Inverse Dynamics
Inverse Dynamics
APA 6903
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What is “Inverse Dynamics”?
APA 6903
Fall 2013 3
What is “Inverse Dynamics”?
Motion – kinematics Force – kinetics Applied dynamics
APA 6903
Fall 2013 4
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: a given action creates an equal and opposite reaction
amF
F
0
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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
FForces
F = maEquations of
motion
∫∫Double
integration
xDisplacements
xDisplacements
d2x / dt2
Double differentiation
F = maEquations of
motion
FForces
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Dynamics
Direct method• Forces are known• Motion is calculated by integrating once to obtain
velocity, twice to obtain displacement
FForces
F = maEquations of
motion
∫∫Double
integration
xDisplacement
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Dynamics
Inverse method• Displacements/motion are known• Force is calculated by differentiating once to
obtain velocity, twice to obtain acceleration
xDisplacements
d2x / dt2
Double differentiation
F = maEquations of
motion
FForces
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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
Forces/moments known at foot segment
Using Newton-Euler formulas, calculation begins at foot, then to ankle
Proceed from distal to proximalKNOWNS
UNKNOWNS
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FBD of Foot
$%#&?! Multiple unknowns Centre of
gravity
Fg
Centre of pressure
Triceps sural force
Anterior tibial muscle force
Bone force
Ligament force
Joint moment
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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
Displace forces to joint centre
Force equal and opposite
Centre gravity
Fr
Centre of pressure
FFoot muscle
forces
F*Force at joint
centre
-F*Force equal and opposite
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Simplification
Replace coupled forces with moment
Centre of gravity
FFoot muscle
force
F*Force at joint
centre
-F*Force equal and opposite
M
Moment
Fr
Centre of pressure
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Simplification
Representation net moments and forces at ankle
Freaction
xreaction, yreaction
mfootg
Fankle
xankle, yankle
Mankle
rcm,dist
rcm,prox
cm = centre of mass prox = proximal dist = distal
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3rd Step
f = foot a = ankle r = reaction prox = proximal dist = distal
Fa
Ma
Tr
Fr
mfafmfg
Ifαf
Force/moment known(force plate)
Unknown forces/moments at
ankle
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3rd Step
Therefore, ankle joint expressed by:
gmFamF
amgmFF
amF
frffa
fffra
ff
ffrdistcmaproxcmra
ffrdistcmaproxcmra
ff
IFrFrTM
IFrFrTM
IM
,,
,,
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3rd Step
Thus, simply in 2-D :
Much more complicated in 3-D!
gmFamF
FamF
amF
fyryffya
xrxffxa
ff
,,,
,,,
ffrdistcmaproxcma
ff
IFrFrM
IM
,,
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3rd Step
Moment is the vector product of position and force
NOT a direct multiplication
xyyxz
z
FrFrM
FrM
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3rd Step
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)
UNKNOWNS
KNOWNS
h
h
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3rd Step
Knee joint is expressed by:
gmFamF
amgmFF
amF
sassk
sssak
ss
ssadistcmkproxcmak
ssadistcmkproxcmak
ss
IFrFrMM
IFrFrMM
IM
,,
,, 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)
Equal and opposite force at distal extremity of segment (Newton’s 3rd Law)
Next, determine unknowns at proximal extremity of next segment (hip)
UNKNOWNS
