2. (a)From data listed in Table A.3 in Appendix A, calculate the center of
mass of the lower limit for frame 70.
0.8
(1.833, 0.650) cm of thigh
0.7
meters
0.6
0.5
0.4
(1.672, 0.362) cm of shank
0.3
0.2
(1.482, 0.157) cm of foot
0.1
1.4
1.5
1.6
1.7
1.8
1.9
meters
See Power Point presentation for lecture on anthropometry for step-by-step
calculation of the center of mass of the system of the lower limb.
mass of foot = (total body mass) (proportion for foot)
= (68.5 kg) (0.0145) = 0.99325 kg
mass of shank = (total body mass) (proportion for shank)
= (68.5 kg) (0.0465) = 3.18525 kg
mass of thigh = (68.5 kg) (0.100) = 6.85 kg
 mass of lower extremity = 11.0285 kg
2.0

foot as a proportion of lower extremity:
footproportion =
mass
foot
mass
=
lower extremi ty
0.99325kg
11.0285kg
footproportion = 0.09006
shankproportion =
mass
shank
mass
=
3.18525kg
11.0285kg
=
6.85kg
11.028kg
lower extrem ity
shankproportion = 0.288819
thighproportion =
mass
thigh
mass
lower extremi ty
thighproportion = 0.621118

of proportions of lower extremity = 1.0
segment proportion of
x center of x-product
lower extremity mass (cm)
y center of y-product
mass (cm)
Foot
0.090062112
1.482
0.13347205
0.157
0.014139752
Shank
0.288819876
1.672
0.482906832 0.362
0.104552759
Thigh
0.621118012
1.833
1.138509317 0.650
0.403726708
1.754888199
0.522419255
1.000
center of mass of lower extremity 
(1.75, 0.522)
same as book
answer
2. (b) Using data of stride length (problem 2.(f) in section 2.8 and a stride time of
68 frames, create an estimate (assuming symmetrical gait) of the coordinates of the
left half of the body for frame 30.
From problem 2. (f):
First determine stride length: Book suggests that stride length is from foot flat to
foot flat.
heel
The foot remains flat
on the surface for a period
of time.
heel
HCR
foot flat right
Book suggests that the first and second foot flat periods are:
First foot flat – frames 35-40º, x coordinates 1.2384 – 1.2435
Second foot flat – frames 102-106º, x coordinates 2.6524 – 2.6515
Which x coordinate of the heel should be used to determine stride length?
Possibilities: x35  x102  2.6524 – 1.2384 = 1.414m
or
X40  x106  2.6515 – 1.2435 = 1.408m
* There are other possibilities – for example mid-foot flat to mid-foot flat.
** Note that I would have calculated stride length using first HCR to second HCR
(frame 28 to frame 97; 2.6422 – 1.2280 = 1.4142m) because it is one frame to
another frame versus a period to another period. Note that it is close in value to
x35  x102 .
Next, let’s determine where frame 30 occurs within the complete stride:
69 intervals
of time
Event
TOR
Frame #
1
HCR
28
TOR
HCR
69 intervals
of time
30
70
97
*Note that there are 69 intervals
of time. Book is in error
suggesting 68 frames. If we use
book’s suggestions of 68
intervals, 1/2 of stride time would
be 34 intervals.
 right and left sides of body are
out of phase with each other by
34 frames (time intervals).
What frame of the right side of the body, for which we have data corresponds to
the left side?
Answer: frame 30 + 34 = frame 64. The assumption is that we can use information
about the right side of the body and translate it one half stride to make if the left
side. Note that all y coordinates will be unchanged. All x coordinates from frame
64 must have 1/2 stride length subtracted from them. In other words, 1.414m/2 =
0.707 m must be subtracted from each x value of frame 64 from the right side of
the body to create segments of the left side of the body for frame 30, assuring that
the right and left sides are out of phase by 1/2 stride.
From the segment centers of mass (Table A.3  note book error), calculate the
center of mass of the right half of the body (right foot + right leg (or shank) + right
thigh + 1/2 right HAT):
segment
mass proportion
of right half
x cm
x-product
y cm
y-product
R. foot
0.0145/.5 = 0.029 1.329
0.038541
0.084
0.002436
R. shank
0.0465/.5 = 0.093 1.212
0.112716
0.329
0.030597
R. thigh
0.100/.5 = 0.200
1.110
0.222
0.670
0.134
R.
0.339/.5 = 0.678
1.029
0.697662
1.089
0.738342
1
2
HAT
 1
1.070919
 0.905375
Center of mass for right half of body for frame 30 is (1.070919m, 0.905375m)
Calculate the center of mass of the left half of the body:
segment
mass proportion x cm
of left half
x-product y cm
L. foot
0.029
x64 right-0.707=
1.38-0.707=0.673
0.019517
L. shank
0.093
L. thigh
0.200
L. 1 2 HAT 0.678

