ECE569 Fall 2015 Partial Solution to Problem Set 3

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1 ECE569 Fall 215 Partial Solution to Problem Set 3 These problems are from the textbook by Spong et al. 1, which is the textbook for the ECE569 this Fall 215 semester. As such, many of the problem statements are taken verbatim from the text; however, others have been reworded for reasons of efficiency or instruction. Solutions are mine. Any errors are mine and should be reported to me, skoskie@iupui.edu, rather than to the textbook authors Jacobian J 11 of 2-link Elbow Manipulator 26. We know that J 11 = [ z (o c o ) z 1 (o c o 1 ) z 2 (o c o 2 ) ]. The challenge is to find the frame origins and the z i vectors. The first step of our analysis would be to find the frames of reference. However, in this problem we are given frames of reference in the Figure. I am using the given frames in this solution. To determine the Jacobian J 11 we will complete the remaining steps: Determine the DH parameters for the elbow manipulator The link lengths are d 1 along the z for the first link, and a 2 and a 3 along x 2 and x 3 for the second and third links, respectively. Frames F 1 and F 2 have the same orientation, so the parameters α i and θ i are zero for these links. Now, we need a rotation or rotations that take F to F 1 in conjunction with the translation by d 1 along the z axis. Note that the DH convention allows us to rotate only about the x and z axes, so we must find one or more rotations about these axes. Also, the order for the DH matrices is A i = Rot zi,θ i T rans zi,d i T rans xi,a i Rot xi,α i. Thus we must first rotate about z, then about x. First we rotate about z by θ 1 = π/2. Now we have x 1 pointing to the right, y back into the page, and z still pointing up. Next, we rotate by α 1 = π/2 about the current x to obtain z 1 facing out of the page, and y 1 pointing up. We now have all of the information we need to create the table of DH parameters: Determine the A i matrices Having obtained the DH parameters, it is fairly trivial to calculate the A i matrices. In the interest of fitting the matrices on the 1 Spong, M., S. Hutchinson, and M. Vidyasagar, Robot Modeling and Control, John Wiley & Sons,

2 ECE56 Fall 215 Partial Partial Problem Set 3 November 27, Table 1: DH Parameters for the Elbow Manipulator Link a i α i d i θ i 1 π/2 d 1 π/2 + θ 1 2 a 2 θ 2 3 a 3 θ 3 * indicates variables page I will abbreviate as follows: c 1 = cos(π/2 + θ 1 ) s 1 = sin(π/2 + θ 1 ) c 2 = cos θ 2 s 2 = sin θ 2 c 3 = cos θ 3 s 3 = sin θ 3 c 12 = cos(π/2 + θ 1 + θ 2 ) s 12 = sin(π/2 + θ 1 + θ 2 ) The A i matrices are: A 1 = = c 1 s 1 s 1 c 1 1 c 1 s 1 s 1 c 1 1 d d cos π/2 sin π/2 sin π/2 cos π/2 Of course we must remember to rewrite this in the usual notation. Noting that we have A 1 = sin(π/2 + θ) = cos θ cos(π/2 + θ) = sin θ, cos(π/2 + θ 1 ) sin(π/2 + θ 1 ) sin(π/2 + θ 1 ) cos(π/2 + θ 1 ) 1 d 1 = sin θ 1 cos θ 1 cos θ 1 sin θ 1 1 d 1,

3 ECE56 Fall 215 Partial Partial Problem Set 3 November 27, and A 2 = A 3 = cos θ 2 sin θ 2 a 2 cos θ 2 sin θ 2 cos θ 2 a 2 sin θ 2 1 cos θ 3 sin θ 3 a 3 cos θ 3 sin θ 3 cos θ 3 a 3 sin θ 3 1,. Determine the Ti matrices The Ti T 1 = A 1 = T 2 = A 1 A 2 = c 1 s 1 s 1 c 1 1 d 1 T 3 = A 1 A 2 A 3 = matrices are, with some help from Matlab, c 1 c 2 c 1 s 2 s 1 a 2 c 1 s 1 c 2 s 1 s 2 c 1 a 2 s 1 s 2 c 2 d 1 [ c1*c2, -c1*s2, s1, a2*c1] [ c2*s1, -s1*s2, -c1, a2*s1] [ s2, c2,, d1] [,,, 1] s 1 c 23 s 1 s 23 c 1 a 2 s 1 + a 3 s 1 c 2 c 1 c 23 c 1 s 23 s 1 a 2 c 1 a 3 c 1 c 2 s 23 c 23 d 1 + a 3 s 2 [ c2*c3*s1 - s1*s2*s3, - c2*s1*s3 - c3*s1*s2, -c1, a2*s1 + a3*c2*s1] [ c1*s2*s3 - c1*c2*c3, c1*c2*s3 + c1*c3*s2, s1, - a2*c1 - a3*c1*c2] [ c2*s3 + c3*s2, c2*c3 - s2*s3,, d1 + a3*s2] [,,, 1] Determine the origins and the z i From the geometry we determine that o c = s 1 (a 2 c 2 + a 3 c 12 ) c 1 (a 2 s 2 + a 2 s 12 ). The sine and cosine of θ 1 arise from the fact that the d 1 frame F 1 is rotated with respect to F. To see the rest, compare the two-link robot example in the text. So, what is wrong with our T i matrices? The origins are found in the first three elements of the fourth column of the T i matrices:

