1. Inverse Dynamics

2. APA 6903 Fall 2013 2 What is “Inverse Dynamics”?

3. APA 6903 Fall 2013 3 What is “Inverse Dynamics”?  Motion – kinematics  Force – kinetics  Applied dynamics

4. APA 6903 Fall 2013 4 What is “Inverse Dynamics”? KinematicsKinetics “COMPUTATION” Resultant joint loading

5. APA 6903 Fall 2013 5 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

6. APA 6903 Fall 2013 6 Inverse Dynamics  Newton’s Laws of motion 1 st : 2 nd : 3 rd : a given action creates an equal and opposite reaction

7. APA 6903 Fall 2013 7 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)

8. APA 6903 Fall 2013 8 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

9. APA 6903 Fall 2013 9 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

10. APA 6903 Fall 2013 10 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

11. APA 6903 Fall 2013 11 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

12. APA 6903 Fall 2013 12 Objective  Combines Anthropometry: anatomical landmarks, mass, length, centre of mass, inertial moments Kinematics: goniometre, reflective markers, cameras Kinetics: force plates

13. APA 6903 Fall 2013 13 1 st Step Establish a model

14. APA 6903 Fall 2013 14 1 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

15. APA 6903 Fall 2013 15 Segmentation  Assume Each segment is symmetric about its principal axis Angular velocity and longitudinal acceleration of segment are neglected Frictionless

16. APA 6903 Fall 2013 16 2 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

17. APA 6903 Fall 2013 17 Free Body Diagram  Statics – analysis of physical systems  Statically determinant

18. APA 6903 Fall 2013 18 Free Body Diagram

19. APA 6903 Fall 2013 19 3 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

20. APA 6903 Fall 2013 20 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

21. APA 6903 Fall 2013 21 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

22. APA 6903 Fall 2013 22 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

23. APA 6903 Fall 2013 23 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

24. APA 6903 Fall 2013 24 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

25. APA 6903 Fall 2013 25 3 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

26. APA 6903 Fall 2013 26 3 rd Step  Therefore, ankle joint expressed by:

27. APA 6903 Fall 2013 27 3 rd Step  Thus, simply in 2-D :  Much more complicated in 3-D!

28. APA 6903 Fall 2013 28 3 rd Step  Moment is the vector product of position and force  NOT a direct multiplication

29. APA 6903 Fall 2013 29 3 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

30. APA 6903 Fall 2013 30 3 rd Step  Knee joint is expressed by:  k = knee  s = shank  a = ankle  cm = centre of mass  prox = proximal  dist = distal

31. APA 6903 Fall 2013 31 3 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

32. APA 6903 Fall 2013 32 3 rd Step  Hip joint is expressed by:  k= knee  h = hip  t = thigh  cm = centre of mass  prox = proximal  dist = distal

33. APA 6903 Fall 2013 33 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

34. APA 6903 Fall 2013 34 Exercise FootShank m (kg)13 I (kg m 2 )0.00400.0369 a x (m/s 2 )-0.361.56 a y (m/s 2 )-0.56-1.64 α (rad/s 2 ) -3.41-9.39 CM at x,y (m)0.04, 0.090.06, 0.34 AnkleKnee Location in x, y (m)0.10, 0.120.02, 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

35. APA 6903 Fall 2013 35 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

36. APA 6903 Fall 2013 36 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

37. APA 6903 Fall 2013 37 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

38. APA 6903 Fall 2013 38 Exercise – ankle (F) F r,x = 6 N F r,y = 1041 N MaMa F a,x F a,y IαIα mfgmfg -0.56 m/s 2 -0.36 m/s 2 ankle = (0.10, 0.12) CM foot = (0.04, 0.09) CP = (0.0, 0.03)

39. APA 6903 Fall 2013 39 Exercise – ankle (M) F r,x = 6 N F r,y = 1041 N MaMa F a,x F a,y IαIα mfgmfg -0.56 m/s 2 -0.36 m/s 2 ankle = (0.10, 0.12) CM foot = (0.04, 0.09) CP = (0.0, 0.03)

