INTRODUCTION OF GEARS AND GEAR KINEMATICS

PROF. K N WAKCHAURE
SRES COLLEGE OF ENGINEERING,
KOPARGAON
INTRODUCTION OF GEARS AND
GEAR KINEMATICS

Power transmission is the movement of energy from its place of
generation to a location where it is applied to performing useful
work
A gear is a component within a transmission device that
transmits rotational force to another gear or device.
2PROF. K N WAKCHAURE

 The materials used for the manufacture of gears depends upon the strength and service
conditions like wear, noise etc.
 The gears may be manufactured from metallic or non-metallic materials. The metallic gears
with cut teeth are commercially obtainable in cast iron, steel and bronze.
 The non-metallic materials like wood, rawhide, compressed paper and synthetic resins like
nylon are used for gears, especially for reducing noise.
 The cast iron is widely used for the manufacture of gears due to its good wearing properties,
excellent machinability and ease of producing complicated shapes by casting method. The
cast iron gears with cut teeth may be employed, where smooth action is not important.
 The steel is used for high strength gears and steel may be plain carbon steel or alloy steel.
The steel gears are usually heat treated in order to combine properly the toughness and
tooth hardness.
 The phosphor bronze is widely used for worms gears in order to reduce wear of the worms
which will be excessive with cast iron or steel.

1. According to the position of axes of the shafts.
a. Parallel
1.Spur Gear
2.Helical Gear
3.Rack and Pinion
b. Intersecting
Bevel Gear
c. Non-intersecting and Non-parallel
worm and worm gears

 Teeth is parallel to axis of rotation
 Transmit power from one shaft to
another parallel shaft
 Used in Electric screwdriver, oscillating
sprinkler, windup alarm clock, washing
machine and clothes dryer

 Advantages
 They offer constant velocity ratio
 Spur gears are highly reliable
 Spur gears are simplest, hence easiest to design and manufacture
 A spur gear is more efficient if you compare it with helical gear of same size
 Spur gear teeth are parallel to its axis. Hence, spur gear train does not produce axial thrust. So the gear
shafts can be mounted easily using ball bearings.
 They can be used to transmit large amount of power (of the order of 50,000 kW)
 Disadvantages
 Spur gear are slow-speed gears
 Gear teeth experience a large amount of stress
 They cannot transfer power between non-parallel shafts
 They cannot be used for long distance power transmission.
 Spur gears produce a lot of noise when operating at high speeds.
 when compared with other types of gears, they are not as strong as them

 The teeth on helical gears are cut at an angle to the face of the
gear
 This gradual engagement makes helical gears operate much
more smoothly and quietly than spur gears
 One interesting thing about helical gears is that if the angles of
the gear teeth are correct, they can be mounted on
perpendicular shafts, adjusting the rotation angle by 90 degrees

 Advantages
 The angled teeth engage more gradually than do spur gear teeth causing them to run more
smoothly and quietly
 Helical gears are highly durable and are ideal for high load applications.
 At any given time their load is distributed over several teeth, resulting in less wear
 Can transmit motion and power between either parallel or right angle shafts
 Disadvantages
 •An obvious disadvantage of the helical gears is a resultant thrust along the axis of the gear,
which needs to be accommodated by appropriate thrust bearings, and a greater degree of sliding
friction between the meshing teeth, often addressed with additives in the lubricant.
 Thus we can say that helical gears cause losses due to the unique geometry along the axis of the
helical gear’s shaft.
 •Efficiency of helical gear is less because helical gear trains have sliding contacts between the
teeth which in turns produce axial thrust of gear shafts and generate more heat. So, more power
loss and less efficiency

To avoid axial thrust, two
helical gears of opposite hand
can be mounted side by side,
to cancel resulting thrust
forces
Herringbone gears are mostly
used on heavy machinery.

