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

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The screws are placed on the axis of kinematic pairs subsequently. Location of the moving platform can be defined fully if we know the coordinates of three points on it. Since two intersecting screws define a point, we’ll place be able to find the center of the spherical pair using connection point on the moving platform. v E5 ⎡ ⎢ ⎢ ⎢ = ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ v E4 = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ν E5 on the first axis of the spherical pair to ν E4 and v v v sv Cα13 Sα 24 + cv Cα 24 v v v − cv Cα13 Sα 24 + sv Cα 24 v v v sv Cα13 Sα 24 + cv Cα 24 v v svrv Sα13 Sα 24 v v − cvrv Sα13 Sα 24 0 v v v cv Sα 24 − sv Cα13 Cα 24 v v v sv Sα 24 + cv Cα13 Cα 24 v v − Sα Cα ν E 5 . This point coincides with the ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 13 24 v [ ( 2c Cα v 13 Sα v 24 − Cα v v sv 24 ) rv + sva35 ] v ( 2s Cα v Sα v + c Cα v ) r − c a [ v v ] 13 24 − a v 35 24 Cα 13 v v v Sα13 v 35 Sα13 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (4.1) To solve the forward task for the parallel manipulator, we will separately handle each branch as a serial manipulator. Using the recurrent equations (2.28) and (2.30), the expressions for the components of ν E4 and ν E 5 are found as given in (4.1). Note that, to find these expressions, a simple recursive algorithm is used. (Appendix B). 4.1.3 Construction of the Set of Equations to be Solved For two intersecting unit screws ν E4 and ν 5 E in space (a45 = 0), the equations of screw axis in Plücker coordinates can be written as: 40
v v v P4 = yv N 4 − zv M 4 v v v R4 = xvM 4 − yv L4 v v v Q5 = zv L5 − xv N5 v v v Q4 = zv L4 − xv N4 v v v P5 yv N5 − zv M 5 = (4.2) v v v R5 = xvM 5 − yv L5 To define the components of radius vector ρv(xv,yv,zv) in Cartesian coordinates, we need three equations. Solving where v v v v v L12 = M 4 N5 − N4 M 5 x v v v Q4 P4 , P5 Using some of the components of v = v , for xv, yv and zv we get: ( )( ) 1 v v v v v v v − L4P5 + M 5Q4 + N5 R4 L12 ( )( ) 1 v v v v v − = M 4 P5 − M 5 P4 L12 ( )( ) 1 v v v v v − = N P − N P L y (4.3) z v 4 5 5 ν E 4 and 4 12 ν E 5 in equation (4.1), one can find from (4.3) the coordinates xv, yv, zv of the six intersection points of screws, lying on the moving platform. Since we know the distances between the points, we can write the following equalities (Figure 4.2). |A 1 – A 2 | = |rb – ra| |A 3 – A 4 | = |rb – ra| |A 5 – A 6 | = |rb – ra| |A 1 – A 3 | = ra |A 2 – A 4 | = rb 3 |A 3 – A 5 | = 3 ra |A 5 – A 1 | = 3 ra 3 |A 4 – A 6 | = 3 rb |A 6 – A 2 | = 3 rb |A 6 – A 1 | = r a + rb + ra rb 2 2 |A 4 – A 5 | = r a + rb + ra rb 2 2 |A 2 – A 3 | = r a + rb + rarb 2 2 (4.4) 41
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 and 26: LP + MQ + NR = 0 L 1 2 2 + M 2 1 LP
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: Figure 4.1 - The six degree of free
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,
Page 77 and 78: [38] Kotelnikov A. P., Screw calcul
Page 79 and 80: [60] R. I. Alizade, Investigation o
Page 81 and 82: A.1.1 CASSoM Source Code Like all v
Page 83 and 84: in the 'Input Parameters' section.
Page 85 and 86: pty:=roy+dy; ptz:=roz+dz; end; func
Page 87 and 88: Figure A.2 - a Screenshot of iMIDAS
Page 89 and 90: APPENDIX B RECURSIVE SYMBOLIC CALCU
Page 91 and 92: PP := L1 ← 0 M1 ← 0 N1 ← 1 L2
Page 93 and 94: Preleminary function definitions In
Page 95 and 96: C.2 Visual NASTRAN Desktop NASTRAN

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