6.1 synthsis of the mechanism

Sanjivani Rural Education Society’s
Sanjivani College of Engineering, Kopargaon-423603
( An Autonomous Institute Affiliated to Savitribai Phule Pune University, Pune)
NAAC ‘A’ Grade Accredited, ISO 9001:2015 Certified
Subject :- Theory of Machines II
T.E. Mechanical (302043)
Unit 6 SYNTHESIS OF MECHANISM
By
Prof. K. N. Wakchaure(Asst Professor)
Department of Mechanical Engineering
Sanjivani College of Engineering
(An Autonomous Institute)
Kopargaon, Maharashtra
Email: wakchaurekiranmech@Sanjivani.org.in Mobile:- +91-7588025393

INTRODUCTION
• The synthesis of mechanism is the design or creation of a mechanism to produce a desired output
motion for a given input motion.
• In other words, the synthesis of mechanism deals with the determination of proportions of a
mechanism for the given input and output motion.
• Kinematic analysis is the process of determination of velocity and acceleration of the various links of an
existing mechanism.
• On the other hand, 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 position of input and output links are known at
fewer positions as compared to large number of positions.

Steps in the Synthesis
In the application of synthesis, to the design of a mechanism, the problem divides itself into the following three parts:
1. Type synthesis, i.e. the type of mechanism to be used,
• This includes Type of mechanism spatial or Planar Mechanism
• Chain drive, Belt drive, gear drive, 4- mechanism
• Types of Joint Sliding pair, revolute pair spherical Pair etc.
• Higher or Lower Pair
• Types of link- Binary, ternary, quadratic
• Types of Gears, Gear trains, Cam and follower
2. Number synthesis, i.e. the number of links and the number of joints needed to produce the required motion,
• no. of links and joints
• No. teeth and no. of gears
3. Dimensional synthesis, i.e. the proportions or lengths of the links necessary to satisfy the required motion
characteristics.
• Link lengths and angles
• Pitch circle diameter, center distance
• Stroke length and Angle of Ascent and Descent in Cam and Follower

INTRODUCTION
• In designing a mechanism, one factor that must be kept in mind is that of the accuracy required of the mechanism.
Sometimes, it is possible to design a mechanism that will theoretically generate a given motion.
• The difference between the desired motion and the actual motion produced is known as structural error.
• 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+1and 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.
• 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.
• The error resulting from tolerances in the length of links and bearing clearances
is known as mechanical error.
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, tolerances in the length of links and bearing clearances,
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.

Classifications of Synthesis Problem
1. Function generation.
The major classification of the synthesis problems that arises in the design of links in a mechanism is a function
generation. In designing a mechanism, the frequent requirement is that the output link should either rotate, oscillate or
reciprocate according to a specified function of time or function of the motion of input link.
This is known as function generation.
A simple example is that of designing a four bar mechanism to generate the function y = f (x). In this case, x represents
the motion of the input link and the mechanism is to be designed so that the motion of the output link approximates the
function y.

2. Path generation.
In a path generation, the mechanism is required to guide a point (called a tracer point or coupler point) along a path
having a prescribed shape. The common requirements are that a portion of the path be a circular arc, elliptical or a straight
line.

3. Body guidance.
In body guidance, both the position of a point within a moving body and the angular displacement of the body are
specified. The problem may be a simple translation or a combination of translation and rotation.

