University:
Concordia University
Course:
ENGR 6411 | Robotic Manipulators I: Mechanics
Academic year:

2022
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488

Pages:

14
Author:

fynegafn8ps

Chapter 3 Manipu]ator Kinematics Exercises 3.1) о о о _ 2'т 0 ) d; а;-1 «i-1 о L, L, [С, s,0 -S, О О О О 'тт-"т'т - 1 2 3 where: L,] 1 Q + 02 + Оз) + 02 + Оз), О. а, > О(а; > О always) :}'' Chapter 4 Inverse Manjpu]ator Юnematics Exercises 4.1) 4.2) This рrоЫет сап have different solutions depending how it is interpreted. 1 intendcd that а goal is specificd which includes а desired orientation of the last link. ln this case, the solution is fairly easy. ir is givcn, so compute: t:,т =fтtт�т- 1 Now � Т =� Т which we write out as: R11 = [ R" от 3 Rз1 From thc solution of exercise 3 from chaptcr 3 wc have: Equatc elements (1. 3): S, = R13 Equate clements (2, 3): -С1 :. 101 = R23 = atan2(R1з, -R23) 1 lf both R13 = О and R,з = О the goal is unattainaЬJe. Equate elements (1, 4): Р, = C,(C2 L 2 + L,) Equate elements (2,4): Ру = S1(C2L2 + L1) 4.2) (Co111i11ued) lfC1 #Otl1en С2 1 Р, ( - - -L1 L2 С, = Else С, =..!.._ L, ) (!.z. - L,) S, Equate Elements (3,4): Р, = S2L2 Equate elements (3, 1): S23 = R31 Equate eleшents (3. 2): Сн so, [ е3 = Rз2 = atan2(R31, R"J - е, [ Ifboth R31 and R32 are zero, the goal is unattainaЬle. А second interpretatioп of thc proЫcm is that only а desired position is given (по orientation). ln this therc may Ье up 10 four solutions: Assumc3 Pioo1 = LзХз, theo First, S, = Р, L, + L2C2 + LзС13 so. [ е, = atan2(P,.,P,) or atan2(-P,,-P,) [ Since the sign of the "L1 + L2C2 + L3C'2J" term may Ье + or -. Next, define: --L, ifC1 #О 2 - L, { s, Р, And wc have: if S1 "# О а= �1 L2S2 + LзS,з = Р, Square and add these two equations to get: 1 2 +P2 -L2 -L 2) С3 =--(a 2 3 2L2Lз S3 = ±/1 - с;; ' [ Оз = atan2(Sз, Сз) [ 4.2) (Co111i11ued) Fiпally. LзС2з = « - L2C2 LзS2, = Р, - L2S2 so, 182 = atan2(P, - L,S,,a - L,C,)-831 4.3) А fi.:1t disk of zero thick:ness with inner radius equal to minimum extcnsion, and outer radiнs equal 10 rnaximum extcnsion. � 4.4) For an arm Iike this it is reasonable to specify а goal Ьу giving the desired х & у coordinates of the tip and а value for 0з, thc wrist roll. After transforming back 10 Рх & Рт of the wrisr (i.e. remove orrset due to Lз) the solutioп is simple and corresponds to thc conversion between cartesian and роlаг coordinates. 4.5) Tum the results of section 4.7 into а computer algorithm, plus check each solulion to sce if joiпts arc in range. 4,6) То derive the "nearest" solution, we would like to min.imize the rotation of cach joinl. Denote the starting angle of each joint as Bj ,.. (for j-rh joint), and 1he final position or eacl1 joint as 0j F• For each proposed solution, compure: S= r::., IBJF - 8 ,I 1 And choosc the ·'nearest" as the one which minimizes S. Sometimes а wcighting factor is used (to penalize motion of"large" join[s, forexample) апd so tbe score for each proposed solution is: S = I:f;, W;IB;, - B;,I 4.7) RcpeatabШty is affected Ьу: 1) Stepoy s1n1c crror is servo system 2) FlexiЬility of links 3) BackJash in gears 4) Looseness in Ьearings 5) Noise in sensor readings 6) Thermal effects Accuracy is affected Ьу all of the above, plus: 1) lmprecise knowledgc of D.H. Parameters. 