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I I -El I
Professional Background Materials
for War Mathematics
FLORIDA PROGRAM
FOR IMPROVEMENT OF SCHOOLS
Bulletin No. 41
September, 1942
STATE DEPARTMENT OF EDUCATION
TALLAHASSEE, FLORIDA
COLIN ENGLISH, SUPERINTENDENT
37 009759
no. 4/
I
UNIVERSITY
OF FLORIDA
LIBRARY
Professional Background Materials
For War Mathematics
Bulletin No. 41
September, 1942
PREPARED AT
FLORIDA CURRICULUM LABORATORY
UNIVERSITY OF FLORIDA
M. L. STONE, DIRECTOR
W. L. HUTCHINGS, CONSULTANT
DIVISION OF INSTRUCTION
M. W. CAROTHERS, DIRECTOR
STATE DEPARTMENT OF EDUCATION
TALLAHASSEE, FLORIDA
COLIN ENGLISH, SUPERINTENDENT
.f ~:rb
V:. i
VOCATIONAL PRINT SHOP, WEST PALM BEACH
-401-
Contents
Page
Foreword v
Diagram of Bulletin Production vi
Section
I. Notes On Elementary Operations 1
Significant Figures and Accuracy . 7
The Mil 12
Probable Error ... 14
Distance Rate Time Problems 15
II. Map Projections and Map Problems 17
Map Projections 19
Problems Associated with Maps 31
III. Vectors and Vector Problems 40
Projectile Motion 42
Vector Forces Acting on Airplanes 46
IV. Navigation 50
Determination of Direction-The Compass 50
Pilotage Problems . 55
Vector Problems in Aerial Navigation . 59
Time and Celestial Navigation . 71
iii
149362
Goreword
On May 7, 8, and 9, 1942, the State Department of Education
and the American Council on Education jointly sponsored a
conference to consider what reorganization should be made in
the secondary school mathematics program to provide better the
mathematical skills and concepts needed by those entering either
industry or the armed forces. Following an exploration of the
total problem the conference recommended that secondary school
boys and girls with special mathematical aptitude should be
guided into the regular four-year sequence, that a one-year
course should be planned for those who have discontinued mathe-
matics, and that professional materials be planned to assist
teachers in understanding the use of mathematics in industry
and in war. Therefore, committees working at the Florida Cur-
riculum Laboratory have prepared a pupil text, "Mathematics
Essentials for the War Effort," and "Professional Background
Materials for War Mathematics."
"Professional Background Materials for War Mathematics"
is an attempt to give the teachers of "Mathematics Essentials for
the War Effort" some interpretative materials, primarily illus-
trating applications of elementary mathematics to war problems.
No attempt has been made to treat any topic exhaustively. Many
of the problems and illustrations included may be used as illus-
trative material in other secondary mathematics courses.
The State Department of Education appreciates the efforts
of those who participated in the production of these materials.
The membership of the committee was as follows: Dr. W. L.
Hutchings, Consultant, Rollins College, Winter Park, Florida;
Miss Clare Goertz, Florida High School, Tallahassee, Florida, and
Mrs. E. R. Smith, Leon High School, Tallahassee, Florida. Especial
appreciation is due the American Council on Education and par-
ticularly to Dr. Charles E. Prall for the financial assistance and
encouragement in making this publication possible. Thanks are
also due to Lieutenant Commander Norman P. Anderson of the
United States Navy, Major Joe M. Robertson of the United
States Army, and Mr. Robert A. Thompson, Engineering Depart-
'ment, University of Florida, and to various other members of the
Mathematics and Engineering Departments of the University of
Florida for valuable criticism of the materials.
State Superintendent
of Public Instruction
CONTINUING PRODUCTION OF CURRICULUM BULLETINS
FLORIDA PROGRAM FOR IMPROVEMENT OF SCHOOLS
Series Beginning 1938
No. 9 5
A G u i d e to No. 4. Florida's School Health Program (1942, Revised)
P r a c t i c e No. 22K. Teaching Actions and Effects of Alcohol and Other Narcotics (1941)
i n F o ri d a o No. 26. Arithmetic in the Elementary School (1942)
e m e nta ry No. 27. State Adopted Library Books for Florida Schools (1942)
Schools (1940) o-
No. 2 A v e n u e s o f No. 1. Guide to Exploratory Work* (1938)
W a y s t o Understand No. 4. Florida's School Health Program (1942, Revised)
B e t t e r I n ing, A Bulle- No. 5. Physical Education (1942, Revised)
s t r u c t i o n tin for Parents No. 11. Business Education (1940)
i n F I o ri d a a n d L a y No. 12. Industrial Arts (1940)
Schools (1939) Groups (1940) No. 22K. Teaching Actions and Effects of Alcohol and Other Narcotics (1941)
No. 25. Home Economics Books and Other Source Materials (1941)
- No. 27. State Adopted Library Books for Florida Schools (1942)
No. 28. A Teacher's Guide in the Social Studies for the Secondary
No. 10 m Schools of Florida (1942)
A Guide to a 2 No. 29. Everyday Living (1942)
F u n c t i o n a 1I No. 40. Mathematics Essentials for the War Effort (1942)
Program in the I No. 41. Professional Background Materials for War Mathematics (1942)
Second ry 0 No. 42. Wartime Course in Physics (1942)
School
(1940) Technology Series:
Book 1. General Mechanics (1940)
Book 2. Engines (1942)
Book 3. Aeronautics (1941)
Now out of print.
SECTION I
Notes On Elementary Operations
In times like these all education should be pointed toward
what is of immediate use. There is no time to spend on beautiful,
interesting, or otherwise worthwhile material if it cannot be
applied. The teacher of mathematics should clearly recognize the
fact that the great majority of men in the armed forces, as in
civil life, will rarely be required to perform more than the most
elementary mathematical operations. It is in the fundamental
field of arithmetic that the greatest deficiency is found among
men entering the Army and Navy. In many cases our armed
services have had to lower their educational requirements for
specialist branches, due to a general lack of mathematical train-
ing, and have had to slow up their training program in order to
give the men instruction and review in elementary arithmetic.
The above statements are not meant to imply that elemen-
tary instruction in mathematics should stop with arithmetic. All
branches of mathematics provide useful training for our armed
forces. The fact is emphasized that too few of our high school
and college graduates obtain enough practice and drill in the
handling of numbers beyond the 8th or 9th grade level and that
instruction and drill in arithmetical processes should be con-
sidered an integral part of every mathematics course.
In connection with operations with numbers the following
quotations are pertinent. The first is from the speech of a naval
officer at a conference on pre-induction training, the second from
a government aircraft navigation manual.
emphasize accurate solutions-theory is worse than
useless unless the answer is right. A battle can be lost by a
misplaced decimal point or an incorrect addition.
If it takes a pilot 3 minutes to carry a minor calculation
to a correct solution it will take him an hour to make 20
computations and it may well happen that this many will
be involved in a flight of one hour's duration. It should
not take over 10 seconds to handle any single minor compu-
tation and get accurate results.
The commonest operation with numbers required of the
enlisted man is that of receiving, transmitting, recording, or
2 BACKGROUND MATERIALS FOR WAR MATHEMATICS
acting upon an order in the form of a sequence of digits. Orders
for time, direction, speeds, angles in gunnery and many others
are given primarily as a sequence of digits. In transmitting
numbers as in writing them the digits only are given and not
the units. Zero is always read as zero and not "oh" or "ought."
Thus 207.5 would be read two zero seven point five. Time spent
on practice in the quick and accurate reproduction and reten-
tion of numbers given verbally or by means of "flash cards"
would be valuable military training. When numbers are written
as part of orders or reports it is common military practice to
require that zeros have a bar drawn through them and unity be
underlined so as to avoid any possible confusion with letters.
In addition to drill in the retention of numbers, drill in
learning to estimate quickly and accurately the number and
kinds of objects in a group would have distinct value. Military
observers must frequently estimate the number and kind of
hostile troops, airships, ships, etc. A report that there were "a
flock of ships out there" may be true but hardly accurate enough
to base correct military action upon.
Military men must also be able to estimate distances, angles
and speeds quickly and accurately. Thus an aviator who knows
the width of his fingers and hands or the length of his thumb
can scale distances from a map quite accurately without the in-
convenience of removing both hands from the controls. Few
people ever have more than a vague idea of the length of their
stride. Speed can be estimated fairly well by learning to "count
seconds" and judging the distance traveled in feet during the
count. Miles per hour is very close to two-thirds of feet per sec-
ond (accurately 15/22). The face of a clock or watch can serve
as a crude but effective protractor in estimating angles.
Next to simple arithmetic the commonest mathematical
operations are drawing figures to scale, the construction and
reading of graphs, and the use of simple formulae. It is sug-
gested that teachers require approximate scale drawings for
all problems requiring figures. Any pilot or navigator must
constantly draw scale diagrams. It must be noted here that no
elaborate or expensive equipment is needed for elementary work.
A simple ruler or straight edge and a cheap compass and pro-
tractor is all that is required. A folded piece of paper makes an
excellent straight edge. Every figure should be accurately and
completely labeled. In making graphs, points should be marked
NOTES ON ELEMENTARY OPERATIONS
with a fine dot and a small circle drawn around each dotted point.
Graphs should show all pertinent information, including desvrip-
tive title, scale and units on the axes. In pilot training centers
it is even found necessary to give instruction in how to center a
protractor correctly in measuring angles!
