Gears, gear ratios and mechanisms for QCE Engineering Unit 4

Syllabus dot point

“Calculate the gear ratio, output speed and output torque of a simple and a compound gear train, and explain how gears trade rotational speed against torque”

QCEEngineeringUnit 4: Machines and mechanisms6 min readUpdated 29 May 2026

Quick answer

The gear ratio of a meshing pair is the ratio of the driven (output) teeth to the driver (input) teeth, $GR = T_{out}/T_{in}$ . A ratio greater than $1$ is a reduction: it slows the output and multiplies the torque. Output speed is the input speed divided by the gear ratio, $N_{out} = N_{in}/GR$ , and ideal output torque is input torque multiplied by the gear ratio, $\tau_{out} = \tau_{in} \times GR$ . Speed and torque trade against each other because power $P = \tau \omega$ is conserved. Meshing gears rotate in opposite directions; an idler gear reverses direction without changing the overall ratio. Compound trains multiply the ratios of each stage.

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What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

QCAA wants you to analyse gear trains: work out the gear ratio from the tooth counts, find the output speed and output torque, and explain the fundamental trade-off that gears make between rotational speed and turning force. You also need to handle idler gears and compound trains and state the direction of rotation. This sits at the heart of the mechanisms content in Unit 4.

The answer

Gear ratio from tooth counts

When two gears mesh, the number of teeth fixes how their rotations relate. The gear ratio for a single pair is:

GR = \frac{T_{out}}{T_{in}}

where $T_{in}$ is the number of teeth on the driver and $T_{out}$ is the number of teeth on the driven gear. A larger driven gear gives $GR > 1$ , called a gear reduction.

Output speed and torque

Because the teeth mesh at the same linear speed, the rotational speeds are inversely proportional to the tooth counts:

N_{out} = \frac{N_{in}}{GR} = N_{in} \times \frac{T_{in}}{T_{out}}

For an ideal (lossless) gear train, power is conserved, $P = \tau \omega$ , so torque rises as speed falls:

\tau_{out} = \tau_{in} \times GR

This is the core trade-off: a reduction gear slows the output and multiplies its torque by the same factor, while a step-up gear speeds the output and reduces its torque. You cannot increase both speed and torque from a fixed input power.

Direction and idler gears

Two externally meshing gears turn in opposite directions. Placing a third gear between them, an idler, reverses the direction again so the output matches the input direction. Crucially, the idler does not change the overall gear ratio: in the train driver to idler to driven, the idler's tooth count cancels, so only the first and last gears set the ratio.

Compound gear trains

A compound train has two gears fixed on the same shaft, so they share a speed. The overall gear ratio is the product of the individual stage ratios:

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

This lets a small gearbox achieve a large reduction (a hand drill or a clock), far more than a single pair could give in a compact space.

Output speed and torque of a reduction gear

Find the output of a simple gear pair

A motor drives a $20$ -tooth pinion at $1500\,\text{rev/min}$ with an input torque of $2.0\,\text{N m}$ . It meshes with a $60$ -tooth gear. Assuming an ideal train, find the gear ratio, the output speed and the output torque, and state the direction.

Gear ratio:

GR = \frac{T_{out}}{T_{in}} = \frac{60}{20} = 3.0

This is a $3:1$ reduction.

Output speed:

N_{out} = \frac{N_{in}}{GR} = \frac{1500}{3.0} = 500\,\text{rev/min}

Output torque (ideal, power conserved):

\tau_{out} = \tau_{in} \times GR = 2.0 \times 3.0 = 6.0\,\text{N m}

The output shaft turns at $500\,\text{rev/min}$ in the opposite direction to the input, delivering $6.0\,\text{N m}$ . The speed dropped by a factor of three and the torque rose by the same factor, confirming power is conserved: $2.0 \times 1500 = 6.0 \times 500$ .

Why this matters for machine control

Gear trains are how engineers match a power source to a load. An electric motor runs efficiently at high speed and low torque; a wheel, a conveyor or a robot joint usually needs the opposite. The gearbox converts one to the other. Choosing the gear ratio sets the machine's final speed and the force it can deliver, which is a central design decision in the Unit 4 engineered solution.

Common errors

Inverting the gear ratio: Output speed divides by the ratio while output torque multiplies by it. Swapping these makes a reduction gear speed up, which is wrong.
Thinking the idler changes the ratio: An idler gear reverses rotation direction but cancels out of the ratio. Only the driver and the final driven gear set the overall ratio.
Expecting more speed and more torque: A gear train cannot raise both from a fixed input power. One always falls as the other rises.
Ignoring direction: Each external mesh reverses rotation. Forgetting to count the reversals gives the wrong output direction.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

Explain how the gearing system on a bicycle provides a mechanical advantage to a cyclist moving up an incline, and the system's effect on the work done on the pedals. Sketch the mechanical components involved to support your response.

Show worked answer →

Five marks: roughly two for the sketch and three for the explanation linking gear ratio, mechanical advantage and work.

Sketch: the front chainring (driver) connected by a chain to the rear sprocket (driven) on the wheel hub, with the pedal crank on the chainring [marks for a labelled chain drive].

Mechanical advantage going up an incline: the cyclist selects a low gear, a small front chainring driving a large rear sprocket. The gear ratio (driven teeth divided by driver teeth) is greater than 1, a reduction, so the turning force (torque) delivered to the wheel is multiplied. This makes it easier to push the bicycle and rider up the incline against gravity, which is the mechanical advantage.

Effect on work done on the pedals: gears change the trade-off between force and distance, not the total energy. Because torque is multiplied, the wheel turns fewer times for each turn of the pedals, so the cyclist must pedal through more rotations (and a greater distance) to climb the hill. The work done on the pedals equals the work delivered to the wheel (ideally), so the lower gear spreads the same work over more, easier pedal strokes rather than reducing the total work needed.

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Sources & how we know this

Engineering General senior syllabus 2025 — QCAA