A BIOMECHANICAL DYNAMIC MODEL FOR

LIFTING IN THE SAGIHAL PLANE

MOUSTAFA MOHAMED EL-BASSOUSSI, B . S . , M.S

A DISSERTATION
 l\)0.43 ACKNOWLEDGMENTS
I would like to place on record my sincere gratitude to
Dr. M.M. Ayoub, the chairman of my advisory committee, for his
direction and encouragement throughout the entire study. Dr. Ayoub
provided numerous suggestions that wfere of considerable value and
was readily available to provide assistance when needed.
m

Gratitude is also extended to Prof. I. Williams for his

guidance, encouragement and co-reading of the manuscript.
I also wish to express my indebtedness to Prof. W. Sandel,
Dr. B. Lambert and Dr. M. Smith for their helpful advice and
constructive criticism. Thanks also goes to J. Gibbs and C. Mittan
for their technical assistance in construction and maintenance of
the equipments.
In addition, it is necessary to extend a "thank you" to the
graduate students who volunteered to participate in the experiment
as test subjects.

11
 TABLE OF CONTENTS
CHAPTER PAGE
ACKNOWLEDGMENTS 11
LIST OF TABLES v
LIST OF FIGURES x
I. INTRODUCTION 1
1.1 Scope 1
1.2 Objectives 2
II. REVIEW OF LITERATURE 4
2.1 Anatomy of the spine related to lifting 4
2.2 Injuries due to lifting 10
2.3 Biomechanics and lifting experiments 17
III. EQUIPMENT AND METHODOLOGY 27
3.1 The experiment and equipment 27
3.2 Independent and dependent variables 29
3.3 Mathematical development of the biomechanical
dynami c model 34
3.3.A Determination of angular
displacements, velocities and
accelerations 37
3.3.B Computation of linear accelerations 38
3.3.C Calculations of reactive forces
and reactive torques at body
articulations 61

111
 3.3.D Calculations of reactive forces
and reactive torques at the
center of the di scs 77
3.3.E Calculation of compressive and
shearing forces 93
IV. RESULTS AND DISCUSSIONS 97
4.1 The net torques at the hip joint 97
4.2 Compressive forces on the spine 106
4.3 Shearing forces on the spine 134
V. CONCLUSIONS AND RECOMMENDATIONS FOR FUTURE RESEARCH 143
5.1 Conclusions 143
5.2 Recommendations for future research 144
BIBLIOGRAPHY 146
APPENDIX A - STATISTICAL ANALYSIS OF THE MAXIMUM NET
TORQUES AT THE HIP JOINT 152
APPENDIX B - STATISTICAL ANALYSIS OF THE MAXIMUM
COMPRESSIVE FORCES ON THE DIFFERENT SPINAL LEVELS ..160
APPENDIX C - DESCRIPTION OF BODY MOTION 191
APPENDIX D - COMPUTER PROGRAM DOCUMENTATION 205

IV
 LIST OF TABLES
TABLE PAGE
1. Maximum values of compressive forces for vertebral
end-plates (kg.) 16
2. Maximum weights (kg.) advisable when carrying loads 23
3. Maximum acceptable weight of lift for males (lbs.) 25
4. Anthropometric data for the experimental subjects 28
5. Values of the biomechanical lifting equivalents,
pound-inches 34
6. Center of gravity and radius of gyration as a
percentage of the segment 1ength 63
7. Effect of the biomechanical lifting equivalents on the
maximum net torques at the hip for the leg lift 99
8. Effect of the biomechanical lifting equivalents on the
maximum net torques at the hip for the back lift 100
9. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum net torques at hip
for third subject (leg lift) 102
10. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum net torques at hip
for third subject (back lift) 102
11. Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum net
torques at hip joint 103
12. Effect of the biomechanical lifting equivalents on the

V
 maximum compressive forces of the upper surface of S-1
for the leg lift 107
13. Effect of the biomechanical lifting equivalents on the
maximum compressive forces of the upper surface of S-1
for the back lift 108
14. Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the lower surface of L-5
for the leg lift 109
15. Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the lower surface of L-5
for the back lift 110
16. Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the upper surface of L-5
for the leg lift HI
17. Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the upper surface of L-5
for the back lift 112
18. Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the lower surface of L-4
for the leg lift 113
19. Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the lower surface of L-4
for the back lift 114
20. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum compressive forces on

vi
 upper surface of S-1 for third subject (leg lift) 121
21. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum compressive forces on
upper surface of S-1 for third subject (back lift) 121
22. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum compressive forces on
lower surface of L-5 for third subject (leg lift) 122
23. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum compressive forces on
lower surface of L-5 for third subject (back lift) 122
24. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum compressive forces on
upper surface of L-5 for third subject (leg lift) 123
25. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum compressive forces on
upper surface of L-5 for third subject (back lift) 123
26. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum compressive forces on
lower surface of L-4 for third subject (leg lift) 124
27. ANOVA for simple linear regression between biomechanical
lifting equivalents and maximum compressive forces on
lower surface of L-4 for third subject (back lift) 124
28. Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum compressive
forces on the upper surface of S-1 129

vii
29. Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum compressive
forces on the lower surface of L-5 127
30. Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum compressive
forces on the upper surface of L-5 128
31. Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum compressive
forces on the lower surface of L-4 129
32. Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the upper surface of S-1
for the leg lift 135
33. Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the upper surface of S-1
for the back lift 136
34. Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the lower surface of L-5
for the leg lift 137
35. Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the lower surface of L-5
for the back lift 138
36. Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the upper surface of L-5
for the leg lift 139
37. Effect of the biomechanical lifting equivalents on the

••
vin
 maximum shearing forces on the upper surface of L-5
for the back lift 140
38. Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the lower surface of L-4
for the leg lift 141
39. Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the lower surface of L-4
for the back lift 142

IX
 LIST OF FIGURES
FIGURE PAGE
1. Lateral and front views of the spine 5
2. Dynamic analysis of the forces on the upper
arm-forearm-hand segments 20
3. Motion patterns for the different articulations during
the lifting task 30
4. The movie camera and the speed control device used
during the experiment 31
5. The projector used in the analysis of the movie films .... 32
6. Displacement, velocity and acceleration patterns using
SI ote and Stone equati ons 39
7. Analysis of circular motions 40
8. Acceleration notations 40
9. The acceleration components at the center of gravity
of the 1 ower 1 eg 43
10. The acceleration components at the center of gravity
of the upper 1 eg 46
11. The acceleration components at the center of gravity
of the trunk 50
12. The acceleration components at the center of gravity
of the upper arm 53
13. The acceleration components at the center of gravity
of the 1 ower arm 58
14. Free-body diagram showing the forces and torques on the

x
 hand during the dynamic activity 65
15. Forces and torques notations 65
16. Free-body diagram for the lower arm showing the forces and
torques during the dynamic activity 68
17. Free-body diagram for the upper arm showing the forces
and torques during the dynamic activity 68
18. Free-body diagram for the trunk showing the forces and
torques during the dynamic activity 71
19. Free-body diagram for the upper leg showing the forces
and torques during the dynamic activity 71
20. Free-body diagram for the lower leg showing the forces
and torques during the dynamic activity 75
21. Average spinal column as presented by Fisher (1967) 78
22. Spinal dimensions taken from data by Fisher (1967) 80
23. Standing erect position when carrying no load 83
24. Angular changes when performing a dynamic activity 85
25. Free-body diagram for the upper trunk link showing
the forces and torques during the dynamic activity 91
26. Free-body diagram for the middle trunk link showing
the forces and torques during the dynamic activity 91
27. Relationship between the biomechanical lifting equivalents
and the maximum net torques at the hip joint 104
28. Leg lift critical positions using the same weight of
lift and different moment arms 118
29. Back lift critical positions using the same weight of

xi
 11ft and different moment arms 119
30. Relationship between the biomechanical lifting
equivalents and the maximum compressive forces on the
upper surface of S-1 119
31. Relationship between the biomechanical lifting
equivalents and the maximum compressive forces on the
lower surface of L-5 126
32. Relationship between the biomechanical lifting
equivalents and the maximum compressive forces on the
upper surface of L-5 127
33. Relationship between the biomechanical lifting
equivalents and the maximum compressive forces on the
upper surface of L-4 128

Xll
 CHAPTER I
INTRODUCTION
Man 1s an expensive means of transport, for, in manual carrying,
the entire muscles of the body are brought into action. Since the
efficiency of the human motor system is rather low, this makes the
human machine an expensive source of energy. Mechanical devices are
considerably cheaper to operate in terms of overall costs.
Although manual handling of loads is expensive and inefficient,
it is often necessary. This may be due to the type of task
performed, the space available which does not permit mechanical
handling equipments, or other restrictions which may be present.
Therefore, it is necessary to improve the operator's efficiency and
reduce fatigue and injury during manual handling of loads. The
necessity of manual lifting and the resulting associated problems
dictate a need for studies which are directed towards a better
understanding of the physical stresses imposed on the musculo-
skeletal system during lifting activities.

1.1 SCOPE
Various methods have been used in studies to determine stresses
acting upon parts of the human body in work situations. Among these
are the following:
(1) Mathematical models devised to predict the magnitude of
external forces which cause partial or complete destructior
of the vertebral or supporting structures.

1
 (2) Use of electromyography to determine the role of the back
muscles with respect to posture and motion.
(3) Use of a force platform to measure changes in location of
the center of body mass during simulated tasks.
(4) Measurement of physiological responses such as oxygen
consumption, respiration, and heart rate during physical
work in a controlled atmosphere as an indication of body
stress.
All of these methods have some merit, but in general none is
considered the ultimate solution for measuring the forces acting upon
various parts of the human body during lifting tasks.

1.2 OBJECTIVES
The main objective of this research is the development of a
methodology for analyzing the amount of physical stress imposed on a
person's musculoskeletal system by infrequent material handling tasks.
The methodology is based upon the concept that the physical stress
incurred by the musculoskeletal system can be analyzed by applying
the well-known laws of engineering mechanics to the human body.
A need appears to exist for the formulation of a biomechanical
dynamic model for lifting in the sagittal plane. The proposed
model will:
1. Provide data on the mechanical stresses on the body when
performing lifting task in the sagittal plane.
2. Determine whether a lift can be performed by an individual
 based on his or her maximum isometric strength data.
3. Aid in selecting the proper methods for lifting by
providing the stresses for different motion patterns at
different speeds. It can help in the selection of
individuals for jobs by comparing each person's estimated
maximum lifting capacities with the stresses predicted
from job requirements.
Because the lower lumbar region of the spine is most vulnerable
to injury due to the mechanical stresses being highest in that
region, this study will concentrate upon the stresses in the region
of the last two lumbar discs (L-5/S-1 and L-4/L-5) and the
surrounding vertebral bodies.
A biomechanical model will be developed for non-repetitive,
short duration lifting. That is, the lift is performed only a few
times during a working day and takes no longer than 5 seconds to
complete, Fisher (1967). This constraint is necessary because
fatigue and cardiovascular problems may become the limiting factors
for repetitive or sustained lifting motion even though the mechanical
stresses involved do not appear to be excessive.
A biomechanical model will be developed only for symmetrical
motions performed in the sagittal plane, because balanced loading
without rotation of the spine is recommended during heavy lifting.
 CHAPTER II
REVIEW OF LITERATURE
2.1 ANATOMY OF THE SPINE RELATED TO LIFTING
The vertebral column. Figure 1, Is composed of 33 vertebrae
superimposed on one another in series. Considering the vertebral
column from caudal to cranial end, the lower four vertebrae are
fused to form the rather Inconsequential coccyx, the next five are
fused into the sacrum, and the remaining 24 vertebrae form the
presacral spinal column. These 24 vertebrae are separated each from
its neighbor above and below, by a fibrocartilaginous intervertebral
disc, and they are united by articular capsules and ligaments. This
alternating arrangement of the bony vertebrae with a cartilaginous
articulation provides a flexible supporting column for the trunk and
upper extremities. The 24 presacral vertebrae are described
structurally as seven cervical, twelve thoracic, and five lumbar
vertebrae.
The size of the presacral vertebrae increases from above
downward, and the lumbar spine, containing only five vertebrae and
their associated Intervertebral discs comprises almost one third of
the length of the presacral spine.
The vertebral column has four important functions:
1 - it transmits the weight of the upper parts of the body
down to the pelvic girdle,
2 - it provides a stable central attachment for the bones and
muscles of the trunk and extremities.
 A T L A S . . - ^ ^ '
AXIS ^^-^^
(Epistropheus)
CERVICAL VERTEBRAE-- —

Ji\^
VERTEBRA,
(prominent)-- -.-.«»

12
* rHORACIC VERTEBRAE--

31

9
^-9

m' Tl

f.
iL7^'' 1

• LUMBAR VERTEBRAE - - -<

!K_;
^ Ar<2a of Concern- -
- 5

SACRUM
(5 Pieces)
V5Q
—VJ
COCCYX
(4 Pieced

Fig. 1: Lateral and front view of spine

 3 - it contributes to the posture and to all the movements of
the body, and
4 - it provides a flexible protective tube for the spinal cord
and for the spinal nerves in part of their action.
The anterior part of the spine consists of the vertebral
bodies, intervertebral discs, and cartilaginous end-plates located
between each vertebra and disc. This anterior part of the spine
supports the compressive loads due to the weight of the body and any
weight held in the hands.
Armstrong (1965), stated that, the weight borne by each
succeeding vertebral body increases as proceeding caudally, and the
surface area of the anterior articulations between the vertebral
bodies increases commensurate with the increasing load. The
vertebrae are unharmed by loads within the limits of their tolerance
no matter how long maintained. When the load applied exceeds the
strength of the vertebrae they fracture and give way, the amount of
disintegration depends on how excessive the load.
The intervertebral discs lie between the bodies of the presacral
vertebrae and are the only structures other than bone or cartilage
which habitually transmit the body weight. The intervertebral discs
have three functions:
1 - they help bind the vertebral bodies together,
2 - they form an integral part of the intervertebral joints
which permit the movement between the vertebrae,
3 - they transmit the body weight.
 7
An intervertebral disc is described by Armstrong (1965) as
consisting of three parts:
1 - the cartilage end-plates enclosing it above and below,
2 - the annul us fibrosus, and
3 - the nucleus polposus.
The cartilage end-plates cover the articulating surfaces of the
vertebral bodies central to the bony epiphyseal ring anteriorly
and laterally. Posteriorly they extend to the margin of the vertebral
body. The cartilage end-plates have two functions:
1 - they shield the vertebral bodies from the effects of the
habitual transmission of weight, preventing pressure
absorption by the bone, and
2 - they allow fluid exchange between the vertebral bodies
and the discs.
The annulus fibrosus surrounds the central nucleus pulposus of the
disc and partly envelops it above and below. The annulus fibrosus
is composed of concentrically arranged lamellae of fibers and
fibrocartilage which pass obliquely from the cartilage plate or
epiphyseal ring above to that below. The annulus fibrosus have five
functions:
1 - it binds the vertebral bodies together giving stability
and unity to the spine,
2 - it allows movement between adjacent vertebrae,
3 - it acts as a check ligament,
4 - it retains the nucleus pulposus, and
 8
5 - it acts as a shock-absorbing mechanism.
The nucleus pulposus constitutes less than one fourth of the volume
of the lumbar intervertebral discs. It is composed of a
semi gelatinous material containing collagenous fibers and a matrix
of chondromucoid elements in the form of a gel. The fibers of the
nucleus pulposus sink into the cartilage plates at acute angles and
attach firmly. The water content of the nucleus is high and varies
with age. Although the nucleus pulposus is a gel, it behaves much
like a liquid, for it is essentially incompressible, transmits
forces equally in all directions, and alters its shape under the
influence of pressure. The nucleus pulposus has three functions:

1 - it acts as a fulcrum for movement between vertebrae,

2 - it helps equalize the stresses place on it by transmitting
forces equally in all directions to the entire annulus,
the cartilaginous end-plates, and through them to the
vertebral bodies,
3 - it acts as a shock absorber.
According to Cailliet (1962), the posterior part of the spine
is primarily a guiding mechanism, but also supports shearing forces,
and is where the muscles of the spine insert.
The muscles of the trunk and the abdomen brace the spine and
add to its stability. They produce and control spinal movements,
and partially sustain the mechanical forces acting on the vertebrae
and on the intervertebral joints. Morris et.al. (1961) found that
a force of four and one half pounds would collapse a muscleless
spine. It was verified by Armstrong (1965) and Bartelink (1957)
that the muscles of the spine are very active during lifting.
The abdominal pressure affects the lifting action in a number
of different ways, depending on the speed of the lift, body
orientation during the lift, and the amount of weight lifted. It
is believed that the abdominal pressure serves as a countermeasure
to relieve a portion of the compressive spinal forces. Asmussen
and Poulsen (1968) concluded that in static holding of burdens up
to 50 kg. in a 45° tilted position with straight back and normal
breathing, the abdominal pressure plays only a minor or negligible
role in relieving the back muscles. Fisher (1967), on the other
hand, determined that the total compressive force on the spine during
lifting could be reduced by as much as 25% by the abdominal
pressure. Morris et.al. (1961) followed Bartelink (1957) in
suggesting that the abdominal cavity shares the load with the
lumbar spine. They found that when intra-abdominal pressure is
raised by contraction of the abdominal muscles, the abdominal
cavity becomes rigid and can transmit force on a plane, well in
front of the lumbar spine. Pressure is exerted on the diaphragm
and on the pelvic cage when lifting a load.
 10
2.2 INJURIES DUE TO LIFTING
The United States has not enacted federal legislation concerning
the handling of materials, however, some states have regulations
they have adopted for their own use. Most of the regulations are
concerned with loads handled by women and children. Examples of
which as published by the International Labor Organization (1966)
i nclude:
Alaska: The general safety code, 1949, states that it is
prohibited to employ any woman to lift any weight in excess
of 3S% of her body weight. Where sustained or repetitive
lifting is required, the absolute maximum is 25 pounds.
California: The supplement to the 1947 Labor Law states
that no female employee may be requested or permitted to
lift any object weighing 50 pounds or over. However an
Industrial Commission Regulation prohibiting females from
lifting burdens in excess of 25 pounds takes precedence
over the 50 pounds limitation in the Labor Law. It also
provides that no women shall carry loads weighing more
than 10 pounds up or down any stairway rising more than
five feet from its base.
Georgia: Regulation 59 issued by the Commissioner of Labor
prescribes that the maximum weight for women shall be 30
pounds. This limit is also applicable to young persons
under 18.
Maine: A decision of the Commissioner of Labor prohibited
the carrying of heavy loads in the case of young persons
under 18 in the following Industries: bakeries, food
products, clothing, metal, machinery and foundries, moulded
rubber products, tanning, oil cloth and textiles Industries.
Maryland: The Safety Code for the protection of Industrial
workers in Foundries, 1948, prescribes a maximum weight of
25 pounds for women employed in foundries.
Massachusetts: It is prescribed that all receptacles
weighing with their contents 75 pounds or more which are to
be moved by female employees in any manufacturing or mechanical
establishment must be provided with pulleys or castors.
 11
Michigan: The maximum weight that a woman may carry is
fixed at 35 pounds, and at 20 pounds when the journey
includes ascending or descending stairs.
Minnesota: The regulations applicable to foundries fix at
25 pounds the maximum permissible weight to be carried by
one woman.
Ohio: The maximum weight for women is fixed at 25 pounds.
Utah: The Industrial Commission Welfare Regulations
prescribe that a woman shall not be required to lift a load
weighing more than 30 pounds and shall not carry a load
weighing more than 15 pounds.
Tichauer (1971) stated that, the weight/bulk ratio of the
object handled is a major determinant of the severity of a lifting
task. It was also shown by him that many of the current so-called
"rules for safe lifting" are simply inadequate. Likewise, weight
limitations on lifting legislated are recommended by governmental,
intergovernmental, and private persons and are often irrelevant and
valueless because they do not consider weight/bulk ratio of the
object, nor do they consider the protection of individuals with
pre-existing impairments.

Brown (1970) reported that the handling of materials has for

long been the source of injury to workmen in industry. Jones (1971)
reports that more than 20% of the compensation dollar in Ontario is
often spent on lower back pain. Eastman Kodak employees experience
approximately four hours per man per year in lost time from work, due
to lifting. Sixty to sixty five percent of the Swedish population
experience lower back pain. And it is estimated that 56% of the
long-term industrial employees in the United States encounter lower
back pain.
 12
The incidence of lower back pain among different occupations
is listed by Magora and Taustein (1969) to be as follows:
Heavy Industry Workers 21.6%
Nurses 16.8
Farmers 14.5
Light Industry Workers 14.1
Bus Drivers 11.9
Post Office Clerks 10.1
Bank Clerks 10.1
Policemen 6.4
It has been reported by Troup (1965) that a relatively large
portion of industrial injuries (as great as 12%) are back disorders
resulting from a lifting task. Fisher (1967) indicates that disc
degeneration is dependent on the magnitude of day-to-day forces to
which it is subjected. This, translated, means that people associated
with heavy work would be more susceptible to disc injury.
Studies by Armstrong (1965); Morris et.al. (1961); Munchinger
(1962) and Roaf (1960) show that injuries to the spine due to
lifting seems to be concentrated on the lower vertebral bodies and
intervertebral discs.
There is a general agreement by Armstrong (1965); Morris et.al.
(1961), Nachemson (1962); Perey (1957) and Roaf (1960), that excessive
compressive forces on the spine will cause end-plate fractures of
the vertebral bodies. Roaf (1960) verified that hyperflexion and
hypertension also lead to injury of the vertebral body before any
 13
other part of the spine is injured Perey (1957) stated that heavy
lifting can cause vertebral fractures. One example was stated by
Perey (1957) concerning a mechanic who lifted a heavy part of an
automobile and suffered severe pain in the lumbar region of his
back. It was later disclosed that the mechanic suffered an anterior
fracture of the vertebral body. These facts suggest that knowledge
of compressive forces on the vertebral bodies during lifting would
be quite valuable to the lifter.
The types of disc injuries are degeneration, protrusion, and
herniation. Degeneration means that the disc is disintegrating and
losing its mechanical properties: the ability of the nucleus to
transmit pressure equally in all directions and the ability of the
annulus to sustain the pressure. Gordon (1961) and Hirsh (1951)
stated that degeneration of the disc is correlated with age and
starts at age twenty or sooner. The decrease in water content is
a part of the degeneration process. Perey (1957) and Gordon (1961)
believe that degeneration is also dependent on the magnitude of day
to day forces put on the discs.
Armstrong (1965) believes that a disc herniation is a three
stage cycle that usually takes months to complete. The first stage
is a softening of the nucleus and part of the annulus which starts
as a result of degeneration or a trauma that weakens some of the
fibers. The second stage is a displacement of the nucleus through
the annulus. This displacement causes a protrusion and if the
displacement is completely through the annulus a herniation (rupture)
 14
is said to have taken place. The third stage is a formation of
fibrosus tissue or a healing process. Once these changes occur the
disc never completely recovers.
The nucleus pulposus may herniate through the annulus fibrosus.
This type of herniation occurs most commonly in the fourth and
fifth lumbar intervertebral discs. In addition, herniation of the
fourth and fifth lumbar discs is most frequently in a posterior or
posterior-lateral direction. That the fourth and fifth lumbar discs
are the most common sites of herniation of the nucleus through the
annulus is amply demonstrated by the following studies reported in
the literature.
Barr and Mixter (1941), with a consecutive series of patients
having herniated discs, observed:
2 patients with herniated 3rd lumbar discs
81 patients with herniated 4th lumbar discs
57 patients with herniated 5th lumbar discs
1 patient with herniated 1st sacral disc
Smith, et.al., (1944) reported a similar series in which
they found:
1 patient with herniated 3rd lumbar disc
62 patients with herniated 4th lumbar discs
40 patients with herniated 5th lumbar discs
Eckert and Decker (1947) reported that in a series of surgical
procedures which they performed, 97% of the discs removed were from
the L-4 level.
 15

Morris et.al. (1961) found that vertebral failure occurred in

subjects under 40 years of age in the range from 1,000 to 1,710
pounds compressive force and as low as 300 pounds for older subjects.
Perey (1957) notes in elderly people the breaking point of the
lumbo-sacral discs has been found to lie around 1,000 pounds and
around 1,760 pounds in people younger than 40 years.
Almost all the data on strength of the back consists of load
limits on the vertebral bodies and intervertebral discs. The results
presented in Table 1. are all compression tests on vertebral bodies.
Usually the procedure was to use two adjoining vertebrae from a
cadaver with the intervertebral disc and compress to fracture.
Sometimes more than two vertebral bodies were used. The parts that
fracture in these tests are the vertebral end-plates.
The strength of the vertebral bodies vary from about 125 to
1,200 kg. This range is due to age, sex, location of the vertebrae,
differences due to measuring procedures, and, individual differences.
The best available data is by Sonoda (1962), of 730 kg. for
the lumbar region for males, and 5/6 of that for females.
The data on intervertebral discs Is not as extensive as on the
vertebral bodies, because compression tests are usually made on
vertebra-disc-vertebra which is compressed until one component
yields, which is usually the end-plate.
 16

TABLE (1)

MAXIMUM VALUES OF COMPRESSIVE FORCES FOR VERTEBRAL END-PLATE (kg.)

LOCATION AND AUTHOR AGE

Under 40 40-50 50-60 Over 60

1 - SONODA (1962)

cervical 418 326 326 205

upper thoracic 370 288 288 181

middle thoracic 431 336 336 211

lower thoracic 644 502 502 315

1umbar 730 570 570 358

2 - EVANS (1966)

L-1 450 354 351

L-2 444 346 335

L-3 497 374 463

L-4 384 364 379

L-5 389 482 200

3 - PEREY (1957)

L-1 522 522 546 283

L-2 570 570 652 261

L-3 575 575 697 248

L-4 583 583 723 540

L-5 685 685 527 238

 17
A study by Sonoda (1962), who tested only discs, gave the
following results for the 40 to 60 group:
Cervical 320 kg.
Upper thoracic 450
Lower thoracic 1,150
Lumbar 1,500
Chaffin (1969) found that the magnitude of shearing forces on
all the segmental levels of lumbar spine, never reach greater
than 50 kg.

2.3 BIOMECHANICS AND LIFTING EXPERIMENTS

Fundamental to the understanding of man's movements is the
realization that he is a living organism and that his structure and
behavior have been shaped by his anthropological ancestors. The
first systematic observations on human movements were made by
Leonardo da Vinci (1452-1519), and appeared in his "Notes on the
Human Body". During the Renaissance, Galileo and Newton established
the experimental and theoretical bases for the analysis of movements.
It was Borelli (1608-1697), one of Galileo's students, who combined
the sciences of mathematics, physics, and anatomy in the first treatise
dealing with biomechanics. "De Motu Animal1 urn" was published in 1680,
and evoked sufficient interest for a second edition in 1710. Bernoulli,
Euler, and Coulomb tried to develop a rational mathematical formula
to express the maximum and optimum capacity of human work as a function
of the acting force, velocity, and duration of the activity.
 18
Biomechanics fuses the engineer's knowledge of analytical
methods of structures, and functions of components with the biologist's
and physiologist's knowledge of the human body. It investigates the
effects of Internal and external forces on human and animal bodies,
in motion and at rest. Therefore, it is the application of engineer-
ing principles to biological and physiological systems. The primary
objective of biomechanics is to increase the efficiency of human
performance by minimizing the effort required to perform necessary
motor activities, and to reduce body injuries.
The traditional approach for evaluating biomechanical forces,
torques, energy and work has been to isolate these body segments
which are Involved, to estimate their mass and location of mass
center and to make the necessary assumptions concerning the body
articulations which will allow analysis according to the laws of
mechanics. For static analysis one needs only to measure the body
orientation and the external forces acting upon it. For dynamic
analysis, however, the inertial characteristics of the body in
motion create forces which are an integral part of the total kine-
matic system. These forces can be ascertained only be studying the
body locations with respect to time.
It has been 84 years since Braune and Fischer (1890) published
their data regarding the mass distribution for the various body
segments. Since then, fundamental extensions by Dempster (1955)
and Drillis et.al. (1966) have resulted in better estimates of:
(1) the location of the mass centers-of-gravity, (2) the link
 19
lengths, and (3) the magnitudes of the moments-of-inertia of the
various body segments.
However, it was not until the widespread use of the commercial
high speed digital computers that this type of data could be
easily used in developing analytical methods to study the mechanics
of the human body. The digital computer has provided a computational
capacity which, in turn, has fostered the development of several
different types of biomechanical models. Some of these models have
been formulated to determine the whole-body center-of-gravity
location when the body is placed in various configurations (Hanovan,
1964). This type of model is used in determining body movements if
mechanically unrestrained and acted upon by changing momentum or
gravitational forces, such as in a vehicle accident or when falling.
Another type of computerized biomechanical model primarily
estimates the forces and torques at various articulations of the body
during voluntary actions (e.g. lifting, running, or throwing). An
early example of this type is a two-link model of the arm developed
by Pearson et.al. (1961). The intent of this model is to compute
the forces and torques at the elbow and shoulder during a sagittal
plane motion of the arm-forearm-hand aggregate. Figure 2. is an
illustration of the mechanical analogue of the arm used in this type
of model. Stroboscopic photographs of the various arm motions
of interest are taken to determine the instantaneous positions,
velocities, and accelerations of the arm segments. This "activity"
data along with the anthropometric dimensions of each segment's
 20

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 21
length and weight, provides enough input information to compute
the stress levels at the elbow, and shoulder, thus providing a means
to achieve a better understanding of both the complex muscle
actions required for control of the arm, and the resulting strain
at the articulations.