 Rack and pinion gears are used to convert rotation (From the
pinion) into linear motion (of the rack)
 A perfect example of this is the steering system on many cars

 Advantages
 Cheap
 Compact
 Robust
 Easiest way to convert rotation
motion into linear motion
 Rack and pinion gives easier and
more compact control over the
vehicle
 Disadvantages
 Since being the most ancient, the wheel is also the most
convenient and somewhat more extensive in terms of
energy too. Due to the apparent friction, you would
already have guessed just how much of the power being
input gives in terms of output, a lot of the force applied
to the mechanism is burned up in overcoming friction, to
be more precise somewhat around 80% of the overall
force is burned to overcome one.
 The rack and pinion can only work with certain levels of
friction. Too high a friction and the mechanism will be
subject to wear more than usual and will require more
force to operate.
 The most adverse disadvantage of rack and pinion would
also be due to the inherent friction, the same force that
actually makes things work in the mechanism. Due to
the friction, it is under a constant wear, possibly needing
replacement after a certain time

 Bevel gears are useful when the direction of a shaft's rotation needs to
be changed
 They are usually mounted on shafts that are 90 degrees apart, but can
be designed to work at other angles as well
 The teeth on bevel gears can be straight, spiral or hypoid
 locomotives, marine applications, automobiles, printing presses,
cooling towers, power plants, steel plants, railway track inspection
machines, etc.

 Advantages
 Worm gear drives operate silently and smoothly.
 They are self-locking.
 They occupy less space.
 They have good meshing effectiveness.
 They can be used for reducing speed and increasing torque.
 High velocity ratio of the order of 100 can be obtained in a single step
 Disadvantages
 Worm gear materials are expensive.
 Worm drives have high power losses
 A disadvantage is the potential for considerable sliding action, leading to low efficiency
 They produce a lot of heat.

 Worm gears are used when large gear reductions are needed. It is common for
worm gears to have reductions of 20:1, and even up to 300:1 or greater
 Many worm gears have an interesting property that no other gear set has: the worm
can easily turn the gear, but the gear cannot turn the worm
 Worm gears are used widely in material handling and transportation machinery,
machine tools, automobiles etc

 Pitch circle. It is an imaginary circle which by pure rolling action would give the
same motion as the actual gear.
 Pitch circle diameter. It is the diameter of the pitch circle. The size of the gear is
usually specified by the pitch circle diameter. It is also known as pitch diameter.
 Pitch point. It is a common point of contact between two pitch circles.
 Pitch surface. It is the surface of the rolling discs which the meshing gears have
replaced at the pitch circle.
 Pressure angle or angle of obliquity. It is the angle between the common normal
to two gear teeth at the point of contact and the common tangent at the pitch
point. It is usually denoted by φ. The standard pressure angles are 14 1/2 ° and
20°.

 Addendum. It is the radial distance of a tooth from the pitch circle to the top of
the tooth.
 Dedendum. It is the radial distance of a tooth from the pitch circle to the bottom
of the tooth.
 Addendum circle. It is the circle drawn through the top of the teeth and is
concentric with the pitch circle.
 Dedendum circle. It is the circle drawn through the bottom of the teeth. It is also
called root circle.
Note : Root circle diameter =
Pitch circle diameter × cos φ ,
where φ is the pressure angle.

Circular pitch. It is the distance measured on the
circumference of the pitch circle from a point of one tooth to
the corresponding point on the next tooth. It is usually
denoted by Pc ,Mathematically,
 A little consideration will show that the two gears will mesh
together correctly, if the two wheels have the same circular
pitch.
Note : If D1 and D2 are the diameters of the two meshing gears
having the teeth T1 and T2 respectively, then for them to
mesh correctly,

Diametral pitch. It is the ratio of number of teeth to the pitch
circle diameter in millimetres. It is denoted by pd.
Mathematically,
Module. It is the ratio of the pitch circle diameter in
millimeters to the number of teeth. It is usually denoted by m.
Mathematically,
Clearance. It is the radial distance from the top of the tooth to
the bottom of the tooth, in a meshing gear. A circle passing
through the top of the meshing gear is known as clearance
circle.
Total depth. It is the radial distance between the addendum
and the dedendum circles of a gear. It is equal to the sum of

 Working depth. It is the radial distance from the addendum circle to the
clearance circle. It is equal to the sum of the addendum of the two meshing
gears.
 Tooth thickness. It is the width of the tooth measured along the pitch circle.
 Tooth space . It is the width of space between the two adjacent teeth measured
along the pitch circle.
 Backlash. It is the difference between the tooth space and the tooth thickness, as
measured along the pitch circle. Theoretically, the backlash should be zero, but
in actual practice some backlash must be allowed to prevent jamming of the
teeth due to tooth errors and thermal expansion.

