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
I.
PAGE
ACKNOWLEDGMENTS
11
LIST OF TABLES
v
LIST OF FIGURES
x
INTRODUCTION
1
1.1
Scope
1
1.2
Objectives
2
II. REVIEW OF LITERATURE
4
III.
2.1
Anatomy of the spine related to lifting
2.2
Injuries due to lifting
10
2.3
Biomechanics and lifting experiments
17
EQUIPMENT AND METHODOLOGY
4
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
3.3.A
34
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
111
61
3.3.D
Calculations of reactive forces
and reactive torques at the
center of the di scs
3.3.E
77
Calculation of compressive and
shearing forces
IV. RESULTS AND DISCUSSIONS
93
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
6.
34
Center of gravity and radius of gyration as a
percentage of the segment 1ength
7.
Effect of the biomechanical lifting equivalents on the
maximum net torques at the hip for the leg lift
8.
63
99
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)
11.
102
Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum net
torques at hip joint
12.
103
Effect of the biomechanical lifting equivalents on the
V
maximum compressive forces of the upper surface of S-1
for the leg lift
13.
107
Effect of the biomechanical lifting equivalents on the
maximum compressive forces of the upper surface of S-1
for the back lift
14.
108
Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the lower surface of L-5
for the leg lift
15.
109
Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the lower surface of L-5
for the back lift
16.
110
Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the upper surface of L-5
for the leg lift
17.
HI
Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the upper surface of L-5
for the back lift
18.
112
Effect of the biomechanical lifting equivalents on the
maximum compressive forces on the lower surface of L-4
for the leg lift
19.
113
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)
28.
124
Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum compressive
forces on the upper surface of S-1
vii
129
29.
Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum compressive
forces on the lower surface of L-5
30.
127
Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum compressive
forces on the upper surface of L-5
31.
128
Regression analysis for the linear relation between
biomechanical lifting equivalents and maximum compressive
forces on the lower surface of L-4
32.
129
Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the upper surface of S-1
for the leg lift
33.
135
Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the upper surface of S-1
for the back lift
34.
136
Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the lower surface of L-5
for the leg lift
35.
137
Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the lower surface of L-5
for the back lift
36.
138
Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the upper surface of L-5
for the leg lift
37.
139
Effect of the biomechanical lifting equivalents on the
••
vin
maximum shearing forces on the upper surface of L-5
for the back lift
38.
140
Effect of the biomechanical lifting equivalents on the
maximum shearing forces on the lower surface of L-4
for the leg
39.
lift
141
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
14.
58
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
17.
Free-body diagram for the upper arm showing the forces
and torques during the dynamic activity
18.
68
68
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
20.
71
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
26.
Free-body diagram for the middle trunk link showing
the forces and torques during the dynamic activity
27.
104
Leg lift critical positions using the same weight of
lift and different moment arms
29.
91
Relationship between the biomechanical lifting equivalents
and the maximum net torques at the hip joint
28.
91
Back lift critical positions using the same weight of
xi
118
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
31.
119
Relationship between the biomechanical lifting
equivalents and the maximum compressive forces on the
lower surface of L-5
32.
126
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 musculoskeletal 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)
VERTEBRA,
(prominent)--
^^-^^
CERVICAL VERTEBRAE-- —
Ji\^
-.-.«»
*
12
rHORACIC VERTEBRAE--
31
9
^-9
m'
Tl
f.
iL7^'' 1
•
!K_;
^
LUMBAR VERTEBRAE - - -<
Ar<2a of Concern- - 5
SACRUM
(5 Pieces)
V5Q
COCCYX
(4 Pieced
Fig. 1: Lateral and front view of spine
—VJ
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,
prohibited to employ any woman to
of 3S% of her body weight. Where
lifting is required, the absolute
1949, states that it is
lift any weight in excess
sustained or repetitive
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.)
AGE
LOCATION AND AUTHOR
Under 40
40-50
50-60
Over 60
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
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
1 - SONODA (1962)
2 - EVANS (1966)
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 engineering 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 kinematic 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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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
YOUTHS
ADULTS
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
Knuckle Height
52
59
66
73
80
Knuckle Height to
Shoulder Height
51
56
62
68
73
Shoulder Height to
Arm Reach
48
43
60
67
72
formulation of the model, and then it was tested using five
additional males and females. Both male and female subjects were
students.
considered.
Physiological and anthropometric variables were
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-toinput 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)
27
(1)
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
no.
