KOM 2131906 GTU Lecture notes

Kinematics Of Machines
2131906

Syllabus
Sr No. Topics Slide No.
1 Introduction to Machines &
Mechanisms
3
2 Special Mechanisms 36
3 Synthesis
4 Velocity Analysis 95
5 Acceleration Analysis 119
6 Gear 136
7 Gear Trains
8 Cam and Follower 207
2

Introduction to Machines &
Mechanisms
 What is Kinematic Link and its types;
 Types of Motion;
 Kinematic Pair and its Types;
 Kinematic Chain;
 Types of Joints;
 Structure, Machine & Mechanisms.
 Degree of Freedom;
 Grubler’s Criterion;
 Kutzback’s Criterion.
 Mechanisms with Lower Pair;
 Exact and Approximate Straight line Mechanism;
 Inversion of Mechanism;
 Four Bar / Quadric Cyclic Chain Mechanism;
 Single Slider Crank Mechanism;
 Double Slider Crank Mechanism.

Kinematics
Study of motion, without
considering the force causing
that motion.
4

Kinematic Link :::
•“Each part of Machine, which moves
relative to some other part.”
•A link may consists of several parts,
which are rigidly fastened together, so
that they do not move relative to one
another.

Types of Link :::
RIGID LINK FLEXIBLE LINK FLUID LINK
• Link which doesn’t
undergo any deformation
while transmitting the
motion.
• Ex. Piston & Connecting
Rod.
• Classified as :::
1. Binary Link ::
• Link with 2 nodes.
2. Ternary Link ::
3. Quaternary Link ::
Link which are partly
deformed while
transmitting the motion
such that it doesn’t affect
the transmission of
motion.
Ex. Belt, Rope & Chain
Drives.
Link which transmit the
motion by fluid pressure
as compression.
Ex. Hydraulic Jack,
Automobile Brakes.
6

Types of Motion :::
COMPLETELY
CONSTRAINED MOTION
INCOMPLETELY
CONSTRAINED MOTION
TYPE - 5
• In which the relative motion
between the links can be possible only
in a definite direction & can be
predicted.
• Ex. 1) Piston and cylinder in a steam
engine
• Motion of piston is limited to
reciprocate .
• Ex. 2) Motion of a square bar in a
square hole.
between the links can be possible in
more than one direction & can’t be
predicted.
• Ex. 1) Motion of a Round bar in a
fixed block.
7

Types of Motion :::
SUCCESSFULLY
CONSTRAINED MOTION
TYPE - 5
between the links is not completely
constrained by itself but it is made so
by some other means.
• Ex. 1) Upward Motion of shaft in foot
step bearing is constrained by applying
Load.
8

Kinematic Pair :::
• “When two kinematic links are connected in such a manner
that their motion is either completely or successfully
constrained.”
According to the Nature of Relative
Motion between the Links
1. Sliding / Prismatic Pair ::
‒ “It’s formed by two links connected in such a manner
that one link is constrained to have a sliding motion
relative to the other link.”
‒ Ex.) Piston and Cylinder arrangement in Steam Engine;
‒ Ex.) Tail – stock of the Lathe.
9

Kinematic Pair :::
constrained.”
2. Turning / Revolute Pair ::
that one link is constrained to turn / revolute relative
to the other link.”
‒ Ex.) Cycle Wheels turning over the Axel.
10

Kinematic Pair :::
constrained.”
3. Screw / Helical Pair ::
‒ “It’s formed by two links when nature of contact
between them is in such a manner that one link can
turn about the other link by screw threads.”
‒ Ex.) Motion of bolt in screw;
‒ Ex.) Motion of lead screw in the nut.
11

Kinematic Pair :::
constrained.”
4. Rolling Pair ::
that one link rolls over the other link.”
‒ Ex.) Ball & Roller Bearing;
‒ Ex.) Shaft in ball bearing.
12

Kinematic Pair :::
constrained.”
5. Spherical / Globular Pair ::
that one link turns / swivel about the other link.”
‒ Ex.) Ball & Socket Joint;
‒ Ex.) Pen Stand;
‒ Ex.) Attachment of Car Mirror.
13

Kinematic Pair :::
constrained.”
According to the Nature of Contact
Between the Links
1. Lower Pair ::
‒ “It’s formed when two links have a surface / area
contact while transmission of motion.”
‒ Ex.) All Sliding , Turning , Screw and Spherical Pair;
‒ Ex.) Piston – Cylinder Arrangement;
‒ Ex.) Shaft in bearing.
14

Kinematic Pair :::
constrained.”
According to the Nature of Contact
Between the Links
2. Higher Pair ::
‒ “It’s formed when two links have a line / point contact
while transmission of motion.”
‒ Ex.) All Rolling Pair;
‒ Ex.) Cam and Follower;
‒ Ex.) Gear Transmission.
15

Kinematic Pair :::
constrained.”
According to the Mechanical
Arrangement
1. Self Closed / Closed Pair ::
‒ “It’s formed when two links are held together
mechanically itself.”
‒ Ex.) All Lower Pair;
18

Kinematic Pair :::
constrained.”
According to the Mechanical
Arrangement
2. Forced Closed / Unclosed / Open Pair ::
‒ “It’s formed when two links are not held together
mechanically itself but by action of external force.”
‒ Ex.) Cam and Follower.
19

Kinematic Chain :::
• “When the kinematic pairs are connected in
such a manner that the last link is joined to first
link to transmit the definite motion.”
OR
• “A combination of kinematic pairs in which each
two link forms a part of the kinematic pair if the
relative motion between the links either
Completely or Successfully Constrained.”
20

11
2
23
3
4
4
Crank shaft of engine with bearing form a Kinematic Pair
Crank with Connecting Rod form a Kinematic Pair
Piston with Connecting Rod form a Kinematic Pair
Piston with Cylinder forms a Kinematic Pair.
Total combination of this pairs
is known as ‘Kinematic Chain.’ 21

Types of Joints in Chain :::
Types of Joints
Binary Joint
“When two links
are joined at the
same connection”
Ternary Joint
“When three links
are joined at the
same connection”
Quaternary Joint
“When four links
are joined at the
same connection”
Link - 1
Link - 2
B1 B2
B3
B4
Link - 1
Link - 2
Link - 5
Link - 6
B1 B2
B3
T1
T2
Link - 5
Link - 1
Link - 2
T1 B1
B2
T2
Q1
Q2
22

BINARY JOINT
Link - 1
Link - 2
B1 B2
B3
B4
All joints = Binary Joints.
Joints = B1, B2, B3 and B4.
Total No. of Joints ::
= B = 4.
TERNARY JOINT QUATERNARY JOINT
Link - 1
Link - 2
Link - 5
Link - 6
B1 B2
B3
T1
T2
All joints = Binary
and ternary Joints.
Binary Joints = B1, B2 and B3.
Ternary Joints = T1 and T1.
1 Ternary Joints = 2 Binary Joints
= B + T = B + 2(T) =3 + 2(2) = 7.
Link - 5
Link - 1
Link - 2
T1 B1
B2
T2
Q1
Q2
All joints = Binary;
ternary & Quaternary Joints.
Binary Joints = B1 and B2.
Ternary Joints = T1 and T1.
Quaternary Joints = Q1 and Q1.
1 Ternary Joints = 2 Binary Joints
1 Quaternary Joints = 3 Binary Joints
= B + T + Q = B + 2(T) + 3(Q)
=2 + 2(2) + 3(2) = 2 + 4 + 6 = 12.
23

Structure and Machine
• Structure: “It’s an assemblage of a no. of resisting bodies
having no relative motion between them & meant for
carrying loads having straining action.”
• Ex. Railway Bridge, Machine Frame.
• Machine: “An assemblage of parts that transmit the forces,
motion and energy in a predetermined manner.”
• Ex. Steam Engine, Lathe and Milling Machine.
STRUCTURE
1. No Relative Motion Betn the links.
2. No Energy is Transformed into
some useful work.
3. Transmits the forces only.
4. Ex. Railway Bridge.
MACHINE
1. Relative Motion Betn the links.
2. Available Energy is Transformed
into some useful work.
3. Transmits the motion & power.
4. Ex. Steam Engine.
24

Machine and Mechanism
• Mechanism: “Kinematic Chain which is used to transmit
the motion with any one link is fixed.”
• Mechanism with 4 Links = Simple Mechanism.
• Mechanism with more than 4 Links = Complex Mechanism.
MECHANISM
1. Kinematic Chain with 1 link fixed.
2. Used to Transmit or transform the
motion.
3. All mechanisms are not machines.
4. Ex. Four Bar Chain.
MACHINE
1. An assemblage of Mechanisms.
2. Used to transmit the motion and
force.
3. All machines are mechanisms.
4. Ex. I.C. Engine.
• Machines and Mechanisms both are combinations of rigid bodies.
• Relative motions along rigid bodies are definite.
25

Degree of Freedom :::
For a Point in a Space :
P(x, y, z)
Y
X
Z
DOF = 3
For a Body in a Space :
Y
X
Z
DOF = 6
• Unconstrained Rigid Bogy moving in a
space can have 6 DOF.
1. Three translation motion among x, y, z;
2. Three Rotational Motion about x, y, z.
26
“Number of independent co-ordinates required to define the
position and orientation of the point.”