 Face of tooth. It is the surface of the gear tooth above the pitch surface.
 Flank of tooth. It is the surface of the gear tooth below the pitch surface.
 Top land. It is the surface of the top of the tooth.
 Face width. It is the width of the gear tooth measured parallel to its axis.
 Profile. It is the curve formed by the face and flank of the tooth.
 Fillet radius. It is the radius that connects the root circle to the profile of the
tooth.

 The common normal at the point
of contact between a pair of teeth
must always pass through the
pitch point.
 The angular velocity ratio
between two gears of a gear set
must remain constant
throughout the mesh.

N
O2
O1
M
K
P
L
G H
ⱷ
Ɵ
ⱷ
ⱷ
rb
r
ra
RA
Rb

PATH OF
CONTACT
 The length of path of contact is the
length of common normal cut off by
the addendum circles of the wheel
and the pinion.
 Thus the length of path of contact is
KL which is the sum of the parts of
the path of contacts KP and PL.
 The part of the path of contact KP is
known as path of approach and the
part of the path of contact PL is
known as path of recess.
Path of
contact

PATH OF
CONTACT radius of the base circle of pinion,
 radius of the base circle of wheel,
 Now from right angled triangle O2KN,
Hence the path of approach
N
O2
O1
M
K
P
L
G H
ⱷ
Ɵ
ⱷ
ⱷ
rb
r
ra
RA
Rb

PATH OF
CONTACT radius of the base circle of pinion,
 radius of the base circle of wheel,
 Similarly from right angled triangle
O1ML
Hence the path of recess
N
O2
O1
M
K
P
L
G H
ⱷ
Ɵ
ⱷ
ⱷ
rb
r
ra
RA
Rb

PATH OF
CONTACT
the path of recess
path of approach
Length of the path of contact,
N
O2
O1
M
K
P
L
G H
ⱷ
Ɵ
ⱷ
ⱷ
rb
r
ra
RA
Rb

ARC OF
CONTACT
• The arc of contact is EPF or GPH.
• Considering the arc of contact GPH,
it is divided into two parts i.e. arc GP
and arc
• PH.
• The arc GP is known as arc of
approach and the arc PH is called arc
of recess.

ARC OF
CONTACT
• the length of the arc of approach (arc GP)
the length of the arc of recess (arc PH)
Since the length of the arc of contact GPH is equal to the sum of the length of arc
of approach and arc of recess, therefore, Length of the arc of contact

• The contact ratio or the number of pairs of
teeth in contact is defined as the ratio of the
length of the arc of contact to the circular
pitch.
• Mathematically, Contact ratio or number
of pairs of teeth in contact

 The number of teeth on each of the two equal spur gears in mesh are 40. The
teeth have 20° involute profile and the module is 6 mm. If the arc of contact is
1.75 times the circular pitch, find the addendum.
 Given : T = t = 40 ; φ = 20° ; m = 6 mm
 Adendum=6.12 mm

INTERFERENCE
 A little consideration will show, that if the radius of the addendum circle
of pinion is increased to O1N, the point of contact L will move from L to N.
 When this radius is further increased, the point of contact L will be on the
inside of base circle of wheel and not on the involute profile of tooth on
wheel.
 The tip of tooth on the pinion will then undercut the tooth on the wheel at
the root and remove part of the involute profile of tooth on the wheel.
 This effect is known as interference, and occurs when the teeth are being
cut. In brief, the phenomenon when the tip of tooth undercuts the root on
its mating gear is known as interference.

N
O2
O1
M
K
P
L
G H
ⱷ
Ɵ
ⱷ
ⱷ
rb
r
ra
RA
Rb
R
r

INTERFERENCE
 The points M and N are called interference points.
 Obviously, interference may be avoided if the path of contact does not extend
beyond interference points.
 The limiting value of the radius of the addendum circle of the pinion is O1N and
of the wheel is O2M.
 we conclude that the interference may only be avoided, if the point of contact
between the two teeth is always on the involute profiles of both the teeth.
 In other words, interference may only be prevented, if the addendum circles of
the two mating gears cut the common tangent to the base circles between the
points of tangency.

