A gear train is a power transmission system made up of two or more gears. The gear to which the force is first applied is called the driver and the final gear on the train to which the force is transmitted is called the driven gear. Any gears between the driver and the driven gears are called the idlers. Conventionally, the smaller gear is the Pinion and the larger one is the Gear. In most applications, the pinion is the driver; this reduces speed but increases torque.

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Types of gear trains

  1. Simple gear train

  2. Compound gear train

  3. Planetary gear train*

Simple Gear Train - Simple gear trains have only one gear per shaft. The simple gear train is used where there is a large distance to be covered between the input shaft and the output shaft.

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Compound Gear Train -

In a compound gear train at least one of the shafts in the train must hold two gears. Compound gear trains are used when large changes in speed or power output are needed and there is only a small space between the input and output shaft.

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Planetary Gear Train -

A planetary transmission system (or Epicyclic system as it is also known), consists normally of a centrally pivoted sun gear, a ring gear and several planet gears which rotate between these. This assembly concept explains the term planetary transmission, as the planet gears rotate around the sun gear as in the astronomical sense the planets rotate around our sun.

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Planetary gearing or epicyclic gearing provides an efficient means to transfer high torques utilizing a compact design.


We have learnt in the previous section that “If two gears are in mesh, then the product of speed (revolutions) and teeth must be conserved”. Let’s check this simple rule with a help of an example.

If you turn a gear with 6 teeth 3 times and is meshed with a second gear having 18 teeth, than the driving gear 18 teeth (6 x 3) will move through the meshed area. This means that the 18 teeth from the second gear also move through the meshed area. If the second gear has 18 teeth, then it only has to rotate once because 18 x 1=18.

Also, the second gear will be turning slower than the first because it is larger, and larger gears turn slower than smaller gears because they have more teeth.

Gear Ratio for Simple Gear Train

Consider a simple gear train shown below. Notice that the arrows show how the gears are turning. When the driver is turning clockwise the driven gear is anti-clockwise.

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Compound Gear Train

The figure below shows a set of compound gears with the two gears, 2 and 3, mounted on the middle shaft b. Both of these gears will turn at the same speed because they are fastened together, i.e. $N_b = N_2 = N_3$

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When gear 1 and gear 2 are in mesh:

$N_1 \times Z_1 = N_2 \times Z_2$

It’s still true that:

$N_1 \times Z_1 = N_b \times Z_2$

$N_b = \frac{Z_1}{Z_2} \times N_1$

If gears 3 and 4 are in mesh: $N_b \times Z_3 = N_4 \times Z_4$

Therefore, $N_4 = \frac{Z3}{Z4} \times N_b = \frac{Z3}{Z4}\times\frac{Z_1}{Z_2} \times N_1$

So the end-to-end gear ratio is $\frac{(Z1\times Z3)}{ (Z2\times Z4)}$ and it does depend on the intermediate gears; unlike the previous case when each gear could turn on its own separate axis. Note that the resultant gear ratio is just the product of the two separate gear ratios:$\frac{Z3}{Z4}\times\frac{Z_1}{Z_2}$


In the figure below, Gears B and C represent a compound gear and have the following details:

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Gear A -120 teeth ,Gear B -40 teeth,Gear C - 80 teeth,Gear D - 20 teeth What is the output in revs/min at D, and what is the direction of rotation if Gear A rotates in a clockwise direction at 30 revs/min?


When answering a question like this, split it into two parts. Treat Gears A and B as the first part of the question. Treat Gears C and D as the second part.

Gear ratio $AB$ = driven/driving = 40/120 = 1/3

Gear ratio $CD$ = driven /driving = 20/80 = 1/4

Since the driving Gear A rotates 30 RPM and the Gear B is smaller than Gear A, we can conclude that the RPMs for Gear B is 30*3 = 90 RPM

Since Gears B and C represent a compound gear, they have the same rotational speed.

Therefore, Gear D speed is obtained by multiplying 4 to Gear C speed of 90 RPM.

Thus, Gear D moves at 90*4 = 360 rev/min OR 1/3 x ¼ = 1/12

Since Gear A moves at 30 RPM and Gear D is smaller, we multiply by 12: 30 x 12 = 360 RPM


In epicyclic gear train, the axis of rotation of one or more of the wheels is carried on an arm which is free to revolve about the axis of rotation of one of the other wheels in the train. The diagram shows a Gear B on the end of an arm A. Gear B meshes with Gear C and revolves around it when the arm is rotated. B is called the planet gear and C the sun.

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Now let’s see what happens when the planet gear orbits the sun ge

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Observe point p and you will see that Gear B also revolves once on its own axis. Any object orbiting around a center must rotate once. Now consider that B is free to rotate on its shaft and meshes with C. Suppose the arm is held stationary and Gear C is rotated once. B spins about its own center and the number of revolutions it makes is the ratio $\frac{N_C}{ N_B}$. B will rotate by this number for every complete revolution of C.

Now consider that C is unable to rotate and the Arm A is revolved once. Gear B will revolve $1 + \frac{N_C}{N_B}$ because of the orbit. It is the extra rotation that causes confusion. One way to get around this is to imagine that the whole system is revolved once. Then identify the gear that is fixed and revolve it back one revolution. Work out the revolutions of the other gears and add them up. The following tabular method makes it easy.

Method 1

Suppose Gear C is fixed and the Arm A makes one revolution. Determine how many revolutions the planet Gear B makes.

Step 1 is to revolve everything once about the centre.

Step 2 is to identify that C should be fixed and rotate it backwards one revolution keeping the arm fixed as it should only do one revolution in total. Work out the revolutions for B.

Step 3 is to simply add them up and find that the total revs of C is zero and the arm is 1.

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The number of revolutions made by B is $(1+ \frac{t_C}{t_B})$. Note that if C revolves -1, then the direction of B is opposite so $+\frac{t_C}{t_B}$

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