Degree of Freedom of a Mechanism:::
• “In which Complete
Motion can’t be
represented on a
Single Plane.”
• Ex. Robot Arm and
Hoisting Machine (i.e.,
Cranes)
• DOF ≤ 6.
Spatial Mechanisms :: Planar Mechanisms ::
• “In which Complete
Motion can be
represented on a Single
Plane.”
• Ex. Four Bar Chain
Mechanisms.
• DOF ≤ 3.
27

DOF For Planar Mechanisms :::
• If Link AB is fixed, position of CD can be completely specified by 3 Variables.
• Translation among X and Y; Rotation about Z.
• So, each link of mechanism have 3 DOF before connecting to the other link.
• If, we are connecting CD to AB at C & A, it resists 2 Translation (X & Y) Motion
and allow only 1 Rotational (Z) Motion OR it resists only 1 Translation (Y)
Motion and allow 1 Translation (Y) Motion and 1 Rotational (Z) Motion.
• Lower Pair decreases 2 DOF.
• Higher Pair decreases Only 1 DOF.
Y
X
Z
C
A B
D
Y
X
Z
C
A B
D
⤫
⤫
√
Y
X
Z
C
A B
D
⤫
√
√
General Relationship for DOF
For Lower Pair For Higher Pair
28

DOF For Planar Mechanisms :::
• This Equation is known as “GRUBLER’S CRITERION.” for finding DOF.
)(1)(2)1(3 HJLN 
θ
θ1 θ2
Four Bar Chain Mechanism
Five Bar Chain Mechanism
• DOF for the Planar Mechanism can also be defined as ::
‒ “No. of Independent Input Parameters required to bring mechanism into
useful motion.”
29

θ
Four Bar Chain Mechanism
Five Bar Chain Mechanism
θ1 θ2
• Here, in Four Bar Chain Mechanism, only 1 Variable (i.e., )
is required to be defined.
‒ So, the DOF for Four Bar Chain Mechanism is 1.
• While in Five Bar Chain Mechanism, 2 parameters are
required to be defined the completely relative motion of
the Mechanism.
‒ So, the DOF for Five Bar Chain Mechanism is 2.
30

Examples for finding DOF :::
⇒ N = 3 ( L – 1 ) – 2 ( J ) – 1 ( H )
∴ N = 3 ( 3 – 1 ) – 2 ( 3 ) – 1 ( 0 )
∴ N = 0.
⇒ called “STRUCTURE”
⇒ N = 3 ( L – 1 ) – 2 ( J ) – 1 ( H )
∴ N = 3 ( 4 – 1 ) – 2 ( 4 ) – 1 ( 0 )
∴ N = 1.
⇒ called “CONSTRAINED
MECHANISM”
⇒ 1 Parameter required.
Link - 1
Link - 2Link - 3
B
A
C
Link - 1
Link - 2
BA
C
D
31

⇒ N = 3 ( L – 1 ) – 2 ( J ) – 1 ( H )
∴ N = 3 ( 5 – 1 ) – 2 ( 5 ) – 1 ( 0 )
∴ N = 2.
⇒ called “UNCONSTRAINED
MECHANISM”
⇒ 2 Parameters Required.
⇒ N = 3 ( L – 1 ) – 2 ( J ) – 1 ( H )
∴ N = 3 ( 5 – 1 ) – 2 ( 6 ) – 1 ( 0 )
∴ N = 0.
⇒ called “STRUCTURE”
Link - 1
Link - 2
BA
C
E
Link - 3
D
Link - 1
Link - 2
Link - 3
A B
C
D
32

Link - 1
Link - 2Link - 4
BA
C
Link - 3
D
⇒ N = 3 ( L – 1 ) – 2 ( J ) – 1 ( H )
∴ N = 3 ( 6 – 1 ) – 2 ( 8 ) – 1 ( 0 )
∴ N = – 1.
⇒ called “UNCONSTRAINED MECHANISM”
33

⇒ N = 3 ( L – 1 ) – 2 ( J ) – 1 ( H )
∴ N = 3 ( 3 – 1 ) – 2 ( 2 ) – 1 ( 1 )
∴ N = 1.
1 & 2 = Resists X & Y Trans. & Provides Z Rotation;
2 & 3 = Resists Z Rotational & Provides X & Y Trans.
⇒ N = 3 ( L – 1 ) – 2 ( J ) – 1 ( H )
∴ N = 3 ( 4 – 1 ) – 2 ( 3 ) – 1 ( 1 )
∴ N = 2.
4 & 1 = Resists Y Translation & Provides X
Translation & Z Rotational.
Link - 1
C
A
B
Link - 2
Link - 3
C
A
B
D
Link - 1
Link - 2
Link - 4
34

DOF For Spatial Mechanisms :::
• This Equation is known as “KUTZBACK’S CRITERION.” for finding DOF.
• DOF for the Spatial Mechanism can be defined as ::
N = 6 ( L – 1 ) – 5 J1 – 4 J2 – 3 J3 – 2 J4 – 1 J5 – 0J6.
Where,
L = No. of Links;
J1 = No. of Pairs having 1 degree of freedom;
J6 = No. of Pairs having 6 degree of freedom.
35

Mechanisms with Lower Pair :::
PANTOGRAPH
( Constructed by Mr. Christophe Steiner )
• “Mechanical linkage connected in a special manner based
on Parallelogram in such a way that the movement of one
specified point is an amplification of another point.”
‒ It means that, if a line drawn is traced by the first point , an
enlarged copy will be drawn by another point.
⇒ APPLICATION ::
o Use to Copy & Scale diagram;
o Guide the Cutting Tool;
o Reproduction of Maps;
o Defined displacement of piston in
Steam Engine.
36

PANTOGRAPH
( Constructed by Mr. Christophe Steiner )
37

RATCHET MECHANISM ESCAPMENT MECHANISM
38

GENEVA MECHANISM or INDEXING MECHANISM
39

Straight Line Mechanisms :::
• In a straight line Mechanism, one point is always moves
in a straight path, which generates STRAIGHT LINE.
STRAIGHT LINE MECHANISMS
EXACT STRAIGHT LINE MECHANISM
Peaucellier Mechanism;
Scott Russell Mechanism;
Hart’s Mechanism.
41

Exact Straight Line Mechanisms :::
• PRINCIPLE :
‒ “If O be the point on the circumference of a circle having
diameter OP, let OA be any chord and B is any point on line of
OA such that,
OA ⤫ OB = Constant
Then locus of point B is a straight line ⏊ to OP.”
O
P
A
B
Q
42

O
P
A
B
Q
C
• On Extension of OP, draw ⏊
passing through point B,
which intersects OP at Q and
is on extension of OA.
• As point O is Fixed in Circle, if
OQ = Constant then B will
move along straight line
passing through Q and ⏊ to
OP.
• In Δ OAP and Δ OQB,
OP
OBOA
OQ
OQ
OB
OA
OP



• Here, OP = Constant as it’s Diameter.
• Now, if ( OA ⤫ OB ) = Constant then OQ = Constant.
• And so, point B will move along a straight line BQ ⏊ to OP.
43

Peaucellier’s Straight Line Mechanisms :::
• Let, OM be the fixed link;
• Links MA, AC, CB, BD, DA, OC & OD
are connected by Turning Pair.
• Point A is Constrained to move
along the circumference of the
circle having Diameter OP by the
link MA.
• Relative Dimensions are ::
.
.
.
ODOC
DABDCBAC
MAOM



44

• Join CD which intersects AB at point R.
• In Δ ORC and Δ BRC,
OC2 = OR2 + RC2 &
BC2 = BR2 + RC2.
• So,
• So, point B will always move along a straight path BQ ⏊ OP.
OC2 – BC2 = (OR2 + RC2 ) – (BR2 + RC2 )
= OR2 – BR2
= ( OR - BR )( OR + BR )
= OA ⤫ OB
• Since, OC and BC are links whose
length is constant. So, (OA ⤫ OB )
also equals to Constant.
45

46

Robert’s Approximate Straight Line
Mechanisms :::
• Here, Links AC, CD, BD, CP, PD all
are joined by a Turning Pairs.
• Mechanism is Fixed at Point A and
B.
ABCD
BDAC
2
1


• If Point P is Made free to move during the operating condition,
it will trace out an approximate straight line ‖ to Line AB (as
shown by center line in figure).
47

Robert’s Approximate Straight Line
Mechanisms :::
48

Watt’s Approximate Straight Line
Mechanisms :::
• Here, Point O and Q is Fixed.
• Links OA, AB, BQ are connected
in Turning Pairs.
• Point P is a Point on Link AB.
AO
BQ
PB
PA