An extension of the Pearson Arm Model was developed by

Plagenhoef (1966). Again, photographic data is used to describe
the body configurations during the relevant task. This spatial
information, combined with additional anthropometric dimensions
regarding the length of arm, trunk, and leg segments and the total
weight of the subject, provide adequate Information to compute the
forces and torques at the elbow, shoulder, hip, and knee, during
various physical activities performed in the sagittal plane.
The SSP (Static Sagittal Plane) model was introduced by
Chaffin (1969). The efforts of this author have been to extend
the biomechanical model of Plagenhoef to include:
(1) an estimate of the stress in the lower lumbar spine,
(2) the addition of external loads on the hands (e.g. in a
materials handling task), and
(3) an evaluation of the effects of various muscle group
strengths on human performance.
As the name infers, this particular model has been developed
to evaluate various "static" situations, such as when one is
holding a weight, or pushing or pulling on a non-moving container.
In addition to these applications, the model can be used to
 22
analyze "slow" moves by formulating the input data to describe a
sequence of static positions with very small changes in each
successive position. In making this type of pseudo-dynamic
analysis, it must also be assumed that the effects of acceleration
and momentum are negligible, which is not Intrinsic.
Essentially, two differences occur between the static and
dynamic activities. First, in the dynamic case the external forces
exerted on the segments of the body must include, in addition to
gravity, the effects of inertia. Therefore, the forces acting on
each segment do not necessarily act in the same direction. This
requires vector addition of the forces, which is not required in the
static lifting case where all forces act in a parallel direction to
gravity. Second, the additional requirement to transform the
static model into a dynamic one is the inclusion of an inertial
torque at the center of gravity of each segment. This characterizes
the force created by rotational acceleration of the segments about
their gravity centers.
Lifting areas which have been investigated experimentally
include the maximum weight of lift, maximum frequency of lift,
rhythm lifting, and maximum acceptable weights of lift for various
heights.
Grandjean (1969) indicated that, to avoid accidents, it is
advisable when carrying loads not to exceed the maximum weights
shown in Table 2.
McFarland (1969) stated, in general, objects become "heavy" at
 23

TABLE (2)
MAXIMUM WEIGHTS (Kg) ADVISABLE WHEN CARRYING LOADS

ADULTS YOUTHS
Men 1Women Boys 1 Girls
Occasional Lifts 50 20 20 15
Frequent or Continuous Lifts 18 12 11-16 7-11

about 35% of body weight. Cathcart et.al. (1967) believe a man

should not lift more than 50% of body weight occasionally and 40%
continuously. Some of the variables one must consider in limiting
the amount of weight which can be safely lifted include age, lifting
configuration, size and shape of the load, weight of the lifter,
the back strength (usually assumed the limiting factor) of the
lifter, etc. Poulsen (1970) Investigated the maximum weight that
should be lifted for two different height ranges. These were from
the floor to table height and from the table height to a shelf
placed at head height.
Van Wely (1961) reported that it appears, as a result of his
investigation, to be statistically significant that the physiological
efficiency not only depends on the weight of the loads but also on
the number of lifts per unit time. Snook and Irvine (1968) performed
an experiment to determine the most acceptable frequencies of lift
for two different weights (35 and 50 pounds) and three different
lifting ranges (floor to knuckle height, knuckle height to shoulder
 24
height, and shoulder height to full arm reach). Eight subjects
(male industrial employees) were asked to lift as frequently as
they could without becoming overheated or tired. The results
indicated that, in general, the subjects preferred to work the
fastest for the knuckle to shoulder height, followed by the shoulder
to arm reach range, with the slowest being the floor to knuckle
height range.
The effects of rhythm lifting have been studied by Ronnholm
(1962) and Ronnholm et.al. (1962). The results of these
experiments showed definitely that lifting performed with a certain
rhythm was much more economical. It also was concluded that, with
a suitable rhythm, it is possible to considerably improve the
mechanical efficiency.
Snook and Irvine (1967) performed an experiment to determine
the maximum acceptable weight of lift using nine male Industrial
employees as their subjects. The subjects were instructed to adjust
their workload to the maximum amount that they could perform
without strain or discomfort and without becoming tired, weakened,
overheated, or out of breath. The frequency of lift was set at four
lifts per minute for trials varying in length from five to thirty
minutes. The results of this experiment are shown in Table 3.
McDaniel (1972) performed an experiment that resulted in the
formulation of a predictive model for the maximum acceptable weights
of lift for males, females, and a model based on both sexes.
Fifteen male subjects and fifteen female subjects were used in the
 25

TABLE (3)
MAXIMUM ACCEPTABLE WEIGHT OF LIFT FOR MALES (Lbs)

Lift Range % of Population

90 75 50 25 10

Floor Level to 52 59 66 73 80
Knuckle Height

Knuckle Height to 51 56 62 68 73
Shoulder Height

Shoulder Height to 48 43 60 67 72
Arm Reach

formulation of the model, and then it was tested using five

additional males and females. Both male and female subjects were
students. Physiological and anthropometric variables were
considered. These variables were subjected to evaluation with a
stepwise regression technique. The results indicated that the
prediction equalling 14.9% for the combined male-female model,
8.78% for the male model, and 6.83% for the female model. It was
also concluded that males and females should not be used together
in the same regression model when predicting acceptable weight of
lift.
Dryden (1973) performed an experiment to predict the maximum
permissible weight of lift from knuckle height to shoulder height.
The subjects were recruited from employment in which lifting
constitutes a portion of the assigned task. It was also the
objective of that research to determine the effects of body
 26
composition on the permissible weight of lift. The model when
tested generated an average error ratio of prediction of 15.4%.
 CHAPTER III
EQUIPMENT AND METHODOLOGY
The objective of industrial "job Improvement" traditionally
has been economically oriented. Thus when one speaks of "Increasing
the efficiency", he is usually referring to raising the output-to-
input ratio of a person or productive employment. In a manual job
this means less man-hours per part produced, or improved quality per
man-hour. The motivation to achieve greater efficiency in a job
is based on time and motion data analysis by engineers and economists,
especially the more time consuming or highly repetitious jobs.
Time is more closely scheduled and required motions are reduced, not
to a minimum but to an optimum. Because of this allocation of
effort, little or no attention has been given to the types of work,
i.e., elements that occur infrequently in a job. Yet when these
non-repetitious elements cause a high amount of physical stress,
they often lead to lost time, which increases indirect costs and
thus decreases the total economic efficiency of the organization.

3.1 THE EXPERIMENT AND EQUIPMENT

Tichauer (1971) introduced the term "Biomechanical Lifting
Equivalent". This term is defined as the moment in pound inches
which would be applied to the lumbar spine if the load were held
stationary, touching the abdomen with the trunk in erect position.
The magnitude of the biomechanical lifting equivalent is calculated
by Tichauer as follows:
M = (8 + L/2) (W) (1)
27
 28
where:

M = the biomechanical lifting equivalent, lbs. in.

8 = approximate distance in inches from the joints of
lumbar spine to front of abdomen.
L = length in inches of one side of a cube of uniform
density lifted during the standard task.
W = the weight in pounds of the cube handled.
Four male volunteers, graduate students, served as experimental
subjects. The anthropometric data of the subjects are shown in
Table 4.

TABLE (4)
ANTHROPOMETRIC DATA FOR THE EXPERIMENTAL SUBJECTS

Subject Total Total Distance in cm. from

no. weight height
(cm) Ankle Knee Hip Shoulder Elbow
(kg) to to to to to
Knee Hip Shoulder Elbow Wrist
1 61.4 179 47 53 42 30 27
2 70.2 171 44 47 42 30 28
3 84.1 183 48 53 43 32 30
4 64.8 166 40 42 47 28 25
 29
Each subject was instructed to do two types of lift, leg lift
and back lift. Three load levels and three boxes with different
sizes were used with each type of lift. Therefore, with each type
of lift for each subject nine biomechanical lifting equivalents
were used.
EQUIPMENT:
Six wheat sized light bulbs were attached to the subject's
articulations: ankle, knee, hip, shoulder, elbow and wrist. All the
light bulbs were on during the experimental lifts.
A 16-mm movie camera with black and white film (4-S Reversal
Film 7277) was used to record the paths of motion of the six
articulations during the lifting tasks. Figure 3. shows the paths
of motion. The movie camera was attached to a control device to
keep its speed constant during the experimental time, which was -5
1000 frames per minute. Figure 4. shows the movie camera and the
control device. During recording the paths of motion the room light
was off and a black background was utilized.
The analysis of the movie film was performed using a projector
that projects frame by frame, elevated above a drawing table, and
focused on drawing paper, as shown in Figure 5.

3.2 INDEPENDENT AND DEPENDENT VARIABLES

Three independent variables are considered:
1 - The subjects (4 subjects)
2 - The methods of lift (2 methods): leg and back lifts
 30

Fig. 3 : Motion patterns for the different

articulations during the lifting task
 31

Fig. 4 : The movie camera and the speed control

device used during the experiment
 32

Fig. 5 : The projector used in the analysis

of the movie films
 33
3 - Weight/bulk ratios:
Three load levels are used, 10, 20 and 30 pounds. These levels
are chosen on the basis of the mean weight of objects commonly handled
in the activities of daily life and in light and moderate work in
Industry as shown by Tichauer (1971). Three 1/4" plywood boxes
were constructed to the following dimensions:
1. 12" X 12" X 6"
2. 18" X 12" X 6"
3. 24" X 12" X 6:

For the three boxes, two dimensions are fixed, while the third one
is variable. The variable dimensions determine the moment arms for
the biomechanical lifting equivalents. The moment arm for the
first box is 14" (8 + 12/2), 17" (8 + 18/2) for the second box, and
20" (8 + 24/2) for the third box. Therefore, nine different
weight/bulk ratios are used with nine different biomechanical
lifting equivalents. The values of the biomechanical lifting
equivalent corresponding to each weight/bulk ratio are shown in
Table 5. The range of the biomechanical lifting equivalents is
from 140 to 600 pound inches.
Two repetitions were recorded for each biomechanical lifting
equivalent, making a total of 144 experimental lifting cycles;
(4 subjects X 2 methods of lift X 9 biomechanical lifting equivalents
X 2 repetitions).
The dependent variables are the maximum compressive and shearing
forces on:
 34
1 - upper surface of S-1
2 - lower surface of L-5
3 - upper surface of L-5
4 - lower surface of L-4

TABLE (5)
VALUES OF THE BIOMECHANICAL LIFTING EQUIVALENTS, POUND-INCHES

Moment items, inches Load Levels, pounds

10 20 30
14 140 280 420
17 170 340 510
20 200 400 600

3.3 MATHEMATICAL DEVELOPMENT OF THE BIOMECHANICAL DYNAMIC MODEL

Assumptions fundamental to model development are:
1 - The human body is made up of rigid links joined at its
articulation points (wrists, elbows, shoulders, hips, knees
and ankles).
2 - It was reported by Pearson (1961) that the semiprone hand and
the lower arm remain aligned during motions which involve the
whole arm in the sagittal plane, therefore, the lower arm and
hand will be treated as one part in motion.
3 - The human body is considered to be made up of six links, which

are:
a. hand b. lower arm c. upper arm

d. trunk e. upper leg f. lower leg

 35
4 - The ankle is assumed to remain in a fixed planar position, which
provides a constant positional reference.
The following steps are used to develop the model:
3.3.A - Determination of angular displacement, velocity and
acceleration.
3.3.B - Computation of linear acceleration at:
1 - The center of gravity of lower leg due to rotation
about ankle joint.
2 - The knee joint due to rotation about ankle joint.
3 - The center of gravity of upper leg due to rotation
about knee and ankle joints.
4 - The hip joint due to rotation about knee and ankle
joints.
5 - The center of gravity of trunk due to rotation about
hip, knee and ankle joints.
6 - The shoulder joint due to rotation about hip, knee
and ankle joints.
7 - The center of gravity of upper arm due rotation about
shoulder, hip, knee and ankle joints.
8 - The elbow joint due to rotation about shoulder, hip,
knee and ankle joints.
9 - The center of gravity of lower arm due to rotation
about elbow, shoulder, hip, knee and ankle joints.
10 - The wrist joint due rotation about elbow, shoulder,
hip, knee and ankle joints.
 36

11 - The center of gravity of the hand due to rotation about

wrist, elbow, shoulder, hip, knee and ankle joints.
3.3.C - Calculations of reactive forces and reactive torques at
body articulations:
1 - The wrist joint.
2 - The elbow joint.
3 - The shoulder joint.
4 - The hip joint.
5 - The knee joint.
6 - The ankle joint.
3.3.D - Calculations of reactive forces and reactive torques at
the centers of the discs:
1 - general calculations of trunk links dimensions
2 - calculation of spinal angles
3 - calculation of abdominal pressure, force and torque
4 - calculation of reactive forces and reactive torques
at the center of L-4/L-5 disc
5 - calculation of reactive forces and reactive torques
at the center 6f L-5/S-1 disc
3.3.E - Calculation of Compressive and shearing forces on:
1 - The upper surface of S-1
2 - The lower surface of L-5
3 - The upper surface of L-5
4 - The lower surface of L-4
 37
3.3.A - DETERMINATION OF ANGULAR DISPLACEMENTS,
VELOCITIES AND ACCELERATIONS
The first phase In the development of the biomechanical
dynamic model requires the resolution of the Instantaneous
angular displacements, velocities and accelerations for each
articulation.
Pearson (1961) used a film analysis to determine discrete
angular displacements at the shoulder and elbow during arm motion
and then computed the angular velocities and angular
accelerations using repeated finite differencing of a Taylor series
of the discrete displacement data. The same technique was attempted
by Chaffin (1967) but due to the photographic technique employed,
which limited the minimum time Intervals to 0.05 seconds, an
unacceptable "round off" error was found with the acceleration
patterns.
SIote and Stone (1963) used interrupted light photographs of
a discrete forearm flexion. It was determined that the space-time
relationship for this motion could be best expressed by the
following equation:

D-, = e ^ [ 2 f i - -sin ?f^] (2)

where:
D{ - angular displacement, radians
Dmax = maximum angular displacement, radians
T = displacement period, seconds
tl = time, seconds
Mathematically, the expression for angular velocity (V^) and
 38

angular acceleration (Ai*) can be derived from equation 2 by taking

the first and second derivatives with respect to time ti.
This result:
Vi- = Dmax |- ^ _ ^^^ 2 ^ T ^ ^3^

Ai = M m a x 3 - ^ 2 ^ 1 ^^j
T2"
Figure 6. presents the basic kinematic equations of a discrete
forearm flexion using the values of T = 0.32 seconds and Dmax =2.32
radians obtained from the actual experimental data of SIote and
Stone. The angular velocity increases from zero to a maximum and
then decreases to zero. A period of acceleration and finally a
period of deceleration as the forearm is brought to rest correspond
to these velocity changes.
SIote and Stone space-time relationship is used in the
development of the model and it is valid for the determination of
angular displacement for each limb's motion. This will be discussed
in "APPENDIX C". Equation (3) is used for calculation of angular
velocities and equation (4) is used for the calculation of angular
accelerations of each limb's motion.

3.3.B - COMPUTATION OF LINEAR ACCELERATIONS

Lifting tasks include circular motions in which the radius of
rotation is constant. Figure 7. The analysis of such motions is
performed as follows:
Let: a = angular displacement, radians.
 39

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c
s.
OJ
+J
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o o
4->
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m CVJ
 40

Fig. 7 : Analysis of circular motions

+ HORIZONTAL
AXIS

<
Ul
y X.
H <
UJ
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Fig. 8 : ACCELERATION NOTATIONS

 41
a = angular velocity, radians/second.
a = angular acceleration, radians/second^.
R = radius of curvature, cm.
Therefore,
N = R (a)2 (5)
T = R (a) (6)
where N and T are the normal and tangential components of the
instantaneous acceleration, cm/sec?^, respectively.
Furthermore, each component of the instantaneous acceleration
can be resolved in the horizontal and vertical planes, thus,
N/X = + N cos (a)
= + R (a)2 cos (a) (7)

N/Y = - N sin (a)

= - R (a)2 sin (a) (8)

T/X = + T sin (a)

= + R (a) sin (a) (9)

T/Y = + T cos (a)

= + R (a) cos (a) (10)

where the (+) and (-) signs describe the direction of acceleration
in the horizontal and vertical planes according to the notations of
Figure 8.
The total acceleration in the horizontal and vertical planes
can be expressed as:
X = + R [(a) sin (a) + (a)2 cos (a)] (11)

Y = - R [(a) cos (a) -'(a)2 sin (a)] (12)

4-
 \2

1 - Linear acceleration at the center of gravity of lower leg due

to rotation about ankle joint.
Let:
(Ul)-i = angular displacement In radians at a time interval 1 of any
point on the lower leg due to its rotation about ankle joint.
This angle is measured counter-clockwise from the negative
horizontal axis to the lower leg. Figure 9.
(Ul)i = angular velocity in radians per second at a time interval i
of any point on the lower leg due to its rotation about
the ankle joint.
(Ul)i = angular acceleration In radians per second squared at a
time interval 1 of any point on the lower leg due to its
rotation about the ankle joint,
r-j = distance in centimeters between the ankle joint and the
center of gravity of lower leg.
Ni (LL) and Ti (LL) = normal and tangential components of
acceleration in cm. per second squared, acting at the center
of gravity of lower leg, at a time interval i, due to
rotation about the ankle joint respectively.
Ni (LL)/X and Ni (LL)/Y = horizontal and vertical components of
Ni (LL), respectively.
Ti (LL)/X and Ti (LL)/Y = horizontal and vertical components of
Ti (LL), respectively.
Xi (LL) and Yi (LL) = horizontal and vertical components of linear
acceleration in cm. per second squared, acting at the center
 43

A = ankle joint
K = knee joint

Fig. 9 : The acceleration components at the

center of gravity of the lower leg
 44

of gravity of lower leg, at a time interval i, due to

rotation about the ankle joint, respectively.
Therefore,

Ti (LL) = r^ (iiDi (13)

Ni (LL) = r^ (Ul)i2 (14)

Ti (LL)/X = Ti (LL) sin (Ul)i

= r (Ul)i sin (Ul)i (15)

Ti (LL)/Y = Ti (LL) cos (Ul)i

= + r^ {Ul)i cos(Ul)i (16)

Ni (LL)/X = + Ni (LL) cos (Ul)i .

= + r^ (Ul)i2 sin (Ul)i (17)

Ni (LL)/Y = - Ni (LL) sin (Ul)i

= - r^ (Ul)i2 sin (Ul)i (18)

Xi (LL) = Ti (LL)/X + Ni (LL)/X

= + ri [(Ul)i sin (Ul)i
+ (Ul)i2 cos (Ul)i] (19)

Yi (LL) = + rj [(Ul)i cos (Ul)i

- (UDi^ sin (Ul)i] (20)
2. Linear acceleration at the knee joint due to rotation about
the ankle joint:
Let:
Rj = distance in cm. between the ankle and the knee joints.
Xi (K) and Yi (K) = horizontal and vertical components respectively
of linear acceleration in cm. per second squared acting at the
knee joint, at a time interval i, due to rotation about the ankle
joint.
 45
Therefore,
Xi (K) = + R^ [(Ul)i sin (Ul)i
+ (UDi^ cos (Ul)i] (21)

Yi (K) = + R^ [(iiDi cos (Ul)i

- (Ul)i2 sin (Ul)i] (22)

3. Linear acceleration at the center of gravity of upper leg due

to its rotation about the knee and ankle joints:
Let:
(U2)i = angular displacement in radians at a time interval i of
any point on the upper leg due to its rotation about the
knee joint. This angle is measured from the upper leg
axis to the lower leg axis, Figure 10.
(U2)i = angular velocity in radians per second at a time interval i
of any point on the upper leg due to its rotation about
the knee joint.
(U2)i = angular acceleration in radians per second squared at a
time interval 1 of any point on the upper leg due to its
rotation about the knee joint.
(Zl)i = angle in radians at a time interval 1 between the positive
horizontal axis and the upper leg.
r^ = distance in cm. between the knee joint and the center of
gravity of the upper leg.
Ni (UL) and Ti (UL) = normal and tangential components
respectively of acceleration in cm. per second squared
acting at the center of gravity of the upper leg, at a
 46

A = ankle joint
K = knee joint
H = hip joint

Fig. 10 : The acceleration components at the

center of gravity of the upper leg
 47
time interval 1, due to rotation about the knee joint,
respectively.
XXi (UL) and YYi (UL) = horizontal and vertical components of
linear acceleration in cm. per second squared, acting at
the center of gravity of the upper leg, at a time interval
1, due to rotation about the knee and ankle joints,
respectively.
Therefore,

(Zl)i = (U2)i - (Ul)i (23)

Ti (UL) = r2 (U2)i (24)

Ni (UL) = r2 (U2)^ (25)

Ti (UL)/X = - Ti (UL) sin (Zl)i

= - r2 (U2)i sin (Zl)i (26)

Ti (UL)/Y = + Ti (UL) cos (Zl)i

= + r2 (U2)i cos (Zl)i (27)

Ni (UL)/X = - Ni (UL) cos (Zl)i

= - r2 (U2)i2 cos (Zl)i (28)

Ni (UL)/Y = - Ni (UL) sin (Zl)i

= - r2 (U2)i2 sin (Zl)i (29)

Xi (UL) = Ti (UL)/X + Ni (UL)/X

= - r2[(U2)i sin (Zl)i
+ (U2)i2 cos (Zl)i] (30)

Yi (UL) = Ti (UL)/Y + Ni (UL)/Y

= + r2 [(U2)i cos (Zl)i
- (U2)i2 sin (Zl)i] (31)
 48

XXi (UL) = Xi (UL) + Xi (K) (32)

YYi (UL) = Yi (UL) + Yi (K) (33)
^- Linear acceleration of the hip joint due to its rotation about
the knee and ankle joints:
Let:

R2 = distance in cm. between the knee and hip joints.

Xi (H) and Yi (H) = horizontal and vertical components respectively
of the linear acceleration in cm. per second squared, acting
at the hip joint, at a time Interval i, due to its rotation
about the knee joint.
XXi (H) 3ind YYi W = horizontal and vertical components
respectively of the linear acceleration in cm. per second
squared, acting at the hip joint, at a time interval i, due
to its rotation about the knee and ankle joints.
Therefore,

Xi (H) = - R2 [(U2)i sin (Zl)i + (U2)i2 cos (Zl)i]....(34)

Yi (H) = + R2 [(U2)i cos (Zl)i - (U2)i2 sin (Zl)i]....(35)

XXi (H) = Xi (H) + Xi (K) (36)

YYi (H) = Yi (H) + Yi (K) (37)

5. Linear acceleration at the center of gravity of trunk due to

its rotation about the hip, knee and ankle joints:
Let:
(U3).j = angular displacement in radians at a time interval i of
any point on the trunk due to its rotation about the hip
 49

joint. This angle Is measured from the trunk axis to the

upper leg axis as shown In Figure 11.
(U3)i = angular velocity in radians per second at a time interval
1 of any point on the trunk due to its rotation about the
hip joint.
(U3)i = angular acceleration in radians per second squared at a
time interval 1 of any point on the trunk due to its
rotation about the hip joint.
(Z2)i = angle in radians, at a time Interval 1, between the
negative horizontal axis and the trunk,
ra = distance in cm. between the hip joint and the center of
gravity of trunk.
Ni (TK) and Ti (TK) = normal and tangential components respectively
of the acceleration in cm. per second squared acting at
the center of gravity of the trunk, at a time interval i,
due to Its rotation about the hip joint.
Ni (TK)/X and Ni (TK)/Y = horizontal and vertical components of
Ni (TK), respectively.
Ti (TK)/X and Ti (TK)/Y = horizontal and vertical components of
Ti (TK), respectively.
Xi (TK) and Yi (TK) = horizontal and vertical components of linear
acceleration in cm. per second squared, acting at the
center of gravity of trunk, at a time Interval 1, due to
its rotation about the hip joint, respectively.
XXi (TK) and YYi (TK) = horizontal and vertical components
 50

K = knee joint
H = hip joint
S = shoulder joint

Fig. 11 : The acceleration components at the

center of gravity of the trunk
 51

respectively of the linear acceleration in cm. per second

squared, acting at the center of gravity of the trunk, at
a time Interval i, due to its rotation about the hip, knee
and ankle joints.
Therefore,
(Z2)i = (U3)i - (Z1)i (38)

Ti (TK) = r3 (U3)i (39)

Ni (TK) = ra (U3)i2 (40)

Ti (TK)/X = + Ti (TK) sin (Z2)i

= + rg (U3)i sin (Z2)i (41)

Ti (TK)/Y = + Ti (TK) cos (Z2)i

= + rg (U3)i cos (Z2)i (42)

Ni (TK)/X = + Ni (TK) cos (Z2)i

= + ra (U3)i2 cos (Z2)i (43)

Ni (TK)/Y = - Ni (TK) sin (Z2)i

= - ra (U3)i2 sin (Z2)i (44)

Xi (TK) = Ti (TK)/X + Ni (TK)/X

= + ra [(U3)i sin (Z2)i
+ (U3)i2 cos (Z2)i] (45)

Yi (TK) = Ti (TK)/Y + Ni (TK)/Y

= + ra [(U3)i cos (Z2)i

- (U3)i2 sin (Z2)i] (46)

XXi (TK) = Xi (TK) + XXi (H) (47)

YYi (TK) = Yi (TK) + YYi (H) (48)

 52

6. Linear acceleration at shoulder joint due to rotation about hip,

knee and ankle joints:
Let:
Ra = distance in cm. between the hip and shoulder joints.
X.J (S) and Yi (S) = horizontal and vertical components
respectively of the linear acceleration in cm. per second
squared, at a time Interval 1, acting at the shoulder joint,
due to its rotation about the hip joint.
XXi (S) and YYi (S) = horizontal and vertical components
respectively of the linear acceleration in cm. per second
squared, at a time interval 1, acting at the shoulder joint,
due to its rotation about the hip, knee and ankle joints.
Therefore,
Xi (S) = Ra [(U3)i sin (Z2)i + (IJ3)i2 cos (Z2)i].... (49)

Yi (S) = Ra [(U3)i cos (Z2)i - m)i^ sin (Z2)i].... (50)

XXi (S) = Xi (S) + XXi (H) (51)

YYi (S) = Yi (S) + YYi (H) (^^^

7. Linear acceleration at the center of gravity of the upper arm

due to its rotation about shoulder, hip and ankle joints:
Let:
(U4)i = angular displacement in radians, at a time interval i, of
any point on the upper arm, due to its rotation about the
shoulder joint. This angle is measured from the upper
arm axis to the trunk axis as shown in Figure 12.
 53
(UA)i

H = hip joint
S = shoulder joint
H
E = elbow joint

Fig. 12 : The acceleration components at the

center of gravity of the upper arm
 54
(U4)i = angular velocity in radians per second at a time interval
1, of any point on the upper arm due to its rotation about
the shoulder joint.
(U4)i = angular acceleration in radians per second squared at a
time interval 1, of any point on the upper arm due to its
rotation about the shoulder joint.
(Z3)i = angle in radians, at a time interval i, between the
negative horizontal axis and the upper arm.
r^^ = distance in cm. between the shoulder joint and the center
of gravity of the upper arm.
Ni (UA) and Ti (UA) = normal and tangential components respectively
of acceleration in cm. per second squared, acting at the
center of gravity of upper arm, at a time interval i, due
to its rotation about the shoulder joint.
Ni (UA)/X and Ni (UA)/Y = horizontal and vertical components of
Ni (UA), respectively.
Ti (UA)/X and Ti (UA)/Y = horizontal and vertical components of
Ti (UA), respectively.
Xi (UA) and Yi (UA) = horizontal and vertical components
respectively of linear acceleration in cm. per second
squared, at a time interval i, acting at the center of
gravity of upper arm, due to its rotation about shoulder
joint.
XXi (UA) and YYi (UA) = horizontal and vertical components
respectively of linear acceleration in cm. per second squared.
 55

at a time interval 1, acting at the center of gravity of

upper arm, due to its rotation about shoulder, hip, knee
and ankle joints.
Therefore,
(Z3)i = (U4)i - (Z2)i - ir (53)

Ti (UA) = ri, (U4)i (54)

Ni (UA) = rn (U4)i2 (55)

Ti (UA)/X = + Ti (UA) sin (Z3)i

= + r4 (U4)i sin (Z3)i (56)

Ti (UA)/Y = - Ti (UA) cos (Z3)i

= - r^ (U4)i cos (Z3)i (57)

Ni (UA)/X = + Ni (UA) cos (Z3)i

= + ri, (U4)i2 cos (Z3)i (58)

Ni (UA)/Y = + Ni (UA) sin (Z3)i

= + ri, (U4)i2 sin (Z3)i (59)

Xi (UA) = Ti (UA)/X + Ni (UA)/X

= + ri, [(U4)i sin (Z3)i
+ (IJ4)i2 cos (Z3)i] (60)

Yi (UA) = Ti (UA)/Y + Ni (UA)/Y

= - ri, [(U4)i cos (Z3)i

- (U4)i2 sin (Z3)i] (61)

XXi (UA) = Xi (UA) + XXi (S) (62)

YYi (UA) = Yi (UA) + YYi (S) (63)

 56

8- Linear acceleration at the elbow joint due to its rotation about

the shoulder, hip, knee and ankle joints:
Let:

R^ = distance in cm. between the shoulder and elbow joints.

