Inverse Dynamics

APA 6903

Fall 2013 2

What is “Inverse Dynamics”?

APA 6903

Fall 2013 3

What is “Inverse Dynamics”?

Motion – kinematics Force – kinetics Applied dynamics

APA 6903

Fall 2013 4

What is “Inverse Dynamics”?

Kinematics Kinetics

“COMPUTATION”

Resultant joint loading

APA 6903

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Inverse Dynamics

Using Newton’s Laws

• Fundamentals of mechanics• Principles concerning motion and movement• Relates force with motion• Relates moment with angular velocity and angular

acceleration

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Inverse Dynamics

Newton’s Laws of motion

• 1st:

• 2nd:

• 3rd: a given action creates an equal and opposite reaction

amF

F

0

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Inverse Dynamics

If an object is at equilibriated rest = static

If an object is in motion = dynamic

If object accelerates, inertial forces calculated based on Newton’s 2nd Law (ΣF = ma)

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Dynamics

Two approaches to solve for dynamics

FForces

F = maEquations of

motion

∫∫Double

integration

xDisplacements

xDisplacements

d2x / dt2

Double differentiation

F = maEquations of

motion

FForces

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Dynamics

Direct method• Forces are known• Motion is calculated by integrating once to obtain

velocity, twice to obtain displacement

FForces

F = maEquations of

motion

∫∫Double

integration

xDisplacement

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Dynamics

Inverse method• Displacements/motion are known• Force is calculated by differentiating once to

obtain velocity, twice to obtain acceleration

xDisplacements

d2x / dt2

Double differentiation

F = maEquations of

motion

FForces

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Objective

Determine joint loading by computing forces and moments (kinetics) needed to produce motion (kinematics) with inertial properties (mass and inertial moment)

Representative of net forces and moments at joint of interest

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Objective

Combines

• Anthropometry: anatomical landmarks, mass, length, centre of mass, inertial moments

• Kinematics: goniometre, reflective markers, cameras

• Kinetics: force plates

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1st Step

Establish a model

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1st Step

Establish the model

Inertial mass and force often approximated by modelling the leg as a assembly of rigid body segments

Inertial properties for each rigid body segment situated at centre of mass

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Segmentation

Assume

• Each segment is symmetric about its principal axis

• Angular velocity and longitudinal acceleration of segment are neglected

• Frictionless

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2nd Step

Measure ALL external reaction forces

Appoximate inertial properties of members

Locate position of the common centres in space

Free body diagram: • forces/moments at joint articulations• forces/moments/gravitational force at centres of mass

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Free Body Diagram

Statics – analysis of physical systems Statically determinant

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Free Body Diagram

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3rd Step

Static equilibrium of segments

Forces/moments known at foot segment

Using Newton-Euler formulas, calculation begins at foot, then to ankle

Proceed from distal to proximalKNOWNS

UNKNOWNS

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FBD of Foot

$%#&?! Multiple unknowns Centre of

gravity

Fg

Centre of pressure

Triceps sural force

Anterior tibial muscle force

Bone force

Ligament force

Joint moment

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Simplify

Multiple unknown force and moment vectors• Muscles, ligaments, bone, soft tissues, capsules, etc.

Reduction of unknown vectors to:• 3 Newton-Euler equilibrium equations, for 2-D (Fx, Fy, Mz)

• 6 equations, for 3-D (Fx, Fy, Fz, Mx, My, Mz)

Representative of net forces/moment

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Simplification

Displace forces to joint centre

Force equal and opposite

Centre gravity

Fr

Centre of pressure

FFoot muscle

forces

F*Force at joint

centre

-F*Force equal and opposite

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Simplification

Replace coupled forces with moment

Centre of gravity

FFoot muscle

force

F*Force at joint

centre

-F*Force equal and opposite

M

Moment

Fr

Centre of pressure

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Simplification

Representation net moments and forces at ankle

Freaction

xreaction, yreaction

mfootg

Fankle

xankle, yankle

Mankle

rcm,dist

rcm,prox

cm = centre of mass prox = proximal dist = distal

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3rd Step

f = foot a = ankle r = reaction prox = proximal dist = distal

Fa

Ma

Tr

Fr

mfafmfg

Ifαf

Force/moment known(force plate)

Unknown forces/moments at

ankle

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3rd Step

Therefore, ankle joint expressed by:

gmFamF

amgmFF

amF

frffa

fffra

ff

ffrdistcmaproxcmra

ffrdistcmaproxcmra

ff

IFrFrTM

IFrFrTM

IM

,,

,,

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3rd Step

Thus, simply in 2-D :