• When Links OA and QB oscillate about Points O & Q
respectively, Point P will trace out an approximate straight
path.
49

Watt’s Approximate Straight Line
Mechanisms :::
50

Inversion of Mechanism :::
• “The process of choosing different links of chain as fixed.”
OR
• “By fixing each link at a time, we get as many mechanism as
the number of links, then each mechanism is called ‘Inversion’
of the original Kinematic Chain.”
• Most important Kinematic Chain are those which consists of 4
lower pairs, each being Sliding or Turning.
• Following 3 Types of Mechanisms with Four Lower Pairs are
important ::
1. Four Bar / Quadric Cyclic Chain Mechanism;
2. Single Slider Crank Mechanism;
3. Double Slider Crank Mechanism.
51

Inversion of Four Bar Chain Mechanism :::
• It is Simplest Kinematic Chain.
• Consists of 4 Links AB, BC,
CD & DA of different length
and each are forming
Turning Pairs.
Link - 1
Link - 2
BA
C
D
• Imp. thing in designing mechanism is to ensure that input
crank makes complete revolution relative to other links.
• To test whether any link in given Mechanism makes the
complete revolution or not can be tested according to
“GRASHOF’S LAW.”
52

GRASHOFF’S LAW :::
• As shown, the Shortest Link – 4 will make a complete revolution relative to
other links. So, Link – 4 = Crank / Driver.
• Link – 1, which is fixed = Fixed Link / Frame.
• Link – 2,which makes partial rotation/oscillation=Lever/Rocker/Follower.
• Link – 3, which connects Crank and Lever = Coupler / Connecting Rod.
• “For planar 4 bar
mechanism, the Sum of
lengths of Shortest and
Longest Links should not
be greater than the sum
of the Lengths of the
remaining Links if there
is to be Continuous
relative motion between 2
Links.”
Link - 1
Link - 2
BA
C
D
Frame / Fixed Link
Lever
/ Rocker
/ Follower
53

Coupled Wheel of Locomotive :::
• Also Known as “Double
Crank Mechanism.”
• The Mechanism of coupled
wheel of Locomotive as
shown consists of 4 Links
connected by turning Pairs.
Link – 1
Fixed
L – 4
Crank
Rotates
L – 3
Connecting Rod
Oscillates
B
A
C
D
• In this mechanism, Links AB and CD having equal length acts as crank as
making complete revolutions and are connected to the respective wheels.
• Links BC = Connecting Rod & Link AD is Fixed in order to maintain the
Constant Center to Center distance between the Wheels.
• This mechanism meant for transmission of rotary motion from one wheel to
the other wheel.
54

Coupled Wheel of Locomotive :::
55

Beam Engine :::
• Also Known as “Crank and
Lever Mechanism.”
• The Mechanism of Beam
Engine as shown consists of 4
Links connected by turning
Pairs.
• Link – 2 makes complete revolution. So, Link AB is Crank, rotates about
fixed point A.
• Link – 4 which is CED, acts as Lever, is pivoted at point E Oscillates in such
a manner that it makes Vertical Reciprocating Motion at the end D.
• Crank and Lever is connected by Link BC, which is Connecting Rod.
• This serves the purpose of this mechanism which is to convert Rotary
Motion into the Reciprocating Motion.
A
B
C
DE
L – 2
Crank
Rotates
Vertical
Reciprocating
Motion
L – 1
Fixed
56

Watt Indicator Mechanism :::
A
B
C
D
E
F
58

Inversion of Single Slider Crank Mechanism :::
• It is Modified Four Bar
Chain.
• Consists of 3 Links AB,
BC, CD are forming
Turning Pairs and 1
Link forms Sliding
Pair.
• Link – 1 & 2; Link – 2 & 3 and Link – 3 & 4 are forming Turning Pairs and
Link – 4 & 1 forms a sliding Pair.
• As Link AB = Crank rotates, Link C = Slider Reciprocates in the Cross Head
which ultimately Reciprocates the Piston in the Cylinder.
• Connection of Crank and Guide which is Link BC is Connecting Rod.
B
A
C
D
Cross Head
L – 2
Crank
Rotates
L – 1
Frame
Fixed
L – 4
Slider
Slides
59

Pendulum Pump / Bull Engine :::
• Here, Link – 4 = AD,
Cylinders, which are Fixed.
• When Crank BC = Link – 2
Rotates, Connecting Rod AB
= Link – 3 Oscillates about
Point A.
• During this Piston D
attached to Piston Rod CD =
Link – 1.
L – 3
Connecting Rod
Oscillates
B
A
C
D
L – 2
Crank
Rotates
L – 4
Cylinder
Fixed
L – 1
Piston Rod
Slides
L – 4
Cylinder
Fixed
61

Oscillating Cylinder Engine :::
• Here, Link – 3 = AB,
Connecting Rod, which
is Fixed.
• When Crank BC = Link
– 2 Rotates, Piston Rod
CD = Link – 1 and
Oscillates about Point A
and Slide the end Piston
in the Link – 4 which is
Oscillating Cylinder.
B
A
C
L – 2
Crank
Rotates
L – 3
Connecting Rod
Fixed
L – 4
Oscillating
Cylinder
D
62

Oscillating Cylinder Engine :::
63

Rotary I.C. Engine / Gnome Engine:::
• When more power
required the Multi
Cylinder Engine is used.
• Generally it is having 7
Cylinders in one plane.
• When we are fixing the
link – 2 = Crank, and
Rotating Link – 4 =
Connecting Rod about
Point A, Link – 3 =
Piston Slides or
Reciprocates in the Link
– 1which is Cylinder.
L – 2
Crank
Fixed
L – 1
Cylinders
Rotates
L – 3
Piston
Slides /
Reciprocates
L – 4
Connecting Rod
Rotates
O
A
64

Rotary I.C. Engine / Gnome Engine:::
65

Crank and Slotted Quick Return Mechanism :::
• If Link – 3 = AC which is
Connecting Rod is Fixed.
• Driving Crank CB rotates with
the uniform angular speed about
Point C.
• Sliding Block, attached to Crank
End, pin at B slides along Slotted
Bar AP & cause AP to Oscillate
about Pivoted point A.
• Short Link PR transmits motion
from AP to RAM which carries
tool & reciprocates along the line
of Stroke which is ⏊ to AC.
B
A
R
P
R2R1
P1 P2
B2B1
RAM
C
β
α
90 – (α/2)
CUTTINGRETURN
L – 2
Crank
Rotates
L – 1
Slider
Slides
L – 4
Slotted Bar
Oscillates
L – 3
Connecting Rod
Fixed
66

B
A
R
P
R2R1
P1 P2
B2B1
RAM
C
β
α
90 – (α/2)
CUTTINGRETURN
L – 2
Crank
Rotates
L – 1
Slider
Slides
L – 4
Slotted Bar
Oscillates
L – 3
Connecting Rod
Fixed
• Extreme Positions, which are
tangential to Circle are AP1 and
AP2.
• Cutting Stroke = Crank Rotates
from CB1 to CB2 or through angle
β in CW direction.
• Return Stroke = Crank Rotates
from CB2 to CB1 or through angle
α in CW direction.
• Due to Uniform Velocity of
Crank, Time of Cutting = Time of
Return.
• So, Time of Cutting Stroke
Time of Return Stroke
--------------------------------





 



360
360
67

B
A
RP
R2R1
P1 P2
B2B1
RAM
C
β
α
90 – (α/2)
CUTTINGRETURN
L – 2
Crank
Rotates
L – 1
Slider
Slides
L – 4
Slotted Bar
Oscillates
L – 3
Connecting Rod
Fixed
• Overall Travel = Line of Stroke
= R1R2
= P1P2
= 2P1Q
= 2(AP1sin (90 – α/2))
= 2APcos (α/2)
Q
1
1
)
2
90(
AP
QP
Sin 

AC
CB
Cos 1
2


AC
BC
AP
AC
CB
AP


2
2 1
• Cutting St. Angle β > Return St. Angle α
• Since, Crank Speed is Uniform, Return
Stroke Speed is Higher than Cutting
Stroke. 68

69

Whitworth Quick Return Mechanism :::
• If Link – 2 = CD which is
Connecting Rod is Fixed.
• Driving Crank AC rotates with
the uniform angular speed about
Point C.
• Sliding Block, attached to Crank
End, pin at A slides along Slotted
Bar AP which Oscillates about
Pivoted point D.
• Link PR transmits motion from
AP to RAM which carries tool &
reciprocates along the line of
Stroke which is ⏊ to CD.
R
RAM
β
α
L – 4
Slider
Slides
P
R2R1
P1 P2
A
A2
A1
L – 3
Crank
Rotates
L – 2
Connecting
Rod
Fixed
L – 1
Slotted Bar
Oscillates C
D
70