Xi (E) and Yi (E) = horizontal and vertical components
respectively of the linear acceleration in cm. per second
squared, at a time interval i, acting at the elbow joint, due
to its rotation about the shoulder joint.
XXi (E) and YYi (E) = horizontal and vertical components
respectively of the linear acceleration in cm. per second
squared, at a time interval 1, acting at the elbow joint,
due to its rotation about the shoulder, hip, knee and ankle
joints.
Therefore,
Xi (E) = + R4 [(U4)i sin (Z3)i + (U4)i2 cos (Z3)i]....(64)

Yi (E) = - Ri, [(U4)i cos (Z3)i - (U4)i2 sin (Z3)i].... (65)

XXi (E) = Xi (E) + XXi (S) (66)

YYi (E) = Yi (E) + YYi (S) (67)

9. Linear acceleration at the center of gravity of the lower arm due

to its rotation about the elbow, shoulder, hip, knee and ankle
joints:
Let:
(U5)i = angular displacement in radians, at a time interval i, of
any point on the lower arm, due to its rotation about the
 57

elbow joint. This angle is measured from the lower arm

axis to the upper arm axis as shown in Figure 13.
m

(U5)i = angular velocity in radians per second, at a time interval

1, of any point on the lower arm, due to its rotation
about the elbow joint.
(U5)i = angular acceleration in radians per second squared, at a
time interval i, of any point on the lower arm, due to
its rotation about the elbow joint.
(Z4)i = angle in radians, at a time interval 1, between the
negative horizontal axis and the lower arm.
rs = distance in cm. between the elbow joint and the center of
gravity of the lower arm.
Ni (LA) and Ti (LA) = normal and tangential components respectively
of the acceleration in cm. per second squared, at a time
interval i, acting at the center of gravity of the lower
arm, due to its rotation about the elbow joint.
Ni (LA)/X and Ti (LA)/Y = horizontal and vertical components of
Ni (LA), respectively.
Ti (LA)/X and Ti (LA)/Y = horizontal and vertical components of
Ti (LA), respectively.
Xi (LA) and Yi (LA) = horizontal and vertical components
respectively of the linear acceleration in cm. per second
squared, at a time interval i, acting at the center of
gravity of the lower arm, due to its rotation about the
elbow joint.
 58

Ti(LA)/Y

S = Shoulder joint
E = Elbow joint
W = Wrist joint

Fig. 13 : The acceleration components at the

center of gravity of the lower arm
 59

XXi (LA) and YYi (LA) = horizontal and vertical components

respectively of the linear acceleration in cm. per second
squared, at a time interval 1, acting at the center of
gravity of the lower arm, due to its rotation about the
elbow, shoulder, hip, knee and ankle joints.
Therefore,

(Z4)i = TT + (Z3)i - (U5)i (68)

Ti (LA) = rs (U5)i (69)

Ni (LA) = rs (U5)i2 (70)

Ti (LA)/X = - Ti (LA) sin (Z4)i

= - rs (U5)i sin (Z4)i (71)

Ti (LA)/Y = + Ti (LA) cos (Z4)i

= + rs (U5)i cos (Z4)i (72)

Ni (LA)/X = + Ni (LA) cos (Z4)i

= + rs (U5)i2 cos (Z4)i (73)

Ni (LA)/Y = + Ni (LA) sin (Z4)i

= + rs (IJ5)i2 sin (Z4)i (74)

Xi (LA) = Ti (LA)/X + Ni (LA)/X

= - rs [(U5)i sin (Z4)i
- (U5)i2 cos (Z4)i] (75)

Yi (LA) = Ti (LA)/Y + Ni (LA)/Y

= + rs [(U5)i cos (Z4)i
+ (IJ5)i2 sin (Z4)i] (76)

XXi (LA) = Xi (LA) + XXi (E)

YYi (LA) = Yi (LA) + YYi (E)

 60
lO- Linear acceleration at the wrist joint due to its rotation about
the elbow, shoulder, hip, knee and ankle joints:
Let:

Rs = distance in cm. between the elbow and wrist joints.

Xi (W) and Yi (W) = horizontal and vertical components respectively
of the linear acceleration in cm. per second squared, at a
time interval 1, acting at the wrist joint, due to its
rotation about the elbow joint.
XXi (W) and YYi (W) = horizontal and vertical components
respectively of the linear acceleration in cm. per second
squared, at a time Interval 1, acting at the wrist joint, due
to its rotation about the elbow, shoulder, hip, knee and
ankle joints.
Therefore,
Xi (W) = - Rs [(U5)i sin (Z4)i - (U5)i2 cos (Z4)i]....(77)
Yi (W) = + Rs [(U5)i cos (Z4)i + {US)]^ sin (Z4)i]....(78)

XXi (W) = Xi (W) + XXi (E) (79)

YYi (W) = Yi (W) + YYi (E) (80)

11. Linear acceleration at the center of gravity of the hand due to

its rotation about the wrist, elbow, shoulder, hip, knee and
ankle joints:
Let:
re = distance in cm. between the center of gravity of the hand
and the elbow joint.
 61
Xi (HA) and Yi (HA) = horizontal and vertical components
respectively of the linear acceleration in cm. per second
squared, at a time Interval 1, acting at the center of
gravity of the hand, due to its rotation about the elbow
joint.
XXi (HA) and YYi (HA) = horizontal and vertical components
respettively of the linear acceleration in cm. per second
squared, at a time interval 1, acting at the center of
gravity of the hand, due to its rotation about the wrist,
elbow, shoulder, hip, knee and ankle joints.
Therefore,

Xi (HA) = - re [(U5)i sin (Z4)i - (U5)i2 cos (Z4)i]....(81)

Yi (HA) = + re [(U5)i cos (Z4)i + (U5)i2 sin (Z4)i]....(82)

XXi (HA) = Xi (HA) + XXi (E) (83)

YYi (HA) = Yi (HA) + YYi (E) (84)

3.3.C - CALCULATIONS OF REACTIVE FORCES AND

REACTIVE TORQUES AT BODY ARTICULATIONS
Accelerations of the body result in inertial forces and torques
on the body. Newton's second law states that an accelerating mass
results in a force. Therefore, a weight of (W) kg., has a mass of
(W/980-616) kg. per cm./sec.^ If the mass acceleration is (A) cm./sec.2,
then the inertial force (F) is:

^ = -(WW) W (85)
 62

The acceleration which will be used in the calculation of

Inertial forces for body links are the horizontal and vertical components
of linear acceleration acting at the center of gravity of the links.
The moment of inertia of a body is a measure of the resistance
a body offers to any change in its angular velocity. It is
determined by the distribution of its mass about the axis of rotation.
The radius of gyration of a body is the distance from its axis of
rotation to a point at which the total mass of the body might be
concentrated without changing its moment of inertia. The moment of
inertia about the center of gravity is found as follows:

ICG = ^0-'^^'
= M K2 - M r2
= M (K2 - r^) (86)

where:
IpQ = moment of inertia about center of gravity.
IQ = moment of inertia about articulation.
M = mass of link.
K = radius of gyration.
r = distance from articulation to center of gravity.
The values of K and r (which used also to calculate the linear
accelerations) are taken from data published by Plagenhoef (1963),
Table (6).
The Inertial resistance torque (T) acting on a body of moment
of inertia (I), produces in it an angular acceleration (A), given by:
T = - (I) (A) (87)
 63

TABLE (6)
CENTER OF GRAVITY AND RADIUS OF GYRATION AS A
PERCENTAGE OF THE SEGMENT LENGTH

Center of Gravity Radius of Gyration

Segment Proximal Distal Proximal Distal

Hand 50.6% 49.4% 58.7% 57.7%

Forearm 43.0 57.0 52.6 64.7
Upper arm 43.6 56.4 54.2 64.5
Foot 50.0 50.0 69.0 69.0
Shank 43.3 56.7 52.8 64.3
Thigh 43.3 56.7 54.0 65.3
Forearm and hand(a) 68.2 31.8 82.7 56.5
Whole upper limb(b) 53.0 47.0 64.5 59.6
Foot and shank(c) 60.6 39.4 73.5 57.2
Whole lower limb(d) 44.7 55.3 56.0 65.0
Trunk 39.6 60.4(^5 49.7 67.5(e)

(a) Forearm and Hand: - From elbow to ulnar styloid

(b) Whole upper limb: - From shoulder joint to ulnar styloid
(c) Foot and shank: - From knee joint to medial malleolus
(d) Whole lower limb: - From hip to medial malleolus
(e) Percentage is taken from top of head
The inertial resistance torque, gives the additional torque
about the axis of rotation due to a shift in the concentration of
the mass, assuming that the moment of inertia stays constant.
The calculation of reactive forces and torques is carried out
 64
using the free-body diagram approach and the corresponding
equilibrium equations:
2: Fx = 0 (88)
2: Fy = 0 (89)
E Mo = 0 (90)
where z Fx and z Fy are respectively the algebraic sums of
horizontal and vertical forces and z Mo is the algebraic sum of the
moments and torques about the joint.
1. Calculation of reactive forces and torque at wrist joint:
Figure 14. is a free-body diagram for the hand link. Let:
W (HA) = weight of the hands in kg. acting at the center of
gravity of the hand. This value is taken as a negative
value in the input of the computer program.
M (HA) = Mass of the hand in kg. per cm./sec.^ acting at the
center of gravity of the hand.
Ke = radius of gyration of the hand in cm.
L = load carried in the hands in kg. acting at the center
of gravity of the hand. This value is taken as a
negative value in the input of the computer program.
M (L) = mass of the load in kg. per cm./sec.2 acting at the
center of gravity of the hand.
Fi (HA)/X = inertial horizontal force in kg. acting at the center
of gravity of the hand.
Fi (HA)/Y = inertial vertical force in kg. acting at the center
of gravity of the hand.
 Mi(W) 65

Ri(W)/Y
Ti(HA)

Fi(HA)/X

Fig. 14 : Free-body diagram showing the forces and

torques on the hand during the dynamic activity

+ VE FORCE
)[

+ VE
- ^ e

tVE FORCE

Fig. 15 : Forces and torques notations

 66

^CGHA " moment of inertia about the center of gravity of the hand.
Ti (HA) = inertial resistance torque in kg. cm. acting at the center
of gravity of the hand.
Ri (W)/X = reactive force in kg. acting at the wrist joint in the
horizontal axis.
Ri (W)/Y = reactive torque in kg. acting at the wrist joint in the
vertical axis.
Mi (W) = reactive torque in kg.cm. acting at the wrist joint.
Therefore:
M (HA) = - W (HA) / 980.616 (91)

M (L) = - L / 980.616 (92)

Fi (HA)/X = - [M (HA)-f M (L)].XXi (HA) (93)

Fi (HA)/Y = - [M (HA) + M (L)].YYi (HA) (94)

^CGHA = ^^ ^^^ "• ^^'•^^ * ^^^^ " ""'^^ ^^^^

Ti (HA) = - IcGHA • (U5)i (96)

Ri (W)/X = - Fi (HA)/X (97)

Ri (W)/Y = - [Fi (HA)/Y + W (HA) + L] (98)

Mi (W) = - Ti (HA) + Fi (HA)/X • re • sin (Z4)i

- [Fi (HA)/Y + W (HA) + L] • re • cos (Z4)i ...(99)

where the (+) and (-) signs follow the notations of Figure 15.
 67
2. Calculation of reactive forces and torque at the elbow joint:
Figure 16. Is a free-body diagram for the lower arm link.
Let:
W (LA) = weight of the lower arms in kg. acting at its center
of gravity. This value is a negative input to the com-
puter program.
M (LA) = mass of the lower arms in kg. per cm./sec. acting at
its center of gravity.
Ks = radius of gyration of the lower arm in cm.
Fi (LA)/X = inertial horizontal force in kg. acting at the center
of gravity of the lower arm.
Fi (LA)/Y = inertial vertical force in kg. acting at the center
of gravity of lower arm.
I^^, . = moment of inertia about the center of gravity of lower
LbLA

arm.
Ti (LA) = inertial resistance torque in kg.cm. acting at the
center of gravity of the lower arm.
Ri (E)/X = reactive force in kg. acting at the elbow joint in the
horizontal axis.
Ri (E)/Y = reactive force in kg. acting at the elbow joint in the
vertical axis.
M^- (E) = reactive torque in kg.cm. acting at the elbow joint.
Therefore:
M (LA) = - W (LA) / 980.616 (TOO)
Fi (LA)/X = - M (LA) . XXi (LA) 00^)
 /^M1(E)

68
^(Z4)i
Ri(E)/|Y \^ VULA)

j-Mi(W)

V w I -Ri(W)/X

-Ri(W)/Y

Fig. 16 : Free-body diagram for the lower arm showing

the forces and torques during the dynamic activity

Ri(S)/Y

Fi(UA)/Y

Ri(E)/Y

Fig. 17 : Free-body diagram for the upper arm showing

the forces and torques during the dynamic activity
 69

Fi (LA)/Y = - M (LA) • YYi (LA) (102)

ICGLA = ^ ^^'^^ ' ("^s" • ^5') (103)

Ti (LA) = - Ij.gLA • (U5)i (104)

Ri (E)/X = Ri (W)/X - Fi (LA)/X (105)

Ri (E)/Y = Ri (W)/Y - Fi (LA)/Y .-. (106)

Mi (E) = - Ti (LA) + Mi (W) - Ri (W)/X • Rs • sin (Z4)i

+ Ri (W)/Y . Rs • cos (Z4)i
+ Fi (LA)/X . rs • sin (Z4)i
- [Fi (LA)/Y + W (LA)] . rs • cos (Z4)i ...(107)

3. Calculation of reactive forces and torque at the shoulder joint:

Figure 17. is a free-body diagram for the upper arm link. Let:
W (UA) = weight of the upper arms in kg. acting at its center of
gravity. This value is a negative input to the computer
program.
M (UA) = mass of the upper arms in kg. per cm./sec. acting at
its center of gravity.
K4 = radius of gyration of the upper arm in cm.
Fi (UA)/X = inertial horizontal force in kg. acting at the center
of gravity of upper arm.
Fi (UA)/Y = inertial vertical force in kg. acting at the center
of gravity of upper arm.
^CGUA "" moment of inertia about the center of gravity of upper arm
Ti (UA) = inertial resistance torque in kg.cm. acting at the center
of gravity of upper arm.
 70
Ri (S)/X = reactive force in kg. acting at the shoulder joint
in the horizontal axis.
Ri (S)/Y = reactive force in kg. acting at the shoulder joint
in the vertical axis.
Mi (S) = reactive torque in kg.cm. acting at the shoulder joint.
Therefore,
M (UA) = - W (UA) / 980.616 108)

Fi (UA)/X = - M (UA) . XXi (UA) 109)

Fi (UA)/Y = - M (UA) • YYi (UA) 110)

= M (UA) . (Kt,2 - n,2) 111)

^CGUA
Ti (UA) = - ICGUA (U4)i 112)

Ri (S)/X = Ri (E)/X - Fi (UA)/X 113)

-^ [AJOA^^^
Ri (S)/Y = Ri (E)/Y - Fi (UA)/Y 114)

Mi (S) - Ti (UA) + Mi (E) - Ri (E)/X • Ri, • sin (Z3)i

+ Ri (E)/Y • Ri, • cos (Z3)i
+ Fi (UA)/X . r^ • sin (Z3)i
- [Fi (UA)/Y + W (UA)] • r^ • cos (Z3)i ...(115)

4. Calculation of reactive forces and torque at the hip joint:

Figure 18. is a free-body diagram for the trunk link. Let:
W (TK) = weight of the trunk in kg. acting at its center of
gravity. This value is a negative input to the computer

program.
 -Mi(S)

71

Ti(TK)

Mi(H)

Fig. 18 : Free-body diagram for the trunk showing

the forces and torques during the dynamic activity

.-Ri(H)/X

Ri(H)/Y

Fig. 19 : Free-body diagram for the upper leg showing

the forces and torques during the dynamic activity
 72

M (TK) = mass of the trunk in kg. per cm./sec. acting at its

center of gravity.
Ka = radius of gyration of the trunk in cm.
Fi (TK)/X = inertial horizontal force in kg. acting at the center
of gravity of the trunk.
Fi (TK)/Y = inertial vertical force in kg. acting at the center
of gravity of the trunk.

^CGTK " '"O'"®"* 0"^ inertia about the center of gravity of the
trunk.
Tj (TK) = inertial resistance torque in kg.cm. acting at the
center of gravity of the trunk.
Ri (H)/X = reactive force in kg. acting at the hip joint in the
horizontal axis.
Ri (H)/Y = reactive force in kg. acting at the hip joint in the
vertical axis.
Mi (H) = reactive torque in kg.cm. acting at the hip joint.
Therefore,
M (TK) = - W (TK) / 980.616 (116)

Fi (TK)/X = - M (TK) . XXi (TK) (117)

Fi (TK)/Y = - M (TK) • XXJ (TK) (118)

^CGTK = ^ (^^^ • ^^'^ • ''^'^ ^^^^^

Ti (TK) = - I^eTK • (^*3)i (120)

Ri (H)/X = Ri (S)/X - Fi (TK)/X (121)

 73

Ri (H)/Y = Ri (S)/Y - Fi (TK)/Y - W (TK) (122)

Mi (H) = - Ti (TK) + Mi (S) + Ri (S)/X • Ra • sin (Z2)i

+ Ri (S)/Y . Ra • cos (Z2)i
- Fi (TK)/X . ra • sin (Z2)i
- [Fi (TK)/Y+ W (TK)] . ra • cos (Z2)i ...(123)

5. Calculation of reactive forces and torque at the knee joint:

Figure 19. is a free-body diagram for the upper leg link. Let:
W (UL) = weight of the upper legs in kg. acting at its center of
gravity. This value is a negative input to the computer
program.
M (UL) = mass of upper legs in kg. per cm./sec.^ acting at the
center of gravity of upper leg.
K2 = radius of gyration of upper leg in cm.
Fi (UL)/X = inertial horizontal force in kg. acting at the center
of gravity of upper leg.
Fi (UL)/Y = inertial vertical force in kg. acting at the center
of gravity of upper leg.
^CGUL "" moment of inertia about the center of gravity of upper
leg.
Ti (UL) = inertial resistance torque in kg.cm. acting at the center
of gravity of upper leg.
Ri (K)/X = reactive force in kg. acting at the knee joint in the

horizontal axis.
Ri (K)/Y = reactive force in kg. acting at the knee joint in the

vertical axis.
 74
Mi (K) = reactive torque in kg.cm. acting at the knee joint.
Therefore,
M (UL) = - W (UL) / 980.616 124)

Fi (UL)/X = - M (UL) . XXi (UL) 125)

Fi (UL)/Y = - M (UL) . YYi (UL) 126)

^CGUL = M (UL) . (K22 - r22) 127)

Ti (UL) = - I CGUL (U2) 128)

Ri (K)/X = Ri (H)/X - Fi (UL)/X 129)

Ri (K)/Y = Ri (H)/Y - Fi (UL)/Y 130)

Mi (K) - Ti (UL) + Mi (H) + Ri (H)/X • R2 • sin (Zl)i

- Ri (H)/Y • R2 • cos (Zl)i
- Fi (UL)/X • r2 • sin (Zl)i
+ [Fi (UL)/Y + W (UL)] . r2 • cos (Zl)i ...(131)

6. Calculation of reactive forces and torque at the ankle joint:

Figure 20. is a free-body diagram for the lower leg link. Let:
W (LL) = weight of the lower legs in kg. acting at its center of
gravity. This value is a negative input to the computer
program.
M (LL) = mass of lower legs in kg. per cm./sec? acting at its cen-
ter of gravity.
Ki = radius of gyration of lower leg in cm.
Fi (LL)/X = inertial horizontal force in kg. acting at the center
of gravity of lower leg.
 -Mi(K) 75

Ti(LL)

Mi(A)

Rt(A)/Y

Fig. 20 : Free-body diagram for the lower leg showing

the forces and torques during the dynamic activity
 76

Fi (LL)/Y = inertial vertical force in kg. acting at the center

of gravity of lower leg.
IQO, I = moment of inertia about the center of gravity of lower
leg.
Ti (LL) = inertial resistance torque in kg.cm. acting at the
center of gravity of lower leg.
Ri (A)/X = reactive force in kg. acting at the ankle joint in the
horizontal axis.
Ri (A)/Y = reactive force in kg. acting at the ankle joint in the
vertical axis.
Mi (A) = reactive torque in kg.cm. acting at the ankle joint.
Therefore,

M (LL) = - W (LL) / 980.616 (132)

Fi (LL)/X = - M (LL) • Xi (LL) (133)

Fi (LL)/Y = - M (LL) • Yi (LL) (134)

ICGLL = ^ (LL) . (Ki^ - ri2) (135)

Ti (LL) = - I^gLL • (Ul)i (136)

Ri (A)/X = Ri (K)/X - Fi (LL)/X (137)

Ri (A)/Y = Ri (K)/Y#Fi (LL)/Y (138)

Mi (A) = - Ti (LL) + Mi (K)

+ Ri (K)/X • Ri • sin (Ul)i
+ Ri (K)/Y • Ri • cos (Ul)i
- Fi (LL)/X . ri • sin (Ul)i
- [Fi (LL)/Y+ W (LL)] • ri • cos (Ul)i ...(139)
 77

3.3.D - CALCULATIONS OF REACTIVE FORCES AND REACTIVE

TORQUES AT THE CENTERS OF THE DISCS
1. General calculations of trunk link dimensions:
Since the main areas of concern in the spine are the L-4/L-5
and L-5/S-1 intervertebral discs and the surrounding vertebral
bodies, it will be assumed that the centers of the two discs are
articulations. The same assumption was used by Fisher (1967) and
by Chaffin (1969). The trunk link will be divided into three links
a - from hip joint to the center of the L-5/S-1 disc
b - from the center of L-5/S-1 disc to the center of L-4/L-5

disc
c - from the center of L-4/L-5 disc to the shoulder joint
The average spinal dimensions. Figure 21., are used in the
model. These dimensions are determined by Fich (1904) and used by
Fisher (1967). The masses of the trunk links were estimated by
Fisher (1967) to be as follows:
a - above L-4/L-5 60% of trunk mass

b - L-4/L-5 to L-5/S-1 5.5% of trunk mass

c - L-5/S-1 to hip joint 34.5% of trunk mass

Fisher (1967) expressed the spinal dimensions as a function of the
hip to shoulder distance "HS" as follows:
a - L-4/L-5 to shoulder joint : 0.730 of HS
b - L-4/L-5 to L-5/S-1 0.075 of HS

c - hip joint to L-5/S-1 0.195 of HS

The centers of gravity of the trunk links were estimated by
 78

DIMENSKDNS ARE
IN MILLIMETERS

Fig. 21 : Average spinal column as presented by Fisher (1969)

 79
Fisher (1967) to be as follows:

a - L-4/L-5 to shoulder joint : the center of gravity is

located at a distance from the center of L-4/L-5 disc equal
to 40% of its length,
b - L-4/L-5 to L-5/S-1 : the center of gravity is located
at a distance from the center of L-5/S-1 disc equal to 40%
of its length,
c - L-5/S-1 to hip joint: : the center of gravity is located
at a distance from the hip joint equal to 40% of its length.
These figures are shown in Figure 22.
Let:
W (TKl) = weight in kg. of the upper trunk limb located between
the shoulder joint and the center of L-4/L-5 disc.
M (TKl) = mass in kg. per cm./sec.^ of the upper trunk limb.
W (TK2) = weight in kg. of the middle trunk limb located between
the center of L-4/L-5 disc and the center of L-5/S-1 disc
M (TK2) = mass in kg. per cm./sec.^ of the middle trunk link
W (TK3) = weight in kg. of the lower trunk limb located between
the center of L-5/S-1 disc and the hip joint.
M (TK3) = mass in kg. per cm./sec. of the lower trunk limb.
Di = length in cm. of upper trunk limb.
D2 = length in cm. of middle trunk limb.
D3 = length in cm. of lower trunk limb,
dj = distance in cm. from center of gravity of upper trunk
limb to center of L-4/L-5 disc.
 80

CG. of Upper
Trunk Link

W(TKi;

CG. of L 4 / L 5

CG.of Middle Trunk

Link

W(TK2)

CG. of L5/S1

C . G . o f L o w * r Trunk Link

W(TK3)

Fig. 22 : Spinal dimensions taken from data by Fisher (1969)

 81
d2 = distance in cm. from the center of gravity of middle trunk
limb to center of L-5/S-1 disc,

da = distance in cm. from the center of gravity of lower trunk

limb to hip joint.
Therefore,
W (TKl) = 0.6 W (TK) 140)

W (TK2) = 0.055 W (TK) 141)

W (TK3) = 0.345 W (TK) 142)

M (TKl) = - W (TKl) / 980.616 143)

M (TK2) = - W (TK2) / 980.616 144)

M (TK3) = - W (TK3) / 980.616 145)

= 0.73 Ra 146)

= 0.075 Ra 147)

D3 = 0.195 Ra 148)

= 0.4 Di = 0.292 Ra 149)

d2 = 0.4 D2 = 0.030 Ra 150)

d3 = 0.4 Da = 0.078 Ra 151)

2. Calculation of Spinal Angles:

Mitchell (1934) measured the spinal angles in the erect position
and found the following:
upper surface of S-1 : 40 degrees (0.698 radians)
lower surface of L-5 : 21 degrees (0.3667 radians)
 82
upper surface of L-5 : 10 degrees (0.1745 radians)
lower surface of L-4 : 5 degrees (0.08725 radians)
These angles are measured from the negative horizontal axis and for
the erect position.

The curvature changes for the column during sagittal rotation

of the hip is assumed from the data of Dempster (1955) which
disclosed that for the first 27 degrees (0.471 radians) of trunk
flexion the pelvis does not rotate, i.e., the rotation is in the
lumbar spine and for each additional degree of trunk rotation the
pelvis contributes about two thirds of a degree.
Chaffin (1974) found that for the first 13** (0.227 radians)
of upper leg flexion, the pelvis does not rotate and for each
additional degree of upper leg flexion the pelvis contributes about
two thirds of a degree.
Also it is assumed that 18/83 and 24/83 of the lumbar rotation
occurs at L-5/S-1 and L-4/L-5 discs respectively, based on data by
Davis (1965), Albrock and Uganda (1957), Lindahl (1966) and
Rolander (1966). The rotation of the pelvic establishes the position
of the sacrum so the angle of the upper surface of S-1 can be deter-
mined.
Let:
El = angle in radians between the horizontal negative axis and
lower leg at the standing erect position in a normal position
without carrying any load. Figure 23.
E2 = angle in radians between the upper leg and lower leg at the
 83

r- \ •

5 \ £ 3 c> V

< <•

r*-' '
D??,K E2
K

-i^
El t) *^A

Fig. 23 : Standing erect position when carrying no load

 84
standing erect position in a normal position without
carrying any load. Figure 23.
E3 = angle in radians between the trunk and upper leg at the
standing erect position in a normal position without
carrying any load. Figure 23.
E4 = angle in radians between the positive horizontal axis and
upper leg in a normal standing erect position without
carrying any load. Figure 23.
E5 = angle In radians between the negative horizontal axis and
trunk in a normal standing erect position without carrying
any load. Figure 23.
(E6)i = angle in radians which represents the deviation between the
upper leg when carrying a load at instant 1 and the normal
standing erect position of the upper leg without carrying
a load. Figure 24.
(E7)i = angle in radians which represents the deviation between the
trunk when lifting a load at Instant i and the normal
standing erect position of the trunk without carrying a
load. Figure 24.
(CHGl)i = the change in the angle of the upper surface of sacrum,
in radians, due to the change of trunk position, at

instant 1.
(CHG2)i = the change in the angle of the upper surface of sacrum,
in radians, due to the change of upper leg position, at

instant 1.
 85

/
/
/
/
/
•*>^i
fe-1
Fig. 24 : Angular changes when performing a dynamic activity
 86

(Al)i = angle in radians between the negative horizontal axis and

upper surface of S-1 at Instant 1.
(A2)i = angle in radians between the negative horizontal axis and
lower surface of L-5 at Instant 1.
(A3)i = angle in radians between the negative horizontal axis and
upper surface of L-5 at Instant i.
(A4)i = angle in radians between the negative horizontal axis and
lower surface of L-4 at Instant 1.
(Bl)i = the complementary angle of (Al)i, in radians.
(B2)i = the complementary angle of (A2)i, in radians.
(B3)i = the complementary angle of (A3)i, in radians.
(B4)i = the complementary angle of (A4)i, in radians.
Therefore,
E4 = E2 - El (152)

E5 = E3 - E4 (153)

(E6)i = E4-(Zl)i (154)

(E7)i = E5 - (Z2)i (155)

(CHGl)i = 0.0 ^. , (E7)i £0.471

= [(E7)i - 0.471] (2/3) , (E7)i > 0.471 ...(156)

(CHG2)i = 0.0 ,;». . (E6)i < 0.227

= [(E6)i - 0.227] (2/3) , (E6)i > 0.227 ....(157)

(Al)i = 0.698^+ (CHGl)i - (CHG2)i (158)

(Bl)i = 1.57 - (Al)i (159)

 87
Let:
(CHG3)i = change in the angle of lower surface of L-5, in radians,
due to change in trunk position, at instant 1.
(CHG4)i = change in the angle of lower surface of L-5, in radians,
due to change in upper leg position, at instant 1.
(CHG5)i = change in the angle of lower surface of L-4, in radians,
due to change in trunk position, at instant 1.
(CHG6)i = change in the angle of lower surface of L-4, in radians,
due to change in upper leg position, at instant 1.
Therefore,
1-i

(CHG3)i = (E7)i (18/83) ,(E7)i < 0.471

= 0^471 (18/83) + [(E7)i - 0.471] (1/3) .,(18/83)
lit, ^r ,(E7)i > 0.471 (160)

(CHG4)i = (E6)i (18/83) ,(E6)i £ 0.227 ^

= 0.227 (18/83) + [(E6)i - 0.227] (1/3) (18/83)
'"^ ,(E6)i > 0.227 (161)