There is almost no training given in any of the traditional
mathematics courses in mental arithmetic. Few of our boys are
able to do more than add or subtract simple digits without re-
course to paper. Without instruction and drill the ability to
calculate mentally comes only with long practice. Mental calcu-
lation as well as most other "short cut" methods depends prima-
rily on the simple properties of our positional number system-
"looking for tens." Thus a beginner would find such an opera-
tion as subtracting 7 x 8 from 163 difficult without a pencil until
it is pointed out that 163 56 = 163 60 + 4 = 107. Valuable
practice in mental arithmetic can be obtained by the use of
arithmetical applications of algebraic identities such as: (a b)
(a + b) = a" b2, (x + a) (x + b) = x2 + (a + b)x -- ab,
(a + b)2 = a2 + b2 + 2ab. Thus 47 x 38 = (40 + 7) (40 2)
= 402 + 5 x 40 14 = 1600 + 200 14 = 1786.
Other devices that are worth mastering to cut down the
labor of computations that may be mentioned are the following:
(1) Cancellation of common factors in the numerator and
denominator. While this process is always taught, it is too often
not insisted on after once being introduced.
(2) The addition of several 2 or 3 digit numbers simultane-
ously, that is, instead of adding column by column and carrying,
begin with the first number, add the units, the tens, the hun-
dreds of the second number in succession and proceed similarly
with each number in the sum.
(3) The use of simple fractions in place of decimal equiva-
lents in finding products and quotients, e. g., 12.5 = 100/8,
25 = 100/4, etc. Thus to divide by 12.5 move the decimal point
two places to the left and multiply by 8.
(4) The use of simple factors of a composite number, e. g.,
if we are required to divide 8632.7 by 56, it is quicker and less
laborious to note that 56 = 7 x 8, divide the dividend by 7 and
the quotient by 8, using simple division; then it is easy to carry
out long division. It is to be observed that no products or sub-
4 BACKGROUND MATERIALS FOR WAR MATHEMATICS
tractions are required when division can be carried out this way.
(5) Multiplication by use of differences, e. g., 97 = 100 3,
69 = 70 1, that is, if we had to multiply a number N by 97
it is far quicker to multiply by 100 and subtract 3 times the
original number than it is to multiply first by 7, then by 90
and add.
An important feature of numerical calculations, almost
wholly overlooked in elementary instruction, is that of esti-
mating in advance the approximate size of the result sought.
Looking for and finding an approximate value is one of the best
checks there is against gross mistakes, particularly in the place-
ment of the decimal point; moreover, the technique of looking
for an approximate answer forces the computer to concentrate
on the steps involved and then to some extent plan the calcula-
tion in advance. In computing with slide rule, and other me-
chanical computers, the computer must find an approximate
answer in order to place the decimal point correctly. The process
of estimating results consists essentially of rounding off the
numbers involved to one or two figures, cancellation of approxi-
mately equal factors from numerator and denominator, and simi-
lar devices. The process can best be shown by example. Thus in
34.27 x 0.7832 4 .7832 800
(.008132) we observe-= 11 (app.), 008 100
7r(0.008132) a .008 8
(app.). Hence the answer is approximately 1100. Since in the
second step we have increased the numerator and decreased the
denominator this approximation is larger than the correct answer.
To four figures the answer is 1051. Again, suppose we are re-
50730 x 2.875,
quired to calculate 1.462 x 34.48 we observe the denominator is
1.462 x 34.48
1.5 x 34 = 50 (app.), 50 into 50,000 = 1,000; 1.000 x 2.875 = 2800
(app.). Alternatively we observe that 2.8/1.4 = 2; 2 into 34
= 17; 51,000 17 = 3,000 (app.). To four figures the answer
is 2893.
One of the commonest operations that must be performed
is taking a mean of a set of observations. In averaging a set of
numbers which do not differ widely, the approximate mean can
usually be obtained by simply averaging the last digits of the
numbers and correcting the approximate mean. The true mean
can always be obtained from an assumed mean by subtracting
the assumed mean from each of the numbers, averaging these
differences (with due regard to the sign of the differences) and
NOTES ON ELEMENTARY OPERATIONS 5
applying this mean as a correction to the assumed mean.
EXAMPLE: Find the average of 4.83, 4.85, 4.87, 4.92, 4.96.
Here the first two figures of the mean are obviously 4.8. Hence
using end figures only the true mean is 4.8 + (.03 + .05 + .07 +
.12 + .16)/5 = 4.886. In practice the decimals would be disre-
garded until the final step.
EXAMPLE: An aviator obtained the following ten observa-
tions for the altitude of a star. Find the average altitude.
360 40.3' 360 52.5'
41.0' 56.8'
43.2' 59.2'
41.7' 370 03.4'
48.4' 08.4'
SOLUTION: Take an assumed mean of 360 50'; neglecting
tenths and subtracting this assumed mean from each of the
numbers we have (-10 -9 -7 -9 -2 +2 +6 +9 +13 +18)/10
= 11/10 = 1.1; averaging the tenths we have (3 + 2 + 7 + 4 +
5 + 8 + 2 + 4 + 4)/10 = .39. Hence the true mean is 36' 50' +
1.1' + .39' = 360 51.5' correct to the nearest tenth of a minute.
The ordinary process of division gives us the figures in the
answer from left to right, consequently we can stop as soon as
the proper number of significant figures are obtained in the
quotient. In the ordinary method of multiplying from right to
left, however, we obtain first the right hand figures in the prod-
uct, figures which must frequently be discarded. This is a waste
of time and labor. It is just as easy to multiply from left to right
as it is from right to left. The following scheme of left to right
multiplication is not well known but is as easily memorized as
the usual process, and with a little practice will cut down the
time required to multiply two numbers by approximately 40
per cent. It is based simply on the technique of algebraic multi-
plication of polynomials and the collection of like powers of 10.
The process is best illustrated by examples.
EXAMPLE: Multiply 732 x 641.
6 BACKGROUND MATERIALS FOR WAR MATHEMATICS
SOLUTION:
We begin by multiplying hundreds by hundreds, then hun-
dreds by tens + tens by hundreds, then hundreds by units +
units by hundreds + tens by tens, etc.
732
641
420000 7 x 6 x 104 732
46000 (7 x 4 + 6 x 3)103 6 41
3100 (7 x 1 + 2 x 6 + 3 x 4)102 42 3 1 1 02
110 (3 x 1 + 4 x 2)10 4 6 1 1
02 1 x 2 46 9 2 12
469212
We now observe that the work can be condensed still further by
omitting the unnecessary zeros and after the second step placing
the significant figures in their proper place in the first two lines.
In each step the last figures of the partial sum should occur one
place to the right of the last figure in the preceding partial sum.
EXAMPLE: Find the product of 98.72 x 68.54 correct to four
figures.
SOLUTION:
98.72 Step 1: 9x6 = 54
68.54 Step 2: 9x8+6x8 = 120
11 Step 3: 9x5+6x7+8x8 = 151
54518xxx Step 4: 9x4+6x2+8x5+8x7 = 144
12044xx Step 5: 8x4+8x2+7x5 = 83 = 80 since
6766.xxx the 3 occurs in the 6th place
The x's mark the place of figures which will not affect the result,
and hence, figures which it is unnecessary to compute. The posi-
tion of the decimal place may be found by the usual method, or
better yet, by observing the power of ten which is involved in
the first product. In step 3 the partial sum is 151. The 51 is placed
in the proper place and the 1 carried above the preceding figure.
Similarly in step 4 is the procedure. It is implied that the calcu-
lations in the individual steps will be performed mentally, other-
wise, there is little saving of time effected. In the example given
both factors have the same number of figures. This is not essen-
tial. The method applies equally well when the number of digits
is different.
NOTES ON ELEMENTARY OPERATIONS 7
SIGNIFICANT FIGURES AND ACCURACY
The numerical data used in solving practical problems is
rarely exact. The numbers used are the result of measurements
and the accuracy of any measurement depends on the instru-
ments or scales used. In using a scale of any kind it is always
possible to read it to the smallest graduated mark on the scale
and with practice to estimate accurately the fractions of the
smallest division to the nearest tenth. Assuming that the ob-
server makes no gross errors in reading the scales, errors in the
measurement will not exceed .05 of the smallest division, and
may, of course, be positive or negative. In reading distances by
scales, particularly on maps or diagrams, it is never good prac-
tice to bring the end of the scale in conjunction with the starting
point. It is easier to judge coincidence between one of the marks
on the scale and points on a diagram than it is the end of the
scale. For this reason all scales used in map work have the zero
point displaced one major unit to the right of the left hand edge
of the scale. Probably the best method of measuring distances
on a diagram is the differential method: Place the rule used along
the line without attempting to adjust the rule so that any par-
ticular mark coincides with a point, read the scale at beginning
and end and take the difference. While this is the most accurate
method it involves extra labor and hence is little used.
The vast majority of problems in elementary texts are
almost wholly expressed in the language of exact numbers.
About the only practice a student obtains in working with ap-
proximate numbers is in using decimal approximations to 7r and
to square roots until he reaches logarithms and the trigonometric
functions. And even in these subjects too little attention is paid
to the significance of rounding off numbers and to the accuracy
of the calculations. The result of such teaching is that a student
gets into the habit of making computations as long as possible,
or else cuts down some of the numbers to 2 or 3 figures, and then
obtains a result to as many figures as possible. Thus from 3 place
data a student will almost invariably get a 6 place result.
