1. Mechanical Engineering Dept, SJSU
Analytical Synthesis
Ken Youssefi
Mechanical Engineering Dept, SJSU

2. Complex Numbers and Polar Notation
x2 + 1 = 0, x2 + x + 1 = 0, x = ?
Euler (1777), i = √ -1
i2 = -1 O A A′
OA′ = - OA
OA′ = i2 OA
i2 represents 180o rotation of a vector
i represents 90o rotation of a vector
Real axis
Imaginary axis r P x iy
Argand Diagram θ
r = x + iy
x = rcos(θ)
y = rsin(θ)
r = rcos(θ) + i rsin(θ)
real part
imaginary part
eiθ = cos(θ) + i sin(θ),
Euler’s Formula
e-iθ = cos(θ) - i sin(θ)
r = r eiθ
Ken Youssefi
Mechanical Engineering Dept, SJSU

3. Closed Loop Vector Equation – Complex Polar NotationB 3 r1 r3 O2 θ2 r4 r2 A O4 B 3 r1 r3 O2 θ4 θ3
θ1 = 0
r2 + r3 = r1 + r4 4 2
r2eiθ2 + r3eiθ3 = r1eiθ1 + r4eiθ4
Positive sign convention - all angles are measured with respect to the horizontal line in counterclockwise direction.
r2 + r3 + r4 + r1 = 0 A O4 2 B 3 4 r2 r1 r4 r3 O2 θ2 θ4 θ3
θ1 = 180
r2eiθ2 + r3eiθ3 + r4eiθ4 + r1eiθ1 = 0
Ken Youssefi
Mechanical Engineering Dept, SJSU

4. Rotational Operator & Stretch Ratioiy Pj θj
rj = rjeiθj Φj
θj = θ1 + Φj P1 θ1 x
r1 = r1eiθ1
rj = rj (r1 / r1 ) ei(θ1
+ Φj)
rj = (rj / r1) r1 eiθ1 eiΦj
rj = r1 ρ eiΦj
Stretch ratio
Rotational operator
Ken Youssefi
Mechanical Engineering Dept, SJSU

5. Analytical Synthesis – Standard Dyad Form
4 Bar mechanism
Left side of the mechanism A1 O2 2 P1 r2
r′3 P r2 r4 r3
r′3
r″3
Left side
Right side B Aj Pj δj A 3
r2 eiβj
r′3 eiαj αj βj
Parallel 4 2 O4 O2
Closed loop vector equation – complex polar notation
r2 + r′3 + δj = r2 eiβj + r′3 eiαj
Design the left side of the 4 bar → r2 & r′3
Design the right side of the 4 bar → r4 & r″3
r2 (eiβj – 1) + r′3 (eiαj – 1) = δj
Standard Dyad form
Ken Youssefi
Mechanical Engineering Dept, SJSU

6. Analytical Synthesis – Standard Dyad Form
Apply the same procedure to obtain the Dyad equation for the right side of the four bar mechanism. P
r4 (eigj – 1) + r″3 (eiαj – 1) = δj
Standard Dyad form for the right side of the mechanism
g → rotation of link 4
α → rotation of link 3
r″3 B r4 4
Rotation of link 4, g O4
r2 (eiβj – 1) + r′3 (eiαj – 1) = δj
Standard Dyad form for the left side of the mechanism
α → rotation of link 3
β → rotation of link 2
Ken Youssefi
Mechanical Engineering Dept, SJSU

7. Analytical Synthesis Two Position Motion & Path Generation Mechanisms
r2 (eiβ2 – 1) + r′3 (eiα2 – 1) = δ2
Dyad equation for the left side of the mechanism. One vector equation or two scalar equations
Left side of the mechanism P2 δ2 A1 O2 2 P1 r2
r′3
r′3 eiα2 α2
Motion generation mechanism, the orientation of link 3 is important (angle alpha)
Parallel A2
Draw the two desired positions accurately. β2
r2 eiβ2
Measure the angle α from the drawing, α2
Measure the length and angle of vector δ2
There are 5 unknowns; r2, r′3 and angle β2 and only two scalar equations (Dyad).
Select three unknowns and solve the equations for the other two unknowns
Given; α2 and δ2
Select; β2 and r′3
Solve for r2
Two position motion gen. Mech.
Three sets of infinite solution
Ken Youssefi
Mechanical Engineering Dept, SJSU

8. Analytical Synthesis Two Position Motion & Path Generation Mechanisms
Apply the same procedure for the right side of the 4-bar mechanism
r4 (eigj – 1) + r″3 (eiαj – 1) = δj
Given; α2 and δ2
Select;, g2 , r″3
Solve for r4
Two position motion gen. Mech.
Given; β2 and δ2
Select; α2 and r′3
Solve for r2
Two position path gen. Mech.
Three sets of infinite solution
Path Generation Mechanism (left side of the mechanism)
Ken Youssefi
Mechanical Engineering Dept, SJSU