Total
weight
(kg)
Total
height
(cm)
1
61.4
70.2
179
171
183
44
48
166
40
2
3
4
84.1
64.8
Ankle
to
Knee
47
Distance in cm. from
Knee
Hip
Shoulder
to
to
to
Hip Shoulder
Elbow
53
47
53
42
42
42
43
47
30
30
32
28
Elbow
to
Wrist
27
28
30
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
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
-5
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
2
3
4
-
upper
lower
upper
lower
surface
surface
surface
surface
of
of
of
of
S-1
L-5
L-5
L-4
TABLE (5)
VALUES OF THE BIOMECHANICAL LIFTING EQUIVALENTS, POUND-INCHES
Load Levels, pounds
10
30
20
Moment items, inches
14
140
170
200
17
20
280
340
400
420
510
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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+J
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Fig. 7 : Analysis of circular motions
A
+
HORIZONTAL
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<
Ul
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H <
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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
K
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) =
(14)
Ti (LL)/X
r^ (Ul)i2
= Ti (LL) sin (Ul)i
= r (Ul)i sin (Ul)i
(15)
= Ti (LL) cos (Ul)i
= + r^ {Ul)i cos(Ul)i
(16)
= + 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)
Ti (LL)/Y
Ni (LL)/X
Xi (LL)
Yi (LL)
= Ti (LL)/X + Ni (LL)/X
= + ri [(Ul)i sin (Ul)i
+ (Ul)i2 cos (Ul)i]
(19)
= + 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
=
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
(U2)i - (Ul)i
(23)
= - 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)
Yi (UL)
= Ti (UL)/X + Ni (UL)/X
= - r2[(U2)i sin (Zl)i
+ (U2)i2 cos (Zl)i]
(30)
= 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
=
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
(U3)i - (Z1)i
(38)
= + 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)
Yi (TK)
= Ti (TK)/X + Ni (TK)/X
= + ra [(U3)i sin (Z2)i
+ (U3)i2 cos (Z2)i]
(45)
= 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
=
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) cos (Z3)i
= - r^ (U4)i cos (Z3)i
(57)
= + Ni (UA) cos (Z3)i
= + ri, (U4)i2 cos (Z3)i
(58)
= + Ni (UA) sin (Z3)i
= + ri, (U4)i2 sin (Z3)i
(59)
= Ti (UA)/X + Ni (UA)/X
= + ri, [(U4)i sin (Z3)i
+ (IJ4)i2 cos (Z3)i]
(60)
Ti (UA)/Y
Ni (UA)/X
Ni (UA)/Y
Xi (UA)
Yi (UA)
(U4)i - (Z2)i - ir
(53)
= 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) cos (Z4)i
= + rs (U5)i cos (Z4)i
(72)
= + Ni (LA) cos (Z4)i
= + rs (U5)i2 cos (Z4)i
(73)
= + Ni (LA) sin (Z4)i
= + rs (IJ5)i2 sin (Z4)i
(74)
= Ti (LA)/X + Ni (LA)/X
= - rs [(U5)i sin (Z4)i
- (U5)i2 cos (Z4)i]
(75)
= Ti (LA)/Y + Ni (LA)/Y
= + rs [(U5)i cos (Z4)i
+ (IJ5)i2 sin (Z4)i]
(76)
Ti (LA)/Y
Ni (LA)/X
Ni (LA)/Y
Xi (LA)
Yi (LA)
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
Segment
Hand
Forearm
Upper arm
Foot
Shank
Thigh
Forearm and hand(a)
Whole upper limb(b)
Foot and shank(c)
Whole lower limb(d)
Trunk
Proximal
50.6%
43.0
43.6
50.0
43.3
43.3
68.2
53.0
60.6
44.7
39.6
Distal
49.4%
57.0
56.4
50.0
56.7
56.7
31.8
47.0
39.4
55.3
60.4(^5
Radius of Gyration
Proximal
Distal
58.7%
52.6
54.2
69.0
52.8
54.0
82.7
64.5
73.5
56.0
49.7
57.7%
64.7
64.5
69.0
64.3
65.3
56.5
59.6
57.2
65.0
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
(90)
= 0
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)
(92)
= - L / 980.616
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
(97)
= - Fi (HA)/X
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 computer 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
Fi (LA)/X = - M (LA) . XXi (LA)
(TOO)
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)
Ti (UA)
= - ICGUA
112)
Ri (S)/X
= Ri (E)/X - Fi (UA)/X
Ri (S)/Y
= Ri (E)/Y - Fi (UA)/Y
^CGUA
Mi (S)
+
+
-
(U4)i
113)
-^
[AJOA^^^
114)
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.
program.
This value is a negative input to the computer
-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)
Mi (H)
= +
-
(122)
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)
Ti (UL)
= - I CGUL
Ri (K)/X
= Ri (H)/X - Fi (UL)/X
129)
Ri (K)/Y
= Ri (H)/Y - Fi (UL)/Y
130)
Mi (K)
+
(U2)
127)
128)
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)
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
75
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)
H
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
D3
146)
= 0.075
Ra
147)
= 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 determined.
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]
(CHG2)i
= 0.0
=
,;».
, (E7)i £0.471
(2/3) , (E7)i > 0.471 ...(156)
. (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)
(CHG4)i
,(E7)i < 0.471
= 0^471 (18/83) + [(E7)i - 0.471] (1/3) .,(18/83)
lit, ^r
,(E7)i > 0.471
(160)
= (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
=
(164)
(B3)i
= 1.57 - (A3)i
(CHG5)i
(CHG6)i
(A2)i + 0.3667 - 0.1745
(165)
= (E7)i (24/83)
,(E7)^| <. 0.471
= 0.471 (24/83) + [(E«)i - 0.2^^] (1/3) (24/83)
,(E7)i > 0.471
(166)
= (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
(B4)i
=
=
(A3)i + 0.08725 - 0.1745
+ (CHG5)i - (CHG6)i
(168)
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
'V
lifting, since it was shown by Bartelink, (1957), that the
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.