Much more complicated in 3-D!

gmFamF

FamF

amF

fyryffya

xrxffxa

ff

,,,

,,,

ffrdistcmaproxcma

ff

IFrFrM

IM

,,

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3rd Step

Moment is the vector product of position and force

NOT a direct multiplication

xyyxz

z

FrFrM

FrM

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3rd Step

Ankle force/moment applied to subsequent segment (shank)

Equal and opposite force at distal extremity of segment (Newton’s 3rd Law)

Next, determine unknowns at proximal extremity of segment (knee)

UNKNOWNS

KNOWNS

h

h

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3rd Step

Knee joint is expressed by:

gmFamF

amgmFF

amF

sassk

sssak

ss

ssadistcmkproxcmak

ssadistcmkproxcmak

ss

IFrFrMM

IFrFrMM

IM

,,

,, k = knee s = shank a = ankle cm = centre of mass prox = proximal dist = distal

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3rd Step

Knee forces/moments applied to subsequent segment (thigh)

Equal and opposite force at distal extremity of segment (Newton’s 3rd Law)

Next, determine unknowns at proximal extremity of next segment (hip)

UNKNOWNS

KNOWNS

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3rd Step Hip joint is expressed by:

gmFamF

amgmFF

amF

tktth

tttkh

tt

tthdistcmhproxcmkh

tthdistcmhproxcmkh

tt

IFrFrMM

IFrFrMM

IM

,,

,, k= knee h = hip t = thigh cm = centre of mass prox = proximal dist = distal

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Exercise

Calculate the intersegment forces and moments at the ankle and knee

Ground reaction forcesFr,x = 6 N

Fr,y = 1041 N

Rigid body diagrams represent the foot, shank, and thigh

Analyse en 2-D

thigh

shank

• x

y Fr,x = 6 N

Fr,y = 1041 N

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Exercise

Foot Shank

m (kg) 1 3

I (kg m2) 0.0040 0.0369

ax (m/s2) -0.36 1.56

ay (m/s2) -0.56 -1.64

α (rad/s2) -3.41 -9.39

CM at x,y (m) 0.04, 0.09 0.06, 0.34

Ankle Knee

Location in x, y (m) 0.10, 0.12 0.02, 0.50

F of horizontal reaction (N) 6

F of vertical reaction (N) 1041

Centre of pressure at x, y (m) 0.0, 0.03

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Exercise

ankle

CMshank

knee

CMfoot

Fr,x

Fr,y

0.5 m

0.04 m

0.34 m

0.12 m

0.03 m0.09 m

0.10 m

0.06 m

0.02 m

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Exercise

thigh

shank

footFr,x = 6 N

Fr,y = 1041 N

thigh

shank

footFr,x = 6 N

Fr,y = 1041 N

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Exercise

shankfoot

Fr,x = 6 N

Fr,y = 1041 N

Ma

Fa,x

Fa,y

Iα

mPg

mfaf,y

mfaf,x Ms,dist

Ms,prox

Fs,dist,x

Fs,dist,y

Fs,prox,y

Fs,prox,x

msas,x

msas,y

msg

Iα

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Exercise – ankle (F)

NF

smkgF

amFF

maF

xa

xa

xffxrxa

xx

36.6

)/36.0(16

,

2,

,,,

NF

smkgsmkgF

amgmFF

maF

ya

ya

yfffyrya

yy

75.1031

)/56.0(1)/81.9(11041

,

22,

,,,

Fr,x = 6 N

Fr,y = 1041 N

Ma

Fa,x

Fa,y

Iα

mfg

-0.56 m/s2

-0.36 m/s2

ankle = (0.10, 0.12)CMfoot = (0.04, 0.09)

CP = (0.0, 0.03)

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Exercise – ankle (M)

mNM

sradmkgmNmN

mNmM

ImFmF

mFmFM

IM

a

a

ffyrxr

yaxaa

996.102

)/41.3)(004.0()04.0)(1041()06.0)(6(

)06.0)(75.1031()03.0)(36.6(

)004.0()03.009.0(

)04.010.0()09.012.0(

22

,,

,,

Fr,x = 6 N

Fr,y = 1041 N

Ma

Fa,x

Fa,y

Iα

mfg

-0.56 m/s2

-0.36 m/s2

ankle = (0.10, 0.12)CMfoot = (0.04, 0.09)

CP = (0.0, 0.03)

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Exercise – knee (F)

