# MCE/EEC 647/747: Robot Dynamics and Control

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1 MCE/EEC 647/747: Robot Dynamics and Control Lecture 4: Velocity Kinematics Jacobian and Singularities Torque/Force Relationship Inverse Velocity Problem Reading: SHV Chapter 4 Mechanical Engineering Hanz Richter, PhD MCE503 p.1/16 Angular Velocity as a Vector wk A point belonging to a rigid body may undergo pure rotational motion about a fixed axis (given by vector k). Then the trajectory of the point is a circle. If θ is the rate of rotation, the angular velocity vector is w = θk θ If a point belonging to the same body has position vector r, then its linear velocity is given by r v = w r v = w r MCE503 p.2/16

2 Skew Symmetric Matrices A skew symmetric matrix S is defined by the property S + S T = 0 For this to occur, the diagonal elements must be zero and we must have s ij = s ji The set of all 3-by-3 skew symmetric matrices is called so(3). To each vector a = [a x, a y, a z ] R 3 we associate a matrix S(a) so(3) of the form S(a) = a z 0 a z a y 0 a x a y a x 0 MCE503 p.3/16 Properties of Skew Symmetric Matrices SHV lists 4 properties. The most remarkable ones are 2 and 4: S(a)p = a p X T SX = 0 Note that any matrix A can be decomposed as the sum of a symmetric component and a skew symmetric component. The last property says that the skew symmetric part does not influence a quadratic function defined using A. MCE503 p.4/16

3 Derivative of Rotation Matrices It is recommended that you familiarize yourself with basic matrix calculus. These results are very useful in many areas of engineering. The Matrix Reference Manual (Imperial College, London) is a very popular document: Following the derivations in SHV we see that dr dθ = SR(θ) where S is a skew symmetric matrix (Eq. 4.13). When θ = θ(t), we regard R as a function of time. So dr dt = S(w)R(t) The w vector can be uniquely extracted from S recalling the earlier association between a skew symmetric matrix and a vector. The w vector is actually the angular velocity, as shown in p.125. MCE503 p.5/16 Addition of Angular Velocities Suppose we have a series of coordinate frames with a fixed, common origin. The rotation matrices Rj i (t) are functions of time, defining the relative orientation of pairs of frames. Let wi,j k denote the angular velocity associated with the time derivative of Rj i, with components expressed in the coordinate frame k. We wish to find the angular velocity of the end effector frame relative to the world frame, expressed in world frame coordinates. That is, we need to find w0,n. 0 The derivation gives that is, w 0 0,n = w 0 0,1 + R 0 1w 1 1,2 + R 0 2w 2 2, R 0 n 1w n 1 n 1,n w 0 0,n = w 0 0,1 + w 0 1,2 + 0w 0 2, w 0 n 1,n If you think of w i,j as a vector, all we are doing is transforming the various relative angular velocities to the world frame and superimposing them. MCE503 p.6/16

4 Linear Velocity Formula Suppose point p is rigidly attached to frame 1, which moves (translation + rotation) w.r.t. frame 0, according to the homogeneous transformation P 0 = H 0 1(t)P 1 Here P = [p T 1] T. A simple derivation shows that the linear velocity of p w.r.t. the world frame is which is the well-known formula. ṗ 0 = S(w)R 0 1p 1 + ȯ = w r + v Example: We solve Problem 4.15 in class. MCE503 p.7/16 The Jacobian In general, the Jacobian of a differentiable function f : R m R n is an n-by-m matrix J where the entries are the first partial derivatives: J ij = f i x j for i = 1..n, j = 1..m. A transformation matrix T 0 n(q) is such a function of several variables (exactly equal to the number of joints) returning a vector of dimension 4 (3D case). In our case, if we form a vector with all joint velocities q, two Jacobians will be found so that v 0 n = J v q w 0 n = J w q An overall Jacobian can be formed as J = [J T v J T w] T, of dimensions 6-by-n (n is the number of links). MCE503 p.8/16