B.
^
If Mi (H) > 0.0
in this case
(Pr)i
= 0.0
Pi
= 0.0
(171)
:
(172)
Fi (AB) = 0.0
(173)
Ti (AB) = 0.0
(174)
If Mi (H) <.0.0
in this case
(U)i
=
(U3)i . (180/Tr)
(Pr)i
= lO-'* [0.6516 - 0.005447 (U)i] [Mi (H)]1.8 ...(176)
Pi
=
(Pr)i . (0.00136)
Di (AB) = 12.7 [sin ((I§ijL+JlZli) + 0.4^1
Fi (AB) =
{?v^i (465)
Ti (AB) = Fi (AB) • Di (AB)
(175)
(177)
(178)
(179)
(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)
(186)
= Mi (H) - Ti (AB)
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
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)
,
(188)
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 relationship between the biomechanical lifting equivalents and the maximum
net torques at the hip joint.
each method of lift.
This was done for each subject for
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
Lifting '
Equivalent
"lbs.in."
140
Maximum Net Torque at Hij) (kg.cm.)
1st subject
1812.592
140
170
1858.538
170
200
1964.740
200
4th subject
2nd subject
3rd subject
.2114.835
2045.107
1525.441
2017.513
2126.761
1527.609
2079.707
2128.877
1603.208
2155.603
2176.382
1594.703
2190.072
2214.036
1666.065
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
2298.264
2322.186
1827.652
340
400
2309.556
2518.327
2388.883
1840.926
400
2364.774
2487.176
2364.248
1857.742
2361.450
1949.000
420
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
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
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
Maximum Net Torque at Hip (kg.cm . )
1790.208
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)
1939 + 1.46614 BLE
(210)
all subjects :
MNT
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
1
367478.625
367478.625
Deviation about Regression
16
20015.691
1250.981
Total
17
387494.316
Due to Regression
MS
293.752
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
1
499069.188
499069.188
Deviation about Regression
15
47966.567
3197.771
Total
16
547035.755
Due to Regression
,..,. __,
F99%
(1,15) = 8.68
MS
F
156.068
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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
Lifting
Equivalent
"lbs.in."
140
Maximum Compressive Force (kg.) on upper
surface of S-1
2nd subject
388.481
419.429
424.135
345.675
404.732
432.802
348.789
425.602
433.569
362.374
415.321
442.877
360.992
431.002
440.439
368.049
432.599
440.684
372.858
140
170
394.410
170
200
3rd subject 4th subject
1st subject
406.971
200
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
453.351
466.234
404.889
340
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
475.827
433.830
420
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
Lifting
Equivalent
"lbs.in."
1
Maximum Compressive Force (kg) on upper
surface of S-1
1st subject
2nd subject
3rd subject
4th subject
140
399.829
450.548
448.130
354.805
140
394.454
450.130
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
510
533.413
519.838
469.074
600
544.027
562.792
570.487
494.437
600
536.299
565.687
547.849
477.046
375.023
-
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
Lifting
Equivalents
"lbs.in."
140
Maximum Compressive Force (kg.) on lower
surface of L-5
1st subject
2nd subject
3rd subject
4th subject
388.148
424.139
432.004
348.769
410.204
440.890
346.599
419.850
441.600
362.337
430.629
450.492
462.116
434.806
448.848
378.609
437.062
448.557
369.617
140
170
394.571
170
200
411.197
200
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
458.444
474.397
408.933
340
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
485.027
420.249
420
-
510
488.399
513.612
498.722
438.564
510
501.690
509.918
494.877
436.944
600
526.793
528.372
505.350
463.153
600
521.293
515.872
514.811
458.579
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
Lifting
Equivalents
Maximum Compressive Force (kg.) on lower
surface of L-5
1
1st subject
2nd subject
3rd subject
4th subject
140
409.196
461.981
458.743
382.903
140
404.462
461.976
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
510
537.311
347.659
480.764
510
546.001
532.960
490.380
600
555.746
576.794
582.805
490.150
600
549.318
550.727
559.805
502.340
"lbs.in."
364.331
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
Equivalents
"lbs.in."
140
Maximum Compressive Force (kg.) on upper
surface of L-5
1st subject
2nd subject
3rd subject
4th subject
386.206
418.770
424.377
346.500
404.386
433.019
343.785
414.593
433.751
359.299
425.060
442.955
359.836
429.687
440.388
371.275
431.478
440.603
366.433
140
170
392.353
170
200
406.575
200
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
453.109
466.514
404.408
340
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
476.686
417.743
420
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
Lifting
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
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
600
536.905
537.731
548.183
493.802
Maximum Compressive Force (kg.) on upper
surface of L-5
354.657
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
Equivalents
"lbs.in."
