Gear ratios and transmission: HSC Engineering Studies Personal and Public Transport

Syllabus dot point

“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”

6 min readUpdated 2 Jun 2026

Quick answer

Gear ratio is the number of teeth on the driven gear divided by the number on the driver. In series, ratios multiply. Higher ratios trade speed for torque (used in first gear); lower ratios trade torque for speed (used at cruise). The engine sits in a narrow power band, so multiple ratios match its output to road speeds from zero to top speed.

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

A gearbox is a trade machine: it swaps speed for turning force (torque), or turning force for speed, but it can never hand you more of both at once, because power in equals power out (minus a bit of friction). Picture pedalling a bike up a steep hill in a low gear: your legs spin fast but the wheel barely turns for each pedal stroke, and that "barely turns" is exactly why the wheel gets so much extra shove. First gear in a car does the same job for launching from a standstill and climbing hills. Once you are cruising, you want the opposite trade: the engine ticking over slowly and economically while the wheels spin much faster, which is what top gear (overdrive) does. A compound gear train, like a car's gearbox plus its final drive, is just several of these trades chained one after another, and the total trade is found by multiplying each individual trade together.

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

Gear ratio, speed and torque

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

Power is conserved in an ideal gear pair: $P = T_{\text{driver}}\,\omega_{\text{driver}} = T_{\text{driven}}\,\omega_{\text{driven}}$ .
Ratios in series MULTIPLY: $GR_{\text{total}} = GR_1 \times GR_2 \times \dots$ , never add.
$GR > 1$ trades speed for torque (used for launch/hills); $GR < 1$ (overdrive) trades torque for speed (used for economical cruising).

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.

Power conservation and drivetrain losses

A gear train cannot create power; in the ideal case input power equals output power, $T_{\text{in}} \omega_{\text{in}} = T_{\text{out}} \omega_{\text{out}}$ . This is why torque is multiplied by exactly the same factor that speed is divided by. In a real transmission, friction at the gear teeth and bearings, plus oil churning (windage), dissipate roughly 5 to 15 percent of the power as heat, so the delivered output torque is slightly below the ideal figure. The efficiency falls further when many gear pairs are in series, because each pair contributes its own loss; the overall efficiency is the product of the individual stage efficiencies. Examiners often ask you to recompute an output torque after applying an efficiency factor, so always check whether a question states a percentage to apply.

Matching the engine to road conditions

The transmission exists because the engine and the road have different needs. The engine produces useful torque only over a narrow band of speed, yet the wheels must turn from zero at a standstill to high speed at cruise. A low gear (high numerical ratio) trades road speed for tractive force, which is what is needed to launch the vehicle and climb hills. A high gear (overdrive, ratio below one) trades force for speed, letting the engine idle along economically at cruising speed. Selecting the right ratio for the situation keeps the engine in its efficient power band, minimises fuel use and reduces wear, which is the engineering purpose of having multiple ratios at all.

Reading compound gear-train questions

Work stage by stage. First find EACH pair's ratio using $GR = N_{\text{driven}} / N_{\text{driver}}$ , then MULTIPLY the ratios together (never add) to get the overall ratio. Apply the overall ratio once to speed (divide) and once to torque (multiply), then apply any given efficiency as a final multiplier on torque only, since speed is a kinematic relationship (fixed by the ratio) while torque is what is actually lost to friction. If a question chains several stages with different efficiencies, the OVERALL efficiency is the product of the stage efficiencies, not their average.

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 speed.

Step 1: Identify driver and driven

On a chain drive the chainring (front, 50 teeth) is the driver and the sprocket (rear, 13 teeth) is the driven gear.

Step 2: Calculate the gear ratio

Using $GR = N_{\text{driven}} / N_{\text{driver}}$ gives $GR = 13 / 50 = 0.26$ (an overdrive, output faster than input).

Step 3: Find the rear wheel speed

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

Common traps

Inverting the ratio: $GR = N_{\text{driven}} / N_{\text{driver}}$ when expressed as the speed reduction (input speed over output speed). Some textbooks invert this. Stick to one convention.
Summing ratios in series: Series ratios multiply, they do not add.
Forgetting drivetrain losses: Real gearboxes are 85 to 95 percent efficient. Output torque is slightly less than the ideal calculation.
Confusing overdrive with high gear: Overdrive means $GR < 1$ . Top gear is the lowest ratio in the gearbox. Both speed up the output relative to the input.

Exam-style questions

Questions in the style of NESA 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.

2020 HSC-style4 marks

A 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).

HSC 2021-style3 marks

Explain why a road vehicle requires a low (high-numerical) first gear and a high (overdrive) top gear.

Show worked answer →

First gear uses a high numerical ratio so that engine torque is multiplied heavily, giving the large tractive force needed to launch the vehicle from rest and climb steep gradients, at the cost of low road speed. Top gear (overdrive, ratio below 1 at the gearbox) lets the wheels turn faster than the engine input, so the engine runs at low, economical rpm at cruising speed, reducing fuel consumption and noise. Markers reward linking high ratio to torque for launch/hills and low ratio to economical high-speed cruise.

