Simple and Planetary gear trains(1)

Introduction

Sometimes there is two or more gears are made to mesh with each other to transmit power from one shaft to another shaft. Such a combination is called gear train or train of toothed wheels. The nature of the train used depends on upon velocity ratio that required and relative position of the axes of shafts. A gear train may consist of the spur, bevel and spiral gears.

Classification of Gear Trains

Following are the different types of gear trains, depending upon the arrangement of wheels.

• Simple gear train
• Compound gear train
• Reverted gear train and
• Epicyclic gear train

In the first three types of gear trains there the axes of the shafts over which the gears are mounted are fixed relative to one other. In the case of epicyclic gear trains the axes of the shafts on which the gears are mounted may move relative to the fixed specified axis.

Simple Gear Train

When only one gear on each shaft as shown in Fig it is known as a simple gear train. The gears are shown by their pitch circles. When the distance between the two shafts is much small than the two gears 1 and 2 are allow to mesh with each other to transmit motion from one shaft to the other that shown in Fig (a) . Since gear 1 drives the gear 2 then, therefore, gear 1 is called the driver and the gear 2 is called the driven or the follower. It is noted that the motion of driven gear is opposite to the motion of driving gear.

Fig: Simple gear train.

Let

N₁ = Speed of driver gear 1 in r.p.m.,

N₂ = Speed of driven gear 2 in r.p.m.,

T₁ = Number of teeth on gear 1,

T₂ = Number of teeth on gear 2,

Since the speed ratio or velocity ratio of the gear train is the ratio of the speed of driver to the speed of driven and ratio of speeds of any pair of gears which in the mesh is the inverse of their number of teeth. Therefore

Speed ratio = N₁/N₂ = T₂/T₁

It may be noted that ratio of the speed of the driven or follower to the speed of the driver is known as train value of the gear train. Mathematically,

Train value = N₂/N₁ = T₁/T₂

From above expression we see that the training value is the reciprocal of the speed ratio. When the distance between the two gears is large. The motion from one gear to another gear. In such a case it may be transmitted by either of following two methods.

1. By providing the large sized gear
2. By providing one or more than one intermediate gear

A little supposition will show that the former method i.e. providing large sized gears is very inconvenient and uneconomical method where the latter method i.e. providing one or more intermediate gear is very convenient and economical.

It is noted that when the number of intermediate gears are odd the motion of both gears i.e. driver and driven is like as shown in Fig (b). But if the number of intermediate gears are even then the motion of the drive will be in the opposite direction of the driver as shown in Fig(c).

Now suppose a simple gears train with one intermediate gear as shown in Fig (b).

Let,

N₁ = Speed of driver in r.p.m

N₂ = Speed of the intermediate gear in r.p.m

N₃ = Speed of driven in r.p.m

T₁ = Number of teeth on driver,

T₂ = Number of teeth on the intermediate gear and

T₃ = Number of teeth on driven or follower

Since the driving gear 1 which is in mesh with the intermediate gear 2. Therefore speed ratio for these two gears is

N₁/N₂ = T₂/T₁

Similarly the intermediate gear 2 is in mesh with driven gear 3. Therefore speed ratio for these two gears is

N₂/N₃ = T₃/T₂

The speed ratio of the gear train as shown in Fig (b) is obtained by multiplying the equations (i) and (ii).

∴ (N₁/N₂)*( N₂/N₃)=(T₂/T₁)*( T₃/T₂) or (N₁/N₃)=(T₃/T₁)

$$Speed{\rm{ }}ratio = \frac{{Speed{\rm{ }}of{\rm{ }}driver}}{{Speed{\rm{ }}of{\rm{ }}driven}} = \frac{{No.{\rm{ }}of{\rm{ }}teeth{\rm{ }}on{\rm{ }}driven}}{{No.{\rm{ }}of{\rm{ }}teeth{\rm{ }}on{\rm{ }}driver}}$$

and $$\;Train{\rm{ }}value = \frac{{Speed{\rm{ }}of{\rm{ }}driven}}{{Speed{\rm{ }}of{\rm{ }}driver}} = \frac{{No.{\rm{ }}of{\rm{ }}teeth{\rm{ }}on{\rm{ }}driver}}{{No.{\rm{ }}of{\rm{ }}teeth{\rm{ }}on{\rm{ }}driven}}$$

Similarly it can be proved that the above equation holds too even if there are any number of intermediate gears. From above we see that the speed ratio and train value in a simple gears train is independent of the size and number of intermediate gears. These intermediate gears are called idle gears as they do not effect the speed ratio or train value of the system. The idle gears are used for two purposes

• To connect gears where a large centre distance is required, and
• To obtain the required direction of motion of driven gear i.e. clockwise or anticlockwise.

