CSCE 689: Forward Kinematics and Inverse Kinematics

1. CSCE 689: Forward Kinematics and Inverse Kinematics Jinxiang Chai

2. Outline Kinematics • Forward kinematics • Inverse kinematics

3. Kinematics The study of movement without the consideration of the masses or forces that bring about the motion

4. Degrees of freedom (Dofs) The set of independent displacements that specify an object’s pose

5. Degrees of freedom (Dofs) The set of independent displacements that specify an object’s pose • How many degrees of freedom when flying?

6. Degrees of freedom (Dofs) The set of independent displacements that specify an object’s pose • How many degrees of freedom when flying? • So the kinematics of this airplane permit movement anywhere in three dimensions • Six • x, y, and z positions • roll, pitch, and yaw

7. Degrees of Freedom • How about this robot arm? • Six again • 2-base, 1-shoulder, 1-elbow, 2-wrist

8. Work Space vs. Configuration Space • Work space • The space in which the object exists • Dimensionality • R3 for most things, R2 for planar arms • Configuration space • The space that defines the possible object configurations • Degrees of Freedom • The number of parameters that are necessary and sufficient to define position in configuration

9. Forward vs. Inverse Kinematics • Forward Kinematics • Compute configuration (pose) given individual DOF values • Inverse Kinematics • Compute individual DOF values that result in specified end effector position

10. End Effector Base Example: Two-link Structure • Two links connected by rotational joints θ2 l2 l1 θ1 X=(x,y)

11. End Effector Base Example: Two-link Structure • Animator specifies the joint angles: θ1θ2 • Computer finds the position of end-effector: x θ2 l2 l1 θ1 X=(x,y) (0,0) x=f(θ1, θ2)

12. End Effector Base Example: Two-link Structure • Animator specifies the joint angles: θ1θ2 • Computer finds the position of end-effector: x θ2 θ2 l2 l1 θ1 X=(x,y) (0,0) x = (l1cosθ1+l2cos(θ2+ θ2) y = l1sinθ1+l2sin(θ2+ θ2))

13. End Effector Base Forward kinematics • Create an animation by specifying the joint angle trajectories θ2 θ2 l2 l1 θ1 X=(x,y) (0,0) θ1 θ2

14. Base Inverse kinematics • What if an animator specifies position of end-effector? θ2 θ2 l2 l1 End Effector θ1 X=(x,y) (0,0)

15. Base Inverse kinematics • Animator specifies the position of end-effector: x • Computer finds joint angles: θ1θ2 θ2 θ2 l2 l1 End Effector θ1 X=(x,y) (0,0) (θ1, θ2)=f-1(x)

16. Base Inverse kinematics • Animator specifies the position of end-effector: x • Computer finds joint angles: θ1θ2 θ2 θ2 l2 l1 End Effector θ1 X=(x,y) (0,0)

17. Why Inverse Kinematics? • Basic tools in character animation - key frame generation - animation control - interactive manipulation • Computer vision (vision based mocap) • Robotics • Bioninfomatics (Protein Inverse Kinematics)

18. Inverse kinematics • Given end effector position, compute required joint angles • In simple case, analytic solution exists • Use trig, geometry, and algebra to solve

19. Inverse kinematics • Analytical solution only works for a fairly simple structure • Iterative approach needed for a complex structure

20. Iterative approach • Inverse kinematics can be formulated as an optimization problem

21. Base Iterative approaches Find the joint angles θ that minimizes the distance between the character position and user specified position θ2 θ2 l2 l1 θ1 C=(Cx,Cy) (0,0)

22. Base Iterative approaches Find the joint angles θ that minimizes the distance between the character position and user specified position θ2 θ2 l2 l1 θ1 C=(c1,c2) (0,0)

23. Iterative approach Mathematically, we can formulate this as an optimization problem: The above problem can be solved by many nonlinear optimization algorithms: - Steepest descent - Gauss-newton - Levenberg-marquardt, etc

24. Gauss-newton approach Step 1: initialize the joint angles with Step 2: update the joint angles:

25. Gauss-newton approach Step 1: initialize the joint angles with Step 2: update the joint angles: How can we decide the amount of update?

26. Gauss-newton approach Step 1: initialize the joint angles with Step 2: update the joint angles:

27. Gauss-newton approach Step 1: initialize the joint angles with Step 2: update the joint angles: Known!

28. Gauss-newton approach Step 1: initialize the joint angles with Step 2: update the joint angles: Taylor series expansion

29. Gauss-newton approach Step 1: initialize the joint angles with Step 2: update the joint angles: Taylor series expansion rearrange

30. Gauss-newton approach Step 1: initialize the joint angles with Step 2: update the joint angles: Taylor series expansion rearrange Can you solve this optimization problem?

31. Gauss-newton approach Step 1: initialize the joint angles with Step 2: update the joint angles: Taylor series expansion rearrange This is a quadratic function of

32. Gauss-newton approach • Optimizing an quadratic function is easy • It has an optimal value when the gradient is zero

33. Gauss-newton approach • Optimizing an quadratic function is easy • It has an optimal value when the gradient is zero b J Δθ Linear equation!

34. Gauss-newton approach • Optimizing an quadratic function is easy • It has an optimal value when the gradient is zero b J Δθ

35. Gauss-newton approach • Optimizing an quadratic function is easy • It has an optimal value when the gradient is zero b J Δθ

36. Jacobian matrix Dofs (N) • Jacobian is a Mby Nmatrix that relates differential changes of θ to changes of C • Jacobian maps the velocity in joint space to velocities in Cartesian space • Jacobian depends on current state The number of constraints (M)

37. Jacobian matrix • Jacobian maps the velocity in joint angle space to velocities in Cartesian space (x,y) θ2 θ1

38. Jacobian matrix • Jacobian maps the velocity in joint angle space to velocities in Cartesian space (x,y) A small change of θ1and θ2results in how much change of end-effector position (x,y) θ2 θ1

39. Jacobian matrix • Jacobian maps the velocity in joint angle space to velocities in Cartesian space (x,y) A small change of θ1and θ2results in how much change of end-effector position (x,y) θ2 θ1

40. Base Jacobian matrix: 2D-link structure θ2 θ2 l2 l1 θ1 C=(c1,c2) (0,0)

41. Base Jacobian matrix: 2D-link structure θ2 θ2 l2 l1 θ1 C=(c1,c2) (0,0) f2 f1

42. Base Jacobian matrix: 2D-link structure θ2 θ2 l2 l1 θ1 C=(c1,c2) (0,0)

43. Base Jacobian matrix: 2D-link structure θ2 θ2 l2 l1 θ1 C=(c1,c2) (0,0)

44. Base Jacobian matrix: 2D-link structure θ2 θ2 l2 l1 θ1 C=(c1,c2) (0,0)

45. Base Jacobian matrix: 2D-link structure θ2 θ2 l2 l1 θ1 C=(c1,c2) (0,0)

46. Base Jacobian matrix: 2D-link structure θ2 θ2 l2 l1 θ1 C=(c1,c2) (0,0)

47. Gauss-newton approach Step 1: initialize the joint angles with Step 2: update the joint angles: Step size: specified by the user

48. Another Example lower arm middle arm Upper arm base How to compute forward kinematics? A 2D lamp with 6 degrees of freedom

49. Another Example lower arm middle arm Upper arm base A 2D lamp with 6 degrees of freedom

50. Another Example base A 2D lamp with 6 degrees of freedom