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

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    Chapter 1 Introduction Exercises 1.1) Here's just an example of a reasonable response: (ref. [8] in Chap. 1) 1955 Denavit & Hartenberg developed methodology for describing linkages. I 961 George Devol patents design of rst robot. 1961 First unimate robot installed. 1968 Shakey Robot developed at S.R.I. 1975 Robot institute of America formed. 1975 Unimation becomes rst Robot Co. to be protable. 1978 First Puma Robot shipped to GM. 1985 Total U.S. market exceeds 500 million dollars (annual revenue). Developments might be split into a technical list and a business list. 1.2) (Based on 1981 numbers) Source: L. Conigliaro, "robotics presentation, institutional investors conf.", May 28, 1981, Bache Newsletter 81-249.: • 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 $0 People Are FIeXible, But More Expensive Every Year 1999 1997 1974 1981 U.S. Automotive 1988 1995

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    1.4) Kinematics is the study of position and derivatives of position without regard to forces which cause the motion. Workspace is the locus of positions and orientations achievable by the end-effector of a manipulator. Trajectory is a time based function which specifies the position (and higher deriva tives) of the robot mechanism for any value of time. 1.5) Frame is a coordinate system, usually specified in position and orientation relative to some imbed ding frame. Degrees of freedom is the number of independent variables which must be specified in order to completely locate all members of a (rigid-body) mechanism. Position control implies the use of a control system, usually in a closed loop manner, to control the position of one or more moving bodies. 1.6) Force control is the use of (usually closed-loop) algorithms to control the forces of contact gener ated when a robot touches its work environment. A robot programming language is a programming language intended for use in specifying manipu lator actions. 1.7) Structural stiffness is the "K" in F = K X (A.K.A "Hooke's law") which describes the rigidity of some structure. Nonlinear control refers to a closed loop control system in which either the system to be controlled, or the control algorithm itself is nonlinear in nature. Off line programming is the process of creating a program for a device without access to that device. 1.8) See references. For example, in 1985 average labor costs of $15 to $20 per hour are reasonable (depending how fringe benefits are calculated). 1.9) Obviously it has increased dramatically. Recently (1988-1990) the ratio doubles or even triples each year. 1.10) See Figure 1.3, but use latest data you can find. 2

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    Chapter 1 Solutions for Intro duction to Rob otics 1. 1 1 do t h e f o l l o w i n g s e v e n t i m e s { p l a y e r C o u n t e r = 1 d o t h e f o l l o w i n g f o u r t i m e s { o p e n g r i p p e r m o v e t o P _ d e c k c l o s e g r i p p e r m o v e t o P _ p l a y e r C o u n t e r p l a y e r C o u n t e r = p l a y e r C o u n t e r + 1 } / / e n d f o u r − t i m e s l o o p } / / e n d s e v e n − t i m e s l o o p 2. Mechanical manipulators: welding robots on automotive assembly lines, wafer-handling robots in semiconductor manufacturing, parallel-platform robots for flight simulators Fixed automation machines: container filling at bottling plant; automatic car wash; printing, cutting, and folding of newspapers 3. A rigid body in space has six degrees of freedom . It’s free to translate in three directions and to independently rotate about each of those three axes. 1.14 . A P 3 = sin ( π / 6) [ | | 3 1 5 ] | | + cos ( π / 3) [ | | 2 6 9 ] | | = [ | | 2 ∙ 5 3 ∙ 5 7 ∙ 0 ] | | 5. Below are some possible considerations when using motors at joints Pros: • Simple design • Low maintenance requirement Cons: • More moving mass → larger motors required • Greater inertia effects

