Advanced Biomechanics of Physical Activity (KIN 831)

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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.
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What is

• Anthropology?
• Anthropometry?

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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

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What functions does anthropometry serve?
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Uses of Anthropometry

• Studies of physical measurements of the human
  body
• Determining differences in individuals and groups
• Age
• Sex
• Race
• Body type (somatotype)

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Emphases of Anthropometric Studies

• Evolution
• Historical
• Performance parameters
• Man-machine interfaces

Past studies
Associated with recent technological developments
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What are some examples of anthropometric
measurements?
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Types of Anthropometric Measurements

• Static measurements
• Length
• Area
• Volume
• Breadths
• Circumference
• Skin fold thickness

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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)

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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?

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Body Segment Lengths

• Vary with body build, sex, racial origin
• Dempsters 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)

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Body Segment Lengths

• Vary with body build, sex, racial origin
• Dempsters data (1955, 1959) segment lengths
  and joint center locations relative to anatomical
  landmarks
• Drillis and Contini (1966) segment lengths/body
  height (see figure)

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What is

• Mass?
• Inertia?
• Density?

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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

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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 lt 1.0

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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.

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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

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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

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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

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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

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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
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Why the same definition ?
Head And Trunk
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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.
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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
86.4
72.1
X
0.433(?X)
0.567(?X)
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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
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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
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  sample table
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.
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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))
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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
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What is inertia?
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According to Newtons first law of motion,
inertia is an objects tendency to resist a
change in velocity. The measure of an objects
inertia is its mass. The more mass an object has
the more inertia it has.
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What is moment of inertia?
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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 objects center of
mass   I ?mixi2 ?m1x12 ?m2x22 ?m3x32 ?
? ? ? ?mnxn2
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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 objects moment of
inertia is determined by its mass and the
distribution of its mass with respect to its axis
of rotation.
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Hypothetical object made up of 5 point masses
m1 m2 m3 m4 m5 0.5 kg
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Calculate the moment of inertia about y-y
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
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Calculate the moment of inertia about x-x
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
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Calculate the moment of inertia about vertical
axis through center of mass
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
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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 segments radius of
  gyration

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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 segments 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
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Radius of Gyration

• Denotes the segments 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(?l)2
• where m mass of segment, l length of
  segment, and ? radius of gyration as a
  proportion of the segment length

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Parallel Axis Theorem

• Moment of inertia can be calculated about any
  parallel axis, given the
• moment of inertia about one axis,
• mass of the segment, and
• perpendicular distance between the parallel axes

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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
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Radius of Gyration/Segment Length in
meters (about a transverse axis)
Segment Center of Gravity Proximal End Distal End
Shank 0.302 0.528 0.643
(see Table 3.1 in class text)
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Moment of inertia varies on the basis of axis of
rotation Proximal end Center of mass Distal end
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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
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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 theoremIk 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. 
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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
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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
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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(?cml)2
3.6kg(0.302)(0.4m)2 0.0525kgm2 Iprox
m(?proxl)2 3.6kg(0.528)(0.4m)2
0.161kgm2 Idist m(?dist)2 3.6kg(0.643)(0.4m)
2 0.238kgm2
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In which twisting movement is the moment of
inertia greater?
Why is a layout flip worth more points than a
tuck flip?
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Why do we bend our knee during the swing phase of
running?
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Lower Extremity
Why do we bend our knee during the swing phase of
running?
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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.
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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
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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 
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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
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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.

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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.).

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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.

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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  
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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.
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?
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Example of PCA of fusiform and pennate muscle
fiber arrangement
Fusiform
Pennate
Shading represents PCA
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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.
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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
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
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Based in PCA, what would you suggest to be the
strongest flexors and extensors?
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PCA Force per unit cross-sectional area -
reports range from 20 100N/cm2 - higher
values reported for pennate muscles
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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.  
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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!!!
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• 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)

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• 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)