Freudenstein’s equation
• Consider a four bar mechanism ABCD, as shown in Fig, in which AB = a, BC = b,
CD = c, and DA = d.
• The link AD is fixed and lies along X-axis.
• Let the links AB (input link), BC (coupler) and DC (output link) make angles 𝜃,𝜙
and 𝛽 respectively along the X-axis or fixed link AD.
For equilibrium of the mechanism, the sum of the components along X-axis and
along
Y-axis must be equal to zero.
First of all, taking the sum of the components along X-axis as shown in Fig
• 𝑎𝑐𝑜𝑠𝜃 + 𝑏𝑐𝑜𝑠𝛽 − 𝑐𝑐𝑜𝑠𝜙 − 𝑑 = 0
• 𝑏𝑐𝑜𝑠𝛽 = 𝑐𝑐𝑜𝑠𝜙 + 𝑑 − 𝑎𝑐𝑜𝑠𝜃
• Squaring both the sides
• 𝑏𝑐𝑜𝑠𝛽 2
= 𝑐𝑐𝑜𝑠𝜙 + 𝑑 − 𝑎𝑐𝑜𝑠𝜃 2
• 𝑏𝑐𝑜𝑠𝛽 2
= 𝑐2
cos2
𝜙 + 𝑑2
+ 2𝑐𝑑𝑐𝑜𝑠𝜙 + 𝑎2
cos2
𝜃 − 2𝑎𝑐𝑐𝑜𝑠𝜙𝑐𝑜𝑠𝜃 − 2𝑎𝑑𝑐𝑜𝑠𝜃 1

• taking the sum of the components along Y-axis as shown in Fig
• 𝑎𝑠𝑖𝑛𝜃 + 𝑏𝑠𝑖𝑛𝛽 − 𝑐𝑐𝑜𝑠𝜙 = 0
• 𝑏𝑠𝑖𝑛𝛽 = 𝑐𝑠𝑖𝑛𝜙 + 𝑎𝑠𝑖𝑛𝜃
• Squaring both the sides
• 𝑏𝑠𝑖𝑛𝛽 2 = 𝑐𝑠𝑖𝑛𝜙 − 𝑎𝑠𝑖𝑛𝜃 2
• 𝑏𝑠𝑖𝑛𝛽 2 = 𝑐2 sin2 𝜙 + 𝑎2 sin2 𝜃 − 2𝑎𝑐𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃 2
adding Equation 1 and 2
𝑏𝑐𝑜𝑠𝛽 2
+ 𝑏𝑠𝑖𝑛𝛽 2
= 𝑐2 cos2 𝜙 + 𝑑2 + 2𝑐𝑑𝑐𝑜𝑠𝜙 + 𝑎2 cos2 𝜃 − 2𝑎𝑐𝑐𝑜𝑠𝜙𝑐𝑜𝑠𝜃 − 2𝑎𝑑𝑐𝑜𝑠𝜃
+ 𝑐2
sin2
𝜙 + 𝑎2
sin2
𝜃 − 2𝑎𝑐𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃
• 𝒃 𝟐(𝐜𝐨𝐬 𝟐 𝜷 + 𝐬𝐢𝐧 𝟐 𝜷) = 𝒄 𝟐(𝒄𝒐𝒔 𝟐 𝝓 + 𝒔𝒊𝒏 𝟐 𝝓) + 𝒅 𝟐 + 𝟐𝒄𝒅𝒄𝒐𝒔𝝓 + 𝐚 𝟐 𝐜𝐨𝐬 𝟐 𝜽 + sin 𝟐 𝜽 −
𝟐𝒂𝒄 𝒔𝒊𝒏𝝓𝒔𝒊𝒏𝜽 + 𝟐𝒂𝒄𝒄𝒐𝒔𝝓𝒄𝒐𝒔𝜽 − 𝟐𝒂𝒅𝒄𝒐𝒔𝜽
• 𝒃 𝟐 = 𝒄 𝟐 + 𝐚 𝟐 + 𝒅 𝟐 + 𝟐𝒄𝒅𝒄𝒐𝒔𝝓 − 𝟐𝒂𝒄 𝒔𝒊𝒏𝝓𝒔𝒊𝒏𝜽 + 𝒄𝒐𝒔𝝓𝒄𝒐𝒔𝜽 − 𝟐𝒂𝒅𝒄𝒐𝒔𝜽