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    Chapter 2 Spatial Descriptions and Transformations Exercises 2.1) R = rot(x, cf>)rot(z, 0) [ 1 0 = 0 Ccp 0 Sep 1 0 0 0 C(f> -S<t> o s<t> c<t> CO -SO 0 50 CO 0 0 0 1 [ C0 -S0 0 ] = Ccf>S0 CcpC0 -Sep Scf>S0 ScpC0 Ccp 2.2) R = rot(x, 45 ° )rot(y, 30 ° ) = [10 .7007 -.701] 66 01 · 5 0 ] 0 .707 .707 -.5 0 .8 66 [ -80 [ .8 66 0 .5 ] = .353 .707 -. 6 12 -.353 .707 . 6 12 2.3) Since rotations are performed about axes of the frame being rotated, these are Euler-Angle style rotations: 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, -0W 1 = roi- 1 (z, -0)roi- 1 (x, - ) = rot(z, 0)rot(x, cf>) Yet another way of viewing the same operation: 1st rotate by rot(z, 0) 2nd rotate by rot(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 -S0 0 ][ 1 c 0 0 0 1 0 0 -0s ] Cq, Sq, Cq, [c -S0Cq, S0Sq, ] = s C0Cq, -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 -.612 0 .707 .707 2.5) If V; is an eigenvector of R, then RV; = 7V; If the eigenvalue associated with V; is 1, then RV;= V; Hence the vector is not changed by the rotation R. So V; is the axis of rotation. 2.6) Imagine a frame {A} whose z axis is aligned with the direction k: Then, the rotation with rotates vectors about k by 0 degrees could be written: R = %Rrot( A z,0)*R [1] We write the description of {A} in {U} as: [A D K x ] UAR = B E K y C F Kz If we multiply out Eq. [1] above, and then simplify using A 2 + B 2 + C 2 = 1, D 2 + E 2 + F 2 = 1, [ABC]· [DEF]= 0, [ABC]© [DEF]= [K x K y Kz] we arrive at Eq. (2.80) in the book. Also, see [R. Paul]* page 25. Ax uz,AZ.i \ Yz Ay 4

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    2.7) Let [ Rn R12 R13 ] R = R21 R22 R23 R31 R32 R33 (1) Compute R11 + R22 + R33 = N (2) If N = 3 , then ( N- 1 ) 0 = Acos - 2 - = ° .Since 0 rotation is zero, K is arbitrary. ( 3 ) If N = -1, then 0 = Acos(-1) = 1 80 ° .In this case Eq. (2.80) becomes: [2K'- I 2K x K y 2 K , K , ] rot(K, 1 8 ° ) = 2I<xK y 2K 2 - 1 2K y K z y 2K x K z 2K y K z 2K;- 1 0 so: 2K; - 1 = R11 =>K x = ±.J(R11 + 1)/2 2K x K y = R21 =>K y = R12/2K x 2K x K z = R31 =>K, = R3t/2K x However, if K x ≅ 0 , then this is ill-defined, so use a different column for solution (not the first column as above). (4) If -1 < N < 3 (so that O < 0 < 1 8 ° ) the 0 use Eq.(2.8 2) in book. 2.8) Procedure RMTOAA is given essentially in the solution to 2.7.However, writing clean code to check the various cases is a good exercise in itself. Procedure AATORM is given by Eq. (2.80) and is easy. 2.9) The subroutines encode Eq. (2.6 4) and equations (2.66 ), (2.67), and (2.68 ). 2.10) The subroutines follow from eq.(2.2) and equations 7 (2.74), (2.7 5), and (2.7 6 ). 2.11) When they represent rotations about the same axis. 2.12) Velocity is a "free vector" and only will be affected by rotation, and not by translation: [ 0 .866 -0.5 ] [ ; 1 ] Ay = ABR B V = 0 .5 0 .866 0 0 0 01 0 0 0 AV = [ -1.34 22.32 3 0 .0 ) 7 5

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    2.13) By just following arrows, and reversing (by inver sion) where needed, we have: BCT = BAT UAy-1 CUy-1 Inverting a transform is done using eq. (2.40) in book. Rest is boring. 2.14) This rotation can be written as: A B T = trans^P, \ A P\)rot(K, 0)trans(- A P t \ A P\) Where rot(K, 0) is written as in eq. (2.77), And trans( A P, \ A P\) = 1 0 0 P x 0 1 0 P y 0 0 1 P z 0 0 0 1 And tran,(-AP, I A PI)=|0 0 r i O 0 -P,l 0 1 0 -P y 0° ! -^ oO o i J Multiplying out we get: • R\\ R\2 R\3 Q x ~ /?21 /?22 /?23 Qy A T= R 3l R 32 /?33 62 . 0 0 0 1 - where the R ij are given be eq. (2.77). And: e x = p x - p,(^vd + c^) - p y (K x KyVe - A: 2 s0) - Pz(K x Kz V0 + K y S0) 6 Q y = P y - Px(K x K y V0 + KzS0) - P y (K2 yV0 + C0) - Pz(K y KzV0 + K x S0) M) (U] (B] [C] Cz = PZ - P X (K X K Z V9 - K y SO) - P y (K y K z V9 + ^5^) -Pi(tf!V0 + C0)

