Solution Manual for Introduction to Robotics: Mechanics and Control, 4th Edition

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Solution Manual for Introduction to Robotics: Mechanics and Control, 4th Edition - Page 1 preview image

Chapter 1IntroductionExercises1.1)Here's just an example of a reasonable response:(ref. [8] in Chap. 1)1955Denavit & Hartenberg developedmethodology for describing linkages.I961George Devol patents design ofrst robot.1961First unimate robot installed.1968Shakey Robot developed at S.R.I.1975Robot institute of America formed.1975Unimation becomesrst Robot Co. to beprotable.1978First Puma Robot shipped to GM.1985Total U.S. market exceeds 500 milliondollars (annual revenue).Developments might be split into a technical listand a business list.1.2)(Based on 1981 numbers)Source:L. Conigliaro, "robotics presentation, institutionalinvestors conf.", May 28, 1981, Bache Newsletter81-249.:• Other - 32.2%• Palletize/Packaging - 2.8%• Machining - 6.8%• Material Handling - 11.8%[ '|Assembly - 22.4%• Welding - 23.9%1.3)$60$50$40$30$20$10$0People Are FIeXible,But More Expensive Every Year1999199719741981U.S. Automotive19881995

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1.4)Kinematicsis the study of position and derivativesof position without regard to forces which causethe motion.Workspaceis the locus of positionsand orientations achievable by the end-effector ofa manipulator.Trajectoryis a time based functionwhich specifies the position (and higher derivatives) of the robot mechanism for any value oftime.1.5)Frameis a coordinate system, usually specified inposition and orientation relative to some imbedding frame.Degrees of freedomis the numberof independent variables which must be specifiedin order to completely locate all members of a(rigid-body) mechanism.Position controlimpliesthe use of a control system, usually in a closedloop manner, to control the position of one ormore moving bodies.1.6)Force controlis the use of (usually closed-loop)algorithms to control the forces of contact generated when a robot touches its work environment.Arobot programming languageis a programminglanguage intended for use in specifying manipulator actions.1.7)Structural stiffnessis the "K" inF =KX(A.K.A"Hooke's law") which describes the rigidity ofsome structure.Nonlinear controlrefers to a closedloop control system in which either the systemto be controlled, or the control algorithm itself isnonlinear in nature.Off line programmingis theprocess of creating a program for a device withoutaccess to that device.1.8)See references. For example, in 1985 average laborcosts of $15 to $20 per hour are reasonable (dependinghow fringe benefits are calculated).1.9)Obviously it has increased dramatically. Recently(1988-1990) the ratio doubles or even triples eachyear.1.10)See Figure 1.3, but use latest data you can find.2

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Chapter1SolutionsforIntroductiontoRobotics1.11dothefollowingseventimes{playerCounter=1dothefollowingfourtimes{opengrippermovetoP_deckclosegrippermovetoP_playerCounterplayerCounter=playerCounter+1}//endfour−timesloop}//endseven−timesloop2.Mechanicalmanipulators:weldingrobotsonautomotiveassemblylines,wafer-handlingrobotsinsemiconductormanufacturing,parallel-platformrobotsforflightsimulatorsFixedautomationmachines:containerfillingatbottlingplant;automaticcarwash;printing,cutting,andfoldingofnewspapers3.Arigidbodyinspacehassixdegreesoffreedom.It’sfreetotranslateinthreedirectionsandtoindependentlyrotateabouteachofthosethreeaxes.1.14.AP3=sin(π/6)[||315]||+cos(π/3)[||269]||=[||2∙53∙57∙0]||5.BelowaresomepossibleconsiderationswhenusingmotorsatjointsPros:•Simpledesign•LowmaintenancerequirementCons:•Moremovingmass→largermotorsrequired•Greaterinertiaeffects