• 𝒃 𝟐(𝐜𝐨𝐬 𝟐 𝜷 + 𝐬𝐢𝐧 𝟐 𝜷) = 𝒄 𝟐(𝒄𝒐𝒔 𝟐 𝝓 + 𝒔𝒊𝒏 𝟐 𝝓) + 𝒅 𝟐 + 𝟐𝒄𝒅𝒄𝒐𝒔𝝓 +
𝐚 𝟐 𝐜𝐨𝐬 𝟐 𝜽 + 𝐬𝐢𝐧 𝟐 𝜽 − 𝟐𝒂𝒄 𝒔𝒊𝒏𝝓𝒔𝒊𝒏𝜽 + 𝟐𝒂𝒄𝒄𝒐𝒔𝝓𝒄𝒐𝒔𝜽 − 𝟐𝒂𝒅𝒄𝒐𝒔𝜽
• 𝒃 𝟐
= 𝒄 𝟐
+ 𝐚 𝟐
+ 𝒅 𝟐
+ 𝟐𝒄𝒅𝒄𝒐𝒔𝝓 − 𝟐𝒂𝒄 𝒔𝒊𝒏𝝓𝒔𝒊𝒏𝜽 + 𝒄𝒐𝒔𝝓𝒄𝒐𝒔𝜽 − 𝟐𝒂𝒅𝒄𝒐𝒔𝜽
• 𝟐𝒂𝒄 𝒔𝒊𝒏𝝓𝒔𝒊𝒏𝜽 + 𝒄𝒐𝒔𝝓𝒄𝒐𝒔𝜽 = 𝐚 𝟐 − 𝒃 𝟐 + 𝒄 𝟐 + 𝒅 𝟐 + 𝟐𝒄𝒅𝒄𝒐𝒔𝝓 −
− 𝟐𝒂𝒅𝒄𝒐𝒔𝜽
• 𝒔𝒊𝒏𝝓𝒔𝒊𝒏𝜽 + 𝒄𝒐𝒔𝝓𝒄𝒐𝒔𝜽 =
(𝐚 𝟐−𝒃 𝟐+𝒄 𝟐+𝒅 𝟐)
𝒂𝒄
+
𝒅
𝒂
𝒄𝒐𝒔𝝓 −
𝒅
𝒄
𝒄𝒐𝒔𝜽
• 𝒔𝒊𝒏𝝓𝒔𝒊𝒏𝜽 + 𝒄𝒐𝒔𝝓𝒄𝒐𝒔𝜽 = 𝒄𝒐𝒔 𝜽 − 𝝓
• 𝒄𝒐𝒔 𝜽 − 𝝓 =
(𝐚 𝟐−𝒃 𝟐+𝒄 𝟐+𝒅 𝟐)
𝒂𝒄
+
𝒅
𝒂
𝒅
𝒄
𝒄𝒐𝒔𝜽
• 𝒄𝒐𝒔 𝜽 − 𝝓 =
𝒅
𝒂
𝒅
𝒄
𝒄𝒐𝒔𝜽 +
(𝐚 𝟐−𝒃 𝟐+𝒄 𝟐+𝒅 𝟐)
𝒂𝒄
• 𝒄𝒐𝒔 𝜽 − 𝝓 = 𝒌 𝟏 𝒄𝒐𝒔𝝓 − 𝒌 𝟐 𝒄𝒐𝒔𝜽 + 𝒌 𝟑
• 𝒌 𝟏 =
𝒅
𝒂
, 𝒌 𝟐 =
𝒅
𝒄
, 𝒌 𝟑 =
(𝐚 𝟐−𝒃 𝟐+𝒄 𝟐+𝒅 𝟐)
𝒂𝒄
• 𝒄𝒐𝒔 𝜽 − 𝝓 = 𝒌 𝟏 𝒄𝒐𝒔𝝓 − 𝒌 𝟐 𝒄𝒐𝒔𝜽 + 𝒌 𝟑 The equation is known as Freudenstein’s equation

