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Gears and gear trains

Gears transmit rotation and torque between shafts with an exact speed ratio, without slipping. This topic explains why the teeth have an involute shape, how they are sized with the module, how the ratio of a train with any number of stages (planetary included) is computed and what forces each tooth carries. The Gears tool computes all of it, with an animation of each mesh.

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Law of gearing and the involute profile

For the speed ratio to be constant at every instant, the common normal to both teeth at the contact point must always pass through a fixed point of the line of centers, the pitch point P. The involute profile (the curve traced by the end of a taut string as it unwinds from a circle, the base circle) satisfies that law and also tolerates small errors in center distance without losing it.

With involutes, the contact point moves along a fixed straight line, the line of action, tangent to both base circles. The angle between that line and the common tangent at P is the pressure angle $ϕ$ , almost always 20°.

Stage 1 of example 1: pitch circles tangent at P, base circles (dashed), outside circles (dotted) and the line of action.

Tooth geometry

Tooth size is set by the module $m = d / z$ (in mm): two gears only mesh if they have the same module and the same pressure angle. For standard 20° full-depth teeth:

Dimension	Formula
Pitch diameter	$d = m z$
Outside diameter	$d_{a} = d + 2 m$
Root diameter	$d_{f} = d - 2.5 m$
Base diameter	$d_{b} = d cos ϕ$
Circular pitch · base pitch	$p = πm$ · $p_{b} = p cos ϕ$
Center distance	$a = \frac{d _{1} + d _{2}}{2}$ (internal: $\frac{d _{2} - d _{1}}{2}$ )

The contact ratio $m_{c}$ is the average number of tooth pairs in contact: the length of the line of action lying between the two outside circles, divided by the base pitch. It must exceed 1 (in practice, 1.2 or more) so that a pair of teeth starts meshing before the previous one lets go:

m_{c} = \frac{r _{a 1}^{2} - r _{b 1}^{2} + r _{a 2}^{2} - r _{b 2}^{2} - a sin ϕ}{p _{b}}

A pinion with very few teeth suffers undercutting: the cutting tool eats into the base of the tooth and weakens it. With 20° full-depth teeth, the minimum to mesh with a rack is

2/ sin^{2} ϕ \approx 17.1

, that is, 18 teeth. The simulator checks it at each stage.

Gear ratio and trains

Since the pitch circles roll without slipping, $ω_{1} r_{1} = ω_{2} r_{2}$ , and the ratio of one stage is

i = \frac{ω _{1}}{ω _{2}} = \frac{n _{1}}{n _{2}} = \frac{z _{2}}{z _{1}} = \frac{d _{2}}{d _{1}}

An external mesh reverses the sense of rotation; an internal one (ring gear) keeps it.
In a simple train (one gear per shaft), the intermediate gears cancel out: the ratio depends only on the first and the last. An idler serves to change the sense or to span a distance.
In a compound train two gears share a shaft, and the total ratio is the product of those of each stage: that is how large ratios are achieved with gears of reasonable size.

Simple train with an idler (the output turns like the input) and a compound two-stage train (54 and 16 turn together on the intermediate shaft).

Type	Shafts	Main feature
Spur	parallel	simple and cheap; somewhat noisy
Helical	parallel	teeth inclined at $ψ$ : smoother and quieter, but with an axial force
Bevel	intersecting (90° typical)	they change the direction of the shaft
Worm and wheel	crossed	very large ratios in one stage; they can be self-locking
Rack and pinion	—	turns rotation into translation: $v = ω d /2$

Torque, power and tooth forces

Power is conserved apart from losses: $P = T ω$ , and with $n$ in rpm, $T [N \cdot m] = 9549 P [kW] / n$ . The output torque is $T_{2} = T_{1} i η$ , with $η$ near 0.98 per spur or helical stage. The force between the teeth acts along the line of action and splits into:

W_{t} = \frac{2 T}{d} W_{r} = W_{t} tan ϕ (helicoidal: W_{a} = W_{t} tan ψ)

$W_{t}$ is the one that transmits the torque; $W_{r}$ pushes the shafts apart and loads the bearings. The tooth works like a small cantilever beam: the Lewis equation gives its bending stress with the form factor $Y$ (which grows with the number of teeth) and Barth's dynamic factor $K_{v} = (6.1 + V) /6.1$ ( $V$ in m/s), which covers the impacts between teeth:

σ = \frac{K _{v} W _{t}}{b m Y}

The surface is checked with the Hertz contact stress between two cylinders. For a final design the full AGMA standard is used, with overload, size and load-distribution factors.

