Kinematic Synthesis

Kinematic Synthesis of Planar Mechanisms
Yatin Kumar Singh Page 1
16.1 Introduction
Kinematic Analysis is the process of determination of velocity and acceleration of the various links of an
existing mechanism. Kinematic Synthesis deals with the determination of the lengths and orientation of the
various lengths of the links so that a mechanism could be evolved to satisfy certain conditions.
Kinematic synthesis of a mechanism requires the determination of lengths of various links that satisfy the
requirements of motion of the mechanism. The usual requirements are related to specified positions of the
input and output links. It is easy to design a planar mechanism, when the positions of input and output links
are known at fewer positions as compared to large number of positions.
16.2 Movability (or Mobility) or Number Synthesis
Movability of a mechanism means the number of degrees of freedom, which is equal to the number of
independent coordinates required to specify its configurations in order to define its motion. This concept is
also known as Number Synthesis.
The Gruebler’s (or Kutzbach) criterion for degrees of freedom of planar mechanisms is given by:
= − − � − .
Where
w�� = + + + ⋯
= number of simple joints or lower pairs having one degree of freedom
n2 = number of binary links
n3 = number of ternary links
n4 = number of quaternary links, and so on
h = number of higher pairs having two degrees of freedom
n = n2 + n3 + n4 + ··· + ni = total number of links.
If h = 0, then
= − − � .
= [ + + + ⋯ − ] − + + + ⋯
Simplifying and re-arranging the equation, we get
= + + [ + + + ⋯ + − ] .
For a fully constrained mechanism, F = 1. Thus
= + [ + + + ⋯ + − ] .
From Eq. (16.4), it is quite evident that the minimum number of binary links is equal to four. Therefore, the
four-bar kinematic chain is the simplest mechanism.
For
F = 1, n2
F = 2, n2
F = 3, n2
············

= �, ≥ � +
For a fully constrained motion, F = 1, so that
= − − � o� � = − .
16.3 Transmission Angle
Transmission angle μ is the interior angle between the coupler and output link as shown in Fig.16.1. If link AB
is the input link, the force applied to the output link DC is transmitted through the coupler BC. For a particular
value of force in the coupler rod, the torque transmitted to the output link (about point D) is maximum when
the transmission angle μ is °. )f links BC and DC become coincident, the transmission angle is zero and the
mechanism would lock or jam. )f μ deviates significantly from °, the torque on the output link decreases.
Sometimes, it may not be sufficient to overcome the friction in the system and the mechanism may be locked
or jammed. (ence, μ is generally kept more than °. The transmission angle is nearly ° for the best
mechanism. To find the positions where the transmission angle is maximum or minimum, apply cosine law to
Δs ABD and BCD.
Fig.16.1 Four-bar mechanism
+ − � = � and + − � = �
From Eqs. (1) and (2), we have
+ − − − � + � =
For maximum or minimum values of μ,
�
�
=
o� � − � ×
�
�
=
�
�
=
�
�
=
o� � =
Since a and d are not zero, so sin θ = or θ = ° or °
Thus, transmission angle is maximum when θ = °, and minimum when θ = °.

16.3.1 Transmission Angle in Slider–Crank Mechanism
The slider-crank mechanism ABC is shown in Fig. . where μ is the transmission angle. To find the positions
of maximum or minimum values of μ, we have
= − = � −
Fig.16.2 Slider-crank mechanism
From Δ BEC, we have
= ° − � = � = � = �
From eq.1 and 2
� − = �
Differentiating Eq. w.r.t. θ and equalizing it to zero, we get
� = − � ×
�
�
�
�
= −
�
�
=
o� � =
As a is not equal to zero, so cos θ =
Thus θ = ° or ° gives the maximum or minimum values of transmission angle μ.
16.4 Limit Positions and Dead Centres of a Four-Bar Mechanism
For the design of four-link mechanism, limit positions and dead centres are essential.
Limit Position: It is the position in which the interior angle between its coupler and input link is either 180°
or 360°. In the limit position of the mechanism, the pivot points A, D and C lie on a straight line as shown in
Fig.16.3(b). In this case, the angle between the input link AD and coupler DC is 180°. A four-link mechanism
can have maximum two limit positions. The second limit position is shown in Fig.16.3(c) where, the angle
between the coupler DC and input link AD is 360°.
The angle of oscillation of output-link BC is given as,
Δϕ = ϕ − ϕ

where ϕ and ϕ are the two extreme positions of output link, BC.
Fig.16.3 Limit positions of four-bar mechanism
Dead Centre Position: It is the position in which the interior angle between the coupler and the output link
(of a four-link mechanism) is either 180° or 360° (refer to Fig.16.4). In dead centre positions, the pivot points
B, C, and D lie in a straight line. There can be two (maximum) dead centre positions. A crank-crank four-bar
link mechanism does not have either a limit position or dead centre positions because both the cranks rotate
through 360°.
Fig.16.4 Dead centre positions of four-bar mechanism
16.5 Dimensional Synthesis
It deals with the determination of actual dimensions of the mechanism to satisfy the specified motion
characteristics. The actual dimensions could be the lengths of the links between adjacent hinge pairs, angles
between the arms of a bell-crank lever, or cam contour dimensions, etc. Synthesis of mechanisms can be
carried out by graphical methods or analytical methods.

16.6 Graphical Method
16.6.1 Pole
Consider a four-bar mechanism O2 ABO4 in two positions O2A1B1O2 and O2A2B2O4 as shown in Fig.16.6. The
coupler link AB has moved from the position A1B1 to A2B2.
Fig.16.6 Pole of a four-bar mechanism
The input link O2A and output link O4B have moved through angles θ12 and ϕ12 respectively in the clockwise
direction. For the motion of the coupler AB from A1B1 to A2B2, P12 is its centre of rotation with respect to the
fixed link. P12 lies at the point of intersection of the perpendicular bisectors of the coupler link AB in its two
positions A1B1 and A2B2. Hence P12 is called the pole. P12 also lies at the point of intersection of the mid-
normals a12 and b12 of the chords A1 A2 and B1 B2 respectively.
Properties of Pole Point
Since AB = A1B1 = A2B2, the perpendicular bisectors of A1 A2 and B1 B2 pass through fixed centres O2 and O4.
Since coupler AB rotates about P12 from position A1B1 to A2 B2, and therefore,
ΔA1 B1 P12 = ΔA2 B2 P12,
∠2 + ∠3 + ∠1 = ∠1 + ∠4 + ∠5
or angle subtended by A1 B1 at P12 = angle subtended by A2 B2 at P12.
Since ∠2 + ∠3 + ∠1 = ∠1 + ∠4 + ∠5
∠2 + ∠3 = ∠4 + ∠5
Thus A1 A2 and B1 B2 subtend equal angle at P12. Since P12 lies on the perpendicular bisectors of A1 A2 and B1
B2,
∠2 = ∠3 and ∠4 = ∠5
Since ∠2 + ∠3 = ∠4 + ∠5
And ∠2 = ∠3 and ∠4 = ∠5
∠2 = ∠4 and ∠3 = ∠5

Therefore, input and output links subtend equal angles at pole P12 as they move from one position to
another.
Since ∠2 + ∠3 + ∠1 = ∠1 + ∠4 + ∠5 = ∠1 + ∠4 + ∠3 ( ∠5 = 3)
i.e., angle subtended by the coupler link AB = angle subtended by the fixed link O2O4.
The triangle A1B1P12 moves as one link about P12 to the position A2 B2 P12. Angular displacement of coupler A1
B1 = angular displacement of P12 B i.e., β12 = ∠4 + ∠5.
16.6.2 Relative Pole
The pole of a moving link is the centre of its rotation with respect to a fixed link. However, if the rotation of
the link is considered relative to another moving link, the pole is called as the relative pole. To determine the
relative pole, fix the link of reference and observe the motion of the other link in the reverse direction.
(a) Determination of Relative Pole for Four-Bar Chain
Consider the four-bar chain O2ABO4 in its two positions O2 AB1O4 and O2 AB2O as shown in Fig.16.7 with links
O2A fixed. The relative pole of AB relative to O2 A is at A. The relative pole of O4B relative to O2A is at R12, that
can be determined as follows:
Let θ12 = angle of rotation of O2A (clockwise)
ϕ12 = angle of rotation of O4B (clockwise)
Fig.16.7 Determination of relative pole of four-bar chain
1.Assume O2 and A as the fixed pivots and rotate O2O4 about O2 through angle θ12 in the counter-clockwise
direction (opposite to the direction of rotation of O2A). Let od4 be the new position after the rotation of O2od4.
2. Locate the point B2 by drawing arc with centres A and od4 and radii equal to AB1 and O4B1 respectively. Then
O2 AB2od4 is inversion of O2AB1od4.
3. Draw mid-normals of O2od4 and B1 B2 passing through O2 and A respectively to intersect at R12, which is the
required relative pole.
ϕ12 − θ12 = angle of rotation of the output link O4B relative to the input link.
The angle will be negative if O4 B > O2A and positive if O4B < O2A.
Angular displacement of R12O4 = angular displacement of O4B1.
∠O4R12O’4 = (∠ϕ12 − ∠θ12) assuming O4B > O2A

2 ∠ = − ∠ϕ12 − ∠θ12)
∠ = ∠ϕ − ∠θ
)n ΔO2 R12 O4,
∠4 = ∠1 + ∠2
∠θ = ∠ϕ − ∠θ + ∠ ∠ = ∠
= ∠ϕ + ∠θ + ∠
∠ = ∠ϕ
Procedure:
1.Join O2O4 and extend it further. Rotate O2O4 about O2 through an angle ½ θ12 in a direction opposite to that of
O2A.
2. Again rotate O2O4 about O4 through an angle ½ ϕ12 in a direction opposite to that of O4B.
3. The point of intersection of these two positions of O2O4 after rotation about O2 and O4, is the required relative
pole R12.
4.The angles subtended by O4 od4 and B1 B2 at R12 are the same.
i.e., ∠O4 R12 od4 = ∠B1 R12 B2
or 2 ∠O4 R12 O2 = 2 ∠B1 R12 A
or ∠O4 R12 O2 = ∠B1 R12A
Angle subtended by the fixed pivots (O2 and O4) at the relative pole = Angle subtended by the coupler AB.
(b) Determination of Relative Pole for Slider-Crank Mechanism
Consider the slider-crank mechanism shown in Fig.16.8. The point B on the slider reciprocates through a
horizontal distance x. Its centre of rotation will lie at infinity on a vertical line where the point O4 can also be
assumed to lie. Then O2 O4 will also be a vertical line through O2. Rotate O2 O4 about O4 through / θ12 in the
counter-clockwise direction. Rotating O2O4 about O4 through / θ12 would mean a vertical line towards left
of O2, at a distance of x/2. The intersections of the two lines locate R12. The procedure to locate the relative
pole of a slider-crank mechanism is as follows:
Fig.16.8 Determination of relative pole of slider-crank mechanism