(A2)i = f ( A l ) i > 0.3667 - 0.698+ (CHG3)i - (CHG4)i (162)

(B2)i = 1.57 - (A2)i (163)

(A3)i = (A2)i + 0.3667 - 0.1745 (164)

(B3)i = 1.57 - (A3)i (165)

(CHG5)i = (E7)i (24/83) ,(E7)^| <. 0.471

= 0.471 (24/83) + [(E«)i - 0.2^^] (1/3) (24/83)
,(E7)i > 0.471 (166)

(CHG6)i = (E6)i (24/83) ,(E6)i £0.227

= 0.227 (24/83) + [(E6)i - 0.227] (1/3) (24/83)
,(E6)i > 0.227 (167)
 88

(A4)i = (A3)i + 0.08725 - 0.1745

+ (CHG5)i - (CHG6)i (168)

(B4)i = 1.57 - (A4)i (169)

^10

3. Calculation of abdominal pressures, forces and torques:

The contribution of intra-abdominal pressure in relieving
compression on the lumbar spine was estimated by Chaffin (1969)
from the data of Morris et.al. (1961). An equation relating the
abdominal pressure with hip torque and the angle between the trunk
and thighs was developed by Chaffin (1969) as follows:
Pr = 10-*+ [0.6516 - 0.005447 (ANGLE)] [Hip torque^-^] ...(170)
where:
ANGLE = angle between trunk and thigh in degrees
Hip torque = torque at the hip in kg.cm
Pr = estimated abdominal pressure in mm Hg. with a maximum of
150 mm Hg.
Experiments by Asmussen (1968) have disclosed that probably
the method of lifting (i.e. quick jerk or slow sustained pull) can
significantly affect the abdominal pressure.
The amount of force created by the abdominal pressure is
estimated by assuming the following four conditions which are
similar to those proposed by Chaffin (1969), Fisher (1967) and
Morris et.al. (1961). These conditions are:
1. The abdominal pressure acts on a diaphram area of 465 cm^.
2. The abdominal muscles do not cause a longitudinal pull in
 lifting, since it was shown by Bartelink, (1957), that the
'V
rectus abdominis was not active during lifting.
3. The line of action of the force acts parallel to the line
of action of the normal compressive force on the lower
lumbar spine.
4. It was shown by Troup (1965) that the distances the
abdominal forces act from the spine are not constant.
Based on the dimensions given by Morris et.al. (1961) an
estimate was made by Fisher (1967) on how the lever arms
vary. At 0** flexion of the trunk, the lever arms decrease
approximately 53% from a 60° flexion, and at 90° of trunk
flexion increases 17%. Chaffin (1969) assumed that the
moment arms vary as the sine of the angle at the hip, with
the erect position having moment arms of 6.7 cm. at the
diaphragm level, and the 90° hip angle position having
14.9 cm. at the diaphram level.
Let:
(Pr)i = abdominal pressure in mm Hg., at instant 1.
Pi = abdominal pressure in kg./sm.^, at instant 1.
Di(AB) = distance the abdominal force acts from the spine, cm., at
Instant 1.
Fi(AB) = abdominal force kg., at instant 1.
Ti(AB) = abdominal torque, kg.cm., at instant i.
(U)i = angle between trunk and thigh, in degrees.
 90
Therefore, there are two possibilities:
A. If Mi (H) > 0.0
in this case
(Pr)i = 0.0 (171)

Pi = 0.0 : (172)

Fi (AB) = 0.0 (173)

Ti (AB) = 0.0 (174)

B. If Mi (H) <.0.0
in this case
(U)i = (U3)i . (180/Tr) (175)

(Pr)i = lO-'* [0.6516 - 0.005447 (U)i] [Mi (H)]1.8 ...(176)

Pi = (Pr)i . (0.00136) (177)

^ Di (AB) = 12.7 [sin ((I§ijL+JlZli) + 0.4^1 (178)

Fi (AB) = {?v^i (465) (179)

Ti (AB) = Fi (AB) • Di (AB) (180)

4. Calculation of reactive forces and reactive torque at center
of L-4/L-5 disc:
Figure 25. is a free-body diagram for the upper trunk limb.
Since the calculation of spinal forces follows the resolution of
forces on the whole body, the inertial forces of the trunk can be
used to estimate the inertial forces on the three links of the
trunk. That is:

F = p% • Fi (TK) (181)
where F is the inertial force on the particular link of the trunk.
 - Mi (S)

-Ri(S)/X 91
^

-Ri(S)/Y
TKTKI)

Mi(L4L5i

CG- of L 4 / L 5 Disc

Rr(L4L6)/Y

Fig. 25 : Free-body diagram for the upper trunk link showing

the forces and torques during the dynamic activity

-Mi(L4L5)

CG. of L4/L5.Disc
-Ri(L4L5yx

-Ri(L4L5)/Y

Ti(TK2X

Mi(L5S1)^.—^
W(TK2)

Ri(L5Sl)/X
.
C G . of L5/S1 Disc

Ri(L5S1)/Y

Fig. 26 : Free-body diagram for the middle trunk link showing

the forces and torques during the dynamic activity
 92
p% is the percent of weight of the particular link of trunk to the
total weight of trunk, and Fi (TK) is the inertial force at the
center of gravity of the trunk. Let:
Fi (TK1)/X = inertial horizontal force in kg. acting at the

center of gravity of upper trunk link.

Fi (TK1)/Y = inertia! vertical force in kg. acting at the center
of gravity of upper trunk link.
Ri (L4L5)/X = reactive horizontal force in kg. acting at the
center of L-4/L-5 disc.
Ri (L4L5)/Y = reactive vertical force in kg. acting at the center
of L-4/L-5 disc.
Mi (L4L5) = reactive torque in kg.cm. acting at the center of
L-4/L-5 disc.
Therefore,

Fi (TK1)/X = 0.6 Fi (TK)/X (182)

Fi (TK1)/Y = 0.6 Fi (TK)/Y (183)

Ri (L4L5)/X = Ri (S)/X - Fi (TK1)/X (184)

Ri (L4L5)/Y = Ri (S)/Y - Fi (TKl )/Y (185)

Mi (L4L5) = Mi (H) - Ti (AB) (186)

5. Calculation of reactive forces and reactive torque at center of

L-5/S-1 disc:
Figure 26. is a free-body diagram for the middle trunk link.
 93
Let:

Fi (TK2)/X = Inertial horizontal force in kg. acting at the

center of gravity of middle trunk link.
Fi (TK2)/Y = Inertial vertical force in kg. acting at the center
of gravity of middle trunk link.
Ri (L5S1)/X = reactive horizontal force in kg. acting at the
center of L-5/S-1 disc.
Ri (L5S1)/Y = reactive vertical force in kg. acting at the center
of L-5/S-1 disc.
Mi (L5S1) = reactive torque in kg.cm. acting at the center of
L-5/S-1 disc.
Therefore,

Fi (TK2)/X = 0.055 Fi (TK)/X (187)

Fi (TK2)/Y = 0.055 Fi (TK)/Y , (188)

Ri (L5S1)/X = Ri (L4L5)/X - Fi (TK2)/X (189)

Ri (L5S1)/Y = Ri (L4L5)/Y - Fi (TK2)/Y (190)

Mi (L5S1) = Mi (H) - Ti (AB) (191)

3.3.E - CALCULATION OF COMPRESSIVE AND SHEARING FORCES

In addition to the normal forces on the links as a result of
body weight and external forces, the erector-spinal muscle exerts a
normal force to counteract the torque around the discs. This force
is also absorbed by the spine and is added to the normal force of
the disc. Assuming one line of action for all the muscles, and
 94
assuming that this line of force acts parallel to the vertebral
bodies, i.e., in a normal disection, it was found from data by
Bartelink (1957), Pearson (1961), Perey (1957) and Thieme (1950),
that the distance acts behind the nucleus of L-4/L-5 disc or
L-5/S-1 disc to be about 5 cms. Therefore, the normal force
exerted by the muscles is equal to the torque action at the center
of the disc in kg.cm. divided by 5.0.
1. Calculation of compressive and shearing forces on upper surface
of S-1:
Let:
Fi (MUS) = normal force exerted by muscles, kg.
Ci (USl) = compressive force on upper surface of S-1, kg.
Si (USl) = shearing force on upper surface of S-1, kg.
Therefore,
Fi (MUS) = Mi (L5Sl)/5.0 (192)

Ci (USl) = Fi (MUS) - Fi (AB)

+ Ri (L5S1)/Y • sin (Bl)i
- Ri (L5S1)/X • cos (Bl)i (193)

S^- (USl) = Ri (L5S1)/Y • cos (Bl)i

+ Ri (L5S1)/X • sin (81 )i (194)

2. Calculation of compressive and shearing forces on lower surface

of L-5:
Let:
Ci (LL5) = compressive force on lower surface of L-5, kg.
Si (LL5) = shearing force on lower surface of L-5, kg.
 95
Therefore,
Ci (LL5) = Fi (MUS) - Fi (AB)
+ Ri (L5S1)/Y • sin (B2)i
- Ri (L5S1)/X • cos (B2)i (195)

Si (LL5) = Ri (L5S1)/Y * cos (B2)i

+ Ri (L5S1)/X . sin (B2)i (196)

3. Calculation of compressive and shearing forces on upper surface

of L-5:
Let:
Ci (UL5) = compressive force on upper surface of L-5, kg.
Si (UL5) = shearing force on upper surface of L-5, kg.
Therefore,
Ci (UL5) = Fi (MUS) - Fi (AB)
+ Ri (L4L5)/Y • sin (B3)i
- Ri (L4L5)/X • cos (B3)i (197)

Si (UL5) = Ri (L4L5)/Y • cos (B3)i

+ Ri (L4L5)/X • sin (B3)i (198)

4. Calculation of compressive and shearing forces on lower surface

of L-4:
Let:
Ci (LL4) = compressive force on lower surface of L-4, kg.
Si (LL4) = shearing force on lower surface of L-4, kg.

Therefore,
Ci (LL4) = Fi (MUS) - Fi (AB)
+ Ri (L4L5)/Y . sin (B4)i
- Ri (L4L5)/X . cos (B4)i (199)
 96
Si (LL4) = Ri (L4L5)/Y • cos (B4)i
+ Ri (L4L5)/X • sin (B4)i (200)

These were the equations used in developing the biomechanical

dynamic model for calculating the forces (compressive and shearing)
on the lumbar spine during sagittal lifting.
 CHAPTER IV
RESULTS AND DISCUSSIONS
4.1 THE NET TORQUES AT THE HIP JOINT
The stresses Induced at the low-back during weight lifting are
due to a combination of weights lifted and the person's method of
lift. Specifically, gravitational forces acting upon the loads
held in the hands and the person's body masses create rotational
moments or torques at the various articulations. The skeletal
muscles are positioned to exert forces at these articulations in such
a manner that they counteract these torques.
The amount or torque at any point is dependent on the amount
of force tending to rotate the segments multiplied by the moment
arm of that force (i.e., the distance from the joint to the force
vector measured normal to the force vector). The moment arm
magnitudes vary with the person's posture. If the load lifted is
held close to the body, the moment arms are small and the resulting
torques at the joints are small. If the load is held away from the
body, the large moment arms will cause large torques. A biomechanical
lifting equivalent takes into account both the weight to be lifted
and the moment arm. Increasing the biomechanical lifting equivalent
by increasing the weight to be lifted and/or the moment arm will
increase the resulting torques at the different articulations.

The torques at the hip are the main components for the
compressive forces on the spine. The abdominal force results in an
abdominal torque which helps the spine carry the load, therefore,

97
 98

the net torque at the hip joint (i.e., the hip torque minus the

abdominal torque) is principally responsible for the compressive

force generated on the spine. Increasing the weight to be lifted

and/or the weight moment arm will increase the net torque at the

hip joint and consequently will increase the compressive force on

the spine. Thus, increasing the biomechanical lifting equivalent

will increase the net torque at the hip joint. Tables 7. and 8.

are the current experimental data for the net torques at the hip

joint at the critical positions (maximum torques) in kg.cm. at the

different biomechanical lifting equivalents and for both the leg and

back methods of lifting.

A regression analysis was carried out to determine the relation-

ship between the biomechanical lifting equivalents and the maximum

net torques at the hip joint. This was done for each subject for

each method of lift. Then a regression analysis was done for the

leg lift of all subjects together, and also for the back lift. In

each case, the relationship between the biomechanical lifting

equivalents and the maximum net torques at the hip joint is a straight

line. The equations of these straight lines are as follows:

For the leg lifts:

1st subject : MNT = 1625 + 1.63752 BLE (201)
2nd subject : MNT = 1903 + 1.34829 BLE (202)
3rd subject : MNT = 1979 + 0.96188 BLE (203)

4th subject : MNT = 1389+ 1.18262 BLE (204)

all subjects : MNT = 1731 + 1.25567 BLE (205)

 99

TABLE (7)

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

NET TORQUES AT THE HIP FOR THE LEG LIFT

Biomechanical Maximum Net Torque at Hij) (kg.cm.)

Lifting '
Equivalent
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 1812.592 .2114.835 2045.107 1525.441

140 2017.513 2126.761 1527.609

170 1858.538 2079.707 2128.877 1603.208

170 2155.603 2176.382 1594.703

200 1964.740 2190.072 2214.036 1666.065

200 2192.073 2176.403 1659.253

280 2105.629 2268.896 2300.163 1762.767

280 2172.467 2333.730 2193.586 1647.843

340 2164.974 2343.123 2308.211 1741.201

340 2298.264 2322.186 1827.652

400 2309.556 2518.327 2388.883 1840.926

400 2364.774 2487.176 2364.248 1857.742

420 2361.450 1949.000

420 2265.865 2528.937 2378.717 1872.653

510 2334.725 2572.544 2477.640 1959.263

510 2454.588 2539.737 2404.673 1976.341

600 2653.785 2749.936 2575.786 2127.964

600 2605.925 2643.930 2571.207 2093.059

 100

TABLE (8)
EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM
NET TORQUES AT THE HIP FOR THE BACK LIFT

Biomechanical Maximum Net Torque at Hip (kg.cm . )

Lifting
Equivalent
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 2020.048 2389.977 2317.707 1888.684

140 1963.224 2415.663 1790.208

170 2122.845 2487.136 2312.452 1848.303

170 2112.003 2474.471 2361.125 1826.242

200 2336.356 2444.260 2430.529 1922.925

200 2185.680 2438.864 2394.832 1896.156

280 2402.134 2555.476 2476.694 2026.509

280 2288.307 2536.365 2519.633 1865.702

340 2420.874 2548.980 2608.208 2132.724

340 2415.104 2559.161 2573.029 2151.660

400 2563.408 2554.693 2792.570 2304.423

400 2542.487 2666.188 2652.515 2210.152

420 2814.306 2747.990 2625.606 2191.559

420 2651.171 2810.410 2628.390 2263.880

510 2807.844 2770.594 2425.836

510 2944.587 2670.596 2449.044

600 2770.387 3016.670 2933.150 2497.611

600 2931.351 2938.576 2830.865 2482.417

 101
For the back lifts
1st subject MNT 1799 + 1.94579 BLE (206)
2nd subject ': MNT 2219 + 1.19280 BLE (207)
3rd subject : MNT 2165 + 1.18597 BLE (208)
4th subject : MNT 1588 + 1.56644 BLE (209)
all subjects : MNT 1939 + 1.46614 BLE (210)
where:
MNT = maximum net torques at the hip joint, kg.cm.
BLE = biomechanical lifting equivalents, lbs.in.
Table 9. is the analysis of variance (ANOVA) for the simple
linear regression between the biomechanical lifting equivalents and
the maximum net torques at the hip joint for the leg lifts of the
third subject, and Table 10. is for his back lifts.
Table 11. is the regression analysis data of all four subjects
for the linear relationship between the biomechanical lifting
equivalents and the maximum net torques at the hip joints.
Figure 27. shows the linear relationship between the biomechanical
lifting equivalents and the maximum net torques at the third
subject's hip for both methods of lift.
From Figure 27. and for both methods of lift, increasing the
biomechanical lifting equivalent by increasing the weight to be
lifted and/or the moment arm will increase the maximum net torques
at the hip joint. Also, for the same biomechanical lifting
equivalent, the back lift produces greater maximum net torques at
the hip joints than the leg lift. The difference in maximum net
 102
TABLE (9)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEFN BIOMECHANICAL LIFTING
EQUIVALENTS AND MAXIMUM NET TORQUES
AT THE HIP FOR THIRD SUBJECT (LEG LIFT)

Source df SS MS

Due to Regression 1 367478.625 367478.625 293.752

Deviation about Regression 16 20015.691 1250.981
Total 17 387494.316

F99% (1,16) = 8.53

TABLE ( 1 0 )

ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL LIFTING

EQUIVALENTS AND MAXIMUM NET TORQUES

AT THE HIP FOR THIRD SUBJECT (BACK LIFT)

Source df SS MS F

Due to Regression 1 499069.188 499069.188 156.068

Deviation about Regression 15 47966.567 3197.771
Total 16 547035.755
,..,. __,

F99% (1,15) = 8.68

 103

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 105
torques at the hip joint between the back and leg methods of lift
increases by increasing the biomechanical lifting equivalent.
ANOVA tables and regression analysis graphs for the first,
second and fourth subjects are presented in Appendix A.
The correlation coefficients in Table 11 (regression analysis)
combining the four subjects reveal low relationships because of the
large variations in the anthropometric data (Table 4 ) .
 106
4.2 COMPRESSIVE FORCES ON THE SPINE
The maximum amount of compression that can be tolerated by the
lumbar region of the spinal column has been estimated from axial
loading compression tests on cadaver columns. Data from separate
studies of this type by such researchers as Evans (1959) and
Sonoda (1962) disclose large biological variations in the ability
of the disc and its weight bearing cartilage endplates to withstand
such stresses. In general, the data of columns from persons under
40 years of age disclose a mean of about 1500 pounds (about 700 kg.)
before the cartilage endplates begin to disclose microfractures.
For each current subject, each method of lift and each
biomechanical lifting equivalent, the maximum compressive forces
on the spine were calculated using the computarized biomechanical
dynamic model. The maximum compressive forces were calculated at
four levels on the spine: upper surface of S-1, lower surface of
L-5, upper surface of L-5 and lower surface of L-4. The maximum
compressive forces on the upper surface of S-1 for the different
biomechanical lifting equivalents for the four subjects are presented
in Table 12. for the leg lifts and in Table 13. for the back lifts.
Current experimental data for the maximum compressive forces on
lower surface of L-5 are presented in Tables 14. and 15., for the
maximum compressive forces on upper surface of L-5 in Tables 16.
and 17., while the maximum compressive forces data on lower surface
of L-4 are presented in Tables 18. and 19.
A regression analysis was carried out to determine the
 107

TABLE ( 1 2 )

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

COMPRESSIVE FORCES OF THE UPPER SURFACE OF S-1 FOR THE LEG LIFT

Biomechanical Maximum Compressive Force (kg.) on upper

Lifting
Equivalent surface of S-1

"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 388.481 419.429 424.135 345.675

140 404.732 432.802 348.789
170 394.410 425.602 433.569 362.374

170 415.321 442.877 360.992

200 406.971 431.002 440.439 368.049

200 432.599 440.684 372.858

280 440.101 456.029 451.691 371.687

280 453.935 447.435 448.816 388.463

340 452.203 457.515 455.678 387.093

340 453.351 466.234 404.889

400 470.410 466.857 478.099 406.807

400 479.787 489.509 477.459 412.158

420 472.680 482.564 468.776 419.230

420 475.827 433.830

510 489.799 509.340 489.708 437.400

510 498.347 504.578 485.865 439.986

600 525.135 522.725 494.988 463.123

600 517.480 511.167 506.628 453.391

 108
TABLE (13)
EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM
COMPRESSIVE FORCES OF THE UPPER SURFACE OF S-1 FOR THE BACK LIFT

Biomechanical Maximum Compressive Force (kg) on upper

Lifting surface of S-1
Equivalent
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 399.829 450.548 448.130 354.805

140 394.454 450.130 375.023

170 404.588 461.426 454.728 379.891

170 407.509 463.526 449.215 367.713

200 421.596 470.609 465.196 386.200

200 451.157 475.733 468.966 385.497

280 454.928 481.305 483.600 399.786

280 431.250 486.624 489.089 368.469

340 464.660 492.669 501.301 421.989

340 456.312 493.052 503.072 419.772

400 487.986 503.668 526.990 440.899

400 490.010 509.205 516.697 449.422

420 517.063 511.890 516.590 450.529

420 519.399 504.625 514.1Q6 425.903

510 523.710 535.255 478.167

533.413 519.838 469.074

510
544.027 562.792 570.487 494.437
600
477.046
1 600 536.299 565.687 547.849
-
 109

TABLE ( 1 4 )

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

COMPRESSIVE FORCES ON THE LOWER SURFACE OF L-5 FOR THE LEG LIFT

Biomechanical Maximum Compressive Force (kg.) on lower

Lifting surface of L-5
Equivalents
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 388.148 424.139 432.004 348.769

140 410.204 440.890 346.599

170 394.571 419.850 441.600 362.337

170 430.629 450.492 462.116

200 411.197 434.806 448.848 378.609

200 437.062 448.557 369.617

280 441.593 454.814 861.104 390.570

280 454.519 461.094 457.128 372.348

340 451.614 364.606 463.976 388.868

340 458.444 474.397 408.933

400 470.597 495.319 486.448 406.751

400 482.765 473.200 485.350 413.186

420 471.499 489.414 477.272 435.347

420 485.027 420.249

488.399 513.612 498.722 438.564

510
501.690 509.918 494.877 436.944
510
526.793 528.372 505.350 463.153
600
521.293 515.872 514.811 458.579
600
-
 no
TABLE ( 1 5 )

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

COMPRESSIVE FORCES ON THE LOWER SURFACE OF L-5 FOR THE BACK LIFT

Biomechanical Maximum Compressive Force (kg.) on lower

Lifting surface of L-5
Equivalents 1

"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 409.196 461.981 458.743 382.903

140 404.462 461.976 364.331

170 418.382 475.187 465.780 376.534

170 414.990 472.589 459.173 386.803

200 463.200 485.626 476.267 394.899

200 431.517 481.355 479.179 395.172

280 466.923 496.750 494.495 410.374

280 442.564 491.810 500.633 378.674

340 476.667 503.012 515.083 430.168

340 467.253 503.649 512.363 432.345

400 499.666 514.362 541.163 461.492

400 501.193 520.447 528.209 450.040

420 532.883 515.129 528.872 461.946

420 529.755 523.271 526.875 437.139

537.311 347.659 480.764

510
546.001 532.960 490.380
510
555.746 576.794 582.805 490.150
600
549.318 550.727 559.805 502.340
600
 Ill

TABLE (16)
EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM
COMPRESSIVE FORCES ON THE UPPER SURFACE OF L-5 FOR THE LEG LIFT

Biomechanical
Lifting Maximum Compressive Force (kg.) on upper
Equivalents surface of L-5

"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 386.206 418.770 424.377 346.500

140 404.386 433.019 343.785

170 392.353 414.593 433.751 359.299

170 425.060 442.955 359.836

200 406.575 429.687 440.388 371.275

200 431.478 440.603 366.433

280 438.745 447.710 452.191 387.327

280 452.201 455.762 449.150 369.966

340 449.973 457.746 455.893 385.824

340 453.109 466.514 404.408

400 468.294 489.444 478.536 404.744

400 478.867 466.830 477.814 410.398

420 470.368 482.875 469.608 432.558

420 476.686 417.743

510 487.375 509.171 490.538 437.575

510 497.779 504.754 486.741 435.396

600 523.973 522.690 496.301 461.249

600 516.988 510.694 507.313 453.450

 112

TABLE ( 1 7 )

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

COMPRESSIVE FORCES ON THE UPPER SURFACE OF L-5 FOR THE BACK LIFT

Biomechanical Maximum Compressive Force (kg.) on upper

Lifting surface of L-5
Equivalents
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 399.818 449.949 448.330 372.902

140 394.971 449.766 354.657

170 407.197 462.704 454.594 367.275

170 404.027 460.474 448.486 379.405

200 451.232 475.448 465.235 385.684

200 421.419 470.476 469.286 386.815

280 454.661 486.052 484.556 399.770

280 431.187 480.743 489.918 368.331

340 364.593 493.429 503.504 419.881

340 455.671 492.690 501.754 421.807

400 488.302 504.567 527.797 450.082

400 490.289 509.335 517.746 440.674

420 519.124 504.595 517.519 450.857

420 517.093 511.548 515.289 525.415

510 523.748 535.865 468.887

510 533.398 520.892 478.763

600 544.282 564.236 571.386 476.947

536.905 537.731 548.183 493.802

600
 113

TABLE (18)
EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM
COMPRESSIVE^-EORCES-OLTHL-LOWEB-SURFACE QI- L-4 FOR THE LEG LIFT

Biomechanical
Lifting Maximum Compressive Force (kg.) on lower
Equivalents surface of L-4
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 385.855 419.532 425.296 346.325

140 405.233 433.880 343.944
170 392.178 415.326 434.504 359.547
170 425.836 443.788 359.671

200 407.445 430.156 440.674 371.637

200 431.919 441.246 366.739

280 438.994 448.407 452.512 387.810

280 452.116 456.737 449.926 370.015

340 449.539 458.578 456.561 386.197

340 454.090 467.248 405.243

400 468.101 490.515 479.414 404.577

400 479.511 467.734 478.769 410.559

420 469.657 483.817 470.611 432.824

420 477.468 417.896

510 486.633 510.107 491.432 436.698

510 498.510 505.832 487.696 435.076

600 524.300 523.796 497.076 461.145

600 517.826 511.609 508.390 454.454

 114

TABLE ( 1 9 )

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

COMPRESSIVE FORCES ON THE LOWER SURFACE OF L-4 FOR THE BACK LIFT

Biomechanical
Lifting Maximum (Compressive Force (kg.) on lower
Equivalents surface of L-4

"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 397.040 445.188 444.256 369.476

140 393.113 443.486 351.325

170 402.483 457.504 449.440 363.832

170 399.233 455.020 442.928 378.335

200 447.349 473.120 460.950 383.306

200 418.004 467.544 467.078 386.005

280 449.565 481.370 482.750 396.211

280 426.593 475.914 486.973 364.465

340 460.775 491.027 499.272 417.093

340 450.438 488.758 498.859 418.273

400 485.737 503.330 523.336 448.157

400 486.745 505.905 514.086 438.149

420 513.410 501.321 514.603 447.984

420 513.235 506.566 512.801 420.410

510 519.057 531.771 464.570

510 529.069 517.953 475.827

600 541.770 561.895 568.583 471.857

600 533.534 532.730 543.840 491.694

 115

relationship between the biomechanical lifting equivalents and the

maximum compressive forces at the four spinal levels. For each

spinal level, the regression analysis was carried out for each

method of lift by each subject. Regression analyses were carried

out for each method of lift for the four subjects combined. In each

case, the relationship between the biomechanical lifting equivalents

and the maximum compressive forces on the different spinal levels Is

a straight line. The equations of these straight lines are as

follows:

Upper Surface of S-1, Leg Lifts:

1st subject USl = 357 + 0.27911 BLE . (211)

2nd subject USl = 382 + 0.23336 BLE.. (212)

3rd subject USl = 409 + 0.15458 BLE . (213)

4th subject USl = 319 + 0.23571 BLE . (214)

all subjects USl = 367 + 0.22373 BLE (215)

Upper surface of S-1. Back Lifts:

1st subject USl = 358 + 0.32710 BLE (216)

2nd subject USl = 421 + 0.22603 BLE .(217)

3rd subject USl = 419 + 0.23289 BLE .(218)

4th subject USl = 325 + 0.27684 BLE .(219)

all subjects USl = 379 + 0.26558 BLE .(220)

Lower Surface of L-5. Leg Lifts:

1st subject LL5 = 357 + 0.28146 BLE .(221)

2nd subject LL5 = 387 + 0.23456 BLE .(222)

3rd subject LL5 = 417 + 0.15730 BLE .(223)

 116

4th subject : LL5 = 319 + 0.23649 BLE (224)

all subjects : LL5 = 371 + 0.22448 BLE (225)
Lower Surface of L-5, Back Lifts:

1st subject LL5 = 367 + 0.33310 BLE (226)

2nd subject LL5 = 436 + 0.20605 BLE (227)
3rd subject LL5 = 429 + 0.23806 BLE (228)
4th subject LL5 = 333+ 0.28328 BLE (229)
all subjects LL5 = 390 + 0.26610 BLE (230)
Upper Surface of L-5, Leg Lifts:

1st subject UL5 = 355 + 0.28051 BLE (231)

2nd subject UL5 = 382 + 0.23496 BLE (232)
3rd subject UL5 = 409 + 0.15665 BLE (233)
4th subject UL5 = 317 + 0.23699 BLE (234)

all subjects UL5 = 366 + 0.22505 BLE (235)

Upper Surface of L-5, Back Lifts:

1st subject UL5 = 357 + 0.32792 BLE (236)

2nd subject UL5 = 425 + 0.20534 BLE (237)

3rd subject UL5 = 419 + 0.23493 BLE (238)

4th subject UL5 = 325 + 0.27784 BLE (239)

all subjects UL5 = 380+ 0.26241 BLE (240)

Lower Surface of L-4, Leg Lifts:

1st subject LL4 = 354 + 0.28123 BLE (241)

2nd subject LL4 = 382 + ).23578 BLE (242)

3rd subject LL4 = 410 + 0.15723 BLE (243)

4th subject LL4 = 317 + 0.23681 BLE (244)

 117
all subjects : LL4 = 366 + 0.22537 BLE (245)
Lower Surface of L-4. Back Lifts:
1st subject LL4 = 353 + 0.32779 BLE (246)
2nd subject LL4 = 4 2 0 + 0.20876 BLE (247)
3rd subject LL4 = 414 + 0.23654 BLE (248)
4th subject LL4 = 322 + 0.27520 BLE (249)
all subjects LL4 = 376 + 0.26289 BLE (250)
where:

USl = maximum compressive force in kg. on upper surface of S-1.

