Solutions Manual - Jacobians, Velocities and Static Forces

Chapter 5 Jacoblans: Velocities and Static Forces Exercises 5.1) The Jacobian in frnme (о) is: :. The same result as when you start ,vith 3 J@, ° namely. the singulэr configurations зге 02 = Q ° or 180 . 5.2) From cxercise 3.3 we have: and: О О 1 О О 1 О О Lз] О О ; 1 о Т= о Т з Т , , , we could then find O J@) quite easily Ьу di fferen­ tiating O PvoRG· Finally. 4 J ® сап Ье calculatcd as ri R 0 J (Ю, This might Ье tedious, so lets try "standard" velocity propagation as done in the tex.t: 5.2) (Conti11ued) 3V, = [ с, s, о] ([ о ] + [ о ]) -Sз О .-.'J@= Сз О О О 1 -L 1 @1 о [ (-L1 - L2C2О - LзС,з) LiJJ. -L,C,lii SзLz c,L, +L, 0 5.3) First. ve\ocity analysis: 1 W1 =bR0 Wo+B1 1 z1 =81i 2 w, = [ -s, оо с,s,] [о]Q + [о]в,о = [s,с,0в,01] с, О -1 О 81 1 О] L, 0 5.3) (Co11tinued) [о] [ s, с, д, + с, s,д, [ Sз 0] [ s, 01] 3W3 = -SСз3 С3 о с,е, + о = -s,s,e, + с,с,0, о о 1 в, 0,+0, о, 4 V, = jR 'V4 = [ .•. ([ - . L,s,e, ] [ S,з61 ] [ L, ]) L,c,e, + с"д, х о О -L,B, L,c,o, о,+ 0, L2 3 , ] L1 C3 0, + Lз(0, + 6з) -L 1 01 - L2 C,01 - LзCnli, S iJ 'J(fl) = [ -L, Next, using forcc analysis: - g L,C, L2Sз L2Сз + Lз о - LзС,з ] 53) (Co111i111,ed) 2N 2 = [с, -S, о][ о ] [L'] Сз О S3 О 'N2 = [ О l + -LзFz LзFr - О О C, Fx-S, Fy х [ SзFx + CзFr ] Fz LзSзFz -L2Fz LзCзFz ] L2(SзFx + C,Fr) + LзFr 1 F1 = [R'F, = [с, -s, О О S2 с, о] -l О C, Fx-S, Fy [ S,Fx + CзFr ] Fz C,(CзFx-SзFr)- S2(SзFx + CзFr) 1F, = [ ] -Fz S2(CзFx -S,Fr) + C,(SзFx + CзFr) 1 N1 = [с, О S, -s, О С, о] [ -l О - L,S,Fz -L2Fz LзCзFz L2(S3F,+CзFr)+LзFr То computc torques, take the z-componcnt of thc 'N,: r, = L,SзFx + (L2Сз + Lз)Fr rз = LзFr Which lcads to same expression as before Гог 'J L1 = ./2.L, 1 . .. Under this cond1tюn S2 = = ±. 707 ./2 and С2 = -.707 :. An isotropic point cx.ists if L 1 = ,./2.L2 and in that case it exists when 0z. = ± 135° ln this configuration, the manipulator looks mornen­ tarily like а Cartesian manipulator. 5.9) U11so\ved. А srnall part of the answer can Ье found in Reference [7] of Chap. 15. з J _ [ L1S2 О ] - L,C, + L, L, so, L1C2+L2J L, -L1C2 - L2] L1S2 L, L, �:;,, 777777 777. 5.11) From (5. 103): Bv = -8RлРХ] "1,1 [вR АО А �R 0.86 0.5 �R A РХ = [ -g.5 0.86 �] [i -5 10 2.5 -4.3 -5.0 [ 2.5 -8.6 ] о = 4ьз 10 0.86 0.5 -0.5 0.86 о •v = [ � о о о о о о l о о о +] -2.5 4.3 5.0 О -4.3 -2.5 8.6 2 о -10 ] ] [ g 1�11 0.86 0.5 -0.5 0.86 О 1.41 о о 1 О 8v = [3.52 - 7.80 - 17.1 l .91 0.51 0] 7 5.12) "Workspace Ьоuпdагу" апу angle set: {8 1 , О,, О} "Workspace interior" апу angle set such that: L1 + L 2 C2 + L 3 C23 = О (01 is arЬitrary) � 5.13) J..