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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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what is inverse dynamics1
What is “Inverse Dynamics”?
  • Motion – kinematics
  • Force – kinetics
  • Applied dynamics
inverse dynamics1
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 dynamics2
Inverse Dynamics
  • Newton’sLaws of motion
    • 1st:
    • 2nd:
    • 3rd: a given action creates an equal and opposite reaction
inverse dynamics3
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
Dynamics
  • Twoapproaches to solve for dynamics
dynamics1
Dynamics
  • Direct method
    • Forces are known
    • Motion iscalculated by integrating once to obtainvelocity, twice to obtaindisplacement
dynamics2
Dynamics
  • Inverse method
    • Displacements/motion are known
    • Force is calculated by differentiating once to obtain velocity, twice to obtain acceleration
objective
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
objective1
Objective
  • Combines
    • Anthropometry: anatomical landmarks, mass, length, centre of mass, inertial moments
    • Kinematics: goniometre, reflective markers, cameras
    • Kinetics: force plates
1 st step
1st Step

Establish a model

1 st step1
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
Segmentation
  • Assume
    • Each segment issymmetric about its principal axis
    • Angularvelocity and longitudinal acceleration of segment are neglected
    • Frictionless
2 nd step
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
Free Body Diagram
  • Statics – analysis of physicalsystems
  • Staticallydeterminant
3 rd step
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
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
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
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

simplification1
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

simplification2
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

3 rd step1
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)

3 rd step2
3rdStep
  • Therefore, ankle joint expressed by:
3 rd step3
3rdStep
  • Thus, simply in 2-D :
  • Much more complicated in 3-D!
3 rd step4
3rdStep
  • Moment is the vectorproduct of position and force
  • NOT a direct multiplication
3 rd step5
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

3 rd step6
3rdStep
  • Knee joint isexpressed by:
  • k = knee
  • s = shank
  • a = ankle
  • cm = centre of mass
  • prox = proximal
  • dist = distal
3 rd step7
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

3 rd step8
3rdStep
  • Hip joint isexpressed by:
  • k= knee
  • h = hip
  • t = thigh
  • cm = centre of mass
  • prox = proximal
  • dist = distal
exercise
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

exercise2
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

exercise3
Exercise

thigh

thigh

shank

shank

foot

foot

Fr,x = 6 N

Fr,x = 6 N

Fr,y = 1041 N

Fr,y = 1041 N

exercise4
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
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
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
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
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
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
slide46
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
slide47
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
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
Global and Local CS

Z = +proximal

GCS = global coordinate system

X = +lateral

LCS = local coordinate system

Y = +anterior

transformation matrix
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 matrix1
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 matrix2
Transformation Matrix
  • In GCS, position vectorr to arbitrary point Pcanbewritten(x,y)
  • In rotated (prime) CS (x’,y’)
transformation matrix3
Transformation Matrix
  • In matrix form

or

transformation matrix4
Transformation Matrix
  • With
  • [T] is orthogonal, therefore
kinematics
Kinematics
  • Global markers in the GCS are numerized and transformed to LCS

pi= position in LCS

Pi= position in GCS

kinetics
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
3-D Calculations
  • Sameprocedures :
    • Distal to proximal
    • Newton-Euler equations
    • Joint reaction forces/moments
    • Transformation from GCS to LCS
3 d calculations1
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
3-D LCS – ankle
  • Thus, ankle joint isexpressed as:

fa

h

ma

h

tr

fr

LCS

3 d lcs ankle1
3-D LCS – ankle

fa

h

ma

h

tr

fr

SCL

lcs foot to gcs
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
LCS to GCS
  • Transformation matrix (transposed)
gcs to lcs shank
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
3-D LCS – knee
  • Thus, knee joint isexpressed as:

SCL

lcs shank to gcs to lcs thigh
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
3-D LCS – hip
  • Thus, hip joint isexpressed as:

mh

fh

mtat

Itαt

Mg

mtg

mk

Fg

fk

SCL

3 d lcs hip1
3-D LCS – hip

mtat

Itαt

Mg

mtg

Fg

SCL

recap1
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
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
interpretations1
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
interpretations2
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
Applications
  • Resultsrepresentvaluable approximations of net joint forces/moments
applications1
Applications
  • Quantifiable results permit the comparison of patient-to-participant’s performance undervarious conditions
    • Diagnostic tool
    • Evaluation of treatment and intervention