LL5 = maximum compressive force in kg. on lower surface of L-5.
UL5 = maximum compressive force in kg. on upper surface of L-5.
LL4 = maximum compressive force in kg. on lower surface of L-4.
BLE = biomechanical lifting equivalent in lbs.in.
From the current experimental data, for all the subjects, the
back lifts produce higher maximum compressive forces on all levels
of the spine than the leg lifts. This is because the back lifts
produce higher maximum net torques at the hip than the leg lifts.
In all the current experimental treatments, the maximum compressive
forces on the different levels of the spine take place in the first
quarter of the lifting time and at the same time that the maximum
net torques occur at the hip. Figures 28. and 29. show the critical
position where the maximum compressive forces occur for leg and back
lifts respectively. The critical positions occur whenever the moment
arms are at their maximum levels. The maximum level of the moment
arms in the back lifts are more than those of the leg lifts. For both
 118

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methods of lift, increasing the box size increases the moment arm
and results in higher compressive force on all levels of the spine.
The difference in maximum compressive force on any spinal
level, between the back and leg methods of lift, increases by
increasing the biomechanical lifting equivalent. Also, a fast leg
or back lift increases the inertial forces due to acceleration,
therefore, a slower lift would decrease the forces on the spine.
Table 20. is the ANOVA for the simple regression between the
biomechanical lifting equivalents and the maximum compressive forces
on the upper surface of S-1 for the leg lifts of the third
subject and Table 21. for his back lifts. ANOVA of the current
experimental data on lower surface of L-5 for the leg lifts of the
third subject are presented in Table 22, and for his back lifts in
Table 23. ANOVA of the current experimental data on upper surface
of L-5 for the leg lifts of the third subject are presented in
Table 24., and for his back lifts in Table 25. ANOVA of the current
experimental data on lower surface of L-4 for the leg lifts of the
third subject are presented in Table 26., and for his back lifts
in Table 27. Figures 30. through 33. show the linear relationship
between the biomechanical lifting equivalents and the maximum
compressive forces on the different spinal levels of the third
subject for both his leg and back lifts. Tables 28. through 31.
are the regression analysis data for the linear relationships between
the biomechanical lifting equivalents and the maximum compressive
forces on the different spinal levels of the four subjects.
 121
TABLE (20)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL LIFTING
EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON UPPER
SURFACE OF S-1 FOR THIRD SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 9480.540 9480.540 389.726

Deviation about Regression 16 389.219 24.326
Total 17 9869.759

F99% (1,16) = 8.53

TABLE (21)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL LIFTING
EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON UPPER
SURFACE OF S-1 FOR THIRD SUBJECT (BACK LIFT)

Source df SS MS F

Due to Regression 1 19238.172 19238.172 282.040

Deviation about Regression 15 1023.162 68.211

Total 16 20261.334

F99% (1,15) = 8.68

 122

TABLE (22)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL LIFTING
EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON THE LOWER
SURFACE OF L-5 FOR THIRD SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 9814.211 9814.211 441.875

Deviation about Regression 16 355.366 22.210
Total 17 10169.577

F99% (1.16) = 8.53

TABLE (23)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL LIFTING
EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON THE LOWER
SURFACE OF L-5 FOR THIRD SUBJECT (BACK LIFT)

Source df SS MS F

Due to Regression 1 20098.852 20098.852 276.513

Deviation about Regression 15 1090.303 72.687

Total 16 21189.155

F99% (1,15) = 8.68

 123
TABLE (24)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL LIFTING
EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON THE UPPER
SURFACE OF L-5 FOR THIRD SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 9741.570 9741.570 421.669

Deviation about Regression 16 369.639 23.102
Total 17 10111.209

F99% (1,16) = 8.53

TABLE (25)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL LIFTING
EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON THE UPPER
SURFACE OF L-5 FOR THIRD SUBJECT (BACK LIFT)

Source df SS MS F

Due to Regression 1 19570.758 19570.758 275.344

Deviation about Regression 15 1066.162 71.077
Total 16 20636.920

F99% (1,15) = 8.68

 124

TABLE ( 2 6 )

ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL LIFTING

EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON THE LOWER

SURFACE OF L-4 FOR THIRD SUBJECT (LEG L I F T )

Source df SS MS

Due To Regression 9808.961 9808.961 414.580

Deviation about Regression 16 378.560 23.660

Total 17 10187.551

F99% (1.16) = 8.53

TABLE ( 2 7 )

ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL LIFTING

EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON THE LOWER

SURFACE OF L-4 FOR THIRD SUBJECT (BACK L I F T )

Source df SS MS F

Due to Regression 1 19850.813 19850. 813 263. 737

Deviation about Regression 15 1129.012 75. 267

Total 16 20979.825

F99% (1,15) = 8.68

 125

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 133

At all spinal levels and for both methods of lift, the

fourth subject had the least maximum compressive forces on his
spine. The fourth subject was the shortest of those used for the
current experimental data. Therefore, the distance the load
travelled was least. The lifting time was almost the same for
all subjects. This means that the velocity and acceleration of
lifting a load by the fourth subject was the least.
Appendix B. shows the statistical analysis of the maximum
compressive forces data at the different spinal levels of the
first, second and fourth subjects.
Appendix C. shows some of the acceleration patterns of the
second and fourth subjects. A description of the body motion is
presented also in Appendix C.
 134

4.3. SHEARING FORCES ON THE SPINE

The shearing forces at any spinal level and for both methods
of lift for all subjects never exceeds 100 kg. Compared to the
compressive forces, these shearing forces would not appear to be
critical. However, definite conclusions can not be made because
the resistance of the spine to shearing forces is not available.
The current experimental data for the maximum shearing forces
at the different biomechanical lifting equvalents are presented as
follows:
leg lifts, upper surface of S-1 Table 32.
back lifts, upper surface of S-1 Table 33.
leg lifts, lower surface of L-5 Table 34.
back lifts, lower surface of L-5 Table 35.
leg lifts, upper surface of L-5 Table 36.
back lifts, upper surface of L-5 Table 37.
leg lifts, lower surface of L-4 Table 38.
back lifts, lower surface of L-4 Table 39.
 135

TABLE (32)

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

SHEARING FORCES ON THE UPPER SURFACE OF S-1 FOR THE LEG LIFT

Biomechanical
Lifting Maximum Shearing Forces (kg.) on upper
Equivalent surface of S-1
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 21.444 27.975 36.713 22.530

140 28.702 37.066 23.992
170 21.444 26.501 36.356 22.530
170 27.749 35.978 32.219
200 21.444 34.171 39.694 22.530
200 35.201 37.645 23.738
280 24.366 34.569 52.352 25.452
280 24.993 27.124 50.062 27.296

340 24.366 26.960 46.432 25.452

340 34.356 46.227 25.452

400 24.572 28.366 39.676 25.452

400 24.366 36.513 37.477 28.395

420 27.287 34.490 42.196 28.373

420 39.045 32.888

510 27.287 29.881 41.545 30.313

510 27.287 29.881 41.179 28.373

600 27.287 30.497 43.868 28.373

600 27.287 29.881 38.745 29.781

 136
TABLE ( 3 3 )

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

SHEARING FORCES ON THE UPPER SURFACE OF S-1 FOR THE BACK LIFT

Biomechanical
Lifting Maximum Shearing Forces (kg.) on upper
Equivalents surface of S-1
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 51.818 62.590 54.437 49.187

140 48.444 62.196 50.089
170 55.921 62.789 57.482 45.779
170 54.441 61.429 54.524 37.641
200 58.802 47.626 55.776 47.214
200 49.365 51.935 54.315 45.265
280 61.526 58.557 56.764 55.363
280 57.354 58.557 58.665 56.408

340 59.054 57.582 61.481 50.218

340 57.911 59.602 56.047 51.697

400 54.352 54.858 67.534 56.662

400 55.331 58.646 57.721 50.162

420 69.295 65.759 61.084 58.837

420 65.532 66.811 61.447 61.020

510 63.760 63.353 69.951

510 67.482 63.106 57.863

600 66.495. 65.231 63.044 66.344

600 61.386 64.996 61.584 54.420

 137

TABLE ( 3 4 )

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

SHEARING FORCES ON THE LOWER SURFACE OF L-5 FOR THE LEG LIFT

Biomechanical
Maximum Shearing Forces (kg.) on lower
Lifting surface of L-5
Equivalents
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 13.588 26.455 24.582 14.277

140 23.217 24.551 21.872

170 13.588 25.490 25.341 14.277

170 24.395 26.001 33.114

200 13.588 38.270 34.697 14.277

200 40.033 32.299 20.426

280 15.440 25.425 45.649 16.128

280 16.522 19.184 42.103 21.108

340 15.440 17.084 40.589 16.128

340 25.291 39.642 16.128

400 22.949 17.084 28.804 18.335

400 15.440 30.640 25.277 24.857

420 17.291 27.458 29.848 17.979

420 26.760 25.761

17.291 18.935 29.827 21.296

510
17.291 18.935 28.735 17.979
510
17.291 31.873 32.141 17.979
600
17.291 18.935 25.107 24.827
600
 138

TABLE (35)

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

SHEARING FORCES ON THE LOWER SURFACE OF L-5 FOR THE BACK LIFT

Biomechanical
Maximum Shearing Forces (kg.) on lower
Lifting
Equivalents surface of L-5

"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 48.573 59.594 53.372 43.977

140 44.077 59.906 45.844

170 52.352 59.831 55.404 40.626

170 50.719 58.971 52.866 36.502

200 53.180 42.884 54.603 42.628

200 43.997 44.717 53.023 37.409

280 57.002 55.964 55.271 50.114

280 53.390 55.964 56.706 52.017

340 51.897 54.402 59.283 42.883

340 53.392 55.754 54.337 51.068

400 49.844 50.127 64.476 49.965

400 46.049 54.042 54.059 43.824

420 63.434 63.223 59.228 50.721

420 57.922 63.475 60.531 55.389

57.955 61.017 63.739

510
60.249 60.641 49.719
510
57.792 58.975 60.863 60.033
600
56.304 56.644 60.476 57.571
600
 139

TABLE (36)

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

SHEARING FORCES ON THE UPPER SURFACE OF L-5 FOR THE LEG LIFT

Biomechanical Maximum Shearing Forces (kg.) on upper

Lifting surface of L-5
Equivalents
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 18.098 25.750 30.853 19.006

140 24.193 31.248 22.049

170 18.098 24.276 30.983 19.006

170 24.631 30.648 30.888

200 18.098 33.082 35.666 19.006

200 34.115 33.833 19.450

280 20.707 29.770 48.367 21.615

280 21.368 22.876 45.858 24.360

340 20.707 22.876 42.263 21.615

340 29.262 41.964 21.615

400 23.071 22.876 34.566 21.615

400 20.707 30.879 31.766 26.235

420 23.316 30.671 36.817 24.224

420 33.858 29.487

510 23.316 25.486 36.297 26.365

510 23.316 25.486 35.381 24.224

600 23.316 29.936 38.843 24.224

600 23.316 25.486 32.615 26.799

 140

TABLE (37)

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

SHEARING FORCES ON THE UPPER SURFACE OF L-5 FOR THE BACK LIFT

Biomechanical Maximum Shearing Forces (kg.) on upper

Lifting surface of L-5
Equivalents
"lbs.in." 1st subject 2nd subject 3rd subject 4th subj«=^ct

140 48.989 59.241 51.420 46.180

140 45.653 59.042 47.112

170 53.095 59.542 54.308 42.891

170 51.610 58.318 51.541 35.725

200 55.480 44.503 52.695 44.325

200 46.348 48.421 51.309 41.553

280 58.697 55.801 53.869 52.321

280 54.599 55.801 55.720 53.466

340 55.763 54.762 58.406 47.004

340 55.210 56.706 53.211 48.703

400 51.546 51.664 64.078 53.239

400 51.484 55.631 54.535 47.002

420 66.461 62.927 58.269 55.281

420 62.291 64.180 58.604 58.110

510 60.969 60.415 66.633

510 64.452 60.175 54.400

600 62.921 62.233 60.144 63.601

600 60.137 61.396 58.749 52.725

 141

TABLE (38)

EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM

SHEARING FORCES ON THE LOWER SURFACE OF L-4 FOR THE LEG LIFT

Biomechanical
Maximum Shearing Forces (kg.) on lower
Lifting surface of L-4
Equivalents
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 17.666 25.917 29.738 18.553

140 24.033 30.271 22.038

170 17.666 24.501 30.272 18.553

170 24.739 30.315 30.699

200 17.666 31.988 36.078 18.553

200 32.787 34.143 19.434

280 20.214 29.349 49.108 21.100

280 20.901 22.331 46.310 24.026

340 20.214 22.331 42.743 21.100

340 29.016 42.228 21.100

400 23.055 22.331 34.183 21.100

400 20.214 30.841 31.078 26.241

420 22.761 30.432 36.068 23.647

420 32.888 29.051

510 22.761 24.878 35.847 25.869

510 22.761 24.878 34.712 23.647

600 22.761 29.918 38.146 23.647

600 22.761 24.878 31.367 1 26.776

 142

TABLE (39)
EFFECT OF THE BIOMECHANICAL LIFTING EQUIVALENTS ON THE MAXIMUM
SHEARING FORCES ON THE LOWER SURFACE OF L-4 FOR THE BACK LIFT

Biomechanical
Lifting Maximum .Shearing Forces (kg.) on lower
Equivalents surface of L-4
"lbs.in." 1st subject 2nd subject 3rd subject 4th subject

140 49.620 59.865 51.165 47.482

140 46.738 59.318 48.142

170 53.946 59.978 54.533 44.211

170 52.591 58.624 51.797 35.859

200 56.743 45.755 53.038 45.385

200 47.685 49.940 51.194 42.133

280 59.861 56.337 54.150 53.569

280 55.549 56.337 55.874 54.549

340 57.456 55.502 58.632 48.542

340 56.546 57.765 53.283 50.320

400 52.690 52.774 64.946 54.475

400 53.125 56.827 55.578 48.434

420 68.073 63.279 58.416 56.785

420 63.992 64.964 59.007 59.654

510 62.196 60.728 68.228

510 66.042 60.486 55.937

600 64.636 63.906 60.352 64.947

600 61.989 63.010 58.686 52.598

 CHAPTER V
CONCLUSIONS AND RECOMMENDATIONS FOR FUTURE RESEARCH
5.1 CONCLUSIONS
The significant conclusions of the current investigation can
be summarized as follows:
1. A biomechanical dynamic model for lifting in the sagittal plane
has been mathematically developed. The model was used for analyzing
the amount of physical stresses on a person's musculoskeletal system
by infrequent material handling tasks. The methodology is based
upon the concept that the physical stresses incurred by the
musculoskeletal system can be analyzed by applying the well-known
laws of engineering mechanics to the human body.
2. Comparison with other biomechanical lifting models is not yet
feasible, because this is the first operative model.
3. Slote and Stone space-time relationship was used in the develop-
ment of the model, and it is validated for the determination of
angular displacements for each joint's motion. For any joint's
motion, the total time to be used in Slote and Stone equation is the
difference in time between the start and end of the motion for that
joint.
4. A biomechanical lifting equivalent takes into account both the
weight to be lifted and the moment arm. Increasing the biomechanical
lifting equivalent by increasing the weight and/or the moment arm
will increase the resulting net torques at the hip joint. The net
torques at the hip joint are the main components of the compressive

143
 144
forces on the spine.

5. For both methods of lift, the relationship between the

biomechanical lifting equivalents and the maximum net torques at the
hip joint is a straight line. For the same biomechanical lifting
equivalent, the back lift produces greater maximum net torques at
the hip joints than the leg lift. The leg and back lift lines
diverge by increasing the biomechanical lifting equivalents.
6. Increasing the biomechanical lifting equivalents will increase
the maximum compressive forces on all spinal levels, for both the
leg and back methods of lift, with straight line relationships.
The leg and back lift lines diverge by increasing the biomechanical
lifting equivalent.
7. The back lift produces greater maximum compressive forces on
all spinal levels than the leg lift, at the same biomechanical
lifting equivalent.
8. The shearing forces at any spinal level and for both methods of
lift never exceeds 100 kg. Compared to the compressive forces,
these shearing forces would not appear to be critical. However,
definite conclusions can not be made because the resistance of the
spine to shearing forces is not available.

5.2 RECOMMENDATIONS FOR FUTURE RESEARCH

A simple lift in the sagittal plane has been used for dynamic
analysis. The basic model designed for current research produced
specific information for industry to use in combatting injuries
 145
of the lower back. The basic model should be expanded so as to be
able to analyze and calculate the stresses as follows:
1. Following the current sagittal plane lift, step forward and place
the weight box 1 , 2 , and 3 feet from the front edge of a bench.
2. Lifting the weight box from the bench and placing it down on the
floor.
3. Turning (twisting) the body while lifting the weight box from the
floor to the elbow height.
4. Twisting the body while lowering the weight box from the bench
to the floor.
5. Carrying the weight box a certain distance to determine the
fatigue effects in these cases:
a - the load in front of the body.
b - the load in one side like single suitcase.
c - the load in two sides like two suitcases.
 BIBLIOGRAPHY

Albrock, D. and Uganda, K. Movement of the Lumbar Spinal Column,

Journal of Bone and Joint Surgery. Vol. 39B, 1957.
Armstrong, J.R. Lumbar Disc Lesions. E.S. Livingstone Ltd., 3rd
edition, 1965.
Asmussen, E. and Poulsen, E. On the Role of Intra-Abdomlnal Pressure
in Relieving the Back Muscles While Holding Weights In a Forward
Inclined Position, No. 28, Danish National Association for
Infantile Paralysis. Hellerup, Denmark, 1968.
Barr, J.S. and Mixter, J. Posterior Protrusion of the Lumbar
Intervertebral Discs, Journal of Bone and Joint Surgery, 1941.
Bartelink, D.L. The role of Abdominal Pressure In Relieving the
Pressure on the Lumbar Intervertebral Discs, Journal of Bone and
Joint Surgery. Vol. 39B, 1957.
Brown, J.R. Lifting. Department of Physiological Hygiene, School
of Hygiene, University of Canada, 1970.
Cailliet, R. Low Back Pain Syndrome, F.A. Davis Company,
Philadelphia, 1962.
Cathcart, E.P., Bedale, E.M., Blair, C , McLeod, K. and Weatherhead, E
The physique of Women In Industry. Industrial Fatigue Research
Board, Report No. 71, cit from Physical Education, Vol. 59, 1957.
Chaffin, D. A computarized Biomechanical Model-Development of and
Use in Studying Gross Body Actions. Journal of Biomechanics,
Vol. 2, 1969.

146
 147

Chaffin, D. An Empirical Investigation of Low-Back Strains and

Vertebrae Geometry, Journal of Biomechanics. Vol. 2, 1969.
Chaffin, D. Personal Communications, January, 1964.
David, P.R., Troup, J. and Burnhard, J. Movements of the Thoracic
and Lumbar Spine When Lifting. Journal of Anatomy. London, Vol
99, 1965.

Dempster, W.T. Space Requirements of the Seated Operator. WADC

Tech. Report, No. 55-159, University of Michigan, 1955.
Drillis, R. and Contini, R. Body segment Parameters, Tech Report
No. 1166-03. New York University, School of Engineering and
Science, Research Division, N.Y., 1966.
Dryden, R.D. A Predictive Model for the Maximum Permissible Weight
of Lift from Knuckle to Shoulder Height. Ph.D. Dissertation,
Texas Tech University, Lubbock, Texas, 1973.
Eckert, C. and Decker, A. Pathological Studies of Intervertebral
Discs, Journal of Bone and Joint Surgery, Vol. 29. 1947.
Evans, F. Studies on the Energy Absorbing Capacity of Human Lumbar
Intervertebral Discs, The Seventh Strapp Car Crash Conference
Proceeding. Springfield, Illinois, 1966.
Fich, R. Handbook der Anatomic und Mechnik der Gelenke. Von Gustav
Fischer, Jena, 1904.
Fisher, B. A Biomechanical Model for the Analysis of Dynamic
Activities. M.S.I.E. Thesis, University of Michigan, Ann Arbor,
Michigan, 1967.
Gordon, E.E. Natural History of the Intervertebral Disc. Archs
 148
Phys. Med.. 1961.
Grandjean, E. Fitting the Task to the Man. London, Taylor and
Francis Ltd., 1969.
Hanovan, E.P. A Mathematical Model for the Human Body. A.D. Rep.
No. 608, 1964.
Hirsh, C. Studies on the Mechanism of Low Back Pain. Acta Orthopoedica,
Vol. 20, No. 4, 1951.
International Labor Organization, Maximum Permissible Weight to be
Carried by One Worker. International Labor Organization Office,
Geneva, 1966.
Jones, D.F. Back Strains: The State of the Art. Journal of Safety
Research, Vol. 3, No. 1, 1971.
Lindahl, 0. Determination of the Sagittal Mobility of the Lumbar
Spine. Acta Orthopoedica, Vol. 37, 1966.
Magora, A. and Taustein, I. An Investigation of the Problem of Sick
Leave in the Patient Suffering from Low Back Pain. Industrial
Medicine and Surgery, Vol. 38, No. 11, 1969.
McDaniel, J. Prediction of Acceptable Lift Capability. Ph.D.
Dissertation, Texas Tech University, Lubbock, Texas, 1972.
McFarland, R.A. Injury - A Major Environmental Problem. Archives
of Environmental Health, Vol. 19, 1969.
Mitchell, M. The Lumbosacral Junction, Journal of Bone and Joint
Surgery, Vol. 16, 1934.
Morris, J.M., Lucas, D.B. and Bresler, B. Role of the trunk in
Stability of the Spine. Journal of Bone and Joint Surgery, Vol.
43A, No. 3, 1961.
 149
Munchinger, R. Manual Lifting and Carrying. International Occupational
Safety and Health Information Sheet, No. 3, 1962, Geneva, Swit.
Nachemson, A. Some Mechanical Properties of the Lumbar Intervertebral
Discs. Bulletin Hospital Joint Disease. Vol. 23, No. 2, 1962.
Pearson, J.R. and McGinley, D.R. Dynamic Analysis of the Upper
Extremity for Planar Motions, Tech. Report No. 04468. University
of Michigan, Medical School, Office of Research Administration,
Ann Arbor, Michigan, 1961.
Perey, 0. Fracture of the Vertebral End-PIate in the Lumbar Spine.
Acta Orthopoedica, Supplement 25, 1957.
,Plagenhoef, S.C. Methods for Obtaining Kinetic Data to Analyze Human /
Motions. Research Q., Vol. 37, 1966. —'
Poulsen, E. Prediction of Maximum Loads in Lifting from Measurements
of Muscular Strength. Communications No. 31, Danish National
Association for Infantile Paralysis, Hellerup, Denmark, 1970.
Roaf, R. A Study of the Mechanics of Spinal Injuries, Journal of
Bone and Joint Surgery, Vol. 42B, No. 4, 1960.
Rolander, S.D. Motion of the Lumbar Spine with Special Reference to
the stability Effect of Posterior Fusion. Acta Orthopoedica,
Supplement 90, 1966.
Ronnholn, N. Physiological Studies on the Optimum Rhythm of Lifting
Work. Ergonomics, Vol. 5, 1962.
Ronnholn, N., Karvonen, M.J. and Lapinleimu, V. Mechanical Efficiency
of Rhythmic and Paced Work of Lifting. Journal of Applied
Physiology, Vol. 17, 1962.
 150

Slote, L. and Stone, G. Biomechanical Power Generated by Forearm

Flexion. Human Factors. Vol. 5, No. 5, 1963.
Smith, A., Deery, E. and Hagman, G. Herniation of the Nucleus
Pulposus. A Study of 100 cases Treated by Operation. Journal
of Bone and Joint Surgery. Vol. 26, 1944.
^.Sonoda, T. Studies on the Compression, Tension and Torsion Strength
of the Human Vertebral Column. Journal of Kyoto Prefect Med.
Iniv., Vol. 71, 1962.
Snook, S.H. and Irvine, C.H. Maximum Acceptable Weight of Lift.
American Industrial Hygiene Association Journal. Vol. 28, 1967.
Snook, S.H. and Irvine, C.H. Maximum Frequency of Lift Acceptable
to Male Industrial Workers. American Industrial Hygiene
Association Journal, Vol. 29, 1968.
Thieme, F.P. Lumbar Breakdown Caused by Erect Posture in Man.
Museum of Anthropology, University of Michigan, No. 4, 1950,
University Press, Ann Arbor, Michigan.
Tichauer, E.R. A Pilot Study of the Biomechanics of Lifting in
Simulated Industrial Work Situations. Journal of Safety Research
Vol. 3, No. 3, 1971.
^ Troup, J.D. Relation of Lumbar Spine Disorders to Heavy Manual
Work and Lifting. The Lancet, April 17, 1965.
Van Wely, P.A. Manual Lifting of Loads. Ergonomic Research Annual
Conference, 1961.
APPENDIX

151
 152
APPENDIX A
STATISTICAL ANALYSIS OF THE MAXIMUM NET TORQUES AT THE HIP JOINT

1. TABLES:

ANOVA for the simple linear regression between thia biomechanical

lifting equivalents and the maximum net torques at the hip joint for:
leg lift of 1st subject Table A-1
back lift of 1st subject Table A-2
leg lift of 2nd subject Table A-3
back lift of 2nd subject Table A-4
leg lift of 4th subject Table A-5
back lift of 4th subject Table A-6
leg lift of the four subjects together Table A-7
back lift of the four subjects together ....Table A-8

2. FIGURES:
The relationships between the biomechanical lifting equivalents and
the maximum net torques at the hip joint for both methods of lift
for:
1st subject Figure A-1
2nd subject Figure A-2
3rd subject Figure A-3
 153
TABLE (A-1)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM NET TORQUES
AT HIP FOR FIRST SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 772447.937 772447.937 208.437

Deviation about Regression 11 40764.914 3705.901
Total 12 813212.851

Fgg^ (1.11) = 9.65

TABLE (A-2)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM NET TORQUES
AT HIP FOR FIRST SUBJECT (BACK LIFT)

Source df SS MS F

Due to Regression 1 1503853.000 1503853.000 158.218

Deviation about Regression 16 152078.688 9504.918

Total 17 1655931.688

F99% (1,16) = 8.53

 154
TABLE (A-3)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM NET TORQUES
AT HIP FOR SECOND SUBJECT(LEG LIFT)

Source df SS MS F

Due to Regression 1 709689.813 709689.813 282.269

Deviation about Regression 15 37717.465 2514.231
Total 16 747403.278

F99% (1,15) = 8.68

TABLE (A-4)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM NET TORQUES
AT HIP FOR SECOND SUBJECT (BACK LIFT)

Source df SS MS. F

Due to Regression 1 472586.000 472586.000 199.507

Deviation about Regression 14 55362.308 3954.450
Total 15 527948.308

F99% (1,14) = 8.86

 155
TABLE (A-5)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM NET TORQUES
AT HIP FOR FOURTH SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 555502.438 555502.438 398.611

Deviation about Regression 16 22297.512 1393.594
Total 17 577799.950

F99% (1,16) = 8.53

TABLE (A-6)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM NET TORQUES
AT HIP FOR FOURTH SUBJECT (BACK LIFT)

Source df SS MS F

Due to Regression 1 974626.938 974626.938 283.701

Deviation about Regression 16 54966.328 3435.396

Total 17 1029593.266

F99% (1,16) = 8.53

 156
TABLE (A-7)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM NET TORQUES AT HIP
FOR THE FOUR SUBJECTS TOGETHER (LEG LIFT)

Source df SS MS F

Due to Regression 1 2341132.000 2341132.000 40.007

Deviation about Regression 64 3745164.000 58518.188
Total 65 6086296.000

F99% (1,64) = 7.08

TABLE (A-8)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM NET TORQUES AT HIP
FOR THE FOUR SUBJECTS TOGETHER (BACK LIFT)

