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What is “Inverse Dynamics”?

- Motion – kinematics
- Force – kinetics
- Applied dynamics

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

Inverse Dynamics

- Newton’sLaws of motion
- 1st:
- 2nd:
- 3rd: a given action creates an equal and opposite reaction

Inverse Dynamics

- If an objectisatequilibriatedrest = static
- If an objectis in motion = dynamic
- If objectaccelerates, inertial forces calculatedbased on Newton’s 2nd Law (ΣF = ma)

Dynamics

- Twoapproaches to solve for dynamics

Dynamics

- Direct method
- Forces are known
- Motion iscalculated by integrating once to obtainvelocity, twice to obtaindisplacement

Dynamics

- Inverse method
- Displacements/motion are known
- Force is calculated by differentiating once to obtain velocity, twice to obtain acceleration

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

Objective

- Combines
- Anthropometry: anatomical landmarks, mass, length, centre of mass, inertial moments
- Kinematics: goniometre, reflective markers, cameras
- Kinetics: force plates

1st Step

Establish a model

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

Segmentation

- Assume
- Each segment issymmetric about its principal axis
- Angularvelocity and longitudinal acceleration of segment are neglected
- Frictionless

2ndStep

- Measure ALL externalreaction forces
- Appoximateinertialproperties 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

Free Body Diagram

- Statics – analysis of physicalsystems
- Staticallydeterminant

3rdStep

- Staticequilibrium of segments
- Forces/moments knownat foot segment
- Using Newton-Euler formulas, calculationbeginsat foot, then to ankle
- Proceedfrom distal to proximal

UNKNOWNS

KNOWNS

FBD of Foot

Triceps sural force

- $%#&?!
- Multiple unknowns

Bone force

Anterior tibial muscle force

Joint moment

Fg

Centre of pressure

Ligament force

Centre of gravity

Simplify

- Multiple unknown force and moment vectors
- Muscles, ligaments, bone, soft tissues, capsules, etc.
- Reduction of unknownvectors to:
- 3 Newton-Euler equilibriumequations, for 2-D (Fx, Fy,Mz)
- 6 equations, for 3-D (Fx, Fy,Fz,Mx,My,Mz)
- Representativeof net forces/moment

Simplification

- Displace forces to joint centre
- Force equal and opposite

F*

Force at joint centre

Fr

Centre of pressure

F

Foot muscle forces

-F*

Force equal and opposite

Centre gravity

Simplification

- Replace coupled forces with moment

F*

Force at joint centre

Fr

Centre of pressure

F

Foot muscle force

-F*

Force equal and opposite

Centre of gravity

M

Moment

Simplification

- Representation net moments and forces atankle

- cm = centre of mass
- prox = proximal
- dist = distal

Mankle

rcm,prox

Fankle

xankle, yankle

rcm,dist

Freaction

xreaction, yreaction

mfootg

3rdStep

Fa

Unknown forces/moments atankle

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

Ma

Ifαf

mfaf

mfg

Tr

Fr

Force/moment known

(force plate)

3rdStep

- Therefore, ankle joint expressed by:

3rdStep

- Thus, simply in 2-D :
- Much more complicated in 3-D!

3rdStep

- Moment is the vectorproduct of position and force
- NOT a direct multiplication

3rdStep

- Ankle force/moment applied to subsequent segment (shank)
- Equal and opposite force at distal extremity of segment (Newton’s 3rd Law)
- Next, determineunknownsat proximal extremity of segment (knee)

UNKNOWNS

h

h

KNOWNS

3rdStep

- Knee joint isexpressed by:

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

3rdStep

- Knee forces/moments applied to subsequent segment (thigh)
- Equal and opposite force at distal extremity of segment (Newton’s 3rd Law)
- Next, determineunknownsat proximal extremity of next segment (hip)

UNKNOWNS

KNOWNS

3rdStep

- Hip joint isexpressed by:

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

Exercise

- Calculate the intersegment forces and moments at the ankle and knee
- Groundreaction forces

