Gear Ratio Calculator
Calculators · Added
A gear train trades speed for torque, and the trade is exact: whatever the ratio does to one it does the reciprocal to the other, because power is conserved. Enter the tooth counts for each stage and this gives the overall ratio, the direction the output turns, and — if you supply a speed and a torque — what comes out the far end after the losses at each mesh are taken into account.
How to use the gear ratio calculator
- 1Enter the tooth counts for the driving and driven gear of the first stage.
- 2Add further stages for a compound train, up to four.
- 3Optionally enter the input speed in rpm and the input torque.
- 4Choose the efficiency per mesh — a spur pair is about 98%, a worm drive can be under 60%.
- 5Read the ratio, whether it is a reduction or an overdrive, and what happens to speed and torque.
Examples
A single reduction stage
- Input
- 12 teeth driving 36 teeth
- Result
- 3:1 reduction — a third of the speed, three times the torque
The output also turns the opposite way, because a single external mesh reverses direction.
A compound train
- Input
- 12→36, then 15→45
- Result
- 9:1 overall
Ratios multiply through the stages. Two modest reductions in series achieve what one impractically large gear pair would.
Speed and torque together
- Input
- 3:1 at 1500 rpm and 2 N·m, 98% per mesh
- Result
- 500 rpm and 5.88 N·m
Torque would be 6 N·m in an ideal gearbox; the missing 2% is friction at the mesh.
About the gear ratio calculator
Why the tooth counts are all that matter
Two meshing gears share the same tooth pitch, so their teeth pass the contact point at the same rate. If the driving gear has 12 teeth and the driven gear has 36, then three turns of the driver pass 36 teeth through the mesh, which is exactly one turn of the driven gear. The physical diameters follow from the tooth counts and the pitch, so they carry no extra information — the ratio is entirely a matter of counting teeth.
This is why gear ratios are exact in a way that belt and friction drives are not. There is no slip to account for and no compliance under load: 36 divided by 12 is 3, precisely and permanently. A toothed belt shares that property, which is why cam timing uses one and why a slipping timing belt is a failure rather than a gradual drift.
For a compound train the stage ratios multiply, because each stage's output is the next one's input. That multiplication is why gearboxes can achieve enormous reductions in a small space: three modest 5:1 stages give 125:1, and four give 625:1, with every individual gear still a reasonable size.
The trade, and where it shows up
An electric motor produces its useful power over a fairly narrow band of speeds, and that band is almost never the speed the driven machine wants. A small DC motor might be efficient at 3000 rpm while the wheel it drives should turn at 60. The gearbox exists to reconcile those two, and the torque multiplication is the reward for accepting the speed reduction — the same reason a cyclist changes down a gear at the bottom of a hill.
Reading the direction of the trade correctly is where the practical error usually lies. A reduction gives more torque and less speed, which is what you want for lifting, driving wheels from a small motor, or moving anything heavy from rest. An overdrive gives more speed and less torque, which suits a propeller, a fan or the top gear of a vehicle already at cruising speed.
The efficiency figure decides whether the trade is worth making at all. Losing 2% per mesh is nothing next to the benefit of running a motor at its efficient speed, so a two-stage spur reduction is close to free. Losing 40% in a worm drive is a real cost, and it is accepted only when its particular virtues — a large ratio in one compact stage, and the fact that the output cannot back-drive the input — are worth paying for.
Frequently asked questions
Which way round should the ratio be written?
Does an idler gear change the ratio?
Why use several stages rather than one big reduction?
Where does the lost power go?
Can a gearbox increase power?
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