• Extreme Positions, which are
tangential to Circle are AP1 and
AP2.
• Cutting Stroke = Crank Rotates
from CA1 to CA2 or through angle
α in CW direction.
• Return Stroke = Crank Rotates
from CA2 to CA1 or through angle
β in CW direction.
• Due to Uniform Velocity of
Crank, Time of Cutting = Time of
Return.
• So,
Time of Cutting Stroke
Time of Return Stroke
--------------------------------





 



360
360
R
RAM
β
α
L – 4
Slider
Slides
P
R2R1
P1 P2
A
A2
A1
L – 3
Crank
Rotates
L – 2
Connecting
Rod
Fixed
L – 1
Slotted Bar
Oscillates C
D
71

• Cutting St. Angle α > Return St. Angle β
• Since, Crank Speed is Uniform, Return
Stroke Speed is Higher than Cutting
Stroke.
R
RAM
β
α
L – 4
Slider
Slides
P
R2R1
P1 P2
A
A2
A1
L – 3
Crank
Rotates
L – 2
Connecting
Rod
Fixed
L – 1
Slotted Bar
Oscillates C
D
72

Inversion of Double Slider Crank Mechanism :::
Elliptical trammel ::
L – 1
Slider
Slides
L – 2
Bar
L – 4
Slotted Plate
Fixed
L – 3
Slider
Slides
A
P
B
• Used for Drawing Ellipse.
• Fixing Link – 4 which is Slotted
Plate have 2 Grooves at Right
Angles to Each Other.
• Link – 1 and 3 are Sliding Pairs
with the Link – 4.
• Link – 2 which is Bar making
Turning Pair with Link – 1 and 3.
• When Link – 1 and 3 Slides along
grooves, Point on Bar, P trace out
an Ellipse on the surface of the
Plate. 73

L – 1
Slider
Slides
L – 2
Bar
L – 4
Slotted Plate
Fixed
L – 3
Slider
Slides
A
P
B
• In Figure Shown Below,
x = PQ & y = PR;
x = AP cosθ & y = BP sinθ
Which is Equation of Ellipse as
AP = Major Axis and
BP = Minor Axis.
B
A
P
x
y
Q
R
θ
.1sincos
sin&cos
22
2
2
2
2




BP
y
AP
x
BP
y
AP
x
74

75

Scotch Yoke Mechanism ::
Link – 1 – Fixed
L – 2
Crank
Rotates
L – 4
Piston
Slides
L – 3
Slider
Slides
• Fixing Link – 1, When we rotate Link – 2 which is Crank about Point A , the
Slider B which is Link – 3 made the Piston ( Link – 4 ) Slides along the
Cylinder.
• It is used to convert Rotary motion of Crank into the Reciprocating Motion
of the Piston and Vice – a – Versa.
B
A
76

Scotch Yoke Mechanism ::
77

Oldham Coupling ::
78

Oldham Coupling ::
79

Devis Steering Gear Mechanism
81

82

83

Ackerman Steering Gear Mechanism
• The whole mechanism of the Ackerman steering gear is on
back of the front wheels; Davis steering gear, it is in front.
• The Ackerman steering gear consists of turning pairs,
whereas Davis steering gear consists of sliding members.
84

Ackerman Steering Gear Mechanism
1. Straight path, links AB and CD are parallel and links BC and AD are
equally inclined to the longitudinal axis of the vehicle.
2. Left & Right, the position of the gear is shown by dotted lines in Fig. In
this position, the lines of the front wheel axle intersect on the back wheel
axle at I, for correct steering.
85

Universal or Hooke’s Joint
86

87

88

89

90

Maximum and Minimum Speeds of Driven Shaft
91

Condition for Equal Speeds of the Driving and Driven Shafts
92

Maximum Fluctuation of Speed
Since α is a small angle, therefore substituting cos α = 1,
and sin α = α radians.
93

Double Hooke’s Joint
Let the driving, intermediate and driven shafts, in the same time, rotate through
angles ,  and 
This shows that the speed of the driving and driven shaft is constant. In other words, this
joint gives a velocity ratio equal to unity, if
1. The axes of the driving and driven shafts are in the same plane, and
2. The driving and driven shafts make equal angles with the intermediate shaft.
94

Velocity analysis of any mechanism can be carried out by
various methods.
1. By graphical method
2. By relative velocity method
3. By instantaneous method
96

 By Graphical Method
The following points are to be considered while solving problems
by this method.
1. Draw the configuration design to a suitable scale.
2. Locate all fixed point in a mechanism as a common point in
velocity diagram.
3. Choose a suitable scale for the vector diagram velocity.
97

4. The velocity vector of each rotating link is r
to the link.
5. Velocity of each link in mechanism has both magnitude and
direction. Start from a point whose magnitude and direction is
known.
6. The points of the velocity diagram are indicated by small letters.
98

To explain the method let us take a few
specific examples.
1. Four – Bar Mechanism:
In a four bar chain ABCD link AD is fixed
and in 15 cm long. The crank AB is 4 cm
long rotates at 180 rpm (cw) while link CD
rotates about D is 8 cm long BC = AD and
BAD| = 60o
. Find angular velocity of link
CD.
Configuration Diagram
60o
wBA
A D
B
C
15 cm
15 cm
8 cm
99

Velocity vector diagram
Vb = r = ba x AB = 4x
60
120x2π
= 50.24 cm/sec
Choose a suitable scale
1 cm = 20 m/s = ab
r
to CD
r
to BC
r
to AB
a, d
b
C Vcb
100

Vcb = bc
Vc = dc = 38 cm/s = Vcd
We know that V = ωR
Vcd = cD x CD
WcD = 75.4
8
38Vcd

CD
rad/s (cw)
101

1. Slider Crank Mechanism:
In a crank and slotted lover mechanism crank rotates
of 300 rpm in a counter clockwise direction. Find
(i) Angular velocity of connecting rod and
(ii) Velocity of slider.
Configuration diagram
60 mm
45o
A
B
150 mm
102

Step 1: Determine the magnitude and velocity of
point A with respect to 0,
VA = O1A x O2A = 60x
60
300x2
= 600  mm/sec
Step 2: Choose a suitable scale to draw velocity vector diagram.
O
Vaa
b
r
to AB r
to OA
Along sides B
Vab = ab
ba =
150BA
Vba
 r/s
Vb = ob velocity of slider
Note: Velocity of slider is along
the line of sliding.
103

3. Shaper Mechanism:
In a crank and slotted lever mechanisms crank O2A rotates
at  r/s in CCW direction. Determine the velocity of slider.
Configuration diagram
4
O1
O2
C
B
3
2
W
5
6
D Scale 1 cm = x m/s
104

Va = 2 x O2A
CO
cO
BO
bO
1
1
1
1

To locate point C 






BO
CO
bOcO
1
1
11
Scale 1 cm = x m/s
d O1O2
VDC
c
a
b
VBA
VAO2 = VA
VBO1
105

 To Determine Velocity of Rubbing
Two links of a mechanism having turning point will be connected
by pins. When the links are motion they rub against pin surface.
The velocity of rubbing of pins depends on the angular velocity of
links relative to each other as well as direction.
106

For example: In a four bar mechanism we have
pins at points A, B, C and D.
 Vra = ab x ratios of pin A (rpa)
+ sign is used  ab is CW and Wbc is CCW i.e. when angular
velocities are in opposite directions use + sign when angular
velocities are in some directions use - ve sign.
VrC = (bc + cd) radius r
VrD = cd rpd
Problems on velocity by velocity vector method (Graphical107

Problems on velocity by velocity vector
method (Graphical solutions)
 Problem 1:
In a four bar mechanism, the dimensions of the links are as given
below:
AB = 50 mm, BC = 66 mm
CD = 56 mm and AD = 100 mm
At a given instant when o
60DAB|  the angular velocity of link
AB is 10.5 r/s in CCW direction.
108

Determine,
i) Velocity of point C
ii) Velocity of point E on link BC when BE = 40 mm
iii) The angular velocity of link BC and CD
iv) The velocity of an offset point F on link BC, if BF = 45 mm, CF
= 30 mm and BCF is read clockwise.
v) The velocity of an offset point G on link CD, if CG = 24 mm,
DG = 44 mm and DCG is read clockwise.
vi) The velocity of rubbing of pins A, B, C and D. The ratio of the
pins are 30 mm, 40 mm, 25 mm and 35 mm respectively.
109

 Solution:
Step -1: Construct the configuration diagram
selecting a suitable scale.
60o
A D
B
C
F
G
Scale: 1 cm = 20 mm
110

Step – 2: Given the angular velocity of link AB and its direction of
rotation determine velocity of point with respect to A (A is fixed
hence, it is zero velocity point).
Vba = BA x BA
= 10.5 x 0.05 = 0.525 m/s
111

Step – 3: To draw velocity vector diagram
choose a suitable scale, say 1 cm = 0.2 m/s.
 First locate zero velocity points.
 Draw a line r
to link AB in the direction of rotation of link AB
(CCW) equal to 0.525 m/s.
C
f
Ved
a, d
e, g
Vba = 0.525 m/s
b
112

