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Link Segment Models
Biomechanical Models
(Anthropometry in Biomechanics)
http://www.sfu.ca/~leyland/Kin201/Anthropometry.pdf
Link Segment Models
Ø  To develop a biomechanical link segment model,
we need to know:
Ø  joint centres of rotation locations (correct
definition of segment)
Ø  segment lengths
Ø  segment masses
Ø  the centre of gravity location and possibly
the segment moment of inertia
Ø  In obtaining this data it is crucial that we know
which bony landmarks to use and how to identify
them (Kin 303)
Ø  The body is modeled
as a series of rigid
segments linked by
hinge joints (in two
dimensions)
Ø  These models can vary
greatly in complexity
Anthropometry
Ø  Anthropometry is the science that deals
with the measure of size (length, breadth,
volume, etc.), mass, shape and inertial
properties of the human body.
Ø  The results of anthropometric
measurements are statistical data
describing human populations in terms of
the above variables.
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Body Segment
Volume, Density & Mass
Why Model?
Ø  Biomechanical models can be used to predict
human muscle force, joint bone-on-bone
force, reach and space requirements.
Ø  Without anthropometry, these models cannot
be developed.
Ø  Focus on a basic understanding of what is
being measured, (lengths, etc. are
straightforward, centre of gravity and moment
of inertia?)
Segment Centre of Mass
The sum of the moments of force (torques)
on all sides of the segments centre of gravity
point must equal zero (note that the mass is
equally distributed but the sum of the
masses is not necessarily equal on both
sides.
60 N
0.5 m
20 N
Ø  Volumes can be determined from living
subjects (e.g. fluid displacement).
Ø  Density has been determined only directly
from cadaver studies or extrapolated from
indirect measurement techniques.
Ø  Density = mass/volume
Centre of Mass (C of g) Location
Forearm
Distance from elbow to forearm centre
of mass is approximately 39-43% of the
segment length.
1.5 m
2
Segmental Centre of Mass
Segment Inertial Properties
Ø measurements based on cadaver
studies
Ø mathematical modelling
Ø radiosotope scanning
Ø  Mass, length and centre of mass location
is sufficient for static analysis of the forces
and moments at each joint for any given
body posture (Kin 201).
Ø  Because the segments rotate during
dynamic activities, the moment of inertia
(I) of the body segments must be known
for dynamic modelling.
Ø  We’ll discuss I later.
Anthropometric Data
Which Population?
Ø  Data for the segments can be obtained
from:
Ø  Population data has
been used to define,
reach, space needs
and other dimensions
for the target use
group.
Ø  Many data sets exist.
Ø  According to Webb Associates, in 1978 the
average stature of female Swedish
civilians was 164.7 cm, and was 153.2 cm
for female Japanese civilians.
Ø  Try to find data relevant to the user group
you are working with.
Ø  When modelling subject specific data is
often required.
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Kin201 Anthropometry
Ø  Obviously segment lengths are related to
overall height (stature) and the mass of
segments is related to the total mass of the
subject.
Ø  However, using a percentage of stature and
total mass is still an estimate.
Ø  We will simplify our anthropometry and just use
data for the 50% male.
Ø  I adapted data from numerous researchers to
develop this set.
Force
Ø  is a push, pull, rub
(friction), or blow Usually drawn as an
arrow indicating direction
(impact)
and magnitude.
Ø  causes or tends to
cause motion or
change in shape
of a body
Anthropometric Data
Human Factors Design Handbook
Woodson, Hillman & Hillman
McGraw Hill, N.Y. 1992
Library 5th floor: call # TA 166 W567
Bodyspace: Anthropometry,
Ergonomics and Design.
Pheasant, S.
Taylor & Francis, London, 1986.
Properties of Forces
Ø  Magnitude
Ø  Direction
Ø  Point of application
Ø  Line of application
Ø  Angle of application
θ
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Vectors and Scalars
Ø  Scalars can be described by magnitude
Ø  e.g., mass, distance, speed, volume
Ø  Scalars such as mass are easy to add
(assuming they have the same units).
