Download unit 1 - biomechanical analysis

UNIT 1 - BIOMECHANICAL ANALYSIS
Dr M J Dolan
Department of Orthopaedic & Trauma Surgery
University of Dundee
Dr T Drew
Department of Orthopaedic & Trauma Surgery
University of Dundee
SECOND EDITION
Edited by
Dr T Drew
Department of Orthopaedic & Trauma Surgery
University of Dundee
Illustrations by
Mr I Christie
Published by
Distance Learning Section
Department of Orthopaedic & Trauma Surgery
University of Dundee
Second Edition published 2005: ISBN 1-903562-46-5
ISBN 978-1-903562-46-8
First Edition published 1994: ISBN 1-899476-49-0
Copyright © 2005 University of Dundee. All rights reserved. No part of this publication may be
reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic,
mechanical, photocopying, and recording or otherwise, without prior permission from the publisher.
The University of Dundee is a Scottish Registered Charity, No SC01509.
UNIT 1 - BIOMECHANICAL ANALYSIS
CONTENTS
1. LINK SEGMENT MODELS
1.1 Inverse Dynamics
1.2 Forces Acting on Body Segments
1.3 Static, Dynamic or Quasi-static?
2. ANTHROPOMETRY
2.1 Segment Length
2.2 Segment Mass and Centre of Mass
2.3 Moment of Inertia and Radius of Gyration
2.4 Muscle Anthropometry
3. FORCES IN JOINTS
3.1 Inter-segment Force
3.2 Joint Force
3.3 Bone-on-bone Force
3.4 Terminology and Definitions
4. MOMENTS ABOUT JOINTS
5. MUSCLE MECHANICS
5.1 Force-length
5.2 Force-velocity
5.3 Physiological Cross-sectional Area and Muscle Architecture
5.4 Electromyography
5.5 The Distribution Problem
SUMMARY
SAQ ANSWERS
END OF UNIT EXERCISE
UNIT 1 - BIOMECHANICAL ANALYSIS
OBJECTIVES
On completing your study of this unit you should be able to:
1. Explain what a link segment model is.
2. List the main assumptions usually used in link segment models.
3. Describe the process of inverse dynamics.
4. List the three classifications of forces that act on body segments.
5. Discuss whether or not a particular problem should be described as static, dynamic
or quasi-static.
6. Define the term anthropometry.
7. Use standardised anthropometric data sets to find segment length, mass, centre of
mass and radius of gyration.
8. Describe the advantages and disadvantages of standardised anthropometric data
sets.
9. List the three parameters which must be considered when the action of muscles are
included on the biomechanical analysis.
10. Define and calculate the inter-segment force.
11. Define and calculate the joint force.
12. Define and calculate the bone-on-bone force.
13. Explain the relationships between inter-segment force, the joint force, and the boneon-bone force.
14. Define the terms net internal joint moment and net external joint moment.
15. Calculate joint moments for simple problems.
16. Describe and explain the relationship between muscle force and length.
17. Describe the relationship between muscle force and velocity.
18. Define the terms pennation angle and physiological cross-sectional area.
19. Calculate the physiological cross-sectional area for muscles.
20. Describe the usefulness of electromyography in biomechanical analysis.
21. Describe the distribution problem and the two techniques used to overcome it.
UNIT 1 - BIOMECHANICAL ANALYSIS
In this first unit of the Skeletal Mechanics module you will be looking at some of the
techniques used for biomechanical analysis of the skeletal system. In particular, you will
be concentrating on how biomechanics is used to estimate the forces and moments
acting at joints. To do this a biomechanical model of the human body must be adopted.
The model should be based on an understanding of the body’s form (i.e.
musculoskeletal anatomy and anthropometrics) and function (i.e. the action of muscles
and ligaments). Since all the variables involved can not be measured then the model
must ultimately rely on a number of assumptions. These assumptions must be clearly
laid out and soundly based.
1. LINK SEGMENT MODELS
The human body is made up of a number of easily identified parts that move relative to
one another. For example, the head can move relative to the thorax. In biomechanics the
different parts are termed body segments or simply segments. The exact definition of
segments will vary according to the problem being studied. For example, if you were
interested in studying walking then you might construct a model similar to the one
shown in Figure 1. In such a model you would probably regard the whole arm as one
body segment, however. However, if you were interested in studying the movement of
the fingers during typing then modelling the whole arm as one segment would be of
little use. Instead you would need to consider each phalanx bone as a separate segment.
BODY SEGMENTS
head
trunk
arm
pelvis
thigh
shank
foot
FIGURE 1. A LINK SEGMENT MODEL CONSISTING OF ELEVEN SEGMENTS: HEAD, TRUNK. PELVIS, AND RIGHT AND LEFT
ARMS, THIGHS, SHANKS AND FEET.
Each body segment has its own mechanical attributes, that is each has its own mass,
length etc. These mechanical attributes are often related to the corresponding whole
body attribute. For example, the length of the thigh is related to height. The mechanical
attributes of the body are explored in detail in the following section on anthropometry.
The body segments which are of interest are usually grouped together in to what is
termed a link segment model. This is a model of the whole body or simply the part of
interest, in which individual segments are linked together. Usually the links correspond
to anatomical joints (Figure 2).
Unit 1 - Biomechanical Analysis
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LINK SEGMENT MODEL
thigh
leg
foot
FIGURE 2. LINK SEGMENT MODEL OF THE LOWER LIMB CONSISTING OF THREE SEGMENTS (THIGH, LEG, FOOT) AND
TWO JOINTS (KNEE AND ANKLE).
All biomechanical models must be based on a number of assumptions. Two of the most
generally applicable assumptions adopted are that:
1. the body segments are rigid.
2. the joints are frictionless.
It follows from the first assumption (the body segments are rigid) that during
movements:
1.
the position of the centre of mass remains fixed relative to the segment.
2.
the moment of inertia of each segment remains constant.
3.
the length of each segment remains constant.
Additional assumptions may also be needed in order to solve a particularly complex
problem or when equipment is not available to make all the required measurements. For
example, a movement may be assumed to take place entirely in one plane. Walking is
often treated this way.
Note that the assumptions given above are not always adopted. In certain cases, an
assumption may influence the result to such a large extent that it cannot be justified.
SAQ 1
(a) What are link segment models made up of?
(b) What are the two main assumptions that are usually adopted for a
link segment model?
1.1 Inverse Dynamics
When all the forces acting on a body segment, or any other object, are known then it is
possible to predict the resulting motion using kinetics. The technique employed to
achieve this is called direct dynamics (Figure 3A). For example, if a force is applied to
a book lying on top of a table it will accelerate and gain velocity. Provided the
magnitude of the force applied, the coefficient of friction between the book and the
table, and the mass of the book are known, then the motion of the book can be predicted
using direct dynamics.
Unit 1 - Biomechanical Analysis
2
In biomechanics it very rare for all the forces acting on a body segment to be known.
Instead it is more usual for the motion resulting from these forces to be known. Using
this knowledge of the motion, the acceleration of a body segment can be calculated.
Using the acceleration and the mass of the segment the magnitude of the resultant force
can be calculated. This process is called inverse dynamics (Figure 3B). This is the
technique most often employed in the biomechanical analysis of the musculoskeletal
system.