NF

smkgNF

amFF

maF

xk

xk

xssxaxk

xx

68.1

)/56.1(336.6

,

2,

,,,

NF

smkgsmkgNF

amgmFF

maF

yk

yk

ysssyayk

yy

24.1007

)/64.1(3)/81.9(375.1031

,

22,

,,,

102.996Nm

Ms,prox

6.36 N

1031.75N

Fs,prox,y

Fs,prox,x

1.56 m/s2

-1.64 m/s2

msg

Iα

ankle = (0.10, 0.12)CMshank = (0.06, 0.34)knee = (0.02, 0.77)

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Exercise – knee (M)

mNM

sradmkgmN

mNmN

mNmNM

ImFmF

mFmFMM

IM

k

k

ssyaxa

ykxkak

42.19

)/39.9)(0369.0()04.0)(75.1031(

)22.0)(36.6()04.0)(24.1007(

)16.0)(68.1(996.102

)04.0()22.0(

)04.0()16.0(

22

,,

,,

102.996Nm

Ms,prox

6.36 N

1031.75N

Fs,prox,y

Fs,prox,x

1.56 m/s2

-1.64 m/s2

msg

Iα

ankle = (0.10, 0.12)CMshank = (0.06, 0.34)knee = (0.02, 0.77)

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Exercise – Results

Joint Force in x (N) Force in y (N) Moment (Nm)

Ankle 6.36 1032 103

Knee 1.68 1007 19.4

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Recap

Establish model/CS GRF and locations Process

• Distal to proximal• Proximal forces/moments

OPPOSITE to distal forces/moments of subsequent segment

• Reaction forces• Repeat

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Principal Calculations

2-D

3-D0

0

0

z

y

x

F

F

F

0

0

0

z

y

x

M

M

M

0

0

y

x

F

F 0zM

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3-D

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3-D

Calculations are more complex – joint forces/moments still from inverse dynamics

Calculations of joint centres – specific marker configurations

Requires direct linear transformation to obtain aspect of 3rd dimension

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3-D

Centre of pressure in X, Y, Z 9 parameters: force components, centre of

pressures, moments about each axis Coordinate system in global and local

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3-D – centre of pressure

direction

XFYFMT

F

dFMY

F

dFMX

yxzz

z

zyzCP

z

zxyCP

M= moment F = reaction force dz = distance between real origin

and force plate origin T = torsion * assuming that ZCP = 0 * assuming that Tx =Ty = 0

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Global and Local CS

Y = +anterior

X = +lateral

Z = +proximal

LCS = local coordinate system

GCS = global coordinate system

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Transformation Matrix

Generate a transformation matrix – transforms markers from GCS to LCS

4 x 4 matrix combines position and rotation vectors

Orientation of LCS is in reference with GCS

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Transformation Matrix

Direct linear transformation used in projective geometry – solves set of variables, given set of relations

Over/under-constrained

Similarity relations equated as linear, homogeneous equations

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Transformation Matrix

In GCS, position vector r to arbitrary point P can be written (x,y)

In rotated (prime) CS (x’,y’)

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Transformation Matrix

In matrix form

or

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Transformation Matrix

With

[T] is orthogonal, therefore

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Kinematics

Global markers in the GCS are numerized and transformed to LCS

pi = position in LCS

Pi = position in GCS

ilocaltoglobali PTp

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Kinematics

zzz

yyy

xxx

zzz

yyy

xxx

zzz

yyy

xxx

PPP

PPP

PPP

A

PPP

PPP

PPP

V

PPP

PPP

PPP

P

321

321

321

321

321

321

321

321

321

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Kinetics

Similar to 2-D, unknown values are joint forces/moments at the proximal end of segment

Calculate reaction forces, then proceed with the joint moments

Transform parameters to LCS

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Kinetics

),,(),,(

),,(),,(

),,(),,(

),,(),0,0(

)0,,()0,,(

),0,0(),0,0(

),,(),,( ,,,,,,

zyxzyx

zyxzyx

cmcmcmcmcmcm

zyx

zz

zryrxrzryrxr

localglobal

vvvVVV

dddDDD

zyxZYX

mgmgmgMG

yxYX

tT

fffFFF

SCSC

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3-D Calculations

Same procedures :• Distal to proximal• Newton-Euler

equations• Joint reaction

forces/moments• Transformation from

GCS to LCS

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3-D Calculations

REMEMBER:• Results at the proximal end of a segment

represent the forces/moments (equal and opposite) of the distal end of the subsequent segment