= O J T @O f. т [-L, S, - L2S12 L,C, + L2C12] [ 10] -= О -L,S12 L2C12 т, = -l0S1L1 - IOL2S12 т2 = -l0L2S12 5.14) So, at singular position: 03 = -ATAN2(dy , а3) lf а,--+ О, thcn 03 -+ -ATAN2(d1 , 0) which is -90'. Х2 а, J' У, 4113 +dy' 5.15) Tl1c ki11ernatics сап Ье done easily to obtai11: 0 Руана Uj = r а• (d2 + L2 + Lз)S1 [ -(d1 + Lб + Lз)С, ] PvoRGX --д0-, = O <) pYORGY д0, д0 PvoRCZ ае, ° 8 1'voRGX а0 PyoRGY 0 a PvoRCZ дО, So, 5,16) You're iп luck! The aпswer is giveп Ьу Equatioп (5.42): 1 W= [� � �i:] [!:] 5.17) For а prismatic joint i. the motion of the tool due to the joint is: 0W; =О Now, so, J(IJ) = [�� _ В, 1 1 i _ _л_,_; __л_,_ -:- -�•: В2 : Вз : В, __ л_ ,_ -!- __л _• ] 85 , 8 , Where cach Ai or Bi is а З х 1 vector givcn Ьу: А·= I 8;= °2;@ (0 Р,�1 - о Piorg) if i revolute 0z; if i linear . о if i revolute i; if; linear 5.18) Only the 4т N column of gт matters. Just dircctly differenliate it w.r.t. 0: 5.19) Ву partiaI differentiation: :. Singular when d, = О 5.20) At а singularily an N-DOF robot only has N-1 dof rcmaining. Hence, it can move freely in some N- l dimensional subspacc. Howe.ver, it is stШ truc that it has п joints. Therefore, we have а device which has опе more joint than the dimcn­ sionality of the sрзсе jt's desclibed in-and rhat's what а redundant manipulator is. Chapter 6 Manipulator Dynamics Exercises 6.1) Use (6.17), but written in polar form is easier. For exainple, /zz: Tzz = J H/Z1Lт1R -Н/2 О Х = Rcos8, fzz = J H/2 О У= -U/21.О (Х' + У2)рг dr d0dZ RsinO, Х 2 2" 1.• О +У'= R 2 (r 2 ) pr3 drdOdZ Tzz = ,r R 4 Нр, Voluшe = "' 2 Н 2 :. Mass = М = р,rг 2 Н :.1 fzz = ½М R 2 1 Similarly (only harder) is: l lxx = lrr = �м R' + �м Н' 1 From symmctry (or integration) 1 lxr = lxz = lrz = О 1 [�мR 2 +2-мн 2 4 12 :. CJ = Q о 6.2) о �МR 2 +2-мн' 4 12 о с, Т given Ьу (6.27).1 с, /zz; = �(Ц + W;') 1 Othcr moments will not mattcr. Derivation follows that of section 6. 7 quitc closcly. The aлswer is: c'w, <8> 1Р2) + 1V,) 2 v, = ;R(ii,z, ® L,x, +0,z, ® (0,z, ® L,x,> + 8z,> '\i2 = jR(L,ii,Y, - L,д;Х, +gZ,) [ с2 О S2] [-L, il;] = -Sь -� сб L�ii, 2 2 \i, = [-L,c,il; + S,g, L,s,il/ + C,g, -L,ii, ] т Vc, = 2W2 <8>'Рс, + 2Wz <8> (21V2 <8> 'Ре,)+ 2 V2 <8> ([ �;;:] <8> [ 2 Lg]) ] Vc. = [ L�ii, • -L,c,ii, + L,s,iJ,i!, +'\i, + [i:::] 0, <8> [-(L,c, + L2C;)iJ; - L,iJf + s,g] 'vc, = (L,s, + L,s,c,>iJI__ + L,ii, +__с,к 2L,S,0,0, - L,0, - L,c,o, lneniaJ Equations: Link / Link 2 [ L�il, -L,c,01 ] + '\i, 6.5) (Co11tim,ed) 2 f2 = [-M,(L, + L2C2JC2lil- м,�,дf + м,s,с] M2(L, + L2C_2)S2fll + ":2L,ё, + м,с,,с 2м,L,See,0,- м,�,е, - м,�2с,в, Backwanl Force lte1·a1ions: Link 2 Link / 1 F, = lR' F, + 'F, (not needed, so skip) 6.5) (Continued) Joim Torque Equations r1 = i п1 - 1 i, + Vi0; Added viscous frictio11 term. = (M1L; + M2(L1 + L2C2) т, - 2(L, 2 )ii, + L,C,)M,L,s,il,il, + v,il, ,, = м,Цii, + (L, + L2C2)M2L,S,ilf + M,gL,c, + v,il, or т = M(0)ii + V(0, В)+ G(0) \vhere: - 2 М(В) _ [ (M,L; + M2(L, + L2 C2) ) о м,ц О ] 6.6) The jacoЬian