Robotics I September, 7 Exercise Consider the rigid body in Fig., a thin rod of length L. The rod will be rotated by an angle α around the z axis, then by an angle β around the resulting x axis, and finally by an angle γ around the resulting y axis. L z x y Figure : Initial placement of a rigid thin rod of length L in an absolute reference frame. Provide the final orientation of the rod, as expressed by a rotation matrix, and the symbolic expression of its angular velocity ω R in terms of the angles (α, β, γ) and their time derivatives ( α, β, γ). Assuming the numerical values L. [m] for the rod, α, β, and γ 6 for the angles, and α β γ /s for their time derivatives, determine: the components of the final unit vector y attached to the rod after the three rotations; the absolute coordinates of the position p of the end-point of the rod after the three rotations; the angular velocity ω of the rod, expressed both in the initial (absolute) reference frame and in the final rotated frame; the absolute velocity ṗ of the end-point of the rod. Exercise Using standard DH angles, a planar R robot with links of lengths l l. [m] is in the initial static configuration q (, 9 ). Plan a rest-to-rest cubic polynomial trajectory in the joint space so as to reach the final end-effector position p goal (.,.66) [m] in minimum time under the following joint velocity limits: q /s, q 9 /s. Provide the minimum time T and the numerical values of the coefficients of the optimal cubic polynomials. If the robot performs the above minimum-time task with a coordinated motion in the joint space, what is the maximum velocity reached by each joint during motion?
Exercise Consider the transmission/reduction assembly of a robot joint driving a single robot link, as sketched in Fig.. The actuator is an electrical motor with maximum rotation speed θ m of its output shaft equal to RPM. The transmission assembly consists of an Harmonic Drive (HD), with its wave generator mounted on the motor shaft (A A ) and whose circular spline has n CS 6 teeth, followed by a two-wheel toothed gear, with a smaller wheel of radius r.9 [cm] mounted on the HD output axis (B B ) and a larger wheel of radius r.7 [cm]. The link rotates with the larger wheel at an angular speed θ l and has a length L.6 [m]. actuator circular spline toothed gear motor shaft A A wave generator harmonic drive B B r r link L Figure : Transmission/reduction assembly of a robot joint. What is the reduction ratio N θ m / θ l of the complete transmission? Does the link rotate in the same or in the opposite direction of the motor shaft? What is the maximum velocity achievable by the end-point of the link? [ minutes, open books but no computer or smartphone]
Solution September, 7 Exercise The sequence of rotations about the first (fixed) axis Z and the following (moving) axes X and Y provides a ZXY Euler representation of orientation. Therefore, the final orientation of the rod is given by R ZXY (α, β, γ) R Z (α) R X (β) R Y (γ) cos α sin α sin α cos α cos β sin β sin β cos β cos γ sin γ sin γ cos γ cos α cos γ sin α sin β sin γ sin α cos β cos α sin γ + sin α sin β cos γ sin α cos γ + cos α sin β sin γ cos α cos β sin α sin γ cos α sin β cos γ cos β sin γ sin β cos β sin γ x (α, β, γ) y (α, β) z (α, β, γ) In particular, the position of the end-point of the rod after the three rotations is p L z. Using the numerical data, the final rotation matrix is ( π R ZXY 6, π 6, π ).69..6..7.69.7.. while the unit vector y y and the position vector p of the end-point of the rod after the rotations are, respectively, y ( π ). 6, π.7, p. z ( π 6 6, π 6, π )..9 [m]...7 The angular velocity ω of the rod associated to the time derivatives ( α, β, γ) in the configuration specified by the angles (α, β, γ) can be computed either from the relation ω z ω y S(ω) ṘZXY (α, β, γ, α, β, γ) R T ZXY (α, β, γ) ω z ω x ω ω y ω x or, more explicitly, as ω z α + x β + y γ α + R Z (α) β + R Z (α) R X (β) γ, ω x ω y ω z.
Performing computations, we obtain ω cos α sin α cos β sin α cos α cos β sin β α β γ cos α β sin α cos β γ. sin α β + cos α cos β γ α + sin β γ This expression is referenced to the initial frame. In the rotated frame, one has cos β sin γ α + cos γ β R ω R T ZXY ω sin β α + γ. cos β cos γ α + sin γ β Finally, the linear velocity ṗ of the end-point of the rod is given by cos α sin β cos γ sin α sin γ ṗ ω p S(ω) p L sin α sin β cos γ + cos α sin γ α Using the numerical data, we obtain ω π..97.97 and Exercise sin α cos β cos γ + cos α cos β cos γ β + sin β cos γ [rad/s], ṗ R ω π.6.9 [m/s]..7 sin α sin β sin γ + cos α cos γ cos α sin β sin γ + sin α cos γ cos β sin γ.96.97 [rad/s],. γ. We first solve the inverse kinematics for the final position p goal (p x,goal, p y,goal ) (., /). Noting that p goal l + l, this point is on the external boundary of the robot workspace. Thus, there is a single configuration with stretched arm as solution, given by ( ) ( ) ( ) ( q,goal ATAN{py,goal, p x,goal } π/ 6 ) q goal [rad]. q,goal As a result the joint displacement is ( ) ( ) q q q goal q 9 q and the associated rest-to-rest cubic trajectories for the two joints, each in time T i >, i,, will take the form ( ) ( ) ) t t q i (t) q,i + q i (, () T i T i
with velocities and accelerations ( ( q i (t) 6 q ) ( ) ) i t t, q i (t) 6 q i T i T i T i Ti ( ( )) t, for i,. T i For each joint, the maximum velocity (in absolute value) will occur at the trajectory midpoint, where the acceleration is zero, i.e., q i (t i ) t i T i q i (T i /) q i T i. Since both displacements q i are positive, we will discard from now on the absolute value. Imposing that the joint velocities both reach their admissible limits, respectively, V /s and V 9 /s, we have { q i (T i /) V i Ti. q i.7, for i, V i., for i. Therefore, the minimum motion time will be T max {T, T } T. s. The associated numerical coefficients of the optimal cubic polynomials in () are q (t) + t 7. t, t [,.7], q (t) 9 + t. t, t [,.]. In order to perform a coordinated motion in the joint space, the fastest joint, namely the first one, should uniformly slow down its motion so that the its final time equals T. Since T T, the needed scaling is by a factor k and, accordingly, the new maximum velocity of joint, reached again at the midpoint of the coordinated trajectory, will be reduced to V /k /s. Indeed, no changes occur in the velocity profile of the second joint. Exercise The considered harmonic drive has n CS 6 teeth on the (internal side of the) circular spline and thus n F S n CS 6 teeth on the external side of the flexspline. Its reduction ratio is thus N HD Since the toothed gear has reduction ratio n F S n F S n CS n F S N T G r r.7.9.9, 6. the reduction ratio of the transmission is θ ṁ N N HD N T G.6. θ l There is a double inversion of rotations in the overall transmission. Thus, the link will rotate in the same direction (clockwise or counterclockwise) of the motor shaft, i.e., sign{ θ m } sign{ θ l }. Finally, the maximum velocity v achievable by the end-point of the link is v L max{ θ l } L max{ θ m } N.6 [m] [RPM] (π/6)[rad/s].6. [m/s].