INVERSE DYNAMICS
(A) direct dynamics
FORCES
RESULTANT FORCE
ACCELERATION
MOTION
(B) inverse dynamics
MOTION
ACCELERATION
RESULTANT FORCE
FIGURE 3. (A) DIRECT DYNAMICS (B) INVERSE DYNAMICS.
SAQ 2 - What is measured and what is calculated in inverse
dynamics?
1.2 Forces Acting on Body Segments
The forces acting on a body segment in a link segment model can be classified as either:
¾ external forces,
¾ internal forces,
¾ inertial forces.
External forces are those forces applied directly to a body segment and those due
to gravity.
EXTERNAL FORCES
External forces result from the interaction with the environment. They take two forms
those resulting from forces applied directly to a body segment, and those due to gravity.
For example, during standing there are two external forces applied directly to the body,
one acting on each foot, and one due to gravity (Figure 4A). Likewise, pushing against
an object would produce an additional external force acting on the hand (Figure 4B).
Most external forces applied directly to the body can be measured using force
transducers (forces transducers are devices that can measure forces). Those due to
gravity can be calculated. Indeed the external forces must be measured in order to
calculate the forces in joints. The gravitational force is equal to the segment’s weight
which is equal to the product of the mass of the segment and the acceleration due to
gravity.
In a link segment model external forces also act on each body segment. These arise due
to the interaction of a body segment with objects external to the link segment model, its
adjacent body segments and the body segment’s weight. For example, Figure 5 shows
the external forces acting on the three body segments that form BiomechAnnie’s upper
limb.
Unit 1 - Biomechanical Analysis
3
external
force
external force
external force
FIGURE 4(A). EXTERNAL FORCES ACTING DURING STANDING.
TANDING.
external force
external force
external force
FIGURE 4(B). EXTERNAL FORCES ACTING DURING PUSHING.
FIGURE 5. EXTERNAL FORCES ON THE BODY SEGMENTS OF BIOMECHANNIE'S UPPER LIMB.
Internal forces are those forces that are internal to the body.
Unit 1 - Biomechanical Analysis
INTERNAL FORCES
4
These forces include those caused by the contraction of muscles and those carried by
ligaments. The forces arising due to the friction in joints are also examples of internal
forces. However, these forces are usually so small that they are negligible and joints are
assumed to be frictionless in most link segment models.
Inertial forces are the forces acting due to the inertia of a body segment as it
accelerates.
INERTIAL FORCES
The inertial force is equal to the product of the mass of a segment and its acceleration.
In many situations, the acceleration of a body segment is so small that the inertial force
is negligible.
SAQ 3 - Name the three types of forces that act on a body segment.
1.3 Static, Dynamic or Quasi-static?
All objects are either in static equilibrium or dynamic equilibrium.
An object that is in static equilibrium has no resultant force or moment acting on it.
The object is therefore either at rest or moving with a constant linear and/or angular
velocity (Figure 6).
STATIC EQUILIBRIUM
F1
Static Equilibrium
sum of forces:
W = mg
ΣF = F2 - F1 - mg = 0
F2
FIGURE 6. STATIC EQUILIBRIUM.
An object that is in dynamic equilibrium has a resultant force or moment acting on it.
The object is therefore undergoing linear acceleration and/or angular acceleration
(Figure 7). The resultant force is equal to the mass of the object multiplied by the
acceleration of the object.
F1
Dynamic Equilibrium
sum of forces:
ΣF = F2 - F1 - mg = ma
a
W = mg
resultant force:
F = ma
F2
FIGURE 7. DYNAMIC EQUILIBRIUM.
Unit 1 - Biomechanical Analysis
5
DYNAMIC EQUILIBRIUM
To calculate accurately the forces acting on a body segment that is accelerating (i.e. one
that is in dynamic equilibrium) inverse dynamics is used. Using inverse dynamics it is
possible to calculate the resultant force acting on a body segment which produced the
observed acceleration. However, when a segment is in static equilibrium (is at rest or
moving with constant velocity), it is not accelerating and there is no resultant force.
Inverse dynamics is therefore not needed. Thus, biomechanical problems involving
body segments in static equilibrium are more easily solved. When a body segment
undergoes only small accelerations the resultant force, which is equivalent to the inertial
force, is often negligible compared to the other forces acting on the segment. The
acceleration of the segment can therefore be ignored. In such cases the link segment
model is said to be quasi-static, that is the segment seems to be static (Figure 8).
QUASI-STATIC
F1
Quasi-static
a is small
a
W = mg
ΣF = F2 - F1 - mg = 0
F2
FIGURE 8. QUASI-STATIC.
For example, during the single stance phase of walking (when only one foot is on the
ground), the inertial effects are negligible when calculating the joint forces and
moments at the ankle joint and it is valid to use a quasi-static model. However, during
the single stance phase of walking, the inertial effects are not negligible when
calculating the forces and moments at the hip joint. This is because the magnitudes of
the accelerations are greater and the whole mass of the lower limb is involved. Thus, if a
quasi-static model was used for the hip significant errors would be introduced and the
results would be invalid.
SAQ 4 - Under what circumstances is it justifiable to treat a link
segment model as quasi-static?
2. ANTHROPOMETRY
Anthropometry is the study of human size and form. Anthropometry is widely
applicable across many subject areas. For example, forensic scientists use it to help
determine the identity of an unknown individual, and dieticians use it to determine
nutritional status.
In biomechanics, a knowledge of the anthropometrics of the subject being analysed is
usually required. It is often too difficult and too time consuming to measure all the
required anthropometric characteristics of each and every subject. Instead recourse is
made to standardised sets of anthropometric data. In biomechanics, the body segment
parameters are of particular importance. These parameters describe the mechanical
attributes of a body segment which are required to solve biomechanical problems. There
are four basic body segment parameters:
¾ length
¾ mass
¾ centre of mass
¾ radius of gyration
Unit 1 - Biomechanical Analysis
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ANTHROPOMETRY
BODY SEGMENT
PARAMETERS
If the actions of muscles are also to be included in the biomechanical analysis then the
origin and insertion sites of the active muscles are also required.
2.1 Segment Length
Using a standardised set of data it is possible to estimate the length of a person’s body
segments from their height. One set of standardised data is shown in Figure 9. The
length of each body segment is expressed as a ratio of body height (H). It is therefore
quite easy to calculate an estimate for the length of a particular segment given the
subject’s height.
FIGURE 9. BODY SEGMENT LENGTHS EXPRESSED AS RATIOS OF BODY HEIGHT. ADAPTED FROM DRILLIS R AND
CONTINI R “BODY SEGMENT PARAMETERS”, REPORT 1163-03, OFFICE OF VOCATIONAL REHABILITATION, DEPARTMENT
OF HEATH, EDUCATION AND WELFARE, NEW YORK, 1966.
The standardised data set given in Figure 9 is applicable to the whole adult population.
However, body dimensions vary with age, body build, sex and racial origin. Therefore
more accurate estimates can be obtained for a particular subject by using a more
appropriate standardised data set. Of course the most accurate way of obtaining body
segment length is to make direct measurements. Standardised sets for body segment
lengths should therefore only be used when it is not possible to make direct
measurements.
Worked Example
At the scene of a crime the police discovered a bare footprint. The foot was measured
and found to be 24.0 cm long and 8.8 cm wide. Estimate the height of the suspect.
Using these values and the standardised data given in Figure 8 we can make two
estimates of the suspect’s height.
The length of the foot is related to height by the equation:
LFOOT = 0.152 × H
where
LFOOT
is the length of the foot, and
H
is the height of the subject.
