Introduction to Robotics (CS223A) Homework #3 Solution ...

1 Introduction to Robotics (CS223A) Homework #3 Solution (Winter 2007/2008) are given that a certain RPR manipulator has the following transformationmatrices, where{E}is the frame of the end c1 s10 0s1c10 000 1 000 0 1 ,03T= c1c3 c1s3 s1L1c1 s1d2s1c3 s1s3c1L1s1+c1d2 s3 c3000001 ,0ET= s1c1s3c1c3L1c1+L2c1c3 s1d2c1s1s3s1c3L1s1+L2s1c3+c1d20c3 s3 L2s30001 Derive the basic Jacobian relating joint velocities to the end-effector s linear andangular velocities in frame{0}.ForJv, we simply differentiate the position of the end-effector expressed in frame{0}, whichis the last column of04T. ForJ we take the z-vectors from01 Tand03T, and since joint 2 isprismatic, it doesn t L1s1 L2s1c3 c1d2 s1 L2c1s3L1c1+L2c1c3 s1d2c1 L2s1s300 L2c3 J = 0 0 s10 0c11 00 the planar PR manipulator shown here:(a)Find the origin of frame{3}expressed in terms of frame{0}, that : you can derive this geometrically, if you want to avoid going throughDH s simplest to do this geometrically.

2 Note: I will often refer to the origin of frame{3}as the tip .In the base frame{0}, the x-coordinate of the tip is found by projecting the linkL2ontothe X0axis and adding toL1. Similarly, the z-coordinate is found by projectingL2ontothe Z0axis and adding to the slider displacementd1. Thus, in frame{0}:0P3org=[xtipztip]=[L1+L2cos( 2)d1+L2sin( 2)](b)Give the2 2 Jacobian that relates the joint velocities to the linear [ xtip d1 xtip 2 ztip d1 ztip 2]=[0 L2sin( 2)1L2cos( 2)](c)For what joint values is the manipulator at a singularity? What motion isrestricted at this singularity?The singularity occurs at a configuration whendet(J) = (J) =J11J22 J12J21=L2sin( 2)So the singularity occurs whenever sin( 2) = 0, which corresponds to: 2= 0oor 180oThese two situations are portrayed below; either the arm is extended out completely orfolded over both cases,the end-effector cannot move instantaneously in the X0direction,ie.

3 The joints cannot produce a velocity component in the the RRR manipulator shown here:X0,1Z0,Z1 out111Z4X4 Note: in the figure, the numbers below the links represent the lengths.(a)Find the DH parameters for this manipulator . Remember to assign theinterior frames of this manipulator using the conventions discussed in : In preparation for part (b), one may add an extra row to the bottom of the tablecorresponding to frame{4}. Since the question didn t specifically ask for this extra row,it s not necessary to have it. After all, frame{4}is fixed wrt to frame{3}and thetransformation between the two can be found (if one so wishes) by i 1ai 1 idi100 102 90o1 20390o1 304 90o100(b)Derive the forward kinematics,04T, of this the above table to compute the DH transformation matrices. In preparation forcomputing the Jacobian in part (c), one may also compute the0iTfor each frame{i}.

4 01T= c1 s10 0s1c10 000 1 000 0 1 12T= c2 s20 100 1 0 s2 c20 000 0 1 02T=01T12T= c1c2 c1s2 s1c1s1c2 s1s2c1s1 s2 c20 0000 1 23T= c3 s30 100 1 0s3c30 000 0 1 03T=02T23T= c1c2c3 s1s3 c1c2s3 s1c3c1s2c1c2+c1s1c2c3+c1s3 s1c2s3+c1c3s1s2s1c2+s1 s2c3s2s3c2 s20001 34T= 1 0 0 10 0 1 00 1 0 00 0 0 1 04T=03T34T= c1c2c3 s1s3 c1s2 c1c2s3 s1c3c1c2c3 s1s3+c1c2+c1s1c2c3+c1s3 s1s2 s1c2s3+c1c3s1c2c3+c1s3+s1c2+s1 s2c3 c2s2s3 s2c3 s20001 (c)Find the basic Jacobian,J0, for this the explicit form: take derivatives of the end-effector position (the last column of04T) forJv, and use the Z-axes of each frame (3rd column of each0iTmatrix calculatedabove) [ 0P4org 1 0P4org 2 0P4org 30Z10Z20Z3]= s1c2c3 c1s3 s1c2 s1 c1s2c3 c1s2 c1c2s3 s1c3c1c2c3 s1s3+c1c2+c1 s1s2c3 s1s2 s1c2s3+c1c30 c2c3 c2s2s30 s1c1s20c1s1s210c2 (d)Find1Jv, the position Jacobian matrix expressed in frame{1}.

5 Compute1 Jvby rotating0Jv(the top half ofJ0) from frame{0}to frame{1}. Notethat this rotation matrix is the inverse (or transpose, in this case) of what appears in01T(computed in part (b)), ie. we need to use10R, c1s10 s1c100 0 1 s1c2c3 c1s3 s1c2 s1 c1s2c3 c1s2 c1c2s3 s1c3c1c2c3 s1s3+c1c2+c1 s1s2c3 s1s2 s1c2s3+c1c30 c2c3 c2s2s3 = s3 s2c3 s2 c2s3c2c3+c2+ 10c30 c2c3 c2s2s3 (e)Use the matrix that you found in part (d) to find the singularities (withrespect to linear velocity) of this singularities occur when the determinant of the Jacobian (inanyframe) is simpler in frame{1}, look at the determinant of1Jv:det(1Jv) = s3 s2c3 s2 c2s3c2c3+c2+ 10c30 c2c3 c2s2s3 = ( s3)(c2c23+c2c3) (c2c3+c2+ 1)( s22s3c3 s22s3 c22s3c3 c22s3)=s3(c2c3+c2+c3+ 1)=s3(1 +c3)(1 +c2)Setting the first term in the above expression to zero gives:sin( 3) = 0 3= 0oor 180oSetting the second term to zero gives the same result as the third term to zero gives.

6 Cos( 2) = 1 2= 180o(f)For each type of singularity that you found in part (e), explain the physicalinterpretation of the singularity, by sketching the arm in a singular configu-ration and describing the resulting limitation on its singularity 3= 0 is when the outer half of the arm is outstretched. In this positionthere is no motion possible in the outThe singularity 3= 180ois when the outer link of the arm is folded in on are two restrictions in this position. First, just as in the outstretched case, thereis no motion possible in the X3direction. Second, because the last two links have thesame length, the end-effector is overlapping with joint 2. As a result there is no motionpossible in the outThe singularity 2= 180ois when half the arm is folded in over itself, causing joints 1and 3 to overlap. In this position, joints 1 and 3 have the same effect on the end-effector as if joint 3 doesn t exist.

7 The ensuing loss of motion can be see in frame{3}: thereis no motion possible in the outY3