1

2
Qualitative synthesis : to create the potential solutions
Some linkage design in the absence of a well-defined algorithm that predicts the solution
Iteration between analysis and synthesis:
With the help of CAD such as Working Model or popular programs
Norton textbook: Chap 3

3
Function generation : correlation of I/P motion with O/P motion in an mechanism ( black box: mechanical analog computer, artillery rangefinder, shipboard gun aiming system); Linkage performs a mathematical function w/ input the independent variable and output the dependent variable
- rocker output
Path generation: control of a Point in the plane such that it follows some prescribed path; The linkage guides some pt. (coupler pt.) along a desired path
Motion generation: control of a Line in the plane such that it follows some prescribed set of sequential positions: bucket control on a bulldozer (rigid body guidance); The linkage guides a body through a specified path with specified orientation
- simple case: coupler output

Traditional
Computer-aided design synthesis
4
- Graphical (chap 3)
- Analytical (chap 5)
- Working model
Synthesis CAD program: Symech http://www.symech.com
More CAD programs: http://www.tenlinks.com/CAE/products/kinematics.htm

5
Rocker output:
- Grashof crank-rocker is desired
- a trivial case of function generator
Coupler output (complex motion)
- more general
- simple case of motion generation
- two positions of a line are defined as O/P

4-bar Grashof crank-rocker synthesis
Design a 4-bar Grashof crank-rocker to give 45° of rocker rotation with equal time forward and back, from a constant speed motor input
6
Norton’s textbook, Example 3-1, p. 93
See Animation\WM\Ch3\fig3_4\ in Norton’s CD-Rom

7

4-bar Grashof crank-rocker synthesis
Step 1
B
1
4
B
2
O
4
8
The linkage in its two toggle positions
Norton’s textbook
Example 3-1, p. 93

9
Step 2, 3
4-bar Grashof crank-rocker synthesis
O
2
B
1
O
4
B
2

10
Step 4a
4-bar Grashof crank-rocker synthesis
O
2
B
1
O
4
B
2

11
4-bar Grashof crank-rocker synthesis
Step 4b a
O
2
A
2
=
1
2
(
O B
2
−
O B
) a
B
1
O
4
B
2

12
4-bar Grashof crank-rocker synthesis
Step 5
Link 2 a
A
1
O
2
A
2
Link 2 a
B
1
O
4
B
2

13
4-bar Grashof crank-rocker synthesis
Step 6
Link 2
Link 3
A
1
O
2
A
2
B
1
O
4
B
2

14
4-bar Grashof crank-rocker synthesis
Step 7
Link 2
Link 3
A
1
O
2
A
2
Link 1
B
1
O
4
B
2
Link 4

4-bar Grashof crank-rocker synthesis
Step 8
Link 2
Link 3
15
A
1
O
2
A
2
B
1
B
2
Link 4
Link 1
O
4
Find the Grashof condition: L+S
<
>
P+Q
If non-Grashof ( > ), then redo Step 3~8 by moving O
2 further from O
4

16
4-bar Grashof crank-rocker synthesis
Step 9
Link 2
O
2
Link 3
B
2
Link 4
Link 1
O
4
- Simulate this mechanism to check its function and transmission angles (Norton’s CD: working model file)

17
4-bar Grashof crank-rocker synthesis

18
In inch
4-bar Grashof crank-rocker synthesis

19
Rocker output- two positions with complex displacement
(motion generation)
Problem: design a 4-bar linkage to move link CD from position C
1
D
1 to C
2
D
2

20
Rocker output- two positions with complex displacement
(motion generation)
Problem: design a 4-bar linkage to move link CD from position C
1
D
1 to C
2
D
2
D
1 Step 1
Draw link CD in two desired positions
C
1
Link 4
C
2
D
2

21
Rocker output- two positions with complex displacement
(motion generation)
Problem: design a 4-bar linkage to move link CD from position C
1
D
1 to C
2
D
2
D
1 Step 2
Draw construction line from C and from
D
1 to D
1
2 to C
2 C
1
Link 4
C
2
D
2

