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

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APPENDIX C SOFTWARE USED FOR NUMERICAL ANALYSIS C.1 MathCAD Solution Using Optimization MathCAD offers a variety of tools for numerical analysis, one of them being the optimization commands. Following the method a single objective function of single or more variables is given and the values of variables to make the function minimum is found with a desired precision. The objective function in our case is assembled in the following way: 1) Transform the functions given in (4.4) as: gn = | fn - cn | = 0 2) Objective function is defined as fobjective= ∑ g n = 0 n= 1 The last step is to give the initial conditions and the constraints for the solver. In the following pages, the actual MathCAD sheet is shown. 12 82
Preleminary function definitions Initial values θ 2 δ 1 ( ) pcss , , r, α3, α4, a5 := δ 2 := 1.42929552 := := θ 1 δ 2 ( ) Q:= 3⋅r1 θ 4 δ 3 := := θ 3 δ 4 ⎛ ⎜ ⎜ ⎜ ⎝ Fixed construction parameters ⎛ ⎝ ⎞ ⎠ θ 6 δ 5 := ( ) B:= r2−r1 := Objective function definitions θ 5 δ 6 ( ) ( ) φ i δ 4 π := 2 ( ) C:= 3⋅r2 ( ) ( ) ( ) sin( α4) := 1 ( ) ( ) ss ⋅ cos α3 ⋅ sin α4 ⋅ a5 + c ⋅ cos α4 ⋅ a5 + r⋅c −c ⋅ cos α3 ⋅ sin α4 ⋅ a5 + ss ⋅ cos α4 ⋅ a5 + r⋅ss sin α 3 π π c1 := cos⎜ 90 ⋅ s 180 1 := sin⎜ 90 ⋅ 180 ⎡ ⎣ ⎤ ⎦ ⎛ ⎝ ⎞ ⎠ ⋅ ⋅ a5 π π c2 := cos⎢( 90 + 120) ⋅ ⎥ s 180 2 := sin⎢( 90 + 120) ⋅ 180 ⎡ ⎣ ⎤ ⎦ π π c3 := cos⎢( 90 + 240) ⋅ ⎥ s 180 3 := sin⎢( 90 + 240) ⋅ 180 i:= 1.. 6 r1 := 1 r2 := 1.5 Input Coordinates θ 1 ⎡ ⎣ ⎡ ⎣ ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ r 2 should be greater then r1 π π π := ( 105) ⋅ θ 3 := ( 87) ⋅ θ 5 := ( 112.61986495) ⋅ 180 180 180 2 2 D:= r1+ r2 + r1 ⋅ r2 δ 6 ⎞ ⎟ ⎠ := 1.42929552 83
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Kinematic and Dynamic Analysis of S
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Research is what I'm doing when I d
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ÖZ Bu tez yeni bir tip uzaysal alt
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Chapter 4 KINEMATIC ANALYSIS ......
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Figure 4.15 - Comparison of results
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Chapter 1 INTRODUCTION The introduc
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Figure 1.4 - The first flight simul
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Generally, the actuators of a seria
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−1 where J = J q J x . . . q = J
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Chapter 2 SCREW KINEMATICS In this
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From equation (2.6) we have; or x z
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Using (2.7) and (2.12), one can fin
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LP + MQ + NR = 0 L 1 2 2 + M 2 1 LP
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Following Crammer’s method, the u
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That’s, the variable angle α ki
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~ ~ ~ ~ ~ ~ Let Ei ( Li , M i, Ni )
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Qk = (NiPj + LjRi - LiRj - NjPi - a
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The methods reported by F. Freudens
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Using equations (3.4) and (3.6), we
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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,
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: PP := L1 ← 0 M1 ← 0 N1 ← 1 L2
Page 95 and 96: C.2 Visual NASTRAN Desktop NASTRAN
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