Transcript cp calculation exercise

Inverse Dynamics
What is “Inverse Dynamics”?
APA 6903
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What is “Inverse Dynamics”?
 Motion – kinematics
 Force – kinetics
 Applied dynamics
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What is “Inverse Dynamics”?
Kinematics
Kinetics
“COMPUTATION”
Resultant joint loading
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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
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Inverse Dynamics
 Newton’s Laws of motion
• 1st:
• 2nd:
• 3rd:

F 0


F  ma
a given action creates an equal and
opposite reaction
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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 2nd Law
(ΣF = ma)
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Dynamics
 Two approaches to solve for dynamics
F
Forces
x
Displacements
F = ma
∫∫
Equations of
motion
Double
integration
d2x / dt2
F = ma
Double
differentiation
Equations of
motion
x
Displacements
F
Forces
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Dynamics
F
Forces
F = ma
∫∫
Equations of
motion
Double
integration
x
Displacement
 Direct method
• Forces are known
• Motion is calculated by integrating once to obtain
velocity, twice to obtain displacement
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Dynamics
x
Displacements
d2x / dt2
F = ma
Double
differentiation
Equations of
motion
F
Forces
 Inverse method
• Displacements/motion are known
• Force is calculated by differentiating once to
obtain velocity, twice to obtain acceleration
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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
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Objective
 Combines
• Anthropometry: anatomical
landmarks, mass, length, centre
of mass, inertial moments
• Kinematics: goniometre,
reflective markers, cameras
• Kinetics: force plates
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1st Step
Establish a model
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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
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Segmentation
 Assume
• Each segment is symmetric
about its principal axis
• Angular velocity and
longitudinal acceleration of
segment are neglected
• Frictionless
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2nd 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
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Free Body Diagram
 Statics – analysis of physical systems
 Statically determinant
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Free Body Diagram
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3rd Step
 Static equilibrium of segments
UNKNOWNS
 Forces/moments known at foot
segment
 Using Newton-Euler formulas,
calculation begins at foot, then to
ankle
KNOWNS
 Proceed from distal to proximal
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FBD of Foot
Triceps
sural force
Ligament
force
Bone force
Anterior tibial
muscle force
Fg
Joint
moment
 $%#&?!
 Multiple unknowns
Centre of
pressure
Centre of
gravity
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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 (Fx, Fy, Mz)
• 6 equations, for 3-D (Fx, Fy, Fz, Mx, My, Mz)
 Representative of net forces/moment
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Simplification
F*
 Displace forces to joint
centre
Force at joint
centre
F
Foot muscle
forces
 Force equal and opposite
Fr
-F*
Centre of
pressure
Force equal
and opposite
Centre
gravity
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Simplification
F*
Force at joint
centre
F
Foot muscle
force
 Replace coupled forces with
moment
Fr
Centre of
pressure
-F*
Force equal
and opposite
M
Moment
Centre of
gravity
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Simplification



Fankle
xankle, yankle
cm = centre of mass
prox = proximal
dist = distal
Mankle
rcm,prox
Freaction
rcm,dist
xreaction, yreaction
mfootg
 Representation net moments and forces at ankle
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3rd Step
Unknown
forces/moments at
ankle
Fa