22
Rocker output- two positions with complex displacement
(motion generation)
Problem: design a 4-bar linkage to move link CD from position C
1
D
1 to C
2
D
2
D
1 Step 3
Bisect C
1
C
2 and D
1
D
2 and extend to intersect at O
4
C
1
Link 4
C
2
D
2
O
4

23
Rocker output- two positions with complex displacement
(motion generation)
Problem: design a 4-bar linkage to move link CD from position C
1
D
1 to C
2
D
2
D
1
Step 4: draw an arc about O intersect O
4
C
1
4 and O
4 to
C
2 at
B
1 and B
2 respectively
C
1
Link 4
C
2
B
1
B
2
D
2
O
4
O
4
C i
D i
= output link, i=1,2

24
Rocker output- two positions with complex displacement
(motion generation)
Problem: design a 4-bar linkage to move link CD from position C
1
D
1 to C
2
D
2
D
1 Step 5:
Follow Steps 2~8 of previous example to complete the linkage
C
1
Link 4
C
2
B
1
B
2
D
2
O
4

25
Rocker output- two positions with complex displacement
(motion generation)
Problem: design a 4-bar linkage to move link CD from position C
1
D
1 to C
2
D
2
Link 4
D
1 a
O
2
A
2
C
1
B
1 a
B
2
C
2
D
2

26
Rocker output- two positions with complex displacement
(motion generation)
Problem: design a 4-bar linkage to move link CD from position C
1
D
1 to C
2
D
2
D
1
Link 4
Link 2
C
1
C
2
Link 3 B
2
A
1
O
2
A
2
B
1
D
2
Link 1 O
4

27
Rocker output- two positions with complex displacement
(motion generation)
Problem: design a 4-bar linkage to move link CD from position C
1
D
1 to C
2
D
2

Three-position Synthesis with specified Fixed Pivots
Problem: design a inverted 4-bar linkage to move link CD from position
C
1
D
1 to C
2
D
2 and then C
3
D
3
. Use specified fixed pivots O
2 and O
4
Step 1
Draw link CD in three desired positions
28
Norton, Section 5.8
Synthesis CAD program: Symech http://www.symech.com

29
Three-position Synthesis with specified Fixed Pivots
Problem: design a inverted 4-bar linkage to move link CD from position
C
1
D
1 to C
2
D
2 and then C
3
D
3
. Use specified fixed pivots O
2 and O
4
Step 2
Draw the ground link O
2
O
4

Three-position Synthesis with specified Fixed Pivots
Step 3
Draw the arcs from C
2 to O
2 and from D
2 to O
2
30
(a) Original coupler three-position problem with specified pivots
(b) Position of the ground plane relative to the second coupler position
(arbitrary).

Three-position Synthesis with specified Fixed Pivots
Step 4
Draw the arcs from C
2 to O
4 and from D
2 to O
4
31
(a) Original coupler three-position problem with specified pivots
(b) Position of the ground plane relative to the second coupler position
(arbitrary).

Three-position Synthesis with specified Fixed Pivots
Step 4
Transferring O
2
O
4 to O
2
’O
4
’
32
(c) Transferring second ground plane position to reference location at first position

Three-position Synthesis with specified Fixed Pivots
Step 5
Repeat the process for the 3 rd coupler position. Transferring O
2
O
4 to O
2
’’O
4
’’
33
O”
2
O”
4
(d) Position of the ground plane relative to the third coupler position (e) transfer 3 rd ground plane position to ref location at 1 st position

Three-position Synthesis with specified Fixed Pivots
34
O”
2
O’
2
O’
4
G
O
2
O”
4
H
O
4
(f) Check the 3 inverted positions of the ground plane
O
2
O
4
O
2
’O
4
’ and O
2
’’O
4
’’
→
E
1
F
1
, E
2
F
2 and E
3
F
3
(g) Find the ground pivot points: G and H

Three-position Synthesis with specified Fixed Pivots
35
(h) The correct inversion of desired linkage

