Gear Trains-1-25.pdf

428 l Theory of Machines
Gear
Gear
Gear
Gear
Gear T
T
T
T
Trains
rains
rains
rains
rains
13
F
F
F
F
Fea
ea
ea
ea
eatur
tur
tur
tur
tures
es
es
es
es
1. Introduction.
2. Types of Gear Trains.
3. Simple Gear Train.
4. Compound Gear Train.
5. Design of Spur Gears.
6. Reverted Gear Train.
7. Epicyclic Gear Train.
8. Velocity Ratio of Epicyclic
Gear Train.
9. Compound Epicyclic Gear
Train (Sun and Planet
Wheel).
10. Epicyclic Gear Train With
Bevel Gears.
11. Torques in Epicyclic Gear
Trains.
13.1.
13.1.
13.1.
13.1.
13.1. Intr
Intr
Intr
Intr
Introduction
oduction
oduction
oduction
oduction
Sometimes, two or more gears are made to mesh with
each other to transmit power from one shaft to another. Such
a combination is called gear train or train of toothed wheels.
The nature of the train used depends upon the velocity ratio
required and the relative position of the axes of shafts. A
gear train may consist of spur, bevel or spiral gears.
13.2.
13.2.
13.2.
13.2.
13.2. T
T
T
T
Types of Gear
ypes of Gear
ypes of Gear
ypes of Gear
ypes of Gear T
T
T
T
Trains
rains
rains
rains
rains
Following are the different types of gear trains, de-
pending upon the arrangement of wheels :
1. Simple gear train, 2. Compound gear train, 3. Re-
verted gear train, and 4. Epicyclic gear train.
In the first three types of gear trains, the axes of the
shafts over which the gears are mounted are fixed relative to
each other. But in case of epicyclic gear trains, the axes of
the shafts on which the gears are mounted may move relative
to a fixed axis.
13.3.
13.3.
13.3.
13.3.
13.3. Simple Gear
Simple Gear
Simple Gear
Simple Gear
Simple Gear T
T
T
T
Train
rain
rain
rain
rain
When there is only one gear on each shaft, as shown
in Fig. 13.1, it is known as simple gear train. The gears are
represented by their pitch circles.
When the distance between the two shafts is small,
the two gears 1 and 2 are made to mesh with each other to
428
CONTENTS
CONTENTS
CONTENTS
CONTENTS

Chapter 13 : Gear Trains l 429
transmit motion from one shaft to the other, as shown in Fig. 13.1 (a). Since the gear 1 drives the gear
2, therefore gear 1 is called the driver and the gear 2 is called the driven or follower. It may be noted
that the motion of the driven gear is opposite to the motion of driving gear.
(a) (b) (c)
Fig. 13.1. Simple gear train.
Let N1 = Speed of gear 1(or driver) in r.p.m.,
N2 = Speed of gear 2 (or driven or follower) in r.p.m.,
T1 = Number of teeth on gear 1, and
T2 = Number of teeth on gear 2.
Since the speed ratio (or velocity ratio) of gear train is the ratio of the speed of the driver to
the speed of the driven or follower and ratio of speeds of any pair of gears in mesh is the inverse of
their number of teeth, therefore
Speed ratio 1 2
2 1
N T
N T
= =
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 2 1
1 2
N T
N T
= =
From above, we see that the train value is the reciprocal of speed ratio.
Sometimes, the distance between the two gears is large. The motion from one gear to another,
in such a case, may be transmitted by either of the following two methods :
1. By providing the large sized gear, or 2. By providing one or more intermediate gears.
A little consideration will show that the former method (i.e. providing large sized gears) is
very inconvenient and uneconomical method ; whereas the latter method (i.e. providing one or more
intermediate gear) is very convenient and economical.
It may be noted that when the number of intermediate gears are odd, the motion of both the
gears (i.e. driver and driven or follower) is like as shown in Fig. 13.1 (b).
But if the number of intermediate gears are even, the motion of the driven or follower will be
in the opposite direction of the driver as shown in Fig. 13.1 (c).
Now consider a simple train of gears with one intermediate gear as shown in Fig. 13.1 (b).
Let N1 = Speed of driver in r.p.m.,
N2 = Speed of intermediate gear in r.p.m.,

N3 = Speed of driven or follower in r.p.m.,
T1 = Number of teeth on driver,
T2 = Number of teeth on intermediate gear, and
T3 = Number of teeth on driven or follower.
Since the driving gear 1 is in mesh with the intermediate gear 2, therefore speed ratio for
these two gears is
1 2
2 1
=
N T
N T
...(i)
Similarly, as the intermediate gear 2 is in mesh with the driven gear 3, therefore speed ratio
for these two gears is
3
2
3 2
=
T
N
N T
...(ii)
The speed ratio of the gear train as shown in Fig. 13.1 (b) is obtained by multiplying the
equations (i) and (ii).
∴ 3
1 2 2
2 3 1 2
× = ×
T
N N T
N N T T
or 3
1
3 1
=
T
N
N T
i.e.
Speed of driver No. of teeth on driven
Speed ratio = =
Speed of driven No. of teeth on driver
and
Speed of driven No. of teeth on driver
Train value = =
Speed of driver No. of teeth on driven
Similarly, it can be proved that the
above equation holds good even if there are
any number of intermediate gears. From
above, we see that the speed ratio and the
train value, in a simple train of gears, is in-
dependent of the size and number of inter-
mediate 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 the following two pur-
poses :
1. To connect gears where a large
centre distance is required, and
2. To obtain the desired direction of
motion of the driven gear (i.e. clockwise or
anticlockwise).
13.4. Compound Gear Train
When there are more than one gear on a shaft, as shown in Fig. 13.2, it is called a compound
train of gear.
We have seen in Art. 13.3 that the idle gears, in a simple train of gears do not effect the speed
ratio of the system. But these gears are useful in bridging over the space between the driver and the
driven.
Gear trains inside a mechanical watch

