机器人学导论(克雷格)第二章作业答案本页仅作为文档封面，使用时可以删除This document is for reference only-rar21year.March2.1 solution:According to the equation of pure transition transformation,the new point after transition is as follows:100235010358(,,)0014711000111transx y z old P Trans d d d P ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⨯==⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦2.3 solution:According to the constraint equations:0;0;01n a n o a o n •=•=•==Thus,the matrix should be like this:00150015100310030102010200010001or --⎡⎤⎡⎤⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥⎢⎥--⎢⎥⎢⎥⎣⎦⎣⎦2.4 Solution:X Y ZP P P ⎛⎫ ⎪ ⎪ ⎪⎝⎭=cos 0sin 010sin 0cos θθθθ⎛⎫ ⎪ ⎪ ⎪-⎝⎭0n a P P P ⎛⎫⎪⎪ ⎪⎝⎭2.7 Solution:According to the equation of pure rotation transformation , the new coordinates are as follows:10022222(,45)03422720222newP rot x P ⎡⎤⎡⎤⎢⎥⎢⎥⎡⎤⎢⎢⎢⎥⎢⎢=⨯==⎢⎥⎢⎢⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦2.9 Solution:Acording to the equations for the combined transformations ,the new coordinates are as follows:B 01005100051100030010310 (,90)(5,3,6)(,90)0010601004900011000111 A BP Rot z Trans Rot x P-⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⨯⨯⨯==⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦Transformations relative to the reference frameTransformations relative to the current frame2.10P=Trans(5,3,6)Rot(x,90)Rot(a,90) PA1 0 0 5 1 0 0 0 0 -1 0 02 = 0 1 03 0 0 -1 0 1 0 0 0 3 0 0 1 6 0 1 0 0 0 0 1 0 5 0 0 0 1 0 0 0 1 0 0 0 1 1 2 = -2 8 1 2.122.14a) For spherical coordinates we have （for posihon )1) r ·cos γ·sin β = 3.1375unitsunits2) r ·sin γ·sin β = 2.195 3) r ·cos β = 3.214I) Assuming sin β is posihve, from a and b → γ=35°from b and c → β=50° from c → r=5II) If sin β were negative. Then γ=35°β=50° r=5 Since orientation is not specified, no more information is available to check the results.b) For case I, substifate corresponding values of sin β , cos β, sin γ, cos γand r in sperical coordinates to get: 0.5265 -0.5735 0.6275 3.1375Tsph(r,β,γ)=Tsph(35,50,5)= 0.3687 0.819 0.439 2.195-0.766 0 0.6428 3.214 0 0 0 12.16 Solution:According to the equations given in the text book, we can get the Euler angles as follows:arctan 2(,)arctan 2(,)y x y x a a or a a Φ=-- Which lead to :21535or Φ=②arctan 2(,)0180x y x y n S n C o S o C or ψ=-Φ+Φ-Φ+Φ= ③arctan 2(,)5050x y z a C a S a or θ=Φ+Φ=- 2.18 Solution:①Since the hand will be placed on the object, we can obtain this:UU U R U R obj H R H R obj T T T T T T ===Thus:10015100101000001UU U H R objT T T --⎡⎤⎢⎥-⎢⎥==⎢⎥⎢⎥⎣⎦②No,it can ’t.If so,the element at the position of the third row and the second column should be 0.However, it isn ’t. ③x=5,y=1,z=0According to the equations of the euler angles:arctan 2(,)arctan 2(,)0180y x y x a a or a a or Φ=--= arctan 2(,)27090x y x y n S n C o S o C or ψ=-Φ+Φ-Φ+Φ=arctan 2(,)27090x y z a C a S a or θ=Φ+Φ=2.21(a)(b) # θd aα 0-1 1θ 0 3d0 1-2 2θ 04d180 2-H 3θ5d(c)O UT =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡+1000100001000121d d 1A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-100001000013111311s d c s c d s c2A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--10001000024222422s d c s c d s c 3A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-100100000053333d c s s c (d)321A A A T T T T O U H O O U H U==2.22(a)(b)(c)O UT = ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡1000010*********l 1A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-1000001000001111s c c s2A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-1000010010000121l l 3A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-10001000001313c s s c 4A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡1000100001000013l (d)4321A A A A T T T T O U H O O U H U==2.23(a)(b)# θ d a α1 1θ0 0 902 2θ''6''150 4 3θ0 1 -904 4θ''180 905 5θ0 0 -906 6θ 5 0 0(c)1A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-1000001001000001 2A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-1000610061.100707.0707.061.100707.0707.0 3A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-1000001001001001 4A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-1000180********1 5A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--100000100707.00707.00707.00707.0 6A =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡1000510000100001(d)H R T =⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡----1000191006010414.1001。
