3131906 - GRAPHICAL AND ANALYTICAL LINKAGE SYNTHESIS

Contents
2.1 Synthesis of Mechanisms ................................................................................................................ 2.2
2.2 Freudenstein’s Equation (Synthesis a four-bar mechanism)......................................................... 2.3
2.3 Two-Position Synthesis of Slider-Crank Mechanisms.................................................................... 2.4
2.4 Two Position Synthesis of Crank and Rocker Mechanism ............................................................ 2.5
2.5 Inversion Method of Synthesis for Four-Bar Mechanism using Three Point................................ 2.7
2.6 Chebychev Spacing for Precision Positions ................................................................................... 2.8
2.7 Problems............................................................................................................................................ 2.9

2.2
Prof. Sunil G. Janiyani, Department of Mechanical Engineering
Kinematics and Theory of Machines (3131906) |
Unit-2 Graphical and Analytical Linkage Synthesis
2.1 Synthesis of Mechanisms
The synthesis of the mechanism is the design or creation of a mechanism to produce the 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.
In the application of synthesis, to the design of a mechanism, the problem divides itself into the following
three parts:
 Type synthesis: Type Synthesis refers to the kind of mechanism selected; it might be a linkage, a
geared system, belts, and pulleys, or even a cam system.
This beginning phase of the total design problem usually involves design factors such as
manufacturing processes, materials, safety space and economics. The study of kinematics is
usually only slightly involved in type synthesis.
 Number synthesis: Number synthesis deals with the number of links, and the number of joints or
pairs that are required to obtain certain mobility. Number synthesis is the second step in design
following type synthesis.
 Dimensional synthesis: The proportions or lengths of the links necessary to satisfy the required
motion characteristics.
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 addition to this, there are errors due to manufacturing. The error resulting from tolerances in the length
of links and bearing clearances is known as mechanical error.
2.1.1 Classifications of Synthesis Problem
a) Function Generation
A frequent requirement in design is that of causing an output member to rotate, oscillate or reciprocate
according to a specified function of time or function of input motion. This is called function generation.
A simple example is that of synthesizing a four-bar linkage to generate the function y=f(x). In this case, x
would represent the motion (crank angle) of input crank, and the linkage would be designed so that the
motion (angle) of the output rocker would approximate the function y.
Other examples of function generation are as follows:
1. In a conveyor line the output member of a mechanism must move at the constant velocity of the
conveyor while performing some operations – Ex. bottle capping, return, pick up the next cap and
repeat the operation.
2. The output member must pause or stop during its motion cycle to provide time for another event.
The second event might be a sealing, stapling, or fastening operation of some kind.
3. The output member must rotate at a specified non-uniform velocity function because it is geared
to another mechanism that requires such a rotating motion.
b) Path Generation:
The second type of synthesis problem is called path generation. This refers to a problem in which a coupler
point is to generate a path having a prescribed shape. Common requirements are that a portion of the path
is a circular arc, elliptical or straight line. Sometimes it is required that the path cross over itself as in a
figure-of-eight.

2.3
c) Body Guidance:
The third general class of synthesis problem is called body guidance. Here we are interested in moving an
object from one position to another.
The problem may call for a simple translation or a combination of translation and rotation (JCB example).
In the construction industry, for example, heavy parts such as scoops and bulldozer blades must be moved
through a series of prescribed positions.
2.2 Freudenstein’s Equation (Synthesis a four-bar mechanism)
Fig.2.1 – Four bar mechanism
 Replace the link of four-bar linkage by position vector and write the vector equation.
𝑟1 + 𝑟2 + 𝑟3 + 𝑟4 = 0
In complex polar notation above equation can be written as
𝑟1𝑒𝑗𝜃1 + 𝑟2𝑒𝑗𝜃2 + 𝑟3𝑒𝑗𝜃3 + 𝑟4𝑒𝑗𝜃4 = 0
Above equation is transformed into complex rectangular form by putting
𝒆𝒋𝜽
= 𝐜𝐨𝐬 𝜽 + 𝒋 ∙ 𝐬𝐢𝐧 𝜽.
∴ 𝑟1( cos 𝜃1 + 𝑗 ∙ sin 𝜃1) + 𝑟2( cos 𝜃2 + 𝑗 ∙ sin 𝜃2) + 𝑟3( cos 𝜃3 + 𝑗 ∙ sin 𝜃3) + 𝑟4( cos 𝜃4 + 𝑗 ∙ sin 𝜃4) = 0
 Now, if the real and imaginary components of the above equation are separated, we obtain the two
algebraic equations
𝑟1 cos 𝜃1 + 𝑟2 cos 𝜃2 + 𝑟3 cos 𝜃3 + 𝑟4 cos 𝜃4 = 0
𝑟1 sin 𝜃1 + 𝑟2 sin 𝜃2 + 𝑟3 sin 𝜃3 + 𝑟4 sin 𝜃4 = 0
In the above equation 𝐬𝐢𝐧 𝜽𝟏 = 𝟎 and 𝐜𝐨𝐬 𝜽𝟏 = −𝟏
∴ −𝑟1 + 𝑟2 cos 𝜃2 + 𝑟3 cos 𝜃3 + 𝑟4 cos 𝜃4 = 0
𝑟2 sin 𝜃2 + 𝑟3 sin 𝜃3 + 𝑟4 sin 𝜃4 = 0