HSC 2023-style6 marks

A vehicle in top gear has a gearbox ratio of 0.80 and a final drive ratio of 3.7. The engine develops 180 N m at 2500 rpm. The driven wheels have a rolling radius of 0.32 m. Determine the wheel rotational speed, the tractive force available at the wheels (assume 90 percent drivetrain efficiency), and the road speed in km/h.

Show worked answer →

Overall ratio.

GR = 0.80 \times 3.7 = 2.96

Wheel speed.

\omega_{\text{wheel}} = \frac{2500}{2.96} = 845 \text{ rpm}

Wheel torque (with efficiency).

T_{\text{wheel}} = T_{\text{eng}} \times GR \times \eta = 180 \times 2.96 \times 0.90 = 479 \text{ N m}

Tractive force.

F = \frac{T_{\text{wheel}}}{r} = \frac{479}{0.32} = 1497 \text{ N}

Road speed. Wheel angular velocity $\omega = 845 \times 2\pi/60 = 88.5$ rad/s, so

v = \omega r = 88.5 \times 0.32 = 28.3 \text{ m/s} = 28.3 \times 3.6 = 102 \text{ km/h}

Markers reward the product of ratios, the wheel speed, the torque multiplied by ratio and efficiency, the tractive force from torque and radius, and the conversion of wheel speed to road speed in km/h.

Practice questions

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

foundation3 marks

A gear pair has a driver gear with 15 teeth and a driven gear with 45 teeth. The driver rotates at 2400 rpm and delivers 60 N m of torque. Calculate the driven gear's ratio, speed and torque, assuming 100 percent efficiency.

Show worked solution →

Ratio.

GR = \frac{N_{\text{driven}}}{N_{\text{driver}}} = \frac{45}{15} = 3.0

Speed.

\omega_{\text{driven}} = \frac{2400}{3.0} = 800\ \text{rpm}

Torque.

T_{\text{driven}} = 60 \times 3.0 = 180\ \text{N m}

Marking criteria: 1 mark for the correct ratio, 1 mark for correct output speed, 1 mark for correct output torque with correct direction of change (speed down, torque up).

foundation3 marks

A compound gear train has a first pair with driver 14 teeth and driven 42 teeth, followed by a second pair with driver 18 teeth and driven 54 teeth. Calculate the overall gear ratio.

Show worked solution →

First pair.

GR_1 = 42/14 = 3.0

Second pair.

GR_2 = 54/18 = 3.0

Overall ratio.

GR_{\text{total}} = GR_1 \times GR_2 = 3.0 \times 3.0 = 9.0

Marking criteria: 1 mark for each correctly calculated pair ratio, 1 mark for correctly multiplying (not adding) the two ratios to get 9.0.

core4 marks

The table below gives the gearbox ratios for a five-speed manual car with a fixed final drive ratio of 3.9. | Gear | Gearbox ratio | |---|---| | 1st | 3.4 | | 2nd | 1.9 | | 3rd | 1.3 | | 4th | 1.0 | | 5th | 0.78 | The engine develops 170 N m at the rpm used in each gear. (a) Calculate the overall ratio and output torque in 2nd gear. (b) Explain, using the table, why 5th gear is called an overdrive.

Show worked solution →

(a) 2nd gear.

GR_{\text{total}} = 1.9 \times 3.9 = 7.41

T_{\text{out}} = 170 \times 7.41 = 1260\ \text{N m}

(b) Overdrive in 5th gear. The gearbox ratio for 5th gear is 0.78, which is less than 1. A ratio below 1 means the output (propeller shaft) turns FASTER than the input from the engine at that stage, before the final drive multiplies torque back up; combined with the final drive, the overall ratio is $0.78 \times 3.9 = 3.04$ , still greater than 1 overall, but well below the 1st gear ratio of $3.4 \times 3.9 = 13.3$ , showing the engine turns far fewer times per wheel revolution in 5th, which is the definition of an overdrive top gear used to keep engine rpm low and economical at cruising speed.

Marking criteria: 1 mark for the correct overall ratio in 2nd gear, 1 mark for the correct output torque, 1 mark for identifying the gearbox ratio below 1 as the overdrive condition, 1 mark for explaining the effect in terms of reduced engine rpm at cruising speed using the table's numbers.

core5 marks

A vehicle in 3rd gear has a gearbox ratio of 1.35 and a final drive ratio of 3.9. The engine develops 165 N m at 3200 rpm. The driven wheels have a rolling radius of 0.30 m. Determine the wheel rotational speed, the tractive force at the wheels (assume 88 percent drivetrain efficiency), and the road speed in km/h.

Show worked solution →

Overall ratio.

GR = 1.35 \times 3.9 = 5.27

Wheel speed.

\omega_{\text{wheel}} = \frac{3200}{5.27} = 607\ \text{rpm}

Wheel torque (with efficiency).