KNOWNS
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3rd Step Hip joint is expressed by:
gmFamF
amgmFF
amF
tktth
tttkh
tt
tthdistcmhproxcmkh
tthdistcmhproxcmkh
tt
IFrFrMM
IFrFrMM
IM
,,
,, 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 forcesFr,x = 6 N
Fr,y = 1041 N
Rigid body diagrams represent the foot, shank, and thigh
Analyse en 2-D
thigh
shank
• x
y Fr,x = 6 N
Fr,y = 1041 N
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Exercise
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
Ankle Knee
Location in x, y (m) 0.10, 0.12 0.02, 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
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Exercise
ankle
CMshank
knee
CMfoot
Fr,x
Fr,y
0.5 m
0.04 m
0.34 m
0.12 m
0.03 m0.09 m
0.10 m
0.06 m
0.02 m
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Exercise
thigh
shank
footFr,x = 6 N
Fr,y = 1041 N
thigh
shank
footFr,x = 6 N
Fr,y = 1041 N
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Exercise
shankfoot
Fr,x = 6 N
Fr,y = 1041 N
Ma
Fa,x
Fa,y
Iα
mPg
mfaf,y
mfaf,x Ms,dist
Ms,prox
Fs,dist,x
Fs,dist,y
Fs,prox,y
Fs,prox,x
msas,x
msas,y
msg
Iα
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Exercise – ankle (F)
NF
smkgF
amFF
maF
xa
xa
xffxrxa
xx
36.6
)/36.0(16
,
2,
,,,
NF
smkgsmkgF
amgmFF
maF
ya
ya
yfffyrya
yy
75.1031
)/56.0(1)/81.9(11041
,
22,
,,,
Fr,x = 6 N
Fr,y = 1041 N
Ma
Fa,x
Fa,y
Iα
mfg
-0.56 m/s2
-0.36 m/s2
ankle = (0.10, 0.12)CMfoot = (0.04, 0.09)
CP = (0.0, 0.03)
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Exercise – ankle (M)
mNM
sradmkgmNmN
mNmM
ImFmF
mFmFM
IM
a
a
ffyrxr
yaxaa
996.102
)/41.3)(004.0()04.0)(1041()06.0)(6(
)06.0)(75.1031()03.0)(36.6(
)004.0()03.009.0(
)04.010.0()09.012.0(
22
,,
,,
Fr,x = 6 N
Fr,y = 1041 N
Ma
Fa,x
Fa,y
Iα
mfg
-0.56 m/s2
-0.36 m/s2
ankle = (0.10, 0.12)CMfoot = (0.04, 0.09)
CP = (0.0, 0.03)
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Exercise – knee (F)
NF
smkgNF
amFF
maF
xk
xk
xssxaxk
xx
68.1
)/56.1(336.6
,
2,
,,,
NF
smkgsmkgNF
amgmFF
maF
yk
yk
ysssyayk
yy
24.1007
)/64.1(3)/81.9(375.1031
,
22,
,,,
102.996Nm
Ms,prox
6.36 N
1031.75N
Fs,prox,y
Fs,prox,x
1.56 m/s2
-1.64 m/s2
msg
Iα
ankle = (0.10, 0.12)CMshank = (0.06, 0.34)knee = (0.02, 0.77)
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Exercise – knee (M)
mNM
sradmkgmN
mNmN
mNmNM
ImFmF
mFmFMM
IM
k
k
ssyaxa
ykxkak
42.19
)/39.9)(0369.0()04.0)(75.1031(
)22.0)(36.6()04.0)(24.1007(
)16.0)(68.1(996.102
)04.0()22.0(
)04.0()16.0(
22
,,
,,
102.996Nm
Ms,prox
6.36 N
1031.75N
Fs,prox,y
Fs,prox,x
1.56 m/s2
-1.64 m/s2
msg
Iα
ankle = (0.10, 0.12)CMshank = (0.06, 0.34)knee = (0.02, 0.77)
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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-D0
0
0
z
y
x
F
F
F
0
0
0
z
y
x
M
M
M
0
0
y
x
F
F 0zM
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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
direction
XFYFMT
F
dFMY
F
dFMX
yxzz
z
zyzCP
z
zxyCP
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
Y = +anterior
X = +lateral
Z = +proximal
LCS = local coordinate system
GCS = global coordinate system
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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)
In rotated (prime) CS (x’,y’)
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Transformation Matrix
In matrix form
or
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Transformation Matrix
With
[T] is orthogonal, therefore
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Kinematics
Global markers in the GCS are numerized and transformed to LCS
pi = position in LCS
Pi = position in GCS
ilocaltoglobali PTp
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Kinematics
zzz
yyy
xxx
zzz
yyy