MINIMUM NUMBER OF TEETH ON THE PINION IN
ORDER TO AVOID
INTERFERENCE

N
O2
O1
M
K
P
L
G H
ⱷ
Ɵ
ⱷ
ⱷ
rb
r
ra
RA
Rb
r
R

 t = Number of teeth on the pinion,,
 T = Number of teeth on the wheel,
 m = Module of the teeth,
 r = Pitch circle radius of pinion = m.t / 2
 G = Gear ratio = T / t = R / r
 φ = Pressure angle or angle of obliquity.
 Let AP.m = Addendum of the pinion, where AP is a fraction
by which the standard addendum of one module for the
pinion should be multiplied in orderto avoid interference.
 From triangle O1NP,
N
O2
O1
M
K
P
L
G H
ⱷ
Ɵ
ⱷ
ⱷ
rb
r
ra
RA
Rb
r
R

This equation gives the minimum number of teeth required on the pinion
in order to avoid interference.

MINIMUM NUMBER OF TEETH ON THE WHEEL IN
ORDER TO AVOID
INTERFERENCE

 T = Minimum number of teeth required on the wheel in order to avoid
interference, And
 AW.m = Addendum of the wheel, where AW is a fraction by which the standard
addendum for the wheel should be multiplied.
 from triangle O2MP

MINIMUM NUMBER OF TEETH TO
AVOID INTERFERENCE
On pinion
On wheel

Determine the minimum number of teeth required on a pinion, in
order to avoid interference which is to gear with,
1. a wheel to give a gear ratio of 3 to 1 ; and 2. an equal wheel.
The pressure angle is 20° and a standard addendum of 1 module for
the wheel may be assumed.
Given : G = T / t = 3 ; φ = 20° ; AW = 1 module
t=16
t=13

 A pair of involute spur gears with 16° pressure angle and pitch of module
6 mm is in mesh. The number of teeth on pinion is 16 and its rotational
speed is 240 r.p.m. When the gear ratio is 1.75, find in order that the
interference is just avoided ; 1. the addenda on pinion and gear wheel ; 2.
the length of path of contact ; and 3. the maximum velocity of sliding of
teeth on either side of the pitch point.
 Given : φ = 16° ; m = 6 mm ; t = 16 ; N1 = 240 r.p.m. or ω1 = 2π × 240/60 = 25.136 rad/s ;
 G = T / t = 1.75 or T = G.t = 1.75 × 16 = 28
Addendum of pinion=10.76mm
Addendum of wheel=4.56mm
 RA = R + Addendum of wheel = 84 + 10.76 = 94.76 mm
 rA = r + Addendum of pinion = 48 + 4.56 = 52.56 mm
 KP=26.45mm; PL=11.94mm;
 KL = KP + PL = 26.45 + 11.94 = 38.39 mm
 W2=14.28rad/sec
 Sliding velocity during engagement= 1043mm/s
 Sliding velocity during disengagement= 471mm/s

 The force between mating teeth can be resolved at the pitch point P into two
 components,
1. Tangential component Ft, which when multiplied by the pitch line velocity, accounts for
the power transmitted.
2. Radial component Fr, which does not work but tends to push the gear apart.
 Fr = Ft tan φ
 Gear pitch line velocity V, in mm/sec, V = πdn/60
 where d is the pitch diameter in mm, gear rotating in n rpm.
 Transmitted power in watts, is in Newton
 P = Ft *V

 A single reduction gear of 120 kW with a pinion 250 mm pitch circle diameter and
speed 650 r.p.m. is supported in bearings on either side. Calculate the total load due to
the power transmitted, the pressure angle being 20°.
 Given : P = 120 kW = 120 × 103 W ; d = 250 mm or r = 125 mm = 0.125 m ;
 N = 650 r.p.m. or ω = 2π × 650/60 = 68 rad/s ; φ = 20°
 T = Torque transmitted in N-m.
 We know that power transmitted (P),=T.ω or
 T = 120 × 103/68 = 1765 N-m
 and tangential load on the pinion,FT = T /r = 14 120 N
 ∴ Total load due to power transmitted, F = FT / cos φ = 15.026 kN Ans.

 Helical gears can be used in a variety of applications since they can be
mounted on either parallel or on 90° non-intersecting shafts.
 Helical gears offer additional benefits relative to spur gears:
o Greater tooth strength due to the helical wrap around
o Increased contact ratio due to the axial tooth overlap
o Higher load carrying capacity than comparable sized spur gears.
 Smoother operating characteristics
 The close concentricity between the pitch diameter and outside diameter
allow for smooth and quiet operation

Automobiles
Presses
Machine tools
Material handling
Feed drives
Marine applications

When two helical gears are engaged as in the,
the helix angle has to be the same on each gear,
but one gear must have a right-hand helix and
the other a left-hand helix.