When the numbers used in calculations are the result of
measurements, they are approximations, true to only a certain
number of figures. Hence the results obtained will be approxi-
mations, the accuracy of which will always depend on the
accuracy of the data. The entries in all numerical tables (except
primes, squares, etc.) and the results of all measurements are
8 BACKGROUND MATERIALS FOR WAR MATHEMATICS
rounded numbers in which the error is not greater than half a
unit of the last significant figure.
A significant figure is any of the digits, 1 to 9. Zero is a
significant figure except when it is used to fix the place of the
decimal point or to fill the place of discarded digits. Thus in
0.00357 the significant figures are 3, 5 and 7. In 380.7 all the
figures are significant. In 58300 there is nothing to indicate
whether the zeros are significant or not. They may or may not
be, depending on how the number was obtained. A good method
of avoiding such ambiguities is to express approximate numbers
in scientific notation. Scientific notation consists simply of mov-
ing the decimal point to a position just after the first non-zero
digit of the number and then multiplying the resulting number
by the proper positive or negative power of 10 to restore it to its
original value, retaining before the power of 10 only the signifi-
cant figures. Thus if 58300 is correct to 3 figures it would be
written 5.83 x 104, if correct to all 5 places, 5.8300 x 104. In
working with approximate numbers the aim of the computer
should be to obtain results consistent with the given data, with
a minimum of labor, and not to carry calculations to the ultimate
decimal point.
Probably no belief about numbers is more widespread, even
in well educated circles, than that the accuracy of a number is
indicated by the number of decimals required to express it. This
is false. Accuracy is indicated by the number of significant
figures. The absolute error in a measurement is the difference
between the true and the measured values; the index of accuracy
of a measurement or calculation is the relative error, that is the
ratio of the absolute error to the true value. It must be stressed
that we never know the actual errors involved; all that can be
obtained is an upper limit which the errors cannot exceed.
The meaning of the above may be clarified by an example.
Thus if a machinist measures the diameter of a shaft as 2", to
a thousandth of an inch, the absolute error will be less than
0.0005 inch; the relative error will be 0.0005/2 = 1/4000, or 1
part in 4000. If an artillery officer measures a base line of 2,000
yards to within one foot, the absolute error will be less than 6
inches; the relative error will be 6/2000 x 36 = 1/12000, or 1
part in 12,000. The accuracy in the latter case is much greater
than the former though the absolute error is 12,000 times as
NOTES ON ELEMENTARY OPERATIONS
large. The relative error of a measurement is independent of the
units used; it is an abstract number.
In calculations with approximate numbers the commonest
operation to be performed is that of rounding the numbers off
to a certain number of significant figures. To round off a number
to k digits, drop all digits to the right of the kth digit (the posi-
tion of the decimal point is immaterial). If the number dropped
is less than one-half unit of the kth place (that is if the k + 1st
digit is less than 5) leave the kth digit unchanged; if the number
dropped is greater than one half a unit of the kth place, increase
the kth digit by one. If the discarded number is exactly one half
a unit of the kth place leave the kth digit unaltered if it is an even
number, but increase it by one if it is an odd number-that is,
round off so as to leave the kth digit an even number in such
cases. If the numbers dropped are to the left of the decimal point
they must be replaced with zeros. Scientific notation employed
when a number has been rounded off according to this rule is
said to be correct to k significant figures.
In rounding off numbers according to this rule, on the aver-
age half the numbers will be increased in value and half will be
decreased and the errors introduced in rounding off tend to cancel
each other and are largely neutralized.
The following figures are each rounded off to 3 significant
figures according to the above rule:
14.8234 is 14.8; 0.4978 is 0.498; 23.65 is 23.6; 23.55 is 23.6
In working with approximate or rounded numbers the fol-
lowing facts are well worth knowing.
From the definition of absolute and relative errors it follows
that absolute error equal relative error times the number.
The absolute error in a number of k significant figures is
less than one half a unit of the kth place; e. g., if an angle is
measured to the nearest minute, the error is not larger than one
half minute. In the sum of m numbers each rounded off to the
same place, the error in the sum may be as great as m/2 units
of the last significant figure.
If the first digit of a number correct to k significant figures
is d and there is more than one significant figure different from
zero, that is, the number is not of the form d(1.000...) x 10n,
10 BACKGROUND MATERIALS FOR WAR MATHEMATICS
then the relative error will be less than 1/2d x 10k-1. If d is
greater than or equal to 5 the relative error is less than 1/10".
Conversely if the relative error is less than 1/ (d + 1) x 10k-1
the number is correct to k significant figures, or at least is in
error by less than one unit in the kth place.
If the relative error is less than 1/2(d + 1) x 101-1 then
the absolute error is less than half a unit of the kth significant
figure and the given number is correct to k figures. Since d may
have any value from 1 to 9, it follows readily that if the relative
error of any number is less than 1/2 x 10k the number is cer-
tainly correct to k significant figures.
If a and b are the first significant figures of two numbers
each correct to k places, and each number has more than one
significant figure different from zero, then their product or
quotient is certainly correct to
k 1 significant figures if a is greater than or equal
to 2, and b is greater than or equal to 2,
k 2 significant figures if a is one or b is one.
The relative error of a product may be as great as the sum
of the relative errors of the factors. The relative error of a
quotient may be as great as the sum of the relative errors of
the divisor and dividend.
The pth power of such a number is certainly correct to
k 1 significant figures if p is less than or equal to d
k 2 significant figures if p is less than or equal to
10d, and its rth root is certainly correct to
k significant figures if rd is greater than or equal to 10,
k 1 significant figures if rd is less than 10.
The relative error of the pth power of a rounded number
equals p times the relative error of the number. The relative
error in the rth root equals the relative error in the number
divided by r.
The absolute error in the common logarithm of a number
is less than one half the relative error of the number.
The converse of this last statement is not true. The error
in an antilog may be many times the error in the logarithm. In
fact if n represents the antilog and 1 the error in the log, then
the error in n is approximately 2.3nl. If the logarithm is not in
NOTES ON ELEMENTARY OPERATIONS 11
error by more than 2 units of the mth place in the mantissa, the
antilog will certainly be correct to m 1 place. The mth place
will rarely be in error by more than 1 or 2 units in a calcula-
tion with m place logarithms.
The result of any calculation with approximate numbers
will not be correct to any more significant figures than the least
accurate of the numbers used. In calculating with approximate
numbers round off all the numbers used to one more significant
figure than there is in the least accurate of the numbers. Keep-
ing this extra place has the effect of making the error introduced
into the answer by these numbers negligibly small as compared
with the error introduced by the least accurate. In carrying out
the steps of the calculation discard meaningless figures and round
off the answer to the proper number of places.
There are two important exceptions to this general rule.
Since accidental errors tend to cancel out, the arithmetic aver-
ages of a series of measurements of the same quantity will
generally be more accurate than the measurements. If a number
of measurements of a quantity to k figures are averaged the
result will in general be accurate to k + 1 figures. In differencing
two rounded numbers which are nearly equal the difference will
not have as many significant figures as either number. Little
reliance can be placed on the difference of two approximate num-
bers in which the first few significant figures are alike. Thus, if
16850 and 16770 are each correct to 4 figures, their difference,
16850 16770 = 80, is correct to only one significant figure,
and even this figure may be in error by one unit. In fact, all that
can be known certainly about this difference is that it lies be-
tween 70 and 90.
EXAMPLE: Find the sum of the approximate numbers 862.32,
387.5, 85.954, and 4.7462, each being correct to its last figure.
SOLUTION:
Since the second number is accurate to only the first decimal
place it would be absurd to retain more than two decimals in
the others. Hence we round them off to two decimals, add, and
give the result to one decimal.
862.32
387.5
85.95
4.75
1340.5
12 BACKGROUND MATERIALS FOR WAR MATHEMATICS
The result is uncertain by one unit in its last figure since the
possible errors add up to more than half a unit of this figure.
EXAMPLE: Find the product of 349.1 x 863.4. How many
figures of the result are trustworthy?
SOLUTION:
The product to six figures is 301413. The relative error in the
0.05 0.05
product is less than or equal to + 0.00020. Hence
349.1 863.4
the absolute error is less than or equal to 301413 x 0.00020 = 60.
The true result therefore lies between 301473 and 301353. The
best we can do is to take the mean of these numbers to four
figures, 349.1 x 863.4 = 3.014 x 106. Even then there is some
uncertainty about the last figure.
5 63
EXAMPLE: How many figures of the quotient are trust-
6.5
worthy assuming the denominator is correct to only two figures?
SOLUTION:
Since we can express 7r to any required degree of accuracy, the
numerator is correct to 3 figures. Hence the only appreciable
error to be considered is the 0.05 in the denominator. To three
figures the quotient is 2.72. The relative error is less than or
equal to 0.05/6.5 <0.0077. Hence the absolute error is less than
or equal to 2.72 x 0.0077 < 0.02. Since the third figure of the
quotient may be in error by two units we are not justified in
calling the result anything but 2.7.
All calculations should be carried only to the limits of ac-
curacy of the measurements involved. In aerial navigation it is
usually sufficient to measure all distances to the nearest mile,
speeds to the nearest mph, angles to the nearest degree. In calcu-
lations involving the position of ships, distances are carried to
the nearest tenth of a mile, angles to the nearest tenth of a
minute. In artillery map work distances can rarely be estimated
to within better than 10 yd. Taking the nautical mile as 2000 yd
is exact enough for most practical problems.
