1. Position synthesis1 Analytic Approach to Mechanism Design ME 324 Fall 2000 http://www.engr.colostate.edu/me/program/courses/ME324/notes/PositionAnalysis.ppt

2. Position synthesis2 Chapter 4 - Analytic Position Analysis 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 Real Axis Imaginary Axis Point A RARA  R cos  jR sin 

3. Position synthesis3 Derivatives, Vector Rotations in the Complex Plane Taking a derivative of a complex number will result in multiplication by j Each multiplication by j rotates a vector 90° CCW in the complex plane Real Imaginary RARA R B = j R R C = j 2 R = -R R D = j 3 R = - j R A B C D

4. Position synthesis4 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 Ground Link Coupler Link 1, length d Pivot 02Pivot 04 A B Link 2, length a Link 3, length b Link 4, length c

5. Position synthesis5 Angle Measurement Convention All angles measured from angle of the ground link Define  1 = 0° One DOF, so can describe all angles in terms of one input, usually  2 1 A B 2 3 4 22 33 44  1 = 0 °

6. Position synthesis6 More on Complex Notation Polar form: re j  Cartesian form: r cos  + j r sin  Euler identity: ±e j  = cos  ± j sin  Differentiation: de j  d   je j 

7. Position synthesis7 The Vector Loop Technique Vector loop equation: R 2 + R 3 - R 4 - R 1 = 0 Alternative notation: R AO2 + R BA - R BO4 - R O4O2 = 0 nomenclature - tip then tail Complex notation: ae j  2 + be j  3 - ce j  4 - de j  1 = 0 Substitute Euler equation: a (cos  2 +j sin  2 ) + b (cos  3 +j sin  3 ) - c (cos  4 +j sin  4 ) - d (cos  1 +j sin  1 ) = 0 d A B a b c 22 33 44 R1R1 R2R2 R3R3 R4R4 O2O2 O4O4

8. Position synthesis8 Vector Loop Technique - continued Separate into real and imaginary parts: Real: a cos  2 + b cos  3 - c cos  4 - d cos  1 = 0 a cos  2 + b cos  3 - c cos  4 - d = 0, since  1 = 0, cos  1 = 1 Imaginary: ja sin  2 + jb sin  3 - jc sin  4 - jd sin  1 = 0 a sin  2 + b sin  3 - c sin  4 = 0, since  1 = 0, sin  1 = 0

9. Position synthesis9 Vector Loop Technique - continued a cos  2 + b cos  3 - c cos  4 - d = 0 a sin  2 + b sin  3 - c sin  4 = 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

10. Position synthesis10 Vector Loop - Summary 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

11. Position synthesis11 Example: Analytic Position Analysis Input position  2 given Solve for  3 &  4  2 =51.3 ° a=1.6 b=2.14 c=2.06 d=3.5  3 =? °  4

12. Position synthesis12 Example: Vector Loop Equation R 2 + R 3 - R 4 - R 1 = 0 ae j  2 + be j  3 - ce j  4 - de j  1 = 0 1.6e j51.3Þ + 2.14e j  3 - 2.06e j  4 - 3.5 e j 0° = 0  2 =51.3° a=1.6 b=2.14 c=2.06 d=3.5  3 =?°  4 R2R2 R3R3 R4R4 R1R1

13. Position synthesis13 Example: Analytic Position Analysis ae j  2 + be j  3 - ce j  4 - de j  1 = 0 a(cos  2 +jsin  2 ) + b(cos  3 +jsin  3 ) - c(cos  4 +jsin  4 ) - d(cos  1 +jsin  1 )=0 Real part: a cos  2 + b cos  3 - c cos  4 - d = 0 1.6 cos  + 2.14 cos  3 - 2.06 cos  4 - 3.5 = 0 Imaginary part: a sin  2 + b sin  3 - c sin  4 = 0 1.6 sin  + 2.14 sin  3 - 2.06 sin  4 = 0  2 =51.3° a=1.6 b=2.14 c=2.06 d=3.5  3 =?°

14. Position synthesis14 Solution: Open Linkage 2 equations from real & imaginary equations 1.6 cos  + 2.14 cos  3 - 2.06 cos  4 - 3.5 = 0 1.6 sin  + 2.14 sin  3 - 2.06 sin  4 = 0 2 unknowns:  3 &  4 Solve simultaneously to yield 2 solutions. Open solution:  3 = 21Þ,  4 = 104°

15. Position synthesis15 Review - Law of Cosines A B C  cos  A 2  B 2  C 2 2AB  arccos A 2  B 2  C 2 2AB      

16. Position synthesis16 Transmission Angles 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° as possible  180  -   acute  180  -  

17. Position synthesis17 Extreme Transmission Angles - Grashof Crank Rocker For a Grashof fourbar, extreme values occur when crank is collinear with ground For the extended position shown:  1 =arccos [ (b 2 +(a+d) 2 - c 2 )/2b (a+d) ]  2 =180° - arccos [ (b 2 +c 2 - (a+d) 2 )/2b c ] a b c d 11 22

18. Position synthesis18 Extreme Transmission Angles - Grashof Crank Rocker For the overlapped case shown:  1 =__________________________  2 =__________________________ a b c d 11 22

19. Position synthesis19 Extreme Transmission Angles - Grashof Double Rocker Remember: coupler makes a full revolution with respect to rockers Transmission angle varies from 0° to 90°

20. Position synthesis20 Extreme Transmission Angles - Non-Grashof Linkage Transmission angle is zero degrees in toggle position: output rocker & coupler Other transmission angle given as:  2 =__________________________ Similar analysis for other toggle position a b c d      

21. Position synthesis21 Calculation of Toggle Angles The input angle,  2, for the first toggle position given as:  2 =__________________________ Similar analysis for the other toggle position a b c d 