ken youssefi mechanical engineering dept, sjsu 1
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Ken Youssefi Mechanical Engineering Dept, SJSU 1
Ken Youssefi Mechanical Engineering Dept, SJSU
Complex Numbers and Polar Notation
x2 + 1 = 0, x2 + x + 1 = 0, x = ? Euler (1777), i = √ -1
i2 = -1
O AA′
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
r = r eiθ
eiθ = cos(θ) + i sin(θ),
Euler’s Formula
e-iθ = cos(θ) - i sin(θ)
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Ken Youssefi Mechanical Engineering Dept, SJSU
Closed Loop Vector Equation – Complex Polar Notation
r2 + r3 = r1 + r4
r2eiθ
2 + r3eiθ
3 = r1eiθ
1 + r4eiθ
4
Positive sign convention - all angles are measured with respect to the horizontal line in counterclockwise direction.
r2eiθ
2 + r3eiθ
3 + r4eiθ
4 + r1eiθ
1 = 0
2
4
θ2
r4r2
A
O4
B
3
r1
r3
O2
θ4
θ3
θ1 = 0
r4r2
A
O4
B
3
r1
r3
O2
r2 + r3 + r4 + r1 = 0A
O4
2
B3
4r2
r1
r4
r3
O2
θ2
θ4
θ3
θ1 = 1803
Ken Youssefi Mechanical Engineering Dept, SJSU
Rotational Operator & Stretch Ratio
P1
θ1
x
iy
r1 = r1eiθ
1
Pj
θj
rj = rjeiθ
j
Φj
θj = θ1 + Φj
rj = (rj / r1) r1 eiθ
1 eiΦj
rj = rj (r1 / r1 ) ei(θ
1 + Φj)
rj = r1 ρ eiΦ
j
Stretch ratioRotational operator
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis – Standard Dyad Form
4 Bar mechanism
A
B
O2O4
2
3
4
P
Left side Right side
r2r4
r3
r′3r″3
Design the left side of the 4 bar → r2 & r′3Design the right side of the 4 bar → r4 & r″3
αj
βj
δj
Closed loop vector equation – complex polar notation
r2 + r′3 + δj = r2 eiβj + r′3 e
iαj
Aj
Pj
r2 eiβj
r′3 eiαj
r2 (eiβj – 1) + r′3 (e
iαj – 1) = δj
Standard Dyad form
Parallel
Left side of the mechanism
A1
O2
2
P1
r2
r′3
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis – Standard Dyad Form
Apply the same procedure to obtain the Dyad equation for the right side of the four bar mechanism.
B
O4
4
P
r4
r″3
Rotation of link 4
r2 (eiβj – 1) + r′3 (e
iαj – 1) = δj
Standard Dyad form for the left side of the mechanism
α → rotation of link 3β → rotation of link 2
r4 (eij – 1) + r″3 (e
iαj – 1) = δj
Standard Dyad form for the right side of the mechanism
→ rotation of link 4 α → rotation of link 3
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis Two Position Motion & Path Generation Mechanisms
α2
β2
δ2
Left side of the mechanism
A1
O2
2
P1
r2
r′3A2
P2
r2 eiβ2
r′3 eiα2
Parallel
r2 (eiβ2 – 1) + r′3 (e
iα2 – 1) = δ2
Dyad equation for the left side of the mechanism. One vector equation or two scalar equations
1. Draw the two desired positions accurately.
Motion generation mechanism, the orientation of link 3 is important (angle alpha)
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
Select; β2 and r′3Solve for r2
Two position motion gen. Mech.
Three sets of infinite solution
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis Two Position Motion & Path Generation Mechanisms
Apply the same procedure for the right side of the 4-bar mechanism
r4 (eij – 1) + r″3 (e
iαj – 1) = δj
Given; α2 and δ2
Select;, 2 , r″3
Solve for r4Two position motion gen. Mech.
Given; β2 and δ2
Select; α2 and r′3Solve for r2
Two position path gen. Mech.
Three sets of infinite solution
Path Generation Mechanism (left side of the mechanism)
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis Three Position Motion & Path Generation Mechanisms
δ2
A1
O2
2
P1
r2
r′3
α2
β2
Parallel
P3
A3
α3
β3
δ3
A2
P2
r2 eiβ2
r′3 eiα2
r2 eiβ3
r′3 eiα3
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis Three Position Motion & Path Generation Mechanisms
Three position motion gen. mech.
