ME 2220_SP15_LEC04 - Position Analysis

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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3

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