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3-D LCS – ankle

Thus, ankle joint is expressed as:

zfzrffza

yfyrffya

xfxrffxa

frffa

fffra

ff

gmfxmf

gmfxmf

gmfxmf

gmfamf

amgmff

amF

,,

,,

,,

h

h

fa

ma

fr

tr

LCS

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h

h

fa

ma

fr

tr

3-D LCS – ankle

xxyyxxyy

zfzzfrdistcmaproxcmrza

zzxxzzxx

yfyyfrdistcmaproxcmrya

yyzzyyzz

xfxxfdistcmaproxcmrxa

ffrdistcmaproxcmra

ffrdistcmaproxcmra

ff

II

Ifrfrtm

II

Ifrfrtm

II

Ifrfrtm

Ifrfrtm

Ifrfrtm

IM

,,,,,

,,,,,

,,,,,

,,

,,

SCL

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LCSfoot to GCS

Resultant forces/moments of the segment are interpreted in LCS of the foot

Next, transform the force/moment vectors (of the ankle) to the GCS, using the appropriate transformation matrix

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LCS to GCS

Transformation matrix (transposed)

Tlocaltoglobalglobaltolocal TT

hglobaltolocalh

hglobaltolocalh

kglobaltolocalk

kglobaltolocalk

aglobaltolocala

aglobaltolocala

mTM

fTF

mTM

fTF

mTM

fTF

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GCS to LCSshank

Determine subsequent segment (shank), using forces/moments obtained from ankle

Transform force/moment global vectors of ankle to LCS of the shank

fk

mk

fa

ma

msas

msg

Isαs

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3-D LCS – knee

Thus, knee joint is expressed as:

zszasszk

ysyassyk

xsxassxk

sassk

sssak

ss

gmfxmf

gmfxmf

gmfxmf

gmfamf

amgmff

amF

,,

,,

,,

SCL

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3-D LCS – knee

xxyyxxyy

zszzskdistcmkproxcmazk

zzxxzzxx

ysyyskdistcmkproxcmayk

yyzzyyzz

xsxxsadistcmkproxcmaxk

ssadistcmkproxcmak

ssadistcmkproxcmak

ss

II

Ifrfrmm

II

Ifrfrmm

II

Ifrfrmm

Ifrfrmm

Ifrfrmm

IM

,,,,,

,,,,,

,,,,,

,,

,,

SCL

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LCSshank to GCS to LCSthigh

Results in reference to LCS of shank

Transform vectors of knee to GCS, using transformation matrix

Then, transform global vectors of the knee to LCS of the thigh

Itαt

mtat

mtg

Fg

Mg

fk

mk

mh fh

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3-D LCS – hip

Thus, hip joint is expressed as:

ztzkttzh

ytykttyh

xtxkttxh

tktth

tttkh

tt

gmfxmf

gmfxmf

gmfxmf

gmfamf

amgmff

amF

,,

,,

,,

Itαt

mtat

mtg

Fg

Mg

fk

mk

mh fh

SCL

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3-D LCS – hip

xxyyxxyy

ztzztkdistcmhproxcmkzh

zzxxzzxx

ytyytkdistcmhproxcmkyh

yyzzyyzz

xtxxtkdistcmhproxcmkxh

ttkdistcmhproxcmkh

ttkdistcmhproxcmkh

tt

II

Ifrfrmm

II

Ifrfrmm

II

Ifrfrmm

Ifrfrmm

Ifrfrmm

IM

,,,,,

,,,,,

,,,,,

,,

,,

Itαt

mtat

mtg

Fg

Mg

SCL

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Recap

Establish model/CS GRF and GCS locations Process

• GCS to LCS• Distal to proximal• Proximal forces/moments• LCS to GCS• Reaction forces/moments of

subsequent distal segment• Repeat

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Interpretations

Representative of intersegmetal joint loading (as opposed to joint contact loading)

Net forces/moments applied to centre of rotation that is assumed (2-D) and approximated (3-D)

Results can vary substantially with the integration of muscle forces and inclusion of soft tissues

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Interpretations

Limitations with inverse dynamics

Knee in extension – no tension (or negligible tension) in the muscles at the joint

With an applied vertical reaction force of 600 N, the bone-on-bone force is equal in magnitude and direction ~600 N

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Interpretations

Knee in flexion, reaction of 600 N produces a bone-on-bone force of ~3000 N (caused by muscle contractions)

Several unknown vectors – statically indeterminant and underconstrained

Require EMG analysis

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Applications

Results represent valuable approximations of net joint forces/moments

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Applications

Quantifiable results permit the comparison of patient-to-participant’s performance under various conditions• Diagnostic tool• Evaluation of treatment and

intervention

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What is “Inverse Dynamics”?

Kinematics Kinetics

“COMPUTATION”

Resultant joint loading

Kinematics Kinetics

Inverse Dynamics

Resultant joint loading

Questions