2.4
 Now,
∴ 𝑟3 cos 𝜃3 = 𝑟1 − 𝑟2 cos 𝜃2 − 𝑟4 cos 𝜃4
∴ 𝑟3 sin 𝜃3 = −𝑟2 sin 𝜃2 − 𝑟4 sin 𝜃4
 Squaring and Adding both the equations
𝑟3
2(cos2
𝜃3 + sin2
𝜃3) = (𝑟1 − 𝑟2 cos 𝜃2 − 𝑟4 cos 𝜃4)2
+ (−𝑟2 sin 𝜃2 − 𝑟4 sin 𝜃4)2
∴ 𝑟3
2
= (𝑟1 − 𝑎)2
+ (−𝑟2 sin 𝜃2 − 𝑟4 sin 𝜃4)2
= 𝑟1
2
− 2𝑎𝑟1 + 𝑎2
+ 𝑟2
2
sin2
𝜃2 + 2𝑟2𝑟4 sin 𝜃2 sin 𝜃4 + 𝑟4
2
sin2
𝜃4
= 𝑟1
2
− 2(𝑟2 cos 𝜃2 + 𝑟4 cos 𝜃4)𝑟1 + (𝑟2 cos 𝜃2 + 𝑟4 cos 𝜃4)2
+ 𝑟2
2
sin2
𝜃2 + 2𝑟2𝑟4 sin 𝜃2 sin 𝜃4
+ 𝑟4
2
sin2
𝜃4
= 𝑟1
2
− 2𝑟1𝑟2 cos 𝜃2 − 2𝑟1𝑟4 cos 𝜃4 + 𝑟2
2
cos2
𝜃2 + 2𝑟2𝑟4 cos 𝜃2 cos 𝜃4 + 𝑟4
2
cos2
𝜃4 + 𝑟2
2
sin2
𝜃2
+ 2𝑟2𝑟4 sin 𝜃2 sin 𝜃4 + 𝑟4
2
sin2
𝜃4
= 𝑟1
2
+ 𝑟2
2
+ 𝑟4
2
− 2𝑟1𝑟2 cos 𝜃2 − 2𝑟1𝑟4 cos 𝜃4 + 2𝑟2𝑟4(cos𝜃2 cos 𝜃4 + sin 𝜃2 sin 𝜃4)
∴ 𝑟3
2
− 𝑟1
2
− 𝑟2
2
− 𝑟4
2
+ 2𝑟1𝑟2 cos 𝜃2 + 2𝑟1𝑟4 cos 𝜃4 = 2𝑟2𝑟4 cos(𝜃2 − 𝜃4)
 Dividing both the sides by 2r2r4
∴
𝑟3
2
− 𝑟1
2
− 𝑟2
2
− 𝑟4
2
2𝑟2𝑟4
+
𝑟1
𝑟4
cos 𝜃2 +
𝑟1
𝑟2
cos 𝜃4 = cos(𝜃2 − 𝜃4)
𝐾1 cos 𝜃2 + 𝐾2 cos 𝜃4 + 𝐾3 = cos(𝜃2 − 𝜃4)
Where
𝐾1 =
𝑟1
𝑟4
, 𝐾2 =
𝑟1
𝑟2
, 𝐾3 =
𝑟3
2
− 𝑟1
2
− 𝑟2
2
− 𝑟4
2
2𝑟2𝑟4
 Freudenstein’s equation enables us to perform this same task by analytical means. Thus suppose
we wish the output lever of a four-bar linkage to occupy the position ∅1, ∅2, and ∅3 corresponding
to the angular positions 𝜓1, 𝜓2, and 𝜓3 of the input lever. We simply replace 𝜃2 with 𝜓𝑖, 𝜃4with ∅𝑖,
and write the equation three times, once for each position.
𝐾1 cos 𝜓1 + 𝐾2 cos 𝜙1 + 𝐾3 = cos(𝜓1 − 𝜙1)
2.3 Two-Position Synthesis of Slider-Crank Mechanisms
The centered slider-crank mechanism has a stroke B1B2 equal to twice the crank radius r2 (B1B2 = 2r2). As
shown, the extreme positions of B1 and B2, also called limiting positions of the slider, are found by
constructing circular arcs through O2 of length (r3 - r2) and (r3 + r2), respectively.
In general, the centered slider-crank mechanism must have r3 > r2. However, the special case of r1 = r2
results in the isosceles slider-crank mechanism, in which the slider reciprocates through O2 and the stroke
4 × r2.