5 Jacobian... As derived in SHV, the J w rows of J are given as J w = [ρ 1 z 0 ρ 2 z 1... ρ n z n 1 ] where ρ i = 1 if joint i is revolute and i = 0 if it is prismatic. z i is the axis of rotation of joint i expressed in the world frame coordinates. The i th column of J v is given as J vi = o0 n q i where o 0 n is the position vector for the origin of the end effector frame, in world coordinates. Note that the velocity ȯ 0 n is obtained by a linear combination of the columns of J v where the scalars are the joint velocities (Eq. 4.49). Therefore J v can be obtained by taking one joint at a time (assuming the others are not moving) and assembling the resulting columns into a matrix. MCE503 p.9/16 Summary of Formulas for the Jacobian Please read Section 4.6 in detail. The upper half of the Jacobian is given as J v = [J v1,...j vn ] where the columns are given by z i 1 (o n o i 1 ) for revolute joint i J vi = for prismatic joint i z i 1 The lower half is given also in columns by z i 1 for revolute joint i J vi = 0 for prismatic joint i MCE503 p.10/16

6 Examples Note: The coordinates of z i w.r.t. the world frame appear as the first three elements of the third column of T 0 i, while o i appears as the first three elements of the fourth column. Therefore we need to find only the third and fourth columns of T. 4.5, 4.6 in class Prob in class Homework 2: 4-17 MCE503 p.11/16 Singularities For a given configuration q, we can view the Jacobian as a linear transformation mapping joint velocities q to end-effector velocities ζ = [v T w T ] T. The velocity vector is then a linear combination of the columns of J: that is, ζ = J(q) q. ζ = J 1 q 1 + J 2 q J n q n For 3D manipulators ζ is a 6-vector. A condition on J(q) so that a q exists that results in any desired ζ is that J(q) be full rank. (Please see Appendix B if needed). When J(q) loses rank for a particular value of q, we encounter a singularity. When the manipulator is in a singular configuration, not all directions of motion are possible for the end effector. We consider the effect of singularities in terms of end effector forces and torques later in the course. MCE503 p.12/16

7 Static Torque/Force Relationship Suppose the end effector is in static equilibrium and subjected to forces and moments (for example pressing against a part). These forces and moments will create torques at revolute joints and forces at prismatic joints. If we organize the end-effector external forces F i and moments n i in a vector F = [F x F y F z n x n y n z ] and the resulting joint torques in a vector τ, the relationship is given by Eq in SHV: τ = J T (q)f It is important to note that this relationship is valid only under static equilibrium. See Example MCE503 p.13/16 Linear Algebra Facts: Moore-Penrose Pseudoinverse Suppose J is an m-by-n matrix with m n. Then the linear system of equations Jx = a may or may not have solutions. If m n, solutions exist only if vector a is a linear combination of the columns of J (a in the range space or column space of J, usually expressed as a col J. To check: a col J if and only if rank(j)=rank([j a]) If m < n and rank(j)=m, JJ T is an m-by-m invertible matrix. We use this to define J + = J T (JJ T ) 1 called the right pseudoinverse of J, since JJ + = I. MCE503 p.14/16

8 Pseudoinverse... As an example, we determine the solvability and find one solution to the system of equations x + y z + w = 3 x y + 2z w = 8 2x + y + z + 3w = 9 using the Matlab commands rank and pinv MCE503 p.15/16 Inverse Velocity Problem We wish to determine the vector of joint velocities q which results in a desired end-effector velocity ζ. The Jacobian relationship ζ = J q can be inverted to give q = J 1 ζ only if J is square and has no singularities. When the robot has more than 6 joints, J is be non-square. If rank J=6, we can use the pseudoinverse to find a solution. Direct substitution shows that q = J + ζ + (I J + J)b is a solution, with b being an arbitrary vector. Note that any vector of the form (I J + J)b is such that J(I J + J)b = 0 which shows that it is possible to have a zero end-effector velocity with non-zero joint velocities! MCE503 p.16/16

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