Source df SS MS F

Due to Regression 1 3202782.000 3202782.000 68.209

Deviation about Regression 67 3146033.000 46955.715
Total 68 6348815.000

F99% (1,67) = 7.08

 157

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 160
APPENDIX B
STATISTICAL ANALYSIS OF THE MAXIMUM COMPRESSIVE
FORCES ON THE DIFFERENT SPINAL LEVELS
1. TABLES:
ANOVA for the simple linear regression between the biomechanical
lifting equivalents and the maximum compressive forces on upper surface
of S-1 for:
leg lift of 1st subject Table B-1
back lift of 1st subject Table B-2
leg lift of 2nd subject Table B-3
back lift of 2nd subject Table B-4
leg lift of 4th subject Table B-5
back lift of 4th subject Table B-6
leg lift of all subjects Table B-7
back lift of all subjects Table B-8
ANOVA for the simple linear regression between the biomechanical
lifting equivalents and the maximum compressive forces on lower
surface of L-5 for:
leg lift of 1st subject Table B-9
back lift of 1st subject Table B-10
leg lift of 2nd subject Table B-11
back lift of 2nd subject Table B-12
leg lift of 4th subject Table B-13
back lift of 4th subject Table B-14
leg lift of all subjects Table B-15
 161
back lift of all subjects Table B-16
ANOVA for the simple linear regression between the biomechanical
lifting equivalents and the maximum compressive forces on upper
surface of L-5 for:

leg lift of 1st subject .Table B-17

back lift of 1st subject Table B-18
leg lift of 2nd subject Table Brl9
back lift of 2nd subject Table B-20
leg lift of 4th subject Table B-21
back lift of 4th subject Table B-22
leg lift of all subjects Table B-23
back lift of all subjects Table B-24
ANOVA for the simple linear regression between the biomechanical
lifting equivalents and the maximum compressive forces on lower surface
of L-4 for:
leg lift of 1st subject Table B-25
back lift of 1st subject Table B-26
leg lift of 2nd subject Table B-27
back lift of 2nd subject Table B-28
leg lift of 4tn subject Table B-29
back lift of 4th subject Table B-30
leg lift of all subjects Table B-31
back lift of all subjects Table B-32
2. FIGURES:
The relationships between the biomechanical lifting equivalents and
 162
the maximum compressive forces on spine for both methods of lift for:
1st subject, upper surface of S-1 Figure B-1
2nd subject, upper surface of S-1 Figure B-2
4th subject, upper surface of S-1 Figure B-3
1st subject, lower surface of L-5 Figure B-4
2nd subject, lower surface of L-5 Figure B-5
4th subject, lower surface of L-5 Figure B-6
1st subject, upper surface of L-5 Figure B-7
2nd subject, upper surface of L-5 Figure B-8
4th subject, upper surface of L-5 Figure B-9
1st subject, lower surface of L-4 Figure B-10
2nd subject, lower surface of L-4 Figure B-11
4th subject, lower surface of L-4 Figure B-12
 163
TABLE (B-1)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE
FORCES ON UPPER SURFACE OF S-1 FOR
FIRST SUBJECT (LEG LIFT)

Source df 3S MS F

Due to Regression 1 22437.062 22437.062 292.445

Deviation about Regression 11 843.946 76.722
Total 12 23281.008

F99% (1.11) = 9.65

TABLE (B-2)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES
ON UPPER SURFACE OF S-1 FOR FIRST SUBJECT (BACK LIFT)

Source df SS MS F

Due to Regression 1 42489.141 42489.141 219.117

Deviation about Regression 16 3102.566 193.910

Total 17 45591.707

F99% (1.16) = 8.53

 164
TABLE (B-3)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF S-1 FOR SECOND SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 21255.953 21255.953 381.740

Deviation about Regression 15 835.226 55.682
Total 16 22091.179

F99% (1,15) = 8.68

TABLE (B-4)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF S-1 FOR SECOND SUBJECT (BACK LIFT)
Li-

Source df SS MS

Due to Regression 1 16963.711 16963.711 439.470

Deviation about Regression 14 540.406 38.600

Total 15 17504.117

F99% (1.14) = 8.86

 165
TABLE (B-5)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF S-1 FOR FOURTH SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 22062.684 22062.684 420.395

Deviation about Regression 16 839.694 52.481
Total 17 22902.378

?gg% (1,16) = 8.53

TABLE (B-6)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF S-1 FOR FOURTH SUBJECT (BACK LIFT)

Source df SS MS

Due to Regression 30433.418 30433.418 200.673

Deviation about Regression 16 2426.503 151.656

Total 17 32859.921

F99% (1.16) = 8.53

 166
TABLE (B-7)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF S-l FOR THE FOUR SUBJECTS (LEG LIFT)

Source df SS MS

Due to Regression 74326.687 74326.687 86.774

Deviation about Regression 64 54819.676 856.557
Total 65 129146.363

F99% (1.64) = 7.08

TABLE ( B - 8 )

ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL

LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON UPPER

SURFACE OF S-1 FOR THE FOUR SUBJECTS TOGETHER (BACK LIFT)

Source df SS MS F

Due to Regression 1 105112.938 105112.938 86.018

Deviation about Regression 67 81873.000 1221.985
Total 68 186985.938
— ' •

F99% ( 1 , 6 7 ) = 7 . 0 8
 167
TABLE (B-9)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-5 FOR FIRST SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 22819.309 22819.309 271.691

Deviation about Regression 11 923.890 83.990
Total 12 23743.199

F99% (1,11) = 9.65

TABLE (B-10)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-5 FOR FIRST SUBJECT (BACK LIFT)

Source df SS MS F

Due to Regression 1 44065.832 44065.832 210.855

Deviation about Regression 16 3343.786 208.987
Total 17 47409.618

F99% (1,16) = 8.53

 168
TABLE (B-11)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-5 FOR SECOND SUBJECT (LEG LIFT)

Source df SS MS TH
Due to Regression 1 21476.180 21476.180 391.997
Deviation about Regression 15 821.800 54.787
Total 16 22297.980
i
F995^ (1.15) = 8.68

TABLE (B-12)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-5 FOR SECOND SUBJECT (BACK LIFT)

Source „ SS MS ,

Due to Regression 1 14102.305 14102.305 328.779

Deviation about Regression 14 600.500 42.893
Total 14702.805

F99% (1.14) = 8.86

 169
TABLE (B-13)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
L9WER SURFACE OF L-5 FOR FOURTH SUBJECT (LEG LIFT)

Source df SS MS

Due to Regression 1 22208.695 22208.695 396.241

Deviation about Regression 16 896.775 56.048
Total 17 23105.470

F99% (1.16) = 8.53

TABLE (B-14)

ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL

LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON

LOWER SURFACE OF L-5 FOR FOURTH SUBJECT (BACK LIFT)

Source df SS MS

Due to Regression 1 31870.094 31870.094 217.150

Deviation about Regression 16 2348.243 146.765

Total 17 34218.337

F99% (1.16) = 8.53

 170
TABLE (B-15)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON LOWER
SURFACE OF L-5 FOR THE FOUR SUBJECTS TOGETHER (LEG LIFT)

Source df SS MS F

Due to Regression 1 74828.313 74828.313 75.774

Deviation about Regression 64 63200.949 987.515
Total 65 138029.262

"99% (1,64) = 7.08

TABLE (B-16)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON LOWER
SURFACE OF L-5 FOR THE FOUR SUBJECTS TOGETHER (BACK LIFT)

Source df SS MS F

Due to Regression 1 105527.875 105527.875 84.207

Deviation about Regression 67 83963.750 1253.190

Total 68 189491.625

F99% (1.67) = 7.08

 171

TABLE (B-17)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF L-5 FOR. FIRST SUBJECT (LEG LIFT)

Source df SS MS

Due to Regression 1 22663.891 22663.891 287.860

Deviation about Regression 11 866.055 78.732

Total 12 23529.946
i
F99% (1.11) = 9-65

TABLE (B-18)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF L-5 FOR FIRST SUBJECT (BACK LIFT)

Source df SS MS

1 42701.902 42701.902 221.043

Due to Regression
16 3090.935 42.680
Deviation about Regression
Total 17 45792.837
1
F99% (1.16) = 8.53
 172
TABLE (B-19)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF L-5 FOR SECOND SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 21544.699 21544.699 384.344

Deviation about Regression 15 840.836 56.056
Total 16 22385.535

F99% (1.15) = 8.86

TABLE (B-20)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF L-5 FOR SECOND SUBJECT (BACK LIFT)

Source df SS MS F

Due to Regression 1 13996.629 13996.629 327.950

Deviation about Regression 14 597.509 42.680

Total 15 14594.138

F99% (1.14) = 8.86

 173

TABLE (B-21)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF L-5 FOR FOURTH SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 22304.801 22304.801 420.526

Deviation about Regression 16 848.644 53.040
Total 17 23153.445

99% (1,16) = 8.53

TABLE (B-22)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF L-5 FOR FOURTH SUBJECT (BACK LIFT)

Source df SS MS

Due to Regression 1 30660.465 30660.465 200.654

Deviation about Regression 16 2444.842 152.803

Total 17 33105.307

F99% (1.16) = 8.53

 174
TABLE (B-23)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF L-5 FOR THE FOUR SUBJECTS TOGETHER (LEG LIFT)

Source df SS MS

Due to Regression 1 75211.188 75211.188 84.274

Deviation about Regression 64 57117.707 892.464
Total 65 132328.895

F99% (1.64) = 7.08

TABLE (B-24)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
UPPER SURFACE OF L-5 FOR THE FOUR SUBJECTS TOGETHER (BACK LIFT)

Source df SS MS F

Due to Regression 1 102612.063 102612.063 84.058

Deviation about Regression 67 81789.125 1220.733
Total 68 184401.188

F99% (1.67) = 7.08

 175
TABLE (B-25)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-4 FOR FIRST SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 22781.316 22781.316 280.185

Deviation about Regression 11 894.387 81.308

Total 12 23675.703

F99% (1.11) = 9.65

TABLE (B-26)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-4 FOR FIRST SUBJECT (BACK LIFT]
^

Source df SS MS

1 42668.219 42668.219 227.164

Due to Regression
Deviation about Regression 16 3005.276 187.830

Total 17 45673.495

Fggcg (1,16) = 8.53

 176

TABLE (B-27)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-4 FOR SECOND SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 21697.098 21697.098 385.239

Deviation about Regression 15 844.817 56.321

Total 16 22541.915

F99% (1.15) = 8.68

TABLE (B-28)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-4 FOR SECOND SUBJECT (BACK LIFT)

Source df SS MS

Due to Regression 1 14470.273 14470.273 280.728

Deviation about Regression 14 721.637 51.545

Total 15 15191.910

F99% (1.14) = 8.86

 177
TABLE (B-29)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-4 FOR FOURTH SUBJECT (LEG LIFT)

Source df SS MS F

Due to Regression 1 22268.691 22268.691 410.403

Deviation about Regression 16 868.169 54.261
Total 17 23136.960

F99% (1.16) = 8.53

TABLE (B-30)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-4 FOR FOURTH SUBJECT (BACK LIFT)

Source df SS MS

Due to Regression 1 30073.872 30073.872 176.428

Deviation about Regression 16 2727.350 170.459

Total 17 32801.222

F99% (1.16) = 8.53

 178
TABLE (B-31)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-4 FOR THE FOUR SUBJECTS TOGETHER (LEG LIFT)

Source df SS MS F

Due to Regression 1 75424.250 75424.250 83.101

Deviation about Regression 64 58088.090 907.626
Total 65 133512.340

99% (1,64) = 7.08

TABLE (B-32)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM COMPRESSIVE FORCES ON
LOWER SURFACE OF L-4 FOR THE FOUR SUBJECTS TOGETHER (BACK LIFT)

Source df SS MS

Due to Regression 102990.313 102990.313 85.385

Deviation about Regression 67 ; 80839.813 1206.564

Total 68 183830.126

F99% (1.67) = 7.08

 179

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APPENDIX C
1. Description of the body motion.
2. Figures:
a. Slote and Stone displacement-time relationship vs. the
current experimental data for:
Knee joint Figure C-1
Hip joint Figure C-2
Shoulder joint Figure C-3
Elbow joint Figure C-4
Wrist joint Figure C-5
b. Acceleration patterns at the center of gravity of the hand
for:
X - direction, 2nd subject Figure C-6
Y - direction, 2nd subject Figure C-7
X - direction, 4th subject Figure C-8
Y - direction, 4th subject Figure C-9
 192
DESCRIPTION OF BODY MOTION
Body motion during a lifting task not only varies from one
individual to another, but also varies for the same person according
to the method of lift. The current analysis of the body motion of the
four human subjects studied reveals more definite information
concerning an articulation and associated body component action and
motion. Each articulation has its own characteristics such as time
of starting and ending motion. For example, in a leg lift the knee
and hip joints start the motion at time zero followed by the ,
shoulder, the wrist and finally the elbow joint.
Leg lift start motion sequence in the average of the four exper-
imental subjects is as follows:
Knee joint at time zero
Hip joint at time zero

Shoulder joint after S% of the total lift time

Wrist joint after 12% of the total lift time

Elbow joint after 15% of the total lift time

Leg lift end motion sequence in the average of the four

experimental subjects:
Knee joint at 80% of the total lift time

Hip joint at 85% of the total lift time

Wrist joint at 88% of the total lift time

Elbow joint at 92% of the total lift time

Shoulder joint at the end of the lift

Therefore, a leg lift starts whenever the knee and hip joints
 193

start their motions and a leg lift ends whenever the shoulder joint
ends its motion.
Back lift start motion sequence in the average of the four
experimental subjects:
Knee joint at time zero
Hip joint after 6% of the total lift time
Shoulder joint after 6% of the total lift time
Elbow joint after 15% of the total lift time
Wrist joint after 15% of the total lift time
Back lift end motion sequence in the average of the four
experimental subjects:
Knee joint after 65% of the total lift time
Hip joint after 75% of the total lift time
Wrist joint after 82% of the total lift time
Elbow joint after 87% of the total lift time
Shoulder joint at the end of the lift
Therefore, a back lift starts whenever the knee joint starts its
motion and a back lift ends whenever the shoulder joint ends its

motion.
The velocity and acceleration of the leg l i f t motion are considered

zero for each articulation in these ranges:

Knee joint the last 20% of total time

Hip joint the last 15% of total l i f t time

Shoulder joint the f i r s t 5% of total l i f t time

 194
Elbow joint the first 15% and the last 8% of
total lift time
Wrist joint the first 12% and the last 12% of
total lift time
The velocity and acceleration of the back lift motion are considered
zero for each articulation in these ranges:
Knee joint : the last 35% of total lift time
Hip joint : the first 6% and the last 24% of
lift time
Shoulder joint the first 6% of total lift time
Elbow joint the first 15% and the last 13% of
total lift time
Wrist joint the first 15% and the last 18% of
total lift time
Slote and Stone displacement-time relationship fits the current
experimental data at every articulation. The total time to be used
in the equation for an articulation is the time difference between
the start and end of motion of this articulation.
For the leg lift, the equation is used in the following average

ranges:
Knee joint from 0% to 80% of total lift time

Hip joint from 0% to 85% of total lift time

Shoulder joint from 5% to 100% of total lift time

Elbow joint from 15% to 92% of total lift time

Wrist joint from 12% to 88% of total lift time

 195

For the back lift, the equation is used in the following average

ranges:
Knee joint from 0% ti 65% of total lift time
Hip joint from 6% to 76% of total lift time
Shoulder joint from 6% to 100% of total lift time
Elbow joint from 15% to 87% of total lift time
Wrist joint from 15% to 82% of total lift time
Figures C-1 through C-5 show the current experimental data with
Slote and Stone equation's data.
 196

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 202

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 204

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 205
APPENDIX D
COMPUTER PROGRAM DOCUMENTATION

The program starts on next page.

 206

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X 3 9 ) , X H A ( 3 9 ) , Y H A ( 3 9 ) , X F A ( 3 9 ) , Y F A ( 3 9 ) , X U A ( 39),YUA( 39),XF
XCGTK( 3 9 ) , Y F C G T K ( 3 9) ,XFT K( 39 ) ,Y FT K ( 39 ) ,XT K ( 3 9 ) , YTK( 39 ) ,
X X R 3 ( 3 9 ) , Y R 3 ( 3 9 ) , T R 3 ( 39) , X F C G U L ( 3 9 ) , Y F C G U L ( 3 9 ) ,BHGH(39)
X,8HGK(39)
D I M E N S I O N X F U L ( 3 9 ) , YFUL ( 3 9 ) , X U L ( 3 9) , YUL ( 39 ) ,XR2 ( 39 ) ,YR
X2 ( 3 9 ) , T R 2 ( 3 9 ) , X F C G L L ( 3 9 ) , YFCGLLC 3 9 ) , X F L L ( 39) , Y F L L ( 3 9 ) ,
> X L L ( 3 9 ) , Y L L ( 3 9) , X R I ( 3 9 ) , Y R 1 ( 3 9 ) , T R 1 ( 3 9 ) , A G 1 ( 3 9 ) , AG2( 39
X ) , A G 3 ( 3 9 ) , A G 4 ( 3 9 ) , A H G H ( 3 9 ) ,AHGK( 3 9)
DIMENSION H I P C H G ( 3 9 ) , D I A D I S { 3 9 ) , A B D Q M ( 3 9 ) , H ( 3 9 ) , A B P R E S
X( 3 9 ) , D F O P C E ( 3 9 ) , A B T n R Q ( 3 9 ) , H I P T H I ( 3 9 ) , T C R N E T ( 3 9 ) , R G l ( 3
X 9 ) , RG2{39 ) , C H G H ( 3 9 )
D I M E N S I O N F M L 5 S 1 ( 3 9 ) , F A B D 0 M { 3 9 ) , C S A C ( 3 9 ) , S S A C ( 3 9 ) , CGMU
X S l ( 3 9 ) , SHUSK 3 9 ) , C L L 5 ( 3 9 ) , S L L 5 ( 39) , C 0 M L L 5 ( 3 9 ) , S H L L 5 ( 3 9
X) , F M L 4 L 5 ( 3 9 ) , C U L 5 ( 3 9 ) , SUL5 ( 39 ) , COMUL 5 ( 39 ) , SHUL 5( 3 9 ) , C L
XL4( 3 9) , S L L 4 { 3 9) , C n M L L 4 ( 3 9 ) , S H L L 4 ( 3 9 )
D I MENS ION A N G S A C ( 3 9 ) , A N G L L 5 ( 3 9 ) , A N G U L 5 ( 39) , A N G L L 4 ( 3 9 ) ,
X A S A C ( 3 9 ) , A L L 5 (3 9 ) , AUL5 ( 39 > , ALL4 ( 39 ) , RS AC ( 39 )» PLL 5( 39 ) ,
X R U L 5 ( 3 9 ) , R L L 4 ( 3 9 ) , X F C G U 3 9 ) , Y F C G 1 ( 3 9 ) , X F C G 2 ( 3 9 ) ,YFCG2 (
X3 9 ) , X L 4 L 5 ( 3 9 ) ,Y L 4 L 5 ( 39 ) , X L5S I ( 39 ) , YL 5S 1( 3 9 ) , V D H ( 39) , vr
XH( 3 9 ) ,CHGK( 3 9 )
READ ( 5 , 9 2 ) N , N A , f M A A , N K , N K K , N H , N H H , N S , N S S , N E , N r E
92 FnRMAT(IlI5)
RE AD( 5 , 8) DMAXA, JMAXK,0'1AXH ,DMA X S , L)MAXE
8 FCRMAT(5FI0 .4 )
R E A O ( 5 , 1 0 ) C O N S T A , C n N S T K , CONSTH,CnNSTS,CONSTF
10 FORMAT(5FI0.4)
RE AD ( 5 , 3 1 ) A K D I S , H K D I S , H S U l S , S t D I S , L W D I b , H - ' \ N n , H H n i S
31 FORMAT( 7F 1 0 . 4 )
 207

RE AD ( 5 , 4 2 ) RAT 1 0 1 , RAT 1 0 2 , RAT 1 0 3 , RAT 1 0 4 , RATIO 5 , R A T I 0 6

42 F0RMAT(6FI0,4)
R E A D ( 5 , 10 1)TOTALM,PWTHA,PWTFA,PWTUA,PWTTK,PWTUL,PWTLL
101 F0RMAT(7F10.4)
RE AD ( 5 , 10 2)GYRAHA,GYRAFA,GYRAUA,GYRATK,GYRAUL,GYRALL
102 F0RMAT(6F10.4)
READ(5,103)WTLOAD
103 F0RMAT(F10.4)
R E A D ( 5 , 5 0 0 0 1 DELTA
5 0 0 0 FORMAT ( F 1 0 . 5 )
READ(5,897) D ISA,DISK,DISH
897 FCRMAT(3F10.4)
TWPI = 6 . 2 8 3 1 8 5 3 0 8
IF(N-l) 201,202,203
201 STOP
202 1=1
T(I)=0,
DDA(I)=CONSTA
DRA( I ) = D D A ( I ) * T W P I / 3 6 0 .
VRA(I)=0.
CRA( I ) = 0 .
. GC TO 5 0 0
203 IF(NA-l) 204,205,206
204 STOP
205 IF(N-NAA) 207,208,209
207 STOP
208 DO 2 1 0 1= 1,NAA
X=I
ZA=NA
Z = NAA
TA(I)=(X-ZA)«DELTA
T( I )=( X-1,)*DELTA
TIMEA=(Z-ZA)«DELTA
EA( I ) = T W P I * T A ( I ) / T I MEA
FA=DMAXA/TWPI
GA=TWPI/TIMEA
RA=DMAXA*TWPI/360.
CCA( I ) = F A * ( E A ( I ) - S I N ( E A ( I ) ) )fCONSTA
DRA(I)=DDA(I)*TWPI/360.
VRA(I )= RA*(GA-(GA*CGS(EA( I) ) ) ) / T W P I
210 C R A ( I ) = R A * G A * G A * S I N ( E A ( I ) )/TWP I
GO TO 5 0 0
209 DC 2 1 1 1= 1,NAA
X^I
Z=NAA
ZA = NA
T A ( I ) = ( X - Z A ) ' : ^ DELTA
T ( I ) = ( X-1.)*L)ELTA
 208

TIMEA=(Z-ZA)*DELTA
E A ( I )=TWPI*TA(1) /TIMEA
FA=0MAXA/TWPI
GA=TWPI/TIMEA
RA=DMAXA*TWPI/360.
OCA(I ) =FA*(EA( I ) - S I N ( E A ( I ) ) ) f C O N S T A
DRA(I)=DDA(I)*TWPI/360.
VRA( I ) = R A * ( G A - ( G A « C O S ( E A { I ) ) ) ) / T W P I
211 CRACI ) = R A * G A « G A * S I N ( E A ( I ) ) / T W P I
DO 2 1 2 I = N A A , N
X=I
T(I)=(X-1.)*DELTA
DDA( I ) = O D A ( N A A )
DRA( I ) = O D A ( n * T W P 1 / 3 6 0 .
VRA( I ) = 0 .
212 CRACD^O.
GG TO 5 0 0
206 IF(N-NAA) 213,214,215
213 STOP
214 DO 2 1 6 1 = 1 , N A
X=I
T(I)=(X-1.)*DELTA
DOA(I)=CONSTA
DRA( I )=DDA( I ) * T W P 1 / 3 6 0 .
VRA(I)=0.
216 CRA(I)=0.
00 217 I=NA,NAA
X-I
ZA=NA
Z=NAA
TA(I )=(X-ZA)*DELTA
T( I ) = ( X - 1 . ) * D E L T A
TIMEA=(Z-ZA)*DELTA
E A ( I ) =T WP I * TA ( I ) / TI ME A
F/i=DMAXA/TWPI
GA=TWPI/TIMEA
RA=DMAXA*TWP I / 3 6 0 .
CCA( I ) = F A * ( E A ( I ) - S I N ( E A ( I ) ) )<-CONSTA
ORA( I )=ODA( I ) « T W P I / 3 6 0 .
VR'\( I ) = R A * ( G A - ( G A * C O S { E A ( I ) ) ) )/TWPI
217 C R A d )=RA*GA'!^GA*SIN(EA( I) )/TWPI
GO TO 500
215 CC 218 1=1,NA
X=I
T ( I )=( X-l.)*DtLTA
CCA( I)=CONSTA
DRA(I)=DDA(I)*TWPI/360.
V R A d ) = 0.
 209
218 CRA(n=:0.
DO 2 1 9 I = N A , N A A
X=I
ZA=NA
Z=NAA
TA(I)=(X-ZA)'<'DELTA
T(I)=(X-1.)*DELTA
TIMEA=(Z~ZA)*DELTA
EA(n=TWPI*TA(I)/TIMEA
FA=DMAXA/TWPI
G A = T W P I / T IMEA
RA=DMAXA*TWPI/360.
ODA( I ) = F A * ( E A ( I ) - S I N ( E A ( I ) ) ) + C O N S T A
DRA( I ) = DDA( I ) * T W P I / 3 6 0 .
VRA( I ) = R A * ( G A - ( G A « C O S ( E A { I ) ) ) ) / T W P I
219 C R A d )=RA*GA*GA*S IN(EA( I ) ) / T W P I
DC 2 2 0 I = N A A , N
X=I
T ( I } = ( X - 1 .)*DELTA
DOAd)=DDA(NAA)
CRA( I ) = D D A ( I ) * T W P I / 3 6 0 .
VRA(I)=0.
220 CRA(I)=0.
GO TO 500
500 DO 32 1 = 1 , N
CGLL = A K D I S * R A T I 0 1
TANCGl d ) = C G L L * C R A ( I )
Q O R C G K I )=CGLL»i'VRA( I ) * V R A ( I )
XCGLL( n = + ( T A N C G l ( I ) * S I N ( D R A ( I ) )i-QGRCGl(I )*COS(DRA(I) )
X)
Y C G L L d )= TANCGl ( I ) * C O S ( D R A ( I ) )-QORCGl ( I ) « S I N ( D R A ( I ) )
T ANK( I J = A K D I S * C R A ( I )
QCRKd )=AKDIS*VRA( I)«VRA( I)
XK( I )=•!-( TANK ( I ) * S I N ( D R A ( I ))4-Q0RK( I ) * C O S ( D R A ( I ) ) )
32 Y K d ) = T A N K ( I)'«=COS (DRA d ) )-QORK( I ) * S I N ( O R A ( I ) )
IF(N-l) 221,222,223
221 STOP
212 1=1
T d )= 0.
CCK( I ) = CONSTK
DRK{ I ) = O D K ( I ) * T W P I / 3 6 0 .
VRK( I ) = 0 .
CRK( I ) = 0 .
GO TO 6 0 0
223 I F ( N K - 1 ) 2 2 4 , 2 2 5 , 2 26
224 STOP
225 IF(N-NKK) 227,228,229
227 STOP
 210
228 DO 2 3 0 1=1,NKK
X=I
ZK = NK
Z=NKK
TK( I )=(X-ZK)*DELTA
Td)=CX~U)*DELTA
TIMEK=(Z-ZK)*OELTA
E K d ) = TWPI*TKd)/TIMEK
FK=DMAXK/TWPI
GK=TWPI/TIMEK
RK=DMAXK*TWPI/360.
ODK( n = F K * ( E K d ) - S I N ( E K ( I ) ) ) < - C O N S T K
DRK( I ) = O D K ( n * T W P I / 3 6 0 .
V R K d ) = R K * ( G K - ( G K * C O S ( E K ( I ) ) ) )/TWPI
230 CRK(I)=RK*GK*GK*SIN(EK( I ) ) / T W P I
GO TO 6 0 0
229 DO 2 3 1 1=1,NKK
X=I
Z=NKK
ZK=NK
TK( I )=(X-ZK)*DELTA
T(I)=(X-1.)*DELTA
TIMEK=(Z-ZK)*DELTA
EK( I ) = T W P I * T K ( D / T I M E K
FK=DMAXK/TWPI
GK=TWPI/TIMEK
RK=0MAXK*TWPI/360.
DDK( I ) = F K * ( E K d ) - S I N ( E K ( I ) ) )<-CONSTK
DRK( I ) = DDK( I ) * T W P I / 3 6 0 .
V R K d )=RK*(GK-(GK*COS(EK( I) ) ) ) / T W P l
231 CRK( I ) = RK*GK*GK*SIN(EK(I)) /TWPI
DO 232 I=NKK,N
X=I
T( I )=(X-1.)*DELTA
DDK( I )=DDK(NKK)
DRK( I )=DDK( I)*TWPI/360.
VRK( I )=0.
232 CRK(I)=0.
GO TO 600
226 IF(N-NKK) 233,234,235
233 STOP
234 DO 2 3 6 1 = 1 , NK
X= I
T( I ) = ( X - 1 . ) * D E L T A
DDK( I ) = CONSTK
DRK( n = D D K ( I ) «TWP 1 / 3 6 0 .
VRK( I ) = 0 .
236 CRK(I)=0.
w

211
DO 2 3 7 I = N K , N K K
X=I
Z=NKK
ZK=NK
TK( I )=(X-ZK)«DELTA
T d ) = (X-l.)*DELTA
TIMEK=(Z-ZK)«DELTA
EKd)=TWPI*TKd)/TIMEK
FK=DMAXK/TWPI
GK=TWPI/TIMEK
RK=DMAXK*TWPI/360.
0 0 K ( I ) = F K * ( E K ( I ) - S I N ( E K ( I ) ) )fCONSTK
DRK( I ) = D D K ( I ) * T W P I / 3 6 0 .
V R K ( I ) = R K * i G K - ( G K * C O S ( E K d ) ) ) )/TWPI
23 7 CRK( I ) = R K * G K * G K * S I N ( E K ( I ) ) / T W P I
GO TO 6 0 0
235 DO 2 3 8 I = 1 , N K
X=I
T(I)=(X-1.)*DELTA
DDK( I ) = CONSTK
DRK( I )=DOK( I ) * T W P I / 3 6 0 .
VRKd)=0.
238 CRK(I)=0.
CO 2 3 9 r = N K , N K K
X=I
Z = NKK
ZK=NK
T K d ) = (X-ZK)*DELTA
T( I ) = ( X - 1 . ) * D E L T A
TIMEK=(Z-ZK)*DELTA
EK( I ) = T W P I * T K ( I ) / T I M E K
FK=DMAXK/TWPI
GK=TWPI/TIMEK
RK=DMAXK*TWPI/360.
CDK( I ) = F K * ( E K ( I ) - S I N ( E K d ) ) )fCONSTK
DPK( I )=DDK( I ) * T W P I / 3 6 0 .
VRK( I ) = R K * ( G K - ( G K * C n S ( E K ( I ) ) ) ) / T W P I
239 C R K d )=RK«GK*GK*SIN(EK{ D J / T W ^ I
DO 2 4 0 l = N K K , N
X=I
T(I)=(X-l.)«DeLTA
DDK( I ) = D D K ( N K K )
DRK( I ) = 0 0 K ( I ) * T W P I / 3 6 0 .
VRK( I ) = 0 .
240 CRK(I)=0.
GO TO 6 0 0
600 DO 4 3 1 = 1 , N
CGUL=HKOIS«RAT 102
 212