But whenever the distance between the driver and the driven or follower has to be bridged
over by intermediate gears and at the same time a great ( or much less ) speed ratio is required, then
the advantage of intermediate gears is intensified by providing compound gears on intermediate shafts.
In this case, each intermediate shaft 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 the other with the driven or follower
attached to the next shaft as shown in Fig.13.2.
Fig. 13.2. Compound gear train.
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 ..., N6 = Speed of respective gears in r.p.m., and
T2 ,T3..., T6 = Number of teeth on respective gears.
Since gear 1 is in mesh with gear 2, therefore its speed ratio is
1 2
2 1
N T
N T
= ...(i)
Similarly, for gears 3 and 4, speed ratio is
3 4
4 3
N T
N T
= ...(ii)
and for gears 5 and 6, speed ratio is
5 6
6 5
N T
N T
= ...(iii)
The speed ratio of compound gear train is obtained by multiplying the equations (i), (ii) and (iii),
∴ 3 5 6
1 2 4
2 4 6 1 3 5
N N T
N T T
N N N T T T
× × = × × or
*
2 4 6
1
6 1 3 5
T T T
N
N T T T
× ×
=
× ×
* Since gears 2 and 3 are mounted on one shaft B, therefore N2 = N3. Similarly gears 4 and 5 are mounted on
shaft C, therefore N4 = N5.

i.e.
Speed of the first driver
Speed ratio =
Speed of the last driven or follower
Product of the number of teeth on the drivens
=
Product of the number of teeth on the drivers
and
Speed of the last driven or follower
Train value =
Speed of the first driver
Product of the number of teeth on the drivers
=
Product of the number of teeth on the drivens
The advantage of a compound train over a simple gear train is that a much larger speed
reduction from the first shaft to the last shaft can be obtained with small gears. If a simple gear train
is used to give a large speed reduction, the last gear has to be very large. Usually for a speed reduction
in excess of 7 to 1, a simple train is not used and a compound train or worm gearing is employed.
Note: 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.
Example 13.1. The gearing of a machine tool is shown
in Fig. 13.3. The motor shaft is connected to gear A and rotates
at 975 r.p.m. The gear wheels B, C, D and E are fixed to parallel
shafts rotating together. The final gear F is fixed on the output
shaft. What is the speed of gear F ? The number of teeth on
each gear are as given below :
Gear A B C D E F
No. of teeth 20 50 25 75 26 65
Solution. Given : NA = 975 r.p.m. ;
TA = 20 ; TB = 50 ; TC = 25 ; TD = 75 ; TE = 26 ;
TF = 65
From Fig. 13.3, we see that gears A, C
and E are drivers while the gears B, D and F are
driven or followers. Let the gear A rotates in
clockwise direction. Since the gears B and C are
mounted on the same shaft, therefore it is a
compound gear and the direction or rotation of
both these gears is same (i.e. anticlockwise).
Similarly, the gears D and E are mounted on the
same shaft, therefore it is also a compound gear
and the direction of rotation of both these gears
is same (i.e. clockwise). The gear Fwill rotate in
anticlockwise direction.
Let NF = Speed of gear F, i.e. last driven or follower.
We know that
Speed of the first driver Product of no. of teeth on drivens
=
Speed of the last driven Product of no. of teeth on drivers
Fig. 13.3
Battery Car: Even though it is run by batteries,
the power transmission, gears, clutches,
brakes, etc. remain mechanical in nature.
Note : This picture is given as additional information
and is not a direct example of the current chapter.

or A B D F
F A C E
50 75 65
18.75
20 25 26
N T T T
N T T T
× × × ×
= = =
× × × ×
∴ A
F
975
52 r. p. m.
18.75 18.75
N
N = = = Ans.
13.5. Design of Spur Gears
Sometimes, the spur gears (i.e. driver and driven) are to be designed for the given velocity
ratio and distance between the centres of their shafts.
Let x = Distance between the centres of two shafts,
N1 = Speed of the driver,
T1 = Number of teeth on the driver,
d1 = Pitch circle diameter of the driver,
N2 , T2 and d2 = Corresponding values for the driven or follower, and
pc = Circular pitch.
We know that the distance between the centres of two shafts,
1 2
2
d d
x
+
= ...(i)
and speed ratio or velocity ratio,
1 2 2
2 1 1
N d T
N d T
= = ...(ii)
From the above equations, we can conveniently find out the values of d1 and d2 (or T1 and T2)
and the circular pitch ( pc ). The values of T1 and T2, as obtained above, may or may not be whole
numbers. But in a gear since the number of its teeth is always a whole number, therefore a slight
alterations must be made in the values of x, d1 and d2, so that the number of teeth in the two gears may
be a complete number.
Example 13.2. Two parallel shafts, about 600 mm apart are to be connected by spur gears.
One shaft is to run at 360 r.p.m. and the other at 120 r.p.m. Design the gears, if the circular pitch is
to be 25 mm.
Solution. Given : x = 600 mm ; N1 = 360 r.p.m. ; N2 = 120 r.p.m. ; pc = 25 mm
Let d1 = Pitch circle diameter of the first gear, and
d2 = Pitch circle diameter of the second gear.
We know that speed ratio,
1 2
2 1
360
3
120
N d
N d
= = = or d2 = 3d1 ...(i)
and centre distance between the shafts (x),
1 2
1
600 ( )
2
d d
= + or d1 + d2 = 1200 ...(ii)
From equations (i) and (ii), we find that
d1 = 300 mm, and d2 = 900 mm
∴ Number of teeth on the first gear,
2
1
300
37.7
25
c
d
T
p
π π ×
= = =

