Download Advanced Biomechanics of Physical Activity (KIN 831)

Advanced Biomechanics of
Physical Activity (KIN 831)
Anthropometry
Material included in this presentation is derived primarily from:
Winter, D.A. (1990). Biomechanical and motor control of human movement.
(2nd ed.). New York: Wiley & Sons
Hamill, J. & Knutzen, K. (2003). Biomechanical Basis of Human Movement
(2nd ed.). Philadelphia: Lippincott Williams & Wilkins
Hamilton, N. & Luttgens, K. (2002). Kinesiology – Scientific Basis of Human
Motion (10th ed.). Boston: McGraw-Hill.
What is
• Anthropology?
• Anthropometry?
Definitions
• Anthropology – the scientific study of the
origin and of the physical, social, and
cultural development and behavior of man
• Anthropometry – the study and technique of
human body measurement for use in
anthropological classification and
comparison
What functions does
anthropometry serve?
Uses of Anthropometry
• Studies of physical measurements of the
human body
• Determining differences in individuals and
groups
–
–
–
–
Age
Sex
Race
Body type (somatotype)
Emphases of Anthropometric Studies
• Evolution
• Historical
Past studies
• Performance parameters
• Man-machine interfaces
Associated with
recent technological
developments
What are some examples of
anthropometric measurements?
Types of Anthropometric
Measurements
• Static measurements
–
–
–
–
–
–
Length
Area
Volume
Breadths
Circumference
Skin fold thickness
Types of Anthropometric
Measurements
• Static measurements (continued)
– Ratios and proportions
•
•
•
•
•
Body mass index – height/stature2
Sitting height/stature
Bicristal/biacromial
Ponderal index – height/weight1/3
Physique (somatotype – endomorphy, mesomorphy,
ectomorphy)
Types of Anthropometric
Measurements (continued)
• Kinetic measurements
– Linear – using mass
• Force = mass x acceleration
F = MA
– Angular – using moment of inertia
• Torque or moment of force = moment of inertia x
angular acceleration
T or M = I
Body Segment Lengths
• Vary with body build, sex, racial origin
• Dempster’s data (1955, 1959) – segment
lengths and joint center locations relative to
anatomical landmarks (see figure)
• Drillis and Contini (1966) – segment
lengths/body height (see figure)
Body Segment Lengths
• Vary with body build, sex, racial origin
• Dempster’s data (1955, 1959) – segment
lengths and joint center locations relative to
anatomical landmarks
• Drillis and Contini (1966) – segment
lengths/body height (see figure)
What is
• Mass?
• Inertia?
• Density?
Density, Mass, and Inertial Properties
• Kinematic and kinetic analyses require data on
mass distribution, mass centers, and moments of
inertia
– Measured directly
• Cadaver
• Segment volume
• Segment density
– Measured indirectly
• Density via MRI
Density, Mass, and Inertial Properties
(continued)
• Whole body density
• Tissue density – varies with type of tissue
• Specific gravity – weight of tissue/weight of
water of same volume
–
–
–
–
Cortical bone = 1.8
Muscle  1.0
Fat  0.7
Lungs < 1.0
Density, Mass, and Inertial Properties
(continued)
• Drillis and Contini (1966) expression for body density
d = 0.69 + 0.0297cE
answer in kilograms/liter
where cE = (height in inches)/(weight in pounds)1/3 ;this is the inverse
ponderal index
d = 0.69 + 0.9cm
answer in kilograms/liter
where cm = (height in meters)/(mass in kilograms)1/3
Note that it can be seen that a tall thin person has a higher
inverse ponderal index than a short fat individual and therefore
greater density.
Proof of conversion from English to metric units in equations:
Since d = 0.69 + 0.0297cE and d = 0.69 + 0.9cm
0.0297cE = 0.9cm
cm/cE = 0.0297/0.9 = 0.033
meters/kilograms1/3 = 0.033
inches/pound1/3
1 meter/1 kilogram1/3
= 0.033 (disregarding units) 
39.37 inches/2.2046 pounds1/3
1
 = 0.033 proof !!!