Rearranging and inserting the value for foot length:
Unit 1 - Biomechanical Analysis
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H=
L FOOT
24
= 158 cm
=
0.152
0152
.
The width of the foot is related to height by the equation:
WFOOT = 0.055 × H
Rearranging and inserting the value for foot length:
H=
WFOOT
8.8
= 160 cm
=
0.055
0.055
Thus the height of the suspect is approximately between 158 and 160 cm.
SAQ 5 - Estimate the length of your upper and lower limbs using your
height and the data in Figure 9. How do these estimates compare with
the actual length of your upper and lower limbs?
2.2 Segment Mass and Centre of Mass
Standardised sets of data have also be produced for the mass and the position of the
centre of mass of each body segment. Both these parameters are required for even the
most simple biomechanical analysis. As these cannot be easily measured recourse is
usually made to standardised data sets. These sets may take several different forms,
each requiring different types and amounts of information. For example, some require
skin fold measurements. The simplest require only the total body mass of the subject
and the length of each segment.
In Table 1 a standard set of anthropometric data for the mass and the position of the
centre of mass of each body segment is presented. The coefficients are in the form of
ratios of segment mass to whole body mass and centre of mass to segment length.
Segment
Hand
Definition
Wrist axis/knuckle II middle finger
Segment
mass/Total
body mass
0.006
Centre of mass/
Segment length
Proximal
Distal
0.560
0.494
Forearm
Elbow axis/ulnar styloid
0.016
0.430
0.570
Upper arm
Glenohumeral axis/elbow axis
0.028
0.436
0.564
Forearm & hand
Elbow axis/ulnar styloid
0.022
0.682
0.318
Total arm
Glenohumeral joint/ulnar styloid
0.050
0.530
0.470
Foot
Lateral malleolus/head metatarsal II
0.0145
0.50
0.50
Leg
Femoral condyles/medial malleolus
0.0465
0.433
0.567
Thigh
Greater trochanter/femoral condyles
0.100
0.433
0.567
Foot & leg
Femoral condyles/medial malleolus
0.061
0.606
0.394
Total leg
Greater trochanter/medial malleolus
0.161
0.447
0.553
TABLE 1. BODY SEGMENT MASSES EXPRESSED AS A RATIO OF TOTAL BODY MASS, AND CENTRE OF MASS POSITIONS
EXPRESSED AS A RATIO OF SEGMENT LENGTH FROM PROXIMAL AND DISTAL ENDS. ADAPTED FROM WINTER DA
“BIOMECHANICS AND MOTOR CONTROL OF HUMAN MOVEMENT, 2ND EDITION”, JOHN WILEY AND SONS LTD,
CHICHESTER, 1990.
For example, Figure 10 shows the whole arm with the masses and positions of the upper
arm and forearm segments. Notice that the position of each centre of mass is closer to
the proximal end of the segment than the distal end. This is because in general the
diameter of each segment increases towards the proximal end.
Unit 1 - Biomechanical Analysis
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0.564 I1
0.436 I1
0.43 I1
0.57 I1
0.028 MBODY
I1
0.016 MBODY
I2
FIGURE 10. MASSES AND POSITIONS OF THE UPPER ARM AND FOREARM SEGMENTS.
Worked Example
For a man with a mass of 75 kg and a forearm of length 25.6 cm calculate the mass and
the position of the centre of mass relative to the proximal end of his forearm.
m = 0.016mBODY = 0.016 × 75 = 1.2 kg
LPROXIMAL = 0.430 LFOREARM = 0.430 × 25.6 = 11.0 cm
Thus the mass of the man’s forearm is 1.2 kg and the centre of mass is 11.0 cm from the
elbow axis.
SAQ 6 - Calculate the mass of the thigh and the position of its centre
of mass relative to its proximal and distal ends, given that the total
body mass of the subject is 80 kg and direct measurements of the
thigh yielded a segment length of 390 mm.
2.3 Moment of Inertia and Radius of Gyration
A rotating body has inertia, similar to that of a body moving in a straight line. The
inertia of a rotating body is called the moment of inertia. It is dependent not only upon
the body’s mass (as with a body moving linearly) but also upon how that mass is
distributed within the body itself relative to the axis of rotation. Formally, the moment
of inertia is defined as the sum of the products of the mass of each particle of the body
and the square of its perpendicular distance from the axis of rotation. Thus, when a body
has its mass concentrated away from the axis about which it is rotating it will have a
large moment of inertia. The moment of inertia is represented by the symbol I and its
units are kg m2.
When considering linear motion we think of all the mass of a body as being
concentrated at its centre of mass. When considering rotary motion, it is useful to think
of all the mass of a body as being concentrated at a particular radius from the axis of
rotation. This radius is termed the radius of gyration of the body. It is represented by
the symbol k and its units are metres. This enables us to describe the moment of inertia
of a body as equivalent to the product of its mass times its radius of gyration squared:
I = mk2
Standardised sets of anthropometric data, such as that in Table 2, usually quote the
radius of gyration of a body segment rather than the moment of inertia since the radius
of gyration is not dependent on the mass of the body segment. The moment of inertia
can therefore be calculated using the calculated radius of gyration and the mass of the
segment, estimated from data such as that in Table 1. The radius of gyration is quoted
about three different axes of rotation as it is dependent on the distribution of mass
relative to the axis of rotation. It will therefore be different for different axes of rotation.
Unit 1 - Biomechanical Analysis
9
MOMENT OF INERTIA
RADIUS OF GYRATION
The three axes of rotation are about the centre of mass, the proximal end and distal end.
It is therefore important that the correct coefficient is chosen.
Segment
Definition
Radius of gyration/Segment length
Centre of
mass
Proximal
Distal
Hand
Wrist axis/knuckle II middle finger
0.297
0.587
0.577
Forearm
Elbow axis/ulnar styloid
0.303
0.526
0.647
Upper arm
Glenohumeral axis/elbow axis
0.322
0.542
0.645
Forearm & hand
Elbow axis/ulnar styloid
0.468
0.827
0.565
Total arm
Glenohumeral joint/ulnar styloid
0.368
0.645
0.596
Foot
Lateral malleolus/head metatarsal II
0.475
0.690
0.690
Leg
Femoral condyles/medial malleolus
0.302
0.528
0.643
Thigh
Greater trochanter/femoral condyles
0.323
0.540
0.653
Foot & leg
Femoral condyles/medial malleolus
0.416
0.735
0.572
Total leg
Greater trochanter/medial malleolus
0.326
0.560
0.650
TABLE 2. BODY SEGMENT RADIUS OF GYRATION EXPRESSED AS A RATIO OF SEGMENT LENGTH RELATIVE TO THE
CENTRE OF MASS, AND PROXIMAL AND DISTAL ENDS. ADAPTED FROM WINTER DA “BIOMECHANICS AND MOTOR
CONTROL OF HUMAN MOVEMENT, 2ND EDITION”, JOHN WILEY AND SONS LTD, CHICHESTER, 1990.
Worked Example
Calculate the radius of gyration and the moment of inertia of the whole arm about its
centre of mass, its proximal end and its distal end for a subject with a whole body mass
of 80 kg and height of 1.80 m.