Three-position Synthesis with specified Fixed Pivots
36
(i) Re-invert to obtain the result (E
1
F
1 pivots)
(j) Re-place the Link CD

37
Three-position Synthesis with specified Fixed Pivots

38
- Used in single point metal cutting machine.
- Forward = cutting

39
Two-Position Motion Generation by Analytical Synthesis
A
2
A
1
P
2
P
1
B
2
B
1
Design a four-bar linkage, which will move a line on its coupler link such that a point P on that line will be first at P
1 and later at
P
2 and will also rotate the line through an angle
α
2 between those two precision positions.
Find the lengths and angles of the four links and the coupler link dimensions A
1
P
1 and B
1
P
1 as shown in Fig 5-1.
o
2
Fig. 5-1 Norton’s textbook o
4

40
Two-Position Motion Generation by Analytical Synthesis
The problem statement is:
Design a four-bar linkage, which will move a line on its coupler link such that a point P on that line will be first at P
1 and later at P
2
2 and will also rotate the line through an angle between those two precision positions. Find the lengths and angles of the four links and the coupler link dimensions
A
1
P
1 and B
1
P
1 as shown in Figure 5-1.

Two-Position Motion Generation by Analytical Synthesis
41
The displacement of the output motion of point
P
P
21
=
R
2
–
R
1
Link 3
V
1
= Z
1
– S
1
Ground link 1
G
1
= W
1
+ V
1
– U
1
(5. 1)
(5. 2a)
(5. 2b)
Thus if we can define the two dyads
W
1
,
Z
1
, and
U
1
,
S
1
, we will have defined a linkage that meets the problem specifications.
W
1
Z
1
Dyads = open kinematic chain of 2 binary links and one joint
V
1
G
1 s
1 u
1

Two-Position Motion Generation by Analytical Synthesis
42
First solve for the left side of the linkage
(vectors around the loop which includes both positions
P
1 and
P
2
W
1 and
Z
1
). A vector loop equation for the left-side dyad .
P
21
W
2
+ Z
2
– P
21
– Z
1
– W
1
= 0 (5.3) z
2 z
1
[ x - component w cos θ ] ( cos β
2
+ z cos φ ( cos α
2
) [ w sin θ ] sin β
) z
φ α
2
2
= p
21 cos δ
2
(5.6a) y - component
[ w sin θ ] ( cos β
+ φ ( cos α
2
2
− 1
) − [ w cos θ ] sin
− 1
) + z
φ α
β
2
2
= p
21 sin δ
2
φ
W
2
β
2
W
1
θ
(5.6b)

Two-Position Motion Generation by Analytical Synthesis
[ w cos θ ] ( cos β
2
+ φ ( cos α
2
) [ w sin θ ] sin β
) z
φ α
2
2
= p
21 cos δ
2
43
[ w
+ sin z
θ ] ( cos β
2 sin φ ( cos α
2
− 1
) − [ w cos θ ] sin
− 1
) + z
φ α
β
2
2
= p
21 sin δ
2
There are and δ
2
. eight
Three variables in these two equations: and δ
2 w , θ , w
, θ , β
2
, z
, φ , α
2 of the eight are defined in the problem statement,
β
2
, z , φ , we are
, p
21 namely α
2
, p
21
. Of the remaining five forced to choose three as “free choices” (assumed values) in order to solve for the other two.
One strategy is to assume :
Values for the three angles, θ , β
2
, φ

44
Two-Position Motion Generation by Analytical Synthesis
A
=
B
=
C
= cos θ cos φ
( cos
( cos
β
2
α
2 p
21 cos δ
2
−
−
1
)
1
)
−
− sin θ sin φ sin β
2 sin α
2
D
E
=
F
=
= sin sin p
21
φ
θ ( cos
( cos sin δ
2
β
2
α
2
−
−
1
)
1
)
+
+ cos θ cos φ sin sin
β
2
α
2
Aw + Bz = C
Dw + Ez = F
Solving simultaneously, w
=
CE
AE
−
−
BF
BD
; z
=
AF
AE
−
−
CD
BD
(5.7a)
(5.7b)
(5.7c)
(5.7d)