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    2.15) Recall that x 2 x 3 e x =1 + x + - + - + ... 2! 3! 1 1 so e K0 =I+ K0 + -K 2 0 2 + -K 3 0 3 + · · · 2! 3! K x K y -K2 x- K2 z K y K z K 2 = --Kl-Kl K X Ky . K X K, K X K, - KyK z -K x - K y _ Writing out the (1,2) element of e K0 (as an example) we have: K0 l 2 l 3 (e )1 2 = 0 + (-K z )0 + -(K x K y )0 + -K 2 0 + · · · ' 2! 3! Recall that: 0 3 5 sin 0 = 0 - - + - - · · · 3! 5! 0 2 04 cos 0 = 1 - -+ - -··· . 2! 4! 0 2 04 06 v = 1 - c = - - - + - - ... 2! 4! 6! We can write: K0 l 3 l 5 (e )1 2 =(-K )0+-K20 --K20 +··· ' z 3! 5! 1 2 1 4 + -(K K )θ - - (K K ) + · · · 2! X y 4 X y Or: (e Ke )\,2 = -K 2 S6 + K x K y Ve Which is the value given in (2.80). Other elements may be checked similarly. 7

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    2.16) In method 1, the multiplication of two 3 x 3 matrices requires 27 multiplications and 18 addi tions. the computation of ABR takes two matrix multiplies, or 57 multiplications, 36 additions the computation of ABR B P is 9 mult. and 6 additions. Hence, in one second this method will require: 30 x computation of A DR = 30 x 54 mult. 30 x 36 add. 100 x computation of A p = 100 x 9 mult. 100 x 6 add. Total= 2520 mult., 1680 Add. In method two, computation of CDR DP requires 9 mult. and 6 add.; likewise the computation of BCR cP and ABR B P, for a total of 27 mult. and 18 add. These must occur 100 times/sec., so in one 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 = R sin Z nθ 2.18) [ Px ] [ R cos ct cos β | AP = P y = R sin α· cos Pz R sm 2.19) In the z-y-z Euler Angle set, the first rotation is: R1 = rot(z, ct) The second rotation expressed in fixed coordi nates is: R 2 = rot(z, ct) rot(y, ) roi- 1 (z, ct) The third is: R3 = (R 2 Ri)rol(z,y)(R 2 R l r t The result is: R = RiR 2 Ri = 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 wherever K ® Q or K · Q occur, collect terms, and you'll get (2.80). 2.21) Just use the given approximations in (2.80) to obtain: -Kz80 1 Kx80 r i R K (S9) = K Z S0 l-KySe K Y se i -K x se More on this in Chapter 5. 2.22) So, given R 1 = R1(a) and R 2 = R K (/3) with a« 1 and f3 « 1; show R 1 R 2 = R 2 R 1 ifwe form the product R 1 R 2 and use a/3 2:: 0 we have: -]za - Kz/3 1 Jxα + Kx/3 Jy a + Kr/3 ] -lxa 1 Kx/3 r i RiR 2 =\ J z a + K z P l-Jya-Kyp We see that j and k, as well as a and f3 appear symmetrically, so R 1 R 2 = R 2 R1. 2.23) By definition u PAoRG = u P 1 . Next, the vector from P 1 TO P 2 is a vector along the positive x axis, so: u XA = u P 2 - u P 1 , which normalized is: uxA = xA/II U XAII- u Now u V = u P 3 - u P 1 . A vector parallel to the positive Y A -axis, can be formed using the Gram Schmidt orthogonalization: VY A = "V-("V."X A )"X A and U Y A = U Y A /\ U Y A \ Finally, the unit vector u Z A can be found by a simple cross-product: U Z A = "X A ®VY A Now: 9 „ r"^ "Y* U Z A "p Mrb -\ ^ = L o o o i J