Ø  Vectors have both magnitude and direction
Ø  e.g., velocity, force, acceleration
Ø  Vectors are represented by arrows
Ø  You should understand vector addition,
subtraction, and multiplication via your prerequisite knowledge.
Vector Components
Dimensions
(fundamental units in brackets)
Ø  Mass (M)
Ø  Length (L)
Ø  Time (T)
Ø  Other fundamental units like temperature
is not a focus of Kin 201
Cartesian Coordinate System
Y
Ø  a = original vector
y ! component
a
x ! component
cos " =
a
a
!
x-component
y-component
sin " =
X
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Quadrants
Directional Conventions
+y
+ve
+ve
Quadrant II (-,+)
Quadrant I (+,+)
y-axis
+x
-x
Quadrant III (-,-)
x-axis
Quadrant IV (+,-)
+ve
-y
Reference Systems
Frames of Reference
Two-Dimensional
Y
Fixed
(XH,YH)
A stationary frame of
reference is used
when we want to
describe the
movement relative to
the environment or
space.
(XK,YK)
(Absolute)
Frame
(XA,YA)
(XT,YT)
(Xheel, Yheel)
X
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Moving (relative) Frame of Reference
Frames of Reference
A moving frame of reference is
usually used when we want to
look at the movement of a
body segment.
Y
Y
X
Three-Dimensional Movement?
For example we may be
interested in the amount of hip,
knee and ankle flexion/
extension in cycling.
Three-Dimensional
Reference System
X, Y, Z axes.
2D sagittal plane
movement
Pages 6 and 7 of Text
X-Axis - Horizontal (sagittal plane)
Y-Axis - Vertical
Z-Axis - Medial/Lateral
I should point out that Z is often used for
the vertical axis and X &Y the horizontal.
3D analysis required
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Cardinal Plane (Principal Axes)
Reference plane of motion where the plane passes through
center of mass when an individual stands in the anatomic
position
Vertical Translation of Barbell
Ø  The text uses a simple
example of pushing a
pencil.
Ø  In the Hang-Squat-Clean
opposite the athlete
ideally wants to move
the bar straight up, via a
vertical pulling force
through the centre of
gravity of the barbell.
Internal Forces (Torques)
Ø  However in this activity
the athletes muscles do
not pull at the centre of
gravity of his limbs.
Ø  So each of the muscle
forces create torques
and the linear upward
movement of the
barbell is achieved by a
complex series of joint
of rotations.
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Force Magnitude & Change in Motion
Components of Torque
Ø  This section of the lecture is really just
emphasizing that F = ma and that the
acceleration of an object will occur in the
direction of the net force on the body.
Ø  I will discuss acceleration, velocity and
displacement in more detail in the another
lecture topic (data acquisition).
Ø  axis of rotation
(fulcrum)
Ø  force (not directed
through axis of
rotation)
Ø  force (moment) arm
force arm
" = F # d$
Torque is a Vector
Ø  Torque has both magnitude (force x
force arm) and direction.
Ø  A counter clockwise torque is positive
and a clockwise torque is negative.
Ø  Make sure that you know which
direction the torque is being applied
!
Muscles Create Torques
Although human motion is
general (translation and
rotation), it is generated by a
series of torques and hence
rotations.
The line of action of muscle
forces do not pass through the
joint axis of rotation
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Force Arm is Perpendicular to
Force Line of Action
Ø  Make sure you use the force arm not the distance
from point of force application to axis of rotation.
Ø  In both these diagram the force is the same
magnitude applied at the same point BUT the
torques are very different!
F
F
Rotation & Leverage
Ø All lever systems
have:
Resistive
Mechanical Advantage
MA
= effort force arm
resistive force arm
Effort (force)
Ø An effort force
Ø a resistive force
and,
Ø an axis of rotation
≠
Mechanical Advantage (MA)
Axis
Classes of Levers
1st Class
MA varies
2nd Class
(MA>1)
Favours the effort force (i.e. a
smaller effort force can balance
a larger resistive force)
3rd Class
(MA<1)
Favours range and speed of
movement.