 From b draw a line r
to BC and from d. Draw d line r
to CD
to interest at C.
 Vcb is given vector bc Vbc = 0.44 m/s
 Vcd is given vector dc Vcd = 0.39 m/s
113

Step – 4: To determine velocity of point E (Absolute
velocity) on link BC, first locate the position of point E
on velocity vector diagram. This can be done by taking
corresponding ratios of lengths of links to vector
distance i.e.
BC
BE
bc
be
  be =
BC
BE
x Vcb =
066.0
04.0
x 0.44 = 0.24 m/s
Join e on velocity vector diagram to zero velocity points a, d
vector de = Ve = 0.415 m/s.
114

Step 5: To determine angular velocity of links BC and CD, we
know Vbc and Vcd.
 Vbc = WBC x BC
 WBC = )(.sec/6.6
066.0
44.0
cwrad
BC
Vbc

Similarly, Vcd = WCD x CD
 WCD = s/r96.6
056.0
39.0
CD
Vcd
 (CCW)
115

Step – 6: To determine velocity of an offset point F
to CF from C on velocity vector diagram.
to BF from b on velocity vector diagram to
intersect the previously drawn line at ‘f’.
 From the point f to zero velocity point a, d and measure
vector fa/fd to get Vf = 0.495 m/s.
116

Step – 7: To determine velocity of an offset point.
to GC from C on velocity vector diagram.
to DG from d on velocity vector diagram to
intersect previously drawn line at g.
 Measure vector dg to get velocity of point G.
Vg = s/m305.0dg 
117

Step – 8: To determine rubbing velocity at pins
 Rubbing velocity at pin A will be
Vpa = ab x rad of pin A = 10.5 x 0.03 = 0.315 m/s
 Rubbing velocity at pin B will be
Vpb = (ab + cb) x rad of point at B.
[ab CCW and cbCW]
Vpb = (10.5 + 6.6) x 0.04 = 0.684 m/s.
 Rubbing velocity at point D will be
 cd x rpd of pin D
= 6.96 x 0.035 = 0.244 m/s
118

va
va'
A
O
A’
δθ
ω
α
ω+δω
ω+αδt
δθ
va'cos δθ
va'sin δθ
• Tangential velocity of point A, va
• During time δt angular distance
travelled δθ
• The Angular velocity at point A’
= ω + αδt
• Tangential velocity at point A’ =
va’ = r ω = r(ω + αδt)

va
va'
A
O
A’
δθ
ω
α
α
ω+δω
ω+αδt
δθ
va'cos δθ
va'sin δθ
Change in velocity perpendicular to OA
= 𝑣 𝑎′ cos 𝛿𝜃 − 𝑣 𝑎
=
𝑣 𝑎′ cos 𝛿𝜃 − 𝑣 𝑎
𝛿𝑡
Acceleration
=
𝑟(𝜔 + 𝛼𝛿𝑡) cos 𝛿𝜃 − 𝑟𝜔
𝛿𝑡
∴ 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑟 𝜔 + 𝑟 𝛼 𝛿𝑡 − 𝑟𝜔
𝛿𝑡
=
𝑟 𝜔 cos 𝛿𝜃 + 𝑟 𝛼 𝛿𝑡 cos 𝛿𝜃 − 𝑟𝜔
𝛿𝑡
As δt → 0 then δθ → 0 & cos δθ → 1
Acceleration of A perpendicular to A = r α

va
va'
A
O
A’
δθ
ω
α
α
ω+δω
ω+αδt
δθ
va'cos δθ
va'sin δθ
Change in velocity parallel to OA = 0
=
𝑣 𝑎′ sin 𝛿𝜃 − 0
𝛿𝑡
Acceleration
=
𝑟(𝜔 + 𝛼𝛿𝑡) sin 𝛿𝜃
𝛿𝑡
∴ 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑟 𝜔 𝛿𝜃 + 𝑟 𝛼 𝛿𝑡 𝛿𝜃
𝛿𝑡
=
𝑟 𝜔 𝛿𝜃
𝛿𝑡
= 𝑟 𝜔 𝜔 = 𝑟 𝜔2 =
𝑣2
𝑟
=
𝑟 𝜔 sin 𝛿𝜃 + 𝑟 𝛼 𝛿𝑡 sin 𝛿𝜃
𝛿𝑡
As δt → 0 then δθ → 0 & sin δθ → δθ
Acceleration of A parallel to A = v2 / r

• Two Components of acceleration
• Centripetal (Parallel to OA), fc = v2 / r
• Tangential (Perpendicular to OA), ft = r α
• When, α = 0, i.e. crank OA rotates with uniform angular
velocity, the value of ft
OA = 0 & Hence fc represents total
acceleration,
o1 oa
a1
o1
a1
oa
fc
OA
ft
OA
fc
OA
ft
OA
OA rotates CW OA rotates CCW

A
o
o1
G
B
O
a
g
b
a1
b1
ab
fc
OB Vob
2 / OB 21.93 m/s2 II to OB → O
fc
AB Vab
2 / AB 2.39 m/s2 II to AB → B
ft
AB L to AB
ft
A II to AO

A D
B
C
G
F
b
a,d
c
ω
α
fc
AB Vab
2 / AB 5.51 m/s2 II to AB → A
ft
AB r α 1.3 m/s2 L to AB
fc
BC Vbc
2 / BC 1.86 m/s2 II to BC → B
ft
BC ------ ----- L to BC
fc
CD Vcd
2 / CD 2.7 m/s2 II to CD → D
ft
CD ------ ----- L to CD

A D
B
C
G
F
b
a,d
c
a1,d1
cd
b1
ab
bc
c1
g1
f1
ω
α
fc
AB Vab
2 / AB 5.51 m/s2 II to AB → A
ft
AB r α 1.3 m/s2 L to AB
fc
BC Vbc
2 / BC 1.86 m/s2 II to BC → B
ft
BC ------ ----- L to BC
fc
CD Vcd
2 / CD 2.7 m/s2 II to CD → D
ft
CD ------ ----- L to CD
1
2
3
4
1
2
3
4

A
B
C
D
E
O
o,c
a
b
d
e
fc
OA Voa
2 / OA 10.975 m/s2 II to OA → O
fc
AB Vab
2 / AB 6.365 m/s2 II to AB → A
ft
AB ----- ----- L to AB
fc
BC Vbc
2 / BC 7.495 m/s2 II to BC → C
ft
BC ----- ----- L to BC
fc
DE Vde
2 / DE 0.386 m/s2 II to DE → D
ft
DE ----- ----- L to DE
ftotal
E ----- ----- Horizontal

fc
OA Voa
2 / OA 10.975 m/s2 II to OA → O
fc
AB Vab
2 / AB 6.365 m/s2 II to AB → A
ft
AB ----- ----- L to AB
fc
BC Vbc
2 / BC 7.495 m/s2 II to BC → C
ft
BC ----- ----- L to BC
fc
DE Vde
2 / DE 0.386 m/s2 II to DE → D
ft
DE ----- ----- L to DE
ftotal
A
B
C
D
E
O
o,c
a
b
d
e
o1,c1
a1
b1
d1
e1
de ab
bc

fc
OC Voc
2 / OC 53.39 m/s2 II to OC → O
ft
OC r α 7.5 m/s2 L to OC
fc
BC Vbc
2 / BC 2.01 m/s2 II to BC → C
ft
BC ----- ----- L to BC
fc
AB Vab
2 / AB 26.17 m/s2 II to AB → A
ft
AB ----- ----- L to AB
fc
BD Vde
2 / DE 8.78 m/s2 II to BD → D
ft
BD ----- ----- L to BD
ftotal
A
C
D
B
O
o,a
b
d
c

fc
OC Voc
2 / OC 53.39 m/s2 II to OC → O
ft
OC r α 7.5 m/s2 L to OC -----
fc
BC Vbc
2 / BC 2.01 m/s2 II to BC → C
ft
BC ----- ----- L to BC -----
fc
AB Vab
2 / AB 26.17 m/s2 II to AB → A
ft
AB ----- ----- L to AB -----
fc
BD Vde
2 / DE 8.78 m/s2 II to BD → B
ft
BD ----- ----- L to BD -----
ftotal
E ----- ----- Horizontal -----
A
C
D
B
O
o,a
b
d
c
o1,a1
bd
b1
d1
c1
oc
ab
bc

fc
OA Voc
2 / OC 53.39 m/s2 II to OC → O
fc
AB Vab
2 / AB 2.01 m/s2 II to AB → A
ft
AB ----- ----- L to AB -----
ftotal
B ----- ----- Horizontal -----
fc
CD Vde
2 / DE 8.78 m/s2 II to CD → D
ft
CD ----- ----- L to CD -----
ftotal
D ----- ----- Vertical -----
A
C
D
BO
ob
d
c
a

fc
OA Voc
2 / OC 19.75 m/s2 II to OC → O
fc
AB Vab
2 / AB 5.55 m/s2 II to AB → A
ft
AB ----- ----- L to AB -----
ftotal
B ----- ----- Horizontal -----
fc
CD Vde
2 / DE 6.32 m/s2 II to CD → C
ft
CD ----- ----- L to CD -----
ftotal
D ----- ----- Vertical -----
A
C
D
BO
ob
d
c
o1
b1
d1
c1
ab
cd
a
a1