Planetary trains

A planetary train has a central gear (sun), several planets that revolve around it mounted on an arm (the carrier) and an internal ring. Its ratio is found by looking at the motion relative to the carrier, where everything is an ordinary train. That is Willis's equation:

\frac{ω _{s} - ω _{c}}{ω _{r} - ω _{c}} = - \frac{Z _{r}}{Z _{s}} ⟹ con la corona fija: i = \frac{ω _{s}}{ω _{c}} = 1 + \frac{Z _{r}}{Z _{s}}

The geometry requires $Z_{r} = Z_{s} + 2 Z_{p}$ .
To fit $N$ equally spaced planets, $(Z_{s} + Z_{r}) / N$ must be an integer.
Fixing another element (the sun or the carrier) gives other ratios with the same gears: that is how automatic gearboxes shift.

Planetary set of example 3: a 24-tooth sun, three 24-tooth planets and a 72-tooth ring.

Worm and wheel

A worm with $z_{1}$ starts driving a wheel with $z_{2}$ teeth gives $i = z_{2} / z_{1}$ : with a 2-start worm and 50 teeth, 25:1 in a single stage. Its lead angle $λ$ ( $tan λ = z_{1} / q$ , with $q$ = worm diameter over the module) sets the efficiency, because the teeth slide a lot:

η = \frac{cos ϕ _{n} - μ tan λ}{cos ϕ _{n} + μ cot λ}

If the wheel tries to drive the worm, the reverse efficiency is $(cos ϕ_{n} - μ cot λ) / (cos ϕ_{n} + μ tan λ)$ . When it comes out zero or negative, the reducer is self-locking: the load cannot drive it backward, which is useful in hoists and hand winches.

Common mistakes

Meshing two gears with different module or pressure angle.
Counting the idler in the ratio (it only changes the sense).
Using the outside diameter instead of the pitch diameter for $W_{t} = 2 T / d$ or for the ratio.
In a planetary set, applying $z_{2} / z_{1}$ directly without moving to the motion relative to the carrier.
Forgetting the efficiency in many-stage trains or in worm drives, where it may be 50 to 90 %.

Worked examples

All four examples are trains from the tool's menu, with a 5 kW motor at 1750 rpm (the worm, 2.2 kW at 1450 rpm), 4140 steel and 20° teeth.

Example 1Two-stage reducer

Stage 1: an 18-tooth pinion and a 54-tooth gear,

m = 3

mm, 30 mm face width. Stage 2: 16 and 64 teeth,

m = 4

mm. Find the total ratio, the output speed and torque (with 98 % per stage) and, for stage 1, the geometry, the forces and the Lewis stress on the pinion.

Solution

Ratio: $i = \frac{54}{18} \cdot \frac{64}{16} = 3 \cdot 4 = 12$ ; $n_{out} = 1750/12 = 145.8$ rpm, in the same sense as the input (two external meshes).
Input torque $T_{1} = 9549 \cdot 5/1750 = 27.28$ N·m; output $T = 27.28 \cdot 12 \cdot 0.9 8^{2} = 314.4$ N·m.
Stage 1 geometry: $d_{1} = 54$ , $d_{2} = 162$ mm, $a = 108$ mm, $d_{a} = 60$ and 168, $d_{b} = 50.74$ and 152.2 mm; contact ratio 1.649.
Forces: $W_{t} = 2 \cdot 27.28/0.054 = 1011$ N and $W_{r} = 1011 tan 20° = 367.8$ N.
Lewis: $V = π \cdot 0.054 \cdot 1750/60 = 4.948$ m/s, $K_{v} = (6.1 + 4.948) /6.1 = 1.811$ , $Y (18) = 0.309$ : $σ = \frac{1.811 \cdot 1011}{30 \cdot 3 \cdot 0.309} = 65.8$ MPa.

i = 12

, 145.8 rpm and 314.4 N·m; stage 1:

a = 108

mm,

W_{t} = 1011

W_{r} = 367.8

N and

σ = 65.8

MPa on the pinion.

Solve it in Simuveliq

In Engineering → Gears, in the examples panel, «Load train» → Two-stage reducer (spur).
The cards give Overall ratio 12 : 1, Output 145.8 rpm, Output torque 314.4 N·m and Overall efficiency 96 %. The Shafts table shows the intermediate shaft at 583.3 rpm turning the other way.
With stage 1 selected, the Geometry, Mesh forces and Strength (Lewis and Hertz) tables give the values above, and the «Step by step» derives them.

