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Robot Motion: Illustrated in Figure I is a robot arm having two "links" connected by two "joints": a shoulder, or base, joint, and an elbow joint There is a motor at each joint and the joint angles are 8 and 62. The (XI. 2) coordinates of the hand at the end of the arm are given by: X1 - Licos +L2 cos(0+0:) x=Lsin 0, +L, sin(0, +8) 12 Hand L2 Elbow Motor $ Base Motor (a) Finish Path of Hand Start (b) Figure 1: Robot arm

The problem is to determine how to control the joint angles by the motors to move the hand from one position to another The arm is to start from rest at a known position and move to a desired position. It must start and stop with zero velocity and acceleration. The following polynomial expressions are to be used for controlling the motion by generating commands to be sent to the joint motor controllers 0,(t) = 8,(0) + azés + azt* + azt* + axt? + azt 82(t) = 620) + bits + b2t+ b3t? + bet? +bst Where 8(t) and 8, (t) are the initial angles at time t = 0 and coefficient vectors a = [a, 22 23 24 as]" and b = [bz bz bzb, bs]' are to be determined to provide the desired motion. The choice of the degree of the polynomials will be explained as the equations of motion are described. For desired initial coordinates (x1, x2) at time t=0 and desired final coordinates at time t=tr, the required values for angles 0:(0), 02(0), 0(te), and (tr) can be found using trigonometry. For given values of 0:(0), 0.(te), and te, matrix equations are to be set up and solved for coefficient vector a. Similarly, for given values of 0,0).02(t), and ts, matrix equations are to be set up and solved for coefficient vector b. These results are to be used to plot the path of the hand. The remaining constraints in the arm motion are that the velocity and acceleration of the links are to be zero at the known starting location and the desired final location. This implies that the angular velocity and angular acceleration of the two angles are to be zero at time t = 0 and t=tr. The angular velocity of the first link at time is the derivative of the angle 0:(t) with respect to time 9, (t) = 5a t* + 4a_t? + 3azt? + 2axt + as The velocity at t=0 is 0,0) = as = 0 and thus, coefficient as = 0. The angular acceleration is the second derivative of the angle 9.(t) with respect to time 0"(t) = 200;t* + 12azt? + 6a3t+204 The acceleration att 0 is 0;"(0) = 204 = 0 and thus, coefficient as = 0. Writing the three constraints on 0:(t) and its derivatives at time t=tein matrix form: ) 5t| 4t 317 ||42 = (200 120 6 8-100700) Similarly, for 0.(t):

C3-106750 t)- 0 (0) 5t|4t|3t (20t 12t; 6) Note that there are three equations and three unknowns for each angle and its two derivatives. This is the reason for choosing a fifth-degree polynomial to control the motion. The lower degree three terms (t',t', tº have coefficients with value zero to meet the constraints at t = 0. This leaves the higher degree three terms (t.14.1) and corresponding three unknown coefficients. Adding additional higher degree terms would require additional constraint equations to be able to compute the values of the corresponding coefficients. Assuming the following initial and final values and link lengths: t=2s 0:(0) = -19 0:(044 0 (1) = 43 02(V) 151 Li - 4 feet L2 = 3 feet which correspond to a starting hand location of (6.5, 0) and a destination location of (0.2). Write a MATLAB script to compute the motion control coefficients and to compute and plot the path of the hand.