Fr,x = 6 N

Fr,y = 1041 N

- Rigid body diagramsrepresent the foot, shank, and thigh
- Analyse en 2-D

thigh

shank

y

Fr,x = 6 N

x

Fr,y = 1041 N

Exercise

Fs,prox,y

Ms,prox

Fs,prox,x

Fa,y

msas,y

mfaf,y

Iα

Ma

msas,x

Fa,x

Iα

msg

mfaf,x

Ms,dist

Fs,dist,x

mPg

Fs,dist,y

Fr,x = 6 N

foot

shank

Fr,y = 1041 N

Exercise– ankle(F)

Fa,y

Ma

-0.56 m/s2

Fa,x

Iα

-0.36 m/s2

mfg

ankle = (0.10, 0.12)

CMfoot = (0.04, 0.09)

CP = (0.0, 0.03)

Fr,x = 6 N

Fr,y = 1041 N

Exercise – ankle(M)

Fa,y

Ma

-0.56 m/s2

Fa,x

Iα

-0.36 m/s2

mfg

ankle = (0.10, 0.12)

CMfoot = (0.04, 0.09)

CP = (0.0, 0.03)

Fr,x = 6 N

Fr,y = 1041 N

Exercise– knee(F)

Fs,prox,y

Ms,prox

Fs,prox,x

-1.64 m/s2

Iα

1.56 m/s2

msg

102.996Nm

6.36 N

1031.75N

ankle = (0.10, 0.12)

CMshank = (0.06, 0.34)

knee = (0.02, 0.77)

Exercise– knee(M)

Fs,prox,y

Ms,prox

Fs,prox,x

-1.64 m/s2

Iα

1.56 m/s2

msg

102.996Nm

6.36 N

1031.75N

ankle = (0.10, 0.12)

CMshank = (0.06, 0.34)

knee = (0.02, 0.77)

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

3-D

- Calculations are more complex – joint forces/moments stillfrom inverse dynamics
- Calculations of joint centres – specific marker configurations
- Requires direct linear transformation to obtain aspect of 3rd dimension

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

3-D – centre of pressure

- M= moment
- F = reaction force
- dz = distance between real origin and force plate origin
- T = torsion
- * assumingthat ZCP= 0
- * assumingthatTx=Ty = 0

direction

Global and Local CS

Z = +proximal

GCS = global coordinate system

X = +lateral

LCS = local coordinate system

Y = +anterior

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 referencewith GCS

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, homogeneousequations

Transformation Matrix

- In GCS, position vectorr to arbitrary point Pcanbewritten(x,y)
- In rotated (prime) CS (x’,y’)

Transformation Matrix

- With
- [T] is orthogonal, therefore

Kinematics

- Global markers in the GCS are numerized and transformed to LCS

pi= position in LCS

Pi= position in GCS

Kinetics

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

3-D Calculations

- Sameprocedures :
- Distal to proximal
- Newton-Euler equations
- Joint reaction forces/moments
- Transformation from GCS to LCS

3-D Calculations

- REMEMBER:
- Resultsat the proximal end of a segment represent the forces/moments (equal and opposite) of the distal end of the subsequent segment

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

LCS to GCS

- Transformation matrix (transposed)

GCS to LCSshank

- Determinesubsequent segment (shank), using forces/moments obtainedfromankle
- Transform force/moment global vectors of ankle to LCS of the shank

fk

mk

Isαs

msas

msg

fa

ma

LCSshank to GCS to LCSthigh

- Resultsin reference to LCS of shank
- Transformvectors of knee to GCS, using transformation matrix
- Then, transform global vectors of the knee to LCS of the thigh

mh

fh

mtat

Itαt

Mg

mtg

mk

Fg

fk

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

Interpretations

- Representative of intersegmetal joint loading(as opposed to joint contact loading)
- Net forces/moments applied to centre of rotation thatisassumed(2-D) and approximated (3-D)
- Resultscanvarysubstantiallywith the integration of muscle forces and inclusion of soft tissues

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 isequal in magnitude and direction ~600 N

Interpretations

- Knee in flexion, reaction of 600 N produces a bone-on-bone force of ~3000 N (caused by muscle contractions)
- Severalunknownvectors – staticallyindeterminant and underconstrained
- Require EMG analysis

Applications

- Resultsrepresentvaluable approximations of net joint forces/moments

Applications

- Quantifiable results permit the comparison of patient-to-participant’s performance undervarious conditions
- Diagnostic tool
- Evaluation of treatment and intervention

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