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Chapter 2Spatial Descriptions and TransformationsExercises2.1)R=rot(x, cf>)rot(z, 0)[ 10=0Ccp0Sep1000C(f>-S<t>os<t>c<t>CO-SO050CO0001[C0-S00]=Ccf>S0CcpC0-SepScf>S0ScpC0Ccp2.2)R=rot(x, 45°)rot(y, 30°)= [10.7007-.701]6601·50]0.707.707-.50.866[ -80[.8660.5]=.353.707-.612-.353.707.6122.3)Since rotations are performed about axes of theframe being rotated, these are Euler-Angle stylerotations:R=rot(z, 0)rot(x, cf>)We might also use the following reasoning:ABR(0, cf>)= BAR-1(0, cf>)=[rot(x, -cf>)rot(z, -0W1=roi-1(z, -0)roi-1(x,-)=rot(z, 0)rot(x, cf>)Yet another way of viewing the same operation:1st rotate byrot(z, 0)2nd rotate byrot(z, 0)rot(x, q,)roi-1(z, 0)3

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2.3)(continued)(This is a similarity transform)Composing these two rotations:=rot(z, 0) rot(x, q,)roe1(z, 0) · rot(z, 0)=rot(z, 0) rot(x, q,)[c=s-S00][1c000100-0s ]Cq,Sq,Cq,[c-S0Cq,S0Sq,]=sC0Cq,-C0Sq,Sq,Cq,2.4)This is the same as 2.3 only with numbers.R=rot(z,30°)rot(x,45)[.866-.353.353]=.50.612-.6120.707.7072.5)IfV;is an eigenvector ofR,thenRV;=7V;Ifthe eigenvalue associated withV;is 1, thenRV;= V;Hence the vector is not changed by the rotationR.SoV;is the axis of rotation.2.6)Imagine a frame {A} whosezaxis is aligned withthe directionk:Then, the rotation with rotates vectors aboutkby0degrees could be written:R = %Rrot(Az,0)*R[1]We write the description of {A} in {U} as:[ADKx]UAR=BEKyCFKzIfwe multiply out Eq. [1] above, and then simplifyusingA2+B2+C2=1, D2+E2+F2=1,[ABC]·[DEF]=0,[ABC]© [DEF]= [KxKyKz]we arriveat Eq. (2.80) in the book. Also, see[R. Paul]*page 25.Axuz,AZ.i\YzAy4

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2.7)Let[RnR12R13]R=R21R22R23R31R32R33(1)ComputeR11+R22+R33=N(2)If N=3, then(N- 1)0=Acos-2-=°.Since0rotation is zero,Kis arbitrary.( 3 )If N=-1, then0=Acos(-1)=180°.Inthis case Eq.(2.80) becomes:[2K'- I2KxKy2K,K, ]rot(K,1 8°)=2I<xKy2K2- 12KyKzy2KxKz2KyKz2K;- 10so:2K; -1=R11 =>Kx=±.J(R11+1)/22KxKy=R21 =>Ky=R12/2Kx2KxKz=R31 =>K,=R3t/2KxHowever, ifKx≅0,then this is ill-defined,so use a different column for solution (notthe first column as above).(4)If -1 < N <3(so thatO < 0 <1 8°)the0use Eq.(2.82) in book.2.8)Procedure RMTOAA is given essentially in thesolution to 2.7.However, writing clean code tocheck the various cases is a good exercise in itself.Procedure AATORM is given by Eq.(2.80) and iseasy.2.9)The subroutines encode Eq.(2.64) and equations(2.66), (2.67), and (2.68).2.10)The subroutines follow from eq.(2.2) and equations7(2.74), (2.7 5), and (2.7 6).2.11)When they represent rotations about the sameaxis.2.12)Velocity is a "free vector" and only will be affectedby rotation, and not by translation:[0.866 -0.5] [;1]Ay = ABRBV =0.50.86600001000AV = [ -1.3422.3230.0 )75

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2.13)By just following arrows, and reversing (by inversion) where needed, we have:BCT=BATUAy-1 CUy-1Inverting a transform is done using eq. (2.40) inbook. Rest is boring.2.14)This rotation can be written as:ABT =trans^P,\AP\)rot(K, 0)trans(-APt\AP\)Whererot(K,0)is written as in eq. (2.77),Andtrans(AP,\AP\) =100Px010Py001Pz0001Andtran,(-AP,IAPI)=|0 0r iO0-P,l010-Py0°!-^oOoi JMultiplying out we get:•R\\R\2R\3Qx~/?21/?22/?23QyAT=R3lR32/?3362.0001 -where theRijare given be eq. (2.77). And:ex=px- p,(^vd + c^) -py(KxKyVe- A:2s0)-Pz(KxKzV0+KyS0)6Qy=Py-Px(KxKyV0+KzS0)-Py(K2 yV0+C0)-Pz(KyKzV0+KxS0)M)(U](B][C]Cz =PZ- PX(KXKZV9-KySO)-Py(KyKzV9+ ^5^)-Pi(tf!V0 + C0)

