Inverse DynamicsInverse Dynamics

D. Gordon E. Robertson, PhD, FCSBD. 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 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

Space Diagram Segments

Ordering of SegmentsOrdering 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 DiagramFree-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?)

1.1.

Draw free-body Draw free-body diagram of terminal diagram of terminal segmentsegment

2.2.

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

centre of gravity(xfoot, yfoot)

weight

3.3.

If ground contact If ground contact add ground add ground reaction force at reaction force at centre of pressurecentre of pressure

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

4.4.

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

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

force fromtriceps surae

force fromtibialis anterior

5.5.

Add bone-on-bone Add bone-on-bone and ligament forces and ligament forces and the frictional and the frictional joint moment of joint moment of forceforce

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

force fromtriceps surae

force fromtibialis anterior

bone-on-bone forces

ligamentforce

Equations are IndeterminateToo many Unknowns, Too few Equations

In In two dimensions two dimensions there are three equations of there are three equations of motion. motion. In In three dimensions three dimensions there are six there are six equations.equations.

But there are more unknown forces But there are more unknown forces (two or (two or more muscles per joint, several ligaments, more muscles per joint, several ligaments, skin, joint capsule, bone-on-bone or cartilage skin, joint capsule, bone-on-bone or cartilage forces, etc.) forces, etc.) then there are equations.then there are equations.

Thus, equations of motion are indeterminate Thus, equations of motion are indeterminate and cannot be solvedand cannot be solved..

Solution:Solution:Reduce number of unknowns to Reduce number of unknowns to

three (2D) or six (3D)three (2D) or six (3D)

The solution is to reduce the number of The solution is to reduce the number of unknowns to three (or six for 3D)unknowns to three (or six for 3D)

These are called the These are called the net force net force ((FFxx, F, Fyy) and the ) and the

net moment of force net moment of force ((MMzz) for 2D ) for 2D

or (or (FFxx, F, Fyy, F, Fzz) and () and (MMxx, M, Myy, M, Mzz) for 3D) for 3D

5a.5a.

Consider a single Consider a single muscle force (muscle force (FF))

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

force fromtriceps surae

force fromtibialis anterior

bone-on-bone forces

ligamentforce

F

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

force fromtriceps surae

force fromtibialis anterior

bone-on-bone forces

ligamentforce

FF

F*

5b.5b.

Move muscle Move muscle forceforce to joint centre to joint centre ((F*F*))

5c.5c.

Add balancing Add balancing force (force (––F*F*))

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

force fromtriceps surae

force fromtibialis anterior

bone-on-bone forces

ligamentforce

FF

F*

–F*

5d.5d.

Force couple Force couple

((FF, , ––FF**) is equal to ) is equal to free moment of free moment of force (force (MMFFkk))

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

force fromtriceps surae

force fromtibialis anterior

bone-on-bone forces

ligamentforce

FF

F*

–F*

MF k

F

–F*

=

5e.5e.

Replace couple Replace couple with free moment with free moment ((MMF F kk))

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

force fromtriceps surae

force fromtibialis anterior

bone-on-bone forces

ligamentforce

FF

F*

–F*

MF k

F

–F*

=

F*

MF k

5.5.

Show all forces Show all forces againagain

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

force fromtriceps surae

force fromtibialis anterior

bone-on-bone forces

ligamentforce

6.6.

Replace muscle Replace muscle forces with forces with equivalent joint equivalent joint forces and free forces and free momentsmoments

centre of gravity(xfoot, yfoot)

weight

Fground

centre of pressure(xground, yground)

force fromtriceps surae

force fromtibialis anterior

bone-on-bone forces

force andmoment fromtriceps surae

force andmoment fromtibialis anterior

ligamentforce

7.7.

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

Mankle k Fankle

Fground

8.8.

Show complete Show complete free-body diagramfree-body diagram

mfoot g jFground

FankleMankle k

centre of pressure(xground, yground)

(xankle, yankle)

9.9.