xxx
zzz
yyy
xxx
PPP
PPP
PPP
A
PPP
PPP
PPP
V
PPP
PPP
PPP
P
321
321
321
321
321
321
321
321
321
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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
),,(),,(
),,(),,(
),,(),,(
),,(),0,0(
)0,,()0,,(
),0,0(),0,0(
),,(),,( ,,,,,,
zyxzyx
zyxzyx
cmcmcmcmcmcm
zyx
zz
zryrxrzryrxr
localglobal
vvvVVV
dddDDD
zyxZYX
mgmgmgMG
yxYX
tT
fffFFF
SCSC
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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:
zfzrffza
yfyrffya
xfxrffxa
frffa
fffra
ff
gmfxmf
gmfxmf
gmfxmf
gmfamf
amgmff
amF
,,
,,
,,
h
h
fa
ma
fr
tr
LCS
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h
h
fa
ma
fr
tr
3-D LCS – ankle
xxyyxxyy
zfzzfrdistcmaproxcmrza
zzxxzzxx
yfyyfrdistcmaproxcmrya
yyzzyyzz
xfxxfdistcmaproxcmrxa
ffrdistcmaproxcmra
ffrdistcmaproxcmra
ff
II
Ifrfrtm
II
Ifrfrtm
II
Ifrfrtm
Ifrfrtm
Ifrfrtm
IM
,,,,,
,,,,,
,,,,,
,,
,,
SCL
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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
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LCS to GCS
Transformation matrix (transposed)
Tlocaltoglobalglobaltolocal TT
hglobaltolocalh
hglobaltolocalh
kglobaltolocalk
kglobaltolocalk
aglobaltolocala
aglobaltolocala
mTM
fTF
mTM
fTF
mTM
fTF
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GCS to LCSshank
Determine subsequent segment (shank), using forces/moments obtained from ankle
Transform force/moment global vectors of ankle to LCS of the shank
fk
mk
fa
ma
msas
msg
Isαs
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3-D LCS – knee
Thus, knee joint is expressed as:
zszasszk
ysyassyk
xsxassxk
sassk
sssak
ss
gmfxmf
gmfxmf
gmfxmf
gmfamf
amgmff
amF
,,
,,
,,
SCL
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3-D LCS – knee
xxyyxxyy
zszzskdistcmkproxcmazk
zzxxzzxx
ysyyskdistcmkproxcmayk
yyzzyyzz
xsxxsadistcmkproxcmaxk
ssadistcmkproxcmak
ssadistcmkproxcmak
ss
II
Ifrfrmm
II
Ifrfrmm
II
Ifrfrmm
Ifrfrmm
Ifrfrmm
IM
,,,,,
,,,,,
,,,,,
,,
,,
SCL
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LCSshank to GCS to LCSthigh
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
Itαt
mtat
mtg
Fg
Mg
fk
mk
mh fh
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3-D LCS – hip
Thus, hip joint is expressed as:
ztzkttzh
ytykttyh
xtxkttxh
tktth
tttkh
tt
gmfxmf
gmfxmf
gmfxmf
gmfamf
amgmff
amF
,,
,,
,,
Itαt
mtat
mtg
Fg
Mg
fk
mk
mh fh
SCL
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3-D LCS – hip
xxyyxxyy
ztzztkdistcmhproxcmkzh
zzxxzzxx
ytyytkdistcmhproxcmkyh
yyzzyyzz
xtxxtkdistcmhproxcmkxh
ttkdistcmhproxcmkh
ttkdistcmhproxcmkh
tt
II
Ifrfrmm
II
Ifrfrmm
II
Ifrfrmm
Ifrfrmm
Ifrfrmm
IM
,,,,,
,,,,,
,,,,,
,,
,,
Itαt
mtat
mtg
Fg
Mg
SCL
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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
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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
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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
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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
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Applications
Results represent valuable approximations of net joint forces/moments
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Applications
Quantifiable results permit the comparison of patient-to-participant’s performance under various conditions• Diagnostic tool• Evaluation of treatment and
intervention
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What is “Inverse Dynamics”?
Kinematics Kinetics
“COMPUTATION”
Resultant joint loading
Kinematics Kinetics
Inverse Dynamics
Resultant joint loading
Questions