Robotics I. April 1, the motion starts and ends with zero Cartesian velocity and acceleration;

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1 Robotics I April, 6 Consider a planar R robot with links of length l = and l =.5. he end-effector should move smoothly from an initial point p in to a final point p fin in the robot workspace so that the motion starts and ends with zero Cartesian velocity and acceleration; at the start, the robot is in the right arm inverse kinematics solution i.e., with positive q, and remains in this type of solution throughout the motion; coordinated motion is enforced to the joints; symmetric limits on joint velocity, acceleration, and jerk are satisfied: q i V i, q i A i, q i J i, i =,. In order to address this motion task, choose a class of trajectories and determine, within the considered class, a minimum time trajectory, given the following position data p in = p. fin =. and joint limits V = [rad/s], A = 3 [rad/s ], J = 3 [rad/s 3 ], V = [rad/s], A = 7.5 [rad/s ], J = 7 [rad/s 3 ]. Provide the minimum feasible time obtained and the maximum absolute values attained by the velocity and the acceleration at the two joints. At the trajectory midpoint, t = /, determine the values of the end-effector Cartesian velocity v and acceleration a, and draw the robot in its current configuration together with the vectors v and a. [8 minutes; open books]

2 Solution April, 6 In view of the nature of the given robot motion limits, it is highly recommended to define the trajectory in the joint space. he direct and inverse kinematics of the R planar robot are given by px l c + l c p = = = fq l s + l s q = with q +/ q +/ = p y AAN{py l + l c p x l s, p x l + l c + p y l s } AAN{s, c } c = p x + p y l l, s = ± c l l, = f p, and where the +/ associated as index to the joint angles q and q mean that for their evaluation the + or, respectively, the sign has been used in the definition of s. Substituting the link lengths and the problem data for p = p in and p = p fin, and picking up the solution with q + > yields q in = 69.5 = [rad], q.3 fin =. = [rad]..784 aking into account the smoothness requirement and the boundary conditions on velocity and acceleration, we choose a polynomial trajectory of degree 5 for each joint. In the double normalized form, its expression is qτ = q in + q τ 3 5τ 4 + 6τ 5.97, q = q fin q in = [rad], τ = t [, ] In order to obtain the maximum values reached along this trajectory by the velocity, acceleration, and jerk, which should satisfy the given limits, we compute the first four time derivatives: q = q 3τ 6τ 3 + 3τ 4 q = q 6τ 8τ + τ 3 q = q 6 36τ + 36τ 3. q = q τ. 4 We analyze the constraints imposed by the joint limits starting with the one with highest differential order. We will work now with scalar quantities, i.e., joint by joint, dropping for simplicity the joint index. he maximum jerk in the closed interval τ [, ] occurs either at the boundaries or where the fourth derivative is zero: q = q = 6 q 3,. q τ τ =.5 q.5 = 3 q 3.

3 hus, the minimum motion time that satisfies the jerk limit is given by 6 q q τ J 3 =: J. J he maximum acceleration occurs where the third derivative is zero no need to check the value at the boundaries, since we have q = q = by construction: 3 q τ = 6τ + 6τ τ =.5 ± qτ = ± q 6. he minimum motion time that satisfies the acceleration limit is given by q qτ A =: A. A Similarly, the maximum velocity occurs where the second derivative is zero again, no need to check the value at the boundaries, since q = q = : qτ = τ 3τ + τ τ = {,.5, } q.5 = 3 q 6, and thus qτ V 3 q 6 V =: V. As a result, the minimum feasible motion time is obtained as { } q q = max { J, A, V } = max J,.48 q,.875. A V Using the data all in radians of the problem at hand, we compute the minimum motion time for the first joint as = max { J,, A,, V, } = max {.579,.9468, 3.696} = [s], where the velocity limit is the most constraining one. Similarly, for the second joint it is = max { J,, A,, V, } = max {.787,.669,.5336} =.787 [s] and the jerk will be the variable reaching first its limit. Since coordinated motion of the joints should be enforced, the common minimum motion time will be = max {, } = = [s], with the second joint traveling much slower than it could in principle. he trajectory profiles of position, velocity, acceleration, and jerk of the two joints are shown in Figs. 3. he peak velocity of the two joints is reached at t = / =.8463 s max q t = q.8463 = [rad/s], max q t = q.8463 =.89 [rad/s], while the peak acceleration in module is attained at t =.5 ± 3/6, namely at t =.783 s max positive acceleration and t =.93 s max negative acceleration = max deceleration max q t = q.783 =.8339 [rad/s ], max q t = q.783 = [rad/s ]. 3

4 3 Joint motion for = Joint velocity for = position rad.5 velocity rad/s time s time s Figure : Position [left] and velocity [right] of joint blue, solid and joint red, dashed Joint acceleration for = Joint jerk for = acceleration rad/s.. jerk rad/s time s time s Figure : Acceleration [left] and jerk [right] of joint blue, solid and joint red, dashed We note also that coordinated motion is symmetric w.r.t. to the total time for all, and thus also for. herefore, the configuration reached at t = / will simply be q := q = q in + q = q in + q fin = 85.8 = [rad] Moreover, it is q := q = 3 q = 6.56 [ /s] =.89 [rad/s], q := q =. he robot analytic Jacobian Jq = fq/ q and its time derivative Hq, q = Jq, l s + l s l s Jq = l c + l c l c l c q + l c q + q l c q + q, Hq, q =, l s q + l s q + q l s q + q 4

5 .6 Cartesian motion of the R robot Figure 3: he planar R robot arm in its initial configuration blue, at the midpoint of the trajectory red, and in the final configuration green, with the Cartesian path traced by its endeffector thin blue line. he two arrows placed at the midpoint represent the end-effector velocity black and acceleration magenta, respectively. heir length has been scaled by a factor to fit. take the numerical values at q = q, q = q J := Jq = , J := Hq, q = herefore, the required Cartesian velocity and acceleration of the robot end-effector at the trajectory midpoint are ṗ = J q = [m/s] p = J q + J q = J q = [m/s ]. 5