1. Draw two parallel lines l1 and l2 at a distance equal to the eccentricity e (if there is any eccentricity).
2.Select a line segment O2C = x/2 on line l1 such that C is measured in a direction opposite to the motion of the
slider.
3. Draw perpendicular lines O2P1 and C P2 to line l1.
4.Make ∠1/2 θ12 at point O2 with the line O2P1 in a direction opposite to the rotation of the input link.
5. The intersection of this line with the line CP2 extended locates the relative pole R12.
16.7 Design of Mechanisms by Relative Pole Method
16.7.1 Four-Bar Mechanism
(a) Two-Position Synthesis
Let for a four-bar mechanism, length of the fixed link O2O4 along with the angular displacement θ12 (between
position 1 and 2) of the input link O2A and angular displacement ϕ12 (between positions 1 and 2) of the
output link O4B, are known.
To design the mechanism (Fig.16.9), first locate the relative pole R12
� = � � �
= ∠θ − ∠ϕ Assu��n� O B ∠O A
= ∠�
Procedure:
1.Construct an angle ψ12 at an arbitrary position R12. Join any two points on the two arms of the angle to obtain
the coupler link AB of the mechanism. Join A with O2 to get the input and output links respectively.
2. Locate point B arbitrarily so that BO4 is the output link. Construct ∠BR12 Z = ψ12. Take any suitable point A or
R12 Z. Join AB and O2A.
3. Instead of locating point B as above, locate point A arbitrarily so that O2A is the input link. Construct ∠AR12 Y
= ψ12. Select any point B on R12 Y. Join A and B to get AB. Join B with O4 to get output link O4B.

Fig.16.9 Two-position method for four-bar mechanism
(b) Three Position Synthesis
Let three positions of input link θ1, θ2 and θ3, and three positions of output link ϕ1, ϕ2 and ϕ3 are known. The
relative poles R12 and R13 can be determined considering,
θ12 = θ2 − θ1 and ϕ12 = ϕ2 − ϕ1 to locate R12
and θ13 = θ3 − θ1 and ϕ13 = ϕ3 − ϕ1 to locate R13
An�� subt�nd�d at R , ∠� = ∠θ − ∠ϕ
An�� subt�nd�d at R , ∠� = ∠θ − ∠ϕ
Fig.16.10 Three-position method for four-bar mechanism
Construct the angles ψ12 and ψ13 at the points R12 and R13 respectively at arbitrary positions such that the
arms of the angles intersect at A and B. Join A and B to get the coupler AB. Join A with O2 and B with O4 to get
input and output links respectively, as shown in Fig.16.10.
16.7.2 Slider–Crank Mechanism
(a) Two Position Synthesis:
Let θ12 = angular displacement of input link O2 A angle between θ1 and θ2) in clockwise direction.

x = linear displacement of slider to the right
e = eccentricity
Fig.16.11 Two position method for slider-crank mechanism
Procedure:
1.Draw two parallel lines l1 and l2 at a distance equal to eccentricity, e.
2. Construct an angle equal to θ12/2 at point R12, chosen arbitrarily and at a convenient position. The
intersection of an arm N of this angle with line l2 gives the position of the slider at point B.
3. Select point A arbitrarily on the other arm M of the angle. Join A with O2 to get the input link (crank) O2A. Join
A and B to get the coupler (connecting rod).
(b) Three-Position Synthesis
Three positions of input link θ1, θ2, and θ3 are known, along with the corresponding slider positions x1, x2 and
x3. Find R12 and R13 as shown in Fig.16.12.
Fig.16.12 Three position method for slider-crank mechanism

Here θ12/2= angle made by the fixed link at R12.
θ13/2= angle made by the fixed link at R13.
Procedure:
1.Construct angle θ12/2= ∠M1 R12 N1 at R12 in an arbitrary position with arm N1 locating point B.
2. Draw angle θ13/2= ∠M2 R13 N2 at R13 with an arm N2 along R13 B.
3.Intersection of the two arms M1 at R12 and M2 at R13 (not through B) of the two angles locates the point A. Join
A with B to get the coupler AB.
16.8 Errors in Kinematic Synthesis of Mechanisms
There are three types of errors present in the design of linkages for function generation. These errors are:
1.Structural,
2.Mechanical, and
3.Graphical.
In function generation, there is correlation between the motion of input and output links. If the motions of
input link is represented as x0, x1, x2, …, xn+1 and the corresponding motions (which is dependent on input
variables x0, x1, …, xn+1) of the output link is represented by y0, y1, y2, …, yn+1, these can be shown on the graph
as indicated in Fig.16.13. We observe that at certain points (P1, P2, and so on) the desired function and
generated function agree well. These points are called Precision Points. The number of such points (from 3
to 6 generally) is equal to the number of design parameters. Except these points the generated function curve
and desired function curve do not agree and are deviating by certain amount of error which is known as
Structural Error. Structural error is the difference between the generated function and the desired
function for a certain value of input variable. So, the precision points are spaced in such a way as to minimize
the structural error of the linkage.
Fig.16.13 Errors in function generation of linkages
Mechanical Errors are caused because of mechanical defects such as improper machining, casting of
components of the linkage, clearance in the components because of rubbing, overloading of linkages, etc.
Graphical Error is caused because of inaccuracy in drawing of perpendicular or parallel lines. It may occur
because of wrong graphical construction and wrong choice of scale. Also, there may be human errors in
drawing work.

16.9 Analytical Method
16.9.1 Function Generation
In function generation, the motion of input (or driver) link is correlated to the motion of output (or follower)
link. Let θ and ϕ be the angles of rotation of input and output links respectively. Let y = f(x) be the function to
be generated. The angle of rotation θ of the input link O2A represents the independent variable x and the
angle of rotation ϕ of the output link O4B represents the dependent variable y, as shown in Fig.16.14. The
relation between x and θ and that between y and ϕ is generally assumed to be linear. Let θi and ϕi be the
initial values of θ and ϕ representing xi and yi respectively.
� − �
−
= � = . =
� − �
−
.
And
ϕ − ϕ
−
= � = . =
ϕ − ϕ
−
.
where the constants rx and ry are called Scale Factors. The subscripts i and f denote the initial and final
values.
Fig.16.14 Function generation method
. . Chebyshev’s Spacing for Precision Points
Let xi and xj be the initial and final values of variable x respectively. A function f(x) is desired to be generated
in the interval xi x xf. Let the generated function be F(x, R1, R2, …, Rn), where R1, R2, …, Rn are the design
parameters. The difference E(x) between the desired function and generated function can be represented by,
= − , � , � … . � .
At precision points, say for x = x1, x2, …, xn, the desired and generated functions agree and E(x) = 0. At other
points E(x) will have some value, called the structural error. It is desirable that E(x) should be minimum.
Therefore, the spacing of precision points is very important. The precision points, according to Chebyshev s
spacing, are given by:
= + [
− �
] , = , , .
w�� =
+
; =
−
; = nu�b�� o� p��c�s�on po�nts

16.9.3 Graphical Method to Locate Precision Points
The Precision points can be obtained by the graphical method from the following steps:
1. Draw a circle of radius b and centre on the x-axis at a distance a from point O.
2. Inscribe a regular polygon of side 2n in this circle such that the two sides are perpendicular to the x-axis.
3.Determine the locations of n accuracy points by projecting the vertices on x-axis as shown in Fig.16.15. It is
sufficient to draw semi-circles only showing inscribed polygon to get the values of precision points.
Fig.16.15 Graphical method to determine precision points
Example: Derive Freudenstein s equation for a four bar linkage.
Solution:
Consider a four-bar mechanism as shown in Fig.16.16 in equilibrium. The magnitudes of AB, BC, CD and DA
are a, b, c and d respectively. θ, β and ϕ are the angles of AB, BC and DC respectively with the x-axis. AD is the
fixed link. AB is the input link and DC the output link.
The displacement along x-axis is, a cos θ + b cos β = d + c cos ϕ
Or b cos β = c cos ϕ − a cos θ + d
� = � − � +
= � + � + − � � − � + �
The displacement along y-axis is,
a sin θ + b sin β = c sin ϕ
or
b sin β = c sin ϕ − a sin θ
� = � − �
= � + � − � �
Adding Eqs. (1) and (2), we get

Fig. . Deriving freudenstein’s equation for four-bar mechanism
b2 = c2 + a2 + d2 − ac sin θ sin ϕ + cos θ cos ϕ − ad cos θ + cd cos ϕ
or 2cd cos ϕ − ad cos θ + a2 − b2 + c2 + d2 = ac sin θ sin ϕ + cos θ cos ϕ)
Dividing throughout by 2ac, we get
� − � +
− + +
= � − � = � − �
� � + � � + � = � − � .
w�� = ; � = − ; � =
− + +
Eq. (16.9) is known as Freudenstein Equation.
16.9.4 Freudenstein’s Equation for the Precision Points
Freudenstein s equation helps to determine the length of links of a four-bar mechanism. The displacement
equation of a four-bar mechanism, shown in Fig.16.17 is given by:
2cd cos ϕ − ad cos θ + a2 − b2 + c2 + d2 = ac cos θ cos ϕ + sin θ sin ϕ)
Fig. . Freudenstein’s equation for the precision points
Dividing throughout by 2ac and rearranging, we get
� − � +
− + +
= � − �
��t � = ; � = − ; � =
− + +
��n � � + � � + � = � − � .
Eq. . is known as the Freudenstein s equation.