First of all we need to calculate the length of the arm and its mass. Using the data in
Figure 9 we can estimate the length of the arm, LA, as follows:
LA = (0.186 + 0.146 + 0.108) × 1.80 = 0.792 m
Using the data in Table 1 we can calculate the mass of the arm, mA, as follows:
mA = 0.05 × 80 = 4.0 kg
Now to calculate the required radii of gyration and moments of inertia the data in Table
2 is needed. For the radius of gyration about the centre of mass, kCM:
kCM = 0.368 LA = 0.368 × 0.792 = 0.291 m
For the moment of inertia about the centre of mass, ICM:
ICM = mAkCM2 = 4.0 × 0.291 × 0.291 = 0.34 kg m2
Similarly for the radius of gyration, kP, and moment of inertia, IP, about the proximal
end:
kP = 0.645 × LA = 0.645 × 0.792 = 0.511 m
IP = mA × kP2 = 4.0 × 0.511 × 0.511 = 1.0 kg m2
Similarly for the radius of gyration, kD, and moment of inertia, ID, about the distal end:
kD = 0.596 × LA = 0.596 × 0.792 = 0.472 m
ID = mA × kD2 = 4.0 × 0.472 × 0.472 = 0.89 kg m2
The relationship between the radii of gyration and the position of the axis of rotation is
illustrated in Figure 11. Notice how the size of the radius of gyration varies.
Unit 1 - Biomechanical Analysis
10
FIGURE 11. RADII OF GYRATION FOR THE ARM.
SAQ 7 - Calculate the moment of inertia of the lower leg and foot
complex about its centre of mass, its proximal end and its distal end,
given that the mass of the subject is 75 kg and the length of the lower
leg and foot complex is 450 mm. Comment on the difference in radius
of gyration about the proximal end and distal end.
2.4 Muscle Anthropometry
When considering the action of muscles around joints there are three parameters to be
considered:
¾ the force produced by the muscle
¾ the insertion point on the bone
¾ the line of action of the force
If only a single muscle is capable of producing a movement or maintaining a position
against a external force then the whole force must be produced by that muscle.
However, the human anatomy is such that several muscles perform the same function. It
is therefore difficult to know the exact contribution of each muscle to the overall force.
If more than one muscle contributes to the force required to produce a movement or
maintain a position then the contribution of each muscle is dependent on the anatomy of
the muscle. This problem is explored further in Section 5 on muscle mechanics.
The insertion point of a muscle on a segment and the line of action determines the effect
that the muscle will have. To some extent the line of action is determined by the origin
and insertion of the muscle. Figure 12 shows three of the forearm flexor muscles (biceps
brachii, brachialis and brachioradialis). Each has a different line of action, produced by
different origins and insertions. Notice how the line of action of each muscle changes as
the arm is extended.
Unit 1 - Biomechanical Analysis
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biceps brachii
brachialis
brachioradialis
FIGURE 12. LINE OF ACTION OF THE FOREARM FLEXORS.
SAQ 8 - When a distal segment moves relative to a proximal segment
what happens to the insertion point and the line of action of a muscle
that connects the two segments?
3. FORCES IN JOINTS
In this section we will be looking at the different types of forces in joints that can be
calculated using biomechanical analysis.
3.1 Inter-segment Force
The inter-segment force, or the inter-segmental force, is the resultant of all the forces
crossing the joint between the two body segments. It is also known as the net joint
force and the resultant joint force. The inter-segment force is comparatively easy to
calculate as no knowledge of the forces produced by muscles and ligaments across a
joint is required. The inter-segment force is calculated in the following worked
examples.
Worked Example
In this example, we will calculate the inter-segment force acting at the ankle joint, i.e.
between the foot and shank segments, during quiet standing. In this case there is no
movement taking place and the body is in static equilibrium. Thus, the only forces
present are those due to gravity and the external forces acting on the feet - the ground
reaction forces. Note that the inter-segment force acting on the foot at the ankle joint,
AF, must be equal and opposite to the inter-segment force acting on the shank at the
ankle joint, AF (Figure 13). Remember that Newton’s third law states that “to every
action there is an equal and opposite reaction”.
shank segment
AF
AS
AF
foot segment
mF g
G
(A)
(B)
FIGURE 13.
Unit 1 - Biomechanical Analysis
12
INTER-SEGMENT
FORCE
Given the mass of the subject, 80 kg, and the mass of the subject’s foot, 1.0 kg, we can
calculate the inter-segment force by considering the foot segment alone. The free body
diagram for the foot is shown in Figure 13(B). The foot segment is in static equilibrium,
the sum of all the forces acting on the segment is equal to zero. There are three forces
acting on the foot: the weight of the segment, the inter-segment force and the ground
reaction force.
If we assume that the body weight is evenly distributed between both lower limbs then
the ground reaction force acting on each foot must be equal to half the body weight:
G = ½mBODYg = 0.5 × 80 × 10 = 400 N
where
G
is the ground reaction force,
mBODY is the body mass of the subject = 80 kg, and
g
is the acceleration due to gravity = 10 m s-1.
Summing all the forces acting on the foot segment:
ΣF = G - mFg - AF = 0
where
mF
is the mass of the foot = 1.0 kg, and
AF
is the inter-segment force acting on the foot at the ankle.
Rearranging for AF:
AF = G - mFg = 400 - 1.0 × 10 = 390 N
Thus, the inter-segment force between the foot and shank has a magnitude of 390 N and
is acting to push the two segments together. As the force is acting to push the two
segments together it must be communicated between the two segments at the joint
surfaces (Figure 14A).
F
F
F
(A)
(B)
FIGURE 14. (A) INTER-SEGMENT FORCE CARRIED BY JOINT (B) INTER-SEGMENT FORCE CARRIED BY LIGAMENTS.
SAQ 9 - Calculate the inter-segment force at the knee joint during
quiet standing. Use the same data given in the above example and 4.0
kg for mass of the shank.
Worked Example
Now, let us calculate the inter-segment force at the ankle joint when the leg is dangling
as shown in Figure 15(A). The free body diagram for this case is shown in Figure 15(B).
Unit 1 - Biomechanical Analysis
13
AF
mF g
(A)
(B)
FIGURE 15. (A) DANGLING FOOT (B) FREE-BODY DIAGRAM OF DANGLING FOOT.
The foot segment is in static equilibrium therefore the sum of all the forces acting on it
is zero:
ΣF = -mFg - AF = 0
Rearranging for AF:
AF = -mFg = -1.0 × 10 = -10 N
Thus, the inter-segment force between the foot and shank has a magnitude of 10 N and
is acting to pull the two segments apart. As the force is acting to pull the two segments
together it must be carried by the ligaments surrounding the joint and the joint capsule
(Figure 14B).
SAQ 10 - If the sign of the inter-segment force is positive does this
mean that it is acting to pull the segments apart or push them
together?
3.2 Joint Force
The joint force is the force acting between two segments at a joint. The joint force
differs from the inter-segment force in that to calculate the joint force the forces arising
due to the activity of the muscles crossing the joint must be included. The joint force is
related to the inter-segment force in that the inter-segment force is equal to the sum of
the joint force and the forces produced by muscles:
inter-segment force = joint force + muscle forces
When there is no muscle activity the joint force is equal to the inter-segment force. Thus
in the worked examples in the previous section on inter-segment forces the joint force is
equal to the inter-segment force as there is no muscle activity.
Worked Example
To illustrate the difference between the joint force and the inter-segment force we will
consider the forces at the shoulder joint when the arm is held out straight to the side as
shown in Figure 16. The mass of the whole arm is 4.0 kg and the position of centre of
mass is from the shoulder joint.