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2.15)Recall thatx2x3ex=1+x+ - + - + ...2!3!11so eK0=I+ K0+-K202+-K303+ · · ·2!3!KxKy-K2 x- K2 zKyKzK2=--Kl-KlKXKy.KXK,KXK,-KyKz-Kx-Ky_Writing out the (1,2) element ofeK0(as an example)we have:K0l2l3(e)1 2=0+(-Kz)0+-(KxKy)0+-K20+ · · ·'2!3!Recall that:035sin 0=0 - -+ - - · · ·3!5!0204cos 0=1 - -+ - -··· .2!4!020406v=1 -c= - - -+ - - ...2!4!6!We can write:K0l3l5(e)12=(-K )0+-K20--K20 +···'z3!5!1214+-(K K)θ- -(K K)+ · · ·2!Xy4XyOr:(eKe)\,2= -K2S6+KxKyVeWhich is the value given in (2.80).Other elementsmay be checked similarly.7

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2.16)In method1, the multiplication of two 3 x 3matrices requires 27 multiplications and 18 additions. the computation ofABRtakes two matrixmultiplies, or 57 multiplications, 36 additions thecomputation ofABRBPis 9 mult. and 6 additions.Hence, in one second this method will require:30 x computation ofA DR= 30 x 54 mult.30 x 36 add.100 x computation of Ap= 100 x 9 mult.100 x 6 add.Total= 2520 mult., 1680 Add.In method two, computation ofCDRDPrequires9 mult. and 6 add.; likewise the computation ofBCR cPandABRBP,for a total of 27 mult. and18 add. These must occur 100 times/sec., so inone second we have:27 x 100 mult.= 2700 mult.18 x 100 add.= 1800 add.Therefore, method 1 is superior, but not by much.2.17)|Px||Rcosθ|Ap=Py Pz=Rsin Znθ2.18)[Px] [Rcos ct cosβ|AP=Py=R sinα· cosPzR sm2.19)In thez-y-zEuler Angle set, the first rotation is:R1=rot(z,ct)The second rotation expressed in fixed coordinates is:R2=rot(z,ct)rot(y,) roi-1(z, ct)The third is:R3 =(R2Ri)rol(z,y)(R2RlrtThe result is:R = RiR2Ri =rot(z, a)rol(y, ft)rot(z, x)Which can be multiplied out to give the result of(2.72).8

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2.20)This is easily derived if you work backwards.i.e, substitute into Rodriquez's formula whereverK®QorK·Qoccur, collect terms, and you'llget (2.80).2.21)Just use the given approximations in (2.80) toobtain:-Kz801Kx80riRK(S9) =KZS0l-KySeKYsei-KxseMore on this in Chapter 5.2.22)So, givenR1=R1(a)andR2=RK(/3)witha«1 andf3«1;showR1R2=R2R1ifwe form theproductR1R2and usea/3 2:: 0we have:-]za - Kz/31Jxα+Kx/3Jya+Kr/3]-lxa1Kx/3riRiR2=\Jza + KzPl-Jya-KypWe see thatjandk,as well asaandf3appearsymmetrically, soR1R2=R2R1.2.23)By definitionuPAoRG=uP1.Next, the vectorfromP1TOP2is a vector along the positive xaxis, so:uXA=uP2- uP1,which normalizedis:uxA=xA/IIUXAII-uNowuV=uP3- uP1.A vector parallel to thepositiveYA-axis,can be formed using the GramSchmidt orthogonalization:VYA="V-("V."XA)"XAandUYA=UYA/\UYA\Finally, the unit vectoruZAcan be found by asimple cross-product:UZA= "XA®VYANow:9„r"^"Y*UZA"pMrb-\^=LoooiJ