Ma
Ifαf
f = foot
a = ankle
r = reaction
prox = proximal
dist = distal
mfaf
mfg
Tr
Fr
Force/moment known
(force plate)
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3rd Step
 Therefore, ankle joint expressed by:
F  m a
f
f
Fa  Fr  m f g  m f a f
Fa  m f a f  Fr  m f g
M  I
f
f
M a  Tr  rcm , prox  Fa   rcm ,dist  Fr   I f  f
M a  Tr  rcm , prox  Fa   rcm ,dist  Fr   I f  f
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3rd Step
 Thus, simply in 2-D :
F  m a
f
f
Fa , x  m f a f , x  Fr , x
Fa , y  m f a f , y  Fr , y  m f g
M  I 
M  r
f
a
f
cm , prox
 Fa   rcm ,dist  Fr   I f  f
 Much more complicated in 3-D!
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3rd Step
 Moment is the vector product of
position and force
M z  r  F 
M z  rx Fy  ry Fx
 NOT a direct multiplication
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3rd Step
UNKNOWNS
 Ankle force/moment applied to subsequent
segment (shank)
 Equal and opposite force at distal extremity
of segment (Newton’s 3rd Law)
 Next, determine unknowns at proximal
extremity of segment (knee)
h
h
KNOWNS
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3rd Step
 Knee joint is expressed by:
F  m a
s
s
Fk  Fa  ms g  ms as
Fk  ms as  Fa  ms g
M  I 
s
s
M k  M a  rcm , prox  Fk   rcm ,dist  Fa   I s s
M k   M a  rcm , prox  Fk   rcm ,dist  Fa   I s s






k = knee
s = shank
a = ankle
cm = centre of mass
prox = proximal
dist = distal
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3rd Step
 Knee forces/moments
applied to subsequent
segment (thigh)
UNKNOWNS
 Equal and opposite force at
distal extremity of segment
(Newton’s 3rd Law)
 Next, determine unknowns
at proximal extremity of
next segment (hip)
KNOWNS
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3rd Step
 Hip joint is expressed by:
F  m a
t
t
Fh  Fk  mt g  mt at
Fh  mt at  Fk  mt g
M  I 
t
t
M h  M k  rcm , prox  Fh   rcm ,dist  Fh   I t t
M h   M k  rcm , prox  Fh   rcm ,dist  Fh   I t t