Two-Position Motion Generation by Analytical Synthesis
Repeat the process for the righthand dyad,
U
1
S
1
U
2
+
S
2
–
P
21
–
S
1
– U
1
= 0 (5.9a)
The x and y component equations become:
S
2
S
1
ψ
μ cos σ (cos
+ s
γ
2 cos
− 1 ) −
ψ (cos
μ
α sin
2
σ sin γ
2
− 1 ) − s sin ψ sin α
2
= p
21 cos δ
2
U
2
γ
2
U
1
(5.10a)
μ sin σ (cos
+ s
γ
2
− 1 ) + μ sin ψ (cos cos
α
2
σ sin γ
2
− 1 ) + s cos ψ sin α
2
= p
21 sin δ
2
45
α
2
, p
21 and
δ
2 Right-hand dyad: three angles
σ
,
γ
2
,
ψ
(5.10b)
Left-hand dyad: three angles
θ
,
β
2
,
φ
σ

46
Two-Position Motion Generation by Analytical Synthesis
A
=
B
=
C
= cos cos ψ p
21
σ cos
( cos
( cos
γ
α
2
2
−
−
1
)
1
)
−
− sin sin
σ
ψ sin sin
γ
δ
2
α
2
2
D
E
=
=
F
= sin sin p
21
σ
ψ
( cos
( cos
γ
α
2
2
−
−
1
)
1
)
+
+ cos cos
σ
ψ sin sin
γ
2
α sin δ
2
2
Au + Bs = C
Du + Es = F
Solving simultaneously, u
=
CE
AE
−
−
BF
BD
; s
=
AF
AE
−
−
CD
BD
(5.11a)
(5.11b)
(5.11c)
(5.11d)

Norton’s FOURBAR program for 2 position synthesis
47
Ex: 5-1, p.219

48
Norton’s FOURBAR program for 2 position synthesis

49
Three-Position Motion Generation by Analytical Synthesis
There are
α
3
, p
21
12 variables
, p
31 namely α
, δ
2
2
, α
3
, p the other FOUR.
and δ
3
21
, p
31 in these
. Six of the 12 are defined in the problem statement,
δ
2 and δ
3
FOUR equations (Eq. 5.22):
.
Of the remaining six left-hand dyad w , θ , w
β
, θ
2
, β
, β
3
,
2
, β
3
, z
, φ , α
2
, z , φ , we are forced to choose three as “free choices” (assumed values) in order to solve for
W
2
+
Z
2
–
P
21
–
Z
1
–
W
1
= 0 (5.19)
W
3
+
Z
3
–
P
31
–
Z
1
–
W
1
= 0 (5.20) right-hand dyad
U
2
+
S
2
–
P
21
–
S
1
– U
1
U
3
+
S
3
–
P
31
–
S
1
– U
1
= 0
= 0
(5.28)

50
Norton’s FOURBAR program for 3 position synthesis

51
Norton’s FOURBAR program for 3 position synthesis

52
Norton’s FOURBAR program for 3 position synthesis with fixed pivot points

53
Norton’s FOURBAR program for 3 position synthesis with fixed pivot points

54
Bulldozer: a tractor-driven machine usually having a broad blunt horizontal blade for moving earth (as in road building) http://www.kenkenkikki.jp/special/no02/e_index.htm
http://maeweb.ucsd.edu/~mae3/spring2002/mae3_19/web/load.html

55
Backhoe: an excavator whose shovel bucket is attached to a hinged boom and is drawn backward to move earth

Mechanism Categories
Function Generation Mechanisms
A function generator is a linkage in which the relative motion between links connected to the ground is of interest.
56
- Correlation of an input motion with an output motion in a mechanism
- Typically, correlation of 2 grounded links

Mechanism Categories
Function Generation Mechanisms
A function generator is a linkage in which the relative motion between links connected to the ground is of interest.
57
A four-bar hand actuated wheelchair brake mechanism

Mechanism Categories
Function Generation Mechanisms
58
A four-bar drive linkage for a lawn sprinkler