=1
y-product
0.099
0.002871
x64 right-0.707=
0.07347
1.497-0.707=0.790
0.347
0.032271
x64 right-0.707=
0.1896
1.655-0.707=0.948
0.663
0.1326
x64 right-0.707=
0.70512
1.083
0.734274
1.747-0.707=1.040
 =0.987707
 =0.902016
y values
unchanged from
frame 64
Center of mass for left half of body for frame 30 is
(0.987707m, 0.902016m)
To determine the center of mass of the entire body:
 add x coordinates of the center of gravity of right and left halves and divide
by 2.
 add y coordinates of the center of gravity of right and left halves and divide
by 2.
X: (1.070919m  0.987707m) 2 = 1.029313m
Y: (0.905375m  0.902016m 2 = 0.9036955m
Close to book answer of (1.025m, 0.904m)

3. (a) Calculate the moment of inertia of the HAT about its center of mass for
the subject described in Appendix A.
Moment of inertia = I0 = m 0 , where I0 = moment of inertia about the center of
mass, m = mass of HAT, and  0 = radius of gyration
2
I0 = [(HAT mass proportion of body) (mass of entire body)]
[(radius of gyration/segment length) (length used for HAT)]2
*Note that three different values are provided in Table 3.1 for HAT (head and
trunk) for radius of gyration per segment length (0.503, 0.496, 0.903) about the
center of mass. Each uses a differently defined length for determining the radius of
gyration. These lengths are: greater trochanter to glenohumeral joint
greater trochanter to mid rib
Appendix A provides data on greater trochanter to base rib cage (mid rib) = 25 cm.
No value is directly provided for length of greater trochanter to glenohumeral joint.
Therefore, we would need to determine this distance indirectly.
Given: Segment length of various segments in Appendix A and body segment
lengths expressed as fractions of body height in figure 3.1. It is possible to
determine length of greater trochanter to glenohumeral joint.
height of greater trochanter = 0.530 total body height
height of glenohumeral joint = 0.818 total body height
 length from greater trochanter to glenohumeral joint is 0.818 H – 0.530 H
= 0.288 H
How can we determine H? We must use some known value of segment length
provided in Appendix A.
For example: greater trochanter height – knee height =thigh length or
0.530 H – 0.285 H= 0.245 H = 31.4 cm
.288H
.245H
=
length from greater
31.4cm
trochanter to
glenohumeral joint
length = 36.91 cm
Another example: knee height – ankle height =
shank length or 0.285 H – 0.039 H =
0.246 H = 42.5 cm
.288H
0.246H
=
length of trochanter
42.5cm
to glenohumeral joint
length = 49.76 cm
*Note difference in length of greater trochanter to glenohumeral joint calculated
from these two methods!!! Big difference!!!
Using values for greater trochanter/mid rib:
I0 = [(0.678)(56.7 kg)] [(0.903)(0.25m)]2 =
(38.4426 kg) (0.22575 m)2 = (38.4426 kg)(0.050963063m2)
= 1.959152626 kg m2
Using values for greater trochanter/glenohumeral joint HAT:
2
I0 = [(0.678)(56.7 kg)] [(0.496)(0.3691m)]2 = 1.288439991kg m
Or
I0 = [(0.678)(56.7 kg)] [(0.496)(0.4976 m)]2 = 2.341730158 kg m2
Using values for greater trochanter/glenohumeral joint trunk head neck:
I0 = [(0.578)(56.7 kg)] [0.503)(0.3691 m)]2 =
Or
1.129626723 kg m2
I0 = [(0.578)(56.7 kg)] [(0.503)(0.4976 m)]2 = 2.053088218 kg m2
Big difference in values!!!
3. (b) Assuming subject is standing erect calculate the moment of inertia of HAT
about hip, knee, and ankle. First, we must determine the distance of the center of
mass of HAT to the hip, knee, and ankle. Note that the center of mass of HAT
must be determined from figure 3.1. HAT length is 0.47 H. If we use thigh length
= 31.4 cm. HAT could be determined.
0.47H
.245H
=
Length of HAT
31.4cm
 length of HAT = 60.24 cm
Note that the center of mass of the HAT is 0.626 (.6024 m) from the proximal end
and 0.374 (.6024 m) from the distal end.
IHAT about hip = IHAT about cm + (massHAT) [(0.374)(0.6024m)]2

