1. No category

Ken Youssefi
Mechanical Engineering Dept, SJSU
1
Complex Numbers and Polar Notation
x2 + 1 = 0, x2 + x + 1 = 0,
Euler (1777), i = √ -1
i2 = -1
x=?
OA′ = - OA
OA′ = i2 OA
O
A′
A
i2 represents 180o rotation of a vector
i represents 90o rotation of a vector
r = x + iy
Argand Diagram
x = rcos(θ)
iy
y = rsin(θ)
Imaginary
axis
P
x
eiθ = cos(θ) + i sin(θ),
Real axis
Ken Youssefi
imaginary part
real part
Euler’s Formula
r
θ
r = rcos(θ) + i rsin(θ)
e-iθ = cos(θ) - i sin(θ)
r = r eiθ
Mechanical Engineering Dept, SJSU
2
Closed Loop Vector Equation – Complex Polar Notation
B
r3
θ3
A
r2 + r3 = r1 + r4
4
r2
iθ2
r2 e
+
iθ3
r3 e
iθ1
= r1 e
+
iθ4
r4 e
3
r4
2
θ4
θ2
θ1 = 0
r1
O2
O4
Positive sign convention - all angles are measured with
respect to the horizontal line in counterclockwise direction.
θ4
A
r2 + r3 + r4 + r1 = 0
r2eiθ2 + r3eiθ3 + r4eiθ4 + r1eiθ1 = 0
3
r3
B
θ3
4
r2
r4
2
θ2
Ken Youssefi
O2
Mechanical Engineering Dept, SJSU
r1
θ1 = 180
O4
3
Rotational Operator & Stretch Ratio
iy
Pj
Φj
θ1
θj
P1
x
r1 = r1eiθ1
rj = rjeiθj
θj = θ1 + Φj
rj = rj (r1 / r1 ) ei(θ1 + Φj)
rj = (rj / r1) r1 eiθ1 eiΦj
rj = r1 ρ eiΦj
Stretch ratio
Ken Youssefi
Rotational operator
Mechanical Engineering Dept, SJSU
4
Analytical Synthesis – Standard Dyad Form
Left side of the mechanism
4 Bar mechanism
P
A
r′3
r″3
B
Pj
r′3 eiαj αj
r3
r2
r4
4
r2 eiβj
O4
O2
P1
Parallel
A1
Aj
2
Left side
δj
3
Right side
βj
r′3
2
r2
O2
Closed loop vector equation – complex polar notation
Design the left side of the 4 bar → r2 & r′3
Design the right side of the 4 bar → r4 & r″3
r2 + r′3 + δj = r2 eiβj + r′3 eiαj
Standard Dyad form
r2 (eiβj – 1) + r′3 (eiαj – 1) = δj
Ken Youssefi
Mechanical Engineering Dept, SJSU
5
Analytical Synthesis – Standard Dyad Form
Apply the same procedure to obtain the Dyad equation for the right side
of the four bar mechanism.
P
r″3
r4
Rotation of link 4,
Standard Dyad form for the right side
of the mechanism
B
r4 (eigj – 1) + r″3 (eiαj – 1) = δj
4
g
g → rotation of link 4
α → rotation of link 3
O4
Standard Dyad form for the left side
of the mechanism
r2 (eiβj – 1) + r′3 (eiαj – 1) = δj
β → rotation of link 2
Ken Youssefi
Mechanical Engineering Dept, SJSU
α → rotation of link 3
6
Analytical Synthesis
Two Position Motion & Path Generation Mechanisms
Left side of the mechanism
P2
δ2
r′3 eiα2 α2
P1
Parallel
A1
A2
r2 eiβ2
β2
2
r2
O2
r′3
Dyad equation for the left side of the mechanism.
One vector equation or two scalar equations
r2 (eiβ2 – 1) + r′3 (eiα2 – 1) = δ2
Motion generation mechanism, the orientation
of link 3 is important (angle alpha)
1. Draw the two desired positions accurately.
2. Measure the angle α from the drawing, α2
3. 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
Two position
motion gen. Mech. Select; β2 and r′3
Ken Youssefi
Mechanical Engineering Dept, SJSU
Solve for r2
Three
sets of
infinite
solution
7
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
Two position
motion gen. Mech.
Given; α2 and δ2
Select;, g2 , r″3
Solve for r4
Path Generation Mechanism (left side of the mechanism)
Two position path
gen. Mech.
Ken Youssefi
Given; β2 and δ2
Select; α2 and r′3
Mechanical Engineering Dept, SJSU
Solve for r2
Three
sets of
infinite
solution
8
Analytical Synthesis
Three Position Motion & Path Generation Mechanisms
P2
r′3 e
P3
r′3 e
iα3
δ2
α2
δ3
A2
P1
Parallel
A1
α3
β3
β2
r2 eiβ2
r′3
2
r2
A3
r2 eiβ3
Ken Youssefi
iα2
O2
Mechanical Engineering Dept, SJSU
9
Analytical Synthesis
Three Position Motion & Path Generation Mechanisms
Three position motion gen. mech.
r2 (eiβ2 – 1) + r′3 (eiα2 – 1) = δ2
Dyad equations
r2 (eiβ3 – 1) + r′3 (eiα3 – 1) = δ3
4 scalar equations
6 unknowns; r2
2 free choices
Three position
motion gen. Mech.
, r′3 , β
2
and β3
Given; α2, α3, δ2, and δ3
Select; β2 and β3
Solve for r2 and r′3
Two sets of infinite solution
Ken Youssefi
Mechanical Engineering Dept, SJSU
10
Analytical Synthesis
Four position motion generation mechanism