Let the input and output are related by some function, such as, y = f(x).
For three specified positions, let
θ1, θ2, θ3 = three positions of input link
ϕ1, ϕ2, ϕ3 = three positions of output link
Then substituting these values in Eq. (16.10), we get
k1 cos ϕ1 + k2 cos θ1 + k3 = cos θ1 − ϕ1)
These equations can be written in the matrix form as:
|
� �
� �
� �
| {
�
�
�
} = {
� − �
� − �
� − �
}
These equations can be solved by any numerical technique. Using Cramer s rule, let
= |
� �
� �
� �
| ; = |
� − � �
� − � �
� − � �
|
= |
� � − �
� � − �
� � − �
| ; = |
� � � − �
� � � − �
� � � − �
|
��n � = ; � = ; � =
Knowing k1, k2 and k3, the values of a, b, c, and d can be calculated. Value of either a or d can be assumed to
be unity to obtain the proportionate values of other parameters.
Example: Design a four-bar mechanism to coordinate three positions of the input and output links given by:
θ1 = 25, ϕ1 = °; θ2 = 35°, ϕ2 = °; θ3 = 50°, ϕ3 = 60°
Solution:
cos θ1 = cos ° = . , cos θ2 = cos ° = . , cos θ3 = cos 50° = 0.6428
cos ϕ1 = cos 30° = 0.8660, cos ϕ2 = cos 40° = 0.7660, cos ϕ3 = cos 60° = 0.5000
cos θ1 − ϕ1 = cos ° − 30°) = 0.9962
cos θ2 − ϕ2 = cos ° − ° = .
cos θ2 − ϕ2 = cos ° − ° = .
� = |
. .
. .
. .
| = . × −
� = |
. .
. .
. .
| = − . × −

� = |
. .
. .
. .
| = − . × −
� = |
. . .
. . .
. . .
| = . × −
� =
�
�
=
− . × −
. × −
= . =
Let d = 1 unit, then a = 5.598 units
� =
�
�
=
− . × −
. × −
= − . = − , = . un�ts
� =
�
�
=
. × −
. × −
= . =
− + +
=
. − + . +
× . × .
= .
The mechanism is shown in Fig.16.18.
Fig.16.18 Four-bar mechanism developed by three position precision points

Unit-2
Introduction
Kinematics deals with the study of relative motion between the various links of a machine ignoring the forces
involved in producing such motion. Thus, kinematics is the study to determine the displacement velocity and
acceleration of the various links of the mechanism. A machine is a mechanism or a combination of
mechanisms that not only imparts definite motion to the various links but also transmits and modifies the
available mechanical energy into some kind of useful energy.
2.2 Velocity Diagrams
Displacement: The displacement of a body is its change of position with reference to a certain fixed point.
Velocity: Velocity is the state of change of displacement of a body with respect to time. It is a vector quantity.
Linear Velocity: It is the rate of change of velocity of a body along a straight line with respect to time. Its
units are m/s.
Angular Velocity: It is the rate of change of angular position of a body with respect to time. Its units are
rad/s. The relationship between velocity v and angular velocity ω is:
= � .
where r = distance of point undergoing displacement from the centre of rotation.
Relative Velocity: The relative velocity of a body A with respect to a body B is obtained by adding to the
velocity of A the reversed velocity of B. If va > vb, then
= ⃗⃗⃗⃗ − ⃗⃗⃗⃗ o� = −
Similarly,
= ⃗⃗⃗⃗ − ⃗⃗⃗⃗ o� = −
Fig.2.1 Relative velocity of a point
Consider two points A and B on a rigid link rotating clockwise about A as shown in Fig.2.1 (a). There can be no
relative motion between A and B as long as the distance between A and B remains the same. Therefore, the
relative motion of B with respect to A must be perpendicular to AB. Hence, the direction of relative velocity of
two points in a rigid link is always along the perpendicular to the straight line joining these points. Let

Unit-2
relative velocity of B with respect to A be represented by vba = ω·AB, then ab is drawn perpendicular to AB to
a convenient scale, as shown in Fig.2.1 (b). Similarly, the linear velocity of any other point C on AB with
respect to A is vca = ω·AC and is represented by vector ac. Hence,
=
�.
�.
=
= =
Hence the point C divides the vector ab in the same ratio as the point C divides the link AB.
2.3 Determination of Link Velocities
There are two methods to determine the velocities of links of mechanisms.
 Relative velocity method
 Instantaneous centre method
2.3.1 Relative Velocity Method
Consider a rigid link AB, as shown in Fig.2.2 (a), such that the velocity of A (va) is vertical and velocity of B (vb)
is horizontal. To construct the velocity diagram, take a point o. Draw oa representing the magnitude and
direction of velocity of A. Draw ob along the direction of vb. From point a draw a line ab perpendicular to AB,
meeting ob in b. Then oab is the velocity triangle, as shown in Fig.2.2 (b). Ob = vb; ab = vba, i.e. the velocity of B
with respect to A. Vector ab is called the velocity image of link AB. The velocity of any point C in AB with
respect to A is given by,
= = . ( )
= + ( )
= + . = + =
Hence vector oc represents the velocity of point C.
Fig.2.2 Relative velocity of points in a kinematic link
Consider a link A1B1 first undergoing rotation by an amount Δθ to A1B 1 and then undergoing translation by
an amount ΔsA to occupy the new position A2B2, as shown in Fig.2.3 (a). Then
∆ = ∆ + ∆

Unit-2
Now let the link A1B1 first undergo linear translation ΔsA to A2B”2 and then angular rotation of Δθ to A2B2, as
shown in Fig.2.3 (b). Then
∆ = ∆ + ∆
Eqs. a and b are same. Dividing by Δt, we get
∆
∆�
=
∆
∆�
+
∆
∆�
Or
= +
Therefore, the velocity of point B is obtained by adding vectorially the relative velocity of point B w.r.t. point
A to the velocity of point A. Now
∆ = . ∆
o�
∆
∆�
= .
∆
∆�
= . �
Also ∠ϕ = 90°
Fig.2.3 Relative velocity of points in a kinematic link
The following conclusions may be drawn from the above analysis:
 The velocity of any point on the kinematic link is given by the vector sum of the velocity of some other point
in the link and the velocity of the first point relative to the other.
 The magnitude of the velocity of any point on the kinematic link relative to the other point in the kinematic
link is the product of the angular velocity of the link and distance between the two points under
consideration.
 The direction of the velocity of any point on a link relative to any other point on the link is perpendicular to
the line joining the two points.
2.3.3 Relative Angular Velocities
Consider two links OA and OB connected by a pin joint at O, as shown in Fig. . . Let ω1 and ω2 be the angular
velocities of the links OA and OB, respectively. Relative angular velocity of OA with respect to OB is,

Unit-2
� = � − �
and relative angular velocity of OB with respect to OA is,
� = � − � = −�
If r = radius of the pin at joint O, then rubbing (or sliding) velocity at the pin joint O,
= � − � �, ��n �� nk� �o�� n �� a�� d��c��on . a
= � + � �, ��n �� nk� �o�� n �� o��o�� d��c��on . b
Fig.2.4 Relative angular velocities
2.3.4 Relative Velocity of Points on the Same Link
Consider a ternary link ABC, as shown in Fig.2.5(a), such that C is any point on the link. Let va and vb be the
velocities of points A and B, respectively. Then
= +
� = =
= + = +
Fig.2.5 Relative velocity of points on the same link
The velocity diagram is shown in Fig.2.5 (b).
Angular velocity of link ABC,
� = = = = = = .

Unit-2
2.3.5 Forces in a Mechanism
Consider a link AB subjected to the action of forces and velocities, as shown in Fig.2.6. Let A be the driving end
and B the driven end. When the direction of the forces and velocities is the same, then
Energy at A = Energy at B
× = ×
= .
Fig.2.6 Force and velocity diagram
Considering the effect of friction, the efficiency of transmission,
=
��
��
=
Or
= .
When the forces are not in the direction of the velocities, then their components along the velocities should
be taken.
2.3.6 Mechanical Advantage
�� , =
��
��
=
�o� a ��c�an��, =
��
��
= =
�
�
.
Considering the effect of friction,
=
�
�
.
2.3.7 Four-Bar Mechanism
(a) Consider the four-bar mechanism, as shown in Fig.2.7 (a), in which the crank O1A is rotating clockwise
with uniform angular speed ω. The linear velocity of point A is va = ω × O1A and it is perpendicular to O1A.
Therefore, draw o1a ⊥ O1A to a convenient scale in Fig.2.7 (b). The velocity of point B is perpendicular to O2B.
Therefore, at point o1, draw o1b ⊥ O2B. The relative velocity of B with respect to A is perpendicular to AB.
Therefore, draw ab ⊥ AB meeting the line drawn perpendicular to O2B at b. Then vb = o1b, and vba = ab. To find
the velocity of joint C, draw ac ⊥ AC and bc ⊥ BC to meet at c. Join o1c. Then vc = o1c.

Unit-2
Now
= + = + =
= + = + =
= + = + =
To find the velocity of any point D in AB, we have
= o� = ( ) .
Fig.2.7 Four-bar mechanism with ternary link
Thus, locate point d in ab and join o1d. Then o1d = vd. To find the relative velocity of C w. r. t. D, join cd. Then
vcd = dc. The velocity image of link ABC is abc.
(b) Now consider the four-bar mechanism, as shown in Fig.2.8 (a), in which the crank AB is rotating at
angular velocity ω in the anticlockwise direction. The absolute linear velocity of B is ω · AB and is
perpendicular to AB. Draw ab ⊥ AB to a convenient scale to represent vba, as shown in Fig.2.8 (b). From b,
draw a line perpendicular to BC and from a another line perpendicular to CD to meet each other at point c.
Then, ac = vca and cb = vbc. To find the velocity of any point E in BC, we have
= o� = ( ) .
Thus, locate point e in cb and join ae. Then ae = vea.
The rubbing velocities at pins of joints are:

Unit-2
: � . ; : � ± � . ; : � ± � . ; : � .
where r is the radius of the pin.
Use the + ve sign when angular velocities are in the opposite directions.
� = and � =
2.3.8 Slider–Crank Mechanism
Consider the slider–crank mechanism, as shown in Fig.2.9 (a), in which the crank OC is rotating clockwise
with angular speed ω. PC is the connecting rod and P is the slider or piston. The linear velocity of C, vc = OC ·
To draw the velocity diagram, draw a line oc = vc from point o, as shown in Fig.2.9 (b), representing the
velocity of point C to a convenient scale. From point c draw a line perpendicular to CP. The velocity of slider P
is horizontal. Therefore, from point o draw a line parallel to OP to intersect the line drawn perpendicular to
CP at p. Then the velocity of the piston, vp = op and the velocity of piston P relative to crank pin C is vpc = cp.
To find the velocity of any point E in CP, we have
Fig.2.9 Slider-crank mechanism
=
= ( ) .
Thus locate point e in cp, join oe. Then ve = oe
Rubbing Velocities are:
: � . ; : � . ; : (� + � ).
2.3.9 Crank and Slotted Lever Mechanism
Consider the crank and slotted lever mechanism, as shown in Fig.2.10 (a). The crank OB is rotating at a
uniform angular speed ω. Let OB = r, AC = l, and OA = d. Linear velocity of B, vb = rω. Draw ob = vb and
perpendicular to OB to a convenient scale, as shown in Fig.2.10 (b). vb is the velocity of point B on the crank
OB. The velocity of the slotted lever is perpendicular to AC. The velocity of the slider is along the slotted lever.
Hence draw a line from b parallel to AC to meet the line perpendicular to AC at p. Now, ap = vpa, and
= = ( ) .
From point c, draw a line perpendicular to CD and from point o draw a line parallel to the tool motion, to
intersect at point d.