and number of teeth on the second gear,
2
2
c
900
113.1
25
d
T
p
π π ×
= = =
Since the number of teeth on both the gears are to be in complete numbers, therefore let us
make the number of teeth on the first gear as 38. Therefore for a speed ratio of 3, the number of teeth
on the second gear should be 38 × 3 = 114.
Now the exact pitch circle diameter of the first gear,
1
1
38 25
302.36 mm
c
T p
d
× ×
′ = = =
π π
and the exact pitch circle diameter of the second gear,
2
2
114 25
907.1 mm
c
T p
d
× ×
′ = = =
π π
∴ Exact distance between the two shafts,
1 2 302.36 907.1
604.73 mm
2 2
d d
x
′ ′
+ +
′ = = =
Hence the number of teeth on the first and second gear must be 38 and 114 and their pitch
circle diameters must be 302.36 mm and 907.1 mm
respectively. The exact distance between the two shafts
must be 604.73 mm. Ans.
13.6. Reverted Gear Train
When the axes of the first gear (i.e. first driver)
and the last gear (i.e. last driven or follower) are co-axial,
then the gear train is known as reverted gear train as
shown in Fig. 13.4.
We see that gear 1 (i.e. first driver) drives the
gear 2 (i.e. first driven or follower) in the opposite direc-
tion. Since the gears 2 and 3 are mounted on the same
shaft, therefore they form a compound gear and the gear
3 will rotate in the same direction as that of gear 2. The
gear 3 (which is now the second driver) drives the gear 4
(i.e. the last driven or follower) in the same direction as
that of gear 1. Thus we see that in a reverted gear train,
the motion of the first gear and the last gear is like.
Let T1 = Number of teeth on gear 1,
r1 = Pitch circle radius of gear 1, and
N1 = Speed of gear 1 in r.p.m.
Similarly,
T2, T3, T4 = Number of teeth on respective gears,
r2, r3, r4 = Pitch circle radii of respective gears, and
N2, N3, N4 = Speed of respective gears in r.p.m.
Fig. 13.4. Reverted gear train.

Since the distance between the centres of the shafts of gears 1 and 2 as well as gears 3 and 4
is same, therefore
r1 + r2 = r3 + r4 ...(i)
Also, the circular pitch or module of all the gears is assumed to be same, therefore number of
teeth on each gear is directly proportional to its circumference or radius.
∴ *T1 + T2 = T3 + T4 ...(ii)
and
Product of number of teeth on drivens
Speed ratio =
Product of number of teeth on drivers
or 1 2 4
4 1 3
×
=
×
N T T
N T T
... (iii)
From equations (i), (ii) and (iii), we can determine the number of teeth on each gear for the
given centre distance, speed ratio and module only when
the number of teeth on one gear is chosen arbitrarily.
The reverted gear trains are used in automotive trans-
missions, lathe back gears, industrial speed reducers, and in
clocks (where the minute and hour hand shafts are co-axial).
Example 13.3. The speed ratio of the reverted gear
train, as shown in Fig. 13.5, is to be 12. The module pitch of
gears A and B is 3.125 mm and of gears C and D is 2.5 mm.
Calculate the suitable numbers of teeth for the gears. No
gear is to have less than 24 teeth.
Solution. Given : Speed ratio, NA/ND = 12 ;
mA = mB = 3.125 mm ; mC = mD = 2.5 mm
Let NA = Speed of gear A,
TA = Number of teeth on gear A,
rA = Pitch circle radius of gear A,
NB, NC , ND = Speed of respective gears,
TB, TC , TD = Number of teeth on respective gears, and
rB, rC , rD = Pitch circle radii of respective gears.
Fig. 13.5
* We know that circular pitch,
2
c
r
p m
T
π
= = π or
.
2
mT
r = , where m is the module.
∴
1
1
.
2
mT
r = ;
2
2
.
2
mT
r = ;
3
3
.
2
mT
r = ;
4
4
.
2
mT
r =
Now from equation (i),
3
1 2 4
.
. . .
2 2 2 2
mT
mT mT mT
+ = +
T1 + T2 = T3 + T4

Since the speed ratio between the gears A and B and between the gears C and D are to be
same, therefore
* C
A
B D
12 3.464
N
N
N N
= = =
Also the speed ratio of any pair of gears in mesh is the inverse of their number of teeth,
therefore
B D
A C
3.464
T T
T T
= = ...(i)
We know that the distance between the shafts
x = rA + rB = rC + rD = 200 mm
or C C
A A B B D D
.
. . .
200
2 2 2 2
m T
m T m T m T
+ = + =
.
...
2
m T
r
 
=
 
 
3
3.125 (TA + TB) = 2.5 (TC + TD) = 400 ...(∵ mA = mB, and mC = mD)
∴ TA + TB = 400 / 3.125 = 128 ...(ii)
and TC + TD = 400 / 2.5 = 160 ...(iii)
From equation (i), TB = 3.464 TA. Substituting this value of TB in equation (ii),
TA + 3.464 TA = 128 or TA = 128 / 4.464 = 28.67 say 28 Ans.
and TB = 128 – 28 = 100 Ans.
Again from equation (i), TD = 3.464 TC. Substituting this value of TD in equation (iii),
TC + 3.464 TC = 160 or TC = 160 / 4.464 = 35.84 say 36 Ans.
and TD = 160 – 36 = 124 Ans.
Note : The speed ratio of the reverted gear train with the calculated values of number of teeth on each gear is
A B D
D A C
100 124
12.3
28 36
N T T
N T T
× ×
= = =
× ×
13.7. Epicyclic Gear Train
We have already discussed that in an epicyclic gear train, the axes of the shafts, over which
the gears are mounted, may move relative to a fixed axis. A simple epicyclic gear train is shown in
Fig. 13.6, where a gear A and the arm C have a common axis at O1 about which they can rotate. The
gear B meshes with gear A and has its axis on the arm at O2, about which the gear B can rotate. If the
* We know that speed ratio A
D
v
Speed of first driver
12
Speed of last dri en
N
N
= = =
Also C
A A
D B D
N
N N
N N N
= × ...(NB = NC, being on the same shaft)
For
A
B
N
N and
C
D
N
N to be same, each speed ratio should be 12 so that
C
A A
D B D
12 12 12
N
N N
N N N
= × = × =