Mechanism Categories
Path Generation Mechanisms
In path generation, we are concerned only with the path of a tracer point and not with the motion (rotation) of the coupler link.
59
Example: film advance mechanism, See Animation/WM/Ch3/camera in Norton’s CD-Rom
Control of a point so that it follows a prescribed path
Doe not specify orientation

60
Mechanism Categories
Path Generation Mechanisms
In path generation, we are concerned only with the path of a tracer point and not with the motion (rotation) of the coupler link.
Crane – straight line motion

Mechanism Categories
Motion Generation Mechanisms
In motion generation, the entire motion of the coupler link is of interest ( rigid body guidance ).
61
- Control of a line in a plane so that it assumes a prescribed set of positions and orientations
- Hard to do with more than 3 positions if ground pivots specified

Mechanism Categories
Motion Generation Mechanisms
In motion generation, the entire motion of the coupler link is of interest ( rigid body guidance ).
62
New Rollerblade brake system

63
Mechanism Categories
Motion Generation Mechanisms
Hood
Four-bar automobile hood linkage design

64
Design a four-bar crank and rocker mechanism to give 45 o of rocker rotation with equal time forward and back, from a constant speed motor input.
1 – Draw the rocker O
4
B in both extreme positions, B
1 and B
2 in any convenient location with angle θ
4
45 o .
=
2 – select a convenient point O
2 line B
1
B
2 extended.
on
3 – Bisect line B
1
B
2
, and draw a circle with that radius about O
2
.
4 – Label the two intersection of the circle with B
1
B
2 extended, A
1 and A
2
.
5 – Measure O
2
A (crank, link2) and
AB (coupler, link3).

65
Design a four-bar crank and rocker mechanism to give 45 o of rocker rotation with equal time forward and back, from a constant speed motor input.

66
Two position synthesis – C
1
D
1 and C
2
D
2
1. Draw the link CD in its two desired positions,
C
1
D
1 and C
2
D
2
2. Connect C
1 to C
2 and D
1 to D
2
.
3. Draw two lines perpendicular to
C
1
C
2 and D
1
(midnormals)
D
2 at the midpoint
4. Select two fixed pivot points, O
2 and O
4
, anywhere on the two midnormals.

Two position synthesis (coupler output) – C
1
D
1 and C
2
D
2
67
The rigid body to be moved from position 1 to 2 is secured to link 3.

Three positions with specified fixed pivot points, coupler as the output
O’
4 1.
Draw the link CD in its three desired positions,
C
1
D
1
, C
2
D
2 and
C
3
D
3 and locate the fixed pivot points O
2 and O
4
.
2.
Draw an arc from
C
1 with radius
O
2
C
2 and another arc from D
1 with radius O
2
D
2
.
Locate the intersection, O’
2
.
68
3.
Draw an arc from
C
1 with radius
O
4
C
2 and another arc from D
1 with radius O
4
D
2
.
Locate the intersection, O’
4
.
C
1
O’
2
C
2
D
1
O ’
2
O
2
C
3
D
2
O ’
4
O
4
D
3

69
Three positions with specified fixed pivot points, coupler as the output
4.
Draw an arc from C
1 with radius O
2
C
3 and another arc from D
1 with radius O
2
D
3
.
Locate the intersection, O”
2
.
5.
Draw an arc from C
1 with radius O
4
C
3 and another arc from D
1 with radius O
4
D
3
.
Locate the intersection, O”
4
.
O”
2
C
1
O ”
2
O”
4
C
2
D
1
O ”
4
D
2
O ’
2
O
2
C
3
O ’
4
O
4
D
3

70
Three positions with specified fixed pivot points, coupler as the output
O ”
4
6.
Connect O
2
O”
2 to O’
2 and O’
2 to
. Draw two midnormals and locate the intersection,
G.
7.
Connect O
4 to O”
4 and O”
4
O’
4
. Draw two midnormals and locate the intersection, to
H.
8.
O
2
G is link 2 and O
4
H is link
4.
9.
Construct a link (3) containing GH and CD.
10. Verify the solution by constructing the mechanism in three position
O ’
4
O ”
2
O ’
2
C
1
C
2
D
1
G
H
O
2
C
3
D
2
O
4
D
3