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Third Class Levers
Ø  The majority of
musculoskeletal
systems are in
third class levers.
Ø  These favour
speed and range
of movement.
Wheel & Axle Arrangements
MA = radius of axle
radius of wheel
only if the effort force is applied to the axle. If
the effort force is on the wheel (resistive on the
axis) reverse this fraction.
Ø  Nearly all musculoskeletal systems in the
human body are third class levers.
Ø  Systems like rotator cuff muscles and other
muscles responsible for longitudinal rotation of
long bones can have MA’s <1. However,
these MA are often quite close to 1.
Sample Midterm Question
a) What class of lever is
shown opposite? [2]
F
b) What are the advantages
and disadvantages to this
type of lever? [2]
R
(i)
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Static Equilibrium
Ø  Unfortunately calculating moments is the
product of two vectors and therefore not
simple multiplication.
Ø  It requires a Cross Product
Ø  This is jumping ahead to
sections of chapter 11
Ø  However, I want to get
started on some torque
calculations as these
appear to give students
the most problems.
Ø  http://en.wikipedia.org/wiki/Cross_product
Ø  FORTUNATELY J as we will be dealing in
two dimensions (and you have a pen in your
hand!) you can determine the direction of the
vector in a practical way.
It is this Simple!
Hold pen at axis
(rotations point)
Vector Products
Push or pull in
direction of
external force
Observe direction of rotation – in our two-dimensional
models it can only be clockwise or anti-clockwise
Static Equilibrium
Ø  I wanted to get started
on some torque
calculations now as
these appear to give
students the most
problems.
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Sum of Moments
Ø  If there is no
acceleration (linear or
angular) then all of the
forces must equal zero
(ΣF = ma = 0)
Ø  Similarly, if there is no
angular acceleration the
sum of the moments
(torques) must also
equal zero (Στ = Iα = 0)
Questions
Ø  Calculate the mass of the total arm (hand, lower
and upper arm).
Ø  Calculate the horizontal distance from the shoulder
to the centre of gravities of the three segments.
Ø  Calculate the moment at the
shoulder in this position ( external
torque due to the arms mass
and location).
Σ = “sum of”
Ø  So if the girl holds her
arm out stationary
(static position) then
all of the forces, and
moments of force
(torques), acting on
the arm must equal
zero.
ΣFx = 0
ΣFy = 0
ΣM = 0
Remember these points J
Ø The only external force acting on this
system will be gravity (g = -9.81 ms-2).
The question would specifically state if any
other forces are to be considered.
Ø The moment of force (torque) is the force
multiplied by the perpendicular distance to
the axis of rotation.
30o
13
Direction Conventions
If I had asked for the
net shoulder muscle
moment the answer
would have been:
Course Recommended Text
Ø  Pages 430 – 442
Ø  I haves discussed levers but you should
be familiar with these terms.
+ 9.51 Nm
Clockwise is negative
Anti-clockwise positive
Static Equilibrium
Ø  If the mass of the forearm/hand segment
and the location of its CG are known, plus
similar information for any external forces
(loads) we can calculate muscle torque.
Ø  What other information do we need if we
want to calculate muscle force?
Ø  Line of action of forces in relation to segment
(insertion point and angle of pull). Basically
we need the moment arm of the muscle.
ΣFx = 0
ΣFy = 0
ΣM = 0
Fm
FR
Ff
Muscle Torque
Fwt
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Static Equilibrium
Static Equilibrium
Calculate the Muscle Force (Fm)?
Fm
Remember you have 2 unknown forces so you
cannot directly solve for Fm using ΣFy = 0.
Fm
ΣFx = 0
ΣFy = 0
ΣM = 0
ΣFx = 0
ΣFy = 0
ΣM = 0
FR
FR
-15N
-50N
Ff
Fwt
Moment arms = 0.03m, 0.15m and 0.3 m respectively
Free Body Diagrams
Ø  The point made on the last slide shows
how the ability to identify the system
and draw a free body diagram is an
ESSENTIAL ability.
