Solutions Manual Trajectory Generation

Chapter 7 Trajectory Generation Exercises 7.1) Three cuЬics arc required to connect а start point. two via points, and а goal point. That is, three for each joint, for а total of 18 cuЬics. fach cuЬic has four cocfficieпts, so 72 coefficients are stored. 7.2) Use (7.6) wilh 80 = -5, 81 = ВО, 11 = 4.0 ао= -5 GJ :::;Q а, = 85 -32 7.3) Using (7.23) we see lhat lhe acceleration musl .. .. 4(85) 85 Ье 8 с: = . lf we choosc exactly 8 = 16 4 � deg / sec2 , then the linear ponion shrinks to 4 zero, and tь = 2 seconds. Lct's instead choose 0 = 85 dcg/ sec 2 , thcn from (7.22) we have 18 = 2 - �- So, the linearponion is 2� scconds long, and the velocity during the linear ponion is ti = (85)(2 - �) deg/ sec. 7 .4) This is basicaLly а programm.ing exercise. The data describlng the path (via points, duration, etc.) should Ье kcpl in а linked list, with one nodc per path segment. Likewise. the planned path (Ьlcnd timcs, vclocitics, etc.) should Ье kept in а second linked list. Тhis linked list is the ошрш of the roшine. 7.5) Use (7.15) to find coefficients. 7.6) Usc (7.11) with the following for the first scg111cnt: 00 = 5 01 = 15 0. = о 01 = 11.5 r1 = 1 This gives: ао = 5а, = О а2 = 12.5 аз= -2.5 And use (7.11) again for the second segment wiLh: 00 = 15 01 =40 80 = 11.5 81=0 t1=I whkh givcs: ао = 15 а,= 17.5 а2 = 40.0 аз= -32.5 Then evaluate (7.3) to plot curvcs. 7.7) Neither segшent is an "intcrior" segment, so (7.26) and (7.28) are used. From (7.26) calculate: IJ, =80 ,, = 1 - /1 - 2(15 - 5) = 1 - ./3 2 80 е,, = • 10 1 -нl- '7) 40 2 +./3 From (7.28): IJз = SGN(l5-40)80 = -80 2 25 tз = 1 -/1 + ( ) = 1 - /1 - � -80 8 li:,, = 25 1 -нl -/1 - �) _ 50 1 + j1 - � FinaHy, t2 is ап interior point so its computed from (7.24): 7.8) Use (7.15) to compute coefficients, and use (7.3) to evaluate splines. 7.9) For first segment, use (7.11) wilh: This gives: ао = 5 а1 = О 15 2 lh= - - For second segn1ent, use. (7.11) with: This gjves: ао = 15 75 а1 = О а2 = -4 аз Then evaluate (7.3) 1О plot curvcs. = 25 4 7.10) This is like exercise 7.7. use (7.26) to calculate: 01 = sgn(l 5 - 5)60 = 60 t, =2- ✓4-'!!!.60 =2- 1/з [ТI_ From (7.28): ё, = sgn(\5- (-10))60= 60 fз = 2-R-=2-n Final\y, from (7.24): 7.11) 'X G =[!O 20 30 О О 30]7 о о о о .707 -.707 .707 .707 -io] 10 7.13) Prograшming exercise. Note: assume sаше tгdjec­ tory for cach joint. 7.14) Programming exercise. Note: assume samc trajec­ tory for each joint. • 1 [ 1/2 1/J] 7.15) let 0, = ---- 3(0, - 0,)- + 3(0G - 0,)2(1f1 +112) ,1, ,1, Then: йJJ =0 3 1 . 012 = ,(0, - 0,) + -0, t11 'п Then, cubic coefficients given Ьу (7.11). 7.17) Ву differentiation: 0(1) = 1801 - 1801 2 ii(t) = 180 - 3601 Thcn, cvaluating at t = О and t = 1, we have: 0(0) = 10 0(1) = 40 0(0) =О 0(1) = О ii(O) = 180 ii(IJ = -180 7.18) Usiпg samc equations as for (7.17), and evaluating at t = 2. 0(2) = -110 8(2) = -360 0(2) = -540 7.19) So, 0(1) = 5 + 140, - 1351 2 0(1) = 140- 2701 0(0) = 10 0(0) = 5 0(1) = 40 0(1) = 10 0(0) = 140 0(1) = -130 7.20) Using equations from 7. l 9, and evaluating а, t = 2: 0(2) = -60 0(2) = -255 0(2) = -400 Chapter 8 Manipulator Mechanism Design Exercises 8.1) The laser beam сап Ье considered а line in space. ln general, it takes 4 D.O.F.S to position а line in space. So, the answer is 4. Note that this is soшewhat "theoretical" in that there is рrоЬаЫу по single 4-DOF design that could perform all s11ch cuHing tasks-however, see prohlem 8.2 below for опе example. 