Inverse Dynamics

1. Inverse Dynamics D. Gordon E. Robertson, PhD, FCSB School of Human Kinetics University of Ottawa

2. Inverse Dynamics(definition) The process of deriving the kinetics (i.e., forces and moments of force) necessary to produce the kinematics (observed motion) of bodies with known inertial properties (i.e., mass and moment of inertia). Typically the process is used to compute internal forces and moments when external forces are known and there are no closed kinematic chains (e.g., batting, shoveling).

3. Two-dimensional Derivation • The following slides outline the derivation of the equations for determining net forces and moments of force for the two-dimensional case. • The three-dimensional case follows the same procedure.

4. Inverse DynamicsKinematic Chains, Segment & Assumptions • First, divide body into kinematic chains • Next, divide chains into segments • Assume that each segment is a “rigid body” • Assume that each joint is rotationally frictionless

5. Space Diagram Segments

6. Ordering of Segments • Start with theterminal segment of a kinematic chain • The ground reaction forces of the terminal segment must be known (i.e., measured) or zero (i.e., free-ended) • If not, start at the other end of the chain (i.e., top-down versus bottom-up) • If external forces are unknown, measure them, otherwise, you cannot proceed

7. Free-body Diagram • Make afree-body diagram (FBD) of the terminal segment • Rules: • Add all known forces that directly influence the free-body • Wherever free-body contacts the environment or another body add unknown force and moment • Simplify unknown forces when possible (i.e., does a force have a known direction, can force be assumed to be zero, is surface frictionless?)

8. 1. • Draw free-body diagram of terminal segment

9. 2. • Add weight vector to free-body diagram at centre of gravity

10. 3. • If ground contact add ground reaction force at centre of pressure

11. 4. • Add all muscle forces at their points of application (insertions)

12. 5. • Add bone-on-bone and ligament forces and the frictional joint moment of force

13. Equations are IndeterminateToo many Unknowns, Too few Equations • In two dimensions there are three equations of motion. In three dimensions there are six equations. • But there are more unknown forces (two or more muscles per joint, several ligaments, skin, joint capsule, bone-on-bone or cartilage forces, etc.) then there are equations. • Thus, equations of motion are indeterminate and cannot be solved.

14. Solution:Reduce number of unknowns to three (2D) or six (3D) • The solution is to reduce the number of unknowns to three (or six for 3D) • These are called the net force (Fx, Fy) and the net moment of force (Mz) for 2D or (Fx, Fy, Fz) and (Mx, My, Mz) for 3D

15. 5a. • Consider a single muscle force (F)

16. 5b. • Move muscle force to joint centre (F*)

17. 5c. • Add balancing force (–F*)

18. 5d. • Force couple (F, –F*) is equal to free moment of force (MFk)

19. 5e. • Replace couple with free moment (MF k)

20. 5. • Show all forces again

21. 6. • Replace muscle forces with equivalent joint forces and free moments

22. 7. Add all ankle forces and moments to obtain net ankle force and moment of force

23. 8. Show complete free-body diagram

24. 9. Show position vectors (rankle , rground)

25. Three Equations of Motion for the Foot Σ Fx = max: Fx(ankle) + Fx(ground) = max(foot) Σ Fy = may: Fy(ankle) + Fy(ground) – mg = may(foot) ΣMz= Ia: Mz(ankle) + [rankle × Fankle] z + [rground × Fground] z = Ifoota (foot)

26. Moment of Force as Cross Product • the moment of a force (M) is defined as the cross-product (x) of a position vector (r) and its force (F). I.e., M = rxF Mz = [r × F]z = rxFy–ryFx rankle = (xankle – xfoot , yankle – yfoot) • [ ... ]z means take the scalar portion in the z- direction

27. Equations of Motion for FootSolve for the Unknowns Σ Fx = max: Fx(ankle) = max(foot) – Fx(ground) Σ Fy = may: Fy(ankle) = may(foot) – Fy(ground) + mg Σ Mz = I a: Mz(ankle) = Ifoota(foot) – [rankle × Fankle] z – [rground × Fground] z Note, moment of inertia (Ifoot) is about the centre of the gravity of the foot, not the proximal or distal end)

28. Apply Newton’s Third Law to Leg: Reaction = – Action • Net force and moment of force at proximal end of ankle causes reaction force and moment of force at distal end of the leg (shank) • Reactions are opposite in direction to actions • I.e., reaction force = – action force reaction moment = – action moment

29. 10. Draw free-body diagram of leg using net forces and moments of force

30. Equations of Motion for Leg Σ Fx = max: Fx(knee)–Fx(ankle) = max(leg) Σ Fy = may: Fy(knee) –Fy(ankle) – mg = may(leg) SMz = I a: Mz(knee) + [rknee × Fknee]z–Mz(ankle) + [rankle × –Fankle]z = Ilega(leg)

31. Equations of Motion for Thigh Σ Fx = max: Fx(hip)–Fx(knee) = max(thigh) Σ Fy = may: Fy(hip) –Fy(knee) – mg = may(thigh) SMz = I a: Mz(hip) + [rhip × Fhip]z–Mz(knee) + [rknee × –Fknee]z = Ithigha(thigh)

32. InterpretationMathematical concepts not anatomical kinetics • These forces and moments are mathematical constructs NOT actual forces and moments. • The actual forces inside joints and the moments across joints are higher because of the cocontractions of antagonists. • Furthermore, there is no certain method to apportion the net forces and moments to the individual anatomical structures.

33. Computerize the Process • Examples: • 2D: Biomech MAS, Ariel PAS, Hu-m-an • 3D: Visual3D, Polygon, KinTools, KinTrak, Kwon3D, Simi