• 𝒌 𝟏 =
𝒅
𝒂
, 𝒌 𝟐 =
𝒅
𝒄
, 𝒌 𝟑 =
(𝐚 𝟐−𝒃 𝟐+𝒄 𝟐+𝒅 𝟐)
𝒂𝒄
• 𝜔1= Angular velocity of the link AB =
𝑑𝜃
𝑑𝑡
,
• 𝜔2= Angular velocity of the link BC =
𝑑𝛽
𝑑𝑡
, and
• 𝜔3= Angular velocity of the link CD =
𝑑𝜙
𝑑𝑡
.
𝜔3 =
−𝑎𝜔1 sin(𝛽 − 𝜃)
csin(𝜙 − 𝛽)
𝜔2 =
−𝑎𝜔1 sin(𝜙 − 𝜃)
csin(𝜙 − 𝛽)
• By differentiating equation 1 & 2

• 𝒌 𝟏 =
𝒅
𝒂
, 𝒌 𝟐 =
𝒅
𝒄
, 𝒌 𝟑 =
(𝐚 𝟐−𝒃 𝟐+𝒄 𝟐+𝒅 𝟐)
𝒂𝒄
• 𝜶1= Angular Acceleration of the link AB =
𝑑𝝎 𝟏
𝑑𝑡
,
• 𝜶2= Angular Acceleration of the link BC =
𝑑𝝎 𝟐
𝑑𝑡
, and
• 𝜶3= Angular Acceleration of the link CD =
𝑑𝝎 𝟑
𝑑𝑡
.
𝛼2 =
−𝑎𝛼1 sin 𝜙 − 𝜃 + 𝑎𝜔1
2
cos 𝜙 − 𝜃 + 𝑏𝜔2
2
cos 𝜙 − 𝛽 − 𝑐𝜔3
2
bsin(𝜙 − 𝛽)
• By differentiating equation 𝜔2 𝑎𝑛𝑑 𝜔3
𝛼3 =
−𝑎𝛼1 sin 𝛽 − 𝜃 + 𝑎𝜔1
2
cos 𝛽 − 𝜃 − 𝑏𝜔3
2
cos 𝜙 − 𝛽 − 𝑏𝜔2
2
csin(𝜙 − 𝛽)
𝜔3 =
−𝑎𝜔1 sin(𝛽 − 𝜃)
csin(𝜙 − 𝛽)
𝜔2 =
−𝑎𝜔1 sin(𝜙 − 𝜃)
csin(𝜙 − 𝛽)

13Subject :- Theory of Machines II
• In designing a mechanism to generate a particular function, it is usually impossible to accurately produce the
function at more than a few points.
• The points at which the generated and desired functions agree are known as precision points or accuracy points
and must be located so as to minimise the error generated between these points.
• The best spacing of the precision points, for the first trial, is called Chebychev spacing.
• According to Freudenstein and Sandor, the Chebychev spacing for n points in the range xs ≥ xi ≥ xf
• (i.e. when x varies between xs and xf ) is given by
𝑥𝑖 =
1
2
𝑥 𝑠 + 𝑥 𝑓 −
1
2
𝑥 𝑓 − 𝑥 𝑠 cos
𝜋 2 × 𝑖 − 1
2 × 𝑛
𝑥𝑖 =
1
2
𝑥 𝑠 + 𝑥 𝑓 −
1
2
× ∆𝑥 × cos
𝜋 2 × 𝑖 − 1
2 × 𝑛
• 𝑥𝑖 = 𝑃𝑟𝑒𝑐𝑖𝑠𝑖𝑜𝑛 𝑃𝑜𝑖𝑛𝑡𝑠
• 𝑥 𝑠 𝑎𝑛𝑑 𝑥 𝑓 𝑆𝑡𝑎𝑛𝑡𝑖𝑛𝑔 𝑎𝑛𝑑 𝑓𝑖𝑛𝑖𝑠ℎ𝑖𝑛𝑔 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦
• ∆𝑥 = 𝑅𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 = 𝑥 𝑓 − 𝑥 𝑠
• i=1,2,…n
Precision Points for Function Generation- Chebychev spacing.

