Process & Mechanical
Gear Ratio & Speed/Torque
Output speed and torque from tooth counts — the ideal kinematic relationship (N₁/N₂ = D₂/D₁ = T₂/T₁), plus mesh efficiency if you want actual delivered torque rather than the theoretical ideal.
Standard method — stated scope limits apply
Gears & Input
Enter tooth counts, input speed and torque.
About This Calculation
A gear (or belt/chain) drive's ratio sets the trade-off between output speed and output torque — conservation of power (ignoring friction losses) means a drive that reduces speed by a given factor increases torque by the same factor, which is the entire reason gearboxes exist: matching a motor's efficient operating speed to whatever speed and torque the driven equipment actually needs. This calculator computes the resulting output speed and torque for a given gear ratio and input conditions.
Frequently Asked Questions
Why does reducing speed through a gearbox increase torque proportionally?
Mechanical power equals torque multiplied by angular speed — for an ideal (100% efficient) gear drive, power in equals power out, so if a gear reduction cuts output speed to a fraction of input speed, output torque must increase by the inverse of that same fraction to keep power balanced.
Does a real gearbox actually preserve power exactly, or is there always a loss?
Real gear drives always have some mechanical loss (friction in gear meshing, bearings, lubricant churning) — typical gearbox efficiency ranges from about 95-98% per reduction stage for well-designed gear sets, meaning actual output torque is slightly less than the ideal speed-ratio calculation alone would predict, and that loss shows up as heat.
How is overall ratio calculated for a multi-stage gear train?
Overall ratio is the PRODUCT of each individual stage’s ratio, not the sum — a two-stage gearbox with a 3:1 first stage and a 4:1 second stage gives a 12:1 overall ratio, which is why compact multi-stage gearboxes can achieve very large overall reductions from modestly-sized individual gear stages.