140
Maximum Compressive Force (kg.) on lower
surface of L-4
1st subject
2nd subject
3rd subject
4th subject
385.855
419.532
425.296
346.325
405.233
433.880
343.944
415.326
434.504
359.547
425.836
443.788
359.671
430.156
440.674
371.637
431.919
441.246
366.739
140
170
392.178
170
200
407.445
200
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
454.090
467.248
405.243
340
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
477.468
417.896
420
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
Equivalents
Maximum (Compressive Force (kg.) on lower
surface of L-4
1st subject
2nd subject
3rd subject
4th subject
140
397.040
445.188
444.256
369.476
140
393.113
443.486
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
"lbs.in."
351.325
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
(215)
BLE
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
=
(244)
317 + 0.23681 BLE
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
Due to Regression
SS
MS
F
1
9480.540
9480.540
Deviation about Regression
16
389.219
24.326
Total
17
9869.759
389.726
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)
df
SS
1
19238.172
Deviation about Regression
15
1023.162
Total
16
20261.334
Source
Due to Regression
F99% (1,15) = 8.68
F
MS
19238.172 282.040
68.211
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
Due to Regression
Deviation about Regression
Total
df
SS
MS
F
1
9814.211
9814.211
441.875
16
355.366
22.210
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)
df
Source
Due to Regression
Deviation about Regression
Total
F99% (1,15) = 8.68
SS
MS
1 20098.852
20098.852
1090.303
72.687
15
16 21189.155
F
276.513
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
Due to Regression
Deviation about Regression
Total
SS
MS
F
1
9741.570
9741.570
421.669
16
369.639
23.102
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)
df
SS
MS
1
19570.758
19570.758
Deviation about Regression
15
1066.162
71.077
Total
16
20636.920
Source
Due to Regression
F99% (1,15) = 8.68
F
275.344
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
Due To Regression
SS
MS
9808.961
9808.961
23.660
Deviation about Regression
16
378.560
Total
17
10187.551
414.580
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
1
19850.813
19850. 813
Deviation about Regression
15
1129.012
75. 267
Total
16
20979.825
Due to Regression
F99% (1,15) = 8.68
F
263. 737
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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
Equivalent
Maximum Shearing Forces (kg.) on upper
surface of S-1
"lbs.in."
1st subject
2nd subject
3rd subject
4th subject
140
21.444
27.975
36.713
22.530
28.702
37.066
23.992
26.501
36.356
22.530
27.749
35.978
32.219
34.171
39.694
22.530
35.201
37.645
23.738
140
170
21.444
170
200
21.444
200
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
34.356
46.227
25.452
340
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
39.045
32.888
420
510
27.287
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41.545
30.313
510
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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
Equivalents
Maximum Shearing Forces (kg.) on upper
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
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
50.089
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
Lifting
Equivalents
Maximum Shearing Forces (kg.) on lower
surface of L-5
"lbs.in."
1st subject
2nd subject
3rd subject
4th subject
140
13.588
26.455
24.582
14.277
23.217
24.551
21.872
25.490
25.341
14.277
24.395
26.001
33.114
38.270
34.697
14.277
40.033
32.299
20.426
140
170
13.588
170
200
13.588
200
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
25.291
39.642
16.128
340
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
26.760
25.761
420
510
17.291
18.935
29.827
21.296
510
17.291
18.935
28.735
17.979
600
17.291
31.873
32.141
17.979
600
17.291
18.935
25.107
24.827
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
Lifting
Equivalents
"lbs.in."
Maximum Shearing Forces (kg.) on lower
surface of L-5
1st subject
2nd subject
3rd subject
53.372
4th subject
43.977
140
48.573
59.594
140
44.077
59.906
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
510
57.955
61.017
63.739
510
60.249
60.641
49.719
600
57.792
58.975
60.863
60.033
600
56.304
56.644
60.476
57.571
45.844
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
Lifting
Equivalents
"lbs.in."
140
Maximum Shearing Forces (kg.) on upper
surface of L-5
1st subject
2nd subject
3rd subject
4th subject
18.098
25.750
30.853
19.006
24.193
31.248
22.049
24.276
30.983
19.006
24.631
30.648
30.888
33.082
35.666
19.006
34.115
33.833
19.450
140
170
18.098
170
200
18.098
200
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
29.262
41.964
21.615
340
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
33.858
29.487
420
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
Lifting
Equivalents
Maximum Shearing Forces (kg.) on upper
surface of L-5
1st subject
2nd subject
3rd subject
4th subj«=^ct
140
48.989
59.241
51.420
46.180
140
45.653
59.042
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
"lbs.in."
47.112
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
Lifting
Equivalents
"lbs.in."
140
Maximum Shearing Forces (kg.) on lower
surface of L-4
1st subject
2nd subject
3rd subject
4th subject
17.666
25.917
29.738
18.553
24.033
30.271
22.038
24.501
30.272
18.553
24.739
30.315
30.699
31.988
36.078
18.553
32.787
34.143
19.434
140
170
17.666
170
200
17.666
200
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
29.016
42.228
21.100
340
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
32.888
29.051
420
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
Equivalents
"lbs.in."