71
Two position Motion synthesis

72
Mechanism used in Mechanical Analog Computer
Type LC-1: WWII Navy PV-1 "Balance Computor
Librascope Incorporation, Burbank, California
Made for the WWII U.S. Navy Lockheed PV-1
Vega "Ventura" antisubmarine patrol bomber. The computer has 19 dials to input fuel load and tanks, bomb load, guns, ammo, torpedos etc. and uses this information to calculates Total Weight and Center of Gravity.
http://dcoward.best.vwh.net/analog/libra.htm

Two-Position Motion Generation by Analytical Synthesis we j ( θ + β
2
)
ze j ( φ + α
2
)
p
21 e j
δ
2
ze j
φ
we j
θ
0 we j
θ e j
β
2
ze j
φ e j
α
2
p
21 e j
δ
2
ze j
φ
we j
θ
0
73 we j
θ ( e j
β
2
1 )
ze j
φ ( e j
α
2
1 )
p
21 e j
δ
2
Euler formula
[ x - component w cos θ ] ( cos β
2
+ z cos φ ( cos α
2
) [ w sin θ ] sin β
) z
φ α
2
2
= p
21 cos δ
2
(5.6a) e j
θ = cos θ + j sin θ , j
= y - component
[ w sin θ ] ( cos
+ φ ( cos
β
2
α
2
− 1
) − [ w cos θ ] sin
− 1
) + z
φ α
β
2
2
=
Slide 42, p
21 sin δ
2
(5.6b)
Textbook p.213
− 1

74
Final Solution to Example: Design a four-bar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion y
O
4
A
1
P
21
A
2
B
1
B
2 x
A
1x
=0.0, B
1x
=1.721, A
2x
=2.656, B
2x
=5.065
A
1y
=0.0, B
1y
=-1.75, A
2y
=-0.751, B
2y
=-0.281
O
2

75
Example: Design a four-bar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion
A
1x
= 0.0
, B
1x
= 1.721, A
2x
= 2.656
, B
2x
= 5.065
A
1y
= 0.0
, B
1y
= -1.75, A
2y
= -0.751
, B
2y
= -0.281
R
1
= (A
1x
, A
1y
), R
2
= (A
2x
, A
2y
), P
21
= R
2
R
1
= (2.656, -0.751), p
21
= 2.76
α
2
= tan − 1
⎛
⎜⎜
A
2 y
A
2 x
−
B
2 y
−
B
2 x
⎞
⎟⎟
− tan − 1
⎛
⎜⎜
A
1 y
A
1 x
−
B
1 y
−
B
1 x
⎞
⎟⎟
= 56 .
519 o δ
2
= tan − 1
⎛
⎜⎜
P
21 y
P
21 x
⎞
⎟⎟
= − 15 .
789 o
From graphical method, determine the I/P for Eq. (5.7)
θ
= 94.394°,
β
2
= -40.366°,
φ
= -45.479°
Solve for ( w z ) dyad using Eq. (5.7) tan − 1 ( y x
) =
ArcTan
⎨
ArcTa n
( y
/
( y x
)
/ x
+
),
π , if x
> 0 otherwise

76
Example: Design a fourbar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion
A
=
B
=
C
= cos θ cos φ
( cos
( cos
β
2
α
2 p
21 cos δ
2
−
−
1
)
1
)
−
− sin θ sin φ sin β
2 sin α
2
(5.7a)
D
E
=
F
=
= sin sin p
21
φ
θ ( cos
( cos sin δ
2
β
2
α
2
−
−
1
)
1
)
+
+ cos θ cos φ sin sin
β
2
α
2
(5.7b)
Aw + Bz = C
Dw + Ez = F w
=
CE
AE
−
−
BF
BD
= 4 .
000
(5.7c) z
=
AF
AE
−
−
CD
BD
= 0 .
000