Unit-2
�� , =
The component of the velocity of the crank perpendicular to the slotted lever is zero at positions B1 and B2.
Thus for these positions of the crank, the slotted lever reverses its direction of motion.
Fig.2.10 Crank and slotted lever mechanism
�� o� cu��n� ��ok�
�� o� ��u�n ��ok�
=
α
° − α
At positions B3 and B4, the component of velocity along the lever is zero, i.e. the velocity of the slider at the
crank pin is zero. Thus the velocity of the lever at crankpin is equal to the velocity of crankpin, i.e. rω. The
velocities of lever and tool at these points are minimum. The maximum cutting velocity occurs at B3 and
maximum return velocity occurs at B4.
�a��u� cu��n� ��oc��y = . � ( ) = � (
+
) .
�a��u� ��u�n ��oc��y = . � ( ) = � (
−
) .
�a��u� cu��n� ��oc��y
�a��u� ��u�n ��oc��y
= (
−
+
) .
2.3.10 Drag Mechanism
The drag mechanism is shown in Fig.2.11 a . Link rotates at constant angular speed ω. Link rotates at a
non-uniform velocity. Ram 6 will move with nearly constant velocity over most of the upward stroke to give a
slow upward stroke and a quick downward stroke when link 2 rotates clockwise.
To determine the velocity diagram, draw o1a = ω · O1A, perpendicular to O1A, as shown in Fig.2.11 (b). From
a draw ab ⊥ AB and from o1 draw o1b ⊥ O2B to intersect at b. Then, o1b = vb and ab = vba. At b draw bc ⊥ BC
and at o1 draw o1c ⊥ O2C, to intersect at c. Then o1c = vc.
Now at o2 draw o2d || O2D and from c draw a line perpendicular to CD to intersect at d. Then o2d = vd, the
velocity of ram 6. o2bc is the velocity image of O2BC.

Unit-2
Fig.2.11 Drag mechanism
2.3.11 Whitworth Quick-Return Motion Mechanism
In the Whitworth mechanism, as shown in Fig.2.12 a , the crank O A rotates with constant angular speed ω.
The link PB oscillates about pin O2. The ram C reciprocates on the guides.
Fig.2.12 Whitworth quick-return motion mechanism
The velocity diagram has been drawn in Fig.2.12 (b). va = o1a = ω · O1A. Draw o1a ⊥ O1A to a convenient scale.
At o2, draw a line perpendicular to PB and at a draw a line parallel to O2P to intersect at p. Then ap = vpa
represents the velocity of slider at P and o2p = vpo2 the velocity of lever PB at P. Produce po2 to b so that
=
Draw bc ⊥ BC and o2c || O2C to intersect at c. Then vc = o2c is the velocity of the ram.
Example: In the mechanism shown in Fig.2.14 (a), the crank O1A rotates at a uniform speed of 650 rpm.
Determine the linear velocity of the slider C and the angular speed of the link BC. O1A = 30 mm, AB = 45 mm,
BC = 50 mm, O2B = 65 mm, and O1O2 = 70 mm.
Solution
Given: N = 650 rpm, lengths of links.

Unit-2
Procedure:
1. Draw the configuration diagram to a convenient scale as shown in Fig.2.14 (a).
2. Calculate the angular velocity of crank O1A.
� = � × = . �ad/�
Fig.2.14
3.Calculate the linear velocity of point A,
va = ω·O1A = 68.07 × 0.03 = 2.04 m/s
4. Draw the velocity diagram as shown in Fig.2.14 (b) to a scale of 1 cm = 0.5 m/s.
5.Draw va = o1a ⊥ O1A such that o1a = 4.08 cm.
6. Draw a line from o2 perpendicular to O2B and another line from a perpendicular to AB to intersect at b. Then
ab = vba and o2b = vb
7. From b, draw a line perpendicular to BC and another line from o2 parallel to the path of motion of the slider to
intersect at c. Then
Velocity of the slider C, vc = o2c
By measurement, we get vc = o2c = 1.4 cm = 0.7 m/s ; vcb = bc = 1.3 cm = 0.65 m/s
An�u�a� ��oc��y o� ��nk BC =
�
��
=
.
.
= �ad �⁄ c�ock�� abou� B
2.4 Instantaneous Centre Method
A link as a whole may be considered to be rotating about an imaginary centre or about a given centre at a
given instant. Such a centre has zero velocity, i.e. the link is at rest at this point. This is known as the

Unit-2
Instantaneous Centre or Centre of Rotation. This centre varies from instant to instant for different
positions of the link. The locus of these centres is termed the Centrode.
2.4.1 Velocity of a Point on a Link
Consider two points A and B on a rigid link, having velocities va and vb, respectively, as shown in Fig.2.50 (a).
From A and B draw lines perpendicular to the directions of motion and let them meet at I. Then I is the
instantaneous centre of rotation of the link AB for its give position.
)f ω is the instantaneous angular velocity of the link AB, then va = ω · )A and vb = ω · )B. Thus
= ( ) . .
The velocity diagram for the link AB has been drawn in Fig.2.50 (b). Triangles IAB and oab are similar. Hence
Fig.2.50 Concept of instantaneous centre
= = .
= = = = �
where C is any point on the link AB.
2.4.2 Properties of Instantaneous Centre
The properties of the instantaneous centre are as follows:
1. At the instantaneous centre of rotation, one rigid link rotates instantaneously relative to another for the
configuration of the mechanism considered.
2. The two rigid links have no linear velocities relative to each other at the instantaneous centre.
3.The two rigid links have the same linear velocity relative to the third rigid link, or any other link.
2.4.3 Number of Instantaneous Centres
The number of instantaneous centres in a mechanism is equal to the number of possible combinations of two
links. The number of instantaneous centres,
=
−
.
where n = number of links.

Unit-2
2.4.4 Types of Instantaneous Centres
The instantaneous centres for a mechanism are of the following types:
1. Fixed Instantaneous Centres.
2. Permanent Instantaneous Centres.
3. Neither Fixed nor Permanent Instantaneous Centres.
Fig.2.51 Instantaneous centres in a four-bar mechanism
Consider a four-bar mechanism shown in Fig.2.51. For this mechanism, n = 4.
Hence N = − / =
The instantaneous centres are: 12, 13, 14, 23, 24, 34.
The instantaneous centres 12 and 14 remain at the same place for the configuration of the mechanism, and
are therefore called fixed instantaneous centres. The instantaneous centres 23 and 34 move when the
mechanism moves. But the joints are permanent; therefore, they are called permanent instantaneous centres.
The instantaneous centres 13 and 24 vary with the configuration of the mechanism and are neither fixed nor
permanent instantaneous centres.
2.4.5 Location of Instantaneous Centres
The following observations are quite helpful in locating the instantaneous centres:
1. For a pivoted or pin joint, the instantaneous centre for the two links lies on the centre of the pin (Fig.2.52 (a)).
2. In a pure rolling contact of the two links, the instantaneous centre lies at their point of contact (Fig.2.52 (b)).
This is because the relative velocity between the two links at the point of contact is zero.
3. In a sliding motion, the instantaneous centre lies at infinity in a direction perpendicular to the path of motion
of the slider. This is because the sliding motion is equivalent to a rotary motion of the links with radius of
curvature equal to infinity (Fig.2.52(c)). If the slider (link 2) moves on a curved surface (link 1), then the
instantaneous centre lies at the centre of curvature of the curved surface (Fig.2.52 (d) and (e)).

Unit-2
Fig.2.52 Locating instantaneous centres
2.4.6 Arnold–Kennedy Theorem
This theorem states that if three-plane bodies have relative motion among themselves their three
instantaneous centres must lie on a straight line.
Consider three rigid links 1, 2, and 3; 1 being a fixed link. 12 and 13 are the instantaneous centres of links 1, 2
and 1, 3 respectively. Let 23 be the instantaneous centre of links 2, 3, lying outside the line joining 12 and 13,
as shown in Fig.2.53. The links 2 and 3 are moving relative to link 1. Therefore, the motion of their
instantaneous centre 23 is to be the same whether it is considered in body 2 or 3. If the point 23 is considered
on link 2, then its velocity v2 is perpendicular to the line joining 12 and 23. If the point 23 lies on link 3, then
its velocity v3 must be perpendicular to the line joining 13 and 23. The velocities v2 and v3 of point 23 are in
different directions, which is not possible. The velocities v2 and v3 of instantaneous centre 23 will be equal
only if it lies on the line joining 12 and 13. Hence all the three instantaneous centres 12, 13 and 23 must lie on
a straight line.
Fig.2.53 Proving three-centres theorem
2.4.7 Method of Locating Instantaneous Centres
The following procedure may be adopted to locate the instantaneous centres:

Unit-2
1.Determine the number of instantaneous centres.
2. Make a list of all the instantaneous centres by writing the link numbers in the first row and instantaneous
centres in ascending order in columns. For example, for a four-bar chain shown in Fig.2.54 (a), we have
3.Locate the fixed and permanent instantaneous centres by inspection. Tick mark (tick) these instantaneous
centres.
4. Locate the remaining neither fixed nor permanent instantaneous centres (circled) by using Arnold–Kennedy s
theorem. This is done by a circle diagram, as shown in Fig.2.54 (b). Mark points on a circle equal to the
number of links in the mechanism. Join the points by solid lines for which instantaneous centres are known
by inspection. Now join the points forming the other instantaneous centres by dotted lines. The instantaneous
centre shall lie at the intersection of the lines joining the instantaneous centres of the two adjacent triangles
of the dotted line. For example, in Fig.2.54 (b), the centre 13 is located at the intersection of lines (produced)
joining the instantaneous centres 12, 23, and 14, 34. Similarly, the centre 24 is located at the intersection of
the lines (produced) joining the centres 23, 34 and 12, 14.
2.4.8 Determination of Angular Velocity of a Link
The angular velocities of two links vary inversely as the distances from their common instantaneous centre to
their respective centres of rotation relative to the frame. For example, for the four-bar mechanism shown in
Fig. . a , if ω2 is the angular velocity of link , then angular velocity ω4 of link 4 will be given by the
following relationship:
�
�
=
−
−
If the respective centres of rotation are on the same side of the common instantaneous centre, then the
direction of angular velocities will be same. However, if the respective centres of rotation are on opposite
sides, then the direction of angular velocities will be opposite.

