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How do gears change the speed and turning force of a rotating drive?

Gear systems and the gear ratio, calculating output speed from the gear ratio, and how gearing trades speed for torque.

An SQA National 5 Engineering Science answer on gear systems, covering the gear ratio as the ratio of driven to driver teeth, calculating output speed, how a gear train trades rotational speed for turning force (torque), and the direction reversal between meshing gears.

Generated by Claude Opus 4.89 min answer

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

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  1. What this key area is asking
  2. Gears and gear trains
  3. The gear ratio
  4. Calculating output speed
  5. Direction and trading speed for force
  6. Why gears matter
  7. Try this

What this key area is asking

The SQA wants you to calculate the gear ratio of a pair of meshing gears, use it to find the output speed, and explain how a gear system trades rotational speed for turning force (torque).

Gears and gear trains

Because the teeth must be the same size to mesh, a gear with more teeth is larger. The teeth never slip, so a gear drive transmits motion exactly, unlike a belt that can slip.

The gear ratio

Calculating output speed

The output (driven) gear speed is found from the input speed and the gear ratio:

output speed=input speedgear ratio.\text{output speed} = \frac{\text{input speed}}{\text{gear ratio}}.

A 4:1 reduction gear driven at 1000 rev/min gives an output of 250 rev/min, four times slower. Conversely, gearing up (ratio less than 1) makes the output faster.

Direction and trading speed for force

Two meshing gears rotate in opposite directions (one clockwise, the other anticlockwise). If you need the output to turn the same way as the input, an idler gear is placed between them; it changes the direction without changing the overall gear ratio.

Crucially, a reduction drive does not give "something for nothing": the output gains turning force only because it loses speed in the same proportion. This is the rotational version of mechanical advantage.

Why gears matter

Gear systems are everywhere a motor's raw speed needs converting into useful motion: in vehicle gearboxes, drills, clocks and robots. They let an engineer match a fast, low-torque motor to a slow, high-torque load, which is one of the most common mechanical design tasks in the assignment.

Try this

Q1. A driver gear of 10 teeth meshes with a driven gear of 50 teeth. State the gear ratio. [1 mark]

  • Cue. 50/10=550/10 = 5, i.e. 5:15:1.

Q2. A 2:1 reduction drive is driven at 600 rev/min600 \text{ rev/min}. Find the output speed. [2 marks]

  • Cue. 600/2=300 rev/min600 / 2 = 300 \text{ rev/min}.

Q3. In which direction does a directly meshing driven gear turn relative to the driver? [1 mark]

  • Cue. The opposite direction.

Exam-style practice questions

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

SQA N5 style3 marksA driver gear has 20 teeth and meshes with a driven gear of 60 teeth. Calculate the gear ratio.
Show worked answer →

Use the gear ratio relationship.

Relationship: gear ratio=teeth on driven gearteeth on driver gear\text{gear ratio} = \dfrac{\text{teeth on driven gear}}{\text{teeth on driver gear}}.

Substitution: gear ratio=6020=3\text{gear ratio} = \dfrac{60}{20} = 3, written as 3:13:1.

Markers reward selecting driven over driver teeth, correct substitution, and stating the ratio (3 or 3:1). This means the driven gear turns once for every three turns of the driver, so it runs slower but with more turning force.

SQA N5 style3 marksA driver gear turns at 900 rev/min and meshes with a driven gear, giving a gear ratio of 3:1. Calculate the output (driven gear) speed.
Show worked answer →

A gear ratio of 3:1 means the output turns three times slower than the input.

Relationship: output speed=input speedgear ratio\text{output speed} = \dfrac{\text{input speed}}{\text{gear ratio}}.

Substitution: output speed=9003=300 rev/min\text{output speed} = \dfrac{900}{3} = 300 \text{ rev/min}.

Markers reward dividing the input speed by the gear ratio and giving the output speed with its unit. A reduction gear ratio always gives a slower output.

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