Unit 1 - Biomechanical Analysis
14
JOINT FORCE
4.0 kg
0.27 m
FIGURE 16. ARM HELD OUT STRAIGHT TO THE SIDE.
First of all we will calculate the inter-segment force at the shoulder joint. For simplicity
we will consider the whole arm to be one segment. There are only two forces acting on
the arm segment: the inter-segment force and the force due to the mass of the arm
segment. These are both shown on the free body diagram in Figure 17.
MA
mAg
IS
FIGURE 17. FREE BODY DIAGRAM OF ARM.
The arm segment is in static equilibrium therefore the sum of all the forces should be
zero. Summing the vertical components, with upwards defined as positive:
ΣFVERTICAL = IS - mAg = 0
where:
IS
is the inter-segment force at the shoulder joint,
MA
is the mass of the whole arm (4.0 kg), and
g
is the acceleration due to gravity (10 m s-1).
Rearranging for IS:
IS = mAg = 4.0 × 10 = 40 N
Thus the inter-segment force at the shoulder joint has a magnitude of 40 N and is acting
upwards.
At first glance this answer appears reasonable, the inter-segment force is acting to
prevent the arm falling down. However, we have not yet considered rotational
equilibrium. If we sum the moments about the joint we find that there must be a
moment, MS, as shown in Figure 18, to counteract the moment produced by the weight
of the segment about the joint.
MS
MA
IS
dCOM
mAg
FIGURE 18. FREE BODY DIAGRAM OF ARM INCLUDING MOMENTS.
Unit 1 - Biomechanical Analysis
15
Thus summing the moments about the shoulder joint:
ΣM = MS - mAgdCOM = 0
Rearranging for MS:
MS - mAgdCOM = 4.0 × 10.0 × 0.27 = 10.8 Nm
where:
MS
is the moment about the shoulder joint, and
dCOM
is the position of the centre of mass from the shoulder joint.
Thus, there must be a moment of magnitude 10.8 Nm to maintain the arm segment in
place. This moment is achieved by internal forces produced by muscle action and
ligaments. We will assume that the moment is produced solely by the deltoid muscle
which is producing a force acting at 15 degrees to the arm segment and is inserted 7 cm
distal and 1 cm superior to the shoulder joint centre (Figure 19).
deltoid muscle force
FD
15°
deltoid
muscle
FIGURE 19. DELTOID MUSCLE FORCE.
The moment produced by the deltoid muscle must be equal to the moment MS and equal
and opposite to the moment produced by the weight of the arm segment. We can
calculate the deltoid muscle force as follows. If we resolve the deltoid muscle force
along the horizontal and vertical, we find that both components produce moments about
the shoulder joint (as shown in Figure 20). This is because of the position of the muscle
insertion relative to the shoulder joint centre.
FV = FD sin 15°
dH
FH =
FD cos 15°
dV
dCOM
mAg
FIGURE 20. FREE BODY DIAGRAM SHOWING FORCES PRODUCING MOMENTS ABOUT SHOULDER JOINT CENTRE.
Applying rotational equilibrium about the shoulder joint centre:
ΣM = FHdV + FVdH - mAgdCOM = 0
Unit 1 - Biomechanical Analysis
16
where:
FH
is the horizontal component of the deltoid muscle force,
dV
is the vertical displacement of the deltoid muscle insertion relative to the
shoulder joint,
FV
is the vertical component of the deltoid muscle force,
dH
is the horizontal displacement of the deltoid muscle insertion relative to the
shoulder joint.
FDcos15°dV + FDsin15°dH - mAgdCOM = 0
FD(cos15°dV + sin15°dH) - mAgdCOM = 0
Rearrange for FD:
FD =
mA gd COM
4.0 × 10 × 0.27
10.8
=
=
cos15° d V + sin15° d H cos15°× 0.01+ sin15°× 0.07 0.0096593 + 0.018117
= 389 N
Calculating the magnitude of the components:
FH = Fcos15° = 389 × cos15° = 376 N
FV = Fsin15° = 389 × sin15° = 101 N
Thus, the force produced by the deltoid muscle must have a magnitude of approximately
390 N to maintain the arm segment in rotational equilibrium.
We have introduced the internal force due to the contraction of the deltoid muscle in
order to account for the moment produced by the weight of the arm segment. This
internal force must be included in any analysis of the joint force. Figure 21 shows the
free body diagram of the arm segment with the vertical and horizontal components of
the deltoid muscle force. The joint force must have a horizontal and a vertical
component to counteract the forces acting on the arm segment and so maintain the arm
segment in static equilibrium.
FV
FH
JH
JV
mAg
FIGURE 21. FREE BODY DIAGRAM INCLUDING DELTOID MUSCLE FORCE COMPONENT AND JOINT FORCE COMPONENTS.
Summing the horizontal forces (defining the forces acting from left to right as positive):
ΣFHORIZ = JH - FH = 0
where:
JH
is the horizontal component of the joint force.
JH = FH = 376 N
Unit 1 - Biomechanical Analysis
17
Summing the vertical forces (defining the forces acting vertically upwards as positive):
ΣFVERTICAL = JV + FV - mAg = 0
JV = -FV + mAg = -101 + 4.0 × 10 = -61 N
2
2
2
J = JH + J V
J
JV
tanθ =
JV
JH
θ
JH
FIGURE 22. COMPONENTS OF THE JOINT FORCE.
The joint force is made up of the two components as shown in Figure 21. We can find
the magnitude of the joint force using Pythagoras’ theorem:
J = J 2H + J 2V = 3762 + 612 = 381 N
Calculating the angle of the joint force, θ, relative to the horizontal:
tan θ =
JV
JH
⎛J ⎞
⎛ 61 ⎞
θ = tan-1 ⎜ V ⎟ = tan-1 ⎜
⎟ = 9.2°
⎝ 376 ⎠
⎝ JH ⎠
Thus the shoulder joint force acting on the arm segment has a magnitude of 380 N and a
direction of 9.2º relative to the horizontal. Comparing this to the inter-segment force of
40 N acting vertically, illustrates that in this there is a considerable difference between
the joint force and inter-segment force.
SAQ 11 - Why is the joint force generally much larger than the intersegment force?
3.3 Bone-on-bone Force
The bone-on-bone force is the force acting between the two bones that form a joint.
The bone-on-bone force is related to the inter-segment force as follows:
BONE-ON-BONE FORCE
inter-segment force = bone-on-bone force + muscle force + ligament force
The ligament force is the force carried by ligaments, joint capsules and muscles that are
not contracting.
In the previous section the joint force was similar expressed as:
inter-segment force = joint force + muscle force
Thus we can say that:
joint force = bone-on-bone force + ligament force
Unit 1 - Biomechanical Analysis
18
LIGAMENT FORCE
The bone-on-bone force is less useful than the joint force as it is usually not possible to
estimate the ligament force. However, in certain circumstances, when the inter-segment
force is negative, i.e. it is acting to pull the segments apart, it is reasonable to say that
the inter-segment force must be carried by the ligaments, joint capsules and muscles that
are not contracting.
For example, in the second worked example in section 3.1 on inter-segment forces, the
inter-segment force at the ankle joint was found to be negative. From this it was
concluded that since no muscles were active the force must be carried by the ligaments,
joint capsule and muscles that are not active, i.e. it is equal to the ligament force. Thus,
the bone-on-bone force must be equal to zero.