Coriolis component of acceleration
δθ
ω
α
ω'
α'
v
rω
v'
r'ω'
v' = v + dv = v + f dt
r' = r + dr
ω' = ω + dω = ω + α dtv'
r'ω' r'ω' sin δθ
r'ω' cos δθ
v' sin δθ
v' cos δθ δθ
δθ

P
r
O
C
Slider - B
A on CD
DR
OA = 60 mm, NOA = 200 RPM clockwise, CD = 300 mm, DR = 400 mm, Angle BOC =
120°, OC = 160 mm, distance between horizontal line from R & point O = 120 mm. Find
out velocity & acceleration of ram R, acceleration of block A along slotted bar CD.
o,c
b
a
d
r1 f c
OB = Vob
2 / OB = 26.33 m/s2 II to OB → O
2 f c
AC = Vac
2 / AC = 4.052 m/s2 II to AC → C
3 f t
AC = unknown ┴ to AC -
4 f cr
AB = 2Vab ωCD = 8.021 m/s2
┴ to CD -
5 f c
AB = unknown II to CD -
6 f c
RD = Vrd
2 / RD = 0.322 m/s2 II to DR → D
7 f t
RD = unknown ┴ to DR -
8 f total
R = unknown - -

O
d1 rd
P
r
C
Slider - B
A on CD
DR
OA = 60 mm, NOA = 200 RPM clockwise, CD = 300 mm, DR = 400 mm, Angle BOC =
120°, OC = 160 mm, distance between horizontal line from R & point O = 120 mm. Find
out velocity & acceleration of ram R, acceleration of block A along slotted bar CD.
o,c
b
a
d
r
o1 , c1
b1
ba
ca
a1
r1
1 f c
OB = Vob
2 / OB = 26.33 m/s2 II to OB → O
2 f c
AC = Vac
2 / AC = 4.052 m/s2 II to AC → C
3 f t
AC = unknown ┴ to AC -
4 f cr
AB = 2Vab ωCD = 8.021 m/s2
┴ to CD -
5 f c
AB = unknown II to CD -
6 f c
RD = Vrd
2 / RD = 0.322 m/s2 II to DR → D
7 f t
RD = unknown ┴ to DR -
8 f total
R = unknown - -

SPUR GEARS
• Gear terminology, law of gearing, Characteristics of
involute action, Path of contact, Arc of contact, Contact
ratio of spur, helical, bevel and worm gears.
• Interference in involute gears.
• Methods of avoiding interference and Back lash.
• Comparison of involute and cycloidal teeth, Profile
modification.
138

Introduction
• Let the wheel A be keyed to the rotating shaft and the wheel B to the shaft, to
be rotated.
• A little consideration will show, that when the wheel A is rotated by a rotating
shaft, it will rotate the wheel B in the opposite direction as shown in Fig. (a).
• The wheel B will be rotated (by the wheel A) so long as the tangential force
exerted by the wheel A does not exceed the maximum frictional resistance
between the two wheels.
• But when the tangential force (P) exceeds the frictional resistance (F),
slipping will take place between the two wheels. Thus the friction drive is
not a positive drive.
In order to avoid the slipping, a
number of projections (called
teeth) as shown in Fig. (b), are
provided on the periphery of the
wheel A, which will fit into the
corresponding recesses on the
periphery of the wheel B. A
friction wheel with the teeth cut
on it is known as toothed
wheel or gear.
139

TYPES OF GEARS
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
140

SPUR GEAR
• 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
141

External and Internal spur Gear…
142

Helical Gear
• 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
143

Herringbone gears
• 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.
144

Rack and pinion
• 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
145

Bevel gears
• 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.
146

Straight and Spiral Bevel Gears
147

WORM AND WORM GEAR
• 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
148

NOMENCLATURE OF SPUR GEARS
149

• 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.
• 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°.
151

• 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.
152

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,
153

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 the
addendum and dedendum.
154

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.
155

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.
156

Path of contact: It is the path traced by the point of contact
of two teeth from the beginning to the end of engagement.
Length of the path of contact: It is the length of the
common normal cut-off by the addendum circles of the
wheel and pinion.
Arc of contact: It is the path traced by a point on the pitch
circle from the beginning to the end of engagement of a
given pair of teeth. The arc of contact consists of two
parts, i.e.
(a) Arc of approach. It is the portion of the arc of contact
from the beginning of the engagement to the pitch point.
(b) Arc of recess: It is the portion of the arc of contact
from the pitch point to the end of the engagement of a
pair of teeth.
157

• The law of gearing states the
condition which must be fulfilled
by the gear tooth profiles to
maintain a constant angular
velocity ratio between two
gears.
• Figure shows two bodies 1 and
2 representing a portion of the
two gears in mesh.
Condition for Constant Velocity Ratio of Toothed Wheels–Law of Gearing
A point C on the tooth profile of the gear 1 is in contact with a point D on the
tooth profile of the gear 2. The two curves in contact at points C or D must
have a common normal at the point. Let it be n - n.
Let,
ω1= instantaneous angular velocity of the gear 1 (CW)
ω2= instantaneous angular velocity of the gear 2 (CCW)
vc = linear velocity of C
vd = linear velocity of D
158

• Then vc = ω1 AC ; in a direction
perpendicular to AC or at an angle
α to n - n,
• vd = ω2 BD ; in a direction
perpendicular to BD or at an angle
β to n - n.
Now, if the curved surfaces of the teeth of two gears are to remain in contact,
one surface may slide relative to the other along the common tangent t-t.
The relative motion between the surfaces along the common normal n - n
must be zero to avoid the separation, or the penetration of the two teeth into
each other.
Component of vc along n - n = vc cos α
Component of vd along n - n = vd cos β
Relative motion along n - n= vc cos α- vd cos β
159

• Draw perpendiculars AE and BF on
n-n from points A and B
respectively.
Then CAE =α and
DBF = β
For proper contact,


vc = ω1 AC vd = ω2 BD
vc cos α- vd cos β =0
ω1 AC cos α- ω2 BD cos β =0
AP
BP
AE
BF
BFAE
BD
BF
BD
AC
AE
AC



2
1
21
21
0
0




160

• Thus, it is seen that the centre line AB is divided at P by the common
normal in the inverse ratio of the angular velocities of the two gears.
• If it is desired that the angular velocities of two gears remain constant, the
common normal at the point of contact of the two teeth should always
pass through a fixed point P which divides the line of centres in the
inverse ratio of angular velocities of two gears.
• As seen earlier, P is also the point of contact of two pitch circles which
divides the line of centres in the inverse ratio of the angular velocities of the
two circles and is the pitch point.
• Thus, for constant angular velocity ratio of the two gears, the common
normal at the point of contact of the two mating teeth must pass
through the pitch point.
Also, as the ∆AEP and ∆ BFP are similar.
EP
FP
AP
BP
AE
BF
EP
FP
AP
BP


2
1


161

VELOCITY OF SLIDING
If the curved surfaces of the two teeth of the gears 1 and 2 are to
remain in contact, one can have a sliding motion relative to the
other along the common tangent t-t at C or D.
Component of vc along t - t = vc sin α
Component of vd along t - t= vd sin β
Relative motion along n – n
= vc sin α - vd sin β
vc = ω1 AC and vd = ω2 BD
EP
FP
AP
BP
AE
BF
EP
FP
AP
BP


2
1


162

• In actual practice following are the two types of teeth commonly
used
1. Cycloidal teeth ; and 2. Involute teeth.
Cycloidal Teeth
• A cycloid is the curve traced by a point on the circumference of
a circle which rolls without slipping on a fixed straight line.
• When a circle rolls without slipping on the outside of a fixed
circle, the curve traced by a point on the circumference of a
circle is known as epi-cycloid.
• On the other hand, if a circle rolls without slipping on the inside
of a fixed circle, then the curve traced by a point on the
circumference of a circle is called hypo-cycloid.
163

Construction of cycloidal teeth for Rack
• In Fig. (a), the fixed line or pitch line of a rack is shown. When the
circle C rolls without slipping above the pitch line in the direction as
indicated in Fig, then the point P on the circle traces epi-cycloid PA.
This represents the face of the cycloidal tooth profile.
• When the circle D rolls without slipping below the pitch line, then the
point P on the circle D traces hypo-cycloid PB, which represents the
flank of the cycloidal tooth. The profile BPA is one side of the
cycloidal rack tooth.
Similarly, the two curves P' A' and
P'B' forming the opposite side of
the tooth profile are traced by the
point P' when the circles C and D
roll in the opposite directions.
167