The helix angleΨ, is always measured on the cylindrical pitch surfaceΨ
value is not standardized. It ranges between 15 and 45. Commonly used
values are 15, 23, 30 or 45. Lower values give less end thrust. Higher
values result in smoother operation and more end thrust. Above 45 is not
Ψ

2. Normal pitch. It is the distance between similar faces
of adjacent teeth, along a helix on the pitch cylinder
normal to the teeth. It is denoted by Pn.
3. Transverse Pitch. It is the distance measured parallel
to the axis, between similar faces of adjacent teeth
denoted by P.
If α is the helix angle, then circular pitch,
1. Helix angle: It is the angle at which
teeth are inclined to the axis of gear. It is also
known as spiral angle.
Φn normal pressure angle

From geometry we have normal pitch
as
Normal module mn is
The pitch diameter (d) of the helical gear is:
The axial pitch (pa) is:
For axial overlap of adjacent teeth, b ≥ pa
In practice b = (1.15 ~2) pa is used.
The relation between normal and
transverse pressure angles is
Pn= Pt. cos()
d= t.mt= t.mn/cos ()
mn= mt.cos
()
tan (Φn)= tan (Φt)/ cos ()
Pa= P/ tan ()

FORCE ANALYSIS
 Fr= Fn. sin(Φn)
 Ft= Fn. cos(Φn). cos()
 Fa= Fn. cos(Φn). sin()
 Fr= Ft. tan(Φ)
 Fn= Ft/ (cos(Φn). sin())



The shape of the tooth in the normal plane is nearly the
same as the shape of a spur gear tooth having a pitch
radius equal to radius Re of the ellipse.
The equivalent number of teeth (also called virtual number of
teeth), te, is defined as the number of teeth in a gear of radius Re:
Re =d/(2.cos2())
te=2Re/mn = d/(mn. cos2())

 A stock helical gear has a normal pressure angle of 20◦, a helix angle of 25◦, and a
 transverse diametral pitch of 6 teeth/in, and has 18 teeth. Find:
 The pitch diameter--- 3 in
 The transverse, the normal, and the axial pitches(P, Pn, Pa)-- 0.52 in, 0.47 in, 1.12
in
 The transverse pressure angle(Φ) --21.88
 Virtual no. of teeth. (Zv)

 spiral gears (also known as skew gears or screw gears) are used
to connect and transmit motion between two non-parallel and
non-intersecting shafts.
 The pitch surfaces of the spiral gears are cylindrical and the
teeth have point contact.
 These gears are only suitable for transmitting small power.
 helical gears, connected on parallel shafts, are of opposite hand.
But spiral gears may be of the same hand or of opposite hand.

A pair of spiral gears 1 and 2, both having left hand
helixes.
The shaft angle θ is the angle through which one of the
shafts must be rotated so that it is parallel to the other
shaft, also the two shafts be rotating in opposite
directions.
α1 and α2 = Spiral angles of gear teeth for gears 1 and 2
Pc1 and Pc2 = Circular pitches of gears 1 and 2,
T1and T2 = Number of teeth on gears 1 and 2,
d1 and d2= Pitch circle diameters of gears 1 and 2,
N1 and N2 = Speed of gears 1 and 2,
Pn = Normal pitch, and
L = Least centre distance between the axes of shafts.

Since the normal pitch is same for both the spiral gears,
therefore

EFFICIENCY OF SPIRAL
GEAR
Ft1 = Force applied tangentially on the driver,
Ft2= Resisting force acting tangentially on the
driven,
Fa1 = Axial or end thrust on the driver,
Fa2 = Axial or end thrust on the driven,
FN = Normal reaction at the point of contact,
Φ = friction angle,
R = Resultant reaction at the point of contact,
and
θ = Shaft angle = α1+ α2
Work input to the driver
= Ft1 × π d1.N1/60 = R cos (α1 - Φ) π d1.N1/60
Work input to the driven gear
= Ft1 × π d2.N2/60 = R cos (α2 + Φ) π d2.N2/60
Ft1
Ft2
Fn
Fn

 Efficiency
Maximum Efficiency
Since the angles θ and φ are constants, therefore the efficiency will be maximum, when cos (α1 – α2 – φ) is
maximum, i.e.