39.37/1.30
Example calculation of density:
Person 70 inches and 170 pounds
d = 0.69 + 0.0297cE
cE = h/w1/3 = 70/1701/3 = 12.64
d = 0.69 + 0.0297 (12.64) = 1.065 kg/l
Density, Mass, and Inertial Properties
(continued)
• Segment densities
– Unique densities for each body segment
– Each segment has a different combination of tissues
(e.g., head vs. shank)
– Individual segments increase in density with increase
in total body density (see figure)
– Upper and lower extremities more dense than whole
body
– Proximal segments less dense than distal segments
Density, Mass, and Inertial Properties
(continued)
• Segment mass and center of mass
– Center of mass and center of gravity used
interchangeably
– Total body mass vs. segment mass
• Increase in total body mass  increase in segmental mass
(proportional increases)
• Possible to express mass of each segment as a proportion of the
total body
• Proportions vary by age, gender, and other factors
• Location of center of mass determined as a percent of segment
length (from proximal or distal end)
• Balance technique used in cadavers to determine center of mass
Total Mass
of a Segment
Mass = mi
mi = diVi
density =
mass/volume
M =  diVi
Note that if d is uniform or assumed to be uniform,
M = d Vi
Note that the center of mass creates the same net moment about any
point along the segment axis as does the original distributed mass
Mx =  mixi
Why the
same
definition ?
Head
And
Trunk
Example: If the greater trochanter has
coordinates (72.1,98.8) and the lateral
femoral condyle has coordinates
(86.4,54.9), calculate the center of mass
of the thigh.
Y
(72.1, 98.9)
98.9
0.433(length)
0.433(Y)
center of mass
0.567(length)
0.567(Y)
(86.4, 54.9)
54.9
72.1
0.433(X)
86.4
0.567(X)
X
Example: If the greater trochanter has coordinates (72.1,98.8) and the lateral
femoral condyle has coordinates (86.4,54.9), calculate the center of mass of the
thigh.
Using information from the proximal end:
x = 72.1 + 0.433(86.4 –72.1) = 78.3
y = 92.8 – 0.433(92.8 – 54.9) = 76.4
Calculate the center of mass of a multisegment system
(e.g., lower extremity, entire body)
- step 1: determine the proportion of mass that each segment is of the
entire multisegment system
- step 2: multiply each segmental proportion times the x coordinate of the
center of mass of that segment
- step 3: multiply each segmental proportion times the y coordinate of the
center of mass of that segment
- step 4: add each of the x products
- step 5: add each of the y products
- step 6: the sums from steps 4 and 5 are the x and y coordinates of the
center of mass of the multisegment system
x0 = m1x1 + m2x2 + m3x3 + + mnxn
M
y0 = m1y1 + m2y2 + m3y3 + + mnyn
M
Segment
Proportion
of total
mass
Segment 1
1
x1
Segment 2
2
Segment 3
3












Segment n
n
 = 1.00
*
sample table
x value of
x product
center of
mass
y value of
center of
mass
y product
 x1
y1
 y1
x2
 x2
y2
 y2
x3
 x3
y3
 y3
xn
 xn
 = x value
of the
center of
mass*
yn
 yn
 = y value
of the
center of
mass*
Note that the calculated center of mass will be relative to the Cartesian coordinate system that was used for the center of masses
used for the individual segments.
Example (not in book): Calculate the
center of mass of the right lower
extremity (foot, shank, and thigh) for
frame 33 of the subject in Appendix A
(use Table A.3(a-c))
Example (not in book): Calculate the center of mass of the right lower extremity (foot, shank, and
thigh) for frame 33 of the subject in Appendix A (use Table A.3(a-c))
- step 1: determine the proportion of mass that each segment is of the entire
multisegment system
mass of subject = 56.7kg
Lower Extremity Mass
mass of foot = 0.0145(56.7kg) = 0.82215kg =
mass of shank =0.0465(56.7kg) = 2.63655kg =
mass of thigh = 0.100(56.7kg) = 5.67kg =
total mass of lower extremity = 9.1287kg
Segmental Proportions of Lower Extremity Mass
foot proportion = 0.8221/9.1287 = 0.09006
shank proportion = 2.6355/9.1287 = 0.28882
thigh proportion = 5.67/9.1287 = 0.62112
total mass proportion = 9.1287/9.1287 = 1.00
- step 2: multiply each segmental proportion times the x coordinate of the
center of mass of that segment
- step 3: multiply each segmental proportion times the y coordinate of the
center of mass of that segment
- step 4: add each of the x products
- step 5: add each of the y products
- step 6: the sums from steps 4 and 5 are the x and y coordinates of the center
of mass of the multisegment system
Segment
Proportion of
total mass
x value of
center of
mass
x product
y value of
center of
mass
y product
Foot
0.09006
1.342
0.12086
0.069
0.00621
Shank
0.28882
1.256
0.36276
0.331
0.0956
Thigh
0.62112
1.175
0.72981
0.678
0.42112
 = 1.0
 = 1.21343
 = 0.5229
What is inertia?