4 ECE56 Fall 215 Partial Partial Problem Set 3 November 27, The z i are found in the first three elements of the third column of Ti (the third c 1 c 1 column of the rotation matrix). Thus we have z 1 = s 1, z 2 = s 1, Compute the cross products to be completed

5 ECE56 Fall 215 Partial Partial Problem Set 3 November 27, Identifying singularities of SCARA manipulator For the SCARA manipulator, (4.12) gives the J 11 portion of the Jacobian, namely a 1 s 1 a 2 s 12 a 1 s 12 J 11 = a 1 c 1 + a 2 c 12 a 1 c 12 1 The determinant is J 11 = 1 a 1s 1 a 2 s 12 a 1 s 12 a 1 c 1 + a 2 c 12 a 1 c 12 = a 2 1s 1 c 12 + a 1 a 2 s 12 c 12 a 2 1c 1 s 12 a 1 a 2 s 12 c 12 ) = a 2 1s 1 c 12 a 2 1c 1 s 12 = a 2 1 (s 1 (c 1 c 2 s 1 s 2 ) c 1 (c 1 s 2 + c 2 s 1 )) = a 2 1(s c 2 1)s 2 = a 2 s 2 Thus J 11 = when either sin θ 2 =, which occurs when θ 2 = ± kπ, for k integer, or when a 2 1 =, which is not true for the SCARA manipulator. Matlab script: >> %%% 4-19 >> clear all >> syms a1 a2 a3 s1 s12 c1 c12 s2 c2 >> alp1 = -a1*s1 -a2*s12 alp1 = - a1*s1 - a2*s12 >> alp2 = a1*c1+a2*c12 alp2 = a1*c1 + a2*c12 >> alp3 = -a1*s12 alp3 = -a1*s12 >> alp4 = a1*c12

6 ECE56 Fall 215 Partial Partial Problem Set 3 November 27, alp4 = a1*c12 >> det(j11) ans = a1^2*c12*s1-11*a2*c12*s12-11*a1*c1*s12 + a1*a2*c12*s12 >> J11 = [alp1 alp3 ; alp2 alp4 ; -1] J11 = [ - a1*s1 - a2*s12, -a1*s12, ] [ a1*c1 + a2*c12, a1*c12, ] [,, -1] >> dj11 = det(j11) dj11 = a1^2*c12*s1 - a1^2*c1*s12 >> sdj11 = subs(dj11, [c12,s12],[c1*c2-s1*s2,c1*s2+c2*s1]) sdj11 = a1^2*s1*(c1*c2 - s1*s2) - a1^2*c1*(c1*s2 + c2*s1) >> simplify(sdj11) ans = -a1^2*s2*(c1^2 + s1^2)

7 ECE56 Fall 215 Partial Partial Problem Set 3 November 27, Jacobian for and Singularities of a three-link Cartesian manipulator The three links of the three-link Cartesian robot in Figure 3-28 have the following directions of prismatic motion: 1. the first block has z 1 directed either into or out of the page, 2. the second robot has z 2 directed left or right, 3. the third has z 3 directed toward the top or bottom of the page. If we take z to be toward the top of the page, x out of the page, and y to the right, then with respect to F, the three z directions are, respectively, 1, 1, and, 1 where I have chosen the positive direction for each. Thus, J v = Because all three joints are prismatic, J ω =, so 1 [ ] 1 Jv = 1 J ω.. Thus, we have three independent columns and det J 11 location is achievable. = 1 indicating that any

8 ECE56 Fall 215 Partial Partial Problem Set 3 November 27, Jacobian for and Singularities of a Three-Link Cartesian Manipulator Solution The three links of the three-link Cartesian robot in Figure 3-28 have the following directions of prismatic motion: 1. the first block has z 1 directed either into or out of the page, 2. the second robot has z 2 directed left or right, 3. the third has z 3 directed toward the top or bottom of the page. If we take z to be toward the top of the page, x out of the page, and y to the right, then with respect to F, the three z directions are, respectively, 1, 1, and, 1 where I have chosen the positive direction for each. Thus, J v = Because all three joints are prismatic, J ω =, so 1 [ ] 1 Jv = 1 J ω.. Thus, we have three independent columns and det J 11 location is achievable. = 1 indicating that any c 215, 217 S. Koskie