SAQ 12 - Write down the definitions for the inter-segment force, the
joint force, the muscle force, the bone-on-bone force and the ligament
force.
3.4 Terminology and Definitions
It is important to note that there is no standard terminology for the two types of forces
that may be calculated at joints and that different authors may use the same terminology
to mean a different force. I have defined five types of force: the inter-segment force,
the joint force, the muscle force, the bone-on-bone force and the ligament force. This
is more than most authors would consider, however, they are necessary to understand
the complexity of the interaction at joints and to allow you to understand the conflicting
use of terminology and different definitions. The two recommended books for this
module, Winter and Nordin & Frankel, use different, conflicting terminology. These are
summarised in Table 3.
Nordin & Frankel
Joint reaction force
Inter-segment force
Joint force
Winter
Joint reaction force
Muscular force
Bone-on-bone force
Bone-on-bone force
Ligament force
TABLE 3. TERMINOLOGY USED FOR THE TWO TYPES OF FORCES AT JOINTS.
If you examine Table 3 you will see that the two books use the term, joint reaction
force, differently. This problem is also prevalent throughout the biomechanics literature.
You must therefore be wary when reading any books or articles on the forces at joints
and be certain of the author’s meaning.
4. MOMENTS ABOUT JOINTS
So far we have only considered the forces acting at joints. The moments acting about
joints, the joint moments, are also important. For example, the joints moments about
the ankle, knee and hip joints are commonly used in clinical and rehabilitation settings
in the assessment of walking.
In static or quasi-static cases, joint moments can be calculated using only the external
forces and moments. No knowledge of the action of muscles is required as in the case of
the joint forces. The overall moment calculated from the external forces and moments is
called the net external joint moment. To maintain equilibrium this must be balanced
by the net internal joint moment, that is the net external joint moment is equal and
opposite to the net internal joint moment:
JOINT MOMENTS
NET EXTERNAL
JOINT MOMENTS
net external joint moment + net internal joint moment = 0
The net internal joint moment is the result of all internal forces and moments, which
are produced by the muscles, the ligaments and the friction at the joints.
Unit 1 - Biomechanical Analysis
19
NET INTERNAL
JOINT MOMENTS
For example, when the foot is flat on the ground during walking there is a net external
flexion moment about the knee joint. To maintain equilibrium, this must be countered
by an equal and opposite net internal moment. This internal extension moment is
produced by the quadriceps muscles which act to extend the knee joint (Figure 23).
quadricep
muscles
internal
extension
moment
external
flexion
moment
FIGURE 23. EXTERNAL AND INTERNAL KNEE JOINT MOMENTS WITH QUADRICEPS ACTIVE.
By considering joint moments it is possible to determine muscle activity. Provided there
is no antagonistic muscle activity, and contribution of the ligaments and joint friction
are negligible, then the net internal moment is equal to the moment produced by the
agonist muscle or muscle group. When there is antagonistic muscle activity, the agonist
muscle must contract more strongly to overcome the moment produced by the
antagonist muscle. In cases when the activity of muscles is unknown, then electromyography can be used to determine the extent of the muscle activity when the muscle
is active. If in the previous example, the hamstrings were active, then they would
produce an additional flexion moment (Figure 24). This would not change the net
internal moment which must be equal to the net external muscle moment. However, the
quadriceps moment must be increased to overcome the moment produced by the
hamstrings.
quadriceps
hamstrings
external
flexion
moment
FIGURE 24. EXTERNAL AND INTERNAL KNEE JOINT MOMENTS WITH ACTIVE HAMSTRINGS AND QUADRICEPS.
SAQ 13 - What is the relationship between the net external joint
moment and the net internal joint moment?
Unit 1 - Biomechanical Analysis
20
Worked Example
Consider the situation shown in Figure 25. The pulley system is exerting a force on the
shank and foot just above the ankle. This force produces a moment about the knee joint
with a moment arm of 38 cm. The knee is being maintained in static equilibrium at 90º
of flexion by the subject. To achieve this he must be contracting his quadriceps muscles
to produce an anti-clockwise moment about the knee joint. What is the magnitude of the
moment produced by the quadriceps to overcome the resistance force?
10 kg
FIGURE 25. PULLEY SYSTEM ACTING ON LEG AT THE ANKLE JOINT.
To find the magnitude of the moment we first of all draw a free body diagram (Figure
26).
MQ
d
mg
F
FIGURE 26. FREE BODY DIAGRAM OF LEG.
The force produced by the pulley system is equal to the product of the attached mass
and the acceleration due to gravity. Thus, using g = 9.8 m s-2:
F = ma = 10.0 × 9.8 = 98 N
Unit 1 - Biomechanical Analysis
21
where:
F
is the pulley force, and
M
is the mass acting on the pulley.
Summing at the moments acting about the knee joint centre (defining clockwise as
positive).
ΣM = Fd - MQ = 0
where:
d
is the moment arm (0.38 m), and
MQ
is the quadriceps moment.
Rearranging:
MQ = Fd = 98 × 0.38 = 37 Nm
Thus the moment produced by the quadriceps is equal to 37 Nm.
SAQ 14 - If in the previous worked example the magnitude of the mass
attached to the pulley system has doubled how would the quadriceps
moment change in order to maintain static equilibrium?
5. MUSCLE MECHANICS
It is possible to calculate inter-segmental forces and joint moments without knowledge
of the action of muscles, but in order to calculate the joint forces knowledge of the
muscle forces is required. However, it is not possible to measure muscle force directly
and accurately predicting muscle forces is extremely difficult as muscles rarely act in
isolation, and several muscles with similar functions cross the same joint.
The magnitude of the force produced by a particular muscle is dependent on a number
of factors including:
¾ force-length relation,
¾ force-velocity relation,
¾ physiological cross-sectional area and muscle architecture.
These can be used to a certain extent to predict the magnitude of force produced by a
muscle. The relationship between these factors and the muscle force is described in the
following three subsections. The subsequent subsections discuss the relationship
between EMG and muscle force and the methods used in biomechanical analysis to
estimate muscle forces.
5.1 Force-length
The relationship between muscle force and length is dependent on the contractile
elements and the passive connective tissue which forms the muscle. The contractile
elements of muscle are the sarcomeres (Figure 27) which consist of the interacting actin
and myosin filaments (also known as the thin and thick myofilaments). The force
produced by each contractile element is dependent on the number of cross bridges
between the actin and myosin filaments which is in turn dependent on the number of
available binding sites.
Unit 1 - Biomechanical Analysis
22
myofibril
(A)
sarcomere
(B)
thin
myofilament
thick
myofilament
cross bridges
FIGURE 27. (A) MYOFIBRIL (B) SARCOMERE.
tension
The relationship between the force produced by the contractile elements, known as the
developed tension, and the length of the muscle is shown in Figure 28. At the resting
length of the muscle, the overlap between actin and myosin filaments is optimal and the
developed tension is maximal. At lengths longer than the resting length, the overlap
between the filaments is reduced and the number of cross bridges between the actin and
myosin filaments is also reduced. Thus, the force that the muscle can produce is also
reduced. At lengths shorter than the resting length, the cross bridges overlap and
interfere, resulting in a reduction in the force that the muscle can produce.
DEVELOPED TENSION
developed tension
50
100
150
length (% of resting length)
resting
length
FIGURE 28. GRAPH OF ACTIVE TENSION AGAINST MUSCLE LENGTH.
