Inverse Dynamics

D. Gordon E. Robertson, Ph.D.School of Human Kinetics

University of Ottawa

PThe process of deriving the kinetics (i.e.,forces and moments of force) necessary toproduce the kinematics (observed motion) ofbodies with known inertial properties (i.e.,mass and moment of inertia).

PTypically the process is used to computeinternal forces and moments when externalforces are known and there are no closedkinematic chains.

Inverse Dynamics

Definition

The following slides outline the derivation of the equations for determining net forces andmoments of force for the two-dimensionalcase.

The three-dimensional case follows the sameprocedure.

Two-dimensional Derivation

Step-by-step explanation of derivation

PFirst divide body into kinematic chains

PNext divide chains into segments

PAssume that each segment is a “rigid body”

PAssume that each joint is rotationallyfrictionless

Inverse Dynamics

Kinematic Chains, Segment & Assumptions

Space Diagram Segments

PStart with the terminal segment of akinematic chain

PThe ground reaction forces of the terminalsegment must be known (i.e., measured) orzero (i.e., free-ended)

P If not, use the other end of the chain

P If external forces are unknown, measurethem, otherwise, you cannot analyze

Inverse Dynamics

Ordering of Segments

PMake a free-body diagram (FBD) of theterminal segment

PRules:< 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 beassumed to be zero, is surface frictionless)

Inverse Dynamics

Free-body Diagram

1.

Draw free-bodydiagram offoot (aterminalsegment)

2.

Addweightvector tofree-bodydiagram atcentre ofgravity centre of gravity

(xfoot, yfoot)

weight

3.

Addgroundreactionforce atcentre ofpressure

Fground

centre of pressure(xground, yground)

4.

Add allmuscleforces attheir pointsofapplication

force fromtriceps surae

force fromtibialis anterior

5.

Add bone-on-bone &ligamentforces &jointmomentof force

bone-on-bone forces

ligamentforce

joint momentof force

P In two dimensions there are three equationsof motion, but there are more than threeunknown forces (two or more muscles perjoint, several ligaments, skin, joint capsule,bone-on-bone (actually cartilage) forces, etc.)

PThus, equations of motion are indeterminantand cannot be solved.

P In three dimensions there are six equationsbut many more than six unknowns.

Equations are Indeterminant

Too many Unknowns, Too few Equations

PThe solution is to reduce the number ofunknowns to three (or six for 3D)

PThese are called the net force (Fx, Fy) andthe net moment of force (Mz) for 2D

or (Fx, Fy, Fz) and (Mx, My, Mz) for 3D

Solution

Reduce number of unknowns to three (2D) or six (3D)

5a.

Considera singlemuscleforce (F)on thefoot

F

5b.

Movemuscleforce tojointcentre(F*)

F

F*

5c.

Addbalancingforce (–F*)

F

F*

–F*

5d.

Forcecouple(F, –F*) isequal tofreemomentof force(MFk)

MF k

F

–F*

=

5e.

Replacecouplewith freemoment

F*

MF k

5.

Show allforcesagain

bone-on-bone forces

ligamentforce

joint momentof force

6.

Replaceeachmuscleforce withequivalentforce andfreemoment offorce

force andmoment fromtriceps surae

force andmoment fromtibialis anterior

7.

Add allankleforces andmomentsto obtainnet ankleforce andmoment offorce

Mankle k Fankle

Fground

8.

Showcompletefree-bodydiagram

mfoot g jFground

FankleMankle k

centre of pressure(xground, yground)

(xankle, yankle)

9.

Showpositionvectors(rankle ,rground)

mfoot g jFground

FankleMankle k

centre of pressure(xground, yground)

(xankle, yankle)

rground

rankle

Ó Fx = max:Fx(ankle) + Fx(ground) = max(foot)

Ó Fy = may:Fy(ankle) + Fy(ground) – mg = may(foot)

Ó Mz = I a:Mz(ankle) + [rankle × Fankle]z

+ [rground × Fground]z = Ifoot a (foot)

Three Equations of Motion forFoot

Newton’s Second Law

P the moment of a force (M) is defined as thecross-product (x) of a position vector (r) andits force (F). I.e., M = r x F

PMz = [ r × F ]z = rx Fy – ry Fx

P rankle = (xankle – xfoot , yankle – yfoot)

P [ ... ]z means take the scalar portion in the z-direction

Moment of Force as Cross Product

Also called Vector Product

Ó Fx = max:Fx(ankle) = max(foot) – Fx(ground)

Ó Fy = may:Fy(ankle) = may(foot) – Fy(ground) + mg

Ó Mz = I a:Mz(ankle) = Ifoot a(foot) – [rankle × Fankle]z

– [rground × Fground]z Note, moment of inertia (Ifoot) is about centre of massof foot, but proximal or distal ends)

Equations of Motion for Foot

Solve for the Unknowns

PNet force and moment of force at proximalend of ankle causes reaction force andmoment of force at distal end of leg

PReactions are opposite in direction to actions

P I.e., reaction force = – action force reaction moment = – action moment

Apply Newton’s Third Law to Leg

Reaction = – Action

10.

Draw free-bodydiagram ofleg

FkneeMknee k

–Mankle k

–Fankle

Ó Fx = max:Fx(knee) – Fx(ankle) = max(leg)

Ó Fy = may:Fy(knee) – Fy(ankle) – mg = may(leg)

Ó Mz = I a:Mz(knee) + [rknee × Fknee]z – Mz(ankle)

+ [rankle × – Fankle]z = Ileg a(leg)

Equations of Motion for Leg

Newton’s Second and Third Laws

Ó Fx = max:Fx(hip) – Fx(knee) = max(thigh)

Ó Fy = may:Fy(hip) – Fy(knee) – mg = may(thigh)

Ó Mz = I a:Mz(hip) + [rhip × Fhip]z – Mz(knee)

+ [rknee × – Fknee]z = Ithigh a (thigh)

Equations of Motion for Thigh

Newton’s Second and Third Laws

PThese forces and moments aremathematical constructs NOT actual forcesand moments.

PThe actual forces inside joints and themoments across joints are higher because ofthe cocontractions of antagonists.

PFurthermore, there is no certain method toapportion the net forces and moments to theindividual anatomical structures.

Interpretation

Mathematical concepts not anatomical kinetics

Computerize the Process

E.g., 2D: Biomech Motion Analysis System;3D: Visual3D, Polygon, KinTools, KinTrak, Kwon3D