Exam (elaborations) TEST BANK FOR Robotics Modelling, Planning and Control 1st Edition By Villani L., Oriolo G., Siciliano B. (Solution Manua)

2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3 Differential Kinematics and Statics . . . . . . . . . . . . . . . . . . . . . . . . 19 4 Trajectory Planning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 5 Actuators and Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 6 Control Architecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 7 Dynamics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 8 Motion Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 9 ForceControl. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 10 Visual servoing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 11 Mobile Robots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 12 Motion Planning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 2 Kinematics Solution to Problem 2.1 Composition of rotation matrices with respect to the current frame gives R(φ) = Rz(ϕ)Rx(ϑ)Rz (ψ). Using the expressions of elementary rotation matrices in (2.6) and (2.8): Rz(ϕ) = ⎡ ⎣ cϕ −sϕ 0 sϕ cϕ 0 0 0 1 ⎤ ⎦ Rx(ϑ) = ⎡ ⎣ 1 0 0 0 cϑ −sϑ 0 sϑ cϑ ⎤ ⎦ Rz (ϕ) = ⎡ ⎣ cψ −sψ 0 sψ cψ 0 0 0 1 ⎤ ⎦ and taking the products gives R(φ) = ⎡ ⎣ cϕcψ − sϕcϑsψ −cϕsψ − sϕcϑcψ sϕsϑ sϕcψ + cϕcϑsψ −sϕsψ + cϕcϑcψ −cϕsϑ sϑsψ sϑcψ cϑ ⎤ ⎦. As for the inverse problem, given a rotation matrix R = ⎡ ⎣ r11 r12 r13 r21 r22 r23 r31 r32 r33 ⎤ ⎦, 2 2 Kinematics the set of Euler angles ZXZ is given by ϕ = Atan2(r13,−r23) ϑ = Atan2  r2 31 + r2 32, r33  ψ = Atan2(r31, r32) when ϑ ∈ (0, π). Otherwise, if ϑ ∈ (−π, 0) then the solution is ϕ = Atan2(−r13, r23) ϑ = Atan2  −  r2 31 + r2 32, r33  ψ = Atan2(−r31,−r32). Solution to Problem 2.2 In the case sϑ = 0, the rotation matrix in (2.18) becomes R(φ) = ⎡ ⎣ cϕ+ψ −sϕ+ψ 0 sϕ+ψ cϕ+ψ 0 0 0 1 ⎤ ⎦ when ϑ = 0. Otherwise, if ϑ = π, then the matrix is R(φ) = ⎡ ⎣ −cϕ−ψ −sϕ−ψ 0 −sϕ−ψ cϕ−ψ 0 0 0 −1 ⎤ ⎦. From the elements [1, 2] and [2, 2] it is possible to compute only the sum or difference of angles ϕ and ψ, i.e., ϕ ± ψ = Atan2(−r12, r22) where the positive sign holds for ϑ = 0 and the negative sign holds for ϑ = π. Solution to Problem 2.3 In the case cϑ = 0, the rotation matrix in (2.21) becomes R(φ) = ⎡ ⎣ 0 sψ−ϕ cψ−ϕ 0 cψ−ϕ −sψ−ϕ −1 0 0 ⎤ ⎦ when ϑ = π/2. Otherwise, if ϑ = −π/2, then the matrix is R(φ) = ⎡ ⎣ 0 −sψ+ϕ −cψ+ϕ 0 cψ+ϕ −sψ+ϕ 1 0 0 ⎤ ⎦. 2 Kinematics 3 From the elements [2, 2] and [2, 3] it is possible to compute only the sum or difference of angles ψ and ϕ, i.e., ψ ± ϕ = Atan2(−r23, r22) where the positive sign holds for ϑ = −π/2 and the negative sign holds for ϑ = π/2. Solution to Problem 2.4 The rotation matrix can be obtained as in (2.24) R(ϑ, r) = Rz(α)Ry(β)Rz(ϑ)Ry(−β)Rz(−α), where the elementary rotation matrices are given as in (2.6) and (2.7): Rz(α) = ⎡ ⎣ cα −sα 0 sα cα 0 0 0 1 ⎤ ⎦ Ry(β) = ⎡ ⎣ cβ 0 sβ 0 1 0 −sβ 0 cβ ⎤ ⎦ Rz(ϑ) = ⎡ ⎣ cϑ −sϑ 0 sϑ cϑ 0 0 0 1 ⎤ ⎦. Taking the first product gives Rz(α)Ry(β) = ⎡ ⎣ cαcβ −sα cαsβ sαcβ cα sαsβ −sβ 0 cβ ⎤ ⎦. The next product gives Rz(α)Ry(β)Rz(ϑ) = ⎡ ⎣ cαcβcϑ − sαsϑ −cαcβsϑ − sαcϑ cαsβ sαcβcϑ + cαsϑ −sαcβsϑ + cαcϑ sαsβ −sβcϑ sβsϑ cβ ⎤ ⎦. Then, by observing that Ry(−β)Rz(−α) = (Rz(α)Ry(β))T , the overall rotation matrix is R(ϑ, r) = ⎡ ⎣ (s2 α + c2 αc2 β)cθ + c2 αs2 β sαcαs2 β(1 − cϑ) − cβsϑ sαcαs2 β(1 − cϑ) + cβsϑ (s2 αc2 β + c2 α)cθ + s2 αs2 β cαsβcβ(1 − cϑ) − sαsβsϑ sαsβcβ(1 − cϑ) + cαsβsϑ cαsβcβ(1 − cϑ) + sαsβsϑ sαsβcβ(1 − cϑ) − cαsβsϑ s2 βcϑ + c2 β ⎤ ⎦. 4 2