Unit-2
Similarly,
�
�
=
−
−
Example: Locate the instantaneous centres of the slider crank mechanism shown in Fig.2.56(a). Find the
velocity of the slider. OA = mm, AB = mm, and OB = mm, ω2 = 12 rad/s cw.
Fig.2.56 Slides-crank mechanism
Solution:
Given: n = , ω2 = 12 rad/s cw, lengths of links.
Number of instantaneous centres = 6
1.Draw the configuration diagram to a convenient scale as shown in Fig.2.56 (a).
2.Locate fixed centres 12 and 34.
3.Locate permanent centre 23. Instantaneous centre 14 is at infinity.
4.Locate neither fixed nor permanent centres 13 and 24 by using Arnold–Kennedy s three centres theorem.
13: 12, 23; 14, 34
24: 12, 14; 23, 34
By measurements, we have
− = mm
− = 390 mm
� = � . �� = × . = . � �⁄ = � . −
Hence,
� = � [
−
−
] = × = . � �⁄
When ω3 is the angular velocity of link 3 about 13.
Velocity of slider,

Unit-2
� = � − = � [
−
−
] − = � . �� [
−
−
] = × . × = . � �⁄
3.1 Introduction
The acceleration may be defined as the rate of change of velocity of a body with respect to time. The
acceleration can be linear or angular. Linear acceleration is the rate of change of linear velocity of a body with
respect to time. Angular acceleration is the rate of change of angular velocity of a body with respect to time.
Acceleration Diagram: It is the graphical representation of the accelerations of the various links of a
mechanism drawn on a suitable scale. It helps us to determine the acceleration of various links of the
mechanism.
The determination of acceleration of various links is important from the point of view of calculating the forces
and torques in the various links to carry out the dynamic analysis.
3.2 Acceleration of a Body Moving in a Circular Path
Consider a body moving in a circular path of radius r with angular speed ω, as shown in Fig.3.1(a). The body
is initially at point A and in time δt moves to point B. Let the velocity change from v to v + δv at B and the
angle covered be δθ.
Fig.3.1 Acceleration for a link
The change in velocity can be determined by drawing the velocity diagram as shown in Fig.3.1(b). In this
diagram, oa = v, ob = v + δv, and ab = change in velocity during time δt. The vector ab is resolved into two
components ax and xb, parallel and perpendicular to oa, respectively.
Now
= − = �� − = + � �� −
= �� = + � ��
The rate of change of velocity is defined as the acceleration. It has two components: tangential and normal.
The tangential component of acceleration ft is the acceleration in the tangential direction. It is defined by,
=
�
=
+ � �� −
�
=
+ � −
�
B�cau�� o� ��a�� an�� co� � ≈
=
�
�
= =
�
= (
�
) = .
Where α = angular acceleration.

Unit-2
The normal component of acceleration fn is the acceleration in the direction normal to the tangent at that
instant. This component is directed towards the centre of the circular path. It is also called the radial or
centripetal acceleration. It is defined by,
=
�
=
+ � ��
�
=
+ � �
�
B�cau�� o� ��a�� an�� n � ≈ �
=
�
�
+
� �
�
≈
�
�
��c��n� ��cond ��, b��n� ��a��
= = � = � = .
Two cases arise regarding the motion of the body.
1. When the body is rotating with uniform angular velocity, then dω/dt = 0, so that ft = 0. The body will have
only normal acceleration, fn = r ω2.
2. When the body is moving on a straight path, r will be infinitely large and v2/r will tend to zero, so that fn = 0.
The body will have only tangential acceleration, ft = r .
3.3 Acceleration Diagrams
3.3.1 Total Acceleration of a Link
Consider two points A and B on a rigid link, as shown in Fig.3.2 (a), such that the point B moves relative to
point A with an angular velocity ω and angular acceleration .
Centripetal (or normal or radial) acceleration of point B with respect to point A is,
= � . = .
Tangential acceleration of point B with respect to point A,
= . .
Total acceleration of B w.r.t. A,
= + ��
= � × + × = � + . .
�� = =
�
.
The acceleration diagram has been represented in Fig.3.2 (b).

Unit-2
Fig.3.2 Total acceleration of a link
3.3.2 Acceleration of a Point on a Link
The accelerations of any point X on the rigid link w.r.t. A as shown in Fig.3.2 (a) are:
= � . � .
= . � .
�� , = +
Therefore, fxa Denoted by ax in the acceleration diagram shown in Fig.3.2 (b) is inclined to XA at the same
angle . Triangles abx and ABX are similar. Thus, point x can be fixed on the acceleration image,
corresponding to point X on the link.
�� , =
�� , =
Acceleration Image: Acceleration images are helpful to find the accelerations of offset points of the links.
The acceleration image is obtained in the same manner as a velocity image.
An easier method of making Δabx similar to ΔABX is by making AB on AB equal to ab and drawing a line
parallel to BX, meeting AX in X . AB X is the exact size of the triangle to be made on ab.
Take ax = AX and bx = B X
Thus the point x is located.
The method is illustrated in Fig.3.2(c).
3.3.3 Absolute Acceleration for a Link
Consider the rigid link AB such that point B is rotating about A with angular velocity ω and angular
acceleration , as shown in Fig. . (a). The point A itself has acceleration fa. The acceleration diagram is
shown in Fig.3.3 (b).
�� , = + + = + � . + . .
Similarly for any other point X,
�� , = + + = + � . + . .

Unit-2
Fig.3.3 Absolute acceleration of a rigid link
3.3.5 Acceleration Diagram for Four-Bar Mechanism
The four-bar mechanism is shown in Fig.3.5 (a). The velocity of point B, vb = ω · AB. The velocity diagram is
shown in Fig.3.5 (b), In this diagram,
Fig.3.5 Acceleration diagram for four-bar mechanism
vb = ab ⊥ AB; dc ⊥ DC and bc ⊥ BC
�� , = =
�� , =
Now we calculate the accelerations of various points and links.
= = ; = = ′
; = = ′′
= +
= ′
+ ′
= +
= ′′
+ ′′

Unit-2
The acceleration diagram is shown in Fig.3.5(c) to a suitable scale. To construct the acceleration diagram,
proceed as follows:
1. Draw ab = fb parallel to AB, which is known in magnitude and direction.
2. Draw bc = fn
cb parallel to BC, which is known in magnitude and direction.
3. Draw cc , representing fn
cb perpendicular to bc , which is known in direction only.
4. Now draw dc = fn
cd parallel to CD, which is known in magnitude and direction.
5. Draw c c, representing ft
cd, perpendicular to dc to intersect c c at point c.
6. Join dc and bc. Then
Acc��a��on o� C ��a�� o B, =
Acc��a��on o� C ��a�� o D, = =
3.3.7 Acceleration Diagram for Slider-Crank Mechanism
(i) Velocity Diagram
For the slider crank mechanism as shown in Fig.3.7 (a). va = ω·OA. The velocity diagram is shown in Fig. .
(b) In this diagram,
Fig.3.7 Acceleration diagram of slider-crank mechanism
va = oa ⊥ OA
ab ⊥ AB
ob || OB
Then
Velocity of slider, vb = ob
Velocity of connecting rod, vba = ab
(ii) Acceleration Diagram
The acceleration diagram is shown in Fig.3.7(c), in which
= � × = =
= = ′
The following steps may be adopted to draw the acceleration diagram:
1. Draw oa = fa parallel to OA on a convenient scale to represent the normal acceleration of the crank OA.

Unit-2
2. Draw ab = fn
ba parallel to AB to represent the normal acceleration of the connecting rod, which is known in
magnitude and direction.
3.Draw a line perpendicular to AB at b to represent the tangential acceleration ft
ba of AB, which is unknown in
magnitude.
4. Draw a line parallel to OB at o to represent the acceleration fb of the slider B. Let these two lines intersect at b.
Join ab. Then
Linear acceleration of slider B, fb = ob
5. To find the acceleration of any point C in AB, we have
=
Locate point c in ab. Join oc. Then
Acceleration of point C, fc = oc
Example: A four-bar mechanism with ternary link is shown in Fig.3.9 (a). The lengths of various links is given
as below: O1O2 = 600 mm, O1A = 300 mm, AB = 400 mm, O2B = 450 mm, AC = 300 mm, BC = 250 mm, AD =
100 mm, and ∠AO1O2 = 75°.
Angular velocity of crank O1A = 20 rad/s
Angular acceleration of crank O1A = 100 rad/s2
Determine (a) acceleration of coupler AB, (b) acceleration of lever O2B, (c) acceleration of points C and D, and
(d) angular acceleration of ternary link.
Solution:
Linear velocity of A, va = ω × O1A = 20 × 0.3 = 6 m/s
1. Draw the configuration diagram as shown in Fig.3.9 (a) to a scale of 1 cm = 100 mm.
2. Draw the velocity diagram as shown in Fig.3.9 (b) to a scale of 1 cm = 1 m/s
3. Measure ab = vba = 3.5 cm = 3.5 m/s; o2b = vb = 4.7 cm = 4.7m/s; o1c = vc = 3.5 cm = 3.5 m/s, vca = ac = 2.7 cm =
2.7 m/s; vcb = bc = 2.1 cm = 2.1 m/s
4. Now
��
��
=
��
��
�� =
. ×
= . c�

Unit-2
5. Locate point d in ab and join o1d. Then vd = o1d = 5.5 cm = 5.5 m/s.
Fig.3.9 Four-bar mechanism with ternary link
6. Calculate the accelerations of various points as follows:
� =
�
� �
=
.
= �/�
��
= � × � � = × . = �/�
� =
�
��
=
.
.
= . �/�
� =
�
� �
=
.
.
= . �/�
� =
�
��
=
.
.
= . �/�
� =
�
��
=
.
.
= . �/�
7. Draw the acceleration diagram on a scale of 1 cm = 10 m/s2, as shown in Fig.3.9(c)
8. Measure the various required lengths on the acceleration diagram.