9. Mechanical Engineering Dept, SJSU
Analytical Synthesis Three Position Motion & Path Generation Mechanisms A2 P2 δ2 A1 O2 2 P1 r2
r′3
r2 eiβ2
r′3 eiα2 α2 β2
Parallel δ3 P3 A3 α3 β3
r2 eiβ3
r′3 eiα3
Ken Youssefi
Mechanical Engineering Dept, SJSU

10. Mechanical Engineering Dept, SJSU
Analytical Synthesis Three Position Motion & Path Generation Mechanisms
Three position motion gen. mech.
Dyad equations
r2 (eiβ2 – 1) + r′3 (eiα2 – 1) = δ2
r2 (eiβ3 – 1) + r′3 (eiα3 – 1) = δ3
4 scalar equations
6 unknowns; r2 , r′3 , β2 and β3
2 free choices
Given; α2, α3, δ2, and δ3
Select; β2 and β3
Solve for r2 and r′3
Three position motion gen. Mech.
Two sets of infinite solution
Ken Youssefi
Mechanical Engineering Dept, SJSU

11. Mechanical Engineering Dept, SJSU
Analytical Synthesis
Four position motion generation mechanism
Dyad equations
r2 (eiβ2 – 1) + r′3 (eiα2 – 1) = δ2
r2 (eiβ3 – 1) + r′3 (eiα3 – 1) = δ3
r2 (eiβ4 – 1) + r′3 (eiα4 – 1) = δ4
Non-linear equations
7 unknowns; r2 , r′3 , β2 , β3 and β4
1 free choices
6 scalar equations
Given; α2, α3, α4 δ2, δ3 and δ4
Four position motion gen. Mech.
Solve for r2 and r′3
Select; one input angle, β2 or β3 or β4
One set of infinite solution
Ken Youssefi
Mechanical Engineering Dept, SJSU

12. Mechanical Engineering Dept, SJSU
Analytical Synthesis
Five position motion generation mechanism
r2 (eiβ2 – 1) + r′3 (eiα2 – 1) = δ2
Non-linear equations
r2 (eiβ3 – 1) + r′3 (eiα3 – 1) = δ3
Dyad equations
r2 (eiβ4 – 1) + r′3 (eiα4 – 1) = δ4
r2 (eiβ5 – 1) + r′3 (eiα5 – 1) = δ5
8 scalar equations
8 unknowns; r2 , r′3 , β2 , β3 , β4 and β5
0 free choice
Given; α2, α3, α4, α5, δ2, δ3, δ4, and δ5
Four position motion gen. Mech.
Solve for r2 and r′3
Select; 0 choice
Unique solution, not desirable
Ken Youssefi
Mechanical Engineering Dept, SJSU

13. Analytical Synthesis –Function Generation Mechanism
Freudenstein’s method r4 r2 A O4 B 3 r1 r3 O2
r1 + r2 + r3 = r4 θ3
r1eiθ1 + r2eiθ2 + r3eiθ3 = r4eiθ4
eiθ = cos(θ) + i sin(θ)
Euler equation θ4 θ2
Real part of the equation
r1 cos(θ1)
+ r2 cos(θ2)
+ r3 cos(θ3)
= r4 cos(θ4)
r1 sin(θ1) + r2 sin(θ2) + r3 sin(θ3) = r4 sin(θ4)
Imaginary part of the equation
Ken Youssefi
Mechanical Engineering Dept, SJSU

14. Analytical Synthesis –Function Generation Mechanism
θ1 = 180
– r1
+ r2 cos(θ2)
+ r3 cos(θ3)
– r4 cos(θ4) = 0
r2 sin(θ2) + r3 sin(θ3) – r4 sin(θ4) = 0
+ r4 cos(θ4)]2
[r1
– r2 cos(θ2)
[r3 cos(θ3)]2 =
[r3 sin(θ3)]2 = [– r2 sin(θ2) + r4 sin(θ4)]2
Add the two equations
r32 = [– r2 sin(θ2) + r4 sin(θ4)]2
+ r4 cos(θ4)]2
+ [r1
– r2 cos(θ2)
r32 = (r1)2 + (r2)2 + (r4)2 – 2r1 r2 cos(θ2) + 2r1 r4 cos(θ4) – 2r2 r4 cos(θ2– θ4 )
Expand and simplify
Ken Youssefi
Mechanical Engineering Dept, SJSU