arm is fixed, the gear train is simple and gear A can drive gear B
or vice- versa, but if gear A is fixed and the arm is rotated about
the axis of gear A (i.e. O1), then the gear B is forced to rotate
upon and around gear A. Such a motion is called epicyclic and
the gear trains arranged in such a manner that one or more of
their members move upon and around another member are
known as epicyclic gear trains (epi. means upon and cyclic
means around). The epicyclic gear trains may be simple or com-
pound.
The epicyclic gear trains are useful for transmitting
high velocity ratios with gears of moderate size in a compara-
tively lesser space. The epicyclic gear trains are used in the
back gear of lathe, differential gears of the automobiles, hoists,
pulley blocks, wrist watches etc.
13.8. Velocity Ratioz of Epicyclic Gear Train
The following two methods may be used for finding out the velocity ratio of an epicyclic
gear train.
1. Tabular method, and 2. Algebraic method.
These methods are discussed, in detail, as follows :
1. Tabular method. Consider an epicyclic gear train as shown in Fig. 13.6.
Let TA = Number of teeth on gear A, and
TB = Number of teeth on gear B.
First of all, let us suppose that
the arm is fixed. Therefore the axes of
both the gears are also fixed relative to
each other. When the gear A makes one
revolution anticlockwise, the gear B will
make *TA / TB revolutions, clockwise.
Assuming the anticlockwise rotation as
positive and clockwise as negative, we
may say that when gear A makes + 1
revolution, then the gear B will make
(– TA / TB) revolutions. This statement
of relative motion is entered in the first
row of the table (see Table 13.1).
Secondly, if the gear A makes
+ x revolutions, then the gear B will
make – x × TA / TB revolutions. This
statement is entered in the second row
of the table. In other words, multiply
the each motion (entered in the first row) by x.
Thirdly, each element of an epicyclic train is given + y revolutions and entered in the third
row. Finally, the motion of each element of the gear train is added up and entered in the fourth row.
* We know that NB
/ NA
= TA
/ TB
. Since NA
= 1 revolution, therefore NB
= TA
/ TB
.
Fig. 13.6. Epicyclic gear train.
Inside view of a car engine.
Note : This picture is given as additional information and is not
a direct example of the current chapter.

Arm fixed-gear A rotates through + 1
revolution i.e. 1 rev. anticlockwise
Arm fixed-gear A rotates through + x
revolutions
Add + y revolutions to all elements
Total motion
Table 13.1. Table of motions
Revolutions of elements
Step No. Conditions of motion Arm C Gear A Gear B
1. 0 + 1
A
B
–
T
T
2. 0 + x
A
B
–
T
x
T
×
3. + y + y + y
4. + y x + y
A
B
–
T
y x
T
×
A little consideration will show that when two conditions about the motion of rotation of any
two elements are known, then the unknown speed of the third element may be obtained by substitut-
ing the given data in the third column of the fourth row.
2. Algebraic method. In this method, the motion of each element of the epicyclic train relative
to the arm is set down in the form of equations. The number of equations depends upon the number of
elements in the gear train. But the two conditions are, usually, supplied in any epicyclic train viz. some
element is fixed and the other has specified motion. These two conditions are sufficient to solve all the
equations ; and hence to determine the motion of any element in the epicyclic gear train.
Let the arm C be fixed in an epicyclic gear train as shown in Fig. 13.6. Therefore speed of the
gear A relative to the arm C
= NA – NC
and speed of the gear B relative to the arm C,
= NB – NC
Since the gears A and B are meshing directly, therefore they will revolve in oppositedirections.
∴ B C A
A C B
–
–
–
N N T
N N T
=
Since the arm C is fixed, therefore its speed, NC = 0.
∴ B A
A B
–
N T
N T
=
If the gear A is fixed, then NA = 0.
B C A
C B
–
–
0 –
N N T
N T
= or B A
C B
1
N T
N T
= +
Note : The tabular method is easier and hence mostly used in solving problems on epicyclic gear train.
Example 13.4. In an epicyclic gear train, an arm carries
two gears A and B having 36 and 45 teeth respectively. If the arm
rotatesat150r.p.m. in the anticlockwise direction about the centre
of the gear A which is fixed, determine the speed of gear B. If the
gear A instead of being fixed, makes 300 r.p.m. in the clockwise
direction, what will be the speed of gear B ?
Solution. Given : TA = 36 ; TB = 45 ; NC = 150 r.p.m.
(anticlockwise)
The gear train is shown in Fig. 13.7. Fig. 13.7

Arm fixed-gear A rotates through + 1
revolution (i.e. 1 rev. anticlockwise)
Arm fixed-gear A rotates through + x
revolutions
Total motion
We shall solve this example, first by tabular method and then by algebraic method.
1. Tabular method
First of all prepare the table of motions as given below :
Table 13.2. Table of motions.
Step No. Conditions of motion Arm C Gear A Gear B
1. 0 + 1
A
B
–
T
T
2. 0 + x
A
B
–
T
x
T
×
3. + y + y + y
4. + y x + y
A
B
–
T
y x
T
×
Speed of gear B when gear A is fixed
Since the speed of arm is 150 r.p.m. anticlockwise, therefore from the fourth row of the table,
y = + 150 r.p.m.
Also the gear A is fixed, therefore
x + y = 0 or x = – y = – 150 r.p.m.
∴ Speed of gear B, A
B
B
36
– 150 150 270 r.p.m.
45
T
N y x
T
= × = + × = +
= 270 r.p.m. (anticlockwise) Ans.
Speed of gear B when gear A makes 300 r.p.m. clockwise
Since the gear A makes 300 r.p.m.clockwise, therefore from the fourth row of the table,
x + y = – 300 or x = – 300 – y = – 300 – 150 = – 450 r.p.m.
∴ Speed of gear B,
A
B
B
36
– 150 450 510 r.p.m.
45
T
N y x
T
= × = + × = +
2. Algebraic method
Let NA = Speed of gear A.
NB = Speed of gear B, and
NC = Speed of arm C.
Assuming the arm C to be fixed, speed of gear A relative to arm C
= NA – NC
and speed of gear B relative to arm C = NB – NC