Note there are many differing
values that can be selected from
previous calculations. We will
pick 2.053 kg m2
= 2.053 kg m2 + (0.678)(56.7 kg) [(0.374)(0.6024m)]2
= 4.004 kgm2
IHAT about knee = IHAT about cm + massHAT [distance of HAT center of mass to knee]2
= 2.053 kg m2 + (38.4426 kg)[(0.225 m + .314m)]2
= 13.22 kg m2
IHAT about ankle = I HAT about cm + mass HAT [distance of HAT center of mass to ankle]2
= 2.053 kg m22 + (38.4426 kg)[0.225 m +0.314 m + 0.425 m]2
=
37.7776 kg m
Note that there is great possibility for variability in these values. It should
also be noted that distance from a joint is a term that is squared. Thus, as
this increases the magnitudes of the moments of inertia increase
substantially.
3.(c) Assuming the center of mass is 1.65 m from the ankle, what
percentage does it contribute to the moment of inertia of HAT about the
ankle?
Ihead about ankle = (masshead) (distance head center of mass to ankle)2
= [(proportion of head and neck)(massbody)] [distance]2
= [(.081)(56.7 kg)][1.65 m]2 = 12.5 kg m2
The contribution of the head to the moment of inertia of HAT about the
ankle is as follows:
2
% contribution of head to IHAT about ankle =
12.5 kgm
37.7776 kgm
= 33.1%
2
X 100
4.(a) Calculate the moment of inertia of the lower limb of the subject in
Appendix A about the hip joint. Assume the knee is not flexed and the foot
is a point mass located 6 cm distal to the ankle.
Ilower limb about hip = Ithigh about hip + Ishank about hip + Ifoot about hip
Calculate Ithigh about hip :
This can be done by either calculating I0 and adding the moment of inertia
associated with parallel axis theorem.
[I0 + (mthigh)(xthigh cm to hip)2] or more easily calculating the moment of inertia
about the proximal (hip) end
Ithigh about hip = (mthigh)(  proximal)2 =
[(0.10)(56.7 kg)][(0.54)(0.314m)]2 = 0.1630 kg m2
Calculate Ishank about hip :
This can be done in a variety of ways. We will calculate the moment of
inertia of the shank about the proximal end and then use the parallel axis
theorem.
Ishank about hip = Ishank about proximal + (mshank)(Xproximal end to hip)2
= [(0.0465)(56.7 kg)][(0.528)(0.425m)]2 + (0.0465)(56.7 kg)[0.314m]2
= 0.3927 kg m2
Calculate Ifoot about hip :
Ifoot about hip = (mfoot)(Xfoot to hip)2
= (0.0145)(56.7 kg)(0.799 m)2
= 0.52486 kgm2
 Ilower limb about hip = 0.1630 kgm2 + 0.3927 kgm2 +0.52486 kgm2
= 1.08056 kgm2
4.(b) Calculate the increase in moment of inertia as calculated in (a) when
a ski boot is worn. The mass of the ski boot is 1.8 kg and assumed to be a
point mass located 1 cm distal to the ankle.
Iboot about hip = (massboot)(Xboot to hip)2
= (1.8 kg)(0.749m)2 =
1.01 kgm2
Ilower limb + boot about hip = Ilower limb about hip + Iboot about hip
= 1.08056 kgm2 + 1.01 kgm2 =
= 2.09 kgm2
% increase in I =
Iboot abouthip
Ilower li m babouthip
1.01kgm
X 100 =
1.08kgm
2
2
X 100
=
93.5% increase