THE MIL
A unit of angular measure that has been employed for years
in Army work but which up to the present has rarely been found
in textbooks is the mil. The mil was designed to be the angle
NOTES ON ELEMENTARY OPERATIONS 13
subtended by a chord, one-thousandths of the distance to the
chord. Neglecting the slight difference between the length of the
arc and chord for this small an angle the mil is 0.001 radian.
Since this would lead to an irrational number of mils in a cir-
cumference with the consequent difficulty of graduating circles
the mil is taken as 1/6400 of a circumference. Hence,
1600 mils = 900
1 mil = 0.05625 = 3.375'
1 = 17.778 mils (app. 18 mils)
The mil is the standard unit of angular measure in all gunnery
work.
With due regard to the degree of approximation involved
we can say that an object which subtends an angle of a small
number of mils (up to 400 in practice) has a width equal to 0.001
of distance of the object. Hence if W be the width, R the dist-
ance, and M the number of mils subtended by W we have the
mil relation
M = 1000W/R
In artillery work where distances are measured in thousands of
yards the 1000 is omitted and the relation written W = MR.
EXAMPLE: An enemy battery, range 4000 yd, subtends an
angle of 15 mils. How many yards of front does it occupy?
SOLUTION:
R = 4000, M = 15, therefore W = 60 yd.
EXAMPLE: The horizontal range to an enemy bomber is 5000
yd. The vertical angle between the horizontal and the line of
sight is 800 mils. The burst of a shell fired at the target is ob-
served to be in the vertical plane of the target and gun, but the
burst is observed to subtend an angle of 8 mils with the plane
to an observer at the gun. What is the deviation in yards of the
burst from the target assuming the range is correct?
SOLUTION:
800 mils = 45. Therefore, the height of the plane is 5000 yd and
the range (airline distance from gun to plane) is 5000 V2 yd.
W RxM 5000 /2 x 8 = 40 V2 yd in the deviation.
1000 1000
14 BACKGROUND MATERIALS FOR WAR MATHEMATICS
EXAMPLE: If the effective range of a machine gun is 1000 yd
and an approaching airplane is known to have a wing speed of
90 ft, how many mils should the plane subtend in the gun sight
before the gunner opens fire?
SOLUTION:
W =90ft = 30yd, R = 1000yd
1000 x 30
Therefore M 30 mils. The gunner should
start firing when the plane subtends an angle of 30 mils.
EXAMPLE: An artillery officer estimates the width of an
approaching tank as 3 yd and measures the angle subtended as
2 mils. What is the range to the tank?
SOLUTION:
W = 3, M = 2. Therefore R = 1500 yd = the range.
PROBABLE ERROR
If a gun is aimed directly at a target and fired under as
nearly identical conditions as possible, the shots will not all strike
the same point but will be dispersed in an elliptical area around a
point called the center of impact. These variations are caused by
small differences in the weight and shape of the bullets, slight
changes in wind and air density, play in the mechanism and
other variable factors. In the case of artillery fire the dispersion
will be much greater along the line of fire (in range) than to
the right or left (deflection). In a large number of shots, the
shots will be grouped more closely toward center of impact, as
many shots will fall beyond the center of impact as fall short; as
many fall to the right as to the left. For practical gunnery the
area in which the shots fall is assumed to be a rectangle, called
the rectangle of dispersion. In most cases it is only necessary to
consider dispersion in range. The measure used to indicate the
spread of the shots and hence the expected accuracy of fire is
the probable error. The probable error is the error that will
be exceeded as often as it is not exceeded.
- Direction of fire
A x
2% 7% 16% 25% 25% 16% 7% 2%,
B y
Center of A B = Center of impact
DISPERSION AND PROBABLE ERROR DIAGRAM
NOTES ON ELEMENTARY OPERATIONS 15
If a line AB be drawn through the center of impact of the
dispersion rectangle, and a line xy be drawn so that 25 per cent
of the shots fall between AB and xy, then the distance Ax is
the probable error of the range. Fifty per cent of the shots will
fall within a distance of one probable error of the center of
impact. If lines are drawn parallel to AB at distances of one
probable error, the percentage of shots falling in each interval
is approximately as shown in the figure. The value of the prob-
able error will change with the range, the kind of gun and the
ammunition used. Its value is determined by firing tests and
tabulated in firing tables.
EXAMPLE: In a series of 24 shots, 18 fell over the target, 6 fell
short. If the range probable error is 30 yd, how much should the
range be changed to place the center of impact at the target?
SOLUTION:
18/24 = 75% overs, 6/24 = 25% shorts. From the dispersion
diagram it is seen that the range should be decreased one prob-
able error or 30 yards.
EXAMPLE: An ammunition dump is 60 yd long. If a gun is
laid with its center of impact at the center of the dump how
many shots must be fired to assure an expectation of 50 hits on
the target if the range probable error is 20 yd?
SOLUTION:
Length of target from center = 30 yd = 1 PE.
We can expect 1 PE = 33% overs on the target
11 PE = 33% shorts on the target
Hence, if 50 shots are to be hits, and x is the number of shots
to be fired, then .66x = 50, whence x = 83 + = shots which
must be fired to assure a reasonable expectation of 50 hits.
Probably no single simple relationship will have more wide-
spread application to military movements than that of distance
equals rate times time.
DISTANCE RATE TIME PROBLEMS
EXAMPLE: In a column of 501 army trucks, each driver is
instructed to follow the truck ahead at a distance in yards equal
to twice the speedometer reading in miles. Assuming that each
driver carries out his instructions and neglecting the length of
16 BACKGROUND MATERIALS FOR WAR MATHEMATICS
the trucks show that the time required for the column to pass a
point on the road is independent of the speed of travel.
SOLUTION:
Let R = speedometer reading in mph. 1 mile = 1760 yd. Length
1000R
of column = 500 x 2R yd =16_0 mi. The time required to pass
1760
a point on the road will be the length of the column divided by
the rate of travel R.
1000R x 1 1000
T = 1-0-- 1 7 0 hr = 34.1 min (app.) which is inde-
pendent of the rate of travel.
EXAMPLE: Suppose the column above travels from A to B,
a distance of 100 mi at 40 mph, the first truck leaving A at 6:00
A. M. At what time will the last truck arrive at B?
SOLUTION:
100
Time for leading vehicle to go from A to B =- 2 hr 30 min.
40
Distance between trucks = 2R = 80 yd
500 x 80
Length of column = 10 22.7 mi (app.)
1760
22 7
Time for last truck to travel 22.7 mi 22 hr = 34 min (app.)
40
Therefore, movement completed at 2 hr 30 min + 34 min after
6:00 A. M. or at 9:04 A. M. In military usage this time would be
written as 0904. Both the Army and Navy now use the 24 hour
clock in recording time. The hours are numbered from 0 to 24
and the designations A. M. and P. M. are dispensed with. Hours
and minutes are written as a sequence of four digits; the first
two digits represent the hour, the last two digits represent the
minutes. Thus 1:05 A. M. would be 0105; 3:34 P. M. would
be 1534.
The system of abbreviation of units used here is that
recommended by American Engineering Societies. Where ab-
breviations of units such as ft for foot or feet, yd for yard or
yards, mph for miles per hour, etc., cannot be mistaken for words,
no period is to be used. Generally no s is added for plurals since
the number involved will imply whether the unit is singular or
plural. When the unit stands alone it should be written in full
and not abbreviated.
SECTION II
Map Projections and Map Problems
In these days of global war, with our men scattered all over
the surface of the earth, everyone is map conscious. Maps of the
earth's land areas and charts of its seas are even more indis-
pensable items of military equipment than they are in civil life.
Planes and ships must have charts to travel by, battles are
planned over maps, troops move and targets are located by the
use of maps. The preparation of maps is a mathematical job and
every mathematics teacher should have some elementary knowl-
edge of the common kinds of maps, how they are derived and
the advantages and disadvantages of the different projections
used.
Before discussing map projections it is necessary to review
briefly a few facts about the earth. The earth is an oblate
spheroid, one of nine planets, revolving about the sun. Through-
out this revolution it rotates, once in twenty-four hours, on an
axis tilted approximately 66.50 to the plane of revolution.
Throughout this revolution the direction of the axis remains prac-
tically unchanged, and all directions on the earth are derived
from this axis. The ends of this axis are the north and south
poles. Though the earth is not a true sphere, having an axial
diameter of approximately 7900 statute miles and an equatorial
diameter of 7927 statute miles, it may be regarded as a sphere
for all except the most precise kinds of measurements.
In order to locate positions, the earth is regarded as covered
with a coordinate network of meridians, and parallels. Meridians,
or longitude circles, are great circles passing through the poles,
a great circle on any sphere being one whose plane passes through
the center. All other circles on a sphere are called, small circles.
Parallels, or latitude circles, are small circles on the earth whose.
planes are perpendicular to the axis of the earth. They are
parallel to the equator, which is the great circle everywhere 90:
distant from the poles.
The prime meridian is the meridian which passes through
the observatory at Greenwich, England. It has been internation-
ally accepted as a reference line for longitude.
The longitude of a plane is the arc of the equator included
between the'prime meridian and the meridian of the place:
* : 1 ' ;' .. ^ '- * *
18 BACKGROUND MATERIALS FOR WAR MATHEMATICS
Longitude is measured along the equator from 0 to 180' east
or west of the prime meridian and expressed in degrees, minutes,
and seconds of arc. Since the earth with reference to the sun
makes one complete rotation of 360 degrees in 24 hours, longi-
tude may also be expressed in units of time as hours, minutes,
and seconds. Thus each degree of longitude equals 1/15 of an
hour or 4 minutes.