Since the gears A and B revolve in opposite directions, therefore
B C A
A C B
–
–
–
N N T
N N T
= ...(i)
Speed of gear B when gear A is fixed
When gear A is fixed, the arm rotates at 150 r.p.m. in the anticlockwise direction, i.e.
NA = 0, and NC = + 150 r.p.m.
∴ B – 150 36
– – 0.8
0 – 150 45
N
= = ...[From equation (i)]
or NB = – 150 × – 0.8 + 150 = 120 + 150 = 270 r.p.m. Ans.
Speed of gear B when gear A makes 300 r.p.m. clockwise
Since the gear A makes 300 r.p.m. clockwise, therefore
NA = – 300 r.p.m.
∴ B – 150 36
– – 0.8
–300 – 150 45
N
= =
or NB = – 450 × – 0.8 + 150 = 360 + 150 = 510 r.p.m. Ans.
Example 13.5. In a reverted epicyclic gear
train, the arm A carries two gears B and C and a
compound gear D - E. The gear B meshes with gear E
and the gear C meshes with gear D. The number of teeth
on gears B, C and D are 75, 30 and 90 respectively.
Find the speed and direction of gear C when gear B is
fixed and the arm A makes 100 r.p.m. clockwise.
Solution. Given : TB = 75 ; TC = 30 ; TD = 90 ;
NA = 100 r.p.m. (clockwise)
The reverted epicyclic gear train is
shown in Fig. 13.8. First of all, let us find the
number of teeth on gear E (TE). Let dB , dC , dD
and dE be the pitch circle diameters of gears B,
C, D and E respectively. From the geometry of
the figure,
dB + dE = dC + dD
Since the number of teeth on each gear,
for the same module, are proportional to their
pitch circle diameters, therefore
TB + TE = TC + TD
∴ TE = TC + TD – TB = 30 + 90 – 75 = 45
The table of motions is drawn as
follows :
Fig. 13.8
A gear-cutting machine is used to cut gears.
Note : This picture is given as additional information
and is not a direct example of the current chapter.

Arm fixed-compound gear D-E
rotated through + 1 revolution ( i.e.
1 rev. anticlockwise)
Arm fixed-compound gear D-E
rotated through + x revolutions
Total motion
Step Conditions of motion Arm A Compound Gear B Gear C
No. gear D-E
1. 0 + 1
E
B
–
T
T
D
C
–
T
T
2. 0 + x
E
B
–
T
x
T
× D
C
–
T
x
T
×
3. + y + y + y + y
4. + y x + y
E
B
–
T
y x
T
× D
C
–
T
y x
T
×
Since the gear B is fixed, therefore from the fourth row of the table,
E
B
– 0
T
y x
T
× = or
45
– 0
75
y x × =
∴ y – 0.6 = 0 ...(i)
Also the arm A makes 100 r.p.m. clockwise, therefore
y = – 100 ...(ii)
Substituting y = – 100 in equation (i), we get
– 100 – 0.6 x = 0 or x = – 100 / 0.6 = – 166.67
Model of sun and planet gears.
INPUT
Spline to Accept
Motor Shaft
Housing OD Designed to meet
RAM Bore Dia, and Share Motor
Coolant Supply
OUTPUT- External Spline to
Spindle
Ratio Detection Switches
Hydraulic or Pneumatic Speed
Change Actuator
Round Housing With O-ring
Seated Cooling Jacket
Motor Flange
Hollow Through Bore for
Drawbar Integration

From the fourth row of the table, speed of gear C,
D
C
C
90
– – 100 166.67 400 r.p.m.
30
T
N y x
T
= × = + × = +
13.9. Compound Epicyclic Gear Train—Sun and Planet Gear
A compound epicyclic
gear train is shown in Fig. 13.9.
It consists of two co-axial shafts
S1 andS2,anannulusgearA which
is fixed, the compound gear (or
planet gear) B-C, the sun gear D
and the arm H. The annulus gear
has internal teeth and the com-
pound gear is carried by the arm
and revolves freely on a pin of the
arm H. The sun gear is co-axial
with the annulus gear and the arm
but independent of them.
The annulus gear A
meshes with the gear B and the
sun gear D meshes with the gear
C. It may be noted that when the
annulus gear is fixed, the sun gear
provides the drive and when the
sun gear is fixed, the annulus gear
provides the drive. In both cases, the arm acts as a follower.
Note : The gear at the centre is called the sun gear and the gears whose axes move are called planet gears.
Fig. 13.9. Compound epicyclic gear train.
Sun and Planet gears.
Speed Change
Shift Axis
Bearing Housing
Output Belt Pulley
Slide Dog
Clutch
Output Sun
Gear
Motor
Flange
Input Sun
Gear
Planet
Gears
Oil
Collector

Arm fixed-gear C rotates through
+ 1 revolution (i.e. 1 rev.
anticlockwise)
Arm fixed-gear C rotates through
+ x revolutions
Add + y revolutions to all
elements
Total motion
Arm fixed-gear D rotates
through + 1 revolution
Arm fixed-gear D rotates
through + x revolutions
elements
Total motion
Let TA , TB , TC , and TD be the teeth and NA, NB, NC and ND be the speeds for the gears A, B,
C and D respectively. A little consideration will show that when the arm is fixed and the sun gear D is
turned anticlockwise, then the compound gear B-C and the annulus gear A will rotate in the clockwise
direction.
The motion of rotations of the various elements are shown in the table below.
Step Conditions of motion Arm Gear D Compound gear Gear A
No. B-C
1. 0 + 1
D
C
–
T
T
D B
C A
–
T T
T T
×
2. 0 + x
D
C
–
T
x
T
× D B
C A
–
T T
x
T T
× ×
3. + y + y + y + y
4. + y x + y
D
C
–
T
y x
T
× D B
C A
–
T T
y x
T T
× ×
Note : If the annulus gear A is rotated through one revolution anticlockwise with the arm fixed, then the
compound gear rotates through TA / TB revolutions in the same sense and the sun gear D rotates through
TA / TB × TC / TD revolutions in clockwise direction.
Example 13.6. An epicyclic gear consists of three gears A, B and C as shown in Fig. 13.10.
The gear A has 72 internal teeth and gear C has 32 external teeth. The gear B meshes with both A
and C and is carried on an arm EF which rotates about the centre of A at 18 r.p.m.. If the gear A is
fixed, determine the speed of gears B and C.
Solution. Given : TA = 72 ; TC = 32 ; Speed of arm EF= 18 r.p.m.
Considering the relative motion of rotation as shown in Table 13.5.
Step No. Conditions of motion Arm EF Gear C Gear B Gear A
1. 0 + 1
C
B
–
T
T
C B C
B A A
– –
T T T
T T T
× =
2. 0 + x
C
B
–
T
x
T
× C
A
–
T
x
T
×
3. + y + y + y + y
4. + y x + y
C
B
–
T
y x
T
× C
A
–
T
y x
T
×