Maximum .Shearing Forces (kg.) on lower
surface of L-4
1st subject
2nd subject
3rd subject
4th subject
140
49.620
59.865
51.165
47.482
140
46.738
59.318
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
48.142
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 development 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.
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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
Due to Regression
SS
F
MS
1
772447.937
772447.937
Deviation about Regression
11
40764.914
3705.901
Total
12
813212.851
208.437
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)
Due to Regression
Deviation about Regression
Total
F99% (1,16) = 8.53
MS
F
1 1503853.000
1503853.000
158.218
152078.688
9504.918
df
Source
16
SS
17 1655931.688
154
TABLE (A-3)
ANOVA FOR SIMPLE LINEAR REGRESSION BETWEEN BIOMECHANICAL
LIFTING EQUIVALENTS AND MAXIMUM NET TORQUES
AT HIP FOR SECOND
Source
SUBJECT(LEG
LIFT)
df
SS
1
709689.813
709689.813
Deviation about Regression
15
37717.465
2514.231
Total
16
747403.278
Due to Regression
MS
F
282.269
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)
df
SS
MS.
1
472586.000
472586.000
Deviation about Regression
14
55362.308
3954.450
Total
15
527948.308
Source
Due to Regression
F99% (1,14) = 8.86
F
199.507
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
1
555502.438
555502.438
Deviation about Regression
16
22297.512
1393.594
Total
17
577799.950
Due to Regression
F
MS
398.611
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
Due to Regression
Deviation about Regression
Total
F99% (1,16) = 8.53
MS
df
SS
1
974626.938
974626.938
16
54966.328
3435.396
17 1029593.266
F
283.701
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
Due to Regression
SS
MS
1 2341132.000 2341132.000
Deviation about Regression
64 3745164.000
Total
65 6086296.000
F
40.007
58518.188
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)
df
Source
SS
MS
1 3202782.000
3202782.000
Deviation about Regression
67 3146033.000
46955.715
Total
68 6348815.000
Due to Regression
F99% (1,67) = 7.08
F
68.209
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
1
22437.062
22437.062
292.445
Deviation about Regression
11
843.946
76.722
Total
12
23281.008
Due to Regression
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)
df
SS
MS
F
1
42489.141
42489.141
219.117
Deviation about Regression
16
3102.566
193.910
Total
17
45591.707
Source
Due to Regression
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
Due to Regression
SS
MS
F
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)
MS
1
16963.711
16963.711
Deviation about Regression
14
540.406
38.600
Total
15
17504.117
df
Due to Regression
F99% (1.14) = 8.86
Li-
SS
Source
439.470
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
Due to Regression
1
22062.684
Deviation about Regression
16
839.694
Total
17
22902.378
SS
F
MS
22062.684 420.395
52.481
?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)
df
Source
Due to Regression
SS
MS
30433.418
30433.418
151.656
Deviation about Regression
16
2426.503
Total
17
32859.921
F99% (1.16) = 8.53
200.673
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
Due to Regression
SS
MS
74326.687
74326.687
856.557
Deviation about Regression
64
54819.676
Total
65
129146.363
86.774
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
Due to Regression
1
SS
105112.938 105112.938
Deviation about Regression
67
81873.000
Total
68
186985.938
—
F99% ( 1 , 6 7 ) = 7 . 0 8
'
•
MS
1221.985
F
86.018
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
1
22819.309
22819.309
271.691
Deviation about Regression
11
923.890
83.990
Total
12
23743.199
Due to Regression
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)
df
SS
MS
F
1
44065.832
44065.832
210.855
Deviation about Regression
16
3343.786
208.987
Total
17
47409.618
Source
Due to Regression
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
Due to Regression
SS
1
21476.180
21476.180
Deviation about Regression
15
821.800
54.787
Total
16
22297.980
i
TH
MS
391.997
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
Due to Regression
Deviation about Regression
Total
F99% (1.14) = 8.86
„
SS
MS
1
14102.305
14102.305
14
600.500
42.893
14702.805
,
328.779
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
1
22208.695
22208.695
Deviation about Regression
16
896.775
56.048
Total
17
23105.470
Due to Regression
396.241
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
Due to Regression
SS
MS
1
31870.094
31870.094
Deviation about Regression
16
2348.243
146.765
Total
17
34218.337
F99%
(1.16) = 8.53
217.150
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
1
74828.313
74828.313
75.774
Deviation about Regression
64
63200.949
987.515
Total
65
138029.262
Due to Regression
"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)
df
SS
MS
F
1
105527.875
105527.875
84.207
Deviation about Regression
67
83963.750
1253.190
Total
68
189491.625
Source
Due to Regression
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)
SS
MS
1
22663.891
22663.891
Deviation about Regression
11
866.055
78.732
Total
12
23529.946
df
Source
Due to Regression
i
287.860