In addition to the active tension produced by the contraction of the muscle fibres there is
also a passive tension. A complete muscle includes a substantial amount of connective
tissue (for example, the sarcolemma, endomysium, perimysium and epimysium). The
behaviour of these connective tissues can be likened to stiff elastic bands, that is they
resist stretching. The passive tension produced by stretching these connective tissues
increases with length. As the length of the muscle increases the contribution of the
active tension to the total muscle force reduces and the contribution of the passive
tension increases dramatically. This is illustrated in Figure 29.
Unit 1 - Biomechanical Analysis
23
PASSIVE TENSION
total tension
tension
passive tension
developed tension
50
100
150
length (% of resting length)
FIGURE 29. GRAPH OF TOTAL TENSION AGAINST MUSCLE LENGTH.
SAQ 15
(a) What are the interacting filaments in sarcomeres called?
(b) At approximately what muscle length is the maximum developed
tension achieved?
5.2 Force-velocity
During all movements muscles are not contracting isometrically (that is they are not
maintaining a constant length) they are either contracting concentrically (shortening) or
eccentrically (lengthening). The relationship between muscle force and velocity is
therefore very important.
During concentric contractions the muscle is shortening (for example the biceps as you
flex your elbow). As the velocity increases the tension in the muscle decreases. This is
shown on the right hand side of the graph in Figure 29 for four different levels of
muscle activation (25%, 50%, 75%, 100%). Note that the velocity is zero where the
curves cross the vertical y-axis. The muscle force at this point is the same as that for an
isometric contraction.
During an eccentric contraction the muscle is lengthening. During an eccentric
contraction the muscle force is greater than during an isometric or concentric
contraction for the same level of muscle activation. This is shown on the left hand side
of the graph in Figure 30.
force
100%
75%
50%
25%
lengthening
0
shortening
velocity
FIGURE 30.
Unit 1 - Biomechanical Analysis
24
Note: the curves in Figure 30 are plotted for a certain muscle length. The forcevelocity curves will of course vary with muscle length; with the maximum forces being
obtained when the muscle is at its optimum length. During an actual limb movement
both the muscle’s velocity and length will be changing so that the amount of force the
muscle produces, for a certain level of activation, will also be changing constantly.
SAQ 16
(a) What is the difference between a concentric and an eccentric
muscle contraction?
(b) Is the muscle force for a certain level of activation greater for
concentric or eccentric contraction?
5.3 Physiological Cross-sectional Area and Muscle Architecture
The maximum force produced by a muscle has been found to be related to its crosssectional area and its muscle architecture (the way in which the muscle fibres are
arranged). The greater the physiological cross-sectional area of a muscle then the better
suited the muscle is for force production.
There is a wide and varied range of muscle architecture. In general the muscle fibres
may be arranged either parallel to the long axis of the muscle or they may be arranged
obliquely. For example, Figure 31A shows a muscle with the muscle fibres arranged in
parallel, and Figure 31B shows a muscle with its muscle fibres arranged like a feather,
this arrangement is called bipennate. An example of a parallel muscle is the rectus
abdominis, and an example of a bipennate muscle is the rectus femoris.
α
(A)
(B)
FIGURE 31. MUSCLE ARCHITECTURES (A) PARALLEL (B) BIPENNATE.
The angle between the long axis of the muscle and the muscle fibres is called the
pennation angle. This angle is shown in Figure 31B. Note that as the muscle changes in
length the pennation angle will also change, being greatest when the muscle is shortest.
Typical values for pennation angle are approximately 5º for the rectus femoris and 30º
for the soleus.
The physiological cross-sectional area is calculated using the following equation:
PCSA =
mcosα
ρ1
Unit 1 - Biomechanical Analysis
25
PENNATION ANGLE
where
m
ρ
l
α
is the muscle mass,
is the muscle density (typically 1.05 g cm-3),
is the muscle fibre length, and
is the pennation angle.
The physiological cross-sectional area is usually expressed in cm2 or m2. Typical values
for physiological cross-sectional areas are approximately 13 cm2 for the rectus femoris
and 60 cm2 for the soleus.
Note: Skeletal muscle tissue is only slightly more dense than water, having a density
of approximately 1.05 g cm-3 compared to the density of water which is 1.00 g cm-3.
Worked Example
Calculate the physiological cross-sectional area for a gastrocnemius specimen given that
its mass is 160 g, its pennation angle is 15º, its muscle fibre length is 50 mm and the
density of skeletal muscle tissue is 1.05 g cm-3.
PCSA =
mcosα 160 × cos15°
= 29 cm2
=
ρ1
1.05 × 5.0
Thus the physiological cross-sectional area for the gastrocnemius specimen is 29 cm2.
SAQ 17 - Calculate the physiological cross-sectional area for the
gastrocnemius specimen in the previous worked example for
pennation angles of 0° and 30° and comment on how the physiological
cross-sectional area is related to pennation angle.
5.4 Electromyography
Electromyography is the study of the electrical signal, called an electromyogram
(EMG), associated with the contraction of a muscle. Under controlled conditions it is
possible to relate the EMG to the magnitude of the muscle force. However, in less
controlled conditions the relationship between EMG and muscle has been found to be
difficult to predict reliably.
Instead of using electromyography to estimate muscle forces it may be used to validate
the results calculated mathematically. For example, if there is agreement between the
occurrence of muscle activity predicted mathematically and that measured using EMG
then the mathematical solution can be considered to be temporally validated (that is the
timing of the activity is valid). However, this does not validate the mathematical
solution quantitatively.
SAQ 18
(a) What is electromyography?
(b) Is there a relationship between EMG and muscle force?
5.5 The Distribution Problem
It is possible to calculate the joint moment and to estimate the force produced by a given
muscle. However, all joints are crossed by several muscles. For example, there are nine
muscles which cross the knee joint. Finding out how the total force is distributed among
these muscles is sometimes referred to as the distribution problem. Such problems are
Unit 1 - Biomechanical Analysis
26
DISTRIBUTION
PROBLEM
also called indeterminant because the number of independent equations available for
their solution is less than the number of unknown values. To solve the distribution
problem and obtain a unique solution, two methods are commonly employed: the
reduction method and the optimisation method.
5.5.1 Reduction method
In the reduction method, the degree of redundancy is reduced until the number of
unknown forces is equal to the number of equations. This can be achieved by
considering the anatomy of the muscles, bones and joints and observing the activation
of the muscles using electromyography. The number of unknown forces can be reduced
by ignoring muscles and/or by grouping muscles with similar function. The number of
equations can be increased, for example by assuming a known force distribution
between muscles.
REDUCTION METHOD
5.5.2 Optimisation method
In the optimisation method, a process or action is either maximised or minimised. This
is based on the underlying assumption that an individual is, either consciously or
unconsciously, adhering to the same optimisation criterion. For example, in many
repetitive movements, such as walking, the optimisation criterion will be to minimise
energy consumption. However, for someone with a degenerative joint disease their
optimisation criterion may be to minimise pain, and if this pain is related to the pressure
between the joint surfaces then the appropriate optimisation criterion may be to
minimise the force carried by the joint surfaces.
SAQ 19 - What is the distribution problem?
Unit 1 - Biomechanical Analysis
27
OPTIMISATION METHOD
SUMMARY
In this unit the basic techniques used in the biomechanical analysis of the skeletal
system have been introduced. For all biomechanical problems a model must be adopted.