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
2 free choices
4 scalar equations
Dyad equationsr2 (e
iβ2 – 1) + r′3 (eiα2 – 1) = δ2
r2 (eiβ3 – 1) + r′3 (e
iα3 – 1) = δ3
6 unknowns; r2 , r′3 , β2 and β3
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis
Four position motion generation mechanism
Given; α2, α3, α4 δ2, δ3 and δ4
Select; one input angle, β2 or β3 or β4 Solve for r2 and r′3
Four position motion gen. Mech.
One set of infinite solution
Dyad equations
r2 (eiβ2 – 1) + r′3 (e
iα2 – 1) = δ2
r2 (eiβ3 – 1) + r′3 (e
iα3 – 1) = δ3
r2 (eiβ4 – 1) + r′3 (e
iα4 – 1) = δ4
Non-linear equations
7 unknowns; r2 , r′3 , β2 , β3 and β4
1 free choices
6 scalar equations
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis
Dyad equations
r2 (eiβ2 – 1) + r′3 (e
iα2 – 1) = δ2
r2 (eiβ3 – 1) + r′3 (e
iα3 – 1) = δ3
r2 (eiβ4 – 1) + r′3 (e
iα4 – 1) = δ4
Non-linear equations
Five position motion generation mechanism
r2 (eiβ5 – 1) + r′3 (e
iα5 – 1) = δ5
8 unknowns; r2 , r′3 , β2 , β3 , β4 and β5
0 free choice
8 scalar equations
Given; α2, α3, α4, α5, δ2, δ3, δ4, and δ5
Select; 0 choice
Four position motion gen. Mech.
Unique solution, not desirable
Solve for r2 and r′3
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis –Function Generation Mechanism
Freudenstein’s method
r4r2
A
O4
B
3
r1
r3
O2
r1 + r2 + r3 = r4
θ2
θ4
θ3
r1eiθ
1 + r2eiθ
2 + r3eiθ
3 = r4eiθ
4
Real part of the equation
eiθ = cos(θ) + i sin(θ)
Euler equation
= r4 cos(θ4)
r1 sin(θ1) + r2 sin(θ2) + r3 sin(θ3) = r4 sin(θ4)
Imaginary part of the equation
r1 cos(θ1) + r2 cos(θ2) + r3 cos(θ3)
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Ken Youssefi Mechanical Engineering Dept, SJSU
Analytical Synthesis –Function Generation Mechanism
θ1 = 180
– r4 cos(θ4) = 0– r1 + r2 cos(θ2) + r3 cos(θ3)
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
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Ken Youssefi Mechanical Engineering Dept, SJSU
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
K2 = – r1
r2
r1
r4
cos(θ2) – r1
r2
cos(θ4) + (r3)
2 – (r1)2 – (r2)
2 – (r4)
2
2r2 r4
= – cos(θ2 – θ4)
(r3)2 – (r1)
2 – (r2)
2 – (r4)
2
2r2 r4
K3 = K1 =r1
r4
Define
K1cos(θ2) + K2 cos(θ4) + K3 = – cos(θ2 – θ4)
Freudenstein’s equation15
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.
Use Chebyshev spacing and three precision points
sj = ½ (so + sn+1) - ½ (sn+1 – so) cos[(2j - 1)π/2n]
r4
r2
A1
O4
B1
3
r1
r3
O2
A5
φ5
φ1
ψ1
B5
ψ5
Select
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
Precision point
so = start point = 1 sn+1 = end point = 10
j = precision point, n = total number of precision points
Determine the precision points
Example – Continuous Function
Ken Youssefi Mechanical Engineering Dept, SJSU 17
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
sj = ½ (so + sn+1) - ½ (sn+1 – so) cos[(2j - 1)π/2n]
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
Boundary condition; x = 1, φ = 45 and x = 10, φ = 105o
φ = (20/3)(x) + 115/3
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; y = log(1) = 0, = 135
and y = log(10) = 1, = 225o
= 90(y) + 135
45 = a(1) + b
105 = a(10) + b
135 = C(0) + d
225 = C(1) + d
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
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
φ = 20/3(x) + 115/3
= 90(y) + 135
Precision points
y1 = log x1 = log (1.6029) = .2049
y2 = log x2 = log (5.5) = .7404
y3 = log x3 = log (9.3971) = .9730
Start
End
Ken Youssefi Mechanical Engineering Dept, SJSU 20
Example – Continuous Function
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)
Freudenstein equationK1cos(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)
Solve for K1, K2, and K3