Unit-2
Acceleration of coupler AB, fba = ab = 7.8 cm = 78 m/s2
Acceleration of lever O2B, fb = o2b = 8.6 cm = 86 m/s2
Acceleration of point C, fc = o1c = 7.9 cm = 79 m/s2
9. Locate point d in ab from the relation,
�� =
�� × ��
��
= . c�
Acceleration of point D, fd = o1d = 11 cm = 110 m/s2.
Tangential acceleration of AB, ft
ba = b b = 7.2 cm = 72 m/s2
An�u�a� acc��a��on o� ��, � =
��
��
=
.
.
= �ad/� c�
3.4 Coriolis Acceleration
The total acceleration of a point with respect to another point in a rigid link is the vector sum of its normal
and tangential components. This holds true when the distance between the two points is fixed and the
relative acceleration of the two points on a moving rigid link has been considered. If the distance between the
two points varies, i.e., the second point which was considered stationary, now slides, the total acceleration
will contain one additional component, called Coriolis Component of Acceleration.
Consider a slider B on a link OA such that when the link OA is rotating clockwise with angular velocity ω the
slider B moves outward with linear velocity v, as shown in Fig. . . Let in time δt the angle turned through by
link OA be δθ to occupy the new position OA and the slider moves to position E. The slider can be considered
to move from B to E as follows:
Fig.3.12 Concept of coriolis acceleration
1.From B to C due to angular velocity ω of link OA.
2. C to D due to outward velocity v of the slider.
3.D to E due to acceleration perpendicular to the rod, i.e., due to Coriolis acceleration. Now
arc DE = arc EF − arc FD = arc EF − arc BC = OF · dθ − OB · dθ
= OF − OB dθ = BF · dθ = CD · dθ
Now
CD = v · dt
and

Unit-2
dθ = ω · dt
Hence arc
DE = (v · dt) · (ω · dt)
Now
arc DE = chord DE, as dθ is very small.
��o�� = �
Bu� = .
Where fcr is the constant Coriolis acceleration of the particle. Hence from (1) and (2), we get
= � .
The direction of the Coriolis acceleration component is such so as to rotate the sliding velocity vector in the
same sense as the angular velocity of OB. This is achieved by turning the sliding velocity vector through 90° in
a manner that the velocity of this vector is the same as that of angular velocity of OA. The method of finding
the direction of Coriolis acceleration is illustrated in Fig.3.13.
Fig.3.13 Finding direction of Coriolis acceleration
Example: In the crank and slotted lever type quick return motion mechanism shown in Fig.3.14 (a), the crank
AB rotates at 120 rpm. Determine (a) velocity of ram at D, (b) magnitude of Coriolis acceleration component,
and (c) acceleration of ram at D · AB = 200 mm, OC = 800 mm, CD = 600 mm, OA = 300 mm.
Solution:
Velocity diagram
� = � × = . �ad/�
vb = vba = ω·AB = . × . = . m/s
1.Draw the configuration diagram to a scale of 1 cm = 100 mm, as shown in Fig.3.14 (a). By measurement, OP =
450 mm.
2.Draw the velocity diagram as shown in Fig.3.14 (b) to a scale of 1 cm = 0.5 m/s, by adopting the following
steps.
3.Draw ba ⊥ AB = 5.02 cm to represent the linear velocity of point B on link AB.
4.Let the coincident point of B on the slotted lever OC be P Draw a line at o perpendicular to OC and another
line at b parallel to OC to meet at point p.
5.By measurement, op = 4.2 cm. Then

Unit-2
�� =
�� × ��
��
=
. ×
= . c�
6.Produce op to oc.
7.Draw a line parallel to XX, the line of stroke of the ram D, and a line at c perpendicular to CD to meet the
previous line at d. Then
Velocity of the slider D, vd = od = 7.8 cm = 3.9 m/s
Velocity of point C, vc = oc = 7.47 cm = 3.74 m/s
An�u�a� ��oc��y o� ��o��d �� C, � =
�
��
=
.
.
= . �ad �⁄ c�
vpo = op = 4.2 cm = 2.1 m/s
vbp = pb = 3 cm = 1.5 m/s
vcd = cd = 3.2 cm = 1.6 m/s
Acceleration Diagram
1.Calculate the accelerations of various points.
2. Coriolis acceleration of B relative to P,
� �
= � × � = × . × . = . �/�
The direction of coriolis component of acceleration is shown in Fig.3.14 (d), which is perpendicular to OC.
� =
�
��
=
.
.
= . �/�
� =
�
��
=
.
.
= . �/�
� =
�
��
=
.
.
= . �/�
3. Draw the acceleration diagram as shown in Fig.3.14(c) to a scale of 1 cm = 2 m/s2, by adopting the following
steps.
4. Draw ab = fn
ba = 15.75 cm parallel to AB.
5. Draw p b = fcr
bp = 7.0 cm perpendicular to OC at point b, and another line perpendicular to OC representing
fs
bp, the sliding acceleration of point B relative to point P. fbp = pb is the total acceleration of B relative to P.
6. Draw op = fn
po = 4.9 cm parallel to OP and another line p p perpendicular to it to meet the previous line at
point p.
7. Join bp and op. By measurement, op = 5.3 cm.
�� =
�� × ��
��
=
. ×
= . c�
8. Extend op to point c so that oc = 9.4 cm.
9. Draw cd = fn
dc = 2.13 cm parallel to CD and draw a line perpendicular to CD at d to represent the tangential
acceleration of CD.
10. Draw a line at o parallel to the line of action of the ram at D meeting previous line at d. Join cd. Then
Acceleration of the ram at D, fd = od = 2.5 cm = 5 m/s2
Velocity of ram = 3.9 m/s

Unit-2
Coriolis acceleration = 14.025 m/s2
Acceleration of ram = 2.5 m/s2
Fig.3.14 Crank and slotted lever mechanism
. Klein’s Construction
Klein s construction is used to draw the velocity and acceleration diagrams for a single slider-crank
mechanism on its configuration diagram. The line that represents the crank in the configuration diagram also
represents the velocity and the acceleration of its moving end to some scale. The following steps may be
adopted to draw the velocity and acceleration diagrams, as shown in Fig.3.24:
1. Draw the configuration diagram OAB of the slider-crank mechanism to a convenient scale.
2. Draw OI ⊥ OB and produce BA to meet O) at C. Then OAC represents the velocity diagram. OA = ωr to some
scale from which the scale of velocity diagram is determined. OC = vbp, CA = vba, OA = vao.

Unit-2
3. With AC as the radius and A as the centre, draw a circle.
4. Locate the mid-point D of AB. With D as centre and DA as radius, draw the circle to intersect the previously
drawn circle at E and F. Join EF intersecting AB at G.
5. EF meets OB at H. If it does not meet OB then produce EF to meet OB at H. Join AH.
6.Then OAGH is the acceleration diagram.
= ; =
Ac��a��on o� ��on ��d�� B , = � .
= � . ; = � .
Fig. . Klein’s construction
7. To find the acceleration of any point X in AB, draw a line XX || OB to intersect HA at X. Join OX. Then
= � . �
= =
� .
4.1 Introduction
The lower pairs are those which have surface contact between them. We study, in this chapter, the various
mechanisms for the generation of straight line motion, both exact and approximate. These mechanisms have a
vital role to play in generating the configurations for machines. The steering gears and (ooke s joint are other
lower pairs that are discussed.
4.2 Pantograph
It is a mechanism to produce the path traced out by a point on enlarged or reduced scale. Fig.4.1 (a) shows
the line diagram of a pantograph in which AB = CD, BC = AD and ABCD is always a parallelogram. OQP is a
straight line. P describes a path similar to that described by Q. It is used as a copying mechanism.

Unit-2
Fig.4.1 Pantograph
Proof: To prove that the path described by P is similar to that described by Q, consider Δs OAQ and OBP which
are similar, because ∠BOP is common.
∠AQO = ∠BPO being corresponding angles as AQ || BP.
��nc�, = = .
In the displaced position shown in Fig.4.1 (b), as all links are rigid,
B1O = BO, D1A1 = DA, A1O = AO
and P1B1 = PB, B1A1 = BA, A1Q1 = AQ
��nc�, =
As A1B1C1D1 is a parallelogram, A1D1 || B1C1, i.e., A1Q1 || B1P1.
OQ1P1 is again a straight line so that Δs QA1Q1 and OB1P1 are similar.
= .
From Eqs. (4.1) and (4.2), we get
=
because OA = OA1 and OB = OB1.
Hence QQ1 is similar to PP1 or they are parallel. Thus path traced by P is similar to that of Q.
The pantograph is used in geometrical instruments, manufacture of irregular objects, to guide cutting
tools, and as indicator rig for cross head.
4.3 Straight Line Motion Mechanisms
Straight line motion can be generated by either sliding pairs or turning pairs. Sliding pairs are bulky and gets
worn out rapidly. Therefore, turning pairs are preferred over sliding pairs for generating straight line motion.
Straight line motion can be generated either accurately or approximately.

