Linkage analysis - Stanford University

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Transcript Linkage analysis - Stanford University

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Analytic Approach to Mechanism Design http://www.engr.colostate.edu/me/program/courses/ME324/notes/PositionAnalysis.ppt

ME 324 Fall 2000

Position synthesis

Chapter 4 Analytic Position Analysis Imaginary Axis

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• A vector can be represented by a complex number • Real part is x-axis • Imaginary part is y axis • Useful when we begin to take derivatives jR sin q R A q R cos q Point A Real Axis

Position synthesis

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Derivatives, Vector Rotations in the Complex Plane • Taking a derivative of a complex number will result in multiplication by j B C • Each multiplication by j rotates a vector 90 °

R C

=

j

2 R = -R CCW in the complex plane

R D

=

j

3 R = -

j

R

R B

=

j

R D Imaginary

R A

A Real

Position synthesis

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Labeling of Links and Link Lengths • Link labeling starts with ground link • Labeling of link lengths starts with link adjacent to ground link • Makes no sense - just go with it

Link 3, length b B Coupler Link 2, length a A Pivot 02 Link 1, length d Ground Link Link 4, length c Pivot 04 Position synthesis

Angle Measurement Convention

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• All angles measured from angle of the ground link • Define q 1 = 0 ° • One DOF, so can describe all angles in terms of one input, usually q 2

2 3

q 3

A

q 2

1

q 1 = 0 °

B 4

q 4

Position synthesis

More on Complex Notation

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• Polar form:

re j

q • Cartesian form:

r

cos q +

j r

sin q • Euler identity: ±

e j

q = cos q ±

j

sin q • Differentiation:

j

q

de d

q =

je j

q

Position synthesis

The Vector Loop Technique • Vector loop equation:

R 2

+

R 3

-

R 4

-

R 1

= 0

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• Alternative notation:

R AO2

+

R BA

-

R BO4

-

R O4O2

= 0

R 3

nomenclature - tip then tail • Complex notation: ae j q 2 + be j q 3 - ce j q 4 - de j q 1 = 0

R 2

O

a A

q 2

d

• Substitute Euler equation: 2 a (cos q 2 +j sin q 2 ) + b (cos q 3 +j sin q 3 ) - c (cos q 4 +j sin q 4 ) - d (cos q 1 +j sin q 1 ) = 0 q 3

b R 1 B c R 4

O 4 q 4

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Vector Loop Technique - continued • Separate into real and imaginary parts: Real: a cos q 2 a cos q 2 since q 1 + b cos q 3 + b cos q 3 = 0, cos q 1 - c cos q 4 - c cos q 4 = 1 - d cos q 1 - d = 0 , = 0

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Imaginary: ja sin q 2 + jb sin q 3 a sin q 2 since q 1 + b sin q = 0, sin 3 q 1 - jc sin = 0 q - c sin q 4 4 - jd sin = 0 , q 1 = 0

Position synthesis

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Vector Loop Technique continued a cos q 2 a sin q 2 + b cos + b sin q 3 q 3 - c cos - c sin q 4 q 4 - d = 0 = 0 • a,b,c,d are known • One of the three angles is given • 2 unknown angles remain • 2 equations given above • Solve simultaneously for remaining angles

Position synthesis

Vector Loop Summary

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• Draw and label vector loop for mechanism • Write vector equations • Substitute Euler identity • Separate into real and imaginary • 2 equations, 2 unknown angles • Solve for 2 unknown angles • Note: there will be two solutions since mechanism can be open or crossed

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Example: Analytic Position Analysis • Input position q 2 • Solve for q 3 & q 4 given b=2.14

q 3 =?

° a=1.6

c=2.06

q 2 =51.3

° d=3.5

q 4 =?

°

Position synthesis

Example: Vector Loop Equation

12 R 2

+

R 3

ae j q 2 + be j

R

q 3

4

-

R 1

- ce j q 4 1.6e

j51.3Þ + 2.14e

j = 0 q 3 - de j q 1 = 0 - 2.06e

j q 4 0 - 3.5

e j 0 ° = b=2.14

q 3 =?

°

R 3 R 2

c=2.06

a=1.6

R 4

q 2 =51.3

° d=3.5

q 4 =?

°

R 1 Position synthesis

Example: Analytic Position Analysis ae j q 2 + be j q 3 - ce j q 4 - de j q 1 = 0 a(cos q 2 +jsin q 2 ) + b(cos q 3 +jsin q 3 ) - c(cos q 4 +jsin q 4 ) - d(cos q 1 +jsin q 1 )=0 Real part: a cos q 2 + b cos q 3 - c cos q 4 1.6 cos 51.3

+ 2.14 cos q 3 - 2.06 cos q 4 - 3.5 = 0 - d = 0 a=1.6

Imaginary part: a sin q 2 + b sin q 3 - c sin q 4

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1.6 sin 51.3

+ 2.14 sin q 3 - 2.06 sin q 4 = 0 = 0 q 2 =51.3

° b=2.14

q 3 =?

° d=3.5

c=2.06

Position synthesis

Solution: Open Linkage 2 equations from real & imaginary equations 1.6 cos 51.3

1.6 sin 51.3

+ 2.14 cos q 3 + 2.14 sin q 3 - 2.06 cos - 2.06 sin q 4 q 4 - 3.5 = 0 = 0 2 unknowns: q 3 & q 4 b=2.14

Solve simultaneously to yield 2 solutions.

q 3 =21Þ a=1.6

Open solution:

c=2.06

q

3 = 21Þ,

q

4 = 104 °

q 2 =51.3Þ d=3.5

q 4 =104Þ

14 Position synthesis

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Review - Law of Cosines cos q =

A

2 +

B

2 -

C

2 2

AB

q = arccos  

A

2 +

B

2 2

AB

-

C

2     A q B C

Position synthesis

Transmission Angles

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• Transmission angle is the angle between the output angle and the coupler • Absolute value of the acute angle • Measure of quality of force transmission • Ideally, as close to 90° 180  m 1 as possible m 1 acute 180  m 2 m 2

Position synthesis

Extreme Transmission Angles Grashof Crank Rocker

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• For a Grashof fourbar, extreme values occur when crank is collinear with ground b m 2 a m 1 For the extended position shown: m 1 =arccos [ (b 2 +(a+d) 2 - c 2 )/2b (a+d) ] m 2 =180 ° - arccos [ (b 2 +c 2 - (a+d) 2 )/2b c ] d c

Position synthesis

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Extreme Transmission Angles Grashof Crank Rocker For the overlapped case shown: m 1 =__________________________ m 2 =__________________________ a b m 1 d m 2 c

Position synthesis

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Extreme Transmission Angles Grashof Double Rocker • Remember: coupler makes a full revolution with respect to rockers • Transmission angle varies from 0 ° to 90°

Position synthesis

Extreme Transmission Angles Non-Grashof Linkage

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• Transmission angle is zero degrees in toggle position: output rocker & coupler a m 2 b • Other transmission angle given as: m 2 =__________________________ • Similar analysis for other toggle position m 1 =0  d c

Position synthesis

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Calculation of Toggle Angles • The input angle, q 2 , for the first toggle position given as: b a q 2 q 2 =__________________________ • Similar analysis for the other toggle position d c

Position synthesis