The big idea: Gear- and belt-driven systems change the direction, speed, power and efficiency of a rotary motion.
The ratio is always driven ÷ driver. Slower means more torque; faster means less. And a compound train multiplies ratios, which is how a large reduction fits in a small housing.
The seven gear systems, each with its shafts — then the ratio rule.
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From a specification to a gear train
Work out the overall ratio needed
Input speed ÷ required output speed. A 12,000 rpm motor driving a 400 rpm output needs 30:1.
Split it into stages
A single pair much above 5:1 needs one gear five times the size of the other. Two stages of about 5.5:1 give 30:1 in a housing a hand can hold.
Choose tooth counts
Pick real numbers: 12 into 66 is 5.5:1. Avoid a whole-number ratio if you can, so the same teeth do not meet every revolution and wear a pattern into each other.
Check the direction
Every external mesh reverses rotation. If the output must turn the same way as the input, either use an even number of meshes or add an idler.
Speed at several points: A 20-tooth driver at 900 rpm meshes with a 60-tooth gear; on the same shaft a 15-tooth gear drives a 45-tooth gear.
After stage one: 900 × 20 ÷ 60 = 300 rpm. After stage two: 300 × 15 ÷ 45 = 100 rpm. Overall 9:1, and the torque is raised about nine times less the losses.
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| Requirement | Arrangement | Consequence |
|---|---|---|
| Reduce speed, raise torque | Small driver into a large driven gear | Slower and stronger. The usual case, because motors are efficient at high speed and products rarely are |
| Increase speed | Large driver into a small driven gear | Faster and weaker. A rotary whisk and a bicycle in its highest gear both do this |
| Keep the direction | Add an idler, or use an even number of meshes | No change to the ratio at all — an idler cancels out of the arithmetic |
| Turn the drive through 90° | Bevel gears, or a worm and wheel for a large reduction as well | A worm also cannot be back-driven, which is a safety property, not a limitation |
| Span a gap quietly and cheaply | A belt drive | It slips under overload, which protects the motor and rules it out where timing matters |
How this is tested — calculating gear and belt ratios and constructing systems that change speed. It comes up two ways:
Paper 1 — multiple choice
- Find the output speed at a stated point in a train.
- Choose the arrangement that gives a required change of speed.
Paper 2 — analysing a product
- Design a gear train to meet a stated speed requirement.
- Calculate the speed at several points and comment on the result.
The trap: Forgetting that an idler leaves the ratio alone. It changes the direction only, and including it in the arithmetic gives a wrong answer confidently.
A motor runs at 3,000 rpm and a mixing paddle must turn at about 100 rpm in the SAME direction. Construct a suitable gear train and justify it.
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