2.5
Fig.2.2 – Centered slider-crank mechanism
Fig.2.3 – General or offset slider crank mechanism
All points on the coupler of the isosceles slider-crank mechanism generate elliptical paths. The paths
generated by the points on the coupler of the slider-crank are not elliptical, but they are always symmetrical
about the axis O2B.
The linkage of general or offset slider-crank mechanism certain special effects can be obtained by
changing the offset distance e. Ex. the stroke B1B2 is always greater than 2 × crank radius r2.
This feature can be used to synthesize a quick return mechanism where a slower working stroke is desired.
Also, the crank angle required to execute the forward stroke is different from that of the return stroke.
2.4 Two Position Synthesis of Crank and Rocker Mechanism
The limiting positions of the rocker in a crank and rocker mechanism are shown as points B1 and B2 (Found
same as slider-crank linkage).
In this particular case, the crank executes the angle Ψ while the rocker moves from B1 to B2. Note on the
return stroke that the rocker swing from B2 to B1 through the same angle but the crank moves through the
angle (360° – Ψ).

2.6
Fig.2.4 - Extreme Position of Crank and Rocker Mechanism
There are many cases in which the crank and rocker mechanism is superior to the cam and follower
system. Among the advantages over the cam, the system is smaller forces involved, the elimination of
retaining spring, and the closer clearance because of the use of revolute pairs.
Cutting stroke B2 to B1 (∅ angle on the rocker) Ψ angle on the crank
Return stroke B1 to B2 (∅ angle on the rocker) 360° – Ψ angle on the crank
𝑄 =
𝜓
2𝜋 − 𝜓
=
180 + 𝛼
180 − 𝛼
{
𝑡1 =
𝜓
𝜔
𝑡2 =
360 − 𝜓
𝜔
=
2𝜋 − 𝜓
𝜔
}
Fig.2.5 - Synthesis of a four-bar linkage to generate rocker angle ∅
To synthesis, a crank and rocker mechanism for a specified value of ∅ and 𝛼, locate the point O4 in the
figure and choose any desired rocker length r4, then draw the two positions O4B1 and O4B2 of link 4
separated by the angle ∅ as given.
Through B1 construct any line X Then through B2 construct the line Y at given angle 𝛼 to line X. The
intersection of these two lines defines the location of the crank pivot O2. Because line X was originally
chosen arbitrarily, there is an infinite number of solutions to this problem.
The distance B2C is 2r2 or twice the crank length. So we bisect this distance to find r2.

2.7
2.5 Inversion Method of Synthesis for Four-Bar Mechanism using Three Point
 In the below figure, the motion of input rocker O2A through the angle 𝜓12 causes the motion of the
output rocker O4B through angle ∅12.
Fig.2.6 - Rotation of input rocker O2A through the angle ψ12 cause rocker O4B to rock through the angle ∅12
 To employ inversion as a technique of synthesis, let us hold O4B stationary and permit the
remaining links, including the frame, to occupy the same relative positions.
 The result is called inverting on the output rocker. Note that A1B1 is positioned the same in the
below figure. Therefore the inversion is made on the O4B1 position. Because O4B1is fixed, the frame
will have to move in order to get the linkage to the A2B2 position. In fact, the frame must move
backward through the angle ∅12. The second position is therefore O2
′
A2
′
B2
′
O4.
Fig.2.7 - Linkage inverted on the O4B position
 The below figure illustrates a problem and the synthesized linkage in which it is desired to
determine the dimensions of a linkage in which the output lever is to occupy three specified
positions corresponding to three given positions of input lever.
 The starting angle of the input lever is 𝜃2; and 𝜓12, 𝜓23, and 𝜓13 are swing angle respectively between
the three design positions 1 and 2, 2 and 3, and 1 and 3. Corresponding angles of swing ∅12, ∅23 and
∅13 are desired for the output lever. The length of link 4 and the starting position O4 are to be
determined.

