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Biomechanical Models

(Anthropometry in Biomechanics)

http://www.sfu.ca/~leyland/Kin201/Anthropometry.pdf

Link Segment Models

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

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)

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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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?)

Body Segment Volume, Density & Mass

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

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 20 N

0.5 m 1.5 m

Centre of Mass (C of g) Location

Forearm

Distance from elbow to forearm centre of mass is approximately 39-43% of the segment length.

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Segmental Centre of Mass Ø  Data for the segments can be obtained

from: Ø measurements based on cadaver

studies Ø mathematical modelling Ø radiosotope scanning

Segment Inertial Properties 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

Ø  Population data has been used to define, reach, space needs and other dimensions for the target use group.

Ø  Many data sets exist.

Which Population?

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.

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.

Force

Ø  is a push, pull, rub (friction), or blow (impact)

Ø  causes or tends to

cause motion or change in shape of a body

Usually drawn as an arrow indicating direction and magnitude.

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 pre-requisite knowledge.

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

Vector Components Ø  a = original vector

acomponentxa

componenty

!=

!=

"

"

cos

sin

!x-component

y-co

mpo

nent

a

Cartesian Coordinate System

Y

X

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Quadrants

Quadrant I (+,+) Quadrant II (-,+)

Quadrant III (-,-) Quadrant IV (+,-)

+y

-y

+x -x

Directional Conventions

+ve

+ve

+ve

x-axis

y-axis

Reference Systems Two-Dimensional

Y

X

(Xheel, Yheel) (XT,YT)

(XK,YK)

(XH,YH)

(XA,YA)

Fixed (Absolute) Frame

Frames of Reference

A stationary frame of reference is used when we want to describe the movement relative to the environment or space.

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Moving (relative) Frame of Reference Y

X

Y

Frames of Reference A moving frame of reference is usually used when we want to look at the movement of a body segment.

For example we may be interested in the amount of hip, knee and ankle flexion/extension in cycling.

Three-Dimensional Movement?

2D sagittal plane movement

3D analysis required

Three-Dimensional Reference System

X, Y, Z axes.

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.

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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

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).

Components of Torque

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: Ø An effort force Ø a resistive force

and, Ø an axis of rotation

Effort (force)

Resistive

Axis

Mechanical Advantage

Mechanical Advantage (MA)

MA = effort force arm resistive force arm

Classes of Levers 1st Class

2nd Class (MA>1) 3rd Class (MA<1)

MA varies

Favours the effort force (i.e. a smaller effort force can balance a larger resistive force)

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]

b) What are the advantages

and disadvantages to this type of lever? [2]

F

R

(i)

200 N

(ii)

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Static Equilibrium

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.

Vector Products

Ø  Unfortunately calculating moments is the product of two vectors and therefore not simple multiplication.

Ø  It requires a Cross Product 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)

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)

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

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).

30o

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.

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Direction Conventions

If I had asked for the net shoulder muscle moment the answer would have been: + 9.51 Nm

Clockwise is negative Anti-clockwise positive

Course Recommended Text

Ø  Pages 430 – 442 Ø  I haves discussed levers but you should

be familiar with these terms.

Static Equilibrium

Fm

Ff Fwt

ΣFx = 0 ΣFy = 0 ΣM = 0

FR

Muscle Torque 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.

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Static Equilibrium Calculate the Muscle Force (Fm)?

Fm

-15N

ΣFx = 0 ΣFy = 0 ΣM = 0

Moment arms = 0.03m, 0.15m and 0.3 m respectively

-50N

FR

Fm

Ff Fwt

ΣFx = 0 ΣFy = 0 ΣM = 0

Static Equilibrium Remember you have 2 unknown forces so you

cannot directly solve for Fm using ΣFy = 0.

FR

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)

-15N

ΣFx = 0 ΣFy = 0 ΣM = 0

Moment arms = 0.03m, 0.15m and 0.3 m respectively

-50N

τm

We can easily calculate τm

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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

-15N

Moment arms = 0.03m, 0.15m and 0.3 m respectively -50N

! = F ! d"

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?)

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Muscle-Joint Complexity

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.

Brachialis

Biceps

Brachioradialis

50 100 150

100 75 50 25 0

Elbow Angle (degrees)

Moment Arm (mm)

Even at the elbow the system is mechanically

indeterminate

L5

L4

6-7 cm

S1

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Stability

Pyramid Quite Stable

Inverted Pyramid Very Unstable

Stability

Pendulum Highly Stable

Ball Neutral Stability

Inverted Pendulum

Abdominals

Quadriceps

Erector Spinae

Gluteus Maximus

Gastrocnemius and Soleus

Anti-Gravity Musculature

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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?

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Buckling

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.

Sample Midterm Question 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]

θ  =  80o

FR FL θ  =  80o

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]