Ø  We must use sum of moments about
the elbow axis. This way we eliminate
FR from the equation as FR passes
through the elbow joint and hence,
does not create a moment of force.
Static Equilibrium
(Kinetic Diagram)
We can easily calculate
τm
ΣFx = 0
ΣFy = 0
ΣM = 0
τm
-15N
-50N
Moment arms = 0.03m, 0.15m and 0.3 m respectively
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Muscle Insertion
Ø  The NET muscle torque calculated previous
is quite accurate (as long as your
anthropometric data is accurate).
Ø  If there is minimal co-contraction the this
would also be an accurate value for the
forearm flexor torque.
Ø  However to calculate muscle force you would
need to know the moment arm for the
muscle.
Ø  If we assume we have one forearm flexor
and we know the moment arm then we can
calculate muscle force.
Static Equilibrium
This was the result of dividing the moment
calculated about the elbow by the moment arm.
575N
! = F ! d"
-15N
-50N
Moment arms = 0.03m, 0.15m and 0.3 m respectively
Simplification #1
Rigid Segments
Ø  Note that we are assuming that the
segments are rigid structures in these
problems.
Ø  From the skeletal lecture you will know
that this is not exact.
Ø  It is a good approximation however.
Simplification #2
Single Equivalent Muscles
Ø  We have assumed there was one muscle
producing the moment (or stabilizing the joint).
Ø  However, this is obviously not accurate.
Ø  For shoulder flexion for example we have two
prime movers and two assistors.
Ø  We often lump such muscle groups together
and term them a “single equivalent muscle”.
Ø  This may appear to be a gross simplification
but you will see we do not make this
assumption at many joints (spine, knee?)
16
100
Muscle-Joint Complexity
Even at the elbow the
system is mechanically
indeterminate
75
Moment
Arm (mm)
Brachialis
50
Biceps
Brachioradialis
Ø  Another important factor is whether the muscles
crossing the joint have a common tendon, or similar
insertion points and similar lines of action.
Ø  Clearly many muscle joint systems are too complex
to model in Kin 201.
25
0
50
100
150
Elbow Angle (degrees)
L4
L5
S1
6-7 cm
17
Stability
Pyramid
Quite Stable
Inverted Pyramid
Very Unstable
Stability
Pendulum
Highly Stable
Ball
Neutral Stability
Anti-Gravity
Musculature
Inverted Pendulum
Erector
Spinae
Abdominals
Gluteus
Maximus
Quadriceps
Gastrocnemius
and Soleus
18
Increased Stability
Ø  Increased base of
support
Ø  Lower centre of gravity
Ø  Higher mass
Ø  Position of centre of
gravity with respect to
base of support (e.g.
leaning into a tackle)
Stability
Ø  Which shape is more stable?
To be confident in how to
answer that question you
would need to ask –
stable in which direction?
Ergonomic Chairs
Ø  The standard is for five legs to increase
base of support.
Chairs?
Ø  Is this a good
idea?
19
Sample Midterm Question
Buckling
Ø  The diagram on the next slide shows our 50th
percentile male in the cross position on the
rings (he is stationary). Assume the long axes of
the arms are horizontal and the force distribution
is the same in both hands (i.e. symmetrical).
The external forces at the hands (FR & FL) act at
the hands centre of mass.
a)  Draw a free body diagram of one total arm
system (upper arm, forearm and hand). Try to
proportion the force vectors somewhat
realistically. Name the forces you put in your
diagram. [4]
Ø  If the work done on
the spine (energy
applied) is greater
than the work the
muscles can do to
stiffen the spine,
then the spine will
buckle.
θ = 80o
FR
θ = 80o
FL
Question continued
b)  What is the value of FR in the diagram
(remember force is a vector)? [4]
c)  What is the muscle moment across his
right shoulder joint? [8]
d)  Name two muscles that would be large
contributors to this muscle moment. [2]
e)  Briefly explain why you cannot accurately
calculate the amount of force these
muscles exert. [2]
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