Inverse Kinematics Kris Hauser CS B659: Principles of Intelligent Robot Motion Spring 2013 1

Articulated Robot Robot: usually a rigid articulated structure Geometric CAD models, relative to reference frames A configuration specifies the placement of those frames (forward kinematics) q1 q2

Inverse Kinematics Problem: given desired position of point on robot (end effector), what joint angles reach it?

Inverse Kinematics Problem: given desired position of point on robot (end effector), what joint angles reach it? q4 q2 q3 q1

Inverse Kinematics Bring point xn on link n to target xT Find q s.t. xT = Tn(q)xn q4 q2 q3 q1

Solving a general equation Solve f(q) = 0 (vector valued nonlinear function) Can include rotation constraints, multiple IK targets q4 q2 q3 q1

Two Approaches for IK Analytical. Write out the equation in terms of q and “invert” it Sometimes there are simple solutions for certain kinematic structures (e.g., industrial robots). If so, computation is fast. Numerical. Iteratively move q to a solution point f(q)=0 General purpose Can incorporate joint limits easily May fall into local minima

Analytical endpoint positioning for a 2R robot arm Without loss of generality, let joint 2 and E.E. position lie on x axis at the reference frame, joint 1 at origin x(q)-xD = 0 L2 L1 xD q2 q1

Analytical endpoint positioning for a 2R robot arm Without loss of generality (why?), let joint 2 and E.E. position lie on x axis at the reference frame, joint 1 at origin x(q)-xD = 0 L2 ||x(q)||2 = ||xD||2 (lhs depends only on q2) ||x(q)||2 = (L1 + c2L2)2 + (s2L2)2 = L12 + 2c2L2L1 + c22L22 + s22L22 = L12 + L22 + 2c2L2L1 L1 xD q2 q1

Analytical endpoint positioning for a 2R robot arm Without loss of generality, let joint 2 and E.E. position lie on x axis at the reference frame, joint 1 at origin x(q)-xD = 0 L2 ||x(q)||2 = ||xD||2 (lhs depends only on q2) ||x(q)||2 = (L1 + c2L2)2 + (s2L2)2 = L12 + 2c2L2L1 + c22L22 + s22L22 = L12 + L22 + 2c2L2L1 c2 = (||xD||2 - L12 - L22)/(2L2L1) If rhs > 1, xD is out of reach If rhs < -1, xD is inside “inner circle” L1 xD q2 q1 c2(q2) Elbow up Elbow down

Analytical endpoint positioning for a 2R robot arm Without loss of generality, let joint 2 and E.E. position lie on x axis at the reference frame, joint 1 at origin x(q)-xD = 0 q2up ||x(q)||2 = ||xD||2 (lhs depends only on q2) ||x(q)||2 = (L1 + c2L2)2 + (s2L2)2 = L12 + 2c2L2L1 + c22L22 + s22L22 = L12 + L22 + 2c2L2L1 c2 = (||xD||2 - L12 - L22)/(2L2L1) If rhs > 1, xD is out of reach If rhs < -1, xD is inside “inner circle” L1 xD q2down q1 c2(q2) Elbow up Elbow down

Analytical endpoint positioning for a 2R robot arm Without loss of generality, let joint 2 and E.E. position lie on x axis at the reference frame, joint 1 at origin x(q)-xD = 0 q2 Once q2 is found, consider angle θD of xD w.r.t origin L1 xD q1

Analytical endpoint positioning for a 2R robot arm Without loss of generality, let joint 2 and E.E. position lie on x axis at the reference frame, joint 1 at origin x(q)-xD = 0 q2 Once q2 is found, consider angle θD of xD w.r.t origin Compute angle θ of E.E. at q=(0,q2) L1 xD q1

Analytical endpoint positioning for a 2R robot arm Without loss of generality, let joint 2 and E.E. position lie on x axis at the reference frame, joint 1 at origin x(q)-xD = 0 Once q2 is found, consider angle θD of xD w.r.t origin Compute angle θ of E.E. at q=(0,q2) Let q1 = θD-θ Done! L1 xD q1 q2

Analytical positioning and orienting for a 3R robot arm Left as an exercise Hint: consider solution in prior slides for choosing the first two joint angles so that the third joint is located at the correct location. Then pick the third.

Drawbacks to Analytical IK Most 6DOF robots encountered in practice are designed to have convenient analytical IK solutions (e.g., 3 intersecting orthogonal axes) General methods for 6DOF robots require solving a high-degree (16) polynomial, which is computationally expensive and suffers from numerical difficulties What about a redundant manipulator (> 6 DOF)?

Papers

Multiplicity issue Let n = # of DOFs, m = # of constraints Roughly, in common cases If n < m, there will be 0 IK solutions If n = m, there may be 1 or more solution If n > m, there will either be 0 or infinite number of solutions Singularities: the uncommon case

Null Space A motion from q->q’ that maintains f(q)=0 is known as a motion in the null space.

Null Space Null space velocity dq must satisfy J(q)dq = 0 => dq lies in the null space of J(q) For any vector y, (I-J+J)y lies in the null space Recall J+ is the pseudoinverse of J A basis of the null space can be found by SVD J = UWVT Let the last k diagonal entries of W be 0, first n-k nonzero WVTdq = 0 First n-k entries of VTdq must be zero Last k entries of VTdq may be non zero => Last k columns of V are a basis for null space

Reminder: Project Proposals 2-3 paragraphs, due 2/1 Include: Title, group members Motivation, significance, and expected demonstration. List of 3-4 milestones Milestone 1 must be completed, or else you will feel like a total failure. Usually complete this before spring break. Milestone 2 should be completed, or else you will feel bad and will deserve a less than stellar grade. Completing milestone 3 will make you (and me) feel pleased with your project. Milestone 4 will make you (and me) very excited, yet is still possible. (Don‘t promise to cure cancer.) Any other details, e.g., implementation, that may be relevant

Recap Two general ways of solving inverse kinematics: analytical and numerical With nonredundant manipulators, there are a finite number of solutions (usually > 1, without joint limits) With redundant manipulators, there are an infinite number of solutions Null space motions