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)
df
Source
SS
MS
1
42701.902
42701.902
Deviation about Regression
16
3090.935
42.680
Total
17
45792.837
Due to Regression
F99% (1.16) = 8.53
1
221.043
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
1
21544.699
21544.699
384.344
Deviation about Regression
15
840.836
56.056
Total
16
22385.535
Due to Regression
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)
df
SS
MS
F
1
13996.629
13996.629
327.950
Deviation about Regression
14
597.509
42.680
Total
15
14594.138
Source
Due to Regression
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
1
22304.801
22304.801
420.526
Deviation about Regression
16
848.644
53.040
Total
17
23153.445
Due to Regression
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)
SS
MS
1
30660.465
30660.465
Deviation about Regression
16
2444.842
152.803
Total
17
33105.307
df
Source
Due to Regression
F99% (1.16) = 8.53
200.654
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
1
75211.188
75211.188
Deviation about Regression
64
57117.707
892.464
Total
65
132328.895
Due to Regression
84.274
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)
df
SS
MS
F
1
102612.063
102612.063
84.058
Deviation about Regression
67
81789.125
1220.733
Total
68
184401.188
Source
Due to Regression
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)
SS
MS
F
1
22781.316
22781.316
280.185
Deviation about Regression
11
894.387
81.308
Total
12
23675.703
df
Source
Due to Regression
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]
^
SS
MS
1
42668.219
42668.219
Deviation about Regression
16
3005.276
187.830
Total
17
45673.495
df
Source
Due to Regression
Fggcg (1,16) = 8.53
227.164
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)
SS
MS
F
1
21697.098
21697.098
385.239
Deviation about Regression
15
844.817
56.321
Total
16
22541.915
df
Source
Due to Regression
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)
SS
MS
1
14470.273
14470.273
Deviation about Regression
14
721.637
51.545
Total
15
15191.910
df
Source
Due to Regression
F99% (1.14) = 8.86
280.728
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
1
22268.691
22268.691
Deviation about Regression
16
868.169
54.261
Total
17
23136.960
Due to Regression
F
MS
410.403
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)
SS
MS
1
30073.872
30073.872
Deviation about Regression
16
2727.350
170.459
Total
17
32801.222
df
Source
Due to Regression
F99% (1.16) = 8.53
176.428
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
1
75424.250
75424.250
Deviation about Regression
64
58088.090
907.626
Total
65
133512.340
Due to Regression
F
83.101
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)
df
Source
Due to Regression
SS
MS
102990.313
102990.313
Deviation about Regression
67
; 80839.813
Total
68
183830.126
F99% (1.67) = 7.08
1206.564
85.385
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 experimental 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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of gravity of the hand for leg and back methods of lift
202
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' Sec.
205
APPENDIX D
COMPUTER PROGRAM DOCUMENTATION
The program starts on next page.
206
y?io^^cu?^o!'^ic?*'^'^^^^^'^^*3^^'^S(39),TE(39),EA(39),EK
viol
,^^^^^!'^^*^^^'^^*^^^'0°'^<^^''DDK(39),DDH(39),DDS(
J A T ^ O ? wi3?l'^'^'^*^^*'°'^'^*^^'*°«*^<39),DRS(39),DRE(39),VR
vro^?hy^^^^*'^^"*^^*'^'^S*^9^'VRE(39),CRA(39),CRK(39),
Ix^nl?!:^^^^^"^^
.CRE(39) ,TANCGU39),Q0RCG1(39),XCGLL(39
V o ^ « ? ^ ! : l ^ ^ ^ ' ^ ' ^ ^ ' ^ ^ ^ ^ * » * ^ ° ^ ' ^ * 3 9 ) , X K ( 3 9 ) t y K ( 3 9 ) ,TANCG2(39)
X,C0RCG2(39),XCGUL(39),YCGUL(39),XXCGUL(39),YYCGUL(39),
XTANH(39),Q0RH(39) ,XH(39) ,YH(39) ,XXH(39),YYH(39),TANCG3
X(39),Q0RCG3(39),XCGTK(39),YCGTK(39),XXCGTK<39),YYCGTK(
X 3 9 ) , T A N S { 3 9 ) , Q n R S { 3 9 ) , X S ( 3 9 ) , Y S ( 3 9 ) , X X S ( 3 9 ) , Y Y S ( 3 9 ) ,TA
XNCG4 ( 3 9 ) , Q0RCG4 ( 3 9 ) ,XCGUA( 39 ) , YCGUA( 3 9 ) , XXCGUA( 3 9 ) , YYC
X G U A ( 3 9 ) t T A N E ( 3 9 ) , Q 0 R E ( 3 9 ) ,X E ( 39 ) , YE ( 3 9 ) ,XX E( 39 ) , YY E( 3 9
X ) , T A N C G 5 ( 3 9 ) , Q 0 R C G 5 ( 3 9 ) , X C G F A ( 3 9 ) ,YCGFA(39) ,XXCGFA(39)
X , Y Y C G F A ( 3 9 ) , T A N W ( 3 9 ) , Q O R W ( 3 9 ) , X W ( 3 9 ) »YW(39 ),XXW( 3 9 ) , Y Y
X W ( 3 9 ) , T A M C G 6 ( 3 9 ) , Q D R C G 6 ( 3 9 ) , X C G H A ( 3 9 ) , Y C G H A ( 3 9 ) ,XXCGHA
X(39) »YYCGHA(39),T(39),XFCGHA(39),YFCGHA(39),XFHA(39),Y
XFHA(39),TINHA(39),TINFA(39),TINUA{39),TINTK(39),TINUL(
X 3 9 ) , T I N L L ( 3 9 ) , X R 6 ( 3 9 ) , Y R 6 ( 3 9 ) , T R 6 ( 3 9 ) , X F C G F A ( 3 9 ) ,YFCGF
XA{39) f X F F A ( 3 9 ) , y F F A ( 3 9 ) , X R 5 ( 3 9 ) , Y R 5 ( 3 9 ) , T R 5 ( 3 9 ) , X F C G U A
X( 3 9 ) , Y F C G U A ( 3 9) , XFUA ( 39 ) , YF UA ( 3 9) , XR4 ( 39 ) , YR4 (39 ) , T R 4 (
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)
92
8
10
31
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
FnRMAT(IlI5)
RE AD( 5 , 8) DMAXA, JMAXK,0'1AXH ,DMA X S , L)MAXE
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
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
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
STOP
202
1=1
.