k= knee
h = hip
t = thigh
cm = centre of mass
prox = proximal
dist = distal
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Exercise
 Calculate the intersegment forces
and moments at the ankle and knee
 Ground reaction forces
thigh
Fr,x = 6 N
Fr,y = 1041 N
shank
 Rigid body diagrams represent the
foot, shank, and thigh
 Analyse en 2-D
Fr,x = 6 N
Fr,y = 1041
N
•
y
x
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Exercise
F of horizontal reaction (N)
6
F of vertical reaction (N)
1041
Centre of pressure at x, y (m)
0.0, 0.03
Foot
Shank
m (kg)
1
3
I (kg m2)
0.0040
0.0369
ax (m/s2)
-0.36
1.56
ay (m/s2)
-0.56
-1.64
α (rad/s2)
-3.41
-9.39
CM at x,y (m)
0.04, 0.09
0.06, 0.34
Location in x, y (m)
Ankle
Knee
0.10, 0.12
0.02, 0.50
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Exercise
0.02 m
0.06 m
0.03 m
0.34 m
CMshank
0.10 m
0.04 m
0.5 m
knee
ankle
CMfoot
0.12 m
0.09
m
Fr,x
Fr,y
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Exercise
thigh
thigh
shank
foot
Fr,x = 6 N
Fr,y = 1041 N
shank
Fr,x = 6 N
foot
Fr,y = 1041 N
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Exercise
Fs,prox,y
Ms,prox
Fs,prox,x
Fa,y
mfaf,y
Fr,x = 6 N
Iα
Ma
Fa,x
Iα
mfaf,x
mPg
Fr,y = 1041 N
foot
msas,y
msas,x
msg
Fs,dist,x
Ms,dist
Fs,dist,y
shank
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Exercise – ankle (F)
F
x
Fa,y
max
 Fa , x  Fr , x  m f a f , x
Fa , x  6  1kg(0.36m / s 2 )
-0.56 m/s2
y
may
 Fa , y  Fr , y  m f g  m f a f , y
Fa , y  1041  1kg(9.81m / s 2 )  1kg(0.56m / s 2 )
Fa , y  1031.75 N
-0.36 m/s2
Fr,x = 6
N
mf g
Fr,y = 1041
N
ankle = (0.10, 0.12)
CMfoot = (0.04, 0.09)
CP = (0.0, 0.03)
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Fa,x
Iα
Fa , x  6.36 N
F
Ma
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Exercise – ankle (M)
Fa,y
 M I
Ma
-0.56 m/s2
M a  Fa , x (0.12  0.09m)  Fa , y (0.10  0.04m) 
Iα
Fr , x (0.09  0.03m)  Fr , y (0.04  0m)  I f  f
M a  (6.36)(0.03m)  (1031.75 N )(0.06m) 
-0.36 m/s2
Fr,x = 6
N
(6 N )(0.06m)  (1041N )(0.04m)  (0.004kg  m 2 )( 3.41rad / s 2 )
M a  102.996 N  m
mf g
Fr,y = 1041
N
ankle = (0.10, 0.12)
CMfoot = (0.04, 0.09)
CP = (0.0, 0.03)
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Fa,x
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Exercise – knee (F)
 Fx max
Fs,prox,y
Ms,prox
 Fk , x  Fa , x  ms as , x
Fs,prox,x
Fk , x  6.36 N  3kg(1.56m / s 2 )
-1.64 m/s2
Fk , x  1.68 N
F
y
Iα
may
msg
 Fk , y  Fa , y  ms g  ms as , y
Fk , y  1031.75 N  3kg(9.81m / s )  3kg(1.64m / s )
2
Fk , y  1007.24 N
1.56 m/s2
2
6.36 N
ankle = (0.10, 0.12)
CMshank = (0.06, 0.34)
knee = (0.02, 0.77)
102.996N
m
1031.75N
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Exercise – knee (M)
 M I
Fs,prox,y
M k  M a  Fk , x (0.16m)  Fk , y (0.04m)
Ms,prox
Fs,prox,x
 Fa , x (0.22m)  Fa , y (0.04m)  I s s
-1.64 m/s2
Iα
M k  102.996 N  m  (1.68 N )(0.16m) 
msg
(1007.24 N )(0.04m)  (6.36 N )(0.22m) 
(1031.75 N )(0.04m)  (0.0369kg  m 2 )( 9.39rad / s 2 )
M k  19.42 N  m
1.56 m/s2
6.36 N
ankle = (0.10, 0.12)
CMshank = (0.06, 0.34)
knee = (0.02, 0.77)
102.996N
m
1031.75N
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Exercise – Results
Joint
Force in x
(N)
Force in y
(N)
Moment
(Nm)
Ankle
6.36
1032
103
Knee
1.68
1007
19.4
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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
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Principal Calculations
2-D
3-D
F
F
F
F
F
x
y
x
y
z
0
0
0
0
0
M
M
M
M
z
0
x
0
y
0
z
0
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3-D
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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 3rd dimension
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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
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3-D – centre of pressure
X CP  
YCP  
M y  Fx  d z
Fz
M z  Fy  d z
Fz
Tz  M z  Fx  Y   Fy  X 
direction
 M= moment
 F = reaction force
 dz = distance between real origin and
force plate origin
 T = torsion
 * assuming that ZCP = 0
 * assuming that Tx =Ty = 0
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Global and Local CS
Z = +proximal
GCS = global
coordinate system
X = +lateral
LCS = local
coordinate
system
Y = +anterior
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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
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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
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Transformation Matrix
 In GCS, position vector r to
arbitrary point P can be
written (x,y)
𝑟 = 𝑟𝑥 𝑖 + 𝑟𝑦 𝑗
 In rotated (prime) CS (x’,y’)
𝑟 = 𝑟𝑥′ 𝑖′ + 𝑟𝑦′ 𝑗′
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Transformation Matrix
𝑟 ∙ 𝑖 = 𝑟𝑥 𝑖 ∙ 𝑖 + 𝑟𝑦 𝑗 ∙ 𝑖 = 𝑟𝑥 + 0
= 𝑟𝑥 ′ 𝑖 ′ ∙ 𝑖 + 𝑟𝑦′ 𝑗 ′ ∙ 𝑖 = 𝑟𝑥 ′ cos(𝑥 ′ , 𝑥) + 𝑟𝑦′ cos(𝑦 ′ , 𝑥)
𝑟 ∙ 𝑗 = 𝑟𝑥 𝑖 ∙ 𝑗 + 𝑟𝑦 𝑗 ∙ 𝑗 = 0 + 𝑟𝑦
= 𝑟𝑥 ′ 𝑖 ′ ∙ 𝑗 + 𝑟𝑦′ 𝑗 ′ ∙ 𝑗 = 𝑟𝑥 ′ cos(𝑥 ′ , 𝑦) + 𝑟𝑦′ cos(𝑦 ′ , 𝑦)
 In matrix form
𝑟𝑥
cos(𝑥 ′ , 𝑥)
𝑟𝑦 = cos(𝑥 ′ , 𝑦)
cos(𝑦 ′ , 𝑥) 𝑟𝑥′
cos(𝑦 ′ , 𝑦) 𝑟𝑦′
or
𝑟 = 𝑇 𝑟′
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Transformation Matrix
𝑟 = 𝑇 𝑟′
 With
cos 𝜃
cos(𝑥 ′ , 𝑥) cos(𝑦 ′ , 𝑥)
𝑇 =
=
𝜋
cos(𝑥 ′ , 𝑦) cos(𝑦 ′ , 𝑦)
cos 𝜃 −
= 𝑠𝑖𝑛𝜃
2
𝜋
cos 𝜃 +
= −𝑠𝑖𝑛𝜃
2
cos(𝜃)
 [T] is orthogonal, therefore
𝑟′ = [𝑇]𝑇 𝑟
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Kinematics
 Global markers in the GCS are numerized and
transformed to LCS
pi  Tglobaltolocal Pi
pi = position in LCS
Pi = position in GCS
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Kinematics
 P1x
P    P1y
 P1z
 P1x
V    P1y
 P1z