77
Example: Design a fourbar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion
From graphical method, determine the I/P for Eq. (5.11)
σ
= -93.449°,
γ
2
=54.330°,
ψ
= 134.521°
A
B
=
=
C
=
D
E
=
=
F
= cos cos ψ p
21
σ cos
( cos
( cos
γ
α
2
2
−
−
1
)
1
)
−
− sin sin
σ
ψ sin sin
γ
α
2
δ
2 sin sin p
21
σ
ψ
( cos
( cos
γ
α
2
2
−
−
1
)
1
)
+
+ cos cos
σ
ψ sin sin
γ
2
α sin δ
2
2
2
(5.11a)
(5.11b)
Au + Bs = C
Du + Es = F
Solving simultaneously, u
=
CE
AE
−
−
BF
BD
= 4 .
000
(5.11c) s
=
AF
AE
−
−
CD
BD
= 2 .
455

78
Example: Design a four-bar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion
Link 3 (5. 2a)
V
1
= Z
1
– S
1
V
1 x
V
1 y
=
= z cos φ − s cos ψ z sin φ − s sin ψ
= 1 .
721
= − 1 .
750
V =2.455
θ
3
= tan − 1
⎛
⎜⎜
V
1 y
V
1 x
⎞
⎟⎟
= − 45 .
479 o
Ground link 1
G
1
=
W
1
+
V
1
–
U
1
(5. 2b)
G
1 x
G
1 y
=
= w cos θ w sin θ
+ v
+ cos θ
3 v sin θ
−
3 u
− cos σ u sin
=
σ
1 .
655
= 6 .
231
θ
1
=
G =6.447
tan − 1
⎛
⎜⎜
G
1 y
G
1 x
⎞
⎟⎟
= 75 .
123 o
Determine the initial & final value of the input crank wrt vector G
θ
2i
=
θ
-
θ
1
=19.271°,
θ
2f
=
θ
2i
+
β
2
= -21.095°
Determine the coupler point wrt point A and the vector V r p
= z = 0.0,
δ p
=
φ
-
θ
3
= 0°

79
Example: Design a fourbar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion
Locate the fixed pivots in the global frame using all the vector definitions
ρ
1
= tan − 1
⎛
⎜⎜
A
1 y
A
1 x
⎞
⎟⎟
= 90 .
0 o
R
1
= (A
1x
, A
1y
) R
1
= 0
O
2 x
O
2 y
=
=
R
1
R
1 cos sin
1
1
−
− z z cos sin
−
− w cos w sin
=
= 0 .
306
− 3 .
988
O
4 x
O
4 y
=
=
R
1
R
1 cos sin
1
1
−
− s s cos
sin
−
− u u cos sin
=
= 1 .
962
2 .
243
θ
O
2
O
4
= tan − 1
⎛
⎜⎜
O
4 y
O
4 x
−
O
2 y
−
O
2 x
⎞
⎟⎟
= 75 .
123 o

80
Example: Design a fourbar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion
Link 1 : g = 6.447,
θ
1
= 75.123°
Link 2 : w = 4.000,
θ
= 94.394°
Link 3 : v = 2.455,
θ
3
= -45.479°
Link 4 : u = 4.000,
σ
= -93.449°
Coupler : r p
= 0,
δ p
= 0°
Crank angle :
θ
2i
= 19.271°
θ
2f
= -21.095°

81
Final Solution to Example: Design a fourbar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion y
O
4
A
1 v
P
21 u
A
2 w
B
1
B
2 x
G
1
=
W
1
+
V
1
–
U
1
O
2

82
Final Solution to Example: Design a fourbar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion y
O
4
A
1 v
P
21 u
A
2 w
B
1
B
2 x
O
2

83
Final Solution to Example: Design a fourbar mechanism to give 2 positions shown in Fig. P3-1 of coupler motion y
O
4 u
A
1 x
P
21
A
2 v
B
2
B
1 w
O
2

84
Lockheed PV-1 Vega "Ventura" antisubmarine patrol bomber