Engineering · Simulator 093
Gears trade speed for turning force. Set how many teeth each gear has, add an idler, or put two gears on one shaft to make a compound train. Watch the gears turn, see which way the last one spins, and work out how fast it goes and how much torque it delivers.
Meshing teeth move past the contact point at the same speed, so for two gears with N₁ and N₂ teeth
ω₂ / ω₁ = −N₁ / N₂
The minus sign means each mesh reverses the direction. With no losses, power P = Tω is the same on both shafts, so torque scales the other way: T₂ = T₁ N₂ / N₁. A small gear driving a large one gives lower speed and higher torque.
In a simple train each shaft carries one gear. The middle gears are idlers: their tooth counts cancel out, so the overall ratio depends only on the first and last gears, but every idler flips the direction once more.
In a compound train two gears share a shaft and turn together, so the stage ratios multiply:
ω_out / ω_in = (N₁ / N₂) × (N₃ / N₄)
which gives large reductions with small gears. Each mesh also loses a little power to friction; the overall efficiency is the per-mesh efficiency raised to the number of meshes.
Ideal spur gears with the same tooth size (module), rigid shafts and no backlash. Torque values assume steady running at the given efficiency.
A motor turns at 3,000 rpm and the output must turn at 300 rpm in the same direction. Design a two-stage compound train using gears with 12 to 60 teeth. Could a simple train of two gears do it?
The total reduction is 3000 / 300 = 10, and the stage ratios multiply, so any pair such as 4 × 2.5 works: 12 → 48 (4:1) and 20 → 50 (2.5:1), or 15 → 60 and 24 → 60. Two meshes reverse the direction twice, so the output turns the same way as the motor. A single pair would need 12 → 120 teeth, outside the range, and would turn the output the opposite way. Check with the compound arrangement and an input of 3000 rpm.