201,202,203
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
207
208
STOP
IF(N-NAA)
207,208,209
STOP
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
C R A ( I ) = R A * G A * G A * S I N ( E A ( I ) )/TWP I
GO TO 5 0 0
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
210
209
204,205,206
208
211
212
206
213
214
216
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
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 .
CRACD^O.
GG TO 5 0 0
IF(N-NAA)
213,214,215
STOP
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.
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
215
C R A d )=RA*GA'!^GA*SIN(EA( I) )/TWPI
GO TO 500
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
219
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
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
212
STOP
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
224
225
227
I F ( N K - 1 ) 2 2 4 , 2 2 5 , 2 26
STOP
IF(N-NKK)
227,228,229
STOP
210
228
DO 2 3 0
X=I
1=1,NKK
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
230
229
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
CRK(I)=RK*GK*GK*SIN(EK( I ) ) / T W P I
GO TO 6 0 0
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
232
226
233
234
236
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.
CRK(I)=0.
GO TO 600
IF(N-NKK)
233,234,235
STOP
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 .
CRK(I)=0.
w
211
23 7
235
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
CRK( I ) = R K * G K * G K * S I N ( E K ( I ) ) / T W P I
GO TO 6 0 0
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
239
240
600
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
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 .
CRK(I)=0.
GO TO 6 0 0
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) } )
43
2A1
242
243
244
245
247
248
250
249
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)
YYH( I )=YH( I )+YK( I )
IFCN-l)
241,242,243
STOP
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
IF(NH-1)244,245,246
STOP
IF(N-NHH)
247,248,249
STOP
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
CRHd)=RH*GH*GH*SIN(EH(I))/TWPI
GO TO 7 0 0
DO 2 5 1 1 = 1 , N H H
X= I
ZH=NH
Z=NHH
213
251
252
246
253
254
256
257
255
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
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.
CRHCI)=0,
GO TO 7 0 0
IF(N-NHH)
253,254,255
STCP
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.
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
CRH( I ) = R H * G H * G H * S I N ( E H d ) ) / T W P I
GC TO 7 0 0
DO 2 5 8 1 = 1,NH
X=I
T d )=(X-1 .)*DELTA
DOH( I )=CONSTH
214
258
DRH(I)=DDH( I ) * T W P I / 3 6 0 .
VRH( I ) = 0 .
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
STOP
342
1=1
T ( I)=0 .
341,342,343
DDS(I)=CONSTS
CRS( I ) = D D S ( I ) * T VP 1 / 3 6 0 .
!?''•
215
343
344
345
347
348
350
349
351
VRS(I)=0.
CRS( I ) = 0 .
GC TO 8 0 0
IF(NS-l)
344,345,346
STOP
IF(N-NSS)
347,348,349
STOP
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
C R S d )=4-RS*GS*GS*SIN(ES(I ) )/TWPI
GO TO 8 0 0
DO 3 5 1
X=I
1 = 1,NSS
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
C R S d )=*-RS«GS*GS*SIN(ES(I ) )/T.^PI
DO 3 5 2 I = N S S , N
X= I
352
346
353
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.
CPS(I)=0.
GO TO 8 0 0
IF(N-NSS)
353,354,355
STGP
m
216
354
356
357
355
358
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 .
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
C R S d )=4-RS*GS*GS*SINCESCI ) ) / T W P I
GO TO 8 0 0
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.
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
3 59
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
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
360
800
38
441
442
443
444
445
447
448
450
DCSC I)=OOS(NSS)
ORSCI ) = O D S ( I ) * T W P I / 3 6 0 .
VRSC I ) = 0 .
CRSd) = 0 .