r2 (eiβ2 – 1) + r′3 (eiα2 – 1) = δ2
Dyad equations
r2 (e
iβ3
– 1) + r′3 (e
iα3
– 1) = δ3
Non-linear
equations
r2 (eiβ4 – 1) + r′3 (eiα4 – 1) = δ4
6 scalar equations
7 unknowns; r2
1 free choices
, r′3 , β
2
, β3 and β4
Given; α2, α3, α4 δ2, δ3 and δ4
Four position
motion gen. Mech. Select; one input angle, β2 or β3 or β4
Solve for r2 and r′3
One set of infinite solution
Ken Youssefi
Mechanical Engineering Dept, SJSU
11
Analytical Synthesis
Five position motion generation mechanism
r2 (eiβ2 – 1) + r′3 (eiα2 – 1) = δ2
Dyad equations
r2 (e
iβ3
– 1) + r′3 (e
iα3
– 1) = δ3
Non-linear
equations
r2 (eiβ4 – 1) + r′3 (eiα4 – 1) = δ4
r2 (eiβ5 – 1) + r′3 (eiα5 – 1) = δ5
8 scalar equations
8 unknowns; r2
0 free choice
, r′3 , β
2
, β3 , β4 and β5
Given; α2, α3, α4, α5, δ2, δ3, δ4, and δ5
Four position
motion gen. Mech. Select; 0 choice
Solve for r2 and r′3
Unique solution, not desirable
Ken Youssefi
Mechanical Engineering Dept, SJSU
12
Analytical Synthesis –Function Generation Mechanism
Freudenstein’s method
B
r3
r1 + r2 + r3 = r4
θ3
A
r1eiθ1 + r2eiθ2 + r3eiθ3 = r4eiθ4
3
r4
r2
θ4
θ2
Euler equation
eiθ = cos(θ) + i sin(θ)
O2
r1
O4
Real part of the equation
r1 cos(θ1) + r2 cos(θ2) + r3 cos(θ3) = r4 cos(θ4)
Imaginary part of the equation
r1 sin(θ1) + r2 sin(θ2) + r3 sin(θ3) = r4 sin(θ4)
Ken Youssefi
Mechanical Engineering Dept, SJSU
13
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
[r3 cos(θ3)]2 = [r1 – r2 cos(θ2) + r4 cos(θ4)]2
[r3 sin(θ3)]2 = [– r2 sin(θ2) + r4 sin(θ4)]2
Add the two equations
r32 = [– r2 sin(θ2) + r4 sin(θ4)]2 + [r1 – r2 cos(θ2) + r4 cos(θ4)]2
Expand and simplify
r32 = (r1)2 + (r2)2 + (r4)2 – 2r1 r2 cos(θ2) + 2r1 r4 cos(θ4) – 2r2 r4 cos(θ2– θ4 )
Ken Youssefi
Mechanical Engineering Dept, SJSU
14
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
r1
cos(θ4) +
cos(θ2) –
r2
r4
Define
r1
K1 =
r4
(r3)2 – (r1)2 – (r2)2 – (r4)2
r1
K2 = –
r2
= – cos(θ2 – θ4)
2r2 r4
K3 =
(r3)2 – (r1)2 – (r2)2 – (r4)2
2r2 r4
K1cos(θ2) + K2 cos(θ4) + K3 = – cos(θ2 – θ4)
Freudenstein’s equation
Ken Youssefi
Mechanical Engineering Dept, SJSU
15
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.
Select
φ1 Input link (crank) start angle = 45o
Precision point
A5
r2
φ5 Input link (crank) end angle = 105o
ψ1 output link (link 4) start angle = 135o
ψ5 output link (link 4) end angle = 225o
r3 3
A1
B1
r4
B5
ψ5
φ5
ψ1
φ1
O2
r1
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]
so = start point = 1 sn+1 = end point = 10
j = precision point, n = total number of precision points
Ken Youssefi
Mechanical Engineering Dept, SJSU
16
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) = 1.6029
x2 = ½ (1 + 10) – ½ (10 – 1) cos(3π/6) = 5.50
x3 = ½ (1+ 10) – ½ (10 – 1) cos(5π/6) = 9.3971
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
17
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,
45 = a(1) + b
105 = a(10) + b
 = c(y) + d,
135 = C(0) + d
225 = C(1) + d
Boundary condition;
x = 1, φ = 45
and
x = 10, φ = 105o
φ = (20/3)(x) + 115/3
Boundary condition;
and
y = log(1) = 0,  = 135
y = log(10) = 1,  = 225o
 = 90(y) + 135
φ1 Input link (crank) start angle = 45o
φ5 Input link (crank) end angle = 105o
Ken Youssefi
ψ1 output link (link 4) start angle = 135o
ψ5 output link
Mechanical
Engineering
Dept, SJSU
(link
4) end
angle = 225o
18
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
Start
Precision
points
End
Ken Youssefi
pts
x

y

1
2
3
4
1.00
1.6029
5.50
9.3971
45.00
49.019
75.0
100.98
0.00
.2049
.7404
.9730
135.00
153.44
201.64
222.57
5
10.00
135.00
1.00
225.00
Mechanical Engineering Dept, SJSU
19
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
135.00
1.00
225.00
K1cos (49.019) + K2 cos (153.44) + K3 = – cos (49.019 – 153.44)
K1cos (75) + K2 cos (201.64) + K3 = – cos (75 – 201.64)
K1cos (100.98) + K2 cos (222.57) + K3 = – cos (100.98 – 222.57)
Ken Youssefi
Solve for K1, K2, and K3
Mechanical Engineering Dept, SJSU
20