4.8) There are an infinite number of solutioпs. Imagine fixing the last link in position and orieпtation: Thcn, the first 3 Li11ks form а "4-bar linkage" which сап take оп an infinity of positions sincc it has а dcgree of freedom. / 4.9) This is slightly trickier lhan it looks at first. Approxinшtely: 4.JO) This subspacc with Ье given in terms of an expres sion for �Т which js а function of three indepen dent vari.1Ыcs х, у, апd В. As these variaЫes run throughout their ranges, а subspace i.s swept out. А, in Example 4.2 the origin of the end-effector framc (hcre, {3}) must have zcro z-component. Also, Zз шust have по z-component and it's direc tio11 is given Ьу the coordinates of 0 PзoRG· So we have D "' Е fJ F О о о wherc: х а= j > x +y у ' andfJ= j ' x +y ' То find expressions for A,B,C,D,E,F as а func tion of iпdependcпt variaЫc, В, consider rotating the vector [001]' aЬout 23 (giveп Ьу [a/!OJ') Ьу amount 0. This serves as our expression for Х 3 and сап Ье Гound from eq. (2.80) Using К = (afJO)', and just taking the З'' column. This yields: kз = [-fJC/laCIISIIJ' Тhen we compute Уз = Zз 181 Хз to get оJ т _ [-{JC/1 аС/1 so о {JSO -aS/1 со о 4.10) (Co11tim,ed) where: х а = Jx 2 + у2 4.11) fJ = у Jх2 + у2 V must Ье rotated into alignmeпt with Zo : Oio =� R 2 (, s, -s,c, [о] [ с,с, о 1 - s,] [ v,] о v,. С, V, Tl1ere arc two solutions: 0, = ata112(-Jv} + V,', V,) О, = atan2(- V,, V,) And 01 = atan2(Jv; + v,;, Vz) О, = atan2(- V1 , V,) + 180' lf V, = V, = О then singu/ar and 02 is arbltrary. 4.12) Find: the set of unattзinaЫe orientatioпs So!шion: Starting with joint З. spinning it, frame 3 can attain any orientation aЬout its i 3 axis, pointiпg upward. The proЫem Lhus reduces to finding restrictions оп the pointing dircction of Zз. Next, spinning aЬout joint 2, Zз сап attain any orientation оп thc surface of а cone of арех angle 2ф: Spinning this ' conc" about joint 1, Zз сап atrain апу orientation except in а сопе shaped zone of арсх angle 360 - 4ф: 1 4.13) 1. Faster coшputatio11ally 2. Finds al/ solutions 3. Sometimes more accurate 4.14) No. Pieper's merhod gives the c]osed. foпn solution ror апу 3-00F manipulalor. (See his thesis for all the cases) 4.15) See references [8] and [9]. 4.16) Since Sa1 = О, easy to use Pieper's method: z = К, (see pages 129-131 of text) or 0 Р,, = са, Fз + Cet1d2 0Р4, = Fз 0 P4 = A3Sa2S3 + DзСа2 (othcr terшs are z�ro) z 1.707 = ./1. + )2 ./1. .П.тsз 2 l.707=Sз+l Sз = 1.707 - 1 = 0.707 since - 180 < 03 < 180, there are 2 sols: l 0з = 45° or 0з = 135° 1 4.17) Since r UJ =0, we use Piepcr's method with = Кз 3 = 3 + 2Fз :. Fз = О 0.707 s, + 0.707 = О Continuing оп yie]ds. 0 = [О, О, -90) 4.18) 2 4.19) 4 .-. s, =-1 .-. 03 = -90 4.20) 1 4.21) 2 4.22) 1 4.23) Just plug in the uш0/2 substitution and collect terms. You'll find you arrive at а quartic in U. You' 11 also see а condition in which the coefficients of U 3 and U become zero- so it Ьecomes quadratic in U 2 . 4.24) Revolшe «;-1 = atan2(-T2з, Тзз) 8;-1 = Т14 di = 2� + Т3� Jт Linear <>1-1 = atan2(-T23, Тз,) а;-1 = Т14 0; = atan2(-T12 , Т11 )