The precision or accuracy points may be easily obtained by using the graphical method as
discussed below.
1. Draw a circle of diameter equal to the range ∆𝑥 = 𝑥 𝑓 − 𝑥 𝑠 .
2. Inscribe a regular polygon having the number of sides equal to twice the number of precision points required, i.e.
for three precision points, draw a regular hexagon inside the circle, as shown in Fig
3. Draw perpendiculars from each corner which intersect the diagonal of a circle at precision points 𝑥1, 𝑥2, 𝑥3.
𝑥1 𝑥2 𝑥3 𝑥 𝑓𝑥𝑖
Precision Points for Function Generation- Chebychev spacing.

Consider a four bar mechanism, as shown in Fig. arranged to generate a function y = f (x)
over a limited range.
Let the range in x is (𝑥 𝑠 − 𝑥 𝑓 ) and the corresponding range in 𝜃 is (𝜃𝑓 − 𝜃𝑠) . Similarly,
let the range in y is (𝑦𝑠 − 𝑦𝑓 ) and the corresponding range in 𝜙 is (𝜙 𝑓 − 𝜙𝑠) .
The linear relationship between x and θ
𝜃 = 𝜃𝑠 +
(𝜃𝑓 − 𝜃𝑠)
𝑥 𝑓 − 𝑥 𝑠
𝑥 − 𝑥 𝑠
linear relationship between y and φ may be written as
𝜙 = 𝜙𝑠 +
(𝜙 𝑓 − 𝜙𝑠)
𝑦𝑓 − 𝑦𝑠
𝑦 − 𝑦𝑠
For n points in the range
𝜃𝑖 = 𝜃𝑠 +
(𝜃𝑓 − 𝜃𝑠)
𝑥 𝑓 − 𝑥 𝑠
𝑥𝑖 − 𝑥 𝑠
𝜙𝑖 = 𝜙𝑠 +
(𝜙 𝑓 − 𝜙𝑠)
𝑦𝑓 − 𝑦𝑠
𝑦𝑖 − 𝑦𝑠
i=1,2,3…n
Angle Relationships for Function Generation

Graphical Synthesis
Inversion Method
4 BAR -2 POSITION
4 BAR -3 POSITION
3R-P – 2 POSITION
3R-P – 3 POSITION
Relative Pole Method
4 BAR -2 POSITION
4 BAR -3 POSITION
3R-P – 2 POSITION
3R-P – 3 POSITION

17
Given Data
A D
𝐵1
′
𝐵2
Θ
Θ12
𝜙12
𝐵2
′
𝐶1
𝜙
1. Draw AD equal to the known length of fixed link, as shown in Fig.
2. At A, draw the input link 1 in its three specified angular positions AB1,
AB2
3. Since we have to invert the mechanism on link 4, therefore draw a line
𝑩 𝟐 𝑫 and and rotate it clockwise (in a direction opposite to the direction
in which link 1 rotates) through an angle 𝝋 𝟏𝟐 (i.e. the angle of the
output link 4 between the first and second position) in order to locate
the point 𝑩 𝟐
′
.
4. Since the mechanism is to be inverted on the first design position,
therefore B1 and B1′ are coincident.
5. Draw the perpendicular bisectors of the lines B1′ B2′ .Assume any point
𝑪 𝟏 on Perpendicular bisector.
6. Join B1′C1 and C1D . The figure AB1′ C1D is the required four bar
mechanism. Now the length of the link 3 and length of the link 4 and its
starting position ( φ ) are determined. 1
2
3
4
𝜃1
𝜃2
𝜙1
𝜙2
𝜙12 = 𝜙2-𝜙1𝜃12 = 𝜃2-𝜃1
Graphical Synthesis INVERSION METHOD
4R-2 POSITION SYNTHESIS

