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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 Opus 4.86 min answerUpdated 2026-07-03

Reviewed by: AI editorial process; not yet individually human-reviewed

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In plain English

Picture a bicycle with a big front chainring and a tiny rear sprocket. Pedal once and the rear wheel spins several times fast, but pedalling gets easy because you are trading speed for effort. Now flip it around: a small front gear driving a huge rear gear makes the wheel turn slower, but with much more twisting force, which is exactly what a crane or hoist gearbox does between the motor and the winch drum, just with several gear pairs stacked one after another so the effect multiplies stage by stage. A worm gear is the odd one out: it is like a screw meshing with a gear, which gives a huge speed reduction in one go, but has an extra party trick, once you stop turning the screw, the gear cannot spin it backwards on its own. That is why warehouse hoists and theatre curtain winches use worm gearboxes, the load just sits there held by friction the moment you let go of the motor, though a proper safety brake is always fitted as well, because "usually holds itself" is not the same as "certified safe".

Note: Lifting Devices is a Preliminary (Year 11) module of NSW Engineering Studies, not an HSC module. The four HSC modules are Civil Structures, Personal and Public Transport, Aeronautical Engineering, and Telecommunications Engineering. This page is kept as Preliminary reference; it is not assessed in the HSC Engineering Studies exam.

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.

Worked example

A goods hoist lifts a 5000 N load at 0.5 m/s on a 0.15 m radius drum. Overall gear efficiency 75 percent. Motor speed 1450 rpm.

Drum torque: $T = 5000 \times 0.15 = 750$ N m.
Drum speed: $\omega = 0.5 / 0.15 = 3.33$ rad/s = 31.8 rpm.
Required GR: $GR = 1450 / 31.8 = 45.6$ .
Motor torque: $T_{\text{motor}} = 750 / (45.6 \times 0.75) = 22$ N m.
Motor power: $P = 22 \times (1450 \times 2 \pi / 60) = 3.34$ kW. A 4 kW motor with margin.

Marker's note: a full-marks answer works through the sequence load and speed to drum torque and drum speed, then to required ratio, then to motor torque, THEN divides by efficiency (not multiplies), before finally sizing the motor power with a sensible margin above the calculated value.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

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.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksA three-stage gear train has individual ratios of 3:1, 4:1 and 2:1. Calculate the overall gear ratio.

Show worked solution →

Series ratios multiply:

GR_{\text{total}} = 3 \times 4 \times 2 = 24

Marking criteria: 1 mark for recognising ratios in series multiply, 1 mark for the correct value (24:1).

foundation3 marksA worm gear pair has a lead angle of 4 degrees and a friction angle of 6 degrees between the worm and gear wheel materials. State, with a reason, whether this pair is self-locking.

Show worked solution →

Self-locking occurs when the friction angle exceeds the lead angle.

Here, friction angle (6 degrees) is greater than the lead angle (4 degrees), so this pair is self-locking: the worm can drive the gear, but the gear cannot back-drive the worm, and the output is held in place with power removed.

Marking criteria: 1 mark for stating the self-locking condition (friction angle > lead angle), 1 mark for comparing the two given angles correctly, 1 mark for the correct conclusion (self-locking) with consequence stated.

core4 marksThe table below gives typical single-stage efficiency ranges for four gear types. | 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 | A hoist designer needs a single gear stage that both gives a 60:1 reduction AND holds the load in place if power is lost, without a separate clutch. (a) Identify which gear type from the table must be used, with a reason. (b) Estimate the overall efficiency lost as heat if 3 kW of input power passes through this stage at the lower end of its efficiency range.

Show worked solution →

(a) Only the worm gear can deliver a 60:1 ratio in a single stage (its typical single-stage range is 40:1 to 100:1) and it is the only gear type in the table with the self-locking property, so it is the only option that meets both requirements without an added clutch.

(b) At the lower efficiency of 0.40:

P_{\text{out}} = \eta \times P_{\text{in}} = 0.40 \times 3\ \text{kW} = 1.2\ \text{kW}

P_{\text{lost as heat}} = 3 - 1.2 = 1.8\ \text{kW}

Marking criteria: 1 mark for correctly identifying worm gear from the table, 1 mark for linking the choice to BOTH the ratio and the self-locking requirement, 1 mark for correct output power calculation, 1 mark for correct heat loss (1.8 kW).

core4 marksA hoist drum of radius 0.18 m must lift a load at 0.40 m/s. The gearbox has an overall ratio of 30:1 and an overall efficiency of 80 percent. Calculate the required motor speed in rpm and the required motor torque if the load is 6000 N.

