Inverse Dynamics

APA 6903 Fall What is “Inverse Dynamics”?

APA 6903 Fall What is “Inverse Dynamics”?  Motion – kinematics  Force – kinetics  Applied dynamics

APA 6903 Fall What is “Inverse Dynamics”? KinematicsKinetics “COMPUTATION” Resultant joint loading

APA 6903 Fall 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

APA 6903 Fall Inverse Dynamics  Newton’s Laws of motion 1 st : 2 nd : 3 rd : a given action creates an equal and opposite reaction

APA 6903 Fall 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 2 nd Law(ΣF = ma)

APA 6903 Fall Dynamics  Two approaches to solve for dynamics F Forces F = ma Equations of motion ∫∫ Double integration x Displacements x Displacements d 2 x / dt 2 Double differentiation F = ma Equations of motion F Forces

APA 6903 Fall Dynamics  Direct method Forces are known Motion is calculated by integrating once to obtain velocity, twice to obtain displacement F Forces F = ma Equations of motion ∫∫ Double integration x Displacement

APA 6903 Fall Dynamics  Inverse method Displacements/motion are known Force is calculated by differentiating once to obtain velocity, twice to obtain acceleration x Displacements d 2 x / dt 2 Double differentiation F = ma Equations of motion F Forces

APA 6903 Fall 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

APA 6903 Fall Objective  Combines Anthropometry: anatomical landmarks, mass, length, centre of mass, inertial moments Kinematics: goniometre, reflective markers, cameras Kinetics: force plates

APA 6903 Fall st Step Establish a model

APA 6903 Fall st 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

APA 6903 Fall Segmentation  Assume Each segment is symmetric about its principal axis Angular velocity and longitudinal acceleration of segment are neglected Frictionless

APA 6903 Fall nd 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

APA 6903 Fall Free Body Diagram  Statics – analysis of physical systems  Statically determinant

APA 6903 Fall Free Body Diagram

APA 6903 Fall rd 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 proximal KNOWNS UNKNOWNS

APA 6903 Fall FBD of Foot  $%#&?!  Multiple unknowns Centre of gravity F g Centre of pressure Triceps sural force Anterior tibial muscle force Bone force Ligament force Joint moment

APA 6903 Fall 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 (F x, F y, M z ) 6 equations, for 3-D (F x, F y, F z, M x, M y, M z )  Representative of net forces/moment

APA 6903 Fall Simplification  Displace forces to joint centre  Force equal and opposite Centre gravity F r Centre of pressure F Foot muscle forces F* Force at joint centre -F* Force equal and opposite

APA 6903 Fall Simplification  Replace coupled forces with moment Centre of gravity F Foot muscle force F* Force at joint centre -F* Force equal and opposite M Moment F r Centre of pressure

APA 6903 Fall Simplification  Representation net moments and forces at ankle F reaction x reaction, y reaction m foot g F ankle x ankle, y ankle M ankle r cm,dist r cm,prox  cm = centre of mass  prox = proximal  dist = distal

APA 6903 Fall rd Step  f = foot  a = ankle  r = reaction  prox = proximal  dist = distal FaFa MaMa TrTr FrFr mfafmfaf mfgmfg IfαfIfαf Force/moment known (force plate) Unknown forces/moments at ankle

APA 6903 Fall rd Step  Therefore, ankle joint expressed by:

APA 6903 Fall rd Step  Thus, simply in 2-D :  Much more complicated in 3-D!

APA 6903 Fall rd Step  Moment is the vector product of position and force  NOT a direct multiplication

APA 6903 Fall rd Step  Ankle force/moment applied to subsequent segment (shank)  Equal and opposite force at distal extremity of segment (Newton’s 3 rd Law)  Next, determine unknowns at proximal extremity of segment (knee) UNKNOWNS KNOWNS h h

APA 6903 Fall rd Step  Knee joint is expressed by:  k = knee  s = shank  a = ankle  cm = centre of mass  prox = proximal  dist = distal

APA 6903 Fall rd Step  Knee forces/moments applied to subsequent segment (thigh)  Equal and opposite force at distal extremity of segment (Newton’s 3 rd Law)  Next, determine unknowns at proximal extremity of next segment (hip) UNKNOWNS KNOWNS

APA 6903 Fall rd Step  Hip joint is expressed by:  k= knee  h = hip  t = thigh  cm = centre of mass  prox = proximal  dist = distal

APA 6903 Fall Exercise  Calculate the intersegment forces and moments at the ankle and knee  Ground reaction forces F r,x = 6 N F r,y = 1041 N  Rigid body diagrams represent the foot, shank, and thigh  Analyse en 2-D thigh shank x y F r,x = 6 N F r,y = 1041 N

APA 6903 Fall Exercise FootShank m (kg)13 I (kg m 2 ) a x (m/s 2 ) a y (m/s 2 ) α (rad/s 2 ) CM at x,y (m)0.04, , 0.34 AnkleKnee Location in x, y (m)0.10, , 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

