Inverse Dynamics

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# Inverse Dynamics - PowerPoint PPT Presentation

Inverse Dynamics. What is “Inverse Dynamics”?. What is “Inverse Dynamics”?. Motion – kinematics Force – kinetics Applied dynamics. What is “Inverse Dynamics”?. Inverse Dynamics. Using Newton’s Laws Fundamentals of mechanics Principles concerning motion and movement

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## Inverse Dynamics

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### Inverse Dynamics

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

0.5 m

0.02 m

knee

0.34 m

0.06 m

CMshank

0.12 m

0.10 m

ankle

0.09 m

0.04 m

0.03 m

CMfoot

Fr,x

Fr,y

Exercise

thigh

thigh

shank

shank

foot

foot

Fr,x = 6 N

Fr,x = 6 N

Fr,y = 1041 N

Fr,y = 1041 N

Exercise

Fs,prox,y

Ms,prox

Fs,prox,x

Fa,y

msas,y

mfaf,y

Ma

msas,x

Fa,x

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

-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

-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

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

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
• In matrix form

or

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
3-D LCS – ankle
• Thus, ankle joint isexpressed as:

fa

h

ma

h

tr

fr

LCS

3-D LCS – ankle

fa

h

ma

h

tr

fr

SCL

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

3-D LCS – knee
• Thus, knee joint isexpressed as:

SCL

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

3-D LCS – hip
• Thus, hip joint isexpressed as:

mh

fh

mtat

Itαt

Mg

mtg

mk

Fg

fk

SCL

3-D LCS – hip

mtat

Itαt

Mg

mtg

Fg

SCL

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
• 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