TANCG2(I )=CGUL*CRK(I)
C 0 R C G 2 d ) = CGUL*VRK( I ) * V R K ( I )
AGK I)=DRK( n - D R A d )
X C G U L d ) = - ( T A N C G 2 ( I ) * S I N ( A G 1 ( I ) ) ) - ( Q0RCG2 ( I ) *COS ( A G l ( I
X) ) )
Y C G U L d ) = <-(TANCG2 ( I ) * C 0 S ( A G 1 ( I ) ) ) - ( Q0RCG2 ( 1 )*S IN ( A G K I
X) } )
XXCGULC ! ) = X C G U L ( D ^ X K d )
YYCGUL( I ) = Y C G U L ( I ) 4 - Y K ( I )
TANH( I ) = H K D I S * C R K d )
QORHd)=HKDIS*VRK( I ) * V R K ( I )
XH( I ) = - ( T A N H ( I ) * S n ( A G l ( I ) ) ) - ( Q O R H ( I ) •COS ( AGl ( I ) ) )
Y H d ) = i - ( T A N H ( I ) * C 0 S ( A G 1 ( I ) ) ) - ( QORH( I ) *S INC AG 1( I ) ) )
XXH(I)=XH(I)<-XKd)
43 YYH( I )=YH( I )+YK( I )
IFCN-l) 241,242,243
2A1 STOP
242 1=1
T ( I )=0 •
ODH(I)=CONSTH
DRH( I ) = D D H ( I ) * T W P I / 3 6 0 .
VRHd)=0.
CRHC I ) = 0 .
GG TO 7 0 0
243 IF(NH-1)244,245,246
244 STOP
245 IF(N-NHH) 247,248,249
247 STOP
248 DC 2 5 0 1 = 1 , NHH
X=I
ZH=NH
Z = NHH
TH(I )=(X-ZH)*DELTA
T { I )=(X-1.)*DELTA
TIMEH = ( Z - Z H ) * D E L T A
EH( I ) = T W P I * T H ( I ) / T I M E H
FH=CMAXH/TWPI
GH=TWPI/TIMEH
RH=0MAXH*TWPI/360. .OK.CTU
D D H d ) = F H * ( E H ( I ) - S I N ( E H d ) ) ) ^-CONSTH
ORH(l)=DDH(I)*TWPI/360.
VRH( I ) = R H * ( G H - ( G H * C O S ( E H { I ) ) ) ) / T W P I
250 CRHd)=RH*GH*GH*SIN(EH(I))/TWPI
GO TO 7 0 0
249 DO 2 5 1 1 = 1 , N H H
X= I
ZH=NH
Z=NHH
 213
TH( I ) = ( X - Z H ) * D E L T A
T d)=(X-U)*DELTA
TIMEH=(Z-ZH)*DELTA
E H d ) = TWPI*TH(I)/TIMEH
FH=0MAXH/TWPI
GH=TWPI/TIMEH
RH=0MAXH*TWPI/360.
ODH ( I J = F H * ( EH ( I ) - S I N ( EH ( I ) ) ) i-CONS T H
DRH( I ) = D D H ( I ) * T W P I / 3 6 0 .
VRHCI ) = R H * ( G H - ( G H * C O S ( e H ( I ) ) ) ) / T W P I
251 CRH(I)=RH*GH*GH*SIN(EHd ))/TWPI
DO 2 5 2 I = N H H , N
X=I
T d )=(X-1.)*DELTA
DDH( I ) = O O H { N H H )
ORH( I ) = D D H ( n * T W P 1 / 3 6 0 .
VRHd )=0.
252 CRHCI)=0,
GO TO 7 0 0
246 IF(N-NHH) 253,254,255
253 STCP
254 DO 2 5 6 1 = 1 , N H
X=I
TCI)=(X-1.)*DELTA
CDH( I ) = CONSTH
O R H ( I ) = ODH( I ) « T W P I / 3 6 0 .
VRH(I)=0.
256 CRH(I)=0.
DC 2 5 7 I = N H , N H H
X=I
ZH=NH
Z = NHH
TH( I ) = ( X - Z H ) * D E L T A
T d ) = (X-1 .)*DELTA
TIMEH=(Z-ZH)*DELTA
EF( I ) = TWP I * T H ( I ) / T I M E H
FH=DMAXH/TWPI
GH=TWPI/TIMEH
RH=DMAXH*TWPI/360.
DOH( I ) = F H * ( E H ( n - S I N ( E H ( d ) ) fCONSTH
DRH( I )=DOH( I ) * T W P I / 3 6 0 .
VRH( I ) = R H * ( G H - ( G H * C O S (EH( I ) ) ) J/T'-,'P I
257 CRH( I ) = R H * G H * G H * S I N ( E H d ) ) / T W P I
GC TO 7 0 0
255 DO 2 5 8 1 = 1,NH
X=I
T d )=(X-1 .)*DELTA
DOH( I )=CONSTH
 214
DRH(I)=DDH( I ) * T W P I / 3 6 0 .
VRH( I ) = 0 .
258 CRH(I)=0.
DC 2 5 9 I = N H , N H H
X=I
ZH=NH
Z=NHH
T H d ) = ( X-ZH)^DELTA
T d ) = ( X - l .)*DELTA
TIMEH=(Z-ZH)*DELTA
E H ( I )=TWPI*TH( D / T I M E H
FH=OMAXH/TWPI
GH=TWPI/TIMEH
RH=DMAXH*TWPI/360.
D D H ( I ) = F H * ( E H ( 1 ) - S I N ( E H ( I)))i-CONSTH
ORHC I )=ODH( I ) * T W P I / 3 6 0 .
VRH{I)=RH*(GH-(GH*COS(EH(I))))/TWPI
259 CRH(I)=RH*GH*GH*SIN(EH(I))/TWPI
DC 2 6 0 I = N H H , N
X= I
T( I ) = ( X - 1 . ) * D E L T A
D C H { I ) = DDH(NHH)
DRH(I)=DDH(I)*TWPI/360.
VRH( I ) = 0 .
2 60 CRH(I)=0.
GO TO 7 0 0
700 CO 5 4 1 = 1 , N
CGTK=HHOI S * R A T I 0 3
TANCG3( I ) = CGTK*CRH( I )
G0RCG3 d ) = CGTK*VRH( I ) * V R H ( I )
AG2(I)=ORH(I)-AG1(I)
XCGTK( I ) = 4 - ( T A N C G 3 ( I ) * S I N ( A G 2 ( I ) ) 4-QORCG3d ) * C 0 S ( A G 2 ( I ) )
X)
Y C G T K d ) = TANCG3( I )*COS( A G 2 d ) ) - Q 0 R C G 3 ( I ) * S I N ( AG2 ( I ) )
XXCGTK( I ) = XCGTK( I ) f X X H ( I )
YYCGTKC I ) = Y C G T K { I)4-YYH( I )
TANS( I ) = H S D I S * C R H ( I )
CORS ( I ) = H S D I S * V R H ( I )*VRH( I )
XS( I )=*•( TANS( I ) * S I N ( A G ? ( I ) ) < - Q 0 R S d ) * C 0 S ( A G 2 d ) ) )
Y S ( I ) = TANS( I ) * C O S ( A G 2 ( I ) ) - 0 0 R S ( I ) * S I N { A G 2 d ) )
XXS( I ) = X S ( D^XXHi I)
54 Y Y S d ) = Y S ( I )+YYH( I )
IF(N-l) 341,342,343
341 STOP
342 1=1
T ( I)=0 .
DDS(I)=CONSTS
CRS( I ) = D D S ( I ) * T VP 1 / 3 6 0 .
!?''•

215
VRS(I)=0.
CRS( I ) = 0 .
GC TO 8 0 0
343 IF(NS-l) 344,345,346
344 STOP
345 IF(N-NSS) 347,348,349
347 STOP
348 CC 3 5 0 1 = 1 , N S S
X=I
ZS=NS
Z = NSS
T S d ) = ( X-ZS)*DELTA
T ( I)={X-1.)*DELTA
TIMES=(Z-ZS)*DELTA
ES( n = T W P I * T S ( D / T I M E S
FS==DMAXS/TWPI
GS=TWPI / T I M E S
RS=0MAXS*TWPI/360.
ODS( I ) = + F S * ( E S ( n - S I N ( E S ( I ) ) )<-CONSTS
DRS( I ) = DDS( I ) * T W P I / 3 6 0 .
VRS ( I ) = * - R S * ( G S - ( G S * C O S ( E S ( I ) ) ) ) / T W P I
350 C R S d )=4-RS*GS*GS*SIN(ES(I ) )/TWPI
GO TO 8 0 0
349 DO 3 5 1 1 = 1,NSS
X=I
ZS=NS
Z = NSS
T S d ) = ( X-ZS)*DELTA
T d)=(X-l.)*DELTA
TIMES=(Z-ZS)*DELTA
ESC I ) = TWP I * T S ( I ) / T I M E S
FS=DMAXS/TWPI
GS=TWPI/TIMES
RS=DMAXS*TWPI/360.
DDS( I ) = 4 - F S * ( E S ( I ) - S I N ( E S ( I ) ) ) fCONSTS
DRS( I ) = DDS( I ) * T W P I / 3 6 0 .
VRS CI )=<-RS*(GS-CGS^CGS( ES{ I ) ) ) ) / T W P I
351 C R S d )=*-RS«GS*GS*SIN(ES(I ) )/T.^PI
DO 3 5 2 I = N S S , N
X= I
T( I ) = ( X - D - D E L T A
DDS( I ) = D D S ( N S S )
DRS( I ) = D D S ( I ) ' . ' = T W P I / 3 6 0 .
V R S d ) = 0.
352 CPS(I)=0.
GO TO 8 0 0
346 IF(N-NSS) 353,354,355
353 STGP
m

216
354 00 356 1=1,NS
X=I
Td)=(X-l.)*DELTA
ODSC I ) = CONSTS
DRSC I ) = DDS( I ) * T W P I / 3 6 0 .
VRSC I ) = 0 .
356 CRS(I)=0,
DO 3 5 7 I = N S , N S S
X=I
ZS = NS
Z = NSS
TSCI ) = C X - Z S ) * D E L T A
T (I)=CX-1.)*0ELTA
TIMES=(Z-ZS)*DELTA
ESC I ) = T W P I * T S ( I ) / T I M E S
FS=DMAXS/TWPI
GS=TWP! / T I M E S
RS=DMAXS*TWPI/360.
DOS(I)=+FS*(ES(I)-SINCES(I)))+CONSTS
ORSC I )=DDSC I ) * T W P I / 3 6 0 .
VRS CI )=4-RS*CGS-CGS*C0S( ESC I ) ) ) )/TWP I
357 C R S d )=4-RS*GS*GS*SINCESCI ) ) / T W P I
GO TO 8 0 0
355 DO 3 5 8 1 = 1 , N S
X=I
T d)=CX-l.)*DELTA
DOSC d = C O N S T S
DRSC I ) = D D S ( I ) « T W P I / 3 6 0 .
VRSd)=0.
358 CRSCI)=0.
DO 3 5 9 I = N S , N S S
X=I
ZS=NS
Z=NSS
TSCI ) = C X - Z S ) * D E L T A
T d )=( X-1.)*DELTA
T I M E S = ( Z - Z S )*DELTA
ES( I ) = T W P I * T S d ) / T I MES
FS=DMAXS/TWPI
GS=TWPI/TIMES
RS=DMAXS*TWPI / 3 6 0 .
PCS ( I ) = 4 - F S * ( E S ( I ) - S IN(ESC I ) ) ) t-CGNSTS
DRSCI)=DDS(I)«TWPl/360.
VRSC I )=4-RS'< CGS-CGS«CGS(ESC I ) ) ) ) /TWPI
3 59 CRSCI ) = f R S * G S * G S * S INCES d ) ) / T W P I
DO 3 6 0 I = N S S , N
X= I
TCI) =CX-l.)*l)ELTA
 217

DCSC I)=OOS(NSS)
ORSCI ) = O D S ( I ) * T W P I / 3 6 0 .
VRSC I ) = 0 .
360 CRSd) = 0 .
GO TO 800
800 DO 38 1=1,N
CGUA=SE0IS*RATI04
TANCG4C I ) =CGUA*CRSC I )
Q0RCG4C I)=CGUA*VRSC I ) * V R S d )
AG3CI)=DRSCI)-AG2 C I ) - C T W P 1 / 2 . )
XCGUAC I ) = TANCG4C n * S I N C A G 3 d ) )<-CQ0RCG4CI )*C0SCAG3CI ) ) )
Y C G U A d ) = - T A N C G 4 d )*C0SCAG3C I ))>Q0RCG4d ) * S I N C A G 3 d ) )
XXCGUAC I ) =XCGUAC I X - X X S C I )
YYCGUAC I)=YCGUAC I)^-YYSC I)
TANECI ) = S E D I S * C R S d )
QORECI)=SeDIS*VRSCI)*VRSCI)
XECI )=TANEC I ) * S I N C A G 3 d ))f(QOREC I )*C0 SC AG 3CI ) ) )
Y E d ) = - T A N E d ) * C 0 S C A G 3 C I))4-Q0REC I )^S INC AG3 CI) )
XXEC I )=XEC I) + XXSC I)
38 YYEC I)=YECI)<-YYSC I)
IFCN-l) 441,442,443
441 STOP
442 1=1
TCI ) = 0.
CDEC I) = CONSTE
DRECI)=DDECI)*TWPI/360.
VREC I ) = 0 .
CREd)=0.
GO TO 900
443 IFCNE-1) 444,445,446
444 STCP
445 IFCN-NEE) 4 4 7 , 4 4 8 , 4 4 9
447 STOP
448 DO 4 5 0 I=1,NEE
X=I
ZE=NE
Z=NEE
TEC I ) = ( X - Z E ) * D E L T A
TCI)=CX-1.)*DELTA
TIMEE=(Z-ZE)*DELTA
EEC I ) = TWPI*TEC l ) / T IMEE
FE=OMAXE/TWPI
GE=TWPI/TIMEE
RE=DMAXE*TWPI/360 .
DDE( I)=^-FE*CEEC I ) - S I N ( P E ( n ) ) f CONST E
DREC I )=ODE( I ) * T / ^ P I / 3 6 0 .
VRECI)=«-RE*CGE-(GE'^COS(EEC I ) ) ) )/TWP I
450 CREC I ) = 4-RE«GE*GE'^SIN(EF( I ) ) /TWPI
 218
GC TO 900
449 DO 4 5 1 1=1,NEE
X=I
ZE = NE
Z=NEE
TEC I )=CX-ZE)^OELTA
TCI)=CX-1,)*DELTA
TIMEE=CZ-ZE)*DELTA
EEC I ) = TWPI*TEC I ) / T I M E E
FE=DMAXE/TWPI
GE=TWPI/TIMEE
RE=DMAXE*TWPI/360.
DDEC I ) = ^-FE*CEEC I)-SINCEEC I ) ) )>CONSTE
DREC I)=DDEC I ) * T W P I / 3 6 0 .
VRE CI ) =^-R E* C GE- C GE*COSC FE CI ) ) ) ) / T WPl
451 CREC I )=+RE*GE*GE*SINCEE( I ) )/TWPI
DC 452 I = N e E , N
X=I
TC I ) = C X - 1 . ) * D E L T A
DDEC I)=DDECNEE)
DREC I ) = DDEC I ) * T W P I / 3 6 0 .
VRECI ) = 0 .
452 CREd)=0.
GO TO 900
446 IFCN-NEE) 4 5 3 , 4 5 4 , 4 5 5
453 STOP
454 DO 456 1=1,NE
X= I
TCI)=C X - 1 . ) * D E L T A
DCEC I ) = CONSTF
DRECI)=DDEC I ) * T W P I / 3 6 0 .
VREC I ) = 0 .
456 CRECI)=0.
DO 4 5 7 I=NE,NEE
X=I
ZE = NE
Z=NEE
TEC I )=(X-ZE)«DELTA
Td)=CX-l.)*D£LTA
TIMEE=CZ-ZE)*DELTA
EEC I)=TWPI*TEC I ) / T I M F E
FE=DMAXE/TWPI
GE = TWP I / T I M E F
RE=0MAXE«TWPI/360 .
DOE( I)=«-FE*CEEC l ) ~ S I N ( E E ( I ) ) ) f CONST E
DREC I ) = DDEC I ) * T W P I / 3 6 0 .
VRECI)=-«-RE* CGE-(GE«CGSC£b( I ) ) ) )/TWP I
457 CREl I ) = f R E * G E * G E * S I N ( E E ( I ) ) / T W P I
 219
GC TO 900
455 00 458 1=1,NE
X=I
Td)=(X-l.)*DELTA
DDEC n = CONSTE
DREC I)=ODEC n * T W P I / 3 6 0 .
VRECn=0.
458 CRECI)=0.
DC 4 5 9 I = N E , N E E
X=I
ZE=NE
Z = NEE
TEd)=CX-ZE)*DELTA
TC I ) = C X - 1 . ) * D E L T A
TIMEE=CZ-ZE)*DELTA
EEC I ) = T W P I * T F d ) / T I M E E
FE=DMAXE/TWPI
GE=TWPI/TIMEE
RE=DMAXE*TWP 1 / 3 6 0 .
DDEC I ) = f F E « C E E C I ) - S INC EEC I ) ) ) 4-CONST E
DRECI)=ODECI)*TWPI/360.
VREC I )=4-RE*CGE-CGE*C0SCEE( I ) ) ) ) / T W P I
459 CRECI)=+RE*GE*GE*SIN(EECI))/TWPI
DO 4 6 0 I = N E E , N
X=I
TCI ) = C X - 1 . ) * D E L T A
DDEC I ) = ODECNEE)
DREC I ) = DDEC D n W P 1 / 3 6 0 .
VRECI)=0.
460 CRECI) = 0.
GC TO 900
900 DO 70 1=1,N
CGFA=EWDIS*RATI05
TANCG5CI)=CGFA*CRECI)
Q0RCG5C I ) = CGFA*VRE( I )*VRE( I )
A G 4 d ) = C T W P I / 2 . ) * - A G 3 C I )-DREC I )
XCGFACI )=-TANCG5C I ) * S I N C A G 4 C I ) ) f ( Q 0 RCG5 ( I )*CGS ( AG4 ( I ) )
X)
YCGFA( I ) = T A N C G 5 ( I ) * C O S ( AG4C I ) ) 4-QORCG5 ( I ) «S IN ( AG4( I ) )
XXCGFAC I ) = XCGFA( I ) 4 - X X E d )
YYCGFA( I ) = Y C G F A ( I )fYYEC I )
TANW(I)=EWDIS*CRE(I)
OORW( I ) = E W D I S * V R E ( I ) * V R E C I )
XWCI ) = -TANWC I )*S INCAG4C I ))<-(Q(lRW( I ) * C G S { AG4( I ) ) )
YWCI )=TANWCI ) * C 0 S C A G 4 C I ) ) f QORW( I ) ^^'S I N ( AG4 ( I ) )
XXW( I ) = XW( I ) + X X E ( I )
70 YYWd ) =YWd)4-YYE( I)
DO 7 1 1 = 1 , N
 220
CGHAND=HAND*RATI06
TANCG6(I)=CGHAN0*CRE(I)
Q0RCG6CI) = CGHAND*VRE(I)*VRE( I )
XCGH A d ) = - T ANCG6 CI ) * S IN C AG4C I ) ) 4-C Q0RCG6 C I ) *COSC AG4C I ) J

Y C G H A ( I ) = TANCG6CI)*C0SCAG4C I ) ) 4-Q0RCG6d ) *SINC AG4( I ) )

XXCGHACn=XCGHA(n4-XXWCI) ^I'NiAb'tiiji
71 YYCGHAC l ) = YCGHAC I ) ^ Y Y W ( I )
WTHA=4-CT0TALW*PWTHA )
^TFA = i-C TOTALW*PWTFA)
WTUA=-i-CTOTALW*PWTUA)
V»TTK=*-CTCTALW*PWTTK)
WTUL = 4-C TOTALW*PWTUL)
WTLL=4-CT0TALW*PWTLL )
SSAMHA=-WTHA/980.616
SSAMFA=-WTFA/980.616
SSAMUA=-WTUA/98 0 , 6 1 6
SSAMTK=-WTTK/980.616
SSAMUL=-WTUL/980.616
SSAMLL=-WTLL/980.616
SSAMLD=-WTLOAD/980.616
FACTR6 = C G Y R A H A * G Y R A H A ) - C R A T I 0 6 * R A T I 0 6 )
FACTR5 = C G Y R A F A * G Y R A F A ) - C R A T I 0 5 * R A T I 0 5 )
FACTR4 = C G Y R A U A * G Y R A U A ) - ( R A T I 0 4 « R A T I 0 4 )
FACTR3 =(GYRATK'!=GYRATK)-(RATI03*RAT 10 3 )
FACTR2 = ( G Y R A U L * G Y R A U L ) - ( R A T I 0 2 * R A T I 0 2 )
FACTRl = ( G Y R A L L * G Y R A L L ) - ( R A T I 0 1 « R A T I 0 l )
CGHAIN = ( SSAMLD+SSAMHA)*FACTR6
CGFAIN=SSAMFA*FACTR5
CGUAIN=SSAMUA*FACTR4
C G T K I N = SSAMTK>:^FACTR3
CGULIN=SS AMUL«FACTR2
CGLLIN=SSAMLL*FACTRl
DO 1 0 4 1= 1,N
XFCGHA( I ) = - ( S S A M H A + SSAMLD)*XXCGHA( I )
YFCGHACI) =-C SSA.MHAfSSAMLD)*YYCGHAd)
XFHAC I ) = XFCGHA( I )
YFHAC I ) = Y F C G H A ( I ) <-WT LOAC^-WT HA
XHAC I )=XFHA( I )
YHAC I ) = Y F H A C I )
TINHACI )=CGHAIN*CRECI ) « ( - ! . )
XR6( I ) = -XFHA( I )
Y R 6 d )=-YFHA( I)
10 4 TR6( I ) = « - T I N H A ( I ) - ( X F H A ( d - CGHANIJ-S I N( AG4 ( I ) ) ) ^ - ( V F H A d )
X*CGHAND*COS(AG4(I)))
DO 7 0 1 1 = 1 , N
X FC GFA( I ) = SSAMFA*XXCGF A d ) * ( - ! . )
YFCGFA( I)=SSAMFA':«YYCGFA( ! ) * ( - ! . )
 221
X F F A ( I ) = XFCGFAC I ) - X R 6 C I )
Y F F A d ) = Y F C G F A C I)+WTFA-YR6C I )
XFAd)=XFCGFAC I )
YFAC n = Y F C G F A C I ) + WTFA
XR5C I ) = - X F F A C I )
Y R 5 C I ) = - Y F F A ( I)
T I N F A d ) = CGFA IN*CREC I ) * C - 1 . )
701 T R 5 d ) = 4 - T I N F A d ) + T R 6 d ) i - C X R 6 C I ) *EWD IS *S IN ( AG4CI ) ) ) - ( X F
XCGFACI)*CGFA*SINCAG4CI)))-CYR6CI)*EWDIS*C0SCAG4CI)))+C
XYFAC I ) * C G F A * C 0 S C A G 4 d ) ) )
00 702 1=1,N
XFCGUACI)=SSAMUA*XXCGUACI)*C-1.)
YFCGUACII=SSAMUA*YYCGUACI)*C-1.)
XFUAC I ) = XFCGUAC I ) - X R 5 C I )
YFUAC I ) = Y F C G U A ( I ) f W T U A - Y R 5 ( I )
XUAC I ) = X F C G U A d )
YUAC I )=YFCGUAC I)i-WTUA
XR4CI)=-XFUAC I)
YR4CI)=-YFUACI)
T I N U A C I )=CGUAIN*CRSC I )
702 T R 4 C I ) = * T I N U A d ) ^ - T R 5 CI)4-CXR5 ( I ) * S E D I S * S INC A G 3 d ) ) ) - C X F
XCGUAC I ) * C G U A * S I N C A G 3 C I ) ) )-C YR5CI ) * SEDI S*CQSC AG3 C I ) ) ) ^-C
XYUACI)*CGUA*COS C A G 3 C I ) ) )
DO 7 0 3 1=1,N
XFCGTKC I ) = S S A M T K * X X C G T K C I ) * C - l . )
YFCGTKCI)=SSAMTK«YYCGTKCI)*C-1.)
XFTKC I ) = XFCGTKC I ) - X R 4 d )
YFTKCI)=YFCGTKC I ) fWTTK-YR4C I )
XTKCI)=XFCGTKCI)
YTKC I )=YFCGTKCI)+WTTK
XR3CI)=-XFTKC I)
YR3C I ) = - Y F T K ( I )
T I N T K d )=CGTK IN*CRH( I ) * ( - ! . )
703 T R 3 d ) = 4 - T I N T K d ) < - T R 4 d ) - C X R 4 ( I ) * H S D I S * S I N ( A G 2 ( I ) ))*-(XT
XKC I ) « C G T K * S I N ( A G 2 ( I ) ) ) - ( YR 4 ( 1 ) * H SDI S*COS ( AG2 ( I ) ) ) + ( YTK
X d )*CGTK*C0SCAG2 ( I ) ) )
DO 1 2 1 1 = 1 , N
X F C G U L C I ) = SSAMUL*XXCGUL( I ) * C - 1 . )
YFCGULCI)=SSAMUL*YYCGUL(I)*(-!.)
XFULC I ) = XFCGUL{ I ) - X R 3 C I )
YFULC I ) = YFCGUL( I ) «-WTUL-YR3( I )
XULC I )=XFCGULCI )
YUL( I )=YFCGUL(I)+WTUL
XR2C I ) = - X F U L ( I )
YR2C I ) = - Y F U L ( I )
T INUL CI )=CGUL IN«CRKC I ) , ,
121 T R 2 d ) = - i - T I N U L d ) 4 - T R ' d I ) - ( X R 3 ( d =<=HKD I S *S IN ( AG 1 ( I ) ) ) f ( X U
X L C I ) * C G U L « S I N ( A G 1 ( I ) ) ) ^ ( YR 3 d ) *HKOI S'^^CGS ( AG I ( I ) ) ) - ( YUL
 222

X( I ) * C G U L * C O S ( A G l d ) ) )
DO 7 0 4 1 = 1 , N
XFCGLLCI)=SSAMLL*XCGLLC I ) * C - 1 . )
YFCGLLCn=SSAMLL*YCGLLC I ) * C - 1 . )
XFLLC I ) = X F C G L L C d - X R 2 d )
YFLLCI)=YFCGLLC I ) •WTLL-YR2 ( I )
XLLCI)=XFCGLL(I)
YLLC I ) = Y F C G L L C I )^-WTLL
XR1CI)=-XFLLC I)
YRIC I ) = - Y F L L C I )
T I N L L C l ) = CGLLIN*CRACI ) * C - 1 . )
704 T R I C I ) = < - T I N L L d ) 4 - T R 2 C I ) - C X R 2 C I ) * A K D I S * S I N C DRACI ) ))«-CXL
XL C I ) * C G L L * S I N C U R A C I ) ) )-C YR2CI ) * A K D I S*COS CDRA CI ) ) ) + CYLL
Xd)*CGLL*COSCDRAC I ) ) )
WT1=.6*WTTK
WT2=.055*WTTK
WT3=.345*WTTK
SSAM1=-WT1/980.616
SSAM2=-WT2/980.616
SSAM3=-WT3/980.616
CG1IN=CGTKIN*.6
CG2IN=CGTKIN*.055
CG3IN=CGTKIN*.345
DIS1=HSDIS*.73
D I S2=HSDI S * . 0 7 5
DIS3=HSDIS*.195
CG1 = . 4 * D I S 1
CG2=.4*DIS2
DC 6 0 1 1 = 1 , N
RGICI)=AG1CI)*360./TWPI
RG2CI ) = AG2C I ) * 3 6 0 . / T W P I
VCHCI)=CDISK-DISA)-RG1( I )
VDHC d = D I S H - C 0 I S K - D I S A ) - R G 2 ( I )
I F ( V O H ( I ) - 2 7 . ) 2 6 0 2 , 2 6 0 2 , 260 3
2 602 C H G H d ) = 0 .
GO TO 2 6 0 4
2 6 0 3 CFGHCI )=CVDHC I ) - 2 7 . ) ' ! ^ 2 . / 3 .
2604 I F C V C H C I ) - 1 3 . ) 2605,2635,2606
2 6 0 5 CFGKC I ) = 0 .
GO TO 2 6 0 7
2606 C H G K C I ) = C V C H C I ) - 1 3 . ) * 2 . / 3 .
2 6 0 7 ANGSACCI ) = 4 0 . K H G H ( I ) - C H G K d )
ASACCI)=90.-ANGSACd)
RSACCI )=ASAC( I ) * T W P I / 3 6 0 .
IFCVDHCI ) - 2 7 . ) 2 9 0 2 , 2 9 0 2 , 2 9 0 3
2902 A H G H C I ) = 2 7 . * 1 8 . / 8 3 .
GO TO 290 4
2903 AFGHCI) = C C C V D H d ) - 2 7 . ) * . ^ 3 3 ) * - 2 7 . ) * l 8 . / 8 3 .
 223

