APPENDIX: INVERSE DYNAMICS ANALYSIS
In this section, we derive the equations of the reaction forces and moments for the ankle and knee
joints. We are mainly interested in the expression of the knee moment. Winter presents a detailed
explanation of the inverse dynamics analysis [46] . Our analyses develop the Newtonian equations of
motion of foot with respect to (w.r.t.) the global coordinate system as shown by X  Y  Z in Fig. S1.
Next, we derive an expression for the moment of the ankle joint w.r.t. the anatomical coordinate frame of
the foot shown by x f  y f  z f in Fig. S1. The anatomical coordinate frame of the foot is established by
placing y f on the axis connecting the toe to the ankle center, z f along with Z and x f along with the
cross-product of y f and z f . The Newtonian equations of motion of the shank give us the reaction force
of the knee in the global axes, using the reaction force of the ankle. Then, we apply the reaction forces of
the ankle and knee in the Euler equations of motion and derive an expression for the moment of the knee
w.r.t. the anatomical coordinate frame of the shank shown by xs  ys  zs . The anatomical coordinate
frame of the shank is established by placing y s on the axis connecting the ankle center to the knee center,
z s along Z and xs along the cross-product of y s and z s . Table S1 lists the parameters that are used in
this text and brief description of each parameter.
The pair of ground reaction force ( FG ) and ground reaction moment ( M G ) is transferred to the distal
joint (i.e. toe) from the center of pressure (COP) to obtain the distal force ( RDf ) and the distal moment (
M Df ) of the foot w.r.t. X  Y  Z . Hence w.r.t. X  Y  Z we get:
RDf  FG
(A-1-a)
M Df  M G  FG  r
(A-1-b)
and w.r.t. x f  y f  z f we get:
Rdf  GA f FG

M df  GA f M G  FG  r
(A-2-a)

(A-2-b)
r is the vector connecting the distal joint to COP and GA f is a proper (i.e. reserves inner product
and has a determinant of 1) rotation from X  Y  Z to x f  y f  z f . The reaction force at the ankle (
RPf ) is derived using the Newtonian equation of motion for the foot:
F
f
 m f a f  RPf  RDf  m f a f  m f geY
(A-3)
where, Ff denotes any force that is applied on the foot segment and a f is the acceleration of the center
of mass of the foot ( COM f ). m f is the mass of the foot segment, eY is a unit vector along the Y-axis,
and g is the acceleration due to gravity. Equation (A-1-a) gives us the proximal force of the foot w.r.t.
X  Y  Z as:
RPf  FG  m f a f  m f geY
(A-4)
which could be transformed to x f  y f  z f through a rotation, as:
R pf  GA f RPf
(A-5)
Now, we develop the Euler equation of motion for the foot segment w.r.t. x f  y f  z f to derive the
expression of the moment at its proximal joint (i.e. ankle):
M
f
  I f   f   f  U f
(A-6)
wherein, M f denotes any moment that is applied on the foot segment and  I f  is the matrix of moment
of inertia.  f the angular velocity,  f the angular acceleration, and U f is the angular momentum of the
foot segment. Expanding the left hand side of the above equation gives us:
M pf  M df  Rdf  d f  R pf  p f   I f   f   f  U f
(A-7)
f
where, M p is the moment at the proximal joint of the foot segment expressed w.r.t. x f  y f  z f . d f is
the vector that connects the center of mass of the foot ( COM f ) to the toe and p f is the vector that
connects COM f to the ankle joint both expressed w.r.t. x f  y f  z f . Here, the tip of the toe is chosen
such that COM f relies on the origin of x f  y f  z f Now, we insert the corresponding terms in
equation (A-7):


M pf  GA f M G  FG  r  GA f FG  d f

(A-8)

 GA f FG  m f a f  m f geY  p f   I f   f   f  U f
Since GA f is a proper rotation, M p could be transformed into X  Y  Z as:
f




