Gear Ratios
Follow tooth counts through pairs, idlers and compound trains; multiply stage efficiencies and choose gears to meet speed and torque limits.
Experiment: trade speed for turning effort; Interactive
Experiment: trade speed for turning effort
Predict first: a 20-tooth gear turns at 120 rpm clockwise and drives a 40-tooth gear. Will the larger gear turn faster or slower? What changes if you add an idler between them?
One mesh: 2:1 reduction
Add an idler: still 2:1
Two stages: 2 × 2 = 4:1
All three examples use the same ideal input: 120 rpm and 2 N·m. Input power is 25.13 W, and ideal output power is also 25.13 W. More output torque comes with less speed.
With each actual mesh at 96% efficiency, the pair delivers 3.84 N·m and 24.13 W; the idler train delivers 3.6864 N·m and 23.16 W; the compound train delivers 7.3728 N·m and 23.16 W. Their output speeds remain 60, 60 and 30 rpm. This comparison charges a loss at both idler meshes.
Count the teeth to find the speed
Count the teeth to find the speed
The driver supplies motion; the driven gear receives it. For one external gear mesh, each tooth passing on the driver advances the driven gear by one tooth space. A gear with twice as many teeth needs twice as many passing teeth to complete a turn.
Output speed = input speed × driver teeth ÷ driven teeth. A 20-tooth driver at 120 rpm driving 40 teeth gives 120 × 20 ÷ 40 = 60 rpm. External meshing reverses the direction.
This lesson uses reduction ratio i = input speed ÷ output speed, with speed magnitudes. A 2:1 reduction means two input turns for one output turn. Ratios above 1 reduce speed; ratios below 1 increase it. At zero speed, use tooth counts to find the ratio; do not divide 0 by 0.
Notation map: this page's original equations use N for teeth and n for speed. The course uses n for teeth, ω for angular velocity and τ for torque. Thus page N = course n; page speed n is written ω in the additions below. GR and i name the same reduction ratio.
In course notation, using magnitudes: GR = nout/nin = dout/din = ωin/ωout = τout/τin for an ideal single pair. Here d is pitch diameter; the torque equality assumes no losses. For a compound train, multiply the stage ratios; the first and last tooth counts alone do not give the overall ratio.
Why tooth counts work, and what the dashed circles mean
The gears here share one tooth size, called module, and a 20° pressure angle. Their pitch diameters are proportional to tooth count. The dashed pitch circles roll together at equal tangential speed, so nA × NA = −nB × NB when clockwise is positive on both shafts. Here n is signed rotational speed and N is tooth count.
Tooth flanks are involute curves. Actual tooth surfaces can slide during contact even though the pitch circles roll. The drawing samples those curves, simplifies the tooth roots and adds a small visual clearance. It is a motion model, not a manufacturing drawing.
An idler and a compound shaft do different jobs
An idler and a compound shaft do different jobs
A → B turns the output opposite to the input, viewed from the same end of each shaft.
A → I → B reverses twice. I affects its own speed and the spacing, but its tooth count cancels from the end-to-end ratio.
A → B, then C → D, with B and C fixed together: i = (NB/NA) × (ND/NC).
Same shaft means the same angular speed. A large B and a small C fixed to one shaft make the same number of turns in the same direction. Their rim speeds differ because their radii differ. They do not mesh with each other.
In the compound example, 20 → 40 teeth halves the speed. A second 20 → 40 stage halves it again: 120 → 60 → 30 rpm. The two external meshes make the final output turn in the same direction as the input.
An idler is not another reduction stage. Drop its tooth count when finding the end-to-end ratio, but count every external mesh for direction: odd means opposite; even means the same, viewed from the same end of the parallel shafts. Do not drop a gear that shares a shaft with a different gear in a compound train.
More torque costs speed; it does not create power
More torque costs speed; it does not create power
Torque is turning effort. In this ideal, lossless train at steady load, output torque magnitude = input torque magnitude × reduction ratio. A 2:1 reduction changes 120 rpm and 2 N·m into 60 rpm and 4 N·m. The 4:1 compound example gives 30 rpm and 8 N·m.
Rotational power is torque × angular speed: P = T × 2πn/60 when T is in N·m and n is in rpm. Both sides of either example carry 25.13 W ideally. Real friction and other losses leave less output power. Tooth strength, bearings, shafts, lubrication and the motor all limit usable torque.
At zero speed, mechanical power is zero. A loaded, stationary gear train can still carry torque. The lab shows ideal static torque balance in that case; it does not model motor stall behavior or promise a safe load.
Primary references and model boundaries
- KHK: Idler gear. The intermediate tooth count cancels from the overall speed ratio.
- KHK: Calculation of Gear Dimensions, section 4.1 and Table 4.1. Standard spur-gear pitch, base, tip and root dimensions and center spacing.
- MIT OpenCourseWare: Gear Trains. Gear bodies, shared shafts, and force and torque balance.
The lab assumes rigid external spur gears, fixed shaft centers, matched tooth size and steady transmitted load. It starts lossless; its efficiency control applies the same specified loss to every actual mesh, including both meshes around an idler. Acceleration, elasticity, backlash dynamics, detailed friction, noise and tooth strength are omitted. Changing tooth counts moves the centers to maintain the correct mesh; it does not represent swapping a gear into a fixed gearbox housing.