18
Given Data
A D
𝐵1
′
𝐵2
𝐵3
′
Θ
Θ12
Θ13
𝜙12
𝐵2
′
𝜙13
𝐵3
′
𝐶1
𝜙
2. At A, draw the input link 1 in its three specified angular positions AB1,
AB2 and AB3.
3. Since we have to invert the mechanism on link 4, therefore draw a line
𝑩 𝟐 𝑫 and and rotate it clockwise (in a direction opposite to the direction
in which link 1 rotates) through an
4. angle 𝝋 𝟏𝟐 (i.e. the angle of the output link 4 between the first and
second position) in order to locate the point 𝑩 𝟐
′
.
5. Similarly, draw another line B3D and rotate it clockwise through an angle
φ13 (i.e. angle of the output link between the first and third position) in
order to locate point 𝑩 𝟑
′
.
6. Since the mechanism is to be inverted on the first design position,
therefore B1 and B1′ are coincident.
7. Draw the perpendicular bisectors of the lines B1′ B2′ and B2′ B3′ . These
bisectors intersect at point C1.
8. Join B1′C1 and C1D . The figure AB1′ C1D is the required four bar
mechanism. Now the length of the link 3 and length of the link 4 and its
starting position ( φ ) are determined.
1
2
3
4
𝜃1
𝜃2
𝜃3
𝜙1
𝜙2
𝜙3
𝜙12 = 𝜙2-𝜙1
𝜙13 = 𝜙3-𝜙1
𝜃12 = 𝜃2-𝜃1
𝜃13 = 𝜃3-𝜃1

19
Given Data 𝜃1
𝜃2
𝑆1
𝑆2
𝑆12 = 𝑆2-𝑆1𝜃12 = 𝜃2-𝜃1
A
e
𝐶1 𝐶2
Θ12
𝐶2
′
B
𝑆12
1. Draw horizontal line l, assume any point A on line.
2. Draw line parallel to line l with offset distance of e.
3. Mark any point 𝑪 𝟏 on offset line also mark 𝑪 𝟐 at a distance of 𝑺 𝟏𝟐.
4. Draw another line 𝑨𝑪 𝟐and rotate it anticlockwise through an angle 𝜽 𝟏𝟐
in order to locate point 𝑪 𝟐
′
.
5. Join 𝑪 𝟏 𝑪 𝟐
′
. Draw the perpendicular bisectors of the lines 𝑪 𝟏 𝑪 𝟐
′
6. Assume any point B on Perpendicular bisectors.
7. Join AB and 𝑩𝑪 𝟏
Graphical Synthesis-INVERSION METHOD
3RP-2 POSITION SYNTHESIS

20
Given Data
1. Draw horizontal line l, assume any point A on line.
2. Draw line parallel to line l with offset distance of e.
3. Mark any point 𝑪 𝟏 on offset line also mark 𝑪 𝟐 and 𝑪 𝟑 at a distance of
𝑺 𝟏𝟐 and 𝑺 𝟏𝟑.
4. Draw another line 𝑨𝑪 𝟐and rotate it anticlockwise through an angle 𝜽 𝟏𝟐
in order to locate point 𝑪 𝟐
′
.
5. Draw another line 𝑨𝑪 𝟑and rotate it anticlockwise through an angle 𝜽 𝟏𝟑
in order to locate point 𝑪 𝟑
′
.
6. Join 𝑪 𝟏 𝑪 𝟐
′
and 𝑪 𝟏 𝑪 𝟑
′
. Draw the perpendicular bisectors of the lines 𝑪 𝟏 𝑪 𝟐
′
and 𝑪 𝟏 𝑪 𝟑
′
. These bisectors intersect at point B.
7. Join AB and 𝑩𝑪 𝟏
𝜃1
𝜃2
𝜃3
𝑆1
𝑆2
𝑆3
𝑆12 = 𝑆2-𝑆1
𝑆13 = 𝑆3-𝑆1
𝜃12 = 𝜃2-𝜃1
𝜃13 = 𝜃3-𝜃1
A
e
𝐶1 𝐶2 𝐶3
Θ12
Θ13
𝐶2
′
𝐶3
′
B
𝑆12
𝑆13
3 POSITION SYNTHESIS