This usually takes the form of a linked segment model which is based on a number of
fundamental assumptions about the body segments and the interaction between them.
Anthropometric data is used to estimate the body segment parameters needed for the
link segment model. Inverse dynamics is employed to calculate the forces and moments
at the joints between the segments using knowledge of the movement of the segments
and of the externally acting forces. Using knowledge of the muscle forces it is possible
to calculate the actual forces carried by the various structures of a joint. However, the
accurate prediction of the magnitude and direction of muscle forces is awash with
difficulties and remains one of the most challenging areas of biomechanics.
Finally it is worth remembering that there are no standardised names for the various
types of forces that can be calculated to describe what is happening at a joint, and that
many authors use the same terminology to describe different forces. It is therefore
important that when reading a book or article you understand the biomechanical
analysis used by the author so that you can correctly interpret the presented results.
Unit 1 - Biomechanical Analysis
28
SAQ ANSWERS
SAQ 1
(a) Linked segment models are made up of body segments.
(b) Two main assumptions usually adopted for a link segment model are that
the body segments are rigid and the joints are frictionless.
SAQ 2
In inverse dynamics the motion is measured and the resultant force is
calculated.
SAQ 3
The three types of forces that act on a body segment are external, internal and
inertial.
SAQ 4
It is justifiable to treat a link segment model as quasi-static only when the
accelerations are small.
SAQ 5
The results of this SAQ are dependent on your own dimensions. As an example
I will use the following dimensions, height = 193 cm, measured upper limb
length = 75 cm, and measured lower limb length = 100 cm.
Length of upper limb is calculated as follows using the ratios given in Figure 9:
LUPPER = (0.186 + 0.146 + 0.108) H = 0.440 × 193 = 85 cm
Length of lower limb is calculated as follows:
LLOWER = (0.186 + 0.146 + 0.108) H = 0.530 × 193 = 102 cm
If the calculated limb lengths are compared to measured limb lengths it can be
seen that the calculated lower limb lengths are reasonable estimates. However,
the calculated upper limb length is 10 cm out. This illustrates the diversity of
human dimensions and why standardised data sets should only be used when
actual measurements are not available.
SAQ 6
To calculate the mass of the thigh, mTHIGH, the ratio reported in Table 1 is used:
mTHIGH = 0.100mBODY = 0.100 × 80 = 8.0 kg
To calculate the position of the centre of mass of the thigh relative to its
proximal end, IPROXIMAL, the ratio, 0.433, reported in Table 2 is used:
IPROXIMAL = 0.433ITHIGH = 0.433 × 390 = 169 mm
To calculate the position of the centre of mass of the thigh relative to its
proximal end, IDISTAL, the ratio, 0.567, reported in Table 2 is used:
Unit 1 - Biomechanical Analysis
29
IDISTAL = 0.567ITHIGH = 0.567 × 390 = 221 mm
The mass of the thigh is 8.0 kg, and its centre of mass is located 169 mm distal
to the greater trochanter and 221 mm proximal to the femoral condyles.
Note that the centre of mass is closer to the proximal end than to the distal end.
This is mainly due to the shape of the thigh which narrows towards its distal end
so that more mass is distributed towards the proximal end. The position of the
centre of mass is also dependent on the densities of the various tissues, such
as bone, muscle and fat, and how they are distributed.
SAQ 7
Calculating the mass of the lower leg and foot complex, m, using the ratio,
0.061, in Table 1:
m = 0.061 × 75 = 4.575 kg
Calculating the radius of gyration about the centre of mass, kCM, using the ratio
0.416, given in Table 2:
kCM = 0.416L = 0.416 × 450 × 10-3 = 0.1872 = 0.19 m
Calculating the moment of inertia about the centre of mass, ICM:
ICM = mAkCM = 4.575 × 0.1872 × 0.1872 = 0.1603 = 0.16 kg m2
Calculating the radius of gyration, kP, using the ratio 0.735, and moment of
inertia, IP, about the proximal end:
kP = 0.735L = 0.735 × 450 × 10-3 = 0.33075 = 0.33 m
IP = mkP = 4.575 × 0.33075 × 0.33075 = 0.50048 = 0.50 kg m2
Calculating the radius of gyration, kD, using the ratio 0.572, and moment of
inertia, ID, about the distal end:
kD = 0.572 × L = 0.572 × 450 × 10-3 = 0.2574 = 0.26 m
ID = mkD = 4.575 × 0.2574 × 0.2574 = 0.3031 = 0.30 kg m2
The radius of gyration about the proximal end is larger than the radius of
gyration about the distal end because more mass is distributed further from the
proximal end than the distal end.
SAQ 8
The insertion point of the muscle which connects the two segments does not
change. However, the line of action of the muscle line changes as the two
segments move relative to one another.
SAQ 9
The free body diagram is shown below:
Unit 1 - Biomechanical Analysis
30
KS
shank and
foot segment
mSg
mF g
G
Summing all the forces acting on the shank and foot segment:
ΣF = G - mFg - mSg - Ks = 0
where
G
is the ground reaction force = 400 N,
mF
is the mass of the foot = 1.0 kg,
mS
is the mass of the shank = 4.0 kg,
g
is the acceleration due to gravity = 10 m s-1, and
KS
is the inter-segment force acting on the knee at the shank.
Rearranging for KS:
KS = G - mSg = 400 - 1.0 × 10 - 4.0 × 10 = 350 N
Thus, the inter-segment force at the knee has a magnitude of 350 N.
SAQ 10
If the sign of the inter-segment force is positive then it means that it is acting to
push the segments together.
SAQ 11
The joint force is generally much larger than the inter-segment force because it
includes muscle forces.
SAQ 12
The inter-segment force is the resultant of all the forces crossing the joint
between two segments.
The joint force is the force acting between two segments at the joint.
The muscle force is the force produced by the muscles crossing the joint.
The bone-on-bone force is the force acting between the bones that form a joint.
The ligament force is the force carried by ligaments, joint capsules and muscles
that are not contracting.
Unit 1 - Biomechanical Analysis
31
SAQ 13
The net external joint moment is equal and opposite to the net internal joint
moment.
SAQ 14
If the mass was doubled then the quadriceps moment would need to be
doubled to maintain static equilibrium.
SAQ 15
(a) The interacting filaments in the sarcomeres are called the actin and myosin
filaments or the thin and thick myofilaments.
(b) The maximum active muscle tension is produced when the muscle length
is approximately equal to the resting length of the muscle.
SAQ 16
(a) In a concentric contraction the muscle is shortening and in an eccentric
contraction the muscle is lengthening.
(b) For a certain level of activation, a muscle will produce a greater force
during an eccentric contraction than during a concentric contraction.
SAQ 17
For a pennation angle of 0°:
PCSA =
mcos0°
160 × 1
= 30.5 cm2
=
ρl
. × 5.0
105
For a pennation angle of 15º:
PCSA =
mcos30° 160 × 0.866
= 26.4 cm2
=
ρl
. × 5.0
105
Thus, as the pennation angle increases the physiological cross-sectional area
decreases according to the cosine function.
SAQ 18
(a) Electromyography is the study of the electrical signal associated with the
contractions of muscles.
(b) There is a relationship between EMG and muscle force. However, it is only
possible to characterise this relationship under controlled conditions.
SAQ 19
The distribution problem is the problem of how the total muscle force is
distributed between all the muscles crossing a joint.
Unit 1 - Biomechanical Analysis
32