Speed of gear C
We know that the speed of the arm is 18 r.p.m. therefore,
y = 18 r.p.m.
and the gear A is fixed, therefore
C
A
– 0
T
y x
T
× = or 32
18 – 0
72
x × =
∴ x = 18 × 72 / 32 = 40.5
∴ Speed of gear C = x + y = 40.5 + 18
= + 58.5 r.p.m.
= 58.5 r.p.m. in the direction
of arm. Ans.
Speed of gear B
Let dA, dB and dC be the pitch circle diameters of gears
A, B and C respectively. Therefore, from the geometry of Fig. 13.10,
C A
B
2 2
d d
d + = or 2 dB + dC = dA
Since the number of teeth are proportional to their pitch circle diameters, therefore
2 TB + TC = TA or 2 TB + 32 = 72 or TB = 20
∴ Speed of gear B C
B
32
– 18 – 40.5 – 46.8 r.p.m.
20
T
y x
T
= × = × =
= 46.8 r.p.m. in the opposite direction of arm. Ans.
Example 13.7. An epicyclic train of gears is arranged as shown in
Fig.13.11. How many revolutions does the arm, to which the pinions B and
C are attached, make :
1. when A makes one revolution clockwise and D makes half a
revolution anticlockwise, and
2. when A makes one revolution clockwise and D is stationary ?
The number of teeth on the gears A and D are 40 and 90
respectively.
Solution. Given : TA = 40 ; TD = 90
First of all, let us find the number of teeth on gears B and C (i.e. TB and TC). Let dA, dB, dC
and dD be the pitch circle diameters of gears A, B, C and D respectively. Therefore from the geometry
of the figure,
dA + dB + dC = dD or dA + 2 dB = dD ...(3 dB = dC)
Since the number of teeth are proportional to their pitch circle diameters, therefore,
TA + 2 TB = TD or 40 + 2 TB = 90
∴ TB = 25, and TC = 25 ...(3 TB = TC)
Fig. 13.10
Fig. 13.11

The table of motions is given below :
Step No. Conditions of motion Arm Gear A Compound Gear D
gear B-C
1. 0 – 1
A
B
T
T
+ A B A
B D D
T T T
T T T
+ × = +
2. 0 – x
A
B
T
x
T
+ × A
D
T
x
T
+ ×
3. – y – y – y – y
4. – y – x – y
A
B
–
T
x y
T
× A
D
–
T
x y
T
×
1. Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise
Since the gear A makes 1 revolution clockwise, therefore from the fourth row of the table,
– x – y = –1 or x + y = 1 ...(i)
Also, the gear D makes half revolution anticlockwise, therefore
A
D
1
–
2
T
x y
T
× = or 40 1
–
90 2
x y
× =
∴ 40 x – 90 y = 45 or x – 2.25 y = 1.125 ...(ii)
From equations (i) and (ii), x = 1.04 and y = – 0.04
∴ Speed of arm = – y = – (– 0.04) = + 0.04
= 0.04 revolution anticlockwise Ans.
2. Speed of arm when A makes 1 revolution clockwise and D is stationary
Since the gear A makes 1 revolution clockwise, therefore from the fourth row of the
table,
– x – y = – 1 or x + y = 1 ...(iii)
Also the gear D is stationary, therefore
A
D
– 0
T
x y
T
× = or
40
– 0
90
x y
× =
∴ 40 x – 90 y = 0 or x – 2.25 y = 0 ...(iv)
From equations (iii) and (iv),
x = 0.692 and y = 0.308
∴ Speed of arm = – y = – 0.308 = 0.308 revolution clockwise Ans.
Arm fixed, gear A rotates
through – 1 revolution (i.e. 1
rev. clockwise)
Arm fixed, gear A rotates
through – x revolutions
Add – y revolutions to all
elements
Total motion

Example 13.8. In an epicyclic gear train, the internal wheels A and B and compound wheels
C and D rotate independently about axis O. The wheels E and F rotate on pins fixed to the arm G. E
gears with A and C and F gears with B and D. All the wheels have
the same module and the number of teeth are : TC = 28; TD = 26;
TE = TF = 18.
1. Sketch the arrangement ; 2. Find the number of teeth on
A and B ; 3. If the arm G makes 100 r.p.m. clockwise and A is fixed,
find the speed of B ; and 4. If the arm G makes 100 r.p.m. clockwise
and wheel A makes 10 r.p.m. counter clockwise ; find the speed of
wheel B.
Solution. Given : TC = 28 ; TD = 26 ; TE = TF = 18
1. Sketch the arrangement
The arrangement is shown in Fig. 13.12.
2. Number of teeth on wheels A and B
Let TA = Number of teeth on wheel A, and
TB = Number of teeth on wheel B.
If dA , dB , dC , dD , dE and dF are the pitch circle diameters of wheels A, B, C, D, E and F
respectively, then from the geometry of Fig. 13.12,
dA = dC + 2 dE
and dB = dD + 2 dF
Since the number of teeth are proportional to their pitch circle diameters, for the same
module, therefore
TA = TC + 2 TE = 28 + 2 × 18 = 64 Ans.
and TB = TD + 2 TF = 26 + 2 × 18 = 62 Ans.
3. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A is fixed
First of all, the table of motions is drawn as given below :
Step Conditions of Arm Wheel Wheel Compound Wheel F Wheel B
No. motion G A E wheel C-D
1. 0 + 1
A
E
T
T
+ A E
E C
–
T T
T T
× A D
C F
T T
T T
+ × A D F
C F B
T T T
T T T
+ × ×
2. 0 + x
A
E
T
x
T
+ × A
C
–
T
x
T
× A D
C F
T T
x
T T
+ × × A D
C B
T T
x
T T
+ × ×
3. + y + y + y + y + y + y
4. + y x + y
A
E
T
y x
T
+ × A
C
–
T
y x
T
× A D
C F
T T
y x
T T
+ × × A D
C B
T T
y x
T T
+ × ×
Fig. 13.12
Arm fixed- wheel A
rotates through + 1
revolution (i.e. 1 rev.
anticlockwise)
Arm fixed-wheel A
rotates through + x
revolutions
Add + y revolutions
to all elements
Total motion
A
C
–
T
T
= A D
C B
T T
T T
= + ×