The latitude of a point on the earth's surface is the angular
distance, measured along the meridian through the point, from
0 at the equator to 90 at the north and south poles. For naviga-
tional purposes the nautical mile is the most convenient unit of
measurement. The nautical mile (6080.27 feet) is approximately
the distance subtended on the earth's surface by an angle of 1'
at the center of the earth. Hence 1 along the equator, or 10 of
latitude is 60 nautical miles.
The direction of a line which passes through a point on the
earth is the inclination of the line to the meridian through the
point. It is measured from 0000 at north, clockwise through 360.
The shortest distance between two places on the earth is the
shorter arc of the great circle passing through the points.
The true bearing of a place on the earth's surface from the
place of an observer is the angle between the arc of the great
circle joining the two places and the meridian of the observer.
Since any great circle, except the equator, cuts each meridian
at a different angle the true bearing of a distant place would
change constantly if one travelled toward it on a great circle
course.
A rhumb line is a line on the earth's surface which intersects
all meridians at the same angle. It is a line of constant direction.
Any two places may be connected by such a line. Parallels and
meridians are special types of rhumb lines. All other rhumb lines
are loxodromic curves, spirals which approach but never reach
the poles.
It is usually much more convenient, in all types of naviga-
tion, to be able to follow to constant course or direction between
two places than it is to follow a great circle route. Unless the
distances involved are large (over a thousand miles) the saving
in distance travelled by following a great circle route which re-
quires constantly changing directions, over that of the rhumb
MAP PROJECTIONS AND MAP PROBLEMS
line course is of no practical importance. Hence courses travelled
are usually rhumb line courses. Even when a great circle route
is to be followed, in practice it is broken down into a series of
rhumb line chords or tangents.
A map or chart is a representation of the earth's surface,
or a portion of it, on a plane area. A map is basically a grid or
lattice of lines representing meridians and parellels. Once the
coordinate lattice is obtained, points and lines representing posi-
tions and outlines may easily be drawn in. The primary aims in
map projections are:
1. To preserve the true shape of physical features, especially
angular relationships
2. To preserve correct proportions of areas
3. To obtain a constant scale of distance
4. To represent certain curves, such as great circles or
rhumb lines as straight lines.
The only kind of surfaces which can be mapped on a plane
without any distortion are developable surfaces. A developable
surface is a ruled surface made up of an assemblage of straight
lines, all of which are either tangents to a space curve, parallel
to each other (cylinders), or passing through a fixed point
(cones).
Since a sphere is a non-developable surface it cannot be
mapped on a plane without some distortion. In any system of
projection, angles, areas or distances, or all three will be dis-
torted away from the origin of projection which may be a point,
a meridian or a parallel. The choice of the system of projection
used to construct a map will depend on the requirements of the
map and the factors that must be considered are: the size of
the area to be mapped; its location on the earth; its shape;
whether angles, distance or areas are to be preserved; the ac-
curacy desired.
MAP PROJECTIONS
Map projections may be divided into two general classes,
geometric and mathematical. In geometric projections, the area
to be mapped is projected onto a plane, cone or cylinder, either
tangent to or cutting the earth, according to a geometric law.
20 BACKGROUND MATERIALS FOR WAR MATHEMATICS
The cone or cylinder is then unrolled onto a plane. Mathematical
projections are based on the geometric projections, but points
on the earth's surface are plotted by means of derived equations,
the equations being derived to correct distortions in the simple
geometric projections.
In perspective geometric projections, points on the earth's
surface are projected onto a plane by lines radiating from a
point. There are three general types in common use.
1. The orthographic, in which the point of projection is at in-
finity and the projecting lines are parallel. This is commonly
used in maps of hemispheres where the plane of projection is
the base of the hemisphere. This kind of map gives a good
general idea of a large area, but the distortion is extremely
great, except in the center of the map. Shape, distance and
area are all distorted.
2. The stereographic, in which an area is projected onto a plane
tangent to the earth at the center of the area, the point
of projection being on the earth's surface diametrically op-
posite the center of the area to be mapped. This projection
is conformal, the shape of figures on the earth's surface are
practically correct, but area and distance are distorted as we
go outward from the point of tangency. The scale of relative
area and distance increase as we go outward from the center.
This projection is the best hemispheric projection for general
purposes, and is widely used for mapping polar areas.
3. The gnomonic, in which an area is projected onto a tangent
plane from the center of the earth. For small areas such as
townships and cities little distortion is introduced and the
gnomonic projection is widely used. It is also used for mapping
polar areas. Its most extensive application is in navigation.
Since the center of projection is the center of the earth, all
great circles appear as straight lines on a gnomonic chart.
Hence charts of this kind are utilized in great circle sailing.
There is so much distortion of area and distance that this
projection is seldom used for mapping large land areas.
MAP PROJECTIONS AND MAP PROBLEMS 21
Fig. 1. Stereographic Projection of
Northern Hemisphere
Fig. 2. Gnomonic Projection on a
Plane Tangent to the Earth at the
North Pole, Showing Distortion
away from the Pole
A cylindrical projection is obtained by projecting points of
the earth from the center onto a cylinder, tangent to the earth
along the equator as shown in figure 3.
I /
P1 /
I /
/I,
1// /
I//f c/ --
656
30C01
Equoaor
Fig. 3. Central Projection of the Earth on a Cylinder Tangent at the Equator
Meridians, which on the earth converge to the poles, become
equally distant parallel vertical lines. Latitude circles become
parallel horizontal lines. The perpendicular relationship of me-
ridians and parallels is preserved. Distances along the equator
appear undistorted on the map. All other distances are distorted.
The amount of distortion increases with distance from the equa-
tor and is not the same in both latitude and longitude. Distances
along the meridians between successive parallels increase rapidly
away from the equator.
4.50 N
I I I
,A
22 BACKGROUND MATERIALS FOR WAR MATHEMATICS
It is easily seen from the figure that the distance on the map
of any parallel from the equator p P p
is proportional to the tangent of
the latitude. Every parallel is ex-
panded to the size of the equator.
Let AB be an arc of a parallel of A B'
latitude L, r the radius of the
arc, R the radius of the earth, L
and CD the corresponding arc of L o c D
AB r
the equator. Then AB r
CD -R -
cos L, or AB = CD cos L. On the Fig. 4. Distortion of Meridians and
cylindrical map the correspond- Parallels in a Cylindrical Projection
ing line A'B' = CD. Hence every parallel is expanded in the ratio
of the secant of its latitude.
Mercator charts are modifications of this simple cylindrical
projection, in which, to avoid local distortion, meridian distances
at each latitude are expanded in the ratio of the secant of the
latitude. The mercator projection can be visualized by letting
the origin of cylindrical projection move up the axis of the
earth at an increasing rate, so determined that the increase in
distortion of the meridians is directly proportional to the increase
in distortion of the parallels of latitude. Small figures on the
map will then be generally similar to figures on the earth but
areas and distances will be distorted, the scale of distortion in-
creasing with latitude.
The parallel of 600 latitude is just half as long as the equa-
tor. A quadrangle subtended by 1 of longitude and latitude at
600 latitude has the same length north and south as a similar
quadrangle at the equator, but the east and west extent is only
half as great. Its area then is approximately half as great as a
square degree quadrangle at the equator. On the mercator pro-
jection the 60 parallel is stretched to double its length, hence
the scale on the meridian is increased an equal amount. The
relative area on the map is thus increased fourfold. At 800 the
relative area is increased 36 times. If the map were extended to
890 latitude the relative area would be increased 3000 times. On
a mercator projection of the world Greenland appears as large as
South America, though it has only one-ninth as great an area.
This distortion must be emphasized since mercator projections
of the world are so commonly used and many people obtain a
MAP PROJECTIONS AND MAP PROBLEMS 23
wholly false idea of geography by failing to realize the great
expansion of area and distance at high latitudes. The mercator
projection is used so frequently due to the ease of construction.
The whole world can be mapped on a continuous rectangular
sheet and it can be prolonged in either an east or west direction
to show part of the map twice.
Since 10 of latitude always represents 60 nautical miles, the
meridians can be subdivided and these subdivisions used as a
scale for measuring distances. Since the scale changes with lati-
tude, in measuring distances between two points, the meridian
scale opposite the mid-latitude between the points should be
used since this represents an average of the changing scale.
The essential and useful feature of mercator charts is that
any straight line joining two points cuts all meridians at the
same angle, and therefore is a rhumb line and indicates the true
direction between the points. For navigational purposes this fea-
ture outweighs the disadvantages introduced by the distortion
of distance and area. Nearly all charts of ocean areas between
latitudes 70 north and south are mercator charts. A great circle
plotted on a mercator chart will appear as a curved line, ap-
parently longer than the rhumb line distance. To plot a great
circle course on a mercator chart, the great circle is first drawn
on a gnomonic chart as a straight line and then points of this line
are transferred to the mercator chart by latitude and longitude.
If the area to be covered does not include more than 3 or
4 degrees of latitude the difference in scale between the parallels
is negligible and a small area mercator chart is easily constructed.
