (brake drum or spring‑applied clutch) can hold the belt in place. If a VFD is used, the drive can be programmed to hold torque during a power‑off event, but a physical brake is still recommended for safety‑critical installations.
How do starting torque and accumulation affect sizing?
Starting torque is often the limiting factor because the motor must accelerate the belt, rollers and the product mass from zero. Compute the equivalent rotating inertia (J) of the moving parts:
J = Σ (m_i × r_i²) where m_i is the mass of each component and r_i its distance from the drive shaft.
For a simple belt‑driven system, the dominant inertia is the belt itself. Approximate belt inertia as:
J_belt ≈ (m_belt × L²) / 12
where L is the belt length. Add the inertia of idlers (treated as point masses at the belt’s midpoint) and the motor rotor inertia (provided by the motor data sheet).
Required start‑up torque (T_start) can be estimated by:
T_start = (J_total × α) + T_friction
α is the angular acceleration needed to reach operating speed in, say, 2 seconds. T_friction is the same 545 N×r value used for steady state, but multiplied by a shock factor of 1.2–1.5 to cover transient spikes.
If the calculated T_start exceeds the motor’s locked‑rotor torque (LRT), you must either:
- Increase the service factor and select a higher‑rated motor,
- Introduce a soft‑starter or VFD to ramp speed, or
- Reduce the belt length or load per start (e.g., by staging product feed).
Accumulation at the head of the conveyor can raise the effective product weight dramatically. A rule of thumb is to add 20 % of the rated product load to the torque calculation if material can pile up more than one belt length.
What checklist confirms a conveyor drive is sized right?
- Gather accurate data: product density, belt speed, belt length, incline angle, component weights.
- Choose the correct friction coefficient based on the support type (slider vs roller).
- Calculate total moving force: W × (μ + sin θ).
- Determine required torque: F_t × drive‑pulley radius.
- Convert torque and speed to mechanical power, then apply efficiency and a service factor.
- Select a gearmotor or motor‑gearbox pair that meets the torque‑speed point with margin.
- Decide if a VFD is needed for soft start, speed control, or energy efficiency.
- Verify starting torque against locked‑rotor torque; add a soft‑starter or redesign the feed if necessary.
- Provide a mechanical backstop or brake for any inclined sections.
- Document the final selection and keep the calculation sheet for future modifications.
Following this sequence removes guesswork and ensures the conveyor runs reliably under the worst‑case load it may encounter.
If you need help locating a suitable gearmotor, gearbox, or VFD, feel free to reach out for a cross‑reference or quotation.
Worth a look next: What Is a V-Belt? Sizes, Sections and Matching; Sheaves, Pulleys and Bushings: QD vs Taper-Lock; What Is a Gearmotor? Types, Ratings and Selection.
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Request a QuoteFrequently asked questions
How do you calculate the torque a conveyor drive needs?
Multiply the total moving weight by the combined resistance factor, which is the friction coefficient plus the sine of the incline angle, to get the tangential force. Then multiply that force by the drive pulley radius. In the worked example, 4 660 N times 0.117 gives about 545 N, and 545 N times 0.15 m gives roughly 82 N·m of torque.
What friction coefficient should I use for a conveyor belt?
The coefficient depends on the support type. A slider flat bed is about 0.15 to 0.20, while roller or ball-bearing support is about 0.02 to 0.05. For a roller-supported belt, 0.03 is a conservative figure. That value is added to the sine of any incline angle, so a 5° incline contributes about 0.087 to the resistance factor.
Do I need a VFD on a conveyor drive?
Not always. When the conveyor runs at constant speed with a predictable load, a gearmotor without a VFD is acceptable. When the line sees frequent start-stop cycles, product buildup or occasional speed changes, a VFD-fed motor is the safer route. A VFD provides soft start, speed control and energy savings, though inclined sections still warrant a physical brake.
What service factor should be used when sizing a conveyor motor?
Industry practice uses a service factor of 1.25 to 1.5 for bulk material conveyors, covering start-up shock, material accumulation and minor overloads. In the worked example, 1.2 kW of electrical power multiplied by 1.4 gives about 1.7 kW of design power. Typical gearmotor efficiency is 0.90, and a VFD-fed motor can reach 0.95.