Compound Gear Train

When there are more than one gear on a shaft which is as shown in Fig it is called a compound gear train.

Fig: Compound gear train

The idle gears in a simple train of gears do not effect the speed ratio of the system. But these gears are applicable in bridging over the space between the driver and driven. But whenever the distance between the driver and the driven has to be bridged over by the intermediate gears and at the same time a great or much less speed ratio is desired then the advantage of intermediate gears is intensified by providing compound gears on intermediate shafts. In such case, each of the intermediate shafts has two gears rigidly fixed to it so that they may have the same speed. One of these two gears meshes with the driver and other with the driven attached to next shaft which is as shown in Fig.

In a compound train of gears, as shown in Fig. 13.2, the gear 1 is the driving gear mounted on shaft A, gears 2 and 3 are compound gears which are mounted on shaft B. The gears 4 and 5 are also compound gears which are mounted on shaft C and the gear 6 is the driven gear mounted on shaft D.

Let

N₁ = Speed of driving gear 1,

T₁ = Number of teeth on driving gear 1,

N₂, N₃, N₄ ..., N₆= Speed of respective gears in r.p.m., and

T₂, T₃, T₄..., T₆ = Number of teeth on respective gears.

Since gear 1 is in mesh with gear 2. Therefore its speed ratio is

N₁/N₂ = T₂/T₁

Similarly, for gears 3 and 4, speed ratio is

N₃/N₄ = T₄/T₃

and for gears 5 and 6, speed ratio is

N₅/N₆ = T₆/T₅

The speed ratio of compound gear train is obtained by multiplying the equations (i), (ii) and (iii) ,

(N₁/N₂)*(N₃/N₄)*( N₅/N₆)=(T₂/T₁)*( T₄/T₃)*( T₆/T₅)

Or, N₁/N₆ = (T₂/T₁)*( T₄/T₃)*( T₆/T₅)

$$\begin{array}{l} Speed{\rm{ }}ratio = \frac{{Speed{\rm{ }}of{\rm{ }}the{\rm{ }}first{\rm{ }}driver}}{{Speed{\rm{ }}of{\rm{ }}last{\rm{ }}driven}} = \frac{{\Pr oduct{\rm{ }}of{\rm{ }}No.{\rm{ }}of{\rm{ }}teeth{\rm{ }}on{\rm{ }}driven}}{{\Pr oduct{\rm{ }}of{\rm{ }}No.{\rm{ }}of{\rm{ }}teeth{\rm{ }}on{\rm{ }}driver}}\\ And\\ \;Train{\rm{ }}value = \frac{{Speed{\rm{ }}of{\rm{ }}last{\rm{ }}driven}}{{Speed{\rm{ }}of{\rm{ }}the{\rm{ }}first{\rm{ }}driver}} = \frac{{\Pr oduct{\rm{ }}of{\rm{ }}No.{\rm{ }}of{\rm{ }}teeth{\rm{ }}on{\rm{ }}driver}}{{\Pr oduct{\rm{ }}of{\rm{ }}No.{\rm{ }}of{\rm{ }}teeth{\rm{ }}on{\rm{ }}driven}} \end{array}$$

The advantage of compound train over simple gear train is that is much larger speed reduction from the first shaft to last shaft that can be obtained with small gears. If a simple gear train is used for a large speed reduction then the last gear has to be very large enough. Mostly for a speed reduction in excess of 7 to 1. A simple train is not used but a compound train or worm gearing is employed. The gears which mesh must have the same circular pitch or module. Thus gears 1 and 2 must have the same module as they mesh together. Similarly gears 3 and 4 and gears 5 and 6 must have the same module.

References:
1. H.H. Mabie and C. F. Reinholtz, “Mechanism and Dynamics of Machinery”, Wiley.
2. J.S. Rao & R.V. Dukkipati Mechanisms and Machine Theory, New Age International (P) Limited..
3. J.E. Shigley and J.J. Uicker, Jr., “ Theory of Machines and Mechanisms”, McGraw Hill.
4. B. Paul, “Kinematics and Dynamics of Planar Machinery”, Prentice Hall.
5. C. E. Wilson, J.P. Sadler and W.J. Michels, “Kinematics and Dynamics of Machinery”, Harper Row.