1
P
x
A   P1y
1
P
 z
P2 x
P2 y
P2 z
P 2 x
P 2 y
P 2
z
2
P
x
2
P
y
2
P
z
P3 x 
P3 y 
P3 z 
P 3 x 

P 3 y 
P 3 z 
3 
P
x
3 
P
y
3 
P
z
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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
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Kinetics
SC global  SC local
( Fr , x , Fr , y , Fr , z )  ( f r , x , f r , y , f r , z )
(0,0, Tz )  (0,0, t z )
( X , Y , 0 )  ( x , y ,0 )
(0,0, MG )  (mg x , mg y , mg z )
( Xcm , Ycm , Zcm )  ( xcm , ycm , zcm )
( Dx , D y , Dz )  ( d x , d y , d z )
(Vx , V y , Vz )  (v x , v y , v z )
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3-D Calculations
 Same procedures :
• Distal to proximal
• Newton-Euler
equations
• Joint reaction
forces/moments
• Transformation from
GCS to LCS
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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
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3-D LCS – ankle
 Thus, ankle joint is expressed as:
F  m
f
af
fa  fr  m f g  m f a f
fa h
mah
fa  m f a f  fr  m f g
f a , x  m f x f  f r , x  m f g x
f a , y  m f x f  f r , y  m f g y
f a , z  m f x f  f r , z  m f g z
tr
fr
LCS
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3-D LCS – ankle
M
 I f
f
ma  t r  rcm , prox  f a   rcm , dist  f r   I f 
f
ma  t r  rcm , prox  f a   rcm , dist  f r   I f 
f
ma , x  t r  rcm , prox  f a   rcm , dist  f   I f , xx
I
zz
 I yy  zz yy
fa h
f ,x
ma , y  t r  rcm , prox  f a   rcm , dist  f r   I f , yy
I xx  I zz  xx zz
ma , z  t r  rcm , prox  f a   rcm , dist  f r   I f , zz
I
yy
 I xx  yy xx
mah

f ,y

tr
f ,z

fr
SCL
APA 6903
Fall 2013
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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
APA 6903
Fall 2013
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LCS to GCS
 Transformation matrix (transposed)
Fa   Tlocalto global  f a 
M a   Tlocalto global ma 
T
localto global
  T
globaltolocal