Construction of cycloidal teeth for gear
• The cycloidal teeth of a gear may be constructed as shown in Fig.
• The circle C is rolled without slipping on the outside of the pitch circle
and the point P on the circle C traces epi-cycloid PA, which represents
the face of the cycloidal tooth.
• The circle D is rolled on the inside of pitch circle and the point P on the
circle D traces hypo-cycloid PB, which represents the flank of the tooth
profile.
The profile BPA is one side of the
cycloidal tooth. The opposite side
of the tooth is traced as explained
above.
168

• An involute of a circle is a plane curve generated by a point on a
tangent, which rolls on the circle without slipping.
169

Involute Teeth
• An involute of a circle is a plane curve generated by a point on a
tangent, which rolls on the circle without slipping as shown in Fig.
• In connection with toothed wheels, the circle is known as base circle.
The involute is traced as follows :
A3, the tangent A3T to the involute is perpendicular
to P3A3 and P3A3 is the normal to the involute.
In other words, normal at any point of an
involute is a tangent to the base circle.
170

Comparison Between Involute and Cycloidal Gears
• In actual practice, the involute gears are more commonly used as compared to cycloidal
gears, due to the following advantages :
Advantages of involute gears
• The most important advantage of the involute gears is that the centre distance for a pair
of involute gears can be varied within limits without changing the velocity ratio. This is
not true for cycloidal gears which requires exact centre distance to be maintained.
• In involute gears, the pressure angle, from the start of the engagement of teeth to the
end of the engagement, remains constant. It is necessary for smooth running and less
wear of gears. But in cycloidal gears, the pressure angle is maximum at the beginning of
engagement, reduces to zero at pitch point, starts decreasing and again becomes maximum
at the end of engagement. This results in less smooth running of gears.
• The face and flank of involute teeth are generated by a single curve where as in cycloidal
gears, double curves (i.e. epi-cycloid and hypo-cycloid) are required for the face and flank
respectively. Thus the involute teeth are easy to manufacture than cycloidal teeth. In
involute system, the basic rack has straight teeth and the same can be cut with simple
tools.
• Note : The only disadvantage of the involute teeth is that the interference occurs with
pinions having smaller number of teeth. This may be avoided by altering the heights of
addendum and dedendum of the mating teeth or the angle of obliquity of the teeth.
171

Advantages of cycloidal gears
Following are the advantages of cycloidal gears :
• Since the cycloidal teeth have wider flanks, therefore the cycloidal
gears are stronger than the involute gears, for the same pitch.
• In cycloidal gears, the contact takes place between a convex flank and
concave surface, whereas in involute gears, the convex surfaces are in
contact. This condition results in less wear in cycloidal gears as
compared to involute gears. However the difference in wear is
negligible.
• In cycloidal gears, the interference does not occur at all. Though
there are advantages of cycloidal gears but they are outweighed by the
greater simplicity and flexibility of the involute gears.
172

• Fig. shows a pinion with centre O1, in mesh with wheel or gear with centre
O2.
• MN is the common tangent to the base circles and KL is the path of contact
between the two mating teeth.
• 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.
• A little consideration will
show, that if the radius of the
addendum circle of pinion is
increased to O 1 N, 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.
173

• Similarly, if the radius of the addendum circle of the wheel increases beyond
O2M, then the tip of tooth on wheel will cause interference with the tooth on
pinion. The points M and N are called interference points.
• From the above discussion, 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.
• When interference is just avoided, the maximum length of path of contact is
MN when the maximum addendum circles for pinion and wheel pass through
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.
174

Beginning of
engagement
End of engagement
175

Pinion (Driver) in CW
Gear
Pinion (Driver) in CCW
Gear
Beginning of
engagement
End of engagement
176

Methods of elimination of Gear tooth
Interference
In certain spur designs if interference exists, it can be
overcome by:
1. Removing the cross hatched tooth tips i.e., using stub
teeth.
2. Increasing the number of teeth on the mating pinion.
3. Increasing the pressure angle
4. Tooth profile modification or profile shifting
5. Increasing the centre distance.
177

Minimum Number of Teeth on the wheel in Order to Avoid
Interference
• We have already discussed that in order to avoid interference, the addendum
circles for the two mating gears must cut the common tangent to the base
circles between the points of tangency. The limiting condition reaches, when
the addendum circles of pinion and wheel pass through points N and M
178

Note: Use this formula while
calculating minimum number
of teeth of wheel and then use
gear ration to calculate the
teeth of pinion
179

Minimum Number of Teeth on the Pinion in Order to
Avoid Interference
• We have already discussed that in order to avoid interference, the addendum
circles for the two mating gears must cut the common tangent to the base
circles between the points of tangency. The limiting condition reaches, when
the addendum circles of pinion and wheel pass through points N and M
180

Two 20° involute spur gears mesh externally and give a velocity
ratio of 3. Module is 3 mm and the addendum is equal to 1.1
module. If the pinion rotates at 120 rpm, determine (i) the
minimum number of teeth on each wheel to avoid interference (ii)
the number of pairs of teeth in contact.
182

Two 20° involute spur gears mesh externally and give a velocity
ratio of 3. Module is 3 mm and the addendum is equal to 1.1
module. If the pinion rotates at 120 rpm, determine (i) the
minimum number of teeth on each wheel to avoid interference (ii)
the number of pairs of teeth in contact.
Thus, 1 pair of teeth will always remain in contact whereas for
78% of the time, 2 pairs of teeth will be in contact.
183

Two 20° involute spur gears have a module of 10 mm. The
addendum is equal to one module. The larger gear has 40 teeth
while the pinion has 20 teeth. Will the gear interfere with the
pinion?
184

A pair of spur gears with involute teeth is to give a gear ratio of 4:
1. The arc of approach is not to be less than the circular pitch and
smaller wheel is the driver. The angle of pressure is 14.5°. Find :
1. the least number of teeth that can be used on each wheel, and
2. the addendum of the wheel in terms of the circular pitch.
185

A pair of spur gears with involute teeth is to give a gear ratio of 4:
1. The arc of approach is not to be less than the circular pitch and
smaller wheel is the driver. The angle of pressure is 14.5°. Find :
1. the least number of teeth that can be used on each wheel, and
2. the addendum of the wheel in terms of the circular pitch.
186

Two 20° involute spur gears have a module of 10 mm. The addendum is one
module. The larger gear has 50 teeth and the pinion has 13 teeth. Does
interference occur? If it occurs, to what value should the pressure angle be
changed to eliminate interference?
187

The following data relate to two meshing involute gears: Number of teeth
on the gear wheel = 60 ; pressure angle = 20o ; Gear ratio = 1.5; Speed of the
gear wheel = 100 rpm; Module = 8 mm; The addendum on each wheel is such
that the path of approach and the path of recess on each side are 40% of the
maximum possible length each. Determine the addendum for the pinion and the
gear and the length of the arc of contact.
188

A pinion of 20o involute teeth rotating at 275 rpm meshes with a gear and
provides a gear ratio of 1.8. The number of teeth on the pinion is 20 and the
module is 8 mm. If the interference is just avoided, determine (i) the addenda
on the wheel and the pinion (ii) the path of contact, and (iii) the maximum
velocity of sliding on both sides of the pitch point.
189

The pinion 1 is the driver and is rotating clockwise.
The wheel 2 is driven in the counter-clockwise
direction. EF is their common tangent to the base
circles.
Contact of the two teeth is made where the
addendum circle of the wheel meets the line of
action EF, i.e., at C and is broken where the
addendum circle of the pinion meets the line of
action, i.e., at D.
CD is then the path of contact.
Let r = pitch circle radius of pinion
R = pitch circle radius of wheel
ra = addendum circle radius of pinion
Ra = addendum circle radius of wheel.
Path of contact = path of approach + path of
recess
CD = CP + PD
= (CF-PF)+(DE-PE)
Observe that the path of approach can be found if the dimensions of the driven
wheel are known. Similarly, the path of recess is known from the dimensions of
    sincossincos 222222
rrrRRR aa 
     sincoscos 222222
rRrrRR aa 
190

CD = CP + PD
= (CF-PF) + (DE-PE)
    sincossincos 222222
rrrRRR aa 
     sincoscos 222222
rRrrRR aa 
Ra C
222
cosRRa 
B
F
222
cosrra 
191

• The arc of contact is the distance travelled by a point on either
pitch circle of the two wheels during the period of contact of a
pair of teeth.
• In Fig., at the beginning of engagement, the driving involute is
shown as GH; when the point of contact is at P it is shown as JK
and when at the end of engagement, it is DL.
• The arc of contact is P'P" and it consists of the arc of approach
P'P and the arc of recess PP".
Let the time to traverse the arc of approach is ta.
Then,
Arc of approach = P'P = Tangential velocity of P’ x Time of approach
192

Arc FK is equal to the path FP as the point P is on the generator
FP that rolls on the base circle FHK to generate the involute PK.
Similarly, arc FH = Path FC.
Arc of recess = PP" = Tang. vel. of P x Time of recess
193

