Gearbox Selection for a Conveyor in a Packaging Cell
A Year 11 team is building a small conveyor to move filled boxes in a simulated packaging cell. The conveyor belt is driven by a DC motor through a gearbox and a chain drive. Requirements: belt speed 0.25 m/s, belt pull 40 N at steady state, and the drive drum is 80 mm in diameter. The chosen motor produces 0.30 N·m of torque at 3000 rpm and draws 2.1 A at 12 V under that load. A 10:1 gearbox is available, plus a 20-tooth driving sprocket and a 40-tooth driven sprocket for the chain stage. A teammate argues the 10:1 gearbox alone is enough and the chain stage is unnecessary complexity. The teacher asks the team to check the numbers before agreeing.
1. Determine the required drum speed and torque, then evaluate whether the proposed drivetrain meets the requirement.
Read the worked answer
Required drum speed Belt speed = π × D × n, so n = 0.25 / (π × 0.080) ≈ 0.995 rev/s ≈ 59.7 rpm.
Required drum torque T = F × r = 40 N × 0.040 m = 1.6 N·m.
Available at the drum Motor torque 0.30 N·m × 10 (gearbox) = 3.0 N·m at 300 rpm. Adding the 2:1 chain stage gives 6.0 N·m at 150 rpm - far more torque than needed but at the wrong speed.
Speed check With gearbox only: 3000/10 = 300 rpm at the drum - about five times too fast. With gearbox plus chain: 150 rpm - still about 2.5 times too fast.
Judgement Neither arrangement matches the required 60 rpm. The drivetrain is torque-rich and speed-poor in the wrong direction: it needs more reduction, not less. Adding the chain stage helps but is insufficient. The team should add a further reduction stage or use a higher-ratio gearbox, and note that a 5× speed error would overload the belt and boxes regardless of torque margin. The teammate's claim is wrong on the numbers.
2. The motor draws 2.1 A at 12 V. Comment on the efficiency of the drivetrain and on what the current figure implies for the power supply.
Read the worked answer
Mechanical output at the drum (if correctly geared) P = F × v = 40 N × 0.25 m/s = 10 W.
Electrical input P = V × I = 12 × 2.1 = 25.2 W.
Overall efficiency 10 / 25.2 ≈ 40%.
Interpretation A 40% figure is plausible for a small DC motor plus two reduction stages, but it is a system number, not a component number. Losses sit in motor copper and brush losses, gearbox friction, and chain friction. If the drivetrain is over-speeding the drum (previous task), the motor will actually unload and draw less than 2.1 A - so the 2.1 A figure only applies once gearing is corrected.
Supply implications
- 2.1 A steady, with a higher stall/inrush current at start - the supply and wiring must be rated for the inrush, not the steady figure.
- At 12 V and 2.1 A, a bench supply of 3 A is marginal; 5 A gives headroom.
- The team should measure actual current at the corrected operating point rather than trust the datasheet value, because belt tension and box load change the figure.
Judgement Efficiency is acceptable for a school build but should be reported honestly as a system value with the measurement conditions stated.