NSWEngineering StudiesSyllabus dot point

Engineering mechanics: How do gear ratios in a vehicle transmission convert engine torque and speed to wheel torque and speed?

Calculate gear ratios in single-pair and compound gear trains, relate input and output speeds and torques, and explain the role of transmission ratios in matching engine output to road conditions

A focused answer to the HSC Engineering Studies Personal and Public Transport dot point on gearing. Single and compound gear ratios, speed and torque relationships, the role of first gear in launch and top gear in cruise, 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 calculate gear ratios in single-pair and compound (series) gear trains, apply the relationships between input and output speed and torque, and explain why a vehicle uses different transmission ratios in different driving conditions.

The answer

Single-pair gear ratio

For a gear pair with a driver gear and a driven gear:

GR = \frac{N_{\text{driven}}}{N_{\text{driver}}}

where $N$ is the number of teeth. Equivalently, $GR$ is the ratio of driver speed to driven speed and the inverse of the diameter ratio.

Speed and torque relations

For an ideal (lossless) gear pair:

\omega_{\text{driven}} = \frac{\omega_{\text{driver}}}{GR} \qquad T_{\text{driven}} = T_{\text{driver}} \times GR

Power is conserved:

P = T_{\text{driver}} \, \omega_{\text{driver}} = T_{\text{driven}} \, \omega_{\text{driven}}

A higher gear ratio means slower output but higher torque. A lower gear ratio (overdrive, with $GR < 1$ ) means faster output but lower torque.

Compound (series) gear trains

When several gear pairs are connected in series (engine, gearbox first stage, gearbox final stage, final drive), the overall ratio is the product of the individual ratios:

GR_{\text{total}} = GR_1 \times GR_2 \times GR_3 \times \dots

A typical six-speed manual gearbox in an Australian family car has gear ratios approximately:

Gear	Ratio
1st	3.5
2nd	2.0
3rd	1.4
4th	1.0
5th	0.85
6th	0.65
Final drive	4.1

The overall reduction in first gear is about $3.5 \times 4.1 = 14.4$ . The engine spins about 14 times for each wheel revolution, multiplying torque by the same factor for hill starts and acceleration.

Why multiple gears are needed

The internal combustion engine produces useful torque only over a narrow speed band (typically 2000 to 5500 rpm for a petrol engine). The wheels need to turn anywhere from zero (at start) to about 1500 rpm (at 130 km/h on standard tyres). The transmission provides the variable reduction so the engine stays in its power band across all road speeds.

In modern vehicles, continuously variable transmissions (CVTs) use a belt and tapered pulleys to vary the ratio continuously. Dual-clutch transmissions use two separate clutches to pre-engage the next gear, reducing shift lag. Electric vehicles typically use a single-speed reduction gear because electric motors produce wide-band torque from zero rpm.

Worked example

A bicycle has 50 teeth on the front chainring and 13 teeth on the rear sprocket. The rider pedals at 80 rpm. Find the rear wheel rotational speed.

$GR = 50 / 13 = 3.85$ (driver $N$ on the chain side is the chainring).

Wait, this is the opposite convention. For a chain drive, the chainring (front, 50 teeth) is the driver and the sprocket (rear, 13 teeth) is the driven. With our definition $GR = N_{\text{driven}} / N_{\text{driver}} = 13 / 50 = 0.26$ .

Rear wheel speed: $\omega_{\text{out}} = 80 / 0.26 = 308$ rpm. At a tyre rolling circumference of 2.1 m, this gives a speed of $308 \times 2.1 / 60 = 10.8$ m/s, or 39 km/h. A reasonable bike top gear.

Past exam questions, worked

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

2020 HSC style4 marksA car's first gear has a pair where the driver gear has 12 teeth and the driven gear has 36 teeth. The final drive has a pinion of 11 teeth and a crown gear of 44 teeth. If the engine produces 200 N m at 3000 rpm, calculate the torque and rotational speed at the driven wheel hub.

Show worked answer →

Gear ratio per pair is the number of teeth on the driven over the driver.

First gear ratio: $GR_1 = 36/12 = 3.0$ .

Final drive ratio: $GR_2 = 44/11 = 4.0$ .

Overall ratio is the product of the two pairs:

GR_{\text{total}} = 3.0 \times 4.0 = 12.0

Output speed (rpm at the wheel hub):

\omega_{\text{out}} = \frac{\omega_{\text{in}}}{GR_{\text{total}}} = \frac{3000}{12} = 250 \text{ rpm}

Output torque (assuming 100 percent efficiency):

T_{\text{out}} = T_{\text{in}} \times GR_{\text{total}} = 200 \times 12 = 2400 \text{ N m}

The transmission multiplies torque by the ratio and divides speed by the ratio. Power ( $P = T \omega$ ) is conserved in an ideal gear train. In practice, 10 to 15 percent of power is lost to friction and oil churning, so output torque is slightly less than this calculation gives.

Markers reward (1) the ratio formula $N_{\text{driven}} / N_{\text{driver}}$ , (2) multiplication of ratios in series, (3) speed division and torque multiplication, and (4) consistent units (rpm and N m).