T
Fk   Tlocalto global  f k 
M k   Tlocalto global mk 
Fh   Tlocalto global  f h 
M h   Tlocalto global mh 
APA 6903
Fall 2013
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GCS to LCSshank
 Determine subsequent segment
(shank), using forces/moments
obtained from ankle
fk
mk
Isαs
 Transform force/moment global
vectors of ankle to LCS of the
shank
m sa s
m sg
fa
ma
APA 6903
Fall 2013
65
3-D LCS – knee
 Thus, knee joint is expressed as:
F
 ms a s
f k  f a  ms g  m s a s
f k  ms a s  f a  ms g
f k , x  ms xs  f a , x  ms g x
f k , y  ms xs  f a , y  ms g y
f k , z  ms xs  f a , z  ms g z
SCL
APA 6903
Fall 2013
66
3-D LCS – knee
M
 I s s
mk  ma  rcm , prox  f k   rcm , dist  f a   I s s
mk   ma  rcm , prox  f k   rcm , dist  f a   I s s
mk , x   ma  rcm , prox  f k   rcm , dist  f a   I s , xx s , x
I
zz
 I yy  zz yy
mk , y   ma  rcm , prox  f k   rcm , dist  f k   I s , yy s , y 
I xx  I zz  xx zz
mk , z   ma  rcm , prox  f k   rcm , dist  f k   I s , zz s , z 
I
yy
 I xx  yy xx
SCL
APA 6903
Fall 2013
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LCSshank to GCS to LCSthigh
 Results in reference to LCS of
shank
mh
fh
mta t
 Transform vectors of knee to
GCS, using transformation
matrix
 Then, transform global vectors
of the knee to LCS of the thigh
Itαt
M
mk
mtg
g
Ffgk
APA 6903
Fall 2013
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3-D LCS – hip
 Thus, hip joint is expressed as:
F
 mt at
mh
f h  f k  mt g  mt at
fh
mta t
f h  mt at  f k  mt g
Itαt
f h , x  mt xt  f k , x  mt g x
f h , y  mt xt  f k , y  mt g y
f h , z  mt xt  f k , z  mt g z
M
mk
mtg
g
Ffgk
SCL
APA 6903
Fall 2013
69
3-D LCS – hip
M
 I t t
mh  mk  rcm , prox  f h   rcm , dist  f k   I t t
mh   mk  rcm , prox  f h   rcm , dist  f k   I t t
mh , x   mk  rcm , prox  f h   rcm , dist  f k   I t , xx t , x
I
zz
mta t
 I yy  zz yy
mh , y   mk  rcm , prox  f h   rcm , dist  f k   I t , yy t , y 
I xx  I zz  xx zz
mh , z   mk  rcm , prox  f h   rcm , dist  f k   I t , zz t , z 
I
yy
 I xx  yy xx
Itαt
M
mtg
g
Fg
SCL
APA 6903
Fall 2013
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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 2013
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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 2013
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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 2013
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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 2013
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Applications
 Results represent valuable
approximations of net joint
forces/moments
APA 6903
Fall 2013
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Applications
 Quantifiable results permit
the comparison of patientto-participant’s performance
under various conditions
• Diagnostic tool
• Evaluation of treatment and
intervention
APA 6903
Fall 2013
76
What is “Inverse Dynamics”?
Kinematics
Kinetics
Inverse
“COMPUTATION”
Dynamics
Resultant joint loading
APA 6903
Fall 2013
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Questions