2.8
Fig.2.8 – Three positions of input lever and output lever
2.6 Chebychev Spacing for Precision Positions
Fig.2.9 - Structural Error
 We need to work with two or three or four positions of the linkage called precision positions and to
find a linkage that exactly satisfies the desired function at a few chosen positions.
 Structural error is defined as the theoretical difference between the function produced by the
synthesized linkage and the function originally prescribed.
 A very good trial for the spacing of these precision positions is called Chebychev spacing. For n
precision position in the range 𝑥0 ≤ 𝑥 ≤ 𝑥𝑛+1, the Chebychev spacing according to Freudenstein
and Sandor, is
𝑥𝑗 =
1
2
(𝑥𝑛+1 + 𝑥0) −
1
2
(𝑥𝑛+1 − 𝑥0)𝑐𝑜𝑠
(2𝑗 − 1)𝜋
2𝑛
Where 𝑗 = 1,2, … 𝑛 And n = No. of precision positions

2.9
2.7 Problems
Ex. 2.1 [GTU; June-2016; 7 Marks] [GTU; Jan.-2016; 7 Marks]
A four-bar mechanism is to be designed, by using three precision points, to generate the
function
y = x1.5
, for the range 1 x 4 .
Assuming 30° starting position and 120° finishing position for the input link and
90°starting position and 180° finishing position for the output link, find the values of x, y,
 and φ corresponding to the three precision points.
Solution: Given Data:
xS = 1; xF = 4; s= 30; F = 120; φS = 90 and φF = 180
The three values of x corresponding to three precision points (i.e. for n = 3) according to
Chebychev’s spacing are given by,
𝒙𝒋 =
𝟏
𝟐
(𝒙𝑭 + 𝒙𝑺) −
𝟏
𝟐
(𝒙𝑭 − 𝒙𝑺) (𝒄𝒐𝒔 [
(𝟐𝒋 − 𝟏)𝝅
𝟐𝒏
])
∴ 𝑥𝑗 =
1
2
(4 + 1) −
1
2
(4 − 1) (𝑐𝑜𝑠 [
(2𝑗 − 1)𝜋
2(3)
])
∴ 𝒙𝒋 = 𝟐. 𝟓 − 𝟏. 𝟓 (𝒄𝒐𝒔 [
(𝟐𝒋 − 𝟏)𝝅
𝟔
])
For j = 1,
𝑥1 = 2.5 − 1.5 (𝑐𝑜𝑠 [
(2(1) − 1)𝜋
6
])
∴ 𝑥1 = 2.5 − 1.5 (𝑐𝑜𝑠 [
𝜋
6
])
∴ 𝑥1 = 2.5 − 1.5 (𝑐𝑜𝑠30°)
∴ 𝒙𝟏 = 𝟏. 𝟐
For j = 2,
𝑥2 = 2.5 − 1.5 (𝑐𝑜𝑠 [
(2(2) − 1)𝜋
6
])
∴ 𝑥2 = 2.5 − 1.5 (𝑐𝑜𝑠 [
3𝜋
6
])
∴ 𝑥2 = 2.5 − 1.5 (𝑐𝑜𝑠90°)
∴ 𝒙𝟐 = 𝟐. 𝟓
For j = 3,
𝑥3 = 2.5 − 1.5 (𝑐𝑜𝑠 [
(2(3) − 1)𝜋
6
])
∴ 𝑥3 = 2.5 − 1.5 (𝑐𝑜𝑠 [
5𝜋
6
])
∴ 𝑥3 = 2.5 − 1.5 (𝑐𝑜𝑠150°)
∴ 𝒙𝟑 = 𝟑. 𝟖