T_{\text{wheel}} = 165 \times 5.27 \times 0.88 = 765\ \text{N m}

Tractive force.

F = \frac{765}{0.30} = 2550\ \text{N}

Road speed. $\omega = 607 \times 2\pi/60 = 63.6\ \text{rad/s}$

v = 63.6 \times 0.30 = 19.1\ \text{m/s} = 19.1 \times 3.6 = 68.8\ \text{km/h}

Marking criteria: 1 mark for the correct overall ratio, 1 mark for correct wheel speed, 1 mark for correct wheel torque including the efficiency factor, 1 mark for correct tractive force, 1 mark for correct road speed in km/h.

exam6 marks

A drivetrain has three stages in series: gearbox (ratio 2.2, stage efficiency 96 percent), transfer case (ratio 1.4, stage efficiency 97 percent), and final drive (ratio 3.7, stage efficiency 95 percent). The engine delivers 210 N m. Calculate the overall gear ratio, the overall drivetrain efficiency, and the actual output torque delivered to the wheels.

Show worked solution →

Overall ratio.

GR_{\text{total}} = 2.2 \times 1.4 \times 3.7 = 11.4

Overall efficiency (the product of the stage efficiencies, since each stage's loss compounds with the next):

\eta_{\text{total}} = 0.96 \times 0.97 \times 0.95 = 0.885 = 88.5\%

Ideal (lossless) output torque.

T_{\text{ideal}} = 210 \times 11.4 = 2394\ \text{N m}

Actual output torque.

T_{\text{actual}} = 2394 \times 0.885 = 2119\ \text{N m}

Markers reward recognising that overall efficiency is the PRODUCT of the individual stage efficiencies (not their sum or average), since each stage's losses act on what has already passed through the previous stage.

Marking criteria: 1 mark for the correct overall ratio, 1 mark for correctly multiplying (not averaging) the three stage efficiencies, 1 mark for the correct overall efficiency, 1 mark for the correct ideal torque, 1 mark for the correct actual torque, 1 mark for a statement that losses compound through stages in series.

exam7 marks

Assess the suitability of a continuously variable transmission (CVT) compared with a conventional stepped manual gearbox and a dual-clutch transmission (DCT), for (a) a small city hatchback used mostly for stop-start commuting, and (b) a high-performance sports car used mostly for spirited driving on winding roads.

Show worked solution →

This is a 7-mark ASSESS: markers reward a judgement for EACH vehicle type supported by contrasted evidence, not a generic description of all three transmissions.

How each transmission works (brief): A stepped manual/automatic has a fixed set of discrete ratios engaged by synchromesh or clutch packs. A CVT uses a belt (or chain) running on two variable-diameter pulleys, allowing the ratio to vary smoothly and continuously, so the engine can be held at its single most efficient rpm regardless of road speed. A DCT uses two clutches, one holding the current gear and one pre-selecting the next, allowing near-instant, smooth shifts under load.
(a) City hatchback, stop-start commuting: The priority here is fuel economy and smoothness at low, constantly varying speeds. A CVT is well suited: it can keep the small engine at its most efficient rpm continuously through the stop-start cycle, avoiding the "shift hunting" a stepped automatic can show in traffic, and its smoothness matters more than outright response in gentle commuting. A manual would work but adds driver workload in heavy traffic; a DCT's advantage (fast shifts under load) is largely wasted at commuter speeds and DCTs can be less smooth at very low, crawling speeds than a CVT.
(b) Sports car, spirited winding-road driving: The priority here is precise, fast, and predictable torque delivery matched to the driver's inputs and to variable engine loads through corners. A DCT is best suited: its near-instant pre-selected shifts under full throttle preserve momentum and driver control, and distinct gear steps give the driver a predictable, engaging relationship between engine note, rpm and road speed. A CVT is poorly suited here despite its efficiency, because the "rubber-band" feel of a continuously varying ratio (engine revs rising without a corresponding, steppy increase in road speed) reduces the precise feedback a performance driver relies on in fast direction changes, and CVT belts have historically struggled with the very high torque loads of high-performance engines.
Judgement: The best transmission choice depends on priority, not a single "best" technology: CVT best serves efficiency and smoothness in low-load, variable-speed commuting, while DCT best serves fast, predictable, high-torque response in performance driving; a stepped manual sits between the two, offering driver engagement and simplicity but neither the seamless efficiency of a CVT nor the shift speed of a DCT.

Marker's note: top-band answers (1) give a distinct, evidence-based recommendation for EACH scenario rather than one blanket answer, (2) explain the underlying mechanism (belt-pulley versus twin-clutch pre-selection) rather than only naming the transmissions, and (3) explicitly weigh the relevant priority (efficiency/smoothness versus response/precision) for each case.

Practise this

Sources & how we know this

Engineering Studies Stage 6 Syllabus (2011) — NESA
Engineering Studies HSC exam papers — NESA