Lay off a horizontal line OA and construct perpendiculars at any
convenient interval to represent meridians. It is usually most
convenient to let each vertical line represent a difference of 10'
of longitude. Lay off a line OC, making an angle with OA equal to
the mid-latitude of the area to be represented. If the north-south
distance of the area is only one or two degrees the initial latitude
may be used without appreciable error. The length of OC inter-
cepted between adjacent meridians is proportional to the secant
of the mid-latitude. Parallels should be drawn at multiples of
this distance from OA. The scale along the parallels represents
arc of longitude only. The scale along the meridians represents
latitude in arc and distance in nautical miles.
24 BACKGROUND MATERIALS FOR WAR MATHEMATICS
B
320
3J0
0 Is 2 _90
61w 630 62 60 60v
Fig. 5. Construction of a Small Area Mercator Chart. The Longitude Scale
represents Arc, the Latitude Scale represents Arc and Distance in
Nautical Miles
Due to the rapid increase in distortion of area and distance
with increase in latitude, mercator projections are poorly suited
to represent land areas. The best maps and land areas are de-
rived from projections of the earth on a cone.
Fig. 6. Projection
on a Cone Tan-
gent to the Earth
Along the Paral-
lel of 450 N Lati-
tude.
MAP PROJECTIONS AND MAP PROBLEMS
Select a standard parallel of latitude near the center of the
area to be mapped and construct a cone tangent to the earth
along this standard parallel. Project the area to be mapped onto
the cone from the center of the earth. If the cone is then unrolled
onto a plane we have a map of the area in which the meridians are
straight lines converging to the apex of the cone; parallels are
arcs of concentric circles with their centers at the apex. It is an
elementary exercise in geometry to show that the angle between
two meridians at the apex is equal to the difference in longitude
of the meridians (in circular measure) times the sine of the
initial latitude. Distances along the standard parallel will be
undistorted and the angles between meridians and parallels are
preserved. But distances and areas are distorted at an increasing
rate as we depart from the standard parallel. There are a number
of modifications of this simple conical projection in widespread
use.
In the Bonne projection the slant height AC of the cone
tangent at a central parallel is computed and reduced to the scale
chosen. This is laid off on the map as the central meridian. With
A as center and AC as radius an arc is struck, representing the
standard parallel. On the initial meridian, distances to scale are
laid off equal to the true meridian distances between parallels on
the earth's surface and through these points concentric arcs with
center at A are drawn. These represent the successive parallels
of latitude. On each of these arcs true distances between merid-
ians at that latitude are laid off starting with the initial meridian.
Through points located in this way meridian curves are drawn.
These curves are concave toward the central meridian and con-
verge to the pole. In this projection distance is preserved along
the central meridian and all parallels. Areas are preserved. Angles
are preserved only along the central meridian. Distortion of angle
and distance, other than along parallels, increases away from the
central line though the distortion is small for distances of several
hundred miles. The Bonne projection is excellent for civil maps
where area is of importance and for maps of countries such as
England or Japan which have a small range of longitude.
One extreme of the Bonne projection is obtained when the
origin is at the equator. The cone becomes a cylinder; parallels
appear as straight lines their correct distance apart; the initial
meridian is a straight line at right angles to the parallels. This
26 BACKGROUND MATERIALS FOR WAR MATHEMATICS
is the sinusoidall projection" and is the basis of many maps
of the tropics.
Probably the best compromise projection that has been de-
veloped in the attempt to preserve area, distance, and angles is
that known as the Lambert conformal. This projection has been
known since 1772 but has only come into widespread use since
the first World War when it was widely used for military maps.
It is the basis for most commercial maps, and is the standard
projection used in all aeronautical charts of the United States.
Like the Bonne, the Lambert projection is based on the cone.
The cone, however, instead of being tangent, intersects the earth
along two standard parallels. Along these parallels the scale of
Fig. 7. Lambert Conformal Conic Projection, showing Portion of
Cone Unrolled
the map will be exact. Between the standard parallels the earth
is projected inward onto the cone and the scale of the map is
smaller than the earth. North and south of the standard parallels
the earth is projected outward, and the scale of the cone is larger.
Generally the intersection parallels are chosen so that two-thirds
of the area to be mapped lies between them, one-sixth lies outside
each standard parallel. When this is done the scale error at the
edges equals the error at the center. In a Lambert map the me-
ridians are converging straight lines, the parallels concentric
circles. The spacing of the parallels is not that obtained by sim-
ple central projection, but it is determined so that the distortions
introduced into latitude and longitude by the projection are equal.
MAP PROJECTIONS AND MAP PROBLEMS
Distortion of area and distance is very small. The Department
of Commerce issues sectional and regional maps of the United
States, primarily for use in aviation, in which the standard paral-
lels are 330N and 450N latitude. The scale error is not greater
than one-half of one per cent anywhere on these maps except
along the extreme north and south boundaries. This projection is
particularly valuable in portraying large longitudinal areas.
Straight lines on a Lambert map are close approximations
to great circles and may be regarded for practical purposes as the
shortest distance between points on the map. Since the merid-
ians converge, a straight line will cut the meridians at different
angles. To find the rhumb line
course between two points it is
only necessary to measure the
angle between the straight line
joining the two points and the
meridian halfway between them.
In following a course determined
in this way an airplane will not -
track the line drawn exactly but
will depart from it as shown by
the dotted line in the figure. If Fig. 8. Course Line and Track (Dot-
the difference in longitude is ted Line) on a Lambert Conformal
greater than 3 or 4 degrees the Map. The Difference is Exaggerated
line should be broken into sections and the rhumb line section
for each course followed.
There is one further type of projection, the Polyconic, widely
used. Instead of using a single cone this projection uses a series
of frustrums of cones tangent to the earth at selected parallels
of latitude. In the developed map the central meridian is a
straight line and is divided into distances proportional to the
actual distance between parallels. Through these points latitude
circles are drawn with their centers on the central meridian and
radii equal to the slant height of the cone tangent at that lati-
tude. Meridians other than the central appear as curved lines.
The distortion of latitude is even less for this projection than
for the Lambert. Since the reference lines are curved this pro-
jection is not suited for navigational purposes, but it is excellent
for general use. It is one of the best projections for representing
areas large in a north-south direction and is widely used in tacti-
cal and ordnance maps and in surveys.
28 BACKGROUND MATERIALS FOR WAR MATHEMATICS
Fig. 9. Polyconic Projection
In addition to the projections given above, the two construc-
tions following are elementary and well illustrate methods of
map drawing.
MAP PROJECTIONS AND MAP PROBLEMS
EXAMPLE: To construct a gnomonic projection on a plane
tangent at the equator.
w r
N
E
S
Fig. 10. Construction of a Gnomonic Chart on a Plane Tangent at the Equator
SOLUTION:
Draw a circle to represent the equator. Construct two per-
pendicular diameters RS and WE and a line tangent to the circle
at R. Divide the arc RE into equal parts; in the figure 100 is used.
Lines OA, OB, OC, etc., represent meridians. Draw the radii of
the arcs to intersect the tangent line in A, B, C, D. These points
of intersection are the points on the equator of the map where
the meridians intersect it. Since the meridians of the sphere are
represented by parallel straight lines perpendicular to the
straight-line equator, the meridians can be drawn when their
points of intersection with the equator are known.
30 BACKGROUND MATERIALS FOR WAR MATHEMATICS
The central meridian is spaced in latitude just as the merid-
ians are spaced on the equator. In this way the points of inter-
section of the parallels with the central meridian are obtained.
To determine the points of intersection, the process is as follows:
In the figure, the meridian 30 out from the central meridian is
selected. Construct CN perpendicular to OC; then CG', which
equals CG, determines G', the intersection of the parallel of 100
north with the meridian of 300 longitude east of the central me-
ridian. Likewise, CH = CH', and so on. Since the projection is
symmetrical with respect to the central meridian and with the
equator, the same values can be transferred to the meridians in
longitude west of the central meridian. After the points of inter-
section of the parallels with the various meridians are deter-
mined, smooth curves should be drawn connecting points at the
same latitude to represent the parallels.
EXAMPLE: To construct a stereographic meridian projection.
In this type of projection a hemisphere is projected onto the
plane of its bounding meridian. Along this meridian the scale
is constant.
A
Fig. 11. Construction
of a Stereographic
Meridian Projection.
MAP PROJECTIONS AND MAP PROBLEMS
SOLUTION:
Represent the central meridian and the equator by two perpen-
dicular diameters PP' and WE of the circle that represents the
boundary of the map. Divide the arc PE into equal parts, in the
figure each being for 10 of latitude. Take arc PF equal to 30
and construct FA tangent to the circle at F. With A as center and
radius AF, draw the arc FR. This arc will represent the parallel
of latitude 60.
Take OB equal to AF. With B as center and radius PB, draw
the arc PSP'. Then this arc will represent the meridian of longi-
tude 60 reckoned from the central meridian PP'. In this way all
the meridians and parallels can be constructed. Hemispheres con-
structed on this projection are very frequently used in geogra-
phies and atlases.
PROBLEMS ASSOCIATED WITH MAPS
On maps of small areas, the distortion introduced by pro-
jecting a portion of the earth's spherical surface onto a plane
is not appreciable and hence any figure on the map may be re-
garded as similar to the corresponding figure on the earth. The
ratio of distance on the map to the corresponding distances on
the ground is approximately constant. This ratio is called the
representative fraction (RF). The RF of various maps differ
widely. Most road maps are on a scale of approximately 1 in.