APA 6903 Fall Exercise ankle CM shank knee CM foot F r,x F r,y 0.5 m 0.04 m 0.34 m 0.12 m 0.03 m 0.09 m 0.10 m 0.06 m 0.02 m

APA 6903 Fall Exercise thigh shank foot F r,x = 6 N F r,y = 1041 N thigh shank foot F r,x = 6 N F r,y = 1041 N

APA 6903 Fall Exercise shankfoot F r,x = 6 N F r,y = 1041 N MaMa F a,x F a,y IαIα mPgmPg m f a f,y m f a f,x M s,dist M s,prox F s,dist,x F s,dist,y F s,prox,y F s,prox,x m s a s,x m s a s,y msgmsg IαIα

APA 6903 Fall Exercise – ankle (F) F r,x = 6 N F r,y = 1041 N MaMa F a,x F a,y IαIα mfgmfg m/s m/s 2 ankle = (0.10, 0.12) CM foot = (0.04, 0.09) CP = (0.0, 0.03)

APA 6903 Fall Exercise – ankle (M) F r,x = 6 N F r,y = 1041 N MaMa F a,x F a,y IαIα mfgmfg m/s m/s 2 ankle = (0.10, 0.12) CM foot = (0.04, 0.09) CP = (0.0, 0.03)

APA 6903 Fall Exercise – knee (F) N m M s,prox 6.36 N N F s,prox,y F s,prox,x 1.56 m/s m/s 2 msgmsg IαIα ankle = (0.10, 0.12) CM shank = (0.06, 0.34) knee = (0.02, 0.77)

APA 6903 Fall Exercise – knee (M) N m M s,prox 6.36 N N F s,prox,y F s,prox,x 1.56 m/s m/s 2 msgmsg IαIα ankle = (0.10, 0.12) CM shank = (0.06, 0.34) knee = (0.02, 0.77)

APA 6903 Fall Exercise – Results Joint Force in x (N) Force in y (N) Moment (Nm) Ankle Knee

APA 6903 Fall 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

APA 6903 Fall Principal Calculations 2-D 3-D

APA 6903 Fall D

APA 6903 Fall 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 3 rd dimension

APA 6903 Fall 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

APA 6903 Fall D – centre of pressure direction  M= moment  F = reaction force  d z = distance between real origin and force plate origin  T = torsion  * assuming that Z CP = 0  * assuming that T x =T y = 0

APA 6903 Fall Global and Local CS Y = +anterior X = +lateral Z = +proximal LCS = local coordinate system GCS = global coordinate system

APA 6903 Fall 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

APA 6903 Fall 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

APA 6903 Fall Transformation Matrix

APA 6903 Fall Transformation Matrix

APA 6903 Fall Transformation Matrix

APA 6903 Fall Kinematics  Global markers in the GCS are numerized and transformed to LCS p i = position in LCS P i = position in GCS

APA 6903 Fall Kinematics

APA 6903 Fall 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

APA 6903 Fall Kinetics

APA 6903 Fall D Calculations  Same procedures : Distal to proximal Newton-Euler equations Joint reaction forces/moments Transformation from GCS to LCS

APA 6903 Fall 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

APA 6903 Fall D LCS – ankle  Thus, ankle joint is expressed as: h h fafa mama frfr trtr LCS

APA 6903 Fall h h fafa mama frfr trtr 3-D LCS – ankle SCL

APA 6903 Fall LCS foot 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

APA 6903 Fall LCS to GCS  Transformation matrix (transposed)

APA 6903 Fall GCS to LCS shank  Determine subsequent segment (shank), using forces/moments obtained from ankle  Transform force/moment global vectors of ankle to LCS of the shank fkfk mkmk fafa mama msasmsas msgmsg IsαsIsαs

APA 6903 Fall D LCS – knee  Thus, knee joint is expressed as: SCL

APA 6903 Fall D LCS – knee SCL

APA 6903 Fall LCS shank to GCS to LCS thigh  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 ItαtItαt mtatmtat mtgmtg FgFg MgMg fkfk mkmk mhmh fhfh

APA 6903 Fall D LCS – hip  Thus, hip joint is expressed as: ItαtItαt mtatmtat mtgmtg FgFg MgMg fkfk mkmk mhmh fhfh SCL

APA 6903 Fall D LCS – hip ItαtItαt mtatmtat mtgmtg FgFg MgMg SCL

APA 6903 Fall 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

APA 6903 Fall 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

APA 6903 Fall 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

APA 6903 Fall 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

APA 6903 Fall Applications  Results represent valuable approximations of net joint forces/moments

APA 6903 Fall Applications  Quantifiable results permit the comparison of patient- to-participant’s performance under various conditions Diagnostic tool Evaluation of treatment and intervention

APA 6903 Fall What is “Inverse Dynamics”? KinematicsKinetics “COMPUTATION” Resultant joint loading KinematicsKinetics Inverse Dynamics Resultant joint loading

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