Calculate a compound gear ratio

Primary Tool

CAS Workspace

The CAS stores the tooth counts as variables, so changing one gear updates every downstream speed and torque. The Gear Ratio Calculator handles compound trains of up to four stages and is the faster tool once you know what each stage contributes.

Problem

A motor turns a 16-tooth gear at $1800$ rpm, delivering $10$ N $\cdot$ m. That gear drives a 48-tooth gear. A 20-tooth gear is fixed to the same shaft as the 48-tooth gear and drives a 60-tooth gear on the output shaft. Find the overall ratio, the output speed, and the output torque.

One stage

For a pair of meshing gears the ratio is driven teeth over driver teeth: $GR = N_{d r i v e n} / N_{d r i v er}$ . A ratio above $1$ slows the output and multiplies torque by the same factor. Two external gears turn opposite ways, so each stage also reverses the direction of rotation.

Two stages on a shared shaft

The 48-tooth and 20-tooth gears are keyed to one shaft, so they turn at the same speed. That shared shaft links the stages, and it is why their ratios multiply.

1Stage 1: $G R_{1} = 48/16 = 3$ .
2Stage 2: $G R_{2} = 60/20 = 3$ .
3Overall: $GR = G R_{1} \times G R_{2} = 3 \times 3 = 9$ , so the train is $9 : 1$ .
4Intermediate shaft speed: $1800/3 = 600$ rpm. Output speed: $600/3 = 200$ rpm, which equals $1800/9$ .
5Ideal torque: the intermediate shaft carries $10 \times 3 = 30$ N $\cdot$ m and the output shaft $30 \times 3 = 90$ N $\cdot$ m, which equals $10 \times 9$ .

Correcting for losses

The $90$ N $\cdot$ m figure assumes lossless meshes. A well-made spur gear mesh transmits roughly $98%$ of the power, and the efficiencies of the two meshes multiply: $0.98 \times 0.98 = 0.9604$ . The expected output torque is then $90 \times 0.9604 = 86.4$ N $\cdot$ m. The output speed does not change, because the tooth counts fix it. Losses show up as torque, not as speed.

Use that torque to size the output shaft. For an allowable shear stress of $40$ MPa, a solid shaft needs $D = (\frac{16 T}{π τ})^{1/3}$ , about $22.5$ mm at $90$ N $\cdot$ m. The Shaft Torsion Calculator returns this directly, and the Torque Calculator covers the force and lever-arm side.

Common Pitfall

The ratio runs through every shaft, not from the first gear to the last. Dividing the last driven gear by the first driver gives $60/16 = 3.75$ , which ignores the 20-tooth gear and understates the reduction by more than half. The correct product is $\frac{48 \cdot 60}{16 \cdot 20} = 9$ . Gears on the same shaft share a speed, so the middle pair never appears as a ratio of its own.

Try a Variation

Swap the roles in stage 2 so the 60-tooth gear drives the 20-tooth gear. What is the new overall ratio, and what happens to output speed and torque? Then compare with adding a third $15 \to 45$ stage to the original train. Which change increases the reduction, and what does it cost in efficiency?

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Worked ExampleEngineeringCAS Workspace

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