QCAA QCAA 2022-style practice: A belt drive has a driver pulley of diameter 120\ mm rotating at 1500\ rev/min and a driven pulley of diameter 300\ mm. Determine the speed ratio and the rotational speed of the driven pulley, assuming no slip.

A 6 mark determine question rewards the ratio then the output speed. Speed ratio (driven to driver) by diameters: N_drivenN_driver = D_driverD_driven = 120300 = 0.4. Driven speed: N_driven = 0.4 × 1500 = 600\ 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\ rev/min.

§-Syllabus dot point

QLDEngineeringSyllabus dot point

How do belt and chain drives transmit motion between shafts that are not touching?

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

A QCE Engineering Unit 4 answer on belt and chain drives. Covers the speed ratio from pulley or sprocket sizes, direction of rotation, slip, and the trade-offs between flat belts, V-belts, toothed belts and chains, with worked drive arithmetic.

Generated by Claude Opus 4.86 min answerUpdated 2026-06-02

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

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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.

Exam-style practice questions

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

QCAA 20226 marksA 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 $600\ \text{rev/min}$ .

QCAA 20234 marksJustify 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.