Open in the simulator →

Example 2Train with an idler

A 20-tooth pinion drives a 30-tooth gear, which in turn drives a 40-tooth one. Find the total ratio and the sense of the output.

Solution

Ratio: $\frac{30}{20} \cdot \frac{40}{30} = \frac{40}{20} = 2$ : the 30-tooth gear cancels out.
Two external meshes: the sense reverses twice, so the output turns like the input. Without the idler it would turn the other way.
$n_{out} = 1750/2 = 875$ rpm; the idler shaft turns at $1750 \cdot 20/30 = 1166.7$ rpm in the opposite sense.

i = 2

, 875 rpm in the same sense as the input.

Solve it in Simuveliq

«Load train» → Train with an idler gear: Overall ratio 2 : 1, «same direction as the input», and the Shafts table with the intermediate one at 1166.7 rpm (↻).

Open in the simulator →

Example 3Planetary with the ring fixed

A planetary set has a 24-tooth sun, a 72-tooth ring and three planets. The sun is the input (1750 rpm) and the ring is fixed. Find the planet teeth, the carrier speed and check that all three fit.

Solution

Planets: $Z_{p} = (72 - 24) /2 = 24$ teeth.
Willis with $ω_{r} = 0$ : $\frac{1750 - ω _{c}}{0 - ω _{c}} = - \frac{72}{24} = - 3 \Rightarrow ω_{c} = 437.5$ rpm, in the same sense as the sun: $i = 1 + 72/24 = 4$ .
Assembly: $(24 + 72) /3 = 32$ , an integer: three planets fit at 120°.
Each planet turns about its own axis at $ω_{p} = ω_{c} (1 - Z_{r} / Z_{p}) = 437.5 \cdot (1 - 3) = - 875$ rpm (backward).

Z_{p} = 24

ω_{c} = 437.5

rpm (

i = 4

), planets at −875 rpm; three can be fitted.

Solve it in Simuveliq

«Load train» → Planetary (ring fixed). The Geometry table gives the teeth 24 · 24 · 72, the speeds «sun 1750 · ring 0 · carrier 437.5» and the spin of each planet, −875 rpm; below, «✓ 3 equally spaced planets can be assembled».
The «Step by step» writes Willis's equation. Change which element is fixed and watch the ratio change.

Open in the simulator →

Example 4Worm reducer

A 2-start worm (

m = 5

mm,

q = 10

μ = 0.05

ϕ_{n} = 20°

) drives a 50-tooth wheel from 2.2 kW at 1450 rpm. Find the ratio, the lead angle, the forward and reverse efficiencies and whether it is self-locking.

Solution

$i = 50/2 = 25$ : the wheel turns at $1450/25 = 58$ rpm.
$tan λ = z_{1} / q = 2/10 \Rightarrow λ = 11.31°$ .
Forward: $η = \frac{0.9397 - 0.05 \cdot 0.2}{0.9397 + 0.05 \cdot 5} = \frac{0.9297}{1.1897} = 78.1 %$ . Reverse: $\frac{0.9397 - 0.05 \cdot 5}{0.9397 + 0.05 \cdot 0.2} = 72.6 %$ .
The reverse one is positive: it is not self-locking. It would need $μ \geq cos ϕ_{n} tan λ = 0.188$ or a smaller lead angle (a single-start worm).
Output torque: $T_{1} = 9549 \cdot 2.2/1450 = 14.49$ N·m and $T_{2} = 14.49 \cdot 25 \cdot 0.781 = 283$ N·m.

i = 25

(58 rpm),

λ = 11.31°

η = 78.1 %

forward and 72.6 % reverse: not self-locking;

T_{2} \approx 283

N·m.

Solve it in Simuveliq

«Load train» → Worm reducer: Overall ratio 25 : 1, Output 58 rpm, Output torque 283.1 N·m and Overall efficiency 78.1 %.
In Geometry: lead angle 11.31° and efficiency «78.1 % · 72.6 %»; below, the warning «Not self-locking». Raise μ to 0.2 and you will see it become self-locking.

Open in the simulator →

Gears and gear trains

Law of gearing and the involute profile

Tooth geometry

Gear ratio and trains

Torque, power and tooth forces

Planetary trains

Worm and wheel

Common mistakes

Worked examples

Example 1Two-stage reducer

Solution

Solve it in Simuveliq

Example 2Train with an idler

Solution

Solve it in Simuveliq

Example 3Planetary with the ring fixed

Solution

Solve it in Simuveliq

Example 4Worm reducer

Solution

Solve it in Simuveliq

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