Since the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the table,
y = – 100 ...(i)
Also, the wheel A is fixed, therefore from the fourth row of the table,
x + y = 0 or x = – y = 100 ...(ii)
∴ Speed of wheel A D
C B
64 26
– 100 100 – 100 95.8 r.p.m.
28 62
T T
B y x
T T
= + × × = + × × = +
= – 4.2 r.p.m. = 4.2 r.p.m. clockwise Ans.
4. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A makes 10 r.p.m. counter
clockwise
Since the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the
table
y = – 100 ...(iii)
Also the wheel A makes 10 r.p.m. counter clockwise, therefore from the fourth row of the
table,
x + y = 10 or x = 10 – y = 10 + 100 = 110 ...(iv)
∴ Speed of wheel A D
C B
64 26
– 100 110 – 100 105.4 r.p.m.
28 62
T T
B y x
T T
= + × × = + × × = +
= + 5.4 r.p.m. = 5.4 r.p.m. counter clockwise Ans.
Example 13.9. In an epicyclic gear of the ‘sun and planet’ type shown
in Fig. 13.13, the pitch circle diameter of the internally toothed ring is to be
224 mm and the module 4 mm. When the ring D is stationary, the spider A,
which carries three planet wheels C of equal size, is to make one revolution in
the same sense as the sunwheel B for every five revolutions of the driving
spindle carrying the sunwheel B. Determine suitable numbers of teeth for all
the wheels.
Solution. Given : dD = 224 mm ; m = 4 mm ; NA = NB / 5
Let TB , TC and TD be the number of teeth on the sun wheel B,
planet wheels C and the internally toothed ring D. The table of motions is given below :
Step No. Conditions of motion Spider A Sun wheel B Planet wheel C Internal gear D
1. 0 + 1
B
C
–
T
T
B C B
C D D
– –
T T T
T T T
× =
2. 0 + x
B
C
–
T
x
T
× B
D
–
T
x
T
×
3. + y + y + y + y
4. + y x + y
B
C
–
T
y x
T
× B
D
–
T
y x
T
×
Fig. 13.13
Spider A fixed, sun wheel
B rotates through + 1
revolution (i.e. 1 rev.
anticlockwise)
Spider A fixed, sun wheel
B rotates through + x
revolutions
elements
Total motion

We know that when the sun
wheel B makes + 5 revolutions, the spi-
der A makes + 1 revolution. Therefore
from the fourth row of the table,
y = + 1 ; and x + y = + 5
∴ x = 5 – y = 5 – 1 = 4
Since the internally toothed ring
D is stationary, therefore from the fourth
row of the table,
B
D
– 0
T
y x
T
× =
or B
D
1 – 4 0
T
T
× =
∴ B
D
1
4
T
T
= or TD = 4 TB ...(i)
We know that TD = dD / m = 224 / 4 = 56 Ans.
∴ TB = TD / 4 = 56 / 4 = 14 Ans. ...[From equation (i)]
Let dB, dC and dD be the pitch circle diameters of sun wheel B, planet wheels C and internally
toothed ring D respectively. Assuming the pitch of all the gears to be same, therefore from the geom-
etry of Fig. 13.13,
dB + 2 dC = dD
Since the number of teeth are proportional to their pitch circle diameters, therefore
TB + 2 TC = TD or 14 + 2 TC = 56
∴ TC = 21 Ans.
Example 13.10. Two shafts A and B are co-axial. A gear C (50 teeth) is rigidly mounted
on shaft A. A compound gear D-E gears with C and an internal gear G. D has 20 teeth and gears
with C and E has 35 teeth and gears with an internal gear G. The gear G is fixed and is concen-
tric with the shaft axis. The compound gear D-E is mounted on a pin which projects from an arm
keyed to the shaft B. Sketch the arrangement and find the number of teeth on internal gear G
assuming that all gears have the same module. If the shaft A rotates at 110 r.p.m., find the speed
of shaft B.
Solution. Given : TC = 50 ; TD = 20 ; TE = 35 ; NA = 110 r.p.m.
Number of teeth on internal gear G
Let dC , dD , dE and dG be the pitch circle diameters of gears C, D, E and G respectively. From
the geometry of the figure,
G C D E
2 2 2 2
d d d d
= + +
or dG = dC + dD + dE
Power transmission in a helicopter is essentially through
gear trains.
Note : This picture is given as additional information and is not a
direct example of the current chapter.
Main rotor Tail rotor
Tail boom
Landing skids Engine, transmis-
sion fuel, etc.
Cockpit
Drive shaft

Let TC , TD , TE and TG be the number of teeth on gears C, D, E and G respectively. Since all
the gears have the same module, therefore number of teeth are proportional to their pitch circle
diameters.
∴ TG = TC + TD + TE = 50 + 20 + 35 = 105 Ans.
Fig. 13.14
Speed of shaft B
Step Conditions of motion Arm Gear C (or Compound Gear G
No. shaft A) gear D-E
1. 0 + 1
C
D
–
T
T
C E
D G
–
T T
T T
×
2. 0 + x
C
D
–
T
x
T
× C E
D G
–
T T
x
T T
× ×
3. + y + y + y + y
4. + y x + y
C
D
–
T
y x
T
× C E
D G
–
T T
y x
T T
× ×
Since the gear G is fixed, therefore from the fourth row of the table,
C E
D G
– 0
T T
y x
T T
× × = or 50 35
– 0
20 105
y x × × =
∴
5
– 0
6
y x = ...(i)
Arm fixed - gear C rotates through + 1
revolution
Arm fixed - gear C rotates through + x
revolutions
Total motion