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2.24)This is a bit tricky here}s most of it:AP=ABRBPdoesn}t change Iength, so.'.Ap.Ap_Bp,Bp=0(A p_BP).(A p+BP)=0}}FG:. F┴ GF=(ABR-1)BPG=(ABR+I)BPBP=(ABR+l)-1G(You can showABR+Inon-shawar):. F=(ABR-I)(ABR+1)-1 GBso,F=BG(Now, one can show that ifX·Y=0 and ifX=AYthen a is skew-symmetric):. B∈ skew-sym matrices(jR-I)=B(jR+I)(I-B)jR=I+ B(now can show (1-B) non-show)IjR=(I-B)-1(1+B)IQ.E.D.2.25)Def. of e-vaI:λiV;=RV;from physicaI insightwe knowV;=RV;whenViis the axis of rota-tion,:. | λ1=1 Ifrom (2.80) one can show (withsome work) that det)(R=1. From Iinear aIgebrawe have:∑iλi=trace(r) πiλi=det(r)/. XiXzA.3 = 1 or A.2^3 = 1[1]And compute the trace(r) from (2.80) to getλ2λ1or λ2λ1++3=+2cosθ+3=2cosθ[2]Now soIve [1] & [2] above forλ2andλ3to get1λ2+ - =2cosθ;λ2λ22 -2cosθ2+1=0λ2=cosθ±√cos2θ-1λ2=cosθ±isinθ:.λ1=1λ2=cos+i=iθθsinθeλ3=cosθ -isinθ=e-iθQ.E.D.10

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2.26)Somebody please send me a simple proof of this.(For any given Euler convention it is not hard.)2.27)[-]003001]ABT=000-1001002.28)tr-[-0010-0.50.8660.8660.500003021]2.29)BCT=0.5-0.866-0.866-0.500000021][-00102.30)0-1C-0.5[0AT=008660.8660-0.50002]3 * 0.5-3 *i°"8662.31)[-]000421]ABT=0000-1-10002.32)|08660.50ACT=O.500-0.86600-100-3421]2.33)[-0.866-0.50BCT=0+10.8660003001]-00 .5 02.34)|08660.50CAT=O.500-0.86600-100-3·.86+2]-4*.8021- 1.511

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2.35)Any R can be given:R=Rx(a)Rr(f3)Rz(y)anddet(R)=det(Rx(a)) · det(Rr(/3)) · det(Rz(y))Using the formulas for rotation about a principleaxis ((2.77) - (2.79)) its easy to showdet(Rx(a))=1Vadet(Ry(/3))=1V/3det(Rz(Y))=1Vy:.det(R)=1 · 1 · 1=1Q.E.D.2.36)From Cayley's formula, the number of parameters needed to specify a rotation is the numberof free parameters in an NXN skew symmetricmatrix, which is½(N2-N).The translationaldegrees of freedom are, of course n, so total is:dof(N)=N+ ½(N2-N)=½(N2+N)Q.E.D.2.37)Form (2,4) element of-ABRTA P00,bTo get: -6.42.38)V{ V2=cos 0,Rpreserves angles, so,12(RVi)T(RVt)=VTv2VfRTRV2= V,rV2/.RTR = I=*.RT=R~l

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2.39)21By=-(1+ru+r22+r33)yB; > epsilon?YesBy=faBJ=(r23 - r32)/4ByBz=(r31 - r13)/4ByB3=(rJ2 - r2J)/4ByNoBy=0Bi=-1/2(r22+r33)Bi > epsilon?YesBJ=[eiiB2=rJ2/2BJB3=r13/2BJNoBJ=0Bi=½(1 -r33)B? > epsilon?Yes-√ɛ22Bz -BzB3=rz3/2BzNoBz=0B3=12.40)Similar to algorithms of Section 2.8.2.41)Similar to algorithms of Section 2.8.13

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Chapter2SolutionsforIntroductiontoRobotics1.a)Use(2.3)toobtainABR=[||1000010−10]||b)Use(2.74)togetα=90degreesβ=90degreesγ=−90degrees2.a)Use(2.64)toobtainABR=[||·330−·770·547·908·418·0396−·259·483·837]||b)Answeristhesameasin(a)accordingto(2.71)3.Use(2.19)toobtainthetransformationmatrices.TherotationisX-Y-Zfixedangles,souse(2.64)forthat3×3submatrix,withanglesγ=0degreesβ=−sin−1tripod_heightdistance_along_optical_axis=−sin−11·55=−107degrees()()αC=0degreesαD=120degreesαE=240degreesThepositionvectorstothecamera-frameoriginsare

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