Show worked solution →

Drum angular speed.

\omega_{\text{drum}} = v/r = 0.40/0.18 = 2.22\ \text{rad/s}

Convert to rpm: $2.22 \times 60/(2\pi) = 21.2$ rpm.

Motor speed.

\omega_{\text{motor}} = \omega_{\text{drum}} \times GR = 21.2 \times 30 = 637\ \text{rpm}

Drum torque.

T_{\text{drum}} = F \times r = 6000 \times 0.18 = 1080\ \text{N m}

Motor torque.

T_{\text{motor}} = \frac{T_{\text{drum}}}{GR \times \eta} = \frac{1080}{30 \times 0.80} = 45\ \text{N m}

Marking criteria: 1 mark for correct drum angular speed, 1 mark for correct motor speed (637 rpm), 1 mark for correct drum torque (1080 N m), 1 mark for correct motor torque (45 N m) including efficiency in the denominator.

exam6 marksA construction hoist for personnel currently uses a worm gearbox for self-locking, backed up by a separate mechanical brake as required by AS1418. A manufacturer proposes replacing the worm gearbox with a higher-efficiency planetary gearbox to reduce running costs, retaining the mechanical brake. Assess this proposal.

Show worked solution →

This is a 6-mark ASSESS: markers reward a judgement weighing efficiency gains against the loss of self-locking, not a one-sided description.

Band 6 plan.

Efficiency case for the change: a planetary gearbox typically runs at 95 to 98 percent efficiency per stage compared with 40 to 85 percent for a worm stage, so replacing the worm gearbox would substantially cut electrical running cost and heat generation over the hoist's operating life.
What is lost: the worm gearbox's self-locking property (friction angle exceeding lead angle) means the load is held in place purely by gear geometry the instant motor torque is removed. A planetary gearbox has no equivalent geometric self-locking; if the mechanical brake fails or is slow to engage, an unlocked planetary drive train could allow the personnel platform to fall under load.
Mitigation already present: because AS1418 already mandates an independent mechanical brake regardless of gearbox type, the proposal does not remove the certified safety device, only the "backup" self-locking behaviour that sat behind it.
Risk judgement: for a load-bearing safety application carrying personnel, removing a passive (always-on, no power required) self-locking mechanism in favour of relying solely on an active brake changes the failure mode from "requires two independent failures" (brake failure AND load slip) to effectively "requires one failure" (brake failure), even though the brake is certified.
Judgement: the efficiency saving is real but should not be adopted without a documented risk assessment and possibly retaining a secondary passive restraint (e.g. a fail-safe spring-applied brake), because for personnel-carrying lifting equipment, removing a passive self-locking layer is a safety regression that outweighs a running-cost saving unless compensated.

Marker's note: top-band answers (1) quantify or at least compare the efficiency difference, (2) correctly explain WHY worm self-locking is a passive/geometric protection while planetary gearboxes lack it, (3) recognise the brake is unchanged but the "layers of protection" have reduced, and (4) reach an explicit, safety-led judgement rather than simply listing pros and cons.

exam5 marksA hoist motor is rated at 40 N m continuous torque at 1450 rpm. It drives a drum of radius 0.25 m through a two-stage gear train: stage 1 is 5:1 helical (efficiency 0.97), stage 2 is 6:1 spur (efficiency 0.98). Calculate the maximum load the hoist can lift and the resulting lifting speed.

Show worked solution →

Overall ratio and efficiency.

GR_{\text{total}} = 5 \times 6 = 30 \qquad \eta_{\text{total}} = 0.97 \times 0.98 = 0.951

Drum torque.

T_{\text{drum}} = T_{\text{motor}} \times GR_{\text{total}} \times \eta_{\text{total}} = 40 \times 30 \times 0.951 = 1141\ \text{N m}

Maximum load.

F_{\text{load}} = \frac{T_{\text{drum}}}{r} = \frac{1141}{0.25} = 4565\ \text{N}

Drum speed.

\omega_{\text{drum}} = \frac{\omega_{\text{motor}}}{GR_{\text{total}}} = \frac{1450}{30} = 48.3\ \text{rpm} = 5.06\ \text{rad/s}

Lifting speed.

v = \omega_{\text{drum}} \times r = 5.06 \times 0.25 = 1.27\ \text{m/s}

Marking criteria: 1 mark for correct combined ratio and efficiency, 1 mark for correct drum torque, 1 mark for correct maximum load (4565 N), 1 mark for correct drum speed conversion, 1 mark for correct final lifting speed (1.27 m/s).