written in frame {О} is: OJ(B) = [-L1 S1 - L2S12 -L,S12] L,C" L,c, +L,c" (5_67) The invcrsc: L2 S12 -L,s, - L,s,, And the timc dcrivativc of O J (0): Тhе rcsl is messy, but follows from (6.99): Vx (B,0) = 0 1-т (B)(V(B,il) - М(В) Gx(0) = 0г7 (0)G(0) 0 1- 1 (fJ)° j(0)B) ] 6.7) То describe the state (0, 19, ii) of а si11gle joi11t \Vhen ea.ch dimension or state is quantized to 16 "Ыns" requircs 16 х 16 х 16 addresses. But \Vhen joint Oi is in any one of thesc "Ьins", joint Oi might Ье in 16 х 16 х 16 different states. So. for general r = /(0, 6, 0)for 3 joints Wc would 11ccd (16 х 16 х 16) 3 locatio11s. Eacl, Jocatio11 stores threc values (r1, r2, rз), so mcmory size required is: 13(16 х 16 х 16) 3 locations 1 6.8) For this manipulator: от1 - 0 \io [с,о s, - S, с, о о = gZoWo = Wo = 0/з = 113 =0 Fonvard Ve/. & Асс. lteralions: Link 1 1 \V1 = [О О ii, ]' 1 \i, = [О О g ]' Li11k 2 (Prismatic) 'W,=;R'W,=[0 -ii, 01' 6.8) (Co11tinued) /nertial Equations: Li11k 1 'N, = [О /zz, О ё, J' Link 2 . 2 Fz=Mz Vcz 2 = -2M2d2fi, - M2d20, ] [ -!tf2g .. -M2d20 1 + M,d, 2N2= c•1,'w,+ 2 W2@ c'l, 2 W2=[0 О O] r Backward Force Iteralions: Link 2 2 /2 = �R 3 Fз + 2F2 = 2F1 ( sec ) last page Uпk 1 1/1 = )R 2 F + 1 F (not needed, so sldp) 2 1 111 = 1 [ о ] + [о] ([' о о][-2м О /zz,0, d2 О @ О О О -1 1 О r1 0 2 2 1 -M2g -M2d20t -M2 d2 01 +м,d, ]) 6.8) (Co11tinued) 6.9) Vcry similar to 6.8. Assume the mass or link З is Мз although this ,vns not stated. 6.10) First, find 2V2: 2 V,= 2 :. J(fJ) = '1-'(fJ) J, [-d, 01] = [-Оd, 01] [d0,,] [-d, о] о 1 =[ -t, �] 'j(fJ) =[ -* gJ Mx(fJ) = 1-т M(0)J-t Мх(0) = [-t, �] [ /zz, "о M,d} i,] Vx(01 0) = J-T(V(fJ1 0)- M(fJ)J-1 jl)) = J-'V(010) - Mx(0)HJ [-t, �] 6.11) Ву gear ratio of )00, I mcan that the motor spins 100 times faster than thc link. The torque оп the Iink is relatcd �о the accel­ eration of the link Ьу: ч = lzz1 WL the gears incrcasc torque Ьу 100, so the motor torque, -см, is; -см= ___.!.__-cL; the motor асс. is Wм = 1ooiVL. 100 1 2 . ) Comblning. we have: rм 1()() /zzt Wм =( Hcncc, inertia appears to motor as: с�) 2 /zz1 6.12} 0, (1) = Bt + С12, so 0, = В+ 2CI, 0 = 2С so, 'Vc, = [2�]@ [g] + [Л@ ([Л@ [g]) = [4�] гgof ] + . [-2(8 + 2С1)2] 1 4С Vc, = 6.14) Tl1e bogus tcrms are: 6. 15) Wcll. lhe ans,ver is lhe same as that shown in Examplc 6.5. 6.16) Upon inspection, this proЬlem quite simple. ** whereO = [�] then: 6.17) See refcrencc [22] from chapter 10, and thc papcrs that it referenccs. See also example 10.6. 6.18) Any reasonaЬle F(0, 6) рrоЬаЫу has the prop­ erty that the friction forcc (or torque) on joint i depends only оп the ve!ocity of joint i, i.e. Also, each f;(0,6;) should Ье "passive", i.e. thc fuпction sl1ou]d lic in the pt & 3rd quadrants. •• solution written Ьу candle light in aftem1ath of 7.0-carthquakc, ! 7.oct.891