8.2) 8.3) Using (8.1) ,ve have: = (0+0)+ (0+0) + (О+ (И -L)) = И -L \V = -,rU3 - -,r L 3 = -л(U 3 - L 3 ) 34 43 4З "Hollow" { Sphere А 8.4) Let Ko.z Ье stiffness of the 0.2 cm diameter shaft, and lч,.з Ье that of the 0.3 cm shaft. Then, using (8.14) and (8.19): 1 1 1 -- = --, + - where 11 = 8 Ko.2n� Ко.з Kto1a1 From (8. 15): = (7.5 х 10 10 )(,r)(0.002)4 ( 32)(0_30) = О . 393 ntmlrad Ко.з = (7.5 х I0 ю)(,r)(0.003)4 (32)(0_30) = 1. 988 ntnJ/rad Ko.z 1 к.,.., 1 1 = (.393)82 + 1.988 = 1 1 = 05427 25.152 + 1.988 1 Кшшl = 1.8426 n(m/rad j 8.5) Torsional loa.d due to accelera(ion of thc ccпter of mass about lhe joint axis is: r = (0.30)(10)(2)(9.8) ntm = 58.8 ntm А torsio11al spring is given Ьу т stiffness is т лв = К ЛВ, so needed 58.8 0.1 К = - = - = 588 ntm/rad From (8. 15), solving for d gives: d= ,/(32)(0.30)(�88) (7.5 Х JOIO)(,r) = 0.01244 m or I d = 1.244 cm 1 8.6) From (8.14): _l_ = _l_ + ..2.._ = 4.333 х 10-3 I ООО 300 КIOWI :. к.,�,= 230.77 nlm/rad 8.7) lf 3 of the attachment points on the top or bottom platc coincide. For example, in fig. 8.22, if q; = q; = q, for any i # j # k, or if р, = Pj = PI for any i # j # k. 8.8) In GruЫer' s formula (8.9) we ha,·e: N = 18 (12 univcrsal joints, 6 prismatic joints) L = 14 (2 parts per activator = 12, top plate, bottom plate) I:.!!, J; = 12 х 3 + 6 = 42 so, F=6(14-18-1)+42=[EJ 8.9) Stiffness of input drive as scen at gear #1: 1 1 1 1 -- = - + - + - = 0.0225; к,.., 1 100 400 100 ooupling ltl J Kgear \ = 44.44 ilha.f) c()Upling 12) Assume that shaft frorn gear 2 to gear З is rigid... 1 1 l __ , - -- = 1.125 х 10-3; = --+ -Kgur 3 2000 6-(Kge:i.r 1) к,.., з = 888.88 1 + __I _ -1- = -_ = 5.3125 х 10-4; К,.,,,• 2000 62 (К,"' з) 1 К,,,,, 8.10) Error is = 1882.35 ntm/md 1 2000 - 1882 35 х IOO=l 6.25% 1 1882.35 8.11) See solution 10 exercise 4. 12 8.12) For i = l, 2, ... 6 1 d; = !/fTq; - р;// 1 8.13) То maximize W = / L I L2S2/ over all configs means maxiшizing W' = L 1 L2. Since L1 + L2 = С (С= some constant); L 2 = С - L, so, W' = L,(C - L,) = CL, - Lf dw' - =c-2L1 dL, .. d2 ш' = -Z dL; Maximum of W' c11.ists whe11 С - 24 = О orL,=C/2 . . 1 L1 = L2 is Oplimal 1 8.14) Samc as ex.ercise 8.13 sincc 3rd dof is orthogonal to first two. 1 L1 = L2 is optimal 1 8.15) LU-decompositio11 allows one to ,vrite the jacoЬian in thc Гorm J(0) = u т riv whcrc" and и are orthogonal matrices. Detem1inзnt of а product equals product of derenninanrs, so DЕТ J = (DETU r )(DEТQ)(DEТV) = 1. DEТQ.I and det Q is the product of the eigenvalues. Q.E.D. 8.16) From (8.15) (0.33 х 7.5 х I0'°)(1r)(0.00l) 4 G1rd4 К=--= 32L (32)(0.40) = 1 0.006135 ntm/rad 1 Vert flimst, Ьccause dinmctcr is I mm! 8.17) From (8.12) 11 = r2/r, = 12/2 = [}] 8.18) From the general case of needing 6, subtract 2 because grasping "along'' and "aЬout" the cylinder's axis is а frcc variaЬlc. 1-Icncc, 0· 8.19) In GrtiЫcr's formula: N = 12 (9 rcal joints, З ball-in-socket contacts) L = 11 (3 links pcr fingcr, basc, ball) r::;., f; = 9 + З х З = 18 so, F = 6(11 - 12 - 1) + 18 = -12 + 18 = 0 8.20) ln GrtlЫers fonnula: N = 6 (3 ball-in-socket, З universal) L = 5 (ground, object, З links) r:r_, f; = 3 х з + з х 2 = 15 so, F =6(5-6- l) + 15 = -12+ 15 =0 8.21) Start witl1 F = K"m(liX, + 8Х2) And substitute in expressions for 8Х1, 8Х2 : Multiply both sides Ьу FKsum Q.E.D. 8.22) From (8.22) we have where L = Lrree + 31 Lconшct So the proЫem is to figure out Lrги and L contaet for the ge11eral 2-pulley setup. where 81 = cos-1 (i) and r1 > r2