• The arc of contact is the length of the pitch circle traversed by
a point on it during the mating of a pair of teeth.
• Thus, all the teeth lying in between the arc of contact will be
meshing with the teeth on the other wheel.
Therefore, the number of teeth in contact =
• As the ratio of the arc of contact to the circular pitch is also the
contact ratio, the number of teeth is also expressed in terms of
contact ratio.
• For continuous transmission of motion, at least one tooth of
one wheel must be in contact with another tooth of the second
wheel.
• Therefore, n must be greater than unity.
cp
1
cos
contactofpath
pitchCircular
contactofArc


Where Pc = πD/T
194

Each of two gears in a mesh has 48 teeth and a module of 8 mm.
The teeth are of 20° involute profile. The arc of contact is 2.25
times the circular pitch. Determine the addendum.
195

A pinion having 30 teeth drives a gear having 80 teeth. The
profile of the gears is involute with 20° pressure angle, 12 mm
module and 10 mm addendum. Find the length of path of
contact, arc of contact and the contact ratio.
196

A pinion having 30 teeth drives a gear having 80 teeth. The profile of the gears
is involute with 20° pressure angle, 12 mm module and 10 mm addendum.
Find the length of path of contact, arc of contact and the contact ratio.
197

Two involute gears in mesh have 20o pressure angle. The gear
ratio is 3 and the number of teeth on the pinion is 24. The teeth
have a module of 6 mm. The pitch line velocity is 1.5 m/s and the
addendum equal to one module. Determine the angle of action of
the pinion (the angle turned by the pinion when one pair of teeth is
in the mesh) and the maximum velocity of sliding.
198

Two involute gears in a mesh have a module of 8 mm and a
pressure angle of 20°. The larger gear has 57 while the pinion has
23 teeth. If the addenda on pinion and gear wheels are equal to
one module, find the (i) contact ratio (ii) angle of action of the
pinion and the gear wheel (iii) ratio of the sliding to rolling velocity
at the (a) beginning of contact (b) pitch point (c) end of contact
199

Two involute gears in a mesh have a module of 8 mm and a pressure angle of
20°. The larger gear has 57 while the pinion has 23 teeth. If the addenda on
pinion and gear wheels are equal to one module, find the (i) contact ratio (ii)
angle of action of the pinion and the gear wheel (iii) ratio of the sliding to rolling
velocity at the (a) beginning of contact (b) pitch point (c) end of contact
200

Two 20° gears have a module pitch of 4 mm. The number of teeth
on gears 1 and 2 are 40 and 24 respectively. If the gear 2 rotates
at 600 rpm, determine the velocity of sliding when the contact
is at the lip of the tooth of gear 2. Take addendum equal to one
module. Also, find the maximum velocity of sliding.
Let pinion (gear 2) be the driver. The
tip of the driving wheel is in contact
with a tooth of the driven wheel at the
end of engagement. Thus, it is
required to find the path of recess
which is obtained from the dimensions
of the driving wheel.
201

(ii) In case the gear wheel is the driver, the tip of the pinion will be in contact
with the flank of a tooth of the gear wheel at the beginning of contact. Thus, it is
required to find the distance of the point of contact from the pitch point, i.e.. path
of approach. The path of approach is found from the dimensions of the driven
wheel which is again pinion.
Thus, path of approach
=9.458mm, same as before and velocity of sliding = 950.8 mm/s
Thus, it is immaterial whether the driver is the gear wheel or the pinion, the
velocity of
sliding is the same when the contact is at the tip of the pinion. The maximum
velocity of sliding will depend upon the larger path considering any of the
wheels to be the driver. Consider pinion to be the driver. Path of recess = 9.458
mm
Path of approach
This is also the path of recess if the wheel becomes the driver.
Maximum velocity of sliding
202

Gears – What are they?
Gears are wheels with teeth. Gears mesh together and
make things turn. Gears are used to transfer motion or power
from one moving part to another.
Sports cars go fast (have speed)
but cannot pull any weight.
Big trucks can pull heavy loads
(have power), but cannot go fast.
Gears cause this.
Gears increase or decrease the
power or speed. 203

Gears are generally used for
following different reasons
To reverse the direction of rotation
To increase or decrease the speed of rotation
To move rotational motion to a different axis
204

TYPES OF GEARS
According to the position of axes of the shafts.
a. Parallel
1.Spur Gear
2.Helical Gear
3.Double Helical Gear
4.Rack and Pinion
b. Intersecting
Bevel Gear
c. Non-intersecting and Non-parallel
1.Worm and worm gears
2.Spiral gears
3.Hypoid gears 205

Classification of Cam
According to
Shape
1. Wedge & Flat Cam
2. Radial or Disc Cam
3. Spiral Cam
4. Cylindrical Cam
5. Conjugate Cam
6. Globoidal Cam
7. Spherical Cam
208
According to
Follower
Movement
Rise-Return-Rise
Dwell-Rise-Return-Dwell
Dwell-Rise-Dwell-Return-
Dwell
According to Manner of constraint
of the Follower
Pre-Loaded Spring Cam
Positive-drive Cam
Gravity Cam

Schematic Diagrams
209
1. Wedge or Flat Cam 2. Radial/ Disc Cam
3. Spiral Cam
4. Cylindrical Cam

Schematic Diagram
210
5. Conjugate Cam
6. Globoidal Cam
7. Spherical Cam

Schematic Diagram
211
According to Follower Movement
(a) R-R-R
(b) D-R-R-D
(c) D-R-D-R-D

Classification of
Follower
212
According to
Shape
a. Knife-edge
b. Roller
c. Mushroom
d. Flat Faced
According to
Movement
Reciprocating
Oscillating
According to
Location of Line
of Movement
Radial
Offset

Schematic Diagrams
213
Radial Followers
Offset Followers
Reciprocating Oscillating

Cam Terminology
• Base Circle: It is the smallest circle tangent to the cam profile
drawn from the centre of rotation of a radial cam.
• Trace Point: It is a reference point on the follower to trace the
cam profile such as the knife-edge of a knife-edged follower and
centre of the roller of a roller follower.
• Pitch Curve: It is the curve drawn by the trace point assuming
that the cam is fixed, and the trace point of the follower rotates
around the cam.
• Pressure Angle: It is the angle between the normal to the pitch
curve at a point and the direction of the follower motion.
• Pitch Point: It is the point on the pitch curve at which the pressure
angle is maximum.
• Pitch Circle: It is the circle passing through the pitch point and
concentric with the base circle.
• Prime Circle: The Smallest Circle drawn tangent to the pitch curve
is known as the Prime Circle
214

Type of Motion of Follower
1. Simple Harmonic Motion
2. Uniform Acceleration & Deceleration
3. Uniform Velocity
4. Cycloidal
216
Cam displacement/Speed calculation Notations
s = Instantaneous Follower Displacement
h = Maximum Follower Displacement
v = Velocity of the Follower
f = Acceleration of the Follwer
θ = Instantaneous Cam Rotation Angle
φ = Cam Rotation Angle for Maximum Follwer Displacement
β = Angle on the Harmonic Circle

1. SHM
217
Distance 𝑠 =
ℎ
2
1 − cos 𝛽
Velocity = 𝑣 =
ℎ
2
𝜋𝜔
𝜑
sin
𝜋𝜃
𝜑
Acceleration = 𝑓 =
ℎ
2
𝜋𝜔
𝜑
2
sin
𝜋𝜃
𝜑
Maximum Velocity = 𝒗 𝒎𝒂𝒙 =
𝒉
𝟐
𝝅𝝎
𝝋
, 𝒂𝒕 𝜽 =
𝝋
𝟐
Maximum Acceleration = 𝒇 𝒎𝒂𝒙 =
𝒉
𝟐
𝝅𝝎
𝝋
𝟐
, 𝒂𝒕 𝜽 = 𝟎

2. Uniform Acceleration &
Deceleration
218
Velocity = 𝑣 =
4ℎ𝜔
𝜑2 𝜃
2𝑠
𝑡2 =
4ℎ𝜔2
𝜑2
𝟐𝒉𝝎
𝝋
, 𝒂𝒕 𝜽 =
𝝋
𝟐

3. Uniform Velocity
219
Velocity = 𝑣 =
ℎ𝜔
𝜑
= 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

4. Cycloidal
220
Velocity = 𝑣 =
ℎ𝜔
𝜑
1 − cos
2𝜋𝜃
𝜑
2ℎ𝜔2
𝜑2 sin
2𝜋𝜃
𝜑
𝟐𝒉𝝎
𝝋
, 𝒂𝒕 𝜽 =
𝝋
𝟐
Maximum Acceleration = 𝒇 𝒎𝒂𝒙 =
𝟐𝒉𝝎 𝟐
𝝋 𝟐 , 𝒂𝒕 𝜽 =
𝝋
𝟒

References:
1. Theory of Machines, Rattan, Tata McGraw-Hill Education,
2009.
2. Theory of Machines, R S Kurmi, Eurasia Publishing House,
2005
221

KOM 2131906 GTU Lecture notes

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KOM 2131906 GTU Lecture notes