21
Given Data
2. At A, draw a line through point A at an angle of
𝜽 𝟏𝟐
𝟐
in opposite
direction of rotation of input link and draw a line through point D at an
angle of
𝝓 𝟏𝟐
𝟐
in opposite direction of rotation of output link
3. These two links are intersects at common point referred as
𝑹 𝟏𝟐 𝒂 𝑷𝒐𝒍𝒆 𝑷𝒐𝒊𝒏𝒕.
4. Measure included angle at 𝑹 𝟏𝟐 mark as 𝝍 𝟏𝟐
5. Draw Input link AB with given length (or Assume) at suitable angle 𝜽.
6. Join 𝑹 𝟏𝟐 𝒕𝒐 𝑩 and draw line and draw line at an angle of 𝝍 𝟏𝟐(Shown in
fig.)
7. Consider any point C on this line.
8. Join BC and CD
9. Mechanism ABCD satisfies the given Conditions.
Θ12
2 𝜙12
2A D
B
𝜓12
C
𝜓12
𝑅12
3
4
1
2
𝜃1
𝜃2
𝜙1
𝜙2
𝜙12
2
=
𝜙2−𝜙1
2
𝜃12
2
=
(𝜃2−𝜃1)
2
Graphical Synthesis RELATIVE POLE METHOD

22
Given Data
𝜽 𝟏𝟐
𝟐
in opposite direction
of rotation of input link and draw a line through point D at an angle of
𝝓 𝟏𝟐
𝟐
in opposite direction of rotation of output link . These two links are
intersects at common point referred as 𝑹 𝟏𝟐 𝒂 𝑷𝒐𝒍𝒆 𝑷𝒐𝒊𝒏𝒕.
𝜽 𝟏𝟑
𝟐
in opposite direction
of rotation of input link and draw a line through point D at an angle of
𝝓 𝟏𝟑
𝟐
in opposite direction of rotation of output link . These two links are
intersects at common point referred as 𝑹 𝟏𝟑 𝒂 𝑷𝒐𝒍𝒆 𝑷𝒐𝒊𝒏𝒕. Measure
included angle at 𝑹 𝟏𝟐 and 𝑹 𝟏𝟑 mark as 𝝍 𝟏𝟐 and 𝝍 𝟏𝟑 respectively.
4. Draw Input link AB with given length (or Assume) at suitable angle 𝜽.
5. Join 𝑹 𝟏𝟐 𝒕𝒐 𝑩 and draw line and draw line at an angle of 𝝍 𝟏𝟐(Shown in fig.)
6. Join 𝑹 𝟏𝟑 𝒕𝒐 𝑩 and draw line and draw line at an angle of 𝝍 𝟏𝟑(Shown in fig.)
7. These two lines intersect at common point C.
8. Join BC and CD
9. Mechanism ABCD satisfies the given Conditions.
Θ12
2 𝜙12
2
𝑅13
A D
Θ13
2
𝜙13
2
B
𝜓13
𝜓12
𝜓13
C
𝜓12
𝑅12
3
4
1
2
3
1
2
𝜃1
𝜃2
𝜃3
𝜙1
𝜙2
𝜙3
𝜙12
2
=
𝜙2−𝜙1
2
𝜙13
2
=
𝜙3−𝜙1
2
𝜃12
2
=
(𝜃2−𝜃1)
2
𝜃13
2
=
(𝜃3−𝜃1)
2

23
Given Data
1. Draw horizontal line and mark any point A on the line.
2. Draw vertical line through point A.
3. Take distance
𝑺 𝟏𝟐
𝟐
from point A and draw line parallel to vertical line.
4. Draw inclined line
𝜽 𝟏𝟐
𝟐
( In the opposite direction of rotation off Crank)
with vertical though point A. Two lines intersects at point 𝑹 𝟏𝟐 and
measure included angle 𝜸
5. Draw crank AB of given length with included angle 𝜽.
6. Join 𝑹 𝟏𝟐 𝑩 and draw line through point 𝑹 𝟏𝟐 at an angle of 𝜸
7. Assume two lines intersects at point 𝑪 𝟏.
8. Join 𝑩𝑪 𝟏 and measure e.
9. Mechanism satisfies the given Conditions.
𝜃1
𝜃2
𝜃3
𝑆1
𝑆2
𝑆3
𝑆12 = 𝑆2-𝑆1
𝑆13 = 𝑆3-𝑆1
𝜃12 = 𝜃2-𝜃1
𝜃13 = 𝜃3-𝜃1
A
𝑆12
2
𝑃12
Θ12
2
𝐵1
γ
γ
𝐶1
e𝜃
3RP-2 POSITION SYNTHESIS

