Robotics I. Figure 1: Initial placement of a rigid thin rod of length L in an absolute reference frame.

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Amanda Arnold
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1 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?

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

2 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]

3 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 ) 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.

4 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 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 [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

5 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].

Robotics I. Test November 29, 2013

Robotics I. Test November 29, 2013 Exercise 1 [6 points] Robotics I Test November 9, 013 A DC motor is used to actuate a single robot link that rotates in the horizontal plane around a joint axis passing through its base. The motor is connected