= 20 mi. The maps most commonly used in artillery work have
a RF of 1/20000 (app. 3 in. = 1 mi). Maps used in aircraft work
are mostly on the scale of 1/500000 (app. 1 in. = 8 mi). Prac-
tically all maps either have a scale printed on them or the R F
printed on the margin.
Since maps are plane areas, various devices are used to indi-
cate differences of elevation on the ground, the most common
being color shading, hachures and contour lines. Color shading
usually consists of a brown wash, the denser the coloring the
greater the elevation. Hachures are shading done with fine lines
running in the direction of the steepest slopes, being heaviest
where the slope is steepest. They are seldom used over a whole
map, but are used extensively to indicate steep banks, cliffs, etc.
A contour is an imaginary line of the earth's surface all points
of which are at the same altitude above mean sea level. They are
the intersections of a series of equally distant horizontal planes
32 BACKGROUND MATERIALS FOR WAR MATHEMATICS
with the earth's surface, (more exactly a series of concentric
spheres). On maps, contours appear as irregular curved lines.
The difference in elevation indicated by successive contours
varies with the scale of the map used and the character of the
ground. The actual elevation is always printed on the contours.
Any contour consists essentially of a series of alternate irregular
V and U shaped arcs, the V's pointing up the valleys with their
points at streams and rivers, the U's directed outward around
the hills between valleys. The principal characteristics of con-
tour lines are the following: All contours are closed curves, either
within or beyond the boundaries of the map. Between critical
points (sudden change in the slope of the ground) contours are
approximately equally spaced and parallel. They never intersect
(except in the case of an overhanging cliff). They are always
perpendicular to the steepest slope and cross ridge lines at right
angles. The more rapidly the slope changes, the closer the contour
lines are. Every contour that closes on a map indicates a summit
or a depression. In addition to the elevation shown on contours
the elevation of peaks is printed.
Practically all maps in use will show meridians of longitude
and parallels of latitude. In addition all military maps will have
a system of grid lines superimposed on the map. These grid
systems are simply a rectangular coordinate system with the
positive Y-axis north, the positive X-axis east. The lines are
uniformly spaced at 1000 yd intervals. The unit of measurement
in practically all military work is 1000 yd. The grid system in
maps covering large areas are all referred to a common origin,
and the lines numbered consecutively, from map to map. The
position of points on the map are given in terms of either co-
ordinates, and a dash is placed between the coordinates. Thus
(863.450 1242.680) indicates that the point is 863,450 yards
to the east, 1,242,680 yards north of the origin. Note that the
decimal point is placed after the thousands. On scales in common
use coordinates can be read easily to the nearest 100 yards and
the nearest 10 yards can be estimated. Some simple illustrations
of the application of elementary mathematics to map work are
given.
EXAMPLE: To construct a movable strip scale for reading
horizontal distances. This is simply the well known problem of
dividing a line segment into a number of equal parts. The scale
should be marked to read actual ground distances. Thus suppose
MAP PROJECTIONS AND MAP PROBLEMS,
the scale is 2" = 1000 yd. Draw a scale and subdivide a 2"
interval into tenths as shown in figure 12.
- 800 600 4 00 I $
Fig. 12. Construction of a Movable Strip Scale for Reading Distances on a Map
Changes in elevation on the ground are expressed in terms
of degrees, gradients or per cent. Degree of slope is simply the
angle measured in degrees, between the horizontal plane and the
plane of the slope. Gradient is the ratio of vertical to horizontal
distances, the former given as unity. A gradient of 1 in 12
means that the ground rises 1 unit in each 12 horizontal units
of distance. The per cent slope is the gradient expressed as a
percentage (i. e. the number of units rise in 100 units of hori-
zontal distance). Thus for a slope of 40, the gradient is 1-14, the
per cent 8.75.
EXAMPLE: To construct a scale for reading slopes from a map.
The distance between contours on a map is called map distance,
(MD), the corresponding horizontal distance on the ground, the
horizontal equivalent (HE), the difference of elevation between
two contour lines the vertical interval (VI). If we know the rep-
resentative fraction (RF) then MD = HE x RF. Also HE = VI
cot 0, where 0 is the slope in degrees. Then MC = VI x RF cot 0.
The map distance will vary inversely as the slope; the greater
the slope the closer the contours and the less the map distance.
In using this formula the MD will be given in terms of the same
units that the VI is given in. For slopes up to 100 it can be safely
assumed that the gradient varies directly with the degree, or
that cot 0 is inversely proportional to 0. Hence to construct a
slope scale determine the MD corresponding to a 1 slope; one-
half of this distance is the MD of a slope of 20, one-third is the
MD of a slope of 3, etc. If these distances are laid off succes-
sively on a straight line we have a map distance slope scale. Thus
suppose the VI is 10 meters, the RF = 1/10000. Then
34 BACKGROUND MATERIALS FOR WAR MATHEMATICS
MD (1 10 x 57.3= 0.0573 meters = 5.73 cm. (cot
10000 10 = 57.3)
MD (2) = 5.73/2 = 2.86 cm
MD (3) = 5.73/3 = 1.91 cm etc., and the scale would
be laid off as shown:
////7/7////// //// //////// / //1 77 t//7
Dbeare.e Sope I 3" 4 1 4- IS- 60 170
Gridient j1-7.3 1-286 /-/9.1 I-4.3 IH-.+ I/-.51-V4
i V = /O MeterS RF= //Io,ooo
Fig. 13. Scale for Reading Slopes from a Contour Map
By sliding this scale over the contours, until a space is found on
the scale equalling the contour spacing on the map, the slope can
be read directly. A knowledge of slopes is indispensable in all
artillery work. The slope of a hill determines whether or not a
gun can be brought to bear on a target beyond the hill, whether
the battery is screened from hostile observation, etc.
EXAMPLE: To construct an elevation profile from a contour
map.
In order to pick locations for guns, observation posts, possi-
ble fields of fire, etc., it is important to be able to construct pro-
files along lines of vision and fire. A profile is simply the irregular
outline traced in a vertical plane by the surface of the earth.
From a contour map, knowing the scale of the map and the con-
tour interval such a profile can be quickly constructed on a sheet
of cross section paper. Take a sheet of cross section paper and
let the horizontal scale be that of the map used. Choose any
convenient vertical scale. The vertical scale should be exagger-
ated in order to bring out detail, since usually changes in eleva-
tion are small compared to horizontal distances. It is most con-
venient to let each horizontal line represent a contour on the
map. The vertical scale is then the ratio of the distance between
the lines and the vertical interval of the contour lines on the
map. Mark the horizontal lines with the elevations of the con-
tours, from the lowest to the highest points which the profile
is to include. Place the edge of the paper along the line to be
profiled. Below each point where a contour line meets the edge
of the paper, mark the horizontal line corresponding to the eleva-
tion of the contour. By joining these points with a smooth curve
we have the required profile.
MAP PROJECTIONS AND MAP PROBLEMS
Fig. 14. Construction of a Profile from a Contour Map
In connection with map problems we include here an elemen-
tary problem of artillery fire using the military grid system of
coordinates. Directions in artillery fire are always measured from
Y-north, the vertical axis, rather than from the X-axis as is
conventional in elementary instruction. Y-north does not neces-
sarily coincide with true north, particularly if the maps used
are conical projections. The angle between Y-north and any line
is called the azimuth of the line. In firing at unseen targets, the
gun sight is usually set on some fixed object called the "aiming
point." The angle between the line of fire of the gun and the line
of sight to the aiming point is called deflection.
EXAMPLE: The positions of a gun and a target have been
plotted on a map. The coordinates of the gun are (715.23 -
1025.61), of the target are (718.34 1027.42). Find the range
and the azimuth of the target from the gun.
36 BACKGROUND MATERIALS FOR WAR MATHEMATICS
Cnr\T TTlT/IT -
OUL U IIU ; IN
Target coordinates
718.34 1027.42
Gun o
coordi-
nates 715.23 1025.61
dx = 3.11, dy = 1.81 .
Sdx 311 r-
Tan Z = --whence
dy 181
'r 716
azimuth = 59 48'02"
Range = (dx)2 + (dy)2 = V12.9482
= 3.597 thousands of yd
= 3600 yd to the nearest 10 yd.
EXAMPLE: An observation post 0 has been located at (813.50
- 1262.14), the coordiates being accurate to within 10 yd. The
distance of a gun from the post -/or MA
has been measured as 2550 yd, i3
its azimuth as 53 40.2'. What
are the coordinates of the gun's
position ?
SOLUTION:
dx = 2550 x sin 530 40.2' =
2054 yd d4
dy = 2550 cos 53 40.2' =
1511 yd
X coordinate of G = 813.50 +
2.05 815.55
Y coordinate of G = 1262.14 + 0
1.51 = 1263.65 Fig. 16.
There are many basic elementary problems of triangulation
connected with mapping. The most important of these are the
problems of intersection and resection. The elementary problem
of intersection is essentially that of finding the distance and di-
rection of an inaccessible point, such as a target, by triangula-
tion from known base points. This type of problem is amply
illustrated in any text on geometry or trigonometry. The problem
of resection is the same as that of intersection except that the
point to be located is that occupied by the observer. The solution
of this problem requires that at least three points, whose loca-
tions on the map are known, are visible to the observer. We give
here two simple geometric methods of solving resection problems.
jd"
__- *_ ..... J
717 718 19
Fig. 15.