Connect the gears to Angular Motion, Rotation and Linear Speed and Torque.
Stage efficiencies multiply; losses accumulate
Stage efficiencies multiply; losses accumulate
Each stage receives only the power that survived the stage before it. Use decimal efficiencies and multiply: ηtotal = η1 × η2 × η3. Never add or average them. See Power and Efficiency for the general meaning of efficiency.
At 98% per stage, three stages give 0.98³ = 0.941192: about 5.88% total loss, not 6%. Fewer, higher-quality stages often beat many cheap stages for efficiency, provided they can meet the ratio and load requirements.
Why adding losses is only an approximation
For a 1000 W input and three 96% stages, the losses are 40 W, 38.4 W and 36.864 W. Each 4% applies to a smaller incoming power. Adding 4% + 4% + 4% incorrectly takes every loss from the original 1000 W.
Positive drives keep the speed ratio; losses reduce torque
Positive drives keep the speed ratio; losses reduce torque
A positive drive keeps input and output linked in a fixed ratio without slipping. Gears, chains and timing belts are positive drives. A slipping flat belt or V-belt is not. Follow Rotation and Linear Speed for the connection from rotation to belt travel.
At a stated steady input speed and load, use magnitudes:
ωout = ωin/i
τout = τin × i × ηtotal
Pout = ηtotal × Pin
The teeth still set the speed exactly in this model. Efficiency reduces output torque from its ideal value. It does not reduce output rpm as well. Use matching torque units on both sides: lbf·ft or N·m. The course conversion is 1 rpm ≈ 0.1047 rad/s when angular speed in rad/s is needed.
Follow both stages of one loaded train
A 24-tooth driver at 1440 rpm and 4 lbf·ft drives 48 teeth. That shaft also carries a 15-tooth driver for a 45-tooth output. Both stages are 96% efficient, with no other losses. After stage 1: 720 rpm and 4 × 2 × 0.96 = 7.68 lbf·ft. The shared shaft carries that same 720 rpm into stage 2. Final output: 1440/(2 × 3) = 240 rpm and 7.68 × 3 × 0.96 = 22.1184 lbf·ft, versus 24 lbf·ft ideally. Overall efficiency is 0.9216.
Count stages for ratio, meshes for direction and stated losses
Count stages for ratio, meshes for direction and stated losses
For a simple train A → I → B, the idler cancels from (NI/NA) × (NB/NI) = NB/NA. The two external meshes still reverse direction twice.
Follow the loss assumption in the problem. Real meshes lose some power, but a problem may explicitly ignore it. If every actual mesh is 96% efficient, A → I → B has ηtotal = 0.96². If a whole stage including its idler is assigned 96%, use that stage value once. If η = 1.00 is stated, ignore losses. Do not infer the loss count from the reduction-stage count when an idler is present.
Choose teeth, then check both output limits; Checklist; Practice
Choose teeth, then check both output limits
- Record input speed and torque, the output speed ceiling and torque floor, available integer tooth counts, maximum ratio per stage, and the loss assumption.
- Use the speed ceiling to find
i ≥ ωin/ωceiling. With at most R:1 per stage, k reduction stages can provide at mostRk:1. - For each candidate stage count k, calculate ηtotal and check the torque bound
i ≥ τfloor/(τin × ηtotal). Use the larger of the speed and torque bounds. - Choose whole-tooth pairs within the range and per-stage limit. Connect successive pairs by shared shafts. Multiply their actual ratios; do not assume a rounded target is achievable.
- Calculate output speed and torque, keeping full precision for both limit checks. Check direction by counting external meshes. An added idler may require recalculating losses.
To choose the faster of feasible designs at the same input speed, compare their actual overall ratios: the smaller one is faster. A feasible design is not automatically the fastest possible design.
Worked design: a winch with a speed ceiling and torque floor
A motor supplies 1680 rpm and 24 lbf·ft. The output must be at most 40 rpm and at least 950 lbf·ft. Available gears have 10–60 teeth, with unlimited copies; each reduction stage is at most 6:1 and 97% efficient. There are no idlers or other losses.
- Speed requires
i ≥ 1680/40 = 42. Two stages give at most6² = 36, so try three. - Three stages give
ηtotal = 0.97³ = 0.912673. Torque requiresi ≥ 950/(24 × 0.912673) ≈ 43.3708. Torque controls. - Choose 12 → 48, 15 → 45, and 16 → 60 teeth:
i = 4 × 3 × 3.75 = 45. Every tooth count and stage ratio meets its limit. ωout = 1680/45 ≈ 37.33 rpm ≤ 40 rpm.τout = 24 × 45 × 0.912673 = 985.68684 lbf·ft ≥ 950 lbf·ft. Both pass. Three external meshes make the output opposite to the input.
This is one feasible train, not a claim of an optimum or a complete winch specification. For load force at a drum, continue to Torque; for travel speed, use Rotation and Linear Speed.
Practice multiplying efficiencies Practice designing to limits