Chapter 5: Complex Number Imaginary

Chapter 5 Complex Number Imaginary Axis (j) • A vector can be represented by a complex number • Real part is x-axis jY=jR sin  • Imaginary part is yaxis • Useful when we begin to take derivatives 1 Point A RA  X=R cos  Real Axis Position synthesis Chapter 5 Complex Number • Vector (R) = Z = x+jy = Rcosθ + j Rsin θ • Complex Conjugate of Z is Imaginary Axis (j) Point A jY=jR sin  RA  Z* = x- jy X=R cos  2 Real Axis Position synthesis Derivatives, Vector Rotations in the Complex Plane Imaginary • Taking a derivative of B a complex number will RB = j R result in multiplication by j C A • Each multiplication by RA RC = j2 R = -R Real j rotates a vector 90° CCW in the complex plane R = j3 R = - j R D D 3 Position synthesis 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 4 Link 3, length b Coupler Link 2, length a B Link 4, length c A Link 1, length d Pivot 02 Ground Link Pivot 04 Position synthesis 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 5 3 2 3 A 2 B 4 4 1 1 = 0 ° Position synthesis More on Complex Notation • Polar form: re j • Cartesian form: r cos + j r sin • Euler identity: ±e j = cos ± j sin • Differentiation: j 6 de j  je d Position synthesis The Vector Loop Technique • Vector loop equation: R2 + R3 - R4 - R1 = 0 • Alternative notation: R3 b RAO2 + RBA - RBO4 - RO4O2 = 0 3 a A nomenclature - tip then tail R2 2 d • Complex notation: R1 aej2 + bej3 - cej4 - dej1 = 0 O 2 • 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 7 B R4 4 c O4 Position synthesis 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 8 Position synthesis 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 9 Position synthesis Vector Loop Summary • • • • • • • 10 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 Position synthesis Example: Analytic Position Analysis • Input position 2 given • Solve for 3 & 4 b=2.14  3=?° a=1.6 2=51.3° d=3.5 11 c=2.06  4 =?° Position synthesis Example: Vector Loop Equation R2 + R3 - R4 - R1 = 0 aej2 + bej3 - cej4 - dej1 = 0 1.6ej51.3Þ + 2.14ej3 - 2.06ej4 - 3.5ej0° = R3 0 R2 b=2.14  3=?° c=2.06 a=1.6 2=51.3° d=3.5 12 R4 R1  4 =?° Position synthesis Example: Analytic Position Analysis aej2 + bej3 - cej4 - dej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 b=2.14 1.6 cos  + 2.14 cos 3  3=?° - 2.06 cos 4 - 3.5 = 0 a=1.6 Imaginary part: c=2.06 a sin 2 + b sin 3 - c sin 4 = 0 2=51.3° d=3.5 1.6 sin  + 2.14 sin 3 - 2.06 sin 4 = 0 13 Position synthesis 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 b=2.14 Solve simultaneously to yield  3=21Þ 2 solutions. a=1.6 Open solution: c=2.06 3 = 21, 4 = 104° 14  2=51.3Þ d=3.5  4=104Þ Position synthesis 15 Position synthesis Review - Law of Cosines 2 2 A B C cos   2AB 2 A2  B2  C 2    arccos   2AB    A B C 16 Position synthesis