Unit-2
4.3.1 Accurate Straight Line Motion Mechanisms
The mechanisms for the generation of accurate straight line motion are as follows:
1.Peaucellier Mechanism
2.Hart Mechanism
3.Scott-Russel Mechanism.
1. Peaucellier Mechanism: A line diagram of the Peaucellier mechanism is shown in Fig.4.2.
OR = OS and OO1 = O1Q.
Point P describes a straight line perpendicular to OO1 produced.
Fig.4.2 Peaucellier straight line mechanism
Proof: Δs ORQ and OSQ are congruent, because OR = OS, OR = QS and OQ is common.
∠ROQ = ∠SOQ. Also ∠OQR = ∠OQS and Δs PRO = Δs PSQ.
∠OQR + ∠RQP = ∠OQS + ∠SQP
But OQ is a straight line.
∠OQR + ∠RQP = ∠OQS + ∠SQP = 180°
Hence OQP is a straight line.
Now OR2 = OT2 + RT2
RP2 = RT2 + TP2
OR2 − RP2 = OT2 − TP2 = (OT + TP)(OT − TP) = OP · OQ
But OR and RP are always constant.
Hence OP · OQ = constant.
Draw PP1 ⊥ OO1 produced and join QQ1. Δs OQQ1 and OPP1 are similar, because ∠OQQ1 = ∠OP1P = 90° and
∠QOQ1 is common. Hence
=
or OQ · OP = OQ1 · OP1 = constant
Now OQ1 = 2OO1 = constant

Unit-2
Hence OP1 = constant
or point P moves in a straight line.
2. Hart Mechanism: The Hart mechanism is shown in Fig.4.3. OO1 is a fixed link and O1Q is the rotating link.
The point Q moves in a circle with centre O1 and radius O1Q. ABCD is a trapezium so that AB = CD, BC = AD and
BD || AC.
= =
P describes a straight line perpendicular to OO1 produced as Q moves in a circle with centre O1.
Proof:
�n � , = ��nc� ||
Fig.4.3 Hart Straight Line Mechanism
Δs ABD and AOP are similar, because ∠AOP = ∠ABD being corresponding angles and ∠DAB is common.
∴ =
o� =
.
.
��a��y a� =
Hence OQ || AC
Now AC || BD, OQ || BD, OP || BD Hence OP || OQ
Since O is a common point, therefore OQP is a straight line.
Now Δs BOQ and BAC are similar.
∴ =
o� =
.
.

Unit-2
. = (
.
) . (
.
) = (
.
) . .
Draw DE ⊥ AC.
= −
= +
. = − + = + − +
= − + = − .
From ΔAED,
AE2 = AD2 − DE2
From ΔGED,
EC2 = CD2 − DE2
− = − .
. = − .
. = (
.
) − = ��
as AO, BO · AB, AD, and CD are fixed.
Hence P describes a straight line perpendicular to OO1 produced as point Q moves in a circle with centre O1.
3. Scott-Russel Mechanism: The Scott-Russel mechanism is shown in Fig.4.4. It consists of a sliding pair and
turning pairs. It can be used to generate approximate and accurate straight lines.
Fig.4.4 Scott-Russel straight line mechanism (CP = CQ)
1. When OC = CP = CQ, Q describes a straight line QO ⊥ OP provided P moves in a straight line along OP.
2. If CP ≠ CQ then Q describes an approximate straight line perpendicular to OP provided P moves along a
straight line OP such that

Unit-2
=
Proof:
(a) As OC = CP = CQ, hence ∠POQ = 90° and OQ ⊥ OP. If P moves along OP then Q moves along a line
perpendicular to OP.
Hence OP = OC cos θ + CP cos θ = PQ cos θ
∠CPI = 90° − θ
Now ∠ICP = ∠COP + ∠CPO = 2θ
Fig.4.5 Scott-Russel mechanism when CP ≠ CQ
Then ∠CIP = ° − ∠CPI − ∠ICP
= ° − ° + θ − θ
= ° − θ
∠CIP = ∠CPI
Or CI = CP
∴ CP = CQ = CO = CI and ∠POQ = 90°
Hence OPIQ is a square. As the path of Q is ⊥ OI, I is the instantaneous centre. Q will move along a line
perpendicular to OP.
(b) When CP ≠ CQ, it will form an elliptical trammel, as shown in Fig.4.5.
+ =
Because x = CQ cos θ and y = PC sin θ
=
�� − ��
�� − ��
=
�
As point C moves in a circle, for Q to move along an approximate straight line,
=
�

Unit-2
Limitations: When OC ⊥ OP, P coincides with O and OQ = 2OC. Here a small displacement of P shall cause a
large displacement of Q, requiring a relatively small displacement of P to give displacement to Q. This requires
highly accurate sliding surfaces.
4.3.2 Approximate Straight Line Motion Mechanisms
The mechanisms for the generation of approximate straight line motion are:
1.Grasshopper Mechanism
2.Watt Mechanism
3.Tchebicheff Mechanism
1. Grasshopper Mechanism: The Grasshopper mechanism is shown in Fig.4.6. The crank OC rotates about a
fixed point O · O1 is a fixed pivot for link O1P. For small angular displacements of O1P, point Q on link PCQ will
trace approximately a straight line path perpendicular to OP if
=
�
Fig.4.6 Grasshopper mechanism
2. Watt Mechanism: The Watt mechanism is shown in Fig.4.7 (a). Links OA and O1B oscillates about O and O1
respectively. AB is a connecting link. P will trace an approximate straight line if
=
In Fig.4.7 (b), θ and ϕ are the amplitudes of oscillation and I is the instantaneous centre of A′B′. P′ is the point
which lies on the approximate straight line described by P.
Proof:
� =
′
and =
′
�
=
′
. (
′
) .

Unit-2
Fig.4.7 Watt mechanism
If ∠B′P′I = 90°, then sin θ = B’P’/B’)
o� =
′ ′
′ ′
; � ≈
′ ′
′
�
≈
′ ′
′
.
′ ′
′ ′
.
′
. (
′
) =
′ ′
′
.
′ ′
′ ′
�o�
′
′
=
′ ′
′
Δs OAA′ and IP′A are approximately similar, as well as Δs O1BB′ and B′P′I are approximately similar.
≈
′ ′
′ ′
≈
Therefore, P divides the coupler AB in the ratio of the lengths of oscillating links. Hence P will describe an
approximate straight line for a certain position of its path.
3. Tchebicheff Mechanism: The Tchebicheff mechanism is shown in Fig.4.8. In this mechanism, OA = O1B
and AP = PB. P is the tracing point. Let AB = 1, OA = O1B = x, and OO1 = y.
�o� = −
− = −
=
+
.
Draw AL ⊥ OO1, then OL = OO1 − O1L
Also O1L = AP1
and = (
−
) = (
−
)
A�a�n = − = −
−
=
+
Further OA2 = AL2 + OL2
= ( − ) + [
+
]
Fig.4.8 Tchebicheff mechanism

Unit-2
o� = + + .
+
= + +
Or y = 2
Substituting in Eq. (4.11), we find that, x = 3.5
i.e., AB : OO1, : OA :: 1 : 2 : 3.5
P moves horizontally because instantaneous centre of AB will lie on point C which is the intersection of OA
and BO1. Thus C lies just below P.
4.4 Intermittent Motion Mechanisms
These mechanisms are used to convert continuous motion into intermittent motion. The mechanisms used
for this purpose are the Geneva wheel and the ratchet mechanism.
1.Geneva Wheel: The Geneva wheel as shown in Fig. 4.10 consists of a plate 1, which rotates continuously and
contains a driving pin P that engages in a slot in the driven member 3. Member 2 is turned ¼ th of a
revolution for each revolution of plate 1. The slot in member 2 must be tangential to the path of pin upon
engagement in order to reduce shock. The angle is half the angle turned through by member 2 during the
indexing period. The locking plate serves to lock member 2 when it is not being indexed. Cut the locking plate
back to provide clearance for member 2 as it swings through the indexing angle. The clearance arc in the
locking plate will be equal to twice the angle α.
2.Ratchet Mechanism: This mechanism is used to produce intermittent circular motion from an oscillating or
reciprocating member. Fig.4.11 shows the details of a ratchet mechanism. Wheel 4 is given intermittent
circular motion by means of arm 2 and driving pawl 3. A second pawl 5 prevents 4 from turning backward
when 2 is rotated clockwise in preparation for another stroke. The line of action PN of the driving pawl and
tooth must pass between centers O and A in order to have the pawl 3 remain in contact with the tooth. This
mechanism is used particularly in counting devices.
Fig.4.10 Geneva wheel Fig.4.11 Ratchet mechanism

Unit-2
4.5 Parallel Linkages
These are the four-bar linkages in which the opposite links are equal in length and always form a
parallelogram. There are three types of parallel linkages: parallel rules, universal drafting machine and lazy
tongs. They are used for producing parallel motion.
1.Parallel Rules: A parallel rule is shown in Fig.4.12, in which AB = CD = EF = GH = IJ and AC = BD, CE = DF, EG
= FH, GI = HJ. Here AB, C, EF, GH and IJ will always be parallel to each other.
2.Universal Drafting Machine: A universal drafting machine is shown in Fig.4.13, in which AB = CD; AC = BD;
EF = GH; and EG = FH. Position of points A and B are fixed. Similarly the positions of points E and F are fixed
with respect to C and D. The positions of scales I and II are fixed with respect to points G and H. Then ABDC is
a parallelogram. Line CD will always be parallel to AB so that the direction of CD is fixed. Therefore, the
direction of EF is fixed. Further EFHG is a parallelogram, so GH is always parallel to EF such that the direction
of GH is also fixed, whatever their actual position may be.
3.Lazy Tongs: Lazy tongs, as shown in Fig.4.14, consists of pin joint A attached to a fixed point. Point B moves
on a roller. All other joints are pin jointed. The vertical movement of the roller affects the movement of the
end C to move in a horizontal direction. It is used to support a telephone or a bulb at point C for horizontal
movements.
Fig.4.12 Parallel rule Fig.4.13 Universal drafting machine
Fig.4.14 Lazy tongs
4.7 Automobile Steering Gear Mechanisms
The mechanisms that are used for changing the direction of motion of an automobile are called steering gears.
Steering is done by changing the direction of motion of front wheels only as the rear wheels have a common
axis which is fixed and moves in a straight line only.