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    2.24) This is a bit tricky here}s most of it: AP = ABR B P doesn}t change Iength, so .'. A p . A p _ B p , B p = 0 ( A p _ BP) . ( A p + BP) = 0 }} F G :. F ┴ G F = (ABR - 1) B P G = (ABR + I) B P B P = (ABR + l)- 1 G (You can show ABR + I non-shawar) :. F = (ABR - I)(ABR + 1)- 1 G B so, F = BG (Now, one can show that if X · Y = 0 and if X = AY then 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) I jR = (I - B)- 1 (1 + B) I Q.E.D. 2.25) Def. of e-vaI: λi V; = RV; from physicaI insight we know V; = RV; when Vi is the axis of rota- tion, :. | λ1 = 1 I from (2.80) one can show (with some work) that det ) (R = 1. From Iinear aIgebra we 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 get λ2 λ 1 or λ2 λ1 + + 3 = + 2cosθ + 3 = 2cosθ [2] Now soIve [1] & [2] above for λ 2 and λ 3 to get 1 λ2 + - = 2cosθ; λ2 λ22 - 2cosθ 2 + 1 = 0 λ2 = cosθ ± √cos 2 θ -1 λ2 = cosθ ± i sinθ :. λ1 = 1 λ2 = cos + i = iθ θ sinθ e λ 3 = cosθ - i sinθ = e- i θ Q.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) [-] 0 0 3001] AB T = 000 -1 0 0 1 0 0 2.28) tr-[-0010 -0.5 0.866 0.866 0.5 0 0 0 0 3021] 2.29) BCT = 0.5 -0.866 -0.866 -0.5 0 0 0 0 0021] [-0010 2.30) 0 -1 C -0.5 [ 0 A T = 0 0 866 0.866 0 -0.5 0 0 0 2 ] 3 * 0.5 -3 * i°" 866 2.31) [-] 0 0 0421] AB T = 000 0 -1 -1 0 0 0 2.32) | 0866 0.5 0 AC T = O.500 -0.866 0 0 -1 0 0 -3421] 2.33) [-0.866 -0.5 0 BC T = 0 +1 0.866 0 0 0 3001] - 00 .5 0 2.34) | 0866 0.5 0 CA T = O.500 -0.866 0 0 -1 0 0 -3·.86+2 ] -4*.8 021 - 1.5 11

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    2.35) Any R can be given: R = Rx(a)Rr(f3)Rz(y) and det(R) = det(Rx(a)) · det(Rr(/3)) · det(Rz(y)) Using the formulas for rotation about a principle axis ((2.77) - (2.79)) its easy to show det(R x (a)) = 1 Va det(R y (/3)) = 1 V/3 det(Rz(Y)) = 1 Vy :. det(R) = 1 · 1 · 1 = 1 Q.E.D. 2.36) From Cayley's formula, the number of parame ters needed to specify a rotation is the number of free parameters in an NXN skew symmetric matrix, which is ½(N 2 - N). The translational degrees of freedom are, of course n, so total is: dof(N) = N + ½(N 2 - N) = ½(N 2 + N) Q.E.D. 2.37) Form (2,4) element of -ABR T A P00,b To get: -6.4 2.38) V{ V2 = cos 0, R preserves angles, so, 12 (RVi) T (RVt)=VTv 2 VfR T RV 2 = V, r V 2 /. R T R = I =*. R T = R~ l

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    2.39) 2 1 B y = -(1 + ru + r22 + r33) y B; > epsilon? Yes B y = fa BJ = (r23 - r32)/4B y Bz = (r31 - r13)/4B y B3 = (rJ2 - r2J)/4B y No B y = 0 Bi = -1 /2(r22 + r33) Bi > epsilon? Yes BJ = [eii B2 = rJ2/2BJ B3 = r13/2BJ No BJ = 0 Bi = ½(1 - r33) B? > epsilon? Yes -√ɛ22 Bz - Bz B3 = rz3/2Bz No Bz = 0 B3 = 1 2.40) Similar to algorithms of Section 2.8. 2.41) Similar to algorithms of Section 2.8. 13

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    Chapter 2 Solutions for Intro duction to Rob otics 1. a) Use (2.3) to obtain A B R = [ | | 1 0 0 0 0 1 0 − 1 0 ] | | b) Use (2.74) to get α = 90 degrees β = 90 degrees γ = − 90 degrees 2. a) Use (2.64) to obtain A B R = [ | | · 330 − · 770 · 547 · 908 · 418 · 0396 − · 259 · 483 · 837 ] | | b) Answer is the same as in (a) according to (2.71) 3. Use (2.19) to obtain the transformation matrices. The rotation is X-Y-Z fixed angles, so use (2.64) for that 3 × 3 submatrix, with angles γ = 0 degrees β = − sin − 1 trip o d_height distance_along_optical_axis = − sin − 1 1 · 5 5 = − 107 degrees ( ) ( ) α C = 0 degrees α D = 120 degrees α E = 240 degrees The p osition vectors to the camera-frame origins are

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