MAP PROJECTIONS AND MAP PROBLEMS
SOLUTION 1:
Let A, B, C be the three visible points. Fasten a sheet of tracing
paper on a horizontal board or table. From a point P on the paper
sight along a straight edge to A, B, C in succession and draw
the rays PA, PB, PC. Now shift the paper over the map until
each ray passes through the point on the map representing the
station to which the ray was drawn. The position of P is then the
observer's location on the map.
SOLUTION 2:
As before let A, B, C be the known points, P the position of
the observer. Measure the angles APB and BPC. On the map
draw the lines AB and BC. On AB as a chord construct the seg-
ment of a circle which contains the angle APB, on BC construct
the segment of a circle which contains angle BPC. The inter-
section of these circles is the required point P. If P is on the
circle through A, B, C the problem cannot be solved.
The trigonometric solution of the three point problem is
long but straightforward. Three possible cases arise. The un-
known point may be within the triangle ABC of known points,
on the same side as C from AB, or on the opposite side. Plot the
points A, B, C and an assumed position of P. Draw the circle
through A, B and P. D is the intersection of CP with this circle.
The three cases are shown in the figure. The sides and angles of
Fig. 17. Diagram for the Trigonometric Solution of the Resection Problem
the triangle ABC are known, the angles X and Y are either the
measured angles BPC, APC or their supplements. The steps in
the solution are now obvious. Solve triangle ABD for AD and BD
o- p
CASE I CASE 2 CASE 3
Fig. 17. Diagram for the Trigonometric Solution of the Resection Problem
the triangle ABC are known, the angles X and Y are either the
measured angles BPC, APC or their supplements. The steps in
the solution are now obvious. Solve triangle ABD for AD and BD
38 BACKGROUND MATERIALS FOR WAR MATHEMATICS
by the law of sines. Then in triangle ADC, sides AD and AC are
known and angle CAD = angle BAC + X. Angle ACD may be
found from the law of tangents. Then in triangle ACP, side AC,
angle APC, and angle ACP are known and AP or CP can be
found from the law of sines. A check is obtained by solving tri-
angle DBC and BPC and finding BP or CP.
Since many military and civil maps are now made by the
use of aerial photography we include a simple problem illustrat-
ing the principles of aerial mapping. An aerial map consists es-
sentially of a mosaic of carefully matched parts of photographic
prints. The scale of the prints is determined by the focal length
of the camera and the height at which the pictures are taken.
Thus if AB is the distance between /
two points on the ground, A'B' the
corresponding distance on the _
print, H the height of the plane, -- ----
f the focal length of the camera,
A'B' f,
we have at once A-- or the
AB H
representative fraction is the ratio
of the focal length of the camera H
to the height of the plane above
the ground. Before the flight a grid
is laid out and the position of prom-
inent land marks indicated. The A 5
problem then reduces to finding the Fig. 18.
number of exposures that must be made for flight at a given
altitude, the time interval between exposures and the number of
trips back and forth over the area to be covered that the pilot
must make. It is customary to allow an overlap on exposures of
60 per cent in the direction of flight and 50 per cent between
strips.
EXAMPLE: A mosaic map is to be made on 6 in. by 8 in. rec-
tangular plates, the longer dimension being on the direction of
flight. The focal length of the camera is 12 in. The plane is to fly
at an altitude of 12,000 ft. Suppose an area to be mapped is 5 mi
wide and 12 mi long. Allowing for overlap, how many exposures
must be made? If the speed of the plane is 100 mph, what is the
time interval between exposures?
MAP PROJECTIONS AND MAP PROBLEMS 39
SOLUTION:
The scale of the prints if f/H = 1/12000
6 = 1 x = 6000 ft = width of area photo-
x 12,000 graphed at each exposure
8 = 1 y = 8000 ft = length of area photo-
y 12,000 graphed at each exposure
Since only 50 per cent of the width is to be used, the number
of strips that must be made is obtained by dividing the width
of the area in feet by 50 per cent of 6000.
5 x 5280
Number of strips- 300 8.8 or 9 strips.
3000
Since only 40 per cent of the length is used, the number of ex-
12 x 5280
posures per strip 0.4 x 8000 19.8 or 20 exposures per strip.
Hence 9 x 20 = 180 exposures must be made.
Ground distance between successive photos is 0.4 x 8000
3200
feet = 3,200 ft. 100 mph = 147 ft sec (app.). Hence 30 = 22
147
sec (app.). An exposure must be made every 22 seconds. It is
assumed that the exposures are made with the camera vertical
and plates parallel with the ground.
SECTION III
Vectors and Vector Problems
One of the most useful concepts in mathematical applica-
tions to physical problems is that of the vector. A vector is a
quantity having both magnitude and direction. Forces, velocities,
accelerations, momentum, etc., are vector quantities. In general
any problem dealing with displacements is a vector problem. The
fundamental idea of the vector is that if two or more vector
quantities such as forces act simultaneously on a body, the re-
sult at the end of any interval of time will be the same as if each
of the vectors involved had acted on the body independently for
the same length of time.
We can replace a set of vectors which act simultaneously by
a single vector, called the resultant of the given vectors. If two
vectors act in the same direction, the resultant will be a vector
in the same direction of magnitude equal to the sum of the mag-
nitude of the two vectors. If they act in opposite directions, the
resultant will be a vector in the direction of the larger, of magni-
tude equal to the difference of the magnitudes of the two vec-
tors. Operations with signed numbers is essentially vector
composition.
If two vectors OA (magnitude a) and OB (magnitude b) act
on a body at O, and the angle between their direction is 0, their
resultant R is obtained by vector composition or addition. That
is, construct OA and OB with lengths proportional to a, b. Lay
/-------- ~--1
o A1
Fig. 19. Vector Compositlon
off AP equal and parallel to OB. Then the resultant R is given in
direction and magnitude by OP. By the use of the law of cosines
VECTORS AND VECTOR PROBLEMS 41
and properties of the right triangle it is readily seen that the
magnitude R and direction of the resultant are given by
R2 = a2 + b2 + 2ab cos 0
S b sin 0
tan =--
a + b cos 0
Angle 0 being the angle which the resultant makes with OA.
If several vectors are involved the resultant may be obtained
quickly and easily by graphical means. Represent the vectors by
arrows pointing in the given directions and of lengths propor-
tional to the given magnitudes. Starting with any one of the vec-
tors place the start of the second arrow at the end of the first,
continue with each vector in turn, preserving the direction in
each case. The line joining the start of the first vector to the
end of the last will represent the resultant.
Analytically the resultant may be found by replacing each
pair of vectors in turn by their resultant, or by the method of
resolutions into components. Not only can the resultant of a set
of vectors always be found, but any vector can be treated as if it
were the resultant of another set of vectors. That is, we can
resolve any vector into components, and this in an infinitude of
ways. In practice vectors are always resolved into components
along and perpendicular to a given direction. If we represent a
vector OP as drawn from the origin of a system of rectangular
coordinates and project it onto the X and Y axes by dropping
perpendiculars from its end point, these projections are the
X and Y (horizontal and verti-
cal) components. Vectorially P
OA + OB = OP. The component
of a vector along a given direc-
tion represents the effective ac- Y
tion of the vector in that direc-
tion. It is to be noted that the
X and Y components of the vec- o X
tor OP are the rectangular co- Fig. 20. Resolution of a Vector
ordinates of the point P, that into Components
the magnitude and direction of the vector OP are the polar co-
ordinates of P. If R is the magnitude of OP and it makes an
angle 0 with the X axis, then the X component of OP, is R cos 0,
the Y component R sin 0. It is readily proved that the X and Y
components of the resultant of any number of vectors is the
respective sum of the X and Y components of the vectors. If the
42 BACKGROUND MATERIALS FOR WAR MATHEMATICS
resultant of a set of vectors is zero then the system of vectors
is in equilibrium. Whenever the net result of several forces acting
on a body is zero, if the vectors representing the forces are added
head to tail in any order, the vector diagram will consist of a
closed polygon.
The independence of the action of vector quantities on a
body may be shown by many simple experiments. Thus if one
ball is projected horizontally at the same instant that another
ball is dropped vertically, they both strike the ground at the
same instant. Gravity accelerates all objects toward the earth
at exactly the same rate, quite independently of any other mo-
tion they may have. The actual motion of the ball thrown hori-
zontally will be the result of two vector components, one horizon-
tal and constant, the other vertical and accelerated, and since
the vertical components of the motion of the two balls are iden-
tical, they drop together.
PROJECTILE MOTION
The motion of any projectile through the air is the resultant
of many vector quantities. Among the most important of these
are the velocity imparted to the projectile by the weapon; grav-
ity, which accelerates the projectile downward; the resistance
of the air which depends on the velocity of the projectile, its size
and shape and the density of the air. If we take all the forces
acting into account the calculation of the curve followed by a
projectile in flight is an extremely difficult problem, which can
be solved only by methods of approximation. If we neglect air
resistance, however, and take into account only the initial veloc-
ity and elevation and the effect of gravity the problem is simple.
EXAMPLE: To find the equation of the trajectory of a pro-
jectile neglecting air resistance.
SOLUTION:
Take the X-axis as horizontal, the Y-axis vertical. Let the origin
be at the gun's muzzle, the elevation of the gun (angle the gun
bore makes with the horizontal) be 0. If V is the initial velocity
of the bullet, its X and Y components are V cos 0, V sin 0, and
if gravity did not act, in time t the position of the bullet would
be X = (V cos 0)t, Y = (V sin 0)t. But in time t an object would
fall the distance gt2/2 where g is the acceleration of gravity
VECTORS AND VECