Transcript
Prof. Albert Espinoza
Mechanical Engineering Department
Polytechnic University of Puerto Rico
Spring 2015
Position Representation: Complex Numbers
To convert from polar to Cartesian form, use the Euler identity:
Polar form: j
A AR R e@
ˆ ˆCartesian form: cos sin + cos sin A A x y A AR i R j R jR R jR
cos sinje j
Polar complex number notation makes vector representation more compact
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Coordinate Conversion (Cartesian to Polar)
Polar complex number notation makes vector representation more compact
2 2
arctan
A x y
y
x
R R R
R
R
ˆ ˆCartesian form: cos sin + cos sin A A x y A AR i R j R jR R jR
Polar form: j
A AR R e@
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Position – Fixed Pivot (Pure Rotation)
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Position in XY coordinates
Position
𝑅𝑃 = 𝑝𝑒𝑗𝜃 = 𝑝 cos 𝜃 + 𝑗 sin 𝜃
Types of Motion
Pure Translation
• All points have same displacement
Pure Rotation
• Different points have different displacement
Complex Motion
• Simultaneous translation and rotation
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Chasles’ Theorem
Any displacement of a rigid body is equivalent to the sum of a translation of any one point on the body and a rotation of the body about an axis through that point.
Total displacement = translation component + rotation component
B A A A B AR R R
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Position Difference (Displacement)
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Position Difference Equation – Two points in the same body
𝑅𝐵𝐴 = 𝑅𝐵 − 𝑅𝐴
𝑅𝐵 = 𝑅𝐴 + 𝑅𝐵𝐴
Same equation represents:
Position Difference
Relative Position
Coordinate Systems
• Coordinate Systems:
GCS = Global Coordinate System, (X, Y)
LNCS = Local Non-Rotating Coordinate System, (x, y)
LRCS = Local Rotating Coordinate System, (x’, y’)
x’
y’
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If the position of A is defined with respect to (wrt) axes x-y, and it is desired to transform its coordinates to the global (inertial) set of axes (X-Y), we use the following equations: Or in matrix form:
Coordinate Transformation
It is often necessary to transform the coordinates of a point defined in one set of axes (x-y) in another (X-Y).
𝑅𝑋 = 𝑅𝑥 cos 𝛿 − 𝑅𝑦 sin 𝛿
𝑅𝑌 = 𝑅𝑥 sin 𝛿 + 𝑅𝑦 cos 𝛿
𝑅𝑋
𝑅𝑌=
cos 𝛿 −sin 𝛿sin 𝛿 cos 𝛿
𝑅𝑥
𝑅𝑦
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Position Analysis of Fourbar Linkage
The fourbar linkage provides 1 DOF (Mobility=1)
Link 2 is normally the input link
Link 4 is normally the output link
All angles are measured w.r.t. +x axis and are positive if CCW
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Possible Solutions of Fourbar Linkage
There are two possible closed-form solutions for θ3 and θ4 for a given θ2 (open and crossed configurations)
Crossed vs Open: Assuming that link 2 is in the first quadrant, a Grashof linkage is defined as crossed if the two links adjacent to the shortest link cross one another and open is they don’t cross.
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Vector Loop – Fourbar Linkage
2 3 4 1 0R R R R
32 4 1 0jj j j
ae be ce de
cos sinje j
2 2 3 3 4 4 1 1cos sin cos sin cos sin cos sin 0 a j b j c j d j
3 2 4
3 2 4
cos cos cos
sin sin sin
b a c d
b a c
2 22 2 2
3 3 2 4 2 4sin cos sin sin cos cosb a c a c d
2 22
2 4 2 4sin sin cos cosb a c a c d
2 2 2 2
2 4 2 4 2 42 cos 2 cos 2 sin sin cos cosb a c d ad cd ac
2 2 2 2
1 2 3 2
d d a b c dK K K
a c ac
1 4 2 2 3 2 4 2 4cos cos cos cos sin sinK K K
2 1 2 2 3
2
1 2 2 3
cos cos
2sin
1 cos
A K K K
B
C K K K
4
42 4
2 tan2
sin ;
1 tan2
2 4
42 4
1 tan2
cos
1 tan2
2 4 4tan tan 02 2
A B C
1,2
2
4
42arctan
2
B B AC
A
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Vector Loop – Fourbar (Continued)
2 2 3 3 4 4 1 1cos sin cos sin cos sin cos sin 0 a j b j c j d j
4 2 3
4 2 3
cos cos cos
sin sin sin
c a b d
c a b
2 22 2 2
4 4 2 3 2 3sin cos sin sin cos cos c a b a b d
2 22
2 3 2 3sin sin cos cos c a b a b d
2 2 2 2
2 3 2 3 2 32 cos 2 cos 2 sin sin cos cos c a b d ad bd ab
2 2 2 2
1 4 5 2
d d c d a bK K K
a b ab
1 3 4 2 5 2 3 2 3cos cos cos cos sin sin K K K
2 1 4 2 5
2
1 4 2 5
cos cos
2sin
1 cos
D K K K
E
F K K K
3
32 3
2 tan2
sin ;
1 tan2
2 3
32 3
1 tan2
cos
1 tan2
2 3 3tan tan 02 2
D E F
1,2
2
3
42arctan
2
E E DF
D
Using a similar approach for θ3:
The ‘minus’ solutions represent the open config. and the ‘plus’ solutions the crossed config. 14
Position Analysis of Fourbar Linkage
Linear position of linkage moving joints:
𝑅𝐴 = 𝑎𝑒𝑗𝜃2 = 𝑎 cos 𝜃2 + 𝑗 sin 𝜃2
𝑅𝐵𝐴 = 𝑏𝑒𝑗𝜃3 = 𝑏 cos 𝜃3 + 𝑗 sin 𝜃3
𝑅𝐵 = 𝑅𝐴 + 𝑅𝐵𝐴 𝑂𝑅 𝑐𝑒𝑗𝜃4 + 𝑅𝑂4= 𝑐 cos 𝜃4 + 𝑗 sin 𝜃4 + 𝑑
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Position of any Point in a Linkage
The position of any point in a linkage requires a simple vector calculations
Position of point P is given as:
Position of point S:
Position of point U:
P A PAR R R
3 32 ( )
jjae pe
2 2 3 3 3 3cos sin cos( ) sin( ) a j p j
PR
PAR
AR
3 Relative angle between vector and PA BAR R
2 2( )jS se R
2 2 2 2cos( ) sin( ) s j
4 4 4 4cos( ) sin( ) u j d
4 4
4
( )jU Oue R R
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Example: Fourbar Linkage
1 2 3 4 2 3Given: 6, 2, 7, 9, 30 , R 6, 30
Determine: The Grashof condition
The possible solutions for both open and crossed configurations
The x and y coord
PL L L L
inates of points , , and (w.r.t. GCS) for the open configurationA B P
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