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

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

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).

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.

Inverse Dynamics Kinematic 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

Space DiagramSegments

Ordering of Segments Start with the terminal 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

Free-body Diagram Make a free-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?)

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

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

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

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

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

Equations are Indeterminate Too many Unknowns, Too few Equations In two dimensions there are three equations of motion. In three dimensions there are six 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. 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. Thus, equations of motion are indeterminate and cannot be solved.

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) The solution is to reduce the number of unknowns to three (or six for 3D) These are called the net force (F x, F y ) and the net moment of force (M z ) for 2D These are called the net force (F x, F y ) and the net moment of force (M z ) for 2D or (F x, F y, F z ) and (M x, M y, M z ) for 3D

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

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

5d. Force couple Force couple (F, –F*) is equal to free moment of force (M F k)

5e. Replace couple with free moment (M F k) Replace couple with free moment (M F k)

5. Show all forces again Show all forces again

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

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

8. Show complete free-body diagram Show complete free-body diagram

9. Show position vectors (r ankle, r ground ) Show position vectors (r ankle, r ground )

Three Equations of Motion for the Foot Σ F x = ma x : F x (ankle) + F x (ground) = ma x (foot) Σ F y = ma y : Σ F y = ma y : F y (ankle) + F y (ground) – mg = ma y (foot) Σ M z = I  : M z (ankle) + [r ankle × F ankle ] z + [r ground × F ground ] z = I foot  (foot)

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 = r x F 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 = r x F M z = [ r × F ] z = r x F y – r y F x r ankle = (x ankle – x foot, y ankle – y foot ) [... ] z means take the scalar portion in the z- direction [... ] z means take the scalar portion in the z- direction

Equations of Motion for Foot Solve for the Unknowns Σ F x = ma x : F x (ankle) = ma x (foot) – F x (ground) Σ F y = ma y : F y (ankle) = ma y (foot) – F y (ground) + mg Σ M z = I  : M z (ankle) = I foot  (foot) – [r ankle × F ankle ] z – [r ground × F ground ] z – [r ground × F ground ] z Note, moment of inertia (I foot ) is about the centre of the gravity of the foot, not the proximal or distal end)

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) 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 Reactions are opposite in direction to actions I.e., reaction force = – action force reaction moment = – action moment I.e., reaction force = – action force reaction moment = – action moment

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

Equations of Motion for Leg Σ F x = ma x : F x (knee) – F x (ankle) = ma x (leg) Σ F y = ma y : F y (knee) – F y (ankle) – mg = ma y (leg)  M z = I a: M z (knee) + [r knee × F knee ] z – M z (ankle) + [r ankle × – F ankle ] z = I leg  (leg) + [r ankle × – F ankle ] z = I leg  (leg)

Equations of Motion for Thigh Σ F x = ma x : F x (hip) – F x (knee) = ma x (thigh) Σ F y = ma y : F y (hip) – F y (knee) – mg = ma y (thigh)  M z = I a: M z (hip) + [r hip × F hip ] z – M z (knee) + [r knee × – F knee ] z = I thigh  (thigh) + [r knee × – F knee ] z = I thigh  (thigh)

Interpretation Mathematical concepts not anatomical kinetics These forces and moments are mathematical constructs NOT actual forces and moments. 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. 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. Furthermore, there is no certain method to apportion the net forces and moments to the individual anatomical structures.

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

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