40. APA 6903 Fall 2013 40 Exercise – knee (F) 102.996N m M s,prox 6.36 N 1031.75N F s,prox,y F s,prox,x 1.56 m/s 2 -1.64 m/s 2 msgmsg IαIα ankle = (0.10, 0.12) CM shank = (0.06, 0.34) knee = (0.02, 0.77)

41. APA 6903 Fall 2013 41 Exercise – knee (M) 102.996N m M s,prox 6.36 N 1031.75N F s,prox,y F s,prox,x 1.56 m/s 2 -1.64 m/s 2 msgmsg IαIα ankle = (0.10, 0.12) CM shank = (0.06, 0.34) knee = (0.02, 0.77)

42. APA 6903 Fall 2013 42 Exercise – Results Joint Force in x (N) Force in y (N) Moment (Nm) Ankle6.361032103 Knee1.68100719.4

43. APA 6903 Fall 2013 43 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

44. APA 6903 Fall 2013 44 Principal Calculations 2-D 3-D

45. APA 6903 Fall 2013 45 3-D

46. APA 6903 Fall 2013 46 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 3 rd dimension

47. APA 6903 Fall 2013 47 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

48. APA 6903 Fall 2013 48 3-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

49. APA 6903 Fall 2013 49 Global and Local CS Y = +anterior X = +lateral Z = +proximal LCS = local coordinate system GCS = global coordinate system

50. APA 6903 Fall 2013 50 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

51. APA 6903 Fall 2013 51 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

52. APA 6903 Fall 2013 52 Transformation Matrix

53. APA 6903 Fall 2013 53 Transformation Matrix

54. APA 6903 Fall 2013 54 Transformation Matrix

55. APA 6903 Fall 2013 55 Kinematics  Global markers in the GCS are numerized and transformed to LCS p i = position in LCS P i = position in GCS

56. APA 6903 Fall 2013 56 Kinematics

57. APA 6903 Fall 2013 57 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

58. APA 6903 Fall 2013 58 Kinetics

59. APA 6903 Fall 2013 59 3-D Calculations  Same procedures : Distal to proximal Newton-Euler equations Joint reaction forces/moments Transformation from GCS to LCS

60. APA 6903 Fall 2013 60 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

61. APA 6903 Fall 2013 61 3-D LCS – ankle  Thus, ankle joint is expressed as: h h fafa mama frfr trtr LCS

62. APA 6903 Fall 2013 62 h h fafa mama frfr trtr 3-D LCS – ankle SCL

63. APA 6903 Fall 2013 63 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

64. APA 6903 Fall 2013 64 LCS to GCS  Transformation matrix (transposed)

65. APA 6903 Fall 2013 65 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

66. APA 6903 Fall 2013 66 3-D LCS – knee  Thus, knee joint is expressed as: SCL

67. APA 6903 Fall 2013 67 3-D LCS – knee SCL

68. APA 6903 Fall 2013 68 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

69. APA 6903 Fall 2013 69 3-D LCS – hip  Thus, hip joint is expressed as: ItαtItαt mtatmtat mtgmtg FgFg MgMg fkfk mkmk mhmh fhfh SCL

70. APA 6903 Fall 2013 70 3-D LCS – hip ItαtItαt mtatmtat mtgmtg FgFg MgMg SCL

71. APA 6903 Fall 2013 71 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

72. APA 6903 Fall 2013 72 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

73. APA 6903 Fall 2013 73 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

74. APA 6903 Fall 2013 74 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

75. APA 6903 Fall 2013 75 Applications  Results represent valuable approximations of net joint forces/moments

76. APA 6903 Fall 2013 76 Applications  Quantifiable results permit the comparison of patient- to-participant’s performance under various conditions Diagnostic tool Evaluation of treatment and intervention

77. APA 6903 Fall 2013 77 What is “Inverse Dynamics”? KinematicsKinetics “COMPUTATION” Resultant joint loading KinematicsKinetics Inverse Dynamics Resultant joint loading

78. Questions