2.10
Since y = x1.5
, therefore the corresponding values of y are
𝒚𝑺 = (𝑥𝑆)1.5
= (1)1.5
= 𝟏
𝒚𝟏 = (𝑥1)1.5
= (1.2)1.5
= 𝟏. 𝟑𝟏𝟔
𝒚𝟐 = (𝑥2)1.5
= (2.5)1.5
= 𝟑. 𝟗𝟓𝟐
𝒚𝟑 = (𝑥3)1.5
= (3.8)1.5
= 𝟕. 𝟒𝟏
𝒚𝑭 = (𝑥𝐹)1.5
= (4)1.5
= 𝟖
The three values of  corresponding to three precision points are given by
𝜃𝑗 = 𝜃𝑆 +
𝜃𝐹 − 𝜃𝑆
𝑥𝐹 − 𝑥𝑆
(𝑥𝑗 − 𝑥𝑆)
∴ 𝜃𝑗 = 30 +
120 − 30
4 − 1
(𝑥𝑗 − 1) = 30 +
90
3
(𝑥𝑗 − 1) = 30 + 30(𝑥𝑗 − 1)
For j = 1,
∴ 𝜽𝟏 = 30 + 30(1.2 − 1) = 𝟑𝟔°
For j = 2,
∴ 𝜽𝟐 = 30 + 30(2.5 − 1) = 𝟕𝟓°
For j = 3,
∴ 𝜽𝟑 = 30 + 30(1.2 − 1) = 𝟏𝟏𝟒°
The three values of φ corresponding to three precision points are given by
𝜑𝑗 = 𝜑𝑆 +
𝜑 − 𝜑𝑆
𝑦𝐹 − 𝑦𝑆
(𝑦𝑗 − 𝑦𝑆)
∴ 𝜑𝑗 = 90 +
180 − 90
8 − 1
(𝑦𝑗 − 1) = 90 +
90
7
(𝑦𝑗 − 1)
For j = 1,
∴ 𝝋𝟏 = 90 +
90
7
(1.316 − 1) = 𝟗𝟒. 𝟎𝟔°
For j = 2,
∴ 𝝋𝟐 = 90 +
90
7
(3.952 − 1) = 𝟏𝟐𝟕. 𝟗𝟓°
For j = 3,
∴ 𝝋𝟑 = 90 +
90
7
(7.41 − 1) = 𝟏𝟕𝟐. 𝟒𝟏°
Ex. 2.2 [GTU; January-2017; 7 Marks] [GTU; December-2014; 7 Marks]
Design a four-bar mechanism to co-ordinate the input and output angles as follows:
Input angles = 15°, 30°, and 45°;
Output angles = 30°, 40°, and 55°.
Solution: Given Data:
1= 15; 2= 30; 3 = 45; φ1 = 30; φ2 = 40 and φ3 = 55

2.11
The Freudenstein’s equation is given by
𝑲𝟏 𝐜𝐨𝐬 𝝋 + 𝑲𝟐 𝐜𝐨𝐬 𝜽 + 𝑲𝟑 = 𝐜𝐨𝐬(𝜽 − 𝝋)
For 1= 15 and φ1 = 30;
𝐾1 cos 30 + 𝐾2 cos 15 + 𝐾3 = cos(15 − 30)
∴ 𝑲𝟏(𝟎. 𝟖𝟔𝟔) + 𝑲𝟐(𝟎. 𝟗𝟔𝟔) + 𝑲𝟑 = 𝟎. 𝟗𝟔𝟔 ⋯ ⋯ ⋯ (𝒊)
For 2= 30 and φ2 = 40;
𝐾1 cos 40 + 𝐾2 cos 30 + 𝐾3 = cos(30 − 40)
∴ 𝑲𝟏(𝟎. 𝟕𝟔𝟔) + 𝑲𝟐(𝟎. 𝟖𝟔𝟔) + 𝑲𝟑 = 𝟎. 𝟗𝟖𝟓 ⋯ ⋯ ⋯ (𝒊𝒊)
For 3= 45 and φ3 = 55;
𝐾1 cos 55 + 𝐾2 cos 45 + 𝐾3 = cos(45 − 55)
∴ 𝑲𝟏(𝟎. 𝟓𝟕𝟒) + 𝑲𝟐(𝟎. 𝟕𝟎𝟕) + 𝑲𝟑 = 𝟎. 𝟗𝟖𝟓 ⋯ ⋯ ⋯ (𝒊𝒊𝒊)
Solving the three simultaneous equations (i), (ii) and (iii), we get
k1 = 0.905 ; k2 = 1.01 and k3 = 1.158
Assuming the length of one of the links, say “a” as one unit, we get the length of the other
links.
Let us assume, a = 1 unit,
𝐾1 =
𝑑
𝑎
∴ 𝒅 = 𝑎 (𝐾1) = 1 (0.905) = 𝟎. 𝟗𝟎𝟓 𝒖𝒏𝒊𝒕𝒔
𝐾2 =
𝑑
𝑐
∴ 𝒄 =
𝑑
𝐾2
=
0.905
1.01
= 𝟎. 𝟖𝟗𝟔 𝒖𝒏𝒊𝒕𝒔
𝐾3 =
𝑎2
− 𝑏2
+ 𝑐2
+ 𝑑2
2𝑎𝑐
∴ 𝐾3(2𝑎𝑐) = 𝑎2
− 𝑏2
+ 𝑐2
+ 𝑑2
∴ 𝑏2
= (𝑎2
+ 𝑐2
+ 𝑑2) − 𝐾3(2𝑎𝑐)
∴ 𝒃 = 𝟎. 𝟕𝟒 𝒖𝒏𝒊𝒕𝒔

