Belt and chain drives for QCE Engineering Unit 4

Syllabus dot point

“Calculate the speed ratio of a belt or chain drive from the pulley or sprocket diameters, and compare belt and chain drives for slip, distance and load capacity”

QCEEngineeringUnit 4: Machines and mechanisms6 min readUpdated 2 Jun 2026

Quick answer

A belt or chain wraps around two pulleys or sprockets and transmits rotation between shafts that are separated. The speed ratio is fixed by the diameters (for belts) or tooth counts (for chains): the driven speed is the driver speed multiplied by the ratio of the driver size to the driven size, $N_2 = N_1 \times (D_1/D_2)$ . A small driver and large driven pulley reduces speed and multiplies torque, just like a gear reduction. An open belt keeps both shafts turning the same way; a crossed belt reverses direction. Belts are quiet, cheap and absorb shock but can slip; toothed (timing) belts and chains do not slip and carry more load, with chains best for heavy power and belts best for smooth, quiet running.

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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 belt and chain drives: work out how they change rotational speed between two shafts using the pulley or sprocket sizes, and compare the two against gears and against each other. These drives transmit power across a gap that gears cannot bridge, so they are everywhere from bicycles to washing machines. Calculating their speed ratio and knowing their trade-offs is core mechanisms content.

The answer

How the speed ratio works

When a belt runs around two pulleys without slipping, the linear speed of the belt is the same everywhere. The rim of each pulley moves at that belt speed, so a larger pulley turns more slowly. The rotational speeds are inversely proportional to the diameters:

N_1 D_1 = N_2 D_2 \quad\Rightarrow\quad N_2 = N_1 \times \frac{D_1}{D_2}

where subscript 1 is the driver and 2 the driven. For a chain drive the same relation holds with tooth counts instead of diameters, $N_2 = N_1 \times (T_1/T_2)$ , because the chain meshes tooth by tooth and cannot slip. A small driver and large driven wheel gives a reduction: lower speed, higher torque.

Direction and configuration

In an open-belt drive the two shafts rotate in the same direction. Crossing the belt in a figure-eight reverses the driven shaft's direction. The distance between shafts can be large, which is the main advantage over gears: a belt or chain spans a gap without a train of intermediate wheels.

Slip and the belt family

A flat or V-belt relies on friction, so under heavy load it can slip, losing a little speed and some efficiency. This is sometimes useful as built-in overload protection. The belt family trades grip against cost:

Flat belts: simple and cheap, but most prone to slip.
V-belts: the wedge shape grips the pulley groove far better, the common choice for machinery.
Toothed (timing) belts: teeth mesh with a grooved pulley, so they cannot slip and keep exact timing, used in engines and printers.

Chains versus belts

A chain runs on toothed sprockets and meshes positively, so it never slips and maintains an exact ratio. Chains transmit much higher power and tolerate heat and oil, which is why bicycles, motorcycles and industrial drives use them. Their cost is more noise, the need for lubrication, and wear that stretches the chain over time. Belts are quieter, cheaper, need no lubrication and absorb shock, but slip and have lower load capacity.

Speed and torque of a belt drive

Find the driven speed and torque

A motor pulley of diameter $D_1 = 80\,\text{mm}$ runs at $N_1 = 1440\,\text{rev/min}$ and delivers a torque of $5.0\,\text{N m}$ . It drives, through an open belt with no slip, a pulley of diameter $D_2 = 240\,\text{mm}$ . Find the driven speed, the direction, and the ideal driven torque.

Speed ratio and driven speed:

N_2 = N_1 \times \frac{D_1}{D_2} = 1440 \times \frac{80}{240} = 1440 \times \frac{1}{3} = 480\,\text{rev/min}

The driven pulley turns at $480\,\text{rev/min}$ , in the same direction as the motor because the belt is open.

Driven torque (ideal, power conserved, $\tau \propto 1/N$ ):

\tau_2 = \tau_1 \times \frac{D_2}{D_1} = 5.0 \times \frac{240}{80} = 5.0 \times 3 = 15\,\text{N m}

So the drive is a $3:1$ reduction: the output runs at one third the speed and three times the torque. Check by conserving power: $\tau_1 \omega_1 = 5.0 \times 1440$ and $\tau_2 \omega_2 = 15 \times 480$ ; both equal $7200$ in rev/min units, confirming the result.

Why this matters for machines and mechanisms

Belt and chain drives let a designer place a motor away from the load and still transmit power, often changing the speed in the process. The choice between a belt and a chain, or between either and a gear train, is a real design decision weighing slip, noise, cost, maintenance and load. In the Unit 4 engineered solution, justifying the drive type and calculating its ratio shows you can match a power source to a mechanism.

Common errors

Inverting the diameter ratio: The driven speed multiplies by driver-over-driven diameter. A large driven pulley slows the output, so its bigger diameter must be on the bottom of the speed equation.
Forgetting belt slip in efficiency: A friction belt can slip under load, so the real output speed and power are slightly below the ideal calculation. Chains and toothed belts do not slip.
Assuming a belt reverses direction: An open belt keeps the same direction; only a crossed belt reverses it. Do not confuse this with meshing gears, which always reverse.
Mixing diameters and tooth counts: Use diameters for plain belts and tooth counts for chains and toothed belts; do not mix the two in one ratio.

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.

Practice6 marks

A belt drive has a driver pulley of diameter

120\ \text{mm}

rotating at

1500\ \text{rev/min}

and a driven pulley of diameter

300\ \text{mm}

. Determine the speed ratio and the rotational speed of the driven pulley, assuming no slip.

Show worked answer →

A 6 mark determine question rewards the ratio then the output speed.

Speed ratio (driven to driver) by diameters: $\dfrac{N_{\text{driven}}}{N_{\text{driver}}} = \dfrac{D_{\text{driver}}}{D_{\text{driven}}} = \dfrac{120}{300} = 0.4$ .

Driven speed: $N_{\text{driven}} = 0.4 \times 1500 = 600\ \text{rev/min}$ .

The larger driven pulley turns more slowly, as expected for a speed reduction. Markers reward the inverse relationship between pulley diameter and speed, the ratio of $0.4$ (a $2.5{:}1$ reduction), and the output speed of $600\ \text{rev/min}$ .

Practice4 marks

Justify the choice of a toothed (timing) belt or a chain over a flat belt where the driven shaft must stay exactly synchronised with the driver.

Show worked answer →

A 4 mark justify answer needs the slip argument.

A flat or V-belt transmits drive by friction, so it can slip under load, meaning the speed ratio is not guaranteed and synchronisation is lost. A toothed belt or a chain engages positively (teeth in grooves, or rollers on sprocket teeth), so there is no slip and the driven shaft stays exactly in step with the driver.

Markers reward identifying friction-driven slip as the flaw of flat belts and positive (form) engagement as why toothed belts and chains maintain synchronisation.

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Sources & how we know this

Engineering General senior syllabus 2025 — QCAA