Robotics & Automation Lecture 17 Manipulability Ellipsoid, Singularities of Serial Arm John T. Wen October 14, 2008
Jacobian Singularity rank(j) = dimension of manipulability ellipsoid = # of independent columns in J = # of independent directions in SE(3) end effector can move. J is singular J loses rank (R(J) reduces in dimension ) manipulability ellipsoid becomes degenerate J gain addition DOF in self motion (N(J) gains dimension). October 14, 2008Copyrighted by John T. Wen Page 1
Manipulability Ellipsoid Jacobian may be interpreted as mapping of a unit ball in the joint velocity space to an ellipsoid in the task space. The principal axes of the ellipsoid are given by the left singular vectors of the Jacobian and the lengths of the principal axes are given by the singular values of the Jacobian. Large singular value indicates the direction that the arm is easy to move, and, conversely, small singular value indicates the direction that the arm is difficult to move (near singularities). For this reason, this ellipsoid is called the manipulability ellipsoid. Dual perspective: the torque induced by the spatial force (force/torque) applied at the end effector is given by τ = J T f or f = J T τ. Again consider a unit ball in τ and the corresponding ellipsoid in f. Then the ellipsoid has the same principal axes but the lengths along these axes are the reciprocal of the manipulability ellipsoid. This is sometimes called the force ellipsoid. In general, a direction that is easy (hard) to manipulate would require large (small) torque to resist spatial force. October 14, 2008Copyrighted by John T. Wen Page 2
Example: 3-DOF wrist (h 1,h 2,h 3 ) = (x,y,z), therefore, the Jacobian is (J) 1 = [h 1,eĥ1θ 1 h 2,eĥ1θ 1 eĥ2θ 2 h 3 ]. Or more conveniently expressed in frame 2: (J) 2 = [h 1,h 2,eĥ2θ 2 h 3 ] = 1 0 s 2 0 1 0 0 0 c 2. Singularity occurs at det(j) = 0 or cos(θ 2 ) = 0 or θ 2 = ± π 2. In coordinate-free perspective, J = [ h 1, h 2, h 3 ], where h 1 h 2 and h 2 h 3. Then J will lose rank if and only if h 1 and h 3 are linearly dependent. October 14, 2008Copyrighted by John T. Wen Page 3
Example: 3-DOF planar arm h 1 = h 2 = h 3 = z, and J = z z z z p 13 z p 23 0 J loses rank (rank less than 3) if and only if p 13 and p 23 are collinear, i.e., θ 2 = 0 or π. In coordinate frame, z z z 0 0 z J 2 = ẑ(e ẑθ 2 p 12 + p 23 ) ẑp 23 0 ẑ(e ẑθ 2 p 12 ) ẑp 23 0 Since p 12 = l 1 x, p 23 = l 2 x, we have the determinant of the (2 2 portion of) Jacobian = l 1 l 2 sinθ 2. October 14, 2008Copyrighted by John T. Wen Page 4
Example: PUMA 560 Choose A at the origin of the spherical wrist joints, then h J A = 1 h 2 h 3 h 4 h 5 h 6 h 1 p 14 h 2 p 24 h 3 p 34 0 0 0 Singularities: [ ] h 4 h 5 h 6 loses rank. [ ] h 1 p 14 h 2 p 24 h 3 p 34 loses rank [ ] First consider h 4 h 5 h 6 losing rank. This is the same as the first example, and singularity occurs at sinθ 5 = 0. October 14, 2008Copyrighted by John T. Wen Page 5
PUMA 560 Singularities [ ] Now consider h 1 p 14 h 2 p 24 h 3 p 34 losing rank. Noting that p 12 is collinear with h 1, h 2 = h 3, and applying elementary column reduction, we obtain [ h 1 p 24 h 2 p 23 h 2 p 34 ]. From before, h 1 = z,h 2 = y,h 3 = y,h 4 = z,h 5 = y,h 6 = z p 01 = 0, p 12 = d 1 z, p 23 = l 2 x d 3 y, p 34 = l 3 x + d 4 z, p 45 = 0, p 56 = 0. Now represent the submatrix in 2-frame, we obtain: d 3 c 2 0 s 3 l 3 c 3 d 4 l 2 c 2 l 3 c 23 d 4 s 23 0 0. d 3 s 2 l 2 c 3 l 3 s 3 d 4 The determinant is l 2 (l 2 c 2 l 3 c 23 d 4 s 23 )(s 3 l 3 c 3 d 4 ). Therefore, the singularities are at: s 3 l 3 c 3 d 4 = 0, or tanθ 3 = d 4 /l 3 (boundary singularity) l 2 c 2 l 3 c 23 d 4 s 23 = 0. This is an interior singularity. October 14, 2008Copyrighted by John T. Wen Page 6
Geometric conditions for singularities A 6-DOF arm is singular if for any two revolute joints, the joint axes are collinear (e.g., spherical wrist). A 6-DOF arm is singular if any three parallel rotational axes lie in a plane (e.g., boundary singularity of an elbow arm). A 6-DOF arm is singular if any four rotational axes intersect at a point or three coplanar revolute axes intersect at a point. October 14, 2008Copyrighted by John T. Wen Page 7
Two Collinear revolute axes Suppose that h i = h j and i, j are both revolute axes. Then the corresponding columns in the Jacobian is h i h j 0 h j h i p ie h j p je h i p i j h j p je Since h i and p i j are collinear, the Jacobian loses rank. October 14, 2008Copyrighted by John T. Wen Page 8
Three Coplanar revolute axes Let i, j,k be three revolute joints and h i = h j = h k = h. The corresponding columns in the Jacobian are: h i h j h k 0 0 h h i p ie h j p je h k p ke h p i j h p jk h p ke Since ( h, p i j, p jk ) are coplanar, the first two columns are linearly dependent. October 14, 2008Copyrighted by John T. Wen Page 9
Four intersecting revolute axes Let (i, j,k,m) be four revolute joints, and the corresponding rotational axes, ( h i, h j, h k, h m ) intersect at a point s. The corresponding columns in the Jacobian are: h i h j h k h m h i p ie h j p je h k p ke h m p me. h i h j h k h m h i p se h j p se h k p se h m p se. Premultiply by rank 3. I 0 p se I, we obtain h i h j h k h m 0 0 0 0, which has maximum October 14, 2008Copyrighted by John T. Wen Page 10
Three coplanar intersecting revolute axes Repeat the same argument as before for the three coplanar axes, { h i, h j, h k }, we obtain the (sub-)jacobian similar to h i h j h k 0 0 0 Since { h i, h j, h k } are coplanar, [ h i, h j, h k ] is of rank 2. October 14, 2008Copyrighted by John T. Wen Page 11