2904 IF(VCHCI)-13.) 2905,2905,2906

2905 AHGK(n=13.*18,/83.
GO TO 2 9 0 7
2906 A H G K d ) = C dVCHC I ) - 1 3 . ) * . 3 3 3 ) + 1 3 . ) * 1 8 . / 8 3 .
2907 A N G L L 5 d ) = A N G S A C C I H-2 1 . - 4 0 . 4-AHGH CI )-AHGKC I )
ALL5CI )=90.-ANGLL5CI )
RLL5 CI ) = A L L 5 d ) * T W P 1 / 3 6 0 .
ANGUL5CI) = A N G L L 5 C I ) i - 2 1 . - 1 0 .
AUL5CI ) = 9 0 . - A N G U L 5 C I )
RUL5 C I ) = A U L 5 ( I ) * T W P I / 3 6 0 .
IFCVDHCI)-2 7.) 7902,7902,7903
7902 BFGHCI ) = 2 7 . * 2 4 . / 8 3 .
GC TO 7 9 0 4
790 3 BHGHC I )=C C CVDHC I ) - 2 7 . ) * . 3 3 3 ) <-27. ) * 2 4 . / 8 3 .
7904 IF(VCHd)-13.) 7905,7905,7906
7905 BHGKd)=13.*24./83.
GO TO 7907
7 9 0 6 BFGKCI )=C C C V C H C I ) - 1 3 . ) * . 3 3 3 ) > 1 3 . ) * 2 4 . / 8 3 .
79C7 A N G L L 4 C I )=ANGUL5( I ) <-5 . - 1 0 . 4-BHGH CI )-BHGKC I )
ALL4C I ) = 9 0 . - A N G L L 4 C I )
601 RLL4CI)=ALL4CI)*TWPI/360.
DO 8 5 0 1= 1,N
H I P C H G d ) = V C H C I )fVOHC I )
HIPTHI CI)=DOHd)
IFCTR3CI ) - 0 . ) 301,302,302
302 DIADISCI)=0.
ABDOMCI)=0.
GO TO 310
301 I F C H I P C H G C n - O . ) 303,303,304
303 OIADISC l ) = 1 2 . 7 * . 4 7
HIOTHI C d = 9 0 .
GC TO 3 1 5
304 IF CHIPCHGCI)-90.) 305,305,306
306 DIAOISC I ) = 1 2 . 7 * 1 . 1 7 7
GO TO 315
305 DIAOISC I )= 12. 7=<^(S INC CHIPCHGC I )*TWPI ) /C360.*2.) )*-.47)
H I P T H I CD = 9 0 .
315 HCI)=-TR3CI) , ^ ,
ABDOMC I ) = 0 . 0 0 0 1 ^ ( 0 . 6 5 1 6 - ( 0 . 0 0 5 4 4 7 * H I P T H I CI ) ) ) * H C I ) * * l .
>8
IFCABDOMC I ) - 1 5 0 . ) 310,310,311
311 ABDOM( I ) = 1 5 0 .
310 A B P R E S d ) =ABOGM( I ) < = 0 . 0 0 1 3 6
CAREA=465.
DFGRCEd)=ABPRES C D^DAREA
ABTORQCI)=DFGRCEC d ^ D I A D I S d )
850 TCRNET C I ) = T R 3 ( I)*-ABTORQ( I )
DC 1 5 1 1=1,N
 224
XFCGIC I )=SSAM1*XXCGTKCI ) * C - 1 . )
YFCGICI)=-(SSAMl«YYCGTK(I))+WTl
X L 4 L 5 d ) = - X F C G l ( I )i-XR4CI)
151 Y L 4 L 5 d ) = - Y F C G l C I )fYR4C I )
OC 1 5 2 1 = 1 , N
XFCG2C I )=SSAM2*XXCGTKCI ) * C - 1 . )
YFCG2( I ) = -CSSAM2*YYCGTKC I ) )+WT2
X L S S I C I )=-XFCG2C I ) < - X L 4 L 5 d )
152 Y L 5 S I C I ) = -YFCG2C I )+YL4L5C I )
OG 1 6 0 1 = 1 , N
FML5S1C I ) = - T O R N E T d ) / 5 .
C S A C ( I ) = Y L 5 S 1 C I ) * S I N C R S A C C l ) ) - X L 5 S l d )*CQS(RSAC( I
SSACd )=YL5S1(I)*C0S(RSAC(I))4-XL5S1(I)*SIN(RSAC( I
COMUSIC I ) = FML5S I d )4-CSACCI ) - D F O R C E ( I )
160 SHUSlCI) = SSACCd
DO 1 6 1 1= 1,N
CLL5 C I ) = Y L 5 S 1 C d * S I N C R L L 5 C I ) ) - X L 5 S l d )*C0S(RLL5C I
SLL5C I ) = Y L 5 S 1 C I ) * C 0 S C R L L 5 C I ) ) f XL5S1 CI )*S INC RLL5 C I
C 0 M L L 5 ( I ) = FML5S1C I ) + C L L 5 d )-DFORCEC I )
161 S H L L 5 d ) = S L L 5 CI)
DO 1 6 2 1= 1,N
F M L 4 L 5 d ) = -T0RNETCI ) / 5 .
C U L S C I ) = Y L 4 L 5 d ) * S I N C RUL5 C D ) - X L 4 L 5 C I )*C0SCRUL5C I
SUL5C I ) = YL4L5C I ) * C 0 S C R U L 5 C I ) )i-XL4L5C I ) * SI NC RUL5 CI
CCMUL5C I ) = F M L 4 L 5 C I ) 1-CUL5 CI )-DFORCEC I )
162 SHUL5C I ) = S U L 5 d )
DO 1 6 3 1 = 1 , N
C L L 4 d ) = Y L 4 L 5 C I ) * SI N C RLL4 C I ) )-X L 4 L 5 C I ) * C 0 S ( R L L 4 C I
SLL4C I ) = YL4L5C I ) * C 0 S C R L L 4 C I ) ) + X L 4 L 5 ( I ) * S I N C R L L 4 C I
CCMLL4C I ) = F M L 4 L 5 C I ) <-CLL4 d )-DFORC EC I )
163 SHLL4C n = S L L 4 C I )
WRITEC6,99)
WRITEC6,1050)
10 50 FORMATC 9 X , « N ' , 8 X , ' N A * , 7 X , » N A A ' , 8 X , • N K « ,7X , •NKK* , 8 X , •NH
X ' , 7 X , ' N H H * , 8 X • N S » , 7 X , 'NSS S B X , ' N E * , 7 X , * N E E » , 5 X , 'DELTA
>• //)
W R I T E C 6 , 10 5 2 ) N , N A , N A A , N K , N K K , N H , N H H , N S , N S S , N E , N E E , D E L T
XA
1 0 5 2 FORMATC H I 1 0 , I F I O . 4 )
WRITEC6,99)

10 53 FORMATC 1 5 X , 'DMA XA • , 1 5 X , • DMA XK* , 13 X , • D MAXH* , 15X , • DN'AXS •

X,15X,« DMAXE'//)
W R I T E C 6 , 1 0 5 5 ) D M A X A , D M A X K , D M A X H , D M AXS , DM AX E
1055 F0RMATC5F20.4)
WRITFC6,99)
WRI TEC 6 , 1 0 5 6 ) ..,MCTUI i A V *r<-r^'
1056 F0RMATC14X,'C0MSTASI4X,'CGNSTKM4X.'.LNSTH',I4X, C .^
 225
XSTS",14X,«C0NSTE«//)
WRITEC6,1058)CONSTA,CONSTK,CONSTH,CONSTS,CON STE
1058 FCRMATC5F20.4)
WRITE(6,99)
WRITEC6,1059)
1 0 5 9 F O R M A T C I O X , ' A K D I S ' , 1 0 X , ' H K D I S * , l O X , • HSDIS • , l O X , ' S E D I S •
X, l O X , ' E W D I S ' , U X , ' H A N D * , 1 0 X , « H H D I S * / / )
WRITEC6,1061)AKDIS,HKDIS,HSDIS,SEDIS,EWDIS,HAND,HHDIS
1061 F0RMATC7F15.2)
WRITEC6,99)
WRITEC6,1062)
1 0 6 2 FORMATC 1 4 X , 'R A T I O 1 • , 14X ,« RA TI 02 • , 1 4 X , • RATI 0 3 ' , 1 4 X , • RAT
XIG4*,14X,'RAT 105*,14X,•RAT 1 0 6 • / / )
WRI TE C 6 , 1 0 6 4 ) RA T I 0 1 , RATI 02 ,R AT I 03 , RAT 104 , RAT 105 , RAT 106
1064 F0RMATC6F20.4)
WRITE(6,99)
WRITEC 6 , 1C65)
1 0 6 5 FORMAT C9X,« T O T A L W S l O X , ' P W T H A ' , l O X , 'PWTFA' , 1 0 X , 'PWTUA*
X,10X,»PWTTK»,10X,>PWTUL« , 1 0 X , • P W T L L » / / )
WRITEC6,1067)T0TALW,PWTHA,PWTFA,PWTUA,PWTTK,PWTUL,PWTL
XL
1067 F0RMATC7F15.4)
WRITEC6,99)
WRITEC6,1068)
1 0 6 8 FORMATC 14 X, 'GYRAHA* , 14X , • G YRAFA • , 14X ,'GYRAUA* , 1 4 X , « G Y R
XATK* , 1 4 X , ' G Y R A U L * , 1 4 X , » GYR ALL • / / )
WRI TEC 6 , 1 0 7 0)GYRAHA,GYRAFA,GYRAUA,GYRATK,GYRAUL,GYRALL
1070 F0RMATC6F20.3)
WRITE(6,99)
WRITEC 6 , 1 0 7 1 )
1071 FORMATC14X,'WTLOAD*//)
WRITEC6,1072)WTLOAD
1072 FORMATCF20.4)
WRITEC6,99)
WRITEC6,1001)
1 0 0 1 FORMAT C9X, d « , 1 6 X , 'TC I ) • , 1 4 X , * D D A ( I ) • , 1 4 X , M)P A CI ) • , 14X
X , « V R A ( I ) » ,14X,«CRAC I )' / / )
DO 1 1 1 = 1 , N
11 WRITEC6,21) I,TC I ) , D D A ( I ) , D R A ( I ) , V R A d ),CPAC I )
21 FORMAT ( I 1 0 , 5 F 2 0 . 3 )
WRITE(6,99)
WRITE(6,1002) . r ^ n ^ l T . . WV
1 0 0 2 F 0 R M A T C 9 X , » I » , 1 6 X , ' T { d ' , I 4 X , ' D D K d ) • ,14 X , • D PK ( I ) • , 14X
X, •VRK( I)« , 1 4 X , ' C R K C ! ) • / / )
12 WRITE(6,22) I , T ( I ) , D D K ( I ) , D R K ( I ) , V R K d ) ,CRK I d
22 FCRNAT dl0,5F2a.3)
WRITEC 6 , 9 9 )
 226

WRITE(6,1003)
1 0 0 3 F 0 R M A T C 9 X , M ' , 1 6 X , ' T C I ) ' , 1 4 X , • DOH CI) • , 1 4 X , • DRHC ! ) • , 14X
X, 'VRHC I ) S 1 4 X , ' C R H C I ) • / / )
DC 13 1 = 1 t N
13 W R I T E C 6 , 2 3) I , T C I ) ,DDHCI) ,DRHCI) , VRHCI),CRHCI )
23 FORMAT dl0,5F20.3)
WRITE(6,99)
WRITEC6, 1004)
1 0 0 4 F C R M A T C 9 X , ' I ' ,16X , • T C I ) • , 1 4 X , 'DOS ( I ) M 4 X , ' D R SC I ) ' , 1 4 X
X,« VRSC I ) ' , 1 4 X , « C R S d ) • / / )
DO 14 1 = 1 , N
14 W R I T E C 6 , 2 4) I , T C I ) , DOS C I ) , D R S C I ) , V R S C I ) , C R S C I )
24 FORMAT CI 1 0 , 5 F 2 0 . 3)
WRITEC6,99)
WRITEC6,1005)
1 0 0 5 F O R M A T C 9 X , • I • , 1 6 X , ' T C I ) « , I 4 X , ' D D E C I ) • , 1 4 X , • D P E ( I ) • , 14X
X,«VRE(I)» , 1 4 X , ' C R E ( I ) ' / / )
DO 15 1 = 1,N
15 WRITE ( 6 , 2 5 ) I , T ( I ) , D D E ( I ) , ORE ( I ) , VRE ( I ) ,CRE CI )
25 FORMAT ( I 1 0 , 5 F 2 0 . 3 )
WRITE ( 6 , 9 9 )
WRITE(6,1006)
100 6 FORMAT ( 9 X , d • , 1 6 X , • T ( I ) • , 1 2 X , • XCGLL( I )» , 12X , • YOG LL ( I ) •
X//)
DC 3 3 1 = 1 , N
33 W R I T E ( 6 , 3 4 ) I , T( I ) , X C G L L ( I ) ,YCGLL( I )
34 FCRMATCI10,3F20.3 )
WRI T E C 6 , 9 9 )
WRITEC6, 1007)
1 0 0 7 F G R M A T C 9 X , ' P ,16X , • T C I ) • , 1 5X , * XK C I ) • , 15X , • YK C I ) • / / )
0 0 39 1 = 1 , N
39 WRITE C 6 , 4 1 ) I , T C I ) , XK ( I ) , YK C I )
41 FORMAT C I 1 0 , 3 F 2 0 . 3 )
WRITEC6,99)
WRITEC6,1008)
1008 FORMAT C 9 X , M « , 1 6 X , ' T C I ) ' , 1 2 X , « X C G U L ( I ) ' , 12X , ' YCGUL CI) •
X//)
DC 35 1 = 1 , N
35 W R I T E ( 6 , 4 5 ) I , TC I ) , X C G U L d ) , Y C G U L C n
45 FCRMATCI10,3F20 . 3 )
WRITE C 6 , 9 9 )

1 0 0 9 F G R M A T C 9 X , M ' ,1 6X , • T ( I ) ' , I I X , • XX CGUL ( I ) M IX , • YYCGJL ( I

X) •//)
DC 46 1=1, N
46 WRITE C6,47) I , T(I ) ,XXCGUL( I ) , YYCGUL(I)
47 FORMATC 110, 3F20.3)
WRITE(6,99)
 227
WRITE(6,1010)
1010 FCRMAT(9X,«P,16X,VT(I)M5X,«XH(I)M5X,«YHd)«//)
DO 4 8 1 = 1 , N
48 WRITE(6,44) I,T( I),XHCI) , Y H d )
44 FORMATdlO,3F20.3)
WRITE C 6 t 9 9 )
WRITEC6,1011)
1 0 1 1 F 0 R M A T C 9 X , ' P »16X , • T C I ) • , 14X , • X X H C I ) • , 14X,'YYHC I ) • / / )
DO 5 1 1 = 1 , N
51 W R I T E C 6 , 5 2 ) I,TC I ) , XXHC I ) , YYHC I )
52 FORMATdlO,3F20.3)
WRITEC6,99)
WRITEC6,1012)
1 0 1 2 FORMATC 9 X , M ' , 1 6 X , » T C I ) ' , 1 2 X , • XCG TKCI ) • ,12X , • YCGTK C I ) •
X//)
DC 5 7 1 = 1 , N
57 WRIT EC 6 , 5 8) I , T ( I ) , X C G T K d ) ,YCGTKC d
58 FCRMATdlO,3F20.3)
WRITEC 6 , 9 9 )
WRITEC6, 1013)
1 0 1 3 FQRMATC9X,* P , 16X , ' T C I ) * , 1 I X , • XXCGTK ( I ) • , 1IX , 'YYCGTKC I
X)«//)
DC 6 4 1 = 1 , N
64 WRITEC6,49)I,TCI),XXCGTKCI),YYCGTKCI)
49 FORMAT CI10,3F20.3)
WRITEC6,99)
WRITEC6, 1014)
1 0 1 4 FORMAT C 9 X , " P , 16X , ' T C I ) • , 1 5 X , 'XSC I ) M 5X, • Y S d ) • / / )
DO 6 2 1 = 1 ,N
62 WRITE(6,56) I,T( I ) , X S C I ) , Y S ( I )
56 FCRMATCI10,3F20.3)
WRITEC6,99)
WRITEC6,1015)
1 0 1 5 F 0 R M A T C 9 X , * P , 1 6X , • T C I ) • ,14X , • XXS d ) • , 14X , ' YYS CI ) • / / )
DO 5 5 1 = 1 , N
55 WRITEC6,59 ) I , T C I ) , X X S ( I ) , Y Y S ( I)
59 FORMATC I 1 0 , 3 F 2 0 . 3 )
WRITEC6,99)
WRITEC6,1016)
1 0 1 6 FORMAT C9X, d • , 1 J X , • TC I ) • , I 2X , • XCG UA CI ) » , 12X , • YCGUA ( I ) •
X//)
DO 5 3 1 = 1 ,N
53 W R I T E C 6 , 6 1) I,TC I ) , X C G U A d ) ,YCGUA( I )
61 FCRMATdl0,3F2C.3)
WRITE C 6 , 9 9 )
WRITEC6,1017)
1 0 1 7 F C R M A T C 9 X , ' P , 1 6 X , ' T ( I ) ' , 1 1 X , • XX CGU A ( I ) • , I I X , • YY CGU A ( I
X) • / / )
 ':->:.T

228
DO 6 3 1 = 1 , N
63 W R I T E C 6 , 6 5 ) I,TC I ) , X X C G U A C I ) ,YYCGUACI)
65 F0RMATdl0,3F20.3 )
WRITEC6,99)
WRITEC6,1018)
1 0 1 8 F 0 R M A T C 9 X , ' P ,16X , • T C n « , 1 5 X , « X EC U M 5 X , 'YEC ! ) • / / )
DO 6 6 1 = 1,N
66 W R I T E C 6 , 6 7 ) I , T C I ) ,XEC I ) , Y E C I )
67 FORMATCI10,3F20.3 )
WRITEC6,99)
WRITEC6,1019)
1 0 1 9 FORMATC 9 X , d » , 1 6 X , • TCI ) • , 14X ,» XXEC I ) • , 1 4 X , » Y Y E C I ) • / / )
DO 6 8 1 = 1 , N
68 W R I T E C 6 , 6 9 ) I , T C I ) ,XXEC I ) , Y Y E C I )
69 FORMATC 1 1 0 , 3 F 2 0 . 3 )
WRITEC6,99 )
WRITEC6,1020)
1 0 2 0 F O R M A T C 9 X , M • , 1 6 X , • T C I ) • , 1 2 X , 'XCGFACI ) ' , 12X,•YCGFACI ) •
X//)
DO 7 2 1 = 1 , N
72 W R I T E C 6 , 7 3 ) I , T C I ) , XCGFAC I ) , YCGFA C I )
73 FCRMATCI10,3F20.3)
WRITEC6,9g)
WPITEC6,102n
1 0 2 1 FORMATC 9 X , M ' ,16X,'Td )• ,11X,"^X XCGFAC I)' , 1IX ,' YY CGFA C I
X)'//)
DO 74 1=1 ,N
74 WRITEC 6 , 7 5 ) 1 , T C I ) ,XXCGFACI),YYCGFACI)
75 FORMAT C 1 1 0 , 3 F 2 0 . 3)
WRITEC6,99)
WRITEC6, 1022)
102 2 FCRMATC9X , ' P , 1 6 X , ' T C I ) ' , 1 5 X , ' X W C D ' , I D X , 'YWC I ) ' / / )
0 0 7 6 1=1 ,N
76 W R I T E C 6 , 7 7 ) I , T C I ),XWC I ) , Y W C I )
77 FCRMATCI10,3F20.3)
WRITEC 6 , 9 9 )
WRITEC6, 1023)
1 0 2 3 F C R M A T C 9 X , ' P , 1 6 X , ' T C I ) ' , 1 4 X , ' XXW d ) • , 14X ,'YYW ( I ) ' / / )
DO 7 8 1 = 1 , N
78 W R I T E C 6 , 7 9 ) I , T ( I ) , X X W C I ),YYW( I )
79 FORMATCI10,3F20.3)
WRITEC6,99)
WRITEC6,1024)
1 0 2 4 F 0 R M A T C 9 X , • I ' 1 6 X , ' T d ) ' , 1 2 X , ' X C G H A ( I ) ' , 12X , • YCGHA ( I ) •
X//)
DO 8 0 1=1 , N
80 W R I T E C 6 , 8 1 ) I,TC I ) , X C G H A C d ,YCGHA( I )
81 FGRMATCI10,3F20 . 3 )
 229
WRITEC6,99)
WRITEC6,1025)
1025 F0RMATC9X,' P , 1 6 X ,
•TCI)'»11X,'XXCGHACI)',11X,'YYCGHACI
X)'//)
CC 8 2 1=1 ,N
82 W R I T E C 6 , 8 3 ) 1 , T C I ) ,XXCGHACI),YYCGHACI)
83 FORMAT C 1 0 , 3 F 2 0 . 3 )
WRITEC6 9 9 )
WRITEC6 1026)
1 0 2 6 FORMAT C9X , ' P , 1 6 X , ' T C I ) ' , 1 4 X , ' X R 6 C I ) • , 1 4 X , ' Y R 6 C I ) ' , 1 4 X
X,«TR6d • / / )
DO 1 0 5 = 1,N
105 WRITFC6 1 0 6 ) I , T C I ) , X R 6 C I ) , Y R 6 C I ) , T R 6 C I )
106 FORMATC 1 0 , 4 F 2 0 . 3 )
WRITEC6 9 9 )
WRITEC 6 1 0 2 7)
1 0 2 7 FORMATC X, • I ' , 1 6 X , 'TC I ) • , 1 4 X , • X R 5 d ) ' , 1 4 X , ' Y R 5 C I ) • ,14X
X,'TRSCI • / / )
DO 1 0 7 = 1,N
107 WRITE(6 1 0 8 ) I , T ( I),XR5( I ) , Y R 5 ( I ),TR5( I)
108 FORMATC 1 0 , 4 F 2 0 . 3 )
WRITEC6 9 9 )
WRITEC6 1 0 2 8 )
1 0 2 8 FORMATC 9 X , ' I • , 1 6 X , ' T d ) • , 1 4 X , ' XR4( I ) ' , 1 4 X , ' Y R 4 ( I )• ,14X
X, • T R 4 d • / / )
DO 1 0 9 = 1,N
109 WRITEC6 1 1 0 ) 1 TC I ) , X R 4 ( I ) , Y R 4 ( I ) , TR4( I)
110 FORMATC 1 0 , 4 F 2 0 . 3 )
WRITEC 6 9 9 )
WRITEC6 1 0 2 9 )
1 0 2 9 FORMATC X , ' P , 1 6 X , ' T ( I ) ' , 1 4 X , ' X R 3 ( I ) ' , 1 4 X , • Y R 3 d ) ' , 14X
X, •TR3C I • / / )
DO 1 1 9 = 1 ,N
119 WRITEC6 1 2 0 ) I T( I ) , X R 3 ( d , Y R 3 ( I ) ,TR3C I )
120 FORMATC 1 0 , 4 F 2 0 . 3 )
WRITEC6 9 9 )
WRITEC6 1 0 3 0 )
1 0 3 0 FORMAT C9 X , ' P I 6 X , 'TC I ) ' , 14X, • X R 2 d ) ' , 14X, 'YR2C I ) ' , 14X
X , ' TR2CI • / / )
DO 1 2 2 = 1,N
122 WRITE(6 1 2 3 ) I , T ( I ) , X R 2 C I ) , Y R 2 ( d , T R 2 ( n
123 FORMATC 1 0 , 4 F 2 0 . 3)
WRITEC6 9 9 )
WRITEC 6 1 03 1)
I 6 X , ' T d ) ' , 14X, » X R I d ) • , 1 4 X , ' Y P l d )• ,14X
1 0 3 1 F0RMATC9 X, ' d
X,'TRlCI • / / )
DO 1 2 4 = 1,N
124 WRIT EC6 1 2 5 ) 1 , T { I ) , X R U I ) , Y R l ( I ) , T R 1 ( d
 230
125 FORMATCI10,4F20.3)
WRITE(6,99)
WPITEC6,2000J
2000 F 0 R M A T C 9 X , » I ' , 2 6 X , ' T C I ) ' , 2 2 X , ' A B D 0 M d ) ' , 2 1 X , • ABPRES CI)
X'//)
DO 3 2 2 1 = 1 , N
322 W R I T E C 6 , 3 2 0 ) I , T C I ) , ABDOMC I ) ,ABPRE SCI)
320 FCRMATCI10,3F30.5)
WRITEC6,99)
WRITE(6,2001)
2 0 0 1 F C R M A T C 9 X , ' P ,21X , ' T C I ) • , 1 6 X , ' D F O R C E C I ) • , 16X , ' ABTORQC I
X)',16X,'TORNETCI)•//)
DC 3 2 3 1=1,N
323 W R I T E C 6 , 3 2 1 ) I , T C I ) , D F G R C E CI) , A B T O R Q C I ) . T O R N E T C I )
321 F0RMATdl0,4F25.5)
WRITEC6,99J
WRITEC6,625)
62 5 F 0 R M A T C 9 X , ' P , 16X , ' T C I ) ' , 1 I X , 'ANGSAC C I ) ' , 1 I X , ' ANGLL 5C I
X) • , l l X , ' A N G U L 5 d ) ' , 1 1 X , ' A N G L L 4 C I ) ' / / )
DO 6 2 6 1 = 1 , N
626 W R I T E C 6 , 6 2 7 ) I , T d ) , ANGSACC I ) , ANGLL5C I ), ANGUL5C I ) , ANGLL
X4CI)
627 FCRMATdlO, 5F20.3)
WRITEC6,99)
WRITEC6,628)
62 8 F C R M A T C 9 X , ' P , 1 6 X , ' T C d ' , 1 3 X , 'ASACCI ) ' , 13X, 'ALL5C I ) ' , 1
>3X,'AUL5CI)',13X,'ALL4CI)'//)
DO 6 2 9 1= 1,N
629 W R I T E C 6 , 6 3 0 ) I , T C I ) , AS AC C I ) , ALL5 ( I )» AUL5( I ) , A L L 4 ( I )
630 F O R M A T d 1 0 , 5 F 2 0 . 3)
WRITE(6,99)
WRITE(6,403)
403 FORM A T ( 9 X , ' I • , 2 I X , ' T ( I ) ' , I 7X, • X L 4 L 5 ( I ) ' , 1 7 X , ' Y L 4 L 5 ( I ) '
X//)
DO 6 5 1 1= 1,N
651 WRITE(6,6 5 2 ) I , T ( I ),XL4L5( I ),YL4L5( I )
652 FORMATdlO,3F25.3)
WRITEC6,99)
WRITEC6,404)
404 FORMATC9X,M',2lX,'Td)',I7X,'XL5SUI)' , I7X , • YL5S1 ( I ) •
X// )
DC 6 5 3 1 = 1 , N
6 53 WRITEC6,6 54)I,TC I ) , X L 5 S I C I ) , Y L 5 S 1 { I )
654 F C R M A T d l O , 3 F 2 5 .?> )
WRI T E C 6 , 9 9 )
WP r T E ( 6 , 4 0 5 ) _ ^. , ^^, , ^ c AT/ T \ I
405 F G P M A T C 9 X , « P ,16X , ' T ( I ) S 11 X , ' F;^L SS I ( I ) M 3X , 'CS AC ( I )
X, U X , 'COMUSK I ) • , 1 2 X , ' S H U S l d ) ' / / )
 231
CC 655 1 = 1 ,N
655 WRI TEC 6 , 6 5 6 ) I ,TC I ) , F M L 5 S 1 C I ) , C S A C C I ) , C O M U S l C I ) , S H U S 1 C I
X)
656 FCRMATdlO,5F20.3)
WRITEC 6 , 9 9 )
WPITEC6,2500)
2 5 0 0 F C R M A T C 9 X , ' P , 1 6 X , • T C I ) ' , 1 I X , ' FML5S 1C I ) • , 13X , • C L L 5 d ) '
X, U X , 'COMLLSC I ) ' , 1 2 X , ' S H L L 5 ( I ) ' / / )
CC 2 5 1 0 1=1 , N
2510 W R I T E ( 6 , 2 5 1 3 ) I , T ( I ) , F M L 5 S I ( I ) , C L L 5 ( I ) ,C0MLL5d ),SHLL5(
XI)
2513 F 0 R M A T ( I 1 0 , 5 F 2 0 . 3 )
WRITE(6,99)
WRITE(6,2501)
2 5 0 1 F 0 R M A T ( 9 X , ' P , 1 6 X , ' T d ) ' , 11 X , • FML4L5 C I ) • , 13X , ' C U L 5 d ) '
X, U X , 'COMUL 5( I ) ' , 1 2 X , ' S H U L 5 C I ) • / / )
DO 2 5 1 1 1 = 1 , N
2 5 1 1 WRI TEC 6 , 2 5 1 4 ) I , T d ) , F M L 4 L 5 ( I ) , CUL5 ( I ) , C0MUL5 d ),SHUL5C
XI )
2514 FGRMATCI10,5F20.3)
WRITEC 6 , 9 9 )
W R I T E ( 6 , 2 50 2 )
2 5 0 2 F C R M A T ( 9 X , ' P ,16X , ' T ( I ) ' , I I X , ' F^^L4L5 ( I ) ' , 13X , ' CLL4 ( I ) '
X, U X , •C0MLL4C I ) ' , 1 2 X , ' S H L L 4 ( I ) ' / / )
CC 2 5 1 2 1 = 1 , N
2512 WRITE(6,2 5 1 5 ) I , T ( I ) ,FML4L5Cl) ,CLL4CI),C0MLL4CI),SHLL4C
xn
2515 F C R W A T d l 0 , 5 F 2 0 . 3 )
99 FORMAT ( I H 1 )
CALL EXIT
END