SPUR GEAR
HELICAL GEAR
Transmit the power between two parallel shaft

No Type Normal Ratio Range
Efficiency
Range
1 Spur 1:1 to 6:1 94-98%
2 Straight Bevel 3:2 to 5:1 93-97%
3 Spiral Bevel 3:2 to 4:1 95-99%
4 Worm 5:1 to 75:1 50-90%
5 Hypoid 10:1 to 200:1 80-95%
6 Helical 3:2 to 10:1 94-98%
7 Cycloid 10:1 to 100:1 75% to 85%

GEAR TRAINS
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.
A gear train may consist of spur, bevel or spiral gears.
Types of Gear Trains
1. Simple gear train,
2. Compound gear train,
 3. Reverted gear train, and
4. Epicyclic gear train.

SIMPLE GEAR TRAIN
 When there is only one gear on each shaft, it is known as simple gear train.
 The gears are represented by their pitch circles.
 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.

 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,
 Ratio of the speed of the driven or follower to the speed of the driver is known as
train value of the gear train.
 speed ratio and the train value, in a simple train of gears, is independent of the
size and number of intermediate gears.

When there are more than one gear on a shaft, it is called a
compound train of gear.
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.

 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 :

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.
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 ...
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

 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.

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.
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 compound.
The epicyclic gear trains are useful for transmitting high velocity
ratios with gears of moderate size in a comparatively 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.

EPICYCLIC GEARMETHOD 1: TABULAR FORM

EPICYCLIC GEAR
 Let the arm C be fixed in an epicyclic gear train as shown in Fig.. 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 opposite directions.
METHOD 2: ALGEBRAIC FORM

 In an epicyclic gear train, an arm carries two gears A and B having 36 and 45
teeth respectively. If the arm rotates at 150 r.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 ?
Given : TA = 36 ; TB = 45 ; NC = 150 r.p.m. (anticlockwise)

 Speed of gear B when gear A is fixed 270 r.p.m. (anticlockwise)
 Speed of gear B whe n gear A makes 300 r.p.m. clockwise = 510 r.p.m. (anticlock)

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.
 TB = 75 ; TC = 30 ; TD = 90 ; NA = 100 r.p.m. (clockwise)
TB + TE = TC + TD
TE = TC + TD – TB = 30 + 90 – 75 = 45

An epicyclic gear consists of three gears A, B and C as shown in
Fig. 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.
Given : TA = 72 ; TC = 32 ; Speed of arm EF = 18 r.p.m.

 An epicyclic train of gears is arranged as shown in Fig.. 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 90respectively.
 Given : TA = 40 ; TD = 90
dA + dB + dC = dD or dA + 2 dB = dD ...( 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

1. Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise 0.04
revolution anticlockwise
2. Speed of arm when A makes 1 revolution clockwise and D is stationary Speed of arm = – y = – 0.308
= 0.308 revolution clockwise

 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.
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
samemodule, 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 4.2 r.p.m. clockwise
4. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A makes 10 r.p.m. counter clockwise 5.4 r.p.m.
counter clockwise

 In an epicyclic gear of the ‘sun and planet’ type shown in Fig. , 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.
Given : dD = 224 mm ; m = 4 mm ; NA = NB / 5

TD = dD / m = 224 / 4 = 56 Ans.
TB = TD / 4 = 56 / 4 = 14 Ans.
TC = 21 Ans.

 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
concentric 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.

Speed of shaft B = Speed of arm = + y = 50 r.p.m. anticlockwise

Fig shows a differential gear used in a motor car. The
pinion A on the propeller shaft has 12 teeth and gears with
the crown gear B which has 60 teeth. The shafts P and Q
form the rear axles to which the road wheels are attached.
If the propeller shaft rotates at 1000 r.p.m. and the road
wheel attached to axle Q has a speed of 210 r.p.m. while
taking a turn, find the speed of road wheel attached to
axle P.

Speed of road wheel attached to axle P = Speed of gear C = x + y
= – 10 + 200 = 190 r.p.m.

x = – 563.8 and y = 563.8
Holding torque on wheel D = 230.6 – 175.7 = 54.9 N-m

Fixing torque required at C = 4416 – 14.7 = 4401.3 N-m

INTRODUCTION OF GEARS AND GEAR KINEMATICS

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INTRODUCTION OF GEARS AND GEAR KINEMATICS