Inverse Kinematics

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

2. q2 q1 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)

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

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

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

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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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 θDof xD w.r.t origin L1 xD q1

13. 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 θDof xD w.r.t origin Compute angle θof E.E. at q=(0,q2) L1 xD q1

14. 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 θDof xD w.r.t origin Compute angle θof E.E. at q=(0,q2) Let q1 = θD-θ Done! L1 xD q1 q2

15. 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.

16. 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)?

17. Papers

18. 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

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

20. 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 VTdqmay be non zero • => Last k columns of V are a basis for null space

21. 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

22. 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