M Pf  M G  FG  r  FG   AG  f d f  FG  m f a f  m f geY   AG  f p f

  AG  f  I f   f   f  U f

where,  AG  f  GA f . One should notice that:
1
(A-9)
FG  L f eYf   FG   AG  f d f  FG   AG  f p f
(A-10)
where, L f is the length of the foot segment and eYf is the unit vector along the y f -axis of the foot
segment expressed w.r.t. Y-axis. Thus, equation (A-9) can be written in the following form:



M Pf  M G  FG  r  FG  L f eYf   m f a f  m f geY   Lpf eYf   AG  f  I f   f   f  U f

(A-11)
f
where, Lp is the distance between COM f and the ankle.
Next, we derive the proximal moment of the shank segment (i.e. knee moment). We exploit the
expressions of the proximal force and moment for the foot segment and use them as the distal force ( RDs )
and moment ( M Ds ) of the shank. In other words w.r.t. X  Y  Z we have:
RDs  RPf
(A-12-a)
M Ds  M Pf
(A-12-b)
which can be transformed to xs  ys  zs using the proper rotation-matrix of GAs :
Rds  GAs RPf
(A-13-a)
M ds  GAs M Pf
(A-13-b)
The proximal reaction force of the shank ( RPs ) is obtained from the Newtonian equation of motion
w.r.t. X  Y  Z , as follows:
F
s
 msas  RPs  RDs  msas  ms geY
(A-14)
where, Fs denotes any force that is applied on the shank segment, ms is the mass and as is the
acceleration of the center of mass ( COM s ) for the shank segment. Applying equations (A-4-a) and (A12-a) in (A-14), we get:
RPs  FG   m f a f  ms as    m f  ms  geY
(A-15)
which could be transformed to xs  ys  zs as:
Rps  GAs RPs
(A-16)
Now, we employ the Euler equation of motion for the shank w.r.t. xs  ys  zs to derive the moment
at the proximal joint (i.e. knee). As such:
 M   I 
s
s
s
 s  U s
(A-17)
where, M s denotes any moment that is applied on the shank.  I s  is the matrix of moment of inertia,  s
the angular velocity,  s is the angular acceleration, and U s is the angular momentum for the shank
segment. Expanding the left hand side of the above equation results in:
M ps  M ds  Rds  d s  Rps  ps   I s  s  s  U s
(A-18)
where, M ps is the moment at the proximal joint of the shank (i.e. knee moment). d s is the vector
connecting the center of mass of the shank ( COM s ) to the ankle and p s the vector connecting COM s
to the knee both expressed in xs  ys  zs . Inserting the corresponding expression of each term in (A-18)
concludes:



M ps  GAs M G  FG  r  FG  L f eYf   m f a f  m f geY   Lpf eYf   AG  f  I f   f   f  U f




 GAs FG  m f a f  m f geY  d s  GAs FG   m f a f  ms as    m f  ms  geY  ps
  I s  s  s  U s

(A-19)
The proximal moment of the shank ( M p ) is transformed to X  Y  Z to reach the following
s
expression for the knee moment ( M Ps ):



M Ps  M G  FG  r  FG  L f eYf   m f a f  m f geY   Lpf eYf   AG  f  I f   f   f U f





 FG  m f a f  m f geY   AG s d s  FG   m f a f  ms as    m f  ms  geY   AG s ps

  AG s  I s  s  s  U s

(A-20)
where,  AG s  GAs . One should notice that:
1
FG  Ls eYs   FG   AG s d s  FG   AG s ps
(A-21)
where, Ls is the length of the shank segment and eYs is a unit vector along y s -axis of the shank segment
expressed w.r.t. X  Y  Z . Hence, we obtain the following expression for the moment of the knee joint:


M Ps  M G  FG   r  L f eYf  Ls eYs  
 m a
f
f

 m f geY    Lpf eYf  Ls eYs    ms as  ms geY   Lsp eYs 
 AG  I    U    AG   I
s
s
s
s
s
s
f
f
  f   f  U f
where, Lp is the distance between COM s and the knee.

(A-22)