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

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From equations (2.13) and (2.14), using the real Plücker coordinates of the screws E1 and E2 we have: L L 2 1 2 2 L L 1 + M + M 2 2 1 2 2 + M 12 + N + N 1 M 2 1 2 2 2 = cosα − w⋅ a + 2w( L P + M Q + N R ) = 1+ w⋅ 0 + N 12 1 N 2 1 2 sinα 1 + 2w( L P + M Q 12 2 1 1 2 + w( P L 2 1 2 1 1 + N 2 2 1 2 + L P + Q M From (2.15) it is clear that unit vectors e1and 2 written using their components as; e e = 0 ⇔ L P + M Q + N R = 0 o 1 1 e e = 1 ⇔ L + M 1 o 2 2 2 1 o 1 1 e e 2 2 e e 2 o 1 2 2 1 12 1 = 1 ⇔ L + e e 2 2 2 1 2 12 2 1 + M 2 2 1 1 2 2 1 + N e e = 0 ⇔ L P + M Q 12 2 1 2 + N 2 2 1 1 = 1 + N 2 = 1 2 1 1 R 2 2 = 0 e e = cosα ⇔ L L + M M + N N = cosα 1 1 2 2 1 12 2 1 R ) = 1+ w⋅ 0 2 1 2 + Q 2 M 1 + R N 1 2 + R 2 N 1 ) (2.15) e , moment vectors o e and o e can be = −a sinα ⇔ PL + L P + Q M + M Q + R N + N R = −a sinα The last equation in (2.16) describes the relative moment of two screws E1 and E2 in screw theory. It can be seen that the relative mutual moment of the two screws equals zero if: • 12 0 = α , Axes of E1 and E2 are parallel • a12 = 0, Axes of E1 and E2 are intersecting That is, from 12 Plücker coordinates of two screws E1 and E2, describing the position of a rigid body in space, just 6 of them are independent. We can find the remaining 6 o coordinates from the constraint equations (2.16). To define the components of e P, Q , R ) 1 2 1 2 12 1 12 2 1 ( 1 1 1 o and e P , Q , R ) from ρ x , y , z ) , ρ x , y , z ) , e L , M , N ) , e L , M , N ) one can 2( 2 2 2 1( 1 1 1 2( 2 2 2 1( 1 1 1 2( 2 2 2 (2.16) use (2.7) for E1 and E2. When the axes of the screws are parallel we will have L = L1 = L2, M = M1 = M2 and N = N1 = N2. So in this case, 9 coordinates is enough instead of 12 coordinates where only 6 of them are independent as we have three constraint equations as given in (2.18) 14
LP + MQ + NR = 0 L 1 2 2 + M 2 1 LP + MQ + NR = 0 2 + N 2 1 2 = 1 2.4 Kinematics of Three Unit Screws in Space Figure 2.4 – Three arbitrary unit screws in space (2.18) Let’s describe three arbitrary unit screws in space; Ei, Ej and Ek as in figure 2.4. We ~ ~ ~ will write the algorithm to find the dual coordinates of unit screw E k ( Lk , M k , Nk ) using ~ ~ ~ ~ ~ ~ known dual coordinates of screws E i( Li , M i, Ni ) and E j ( Lj , M j, N j ) . Dual angles between those three unit screws are defined using (2.14) as A ij = α ij + waij , A jk = α jk + wa jk , ki = α ki waki where α ij α jk , αki A + x z O ~ ~ ~ Ei ( Li , M i, Ni ) aki aij αki , and a ij, a jk , aki are the angles and shortest distances between the corresponding screw axis respectively. Firstly we describe the same task for system of unit vectors e, the real part of a unit screw E. That is, we need to define the direction cosines Lk,Mk,Nk of ek from given unit vectors ei(Li,Mi,Ni) and ej(Lj,Mj,Nj). Now we define a new vector n = ei x ej to determine the orientation of unknown vector ek wrt. the plane defined by ei x ej. n = e ⋅ e i j αij α jk sin α = sinα ij ~ ~ ~ E j ( Lj , M j, N j ) y α jk ij ~ ~ ~ E k ( Lk , M k , Nk ) 15
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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: Using (2.7) and (2.12), one can fin
Page 27 and 28: Following Crammer’s method, the u
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
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[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
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C.2 Visual NASTRAN Desktop NASTRAN
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