15. Analytical Synthesis –Function Generation Mechanism
r32 = (r1)2 + (r2)2 + (r4)2 – 2r1 r2 cos(θ2) + 2r1 r4 cos(θ4) – 2r2 r4 cos(θ2 – θ4)
Divide the above equation by 2r2 r4 r1 r4
cos(θ2) – r2
cos(θ4) +
(r3)2 – (r1)2 – (r2)2 – (r4)2
2r2 r4
= – cos(θ2 – θ4)
K1 = r1 r4
Define
K2 = – r1 r2
(r3)2 – (r1)2 – (r2)2 – (r4)2
2r2 r4
K3 =
K1cos(θ2) + K2 cos(θ4) + K3 = – cos(θ2 – θ4)
Freudenstein’s equation
Ken Youssefi
Mechanical Engineering Dept, SJSU

16. Example – Continuous Function
Synthesize a four bar mechanism to generate a function y = log x in the interval 1  x  10. The input crank length should be 50 mm. B1
Precision point
Select
Input link (crank) start angle = 45o
Input link (crank) end angle = 105o φ1 φ5 r3 3 A1 r4 A5 φ5 r2 B5 ψ5
output link (link 4) start angle = 135o
output link (link 4) end angle = 225o ψ1 ψ5 ψ1 φ1 r1 O2 O4
Determine the precision points
Use Chebyshev spacing and three precision points
sj = ½ (so + sn+1) - ½ (sn+1 – so) cos[(2j - 1)π/2n]
Mon 9/22
so = start point = 1
sn+1 = end point = 10
j = precision point, n = total number of precision points
Ken Youssefi
Mechanical Engineering Dept, SJSU

17. Example – Continuous Function
sj = ½ (so + sn+1) - ½ (sn+1 – so) cos[(2j - 1)π/2n]
x1 = ½ (1 + 10) – ½ (10 – 1) cos(π/6) =
x2 = ½ (1 + 10) – ½ (10 – 1) cos(3π/6) = 5.50
x3 = ½ (1+ 10) – ½ (10 – 1) cos(5π/6) =
Corresponding function values are:
y1 = log x1 = log (1.6029) = .2049
y2 = log x2 = log (5.5) = .7404
y3 = log x3 = log (9.3971) = .9730
Ken Youssefi
Mechanical Engineering Dept, SJSU

18. Example – Continuous Function
Assume linear relationship between φ (input angle) and x, and between  (output angle) and y. Where a, b, c and d are constants.
φ = a(x) + b,
 = c(y) + d,
Boundary condition; x = 1, φ = 45
and x = 10, φ = 105o
45 = a(1) + b
105 = a(10) + b
φ = (20/3)(x) + 115/3
Boundary condition; y = log(1) = 0,  = 135
and y = log(10) = 1,  = 225o
135 = C(0) + d
225 = C(1) + d
 = 90(y) + 135
Input link (crank) start angle = 45o
Input link (crank) end angle = 105o φ1 φ5
output link (link 4) start angle = 135o
output link (link 4) end angle = 225o ψ1 ψ5
Ken Youssefi
Mechanical Engineering Dept, SJSU

19. Example – Continuous Function
y1 = log x1 = log (1.6029) = .2049
φ = 20/3(x) + 115/3
y2 = log x2 = log (5.5) = .7404
 = 90(y) + 135
y3 = log x3 = log (9.3971) = .9730
pts x  y  1
1.00
45.00
0.00
135.00 2
1.6029
49.019
.2049
153.44 3
5.50
75.0
.7404
201.64 4
9.3971
100.98
.9730
222.57 5
10.00
225.00
Start
Precision points
End
Ken Youssefi
Mechanical Engineering Dept, SJSU

20. Example – Continuous Function
Freudenstein equation
K1cos(φ1) + K2 cos(ψ1) + K3 = – cos(φ1 – ψ1)
K1cos(φ2) + K2 cos(ψ2) + K3 = – cos(φ2 – ψ2)
K1cos(φ3) + K2 cos(ψ3) + K3 = – cos(φ3 – ψ3)
pts x  y  1
1.00
45.00
0.00
135.00 2
1.6029
49.019
.2049
153.44 3
5.50
75.0
.7404
201.64 4
9.3971
100.98
.9730
222.57 5
10.00
225.00
K1cos (49.019) + K2 cos (153.44) + K3 = – cos ( – )
K1cos (75) + K2 cos (201.64) + K3 = – cos (75 – )
K1cos (100.98) + K2 cos (222.57) + K3 = – cos ( – )
Solve for K1, K2, and K3
Ken Youssefi
Mechanical Engineering Dept, SJSU