Unit-2
4.7.1 Fundamental Equation for Correct Steering
When a vehicle takes a turn all the wheels should roll on the road smoothly preventing excessive tyre wear.
This is achieved by mounting the two front wheels on two short axles, called stub axles. The stub axles are
pin-jointed with the main front axle which is rigidly attached to the rear axle, on which the back wheels are
attached. Steering is done by front wheels only.
Figure 4.22 shows an automobile taking right turn. When the vehicle takes a turn, the front wheels along with
stub axles turn about their respective pin-joints. The inner front wheel turns through a greater angle as
compared to that of outer front wheel. In order to avoid skidding, for correct steering, the two front wheels
must turn about the same instantaneous centre I, which lies on the axis of the rear wheels. Thus, the condition
for correct steering is that all the four wheels must turn about the same instantaneous centre.
Let AE = l = wheel base; CD = a = wheel track
AB = b = distance between the pivots of front axles
I = common instantaneous centre of all wheels
θ, ϕ = angles turned through by stub-axles of the inner and outer front wheels respectively
Fig.4.22 Automobile steering gear
�o�, = − = �� − ��
�� − �� = .
Eq. (4.12) is called the Fundamental Equation for Correct Steering.
4.7.2 Steering Gears
A steering gear is a mechanism for automatically adjusting values of θ, ϕ for correct steering. The following
steering gears are commonly used in automobiles:
1. Davis Steering Gear
2.Ackermann Steering Gear

Unit-2
1. Davis Steering Gear
Davis steering gear is shown in Fig.4.23. This type of gear has only sliding pairs. Two arms AG and BH are
fixed to the stub axles AC and BD respectively. CAG and DBH form two similar bell-crank levers pivoted at A
and B respectively. KL is a cross-link which is constrained to slide parallel to AB. The ends of the cross-link KL
are pin-jointed to two sliders S1 and S2 as shown. These sliders are free to slide on links AG and BH
respectively. The whole mechanism is in front of the front wheels.
During the straight motion of the vehicle, the gear is in the mid-position, with equal inclination of the arms AG
and BH with the verticals at A and B. The steering is achieved by moving cross-link KL to the right or left of
the mid-position. The steering gear for taking a right turn is shown in Fig.4.24. K′L′ shows the position of the
cross-link KL while taking a right turn.
Determination of Angle α:
Let x = distance moved by KL from mid-position = KK′ − LL′
y = horizontal distance of points K and L from A and B respectively
h = vertical distance between AB and KL
α = angle of inclination of track arms AG and BH with the vertical in mid-position
θ, ϕ = angles turned by stub axles
Fig.4.23 Davis steering gear for straight drive
From Fig.4.24, we have
y = AK sin α = BL sin α
�� − =
−
�� − ��
+ ��
=
−
�o� �� =

Unit-2
Fig.4.24 Davis steering gear taking a right turn
∴
− ��
+ ��
=
−
�� =
− +
.
��a��y, �� + � =
+
�o ��a� �� =
+ +
.
�� − �� = =
�� = =
Generally b/l =0.4 to 0.5, so that α = 11.3° to 14.1°. There will be friction and more wear due to sliding pairs
in the Davis steering gear. Therefore, it becomes inaccurate after some use. On account of these reasons, it is
not used in practice.
2. Ackermann Steering Gear
The Ackermann steering gear has only turning pairs. Fig.4.25 shows the gear for straight drive. The turning
pairs are: AK, KL, LB and AB. The two short arms AK and BL are of equal length and are connected by pin-
joints with front wheel axle AB at A and B respectively. AC and BD are the stub-axles so that CAK and DBL form
bell-crank levers. ABLK form a four - bar linkage. KL is the track rod. For correct steering, we have
�� − �� =
��n��a��y = . �o . ≈ .

Unit-2
Fig.4.25 Ackermann steering gear for straight drive
Determination of Angle α
1. Analytical Method: Consider the Ackermann steering gear, as shown in Fig.4.26 (a), taking a right turn.
The instantaneous centre I lies on a line parallel to the rear axis at a distance of approximately 0.3l above the
rear axis. It may be seen that the whole mechanism of the Ackermann steering gear is on the back of the front
wheels. From Fig.4.26 (b), we have
Projection of arc K’K on AB = Projection of arc L’L on AB
o� ′
= ′
AK [sin α − sin α − ϕ) = BL [sin (α + θ − sin α]
Now AK = BL
sin α − sin α − ϕ) = sin (α + θ − sin α
or sin α − sin α cos ϕ + cos α sin ϕ = sin α cos θ + cos α sin θ − sin α
sin α − cos ϕ − cos θ) = cos α (sin θ − sin ϕ)
�� =
�� − ��
− �� − ��
.
The values of θ and ϕ are known for correct steering. Hence, the value of α can be determined from, Eq.
(4.18).

Unit-2
Fig.4.26 Ackermann steering gear in displaced position
2. Graphical Method to Determine Angle α: Draw a horizontal line OX (Fig.4.27). Make ∠XOQ = θ and ∠XOP
= ϕ. With any radius, draw arc PRQ. Join PR and produce it so that PR = RM. Join MQ and produce. Draw ON ⊥
NQM. Then α = 90° − ∠XON.
Fig.4.27 Graphical method to determine angle α
Proof:
In Fig.4.28, let BL = x so that b = c + 2x sin α.
For correct steering,
b = l(cot ϕ − cot θ) = c cos ψ + x sin (α + θ) + x sin (α − ϕ)
If ψ is small, c cos ψ = c.
c + 2x sin α = c + x sin (α + θ) + x sin (α − ϕ)

Unit-2
or 2 sin α = sin (α + θ) + sin (α + θ)
In Fig.4.29, let r = OP = OR = OQ. In right angled triangle OQN,
∠NOQ = 90° − α − θ = 90 − α − θ)
ON = r cos[90° − α + θ)] = r sin(α + θ)
Draw RL ⊥ ON then RL || SN
i.e. ∠SRQ = ∠α
Also ∠ORL = ∠α
OL = r sin α
Now NL = RS. )n ΔOPR,
��
=
�� ° − . �
=
��
�� . �
= �� . �
From triangle RSM in Fig.4.30, we have

Unit-2
∠MRS = α − 0. ϕ
∠SMR = 90° − α − 0. ϕ)
A��o = �� − . �
Now RM = PR
= �� − . �
o�, = �� . � �� − . � = �� − �� − �
ON = OL + LN = OL + RS
�� + = �� + �� − �� − �
o�, �� = �� + + �� − � .
o� �� =
�� − ��
− �� − ��
.
Hence proved
4.8 (ooke’s Joint or Universal Coupling
It is a device to connect two shafts whose axes are neither coaxial nor parallel but intersect at a point. This is
used to transmit power from the engine to the rear axle of an automobile and similar other applications. It is
also called universal coupling.
The (ooke s joint, as shown in Fig.4.31, consists of two forks connected by a centre piece, having the shape of
a cross or square carrying four trunnions. The ends of the two shafts to be connected together are fitted to the
forks.

Unit-2
Fig.4.31 (ooke s joint
4.8.1 Velocities of Shafts
Let the driving shaft A and the driven shaft B be connected by a joint having two forks KL and MN. The two
forks are connected by a cross KLMN intersecting at O, as shown in Fig.4.32. Let the angle between the axes of
the shafts be α. Let fork KL move through an angle θ in a circle to the position K1L1 in the front view. The fork
MN will also move through the same angle θ. MN being not in the same plane shall move in an ellipse in the
front view to its new position M1N1. To find the true angle, project M1 to the top view, which cuts the
horizontal axis in R and fork MN in R1. Rotate R1 to R2 on the horizontal axis with centre O. Project R2 back in
the front view cutting the circle in M2. Join OM2 Measure angle MOM2, which is the true angle ϕ. Thus when
the driving shaft A revolves through an angle θ, the driven shaft B will revolve through an angle ϕ. Now
�� =
′
′ ; �� =
′
′
��
��
=
′
′ ×
′
′
Bu� ′ = ′
∴
��
��
=
′
′
=
Now OR = OR1 cos α
��
��
=
��
�� = ��
An�u�a� ��oc��y o� d��n ��a��, � =
An�u�a� ��oc��y o� d��n ��a��, � =
�
Differentiating Eq. (4.20), we get
Fig.4.33 Speed polar diagram

Unit-2
��c �.
�
�
= co� �. ��c �
�
�
��
��
��
��
=
co� �
− ��n �. co� �
�
�
=
��
− �� . ��
.
Case 1:
�
�
=
co� � = − ��n �. co� �
�� = ±√
+ ��
.
This condition occurs once in each quadrant, as shown in
Fig.4.33 by points 1, 2, 3 and 4.
Case 2:
ωB/ωA is minimum when the denominator is maximum,
i.e., sin2 α cos2 θ must be minimum, or cos θ = 0°.
Thus ωB/ωA is minimum at θ = 90°, 270°, i.e. at points 6 and 8. Then
�
�
= �� .
Case 3: ωB/ωA is maximum when denominator is minimum,
i.e., when cos2 θ = 1, or cos θ = ±1, θ = 0° or 180°,
i.e. at points 5 and 7. Then
�
�
=
��
.
Case 4:
�a��u� ��uc�ua��on o� ��oc��y o� d��n ��a�� =
� − �
�
�o� � = �
�a��u� ��d ��uc�ua��on =
�
��
− � ��
�
=
− ��
��
= �� .
For α to be small, tan α ≈ α and sin α ≈ α
�� ≈ .
4.8.2 Angular Acceleration of Driven Shaft
�
�
=
��
− �� . ��
�
= � . . �� [
�� . �� . ��
− �� . ��
]
Fig.4.32 (ooke s coupling forks in displaced
position

Unit-2
An�u�a� acc��a��on o� d��n ��a��, =
� �� . ��
− �� . ��
.
�o� acc��a��on �o b� �a��u� o� ��n��u�, =
= � �� . �� [
�� − �� . �� − − �� . �� . ��
− �� . ��
]
= �� + (
��
− ) �� − =
o�, �� ≈
��
− ��
.
4.9 Double Hooke’s Joint
The double (ooke s joint, as shown in Fig.4.34, is used to maintain the speed of driven shaft equal to the
driving shaft at every instant. To achieve this, the driving and the driven shafts should make equal angles with
the intermediate shaft and the forks of the intermediate shaft should lie in the same plane. Let γ be the angle
turned by the intermediate shaft while the angle turned by the driving shaft and the driven shaft be θ and ϕ
respectively. Then
tan θ = cos α tan γ and tan ϕ = cos α tan γ
Therefore, θ = ϕ
Fig.4.34 Double (ooke s joint
However, if the forks of the intermediate shafts lie in perpendicular planes to each other, then the variation of
speed of the driven shaft will be there.
(
�
�
) = ��
(
�
�
) = ��
(
�
�
) = �� .
��a��y (
�
�
) =
��
.

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