Gear Ratio Calculator: Multi-Stage Ratio, Output Speed, Torque and Direction
When you design a gear train, the questions are always the same: what's the overall ratio, how fast does the output turn, how much torque do you get, and which way does it spin? Enter the tooth counts for up to three stages, plus your motor's speed and torque.
The formulas
Each stage's ratio is driven teeth ÷ driver teeth. In a compound train (two gears fixed on the same shaft), the overall ratio is the product of the stages:
Ratio = (Z₂ ÷ Z₁) × (Z₄ ÷ Z₃) × …
- Output speed = input speed ÷ ratio
- Output torque = input torque × ratio × overall efficiency (overall efficiency = per-stage efficiency raised to the number of stages)
With the default values, 10 teeth drive 40 (4 : 1), then a 12-tooth gear on the same shaft drives 36 (3 : 1), for 12 : 1 overall. A 200 rpm motor ends up at about 16.7 rpm, and 0.8 kgf·cm becomes about 0.8 × 12 × 0.9025 ≈ 8.66 kgf·cm at 95% per stage.
Direction of rotation
Two meshing external spur gears turn in opposite directions, so the direction flips at every mesh: an even number of meshes gives the same direction as the input, an odd number gives the opposite. An idler gear between two gears flips the direction once more without changing the ratio. The calculator doesn't count idlers, so adjust the direction yourself if you use one.
What efficiency should you enter?
Precision-machined metal spur gears can exceed 95% per stage. Laser-cut wood or acrylic gears and 3D-printed gears are usually lower, because of tooth accuracy and friction. 95% is only a starting point. To check whether a motor can actually drive your mechanism, try a more conservative value (say 80–90%) and leave a torque margin.
My geared flapping-wing bird is a hand-cranked, laser-cut gear train, a good example of where the lower efficiency of cut-wood gears matters.
Units
Small DC motors and hobby servos usually list torque in kgf·cm: 1 kgf·cm ≈ 0.0981 N·m.
Worked examples
Example 1: one stage, 300 rpm down to 100 rpm
The motor runs at 300 rpm with 0.6 kgf·cm of torque at its working point. The target is 100 rpm, and we'll assume 90% efficiency per stage. The required ratio is 300 ÷ 100 = 3, so an 18-tooth driver meshing with a 54-tooth driven gear gives exactly 54 ÷ 18 = 3.
Using output speed = input speed ÷ total ratio: 300 ÷ 3 = 100 rpm, no speed error from the tooth counts. Output torque = 0.6 × 3 × 0.90 = 1.62 kgf·cm, 10% less than the lossless 1.8 kgf·cm. A single external mesh reverses the direction of rotation.
These numbers come from the input conditions; a heavier load will pull the motor's speed down. Make sure the gears share the same module and pressure angle, and check that the shafts, bearings and teeth can carry the load.
Example 2: two stages, aiming for about 20 rpm
Input is 600 rpm and 0.5 kgf·cm, again at 90% per stage. The target ratio is 600 ÷ 20 = 30. Using gears on hand, stage one is 20 teeth driving 80 (ratio 4), and stage two is 20 teeth driving 140 (ratio 7). Total ratio = 4 × 7 = 28.
Output speed = 600 ÷ 28 = 21.43 rpm, +7.1% from the target. Total efficiency = 0.90² = 0.81, so output torque = 0.5 × 28 × 0.81 = 11.34 kgf·cm, 19% below the ideal 14 kgf·cm.
The 80-tooth gear from stage one and the 20-tooth gear of stage two must be fixed on the same shaft. With two external meshes, the output turns the same way as the input. Keep the load comfortably below the estimated torque, and use the motor's working-point torque. Don't combine stall torque with no-load speed.
Common mistakes
- Inverting the ratio: the reduction ratio is driven teeth ÷ driver teeth.
- Adding stage ratios: multi-stage ratios multiply, and so do the efficiencies.
- Forgetting the compound shaft: if the two gears on the middle shaft aren't locked together, you don't get a two-stage reduction.
- Getting the direction wrong: each external mesh reverses rotation. Add an idler and you have to recount the meshes.
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