24
Given Data
1. Draw horizontal line and mark any point A on the line.
2. Draw vertical line through point A.
3. Take distance
𝑺 𝟏𝟐
𝟐
𝜽 𝟏𝟐
𝟐
with vertical though point A.
5. Two lines intersects at point 𝑹 𝟏𝟐 and measure included angle 𝜸
6. Take distance
𝑺 𝟏𝟑
𝟐
𝜽 𝟏𝟑
𝟐
with vertical though point A. Two lines intersects at point 𝑹 𝟏𝟑. and
measure included angle 𝝍.
8. Draw crank AB of given length with included angle 𝜽.
9. Join 𝑹 𝟏𝟐 𝑩 and draw line through point 𝑹 𝟏𝟐 at an angle of 𝜸
10. Join 𝑹 𝟏𝟑 𝑩 and draw line through point 𝑹 𝟏𝟑 at an angle of 𝝍
11. Above two lines intersects at point 𝑪 𝟏.
12. Join 𝑩𝑪 𝟏 and measure e.
13. Mechanism satisfies the given Conditions.
𝜃1
𝜃2
𝜃3
𝑆1
𝑆2
𝑆3
𝑆12 = 𝑆2-𝑆1
𝑆13 = 𝑆3-𝑆1
𝜃12 = 𝜃2-𝜃1
𝜃13 = 𝜃3-𝜃1
A
𝑆12
2
𝑆13
2
𝑅12
𝑅13
Θ12
2
Θ13
2
𝐵1
γ
γ
𝜓 𝜓
𝐶1
e𝜃
3R-P-2 POSITION SYNTHESIS

25
Given Data 3 𝑃𝑂𝑆𝐼𝑇𝐼𝑂𝑁𝑆 𝑂𝐹 𝐶𝑂𝑈𝑃𝐿𝐸𝑅 𝐿𝐼𝑁𝐾.
x
y
𝐵1(2,0)
𝐴1(2,0)
𝐵2(5,2)
𝐴2(2,0)
50°
𝐴3(12,6)
𝐵3(6,6)
𝑂𝐴
𝑂 𝐵
4
3
1
2
1. Draw XY co-ordinate system.
2. Consider the three positions of the points A and B such as A1, A2, A3 and
B1, B2, B3 as shown in Fig.
3. Find the centre of a circle which passes through three points A1, A2, A3.
This is obtained by drawing the perpendicular bisectors of the line
segments A1 A2 and A2 A3. Let these bisectors intersect at 𝑶 𝑨. It is
evident that a rigid link 𝑨 𝑶 𝑨 pinned to the body at point A and pinned to
the ground at point OA will guide point A through its three positions A1,
A2 and A3.
4. Similarly, find the centre 𝑶 𝑩 of a circle which passes through three points
B1, B2, B3. It is evident that a rigid link 𝑩𝑶 𝑩 pinned to the body at point
B and pinned to the ground at point OB will guide point B through its
three positions B1, B2 and B3.
5. The above construction forms the four bar mechanism 𝑶 𝑨 𝑨𝑩𝑶 𝑩which
guides the body through three specified positions. Fig. shows a four bar
mechanism in these three positions.
6. Measure the co-ordinates of 𝑶 𝑨 𝒂𝒏𝒅 𝑶 𝑩 and Length 𝑶 𝑨 𝑨 𝟏 and 𝑶 𝑩 𝑩 𝟏.
Graphical Synthesis BODY GUIDANCE
3 POSITION SYNTHESIS

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6.1 synthsis of the mechanism

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6.1 synthsis of the mechanism