GO TO 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)
YYEC I)=YECI)<-YYSC I)
IFCN-l) 441,442,443
STOP
1=1
TCI ) = 0.
CDEC I) = CONSTE
DRECI)=DDECI)*TWPI/360.
VREC I ) = 0 .
CREd)=0.
GO TO 900
IFCNE-1) 444,445,446
STCP
IFCN-NEE) 4 4 7 , 4 4 8 , 4 4 9
STOP
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
CREC I ) = 4-RE«GE*GE'^SIN(EF( I ) ) /TWPI
218
449
451
GC TO 900
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
CREC I )=+RE*GE*GE*SINCEE( I ) )/TWPI
DC 452 I = N e E , N
X=I
452
446
453
454
456
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 .
CREd)=0.
GO TO 900
IFCN-NEE) 4 5 3 , 4 5 4 , 4 5 5
STOP
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 .
CRECI)=0.
DO 4 5 7 I=NE,NEE
X=I
457
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
CREl I ) = f R E * G E * G E * S I N ( E E ( I ) ) / T W P I
219
455
458
459
460
900
70
GC TO 900
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.
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
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.
CRECI) = 0.
GC TO 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 )
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
701
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 . )
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
702
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 )
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
2905
2906
2907
7902
790 3
IF(VCHCI)-13.)
2905,2905,2906
AHGK(n=13.*18,/83.
GO TO 2 9 0 7
A H G K d ) = C dVCHC I ) - 1 3 . ) * . 3 3 3 ) + 1 3 . ) * 1 8 . / 8 3 .
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
BFGHCI ) = 2 7 . * 2 4 . / 8 3 .
GC TO 7 9 0 4
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
306
IF CHIPCHGCI)-90.)
305,305,306
DIAOISC I ) = 1 2 . 7 * 1 . 1 7 7
GO TO 315
305
315
311
310
850
DIAOISC I )= 12. 7=<^(S INC CHIPCHGC I )*TWPI ) /C360.*2.) )*-.47)
H I P T H I CD = 9 0 .
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
ABDOM( I ) = 1 5 0 .
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 )
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 )
S H L L 5 d ) = S L L 5 CI)
161
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 )
SHUL5C I ) = S U L 5 d )
162
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
1056
F0RMATC5F20.4)
WRITFC6,99)
WRI TEC 6 , 1 0 5 6 )
..,MCTUI
i A V *r<-r^'
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
12
22
X, •VRK( I)« , 1 4 X , ' C R K C ! ) • / / )
WRITE(6,22)
I , T ( I ) , D D K ( I ) , D R K ( I ) , V R K d ) ,CRK I d
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
48
44
FCRMAT(9X,«P,16X,VT(I)M5X,«XH(I)M5X,«YHd)«//)
DO 4 8 1 = 1 , N
WRITE(6,44) I,T( I),XHCI) , Y H d )
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
75
WRITEC 6 , 7 5 ) 1 , T C I ) ,XXCGFACI),YYCGFACI)
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 • / / )
= 1,N
DO 1 0 9
109
WRITEC6 1 1 0 ) 1 TC I ) , X R 4 ( I ) , Y R 4 ( I ) , TR4( I)
FORMATC 1 0 , 4 F 2 0 . 3 )
110
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
WRITEC6 1 2 0 ) I T( I ) , X R 3 ( d , Y R 3 ( I ) ,TR3C I )
119
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 • / / )
= 1,N
DO 1 2 2
WRITE(6 1 2 3 ) I , T ( I ) , X R 2 C I ) , Y R 2 ( d , T R 2 ( n
122
FORMATC 1 0 , 4 F 2 0 . 3)
123
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
WRIT EC6 1 2 5 ) 1 , T { I ) , X R U I ) , Y R l ( I ) , T R 1 ( d
124
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
323
321
62 5
626
627
62 8
629
630
403
651
652
1=1,N
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 )
F0RMATdl0,4F25.5)
WRITEC6,99J
WRITEC6,625)
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
W R I T E C 6 , 6 2 7 ) I , T d ) , ANGSACC I ) , ANGLL5C I ), ANGUL5C I ) , ANGLL
X4CI)
FCRMATdlO, 5F20.3)
WRITEC6,99)
WRITEC6,628)
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
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 )
F O R M A T d 1 0 , 5 F 2 0 . 3)
WRITE(6,99)
WRITE(6,403)
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
WRITE(6,6 5 2 ) I , T ( I ),XL4L5( I ),YL4L5( I )
FORMATdlO,3F25.3)
WRITEC6,99)
WRITEC6,404)
404
FORMATC9X,M',2lX,'Td)',I7X,'XL5SUI)'
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 )
405
_
, I7X , • YL5S1 ( I ) •
^.
F G P M A T C 9 X , « P ,16X , ' T ( I ) S 11 X , ' F;^L SS I ( I ) M
X, U X , 'COMUSK I ) • , 1 2 X , ' S H U S l d ) ' / / )
, ^^,
, ^ c AT/ T \ I
3X , 'CS AC ( I )
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