Kinematic and Dynamic Analysis of Spatial Six Degree of Freedom ...

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To define the direction of unit vector ek we can use the following equations: e e e k k j ⋅n = Cα ⋅e i ⋅e k kn = Cα ki = Cα kj Sα ij (2.19) In equations (2.19) C stands for cos and S stands for sin. From later on, we will use this convention for the sake of brevity. Writing (2.19) in coordinate form we have: L ( M N − N M ) + M ( N L − L N ) + N ( L M − M L ) = Cα Sα k k k i L L j i L L + M M j k + M k M i j i j + N k + N N k i N j k i = Cα ki = Cα jk j i j k i j i j kn ij (2.20) The system of three linear equations (2.20) in three unknowns Lk, Mk and Nk can be solved using various approaches. In what follows is the application of Crammer’s method to find the solutions. System of equations (2.20) can be written in matrix form as: ( M N − N M ) ( N L − L N ) ( L M M L ) ⎡ i j i j i j i j i j − ⎢ ⎢ Li M i Ni ⎢ ⎣ L j M j N j The discriminant of the coefficients matrix is: ∆ = ∆ = ( M N − N M ) ( N L − L N ) ( L M − M L ) i j L L i j i j i j M M i j i j i j N N i j i i j j ⎤⎡ L ⎥⎢ ⎥⎢ M ⎥ ⎦ ⎢⎣ N k k k ⎤ ⎡Cα kn Sα ij ⎤ ⎥ ⎢ ⎥ ⎥ = ⎢ Cα ki ⎥ ⎥⎦ ⎢ ⎥ ⎣ Cα jk ⎦ i i i i ( M N − N M ) − ( N L − L N ) + ( L M − M L ) i j i j M M j N N j i ( M iN j NiM j ) ( NiL j Li N j ) ( LiM j M iL j ) ij αij 2 2 2 2 2 − + − + − = = S ∆ = n j i j L L j N N j i j i j L L i j M M i j 16
Following Crammer’s method, the unknowns can be found as: Cα Sα jk ( N L − L N ) ( L M − M L ) kn ij i j i j i j i j ∆ L −2 Lk = = Cα ki M i Ni S αij ∆ Cα M j ( M N − N M ) Cα Sα ( L M − M L ) i j i j kn ij i j i j ∆M −2 M k = = Ni Cα ki Ni S αij ∆ N j Cα ( M N − N M ) ( N L − L N ) i j i j i j i j kn ij ∆ N −2 Nk = = Li M i Cα ki S αij ∆ L j jk M j N j N j Cα Sα Cα After expanding the discriminants and simplifying, we have: where k −2 ( LiD 2 + L jD3 + Lij Sαij Cα kn ) S ij −2 ( M iD2 + M jD3 + M ij Sαij Cα kn ) S ij −2 ( NiD2 + N jD3 + Nij Sαij Cα kn ) S ij L = α M = α k k N = α D α D α 2 = Cα ki − Cα jk C ij 3 = Cα jk − Cαij C ki L M N − N M M N L − L N ij = i j i j ij = i j i j ij = i j i j jk N L M − M L From three angles α ki , α jk andα kn , just two of them are independent as we have one constraint equation given as: α kj : 2 2 2 L M + N = 1 (2.21) k + k k Substituting Lk, Mk and Nk we can find the angle α kn as a function of anglesα ki and 17
Page 1 and 2: Kinematic and Dynamic Analysis of S
Page 3 and 4: Research is what I'm doing when I d
Page 5 and 6: ÖZ Bu tez yeni bir tip uzaysal alt
Page 7 and 8: Chapter 4 KINEMATIC ANALYSIS ......
Page 9 and 10: Figure 4.15 - Comparison of results
Page 11 and 12: Chapter 1 INTRODUCTION The introduc
Page 13 and 14: Figure 1.4 - The first flight simul
Page 15 and 16: Generally, the actuators of a seria
Page 17 and 18: −1 where J = J q J x . . . q = J
Page 19 and 20: Chapter 2 SCREW KINEMATICS In this
Page 21 and 22: From equation (2.6) we have; or x z
Page 23 and 24: Using (2.7) and (2.12), one can fin
Page 25: LP + MQ + NR = 0 L 1 2 2 + M 2 1 LP
Page 29 and 30: That’s, the variable angle α ki
Page 31 and 32: ~ ~ ~ ~ ~ ~ Let Ei ( Li , M i, Ni )
Page 33 and 34: Qk = (NiPj + LjRi - LiRj - NjPi - a
Page 35 and 36: The methods reported by F. Freudens
Page 37 and 38: Using equations (3.4) and (3.6), we
Page 39 and 40: Order of a structural group is the
Page 41 and 42: 3.3 Geometrical Structural Synthesi
Page 43 and 44: 3.4 Kinematic Structural Synthesis
Page 45 and 46: Let’s consider a 5 DOF parallel m
Page 47 and 48: Chapter 4 KINEMATIC ANALYSIS Kinema
Page 49 and 50: Figure 4.1 - The six degree of free
Page 51 and 52: v v v P4 = yv N 4 − zv M 4 v v v
Page 53 and 54: The best way of visualizing the dis
Page 55 and 56: 4.2.2 Results for the Actuation of
Page 63 and 64: Figure 4.17 - Corresponding piston
Page 65 and 66: Chapter 5 DYNAMIC ANALYSIS Dynamics
Page 67 and 68: Thus, according to Lagrange, the re
Page 69 and 70: J is the 6x6 Jacobian matrix that
Page 71 and 72: Chapter 6 DISCUSSION AND CONCLUSION
Page 73 and 74: REFERENCES [1] D. Stewart, A platfo
Page 75 and 76: [19] Seung-Bok Choi, Dong-Won Park,
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[38] Kotelnikov A. P., Screw calcul
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[60] R. I. Alizade, Investigation o
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A.1.1 CASSoM Source Code Like all v
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in the 'Input Parameters' section.
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pty:=roy+dy; ptz:=roz+dz; end; func
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Figure A.2 - a Screenshot of iMIDAS
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APPENDIX B RECURSIVE SYMBOLIC CALCU
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PP := L1 ← 0 M1 ← 0 N1 ← 1 L2
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Preleminary function definitions In
Page 95 and 96:
C.2 Visual NASTRAN Desktop NASTRAN
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