NSWEngineering StudiesSyllabus dot point

Engineering mechanics: How are gear trains used to multiply torque in cranes and hoists, and how is the resulting load capacity calculated?

Apply gear ratio and efficiency relationships to multi-stage gear trains in cranes and hoists, calculate motor torque required to lift a given load, and identify the role of worm gears in self-locking lifting applications

A focused answer to the HSC Engineering Studies Lifting Devices dot point on gear trains. Speed and torque in multi-stage gear systems, motor sizing for hoists, worm gears and self-locking, the Port Botany shipping container crane example, and worked HSC-style past exam questions.

Generated by Claude OpusReviewed by Better Tuition Academy6 min answerUpdated 2026-05-18

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What this dot point is asking

NESA wants you to apply gear-ratio and efficiency relationships to multi-stage gear trains in cranes and hoists, size the electric motor for a given lifting capacity and speed, and identify the role of worm gears in providing self-locking behaviour.

The answer

Gear ratio in series

For gear stages in series, the overall ratio is the product:

GR_{\text{total}} = GR_1 \times GR_2 \times \dots

Torque is multiplied; speed is divided:

T_{\text{out}} = T_{\text{in}} \times GR_{\text{total}} \times \eta \qquad \omega_{\text{out}} = \frac{\omega_{\text{in}}}{GR_{\text{total}}}

Each gear stage has its own efficiency. The total efficiency is the product:

\eta_{\text{total}} = \eta_1 \times \eta_2 \times \dots

Typical efficiency per stage:

Gear type	Efficiency per stage
Spur	0.97 to 0.99
Helical	0.96 to 0.98
Bevel	0.94 to 0.97
Worm	0.40 to 0.85 (depending on lead angle)
Chain	0.95 to 0.98

Sizing a motor for a hoist

Given a load $F$ to be lifted at speed $v$ on a drum of radius $r$ :

Force on rope at drum: $F$ .
Required drum torque: $T_{\text{drum}} = F \times r$ .
Required drum speed: $\omega_{\text{drum}} = v / r$ .
Motor torque required: $T_{\text{motor}} = T_{\text{drum}} / (GR \times \eta)$ .
Motor speed required: $\omega_{\text{motor}} = \omega_{\text{drum}} \times GR$ .
Motor power: $P = T_{\text{motor}} \times \omega_{\text{motor}} = F v / \eta$ .

Worm gears and self-locking

A worm gear pair has a screw (worm) meshing with a gear wheel. The lead angle is typically small (under 10 degrees), which gives high gear ratios (40:1 to 100:1 in a single stage) but limits efficiency to about 40 to 70 percent.

Self-locking occurs when the friction angle exceeds the lead angle. In this case, the worm can drive the gear, but the gear cannot back-drive the worm. The output shaft is automatically held in place when the input is removed. Worm gearboxes are standard on warehouse hoists, theatre flies and chair lifts because the load is held in place even if power is lost. The trade-off is that the lower efficiency wastes some of the input energy as heat, and the gearbox usually requires forced lubrication.

Australian context

Ship-to-shore container cranes at Port Botany use multi-stage helical-spur gearboxes between the 200 kW lifting motor and the wire-rope drums to lift 50-tonne containers at 1 to 2 m/s. Tower cranes on Sydney CBD building sites use planetary gearboxes for the hoist and worm gearboxes for the slewing drive. Construction site hoists for personnel use worm gearboxes for self-locking safety; they are also fitted with mechanical brakes as a backup.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC style5 marksA hoist motor produces 50 N m of torque at 1450 rpm. The hoist drives a winch drum of radius 0.20 m through a two-stage gear reduction (first stage 4:1, second stage 5:1) at 85 percent overall efficiency. Calculate the maximum load the hoist can lift in newtons and the lifting speed in metres per second.

Show worked answer →

Overall gear ratio is the product of the two stages.

GR_{\text{total}} = 4 \times 5 = 20

Ideal torque at drum.

T_{\text{drum,ideal}} = T_{\text{motor}} \times GR_{\text{total}} = 50 \times 20 = 1000 \text{ N m}

Actual torque at drum (after efficiency losses).

T_{\text{drum}} = \eta \times T_{\text{drum,ideal}} = 0.85 \times 1000 = 850 \text{ N m}

Maximum load. Force on the rope at the drum:

F_{\text{load}} = \frac{T_{\text{drum}}}{r_{\text{drum}}} = \frac{850}{0.20} = 4250 \text{ N}

Drum speed.

\omega_{\text{drum}} = \frac{\omega_{\text{motor}}}{GR_{\text{total}}} = \frac{1450}{20} = 72.5 \text{ rpm}

Convert to angular velocity in rad/s:

\omega = 72.5 \times \frac{2 \pi}{60} = 7.59 \text{ rad/s}

Lifting speed.

v_{\text{lift}} = \omega \times r_{\text{drum}} = 7.59 \times 0.20 = 1.52 \text{ m/s}

Check power conservation. Input power: $P_{\text{in}} = T \omega = 50 \times (1450 \times 2\pi / 60) = 7.59$ kW. Output power: $F v = 4250 \times 1.52 = 6.46$ kW. Ratio is 0.85, matching the stated efficiency.

Markers reward (1) the product of ratios for series stages, (2) torque multiplication by ratio and reduction by efficiency, (3) the load force from torque and drum radius, and (4) consistent unit conversion when calculating lifting speed.