Since the gear C is rigidly mounted on shaft A, therefore speed of gear C and shaft A is same.
We know that speed of shaft A is 110 r.p.m., therefore from the fourth row of the table,
x + y = 100 ...(ii)
From equations (i) and (ii), x = 60, and y = 50
∴ Speed of shaft B = Speed of arm = + y = 50 r.p.m. anticlockwise Ans.
Example13.11. Fig. 13.15 shows diagrammatically a compound
epicyclic gear train. Wheels A , D and E are free to rotate independently
on spindle O, while B and C are compound and rotate together on spindle
P, on the end of arm OP. All the teeth on different wheels have the same
module. A has 12 teeth, B has 30 teeth and C has 14 teeth cut externally.
Find the number of teeth on wheels D and E which are cut internally.
If the wheel A is driven clockwise at 1 r.p.s. while D is driven
counter clockwise at 5 r.p.s., determine the magnitude and direction of
the angular velocities of arm OP and wheel E.
Solution. Given : TA = 12 ; TB = 30 ;TC = 14 ; NA = 1 r.p.s. ; ND = 5 r.p.s.
Number of teeth on wheels D and E
Let TD and TE be the number of teeth on wheels D and E respectively. Let dA , dB , dC , dD and dE
be the pitch circle diameters of wheels A, B, C, D and E respectively. From the geometry of the figure,
dE = dA + 2dB and dD = dE – (dB – dC)
Since the number of teeth are proportional to their pitch circle diameters for the same module,
therefore
TE = TA + 2TB = 12 + 2 × 30 = 72 Ans.
and TD = TE – (TB – TC) = 72 – (30 – 14) = 56 Ans.
Magnitude and direction of angular velocities of arm OP and wheel E
The table of motions is drawn as follows :
Step Conditions of motion Arm Wheel A Compound Wheel D Wheel E
No. wheel B-C
1. 0 – 1
A
B
T
T
+ A C
B D
T T
T T
+ × A B
B E
T T
T T
+ ×
A
E
T
T
= +
2. 0 – x
A
B
T
x
T
+ × A C
B D
T T
x
T T
+ × × A
E
T
x
T
+ ×
3. – y – y – y – y – y
4. – y – x – y
A
B
–
T
x y
T
× A C
B D
–
T T
x y
T T
× × A
E
–
T
x y
T
×
Fig. 13.15
Arm fixed A rotated through
– 1 revolution (i.e. 1 revolu-
tion clockwise)
Arm fixed-wheel A rotated
through – x revolutions
Add – y revolutions to all ele-
ments
Total motion

Since the wheel A makes 1 r.p.s. clockwise, therefore from the fourth row of the table,
– x – y = – 1 or x + y = 1 ...(i)
Also, the wheel D makes 5 r.p.s. counter clockwise, therefore
C
A
B D
– 5
T
T
x y
T T
× × = or 12 14
– 5
30 56
x y
× × =
∴ 0.1 x – y = 5 ...(ii)
From equations (i) and (ii),
x = 5.45 and y = – 4.45
∴ Angular velocity of arm OP
= – y = –(– 4.45) = 4.45 r.p.s
= 4.45 × 2 π = 27.964 rad/s (counter clockwise) Ans.
and angular velocity of wheel A
E
12
– 5.45 – (– 4.45) 5.36 r.p.s.
72
T
E x y
T
= × = × =
= 5.36 × 2 π = 33.68 rad/s (counter clockwise) Ans.
Example 13.12. An internal wheel B with 80 teeth is keyed to a shaft F. A fixed internal
wheel C with 82 teeth is concentric
with B. A compound wheel D-E
gears with the two internal wheels;
D has 28 teeth and gears with C
while E gears with B. The compound
wheels revolve freely on a pin which
projects from a disc keyed to a shaft
A co-axial with F. If the wheels have
the same pitch and the shaft A makes
800 r.p.m., what is the speed of the
shaft F ? Sketch the arrangement.
Solution. Given : TB = 80 ; TC
= 82 ; TD = 28 ; NA = 500 r.p.m.
Fig. 13.16
First of all, let us find out the number of teeth on wheel E (TE). Let dB , dC , dD and dE be the
pitch circle diameter of wheels B, C, D and E respectively. From the geometry of the figure,
dB = dC – (dD – dE )
Helicopter
Note : This picture is given as additional information and is not a
direct example of the current chapter.

or dE = dB + dD – dC
Since the number of teeth are proportional to their pitch circle diameters for the same pitch,
therefore
TE = TB + TD – TC = 80 + 28 – 82 = 26
Step Conditions of motion Arm (or Wheel B (or Compound Wheel C
No. shaft A) shaft F) gear D-E
1. 0 + 1
B
E
T
T
+ B D
E C
T T
T T
+ ×
2. 0 + x
B
E
T
x
T
+ × B D
E C
T T
x
T T
+ × ×
3. + y + y + y + y
4. + y x + y
B
E
T
y x
T
+ × B D
E C
T T
y x
T T
+ × ×
Since the wheel C is fixed, therefore from the fourth row of the table,
B D
E C
0
T T
y x
T T
+ × × = or
80 28
0
26 82
y x
+ × × =
∴ y + 1.05 x = 0 ...(i)
Also, the shaft A (or the arm) makes 800 r.p.m., therefore from the fourth row of the table,
y = 800 ...(ii)
From equations (i) and (ii),
x = – 762
∴ Speed of shaft F = Speed of wheel B = x + y = – 762 + 800 = + 38 r.p.m.
Example 13.13. Fig. 13.17 shows an epicyclic gear
train known as Ferguson’s paradox. Gear A is fixed to the
frame and is, therefore, stationary. The arm B and gears C
and D are free to rotate on the shaft S. Gears A, C and D have
100, 101 and 99 teeth respectively. The planet gear has 20
teeth. The pitch circle diameters of all are the same so that the
planet gear P meshes with all of them. Determine the
revolutions of gears C and D for one revolution of the arm B.
Solution. Given : TA = 100 ; TC = 101 ; TD = 99 ;
TP = 20 Fig. 13.17
Arm fixed - wheel B rotated
through + 1 revolution (i.e. 1
revolution anticlockwise)
Arm fixed - wheel B rotated
through + x revolutions
elements
Total motion

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Gear Trains-1-25.pdf