According to Newton’s first law of
motion, inertia is an object’s
tendency to resist a change in
velocity. The measure of an object’s
inertia is its mass. The more mass
an object has the more inertia it has.
What is moment of inertia?
Moment of Inertia
 F = ma and  = I, where F = force, m = mass, a = acceleration,
 = torque or moment of force causing angular acceleration,
I = moment of inertia = mixi2, and  = angular acceleration
 I of an object depends upon the point about which it rotates
 I is minimum for rotations about an object’s center of mass
I = mixi2 = m1x12 + m2x22 + m3x32 +     + mnxn2
The angular counterpart to mass is
moment of inertia? It is a quantity that
indicates the resistance of an object to a
change in angular motion. The
magnitude of an object’s moment of
inertia is determined by its mass and the
distribution of its mass with respect to its
axis of rotation.
n
I   mi ri  m1 r1  m2 r 2      mn -1 r n -1  mn r n
i 1
2
2
2
2
2
Hypothetical object made up of 5 point masses
vertical
axis
through
center of
mass
y
m1
0.1m
m2
0.1m
m3
0.1m
m4
0.1m
m5
0.1m
x
0.1m
horizontal
axis
through
center of
mass
x
y
m1 = m2 = m3 = m4 = m5 = 0.5 kg
Calculate the moment of inertia about y-y
vertical
axis
through
center of
mass
y
m1
0.1m
m2
0.1m
m3
0.1m
m4
0.1m
m5
0.1m
0.1m
horizontal
axis
through
center of
mass
x
x
y
5
Iy-y   mi ri  m1 r1  m2 r 2  m3 r3  m4 r4  m5 r5
2
2
2
2
2
2
i 1
Iy-y = (0.5kg)(0.1m)2 + (0.5kg)(0.2m)2 + (0.5kg)(0.3m)2 +
(0.5kg)(0.4m)2 + (0.5kg)(0.5m)2 = 0.275kgm2
Calculate the moment of inertia about x-x
vertical
axis
through
center of
mass
y
m1
0.1m
m2
0.1m
m3
0.1m
m4
0.1m
m5
0.1m
0.1m
horizontal
axis
through
center of
mass
x
x
y
5
I
x -x
  mi ri  m1 r1  m2 r 2  m3 r3  m4 r 4  m5 r5
2
2
2
2
2
2
i 1
Ix-x = (0.5kg)(0.1m)2 + (0.5kg)(0.1m)2 + (0.5kg)(0.1m)2 +
(0.5kg)(0.1m)2 + (0.5kg)(0.1m)2 = 0.025kgm2
Calculate the moment of inertia about vertical axis through
center of mass
vertical
axis
through
center of
mass
y
m1
0.1m
m2
0.1m
m3
0.1m
m4
0.1m
m5
0.1m
0.1m
horizontal
axis
through
center of
mass
x
x
5
y
I
cg
  mi ri  m1 r1  m2 r 2  m3 r3  m4 r 4  m5 r5
2
2
2
2
2
2
i 1
Icg = (0.5kg)(0.2m)2 + (0.5kg)(0.1m)2 + (0.5kg)(0.0m)2 +
(0.5kg)(0.1m)2 + (0.5kg)(0.2m)2 = 0.05kgm2
Moment of Inertia of Segments of
the Human Body
• Segments of body made up of different
tissues that are not evenly distributed or of
uniform shape
• Moment of inertia of body segments
determined experimentally
• Moment of inertia of body segments unique
to individual segments and axes of rotation
• Calculation of moment of inertial of a body
segment is based on the segment’s radius of
gyration
Radius of Gyration
- most techniques that provide values for segment moment of
inertia provide information on the radius of gyration
- moment of inertia can be calculated from the radius of gyration
- radius of gyration denotes the segment’s mass distribution
about an axis of rotation and is the distance from the axis of rotation
to a point at which the mass can be assumed to be concentrated
without changing the inertial characteristics of the segment
I0 = m02
where I0 = the moment of inertia about the center of mass,
m = mass of object and 0 = radius of gyration for rotation
about the center of mass
Radius of Gyration
• Denotes the segment’s mass distribution
about an axis of rotation and is the distance
from the axis of rotation to a point at which
the mass can be assumed to be concentrated
without changing inertial characteristics of
the segment
• Moment of inertia (I) = m(ρl)2
where m = mass of segment, l = length of
segment, and ρ = radius of gyration as a
proportion of the segment length
Parallel Axis Theorem
Moment of inertia can be calculated about any parallel
axis, given the:
1. moment of inertia about one axis,
2. mass of the segment, and
3. perpendicular distance between the parallel axes
Parallel Axis Theorem
I = I0 + mx2
where I = moment of inertia about a parallel axis,
I0 = moment of inertia about the center of mass,
m = mass of object (or segment), and
x = distance from the center of mass and the center of rotation
Radius of Gyration/Segment Length in meters*
(about a transverse axis)
Segment
Shank
Center of
Gravity
Proximal
End
Distal End
0.302
0.528
0.643
*(see Table 3.1 in class text)
Moment of inertia
varies on the basis
of axis of rotation:
Proximal end
Center of mass
Distal end
Example: a) A prosthetic shank has a mass of 3kg and a center of mass at
20cm from the knee joint. The radius of gyration is 14.1cm Calculate the
moment of inertia about the knee joint.