2.12
Ex. 2.3 Synthesize a 4 bar mechanism by the method of inversion for the following specifications.
𝑹𝑨𝑶𝟐
= 𝟐𝟎 𝐦𝐦 𝝍𝟏𝟐 = 𝟒𝟎° 𝝓𝟏𝟐 = 𝟑𝟎° 𝜽𝟐 = 𝟒𝟓°
𝑹𝑶𝟒𝑶𝟐
= 𝟔𝟎 𝐦𝐦 𝝍𝟐𝟑 = 𝟑𝟓° 𝝓𝟐𝟑 = 𝟐𝟓°
Solution:
 The solution to the problem is given in the figure and is based on inverting the linkage
on link 4.
 First, we draw the input rocker O2A in the three specified positions and locate the
desired position for O4.
 Because we will invert on link 4 in the first design position we draw a ray from O4 to
A2 and rotate it backward through the angle 𝜙12 to locate 𝐴2
′
.
 Similarly, we draw another ray O4A3 and rotate it backward through the angle 𝜙13 to
locate 𝐴3
′
.
 Because we are inverting on the first design position, A1 and A1
′
are coincident.
 Now we draw mid normals to the line A1A2
′
and A1A3
′
. These intersect at B1 and
define the length of coupler link 3 and the length of starting position of link 4.
Ex. 2.4 Four bar Crank-Rocker quick return linkage for specified time ratio. Time ratio = 1:1.25
with 45° output rocker motion. Design the synthesis.
Solution:
𝑇𝑅 =
𝛼
𝛽
𝛼 + 𝛽 = 360°
Construction angle 𝛿 = |180 − 𝛼|
= |180 − 𝛽|
1. Draw the output link O4B in both extreme positions, in any convenient location, such that
the desired angle of motion 𝜃4, is subtended.
2. Calculate 𝛼, 𝛽, and 𝛿 using equations. In this example, 𝛼 = 160°, 𝛽 = 200°, 𝛿 = 20°.
3. Draw a construction line through point B1 at any convenient angle.
4. Draw a construction line through point B2 at angle 𝛿 from the first line.
5. Label the intersection of the two construction lines O2.

2.13
6. The line O2O4 now defines the ground link.
7. Calculate the lengths of crank and coupler by measuring O2B1 and O2B2 and solve
simultaneously.
Coupler + crank = 𝑂2𝐵1
Coupler − crank = 𝑂2𝐵2
Or we can construct the crank length by swinging an arc centered at O2 from B1 to cut
line O2B2 extended. Label that intersection B1
′
. The line B2B1
′
is twice the crank length.
Bisect this line segment to measure crank length O2A1.
(a) (b)
(a) Construction of a quick return Grashof crank rocker
(b) The finished linkage in its two toggle positions
References:
1. Theory of Machines, Rattan S S, Tata McGraw-Hill
2. Theory of Machines, Khurmi R. S., Gupta J. K., S. Chand Publication

3131906 - GRAPHICAL AND ANALYTICAL LINKAGE SYNTHESIS

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