Show position Show position vectors (vectors (rrankleankle , ,

rrgroundground))

mfoot g jFground

FankleMankle k

centre of pressure(xground, yground)

(xankle, yankle)

rground

rankle

Three Equations of Motion for the Three Equations of Motion for the FootFoot

Σ Σ FFxx = ma = maxx::

FFxx(ankle) + (ankle) + FFxx(ground) = (ground) = mamaxx (foot)(foot)

Σ Σ FFyy = = mamayy::

FFyy (ankle) + (ankle) + FFyy (ground) – (ground) – mgmg = = mamayy (foot)(foot)

ΣΣ M Mzz= = II ::

MMzz (ankle) + [(ankle) + [rrankleankle × × FFankleankle] ] zz

+ [+ [rrgroundground × × FFgroundground] ] zz = = IIfootfoot (foot)(foot)

Moment of Force as Cross ProductMoment of Force as Cross Product

the the moment of a force (moment of a force (MM) ) is defined as the is defined as the cross-productcross-product ( (xx) of a position vector () of a position vector (rr) and ) and its force (its force (FF). I.e., ). I.e., MM = = rr xx FF

MMzz = [ = [ rr × × FF ]]zz = = rrxx F Fyy –– rryy F Fxx

rrankleankle = ( = (xxankleankle – – xxfootfoot , , yyankleankle – – yyfootfoot))

[ ... ][ ... ]zz means take the scalar portion in the means take the scalar portion in the zz- -

directiondirection

Equations of Motion for Equations of Motion for FootFootSolve for the UnknownsSolve for the Unknowns

Σ Σ FFxx = = mamaxx::

FFxx(ankle) = (ankle) = mamaxx(foot) – (foot) – FFxx(ground)(ground)

Σ Σ FFyy = = mamayy::

FFyy(ankle) = (ankle) = mamayy(foot) – (foot) – FFyy(ground) + (ground) + mgmg

Σ Σ MMzz = = I I ::

MMzz(ankle) = (ankle) = IIfootfoot (foot) – [(foot) – [rrankleankle × × FFankleankle] ] zz

– – [[rrgroundground × × FFgroundground] ] zz

Note, Note, moment of inertiamoment of inertia ( (IIfootfoot) is about the centre of the ) is about the centre of the gravity of the foot, not the proximal or distal end) gravity of the foot, not the proximal or distal end)

Apply Newton’s Third Law to Leg: Apply Newton’s Third Law to Leg: Reaction = – ActionReaction = – Action

Net force and moment of force at proximal Net force and moment of force at proximal end of end of ankleankle causes reaction force and moment causes reaction force and moment of force at distal end of the of force at distal end of the leg (shank)leg (shank)

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

reaction moment = reaction moment = –– action moment action moment

10.10.

Draw free-body Draw free-body diagram of diagram of leg leg using net forces using net forces and moments of and moments of forceforce

FkneeMknee k

–Mankle k

–Fankle

Equations of Motion for Equations of Motion for LegLeg

Σ Σ FFxx = ma = maxx::

FFxx(knee)(knee) –– F Fxx(ankle)(ankle) = ma = maxx(leg)(leg)

Σ Σ FFyy = ma = mayy::

FFyy(knee) (knee) –– FFyy(ankle)(ankle) – mg = ma – mg = mayy(leg)(leg)

MMzz = I a = I a::

MMzz(knee)(knee) + + [[rrkneeknee × F × Fkneeknee]]zz –– M Mzz(ankle)(ankle)

+ [+ [rrankleankle × × –– F Fankleankle]]zz = I = Ilegleg (leg)(leg)

Equations of Motion for Equations of Motion for ThighThigh

Σ Σ FFxx = ma = maxx::

FFxx(hip)(hip) –– F Fxx(knee)(knee) = ma = maxx(thigh)(thigh)

Σ Σ FFyy = ma = mayy::

FFyy(hip) (hip) –– FFyy(knee)(knee) – mg = ma – mg = mayy(thigh)(thigh)

MMzz = I a = I a::

MMzz(hip)(hip) + + [[rrhiphip × F × Fhiphip]]zz –– M Mzz(knee)(knee)

+ [+ [rrkneeknee × × –– F Fkneeknee]]zz = I = Ithighthigh (thigh)(thigh)

InterpretationInterpretationMathematical concepts not anatomical kineticsMathematical concepts not anatomical kinetics

These forces and moments are These forces and moments are mathematical mathematical constructsconstructs NOT actual forces and moments. NOT actual forces and moments.

The actual forces inside joints and the The actual forces inside joints and the moments across joints are moments across joints are higherhigher because of because of the the cocontractionscocontractions of antagonists. of antagonists.

Furthermore, there is no certain method to Furthermore, there is no certain method to apportion the net forces and moments to the apportion the net forces and moments to the individual anatomical structures.individual anatomical structures.

Computerize the ProcessComputerize the Process

Examples:Examples: 2D: Biomech MAS, Ariel PAS, Hu-m-an2D: Biomech MAS, Ariel PAS, Hu-m-an 3D: Visual3D, Polygon, KinTools, KinTrak, 3D: Visual3D, Polygon, KinTools, KinTrak,

Kwon3D, SimiKwon3D, Simi