42 cm
20 cm
62 cm
Example: a) A prosthetic shank has a mass of 3kg and a center of mass at
20cm from the knee joint. The radius of gyration is 14.1cm Calculate the
moment of inertia about the knee joint.
42 cm
20 cm
62 cm
I0 about the center of mass of the shank = m02 = 3kg(0.141meters)2 = 0.06kg meters2
Using the parallel axis theorem:Ik = I0 + mx2 = 0.06kg meters2 + 3kg (0.2meters)2 = 0.18kg
meters2
b) Calculate the moment of inertia for the prosthetic shank about the hip.
Example: a) A prosthetic shank has a mass of 3kg and a center of mass at
20cm from the knee joint. The radius of gyration is 14.1cm Calculate the
moment of inertia about the knee joint.
42 cm
20 cm
62 cm
I0 about the center of mass of the shank = m02 = 3kg(0.141meters)2 = 0.06kg meters2
Using the parallel axis theorem :Ik = I0 + mx2 = 0.06kg meters2 + 3kg (0.2meters)2 = 0.18kg
meters2
b) Calculate the moment of inertia for the prosthetic leg about the hip.
Using the parallel axis theorem:
Ih = I0 + mx2 = 0.06kg meters2 + 3kg (0.62meters)2 = 1.21kg meters2
Note that Ih  20 I0
Calculate the moment of inertial of shank about its
center of gravity and proximal and distal ends
Given: mass of shank = 3.6kg, length of shank = 0.4 meters
Calculate the moment of inertial of shank about its
center of gravity and proximal and distal ends
Given: mass of shank = 3.6kg, length of shank = 0.4 meters
Icm = Io= m(ρcml)2 = 3.6kg[(0.302)(0.4m)]2 = 0.0525kgm2
Iprox = m(ρproxl)2 = 3.6kg[(0.528)(0.4m)]2 = 0.161kgm2
Idist = m(ρdist)2 = 3.6kg[(0.643)(0.4m)]2 = 0.238kgm2
In which twisting movement is the moment of inertia greater?
Why is a layout flip worth more points than a tuck flip?
Why do we bend our knee during the swing phase of running?
Lower Extremity
Why do we bend our knee during the swing phase of running?
Example: Calculate the moment of inertia of the leg about its center of
mass, its distal end, and its proximal end for an 80kg subject who has a
leg (shank) segment = 0.435meters.
Example: Calculate the moment of inertia of the leg about its center of
mass, its distal end, and its proximal end for an 80kg subject who has a
leg (shank) segment = 0.435meters.
mass of the leg is 0.0465 x 80kg = 3.72kg
I0 = 3.72kg(0.435 meters x 0.302)2 = 0.064kg meters2
Ip = 3.72kg(0.435 meters x 0.528)2 = 0.196kg meters2
Id = 3.72kg(0.435 meters x 0.643)2 = 0.291kg meters2
Example: Calculate the moment of
inertia of the leg about its center of
mass, its distal end, and its proximal
end for an 80kg subject who has a leg
(shank) segment = 0.435meters.
Using the Parallel-Axis Theorem:
Example: Calculate the moment of inertia of the leg about its center of
mass, its distal end, and its proximal end for an 80kg subject who has a
leg (shank) segment = 0.435meters.
Using the Parallel-Axis Theorem:
Ip = I0 + mx2 = 0.064kg meters2 + 3.72kg(0.433 x 0.435meters)2 =
0.196kg meters2
Id = I0 + mx2 = 0.064kg meters2 + 3.72kg(0.567 x 0.435meters)2 =
0.2903kg meters2
Muscle Anthropometry
Physiologic Cross-sectional Area (PCA)
• PCA is an experimental measure of
maximum strength of contraction based on
the cross-sectional area of the muscle
• PCA is based on a measure of the number of
sarcomeres that make up the cross-sectional
area of a muscle.
Logic Behind PCA
• The sarcomere is the basic contractile unit of the
muscle. Because they are in series to each other
in making up the length of the myofibril, the
myofibril is limited in strength to the maximum
force that can be generated by the weakest
sarcomere in the series.
• Myofibrils are parallel to each other in a muscle
cell. Therefore, the maximum force of contraction
of the muscle cell is directly related to the sum of
the force of maximum contraction of the each
myofibril (Remember that the maximum force of
contraction of the myofibril is the maximum force
of the weakest sarcomere in its length.).
Logic Behind PCA
• Muscle cells are in parallel with each other.
The maximum force of contraction of a
muscle is directly related to the sum of the
maximum forces of contraction of all the
individual muscle cells that compose the
muscle.
• Therefore, the PCA is directly related to the
number of sarcomeres in this cross-section
and directly related to the maximum force of
contraction of the muscle.
PCA = m/d(l)
where m = mass of the muscle fibers in grams,
d = density of the muscle in grams/cm3 (note that muscle mass 
1.056g/cm3), and
l = length of the muscle fibers in cm
Based on the equation for PCA, in a fusiform muscle, the PCA is
independent of length. Note that changes in length of a muscle are
accompanied by the same proportional changes in muscle mass.
Therefore, PCA is not changed.
l/2
m/2
m
l
PCA = (m/2)/d(l/2)
PCA = m/d(l)
PCA = m/d(l)
Based on the architecture of muscle, why is there no change in PCA?
Example of PCA of fusiform and pennate muscle fiber arrangement:
Fusiform
Pennate
Shading represents PCA
Example of PCA of fusiform and pennate muscle fiber arrangement:
Fusiform
Pennate
Note the difference in the number of muscle fibers that occupy the same volume and the
difference in the cross-sectional area required to cut across all the fibers in the muscle.
In other words the PCA of the pennate is greater than the PCA of the fusiform of equal
volume. Therefore, muscles with pennate fiber arrangements have an advantage for
force of contraction. However, muscles with fusiform fiber arrangement have an
advantage for length of contraction.
Pennation
- angle of pull of the muscle fibers in relationship to the long axis
of the muscle
- as the pennation angle increases, the PCA increases
- as the pennation angle increases, the force of contraction of a single muscle
cell, in the direction of the long axis of the muscle, decreases
- the combination of pennation angle and PCA influence the
maximum force of contraction of a muscle
Fy
Fm
Fy
θ
Fx
Fy = Fm = force of contraction of
each muscle cell in the direction
of the long axis of the
Fy = Fm cos = force of
contract of each muscle cell in
the direction of the long axis
of the muscle
Based in PCA, what would you suggest to be the strongest
flexors and extensors?
PCA
Force per unit cross-sectional area:
- reports range from 20 –100N/cm2
- higher values reported for pennate muscles
Example (not in book): Ignoring any influence pennation
angle might have on force of contraction, calculate the
collective peak force of contraction of the quadriceps (see
Table 3.3 for PCA). Assume that these muscles exert a force
per unit cross- section = 75N/cm2. Calculate the answer in N
and in pounds. Note that 1N = 0.2247lbs.
Example (not in book): Ignoring any influence pennation
angle might have on force of contraction, calculate the
collective peak force of contraction of the quadriceps (see
Table 3.3 for PCA). Assume that these muscles exert a force
per unit cross- section = 75N/cm2. Calculate the answer in N
and in pounds. Note that 1N = 0.2247lbs.
Collective PCA of quadriceps muscle = 12.5cm2 + 30cm2
+26cm2 +25cm2 = 93.5cm2
93.5cm2 x 75N/cm2 = 7012.5N
7012.5N x 0.2247lbs/N = 1575.71 lbs **impressive!!!
Mechanical Advantage of Muscle
-mechanical advantage of a muscle at a joint is determined
by the moment arm length (length of line from the joint
center intersecting the line of long axis of the muscle at a
perpendicular
-mechanical advantage (and moment arm) change with
changes in the angle of the joint
- force of contraction of the muscle also changes with
changes in joint angle (changes in muscle length)
- maximum torque or turning force (moment) at a joint is
therefore the dynamic product of two variables – length
of the muscle (as determined by joint angle) and moment
arm (also determined by joint angle)
Multijoint Muscles
-individual muscles that cross two or more joints
-muscle fiber arrangements of many multijoint muscles may
be insufficient to engage in maximum shortening
(simultaneously affecting multiple joints)
- fusiform muscles have an advantage in length of contraction
in multijoint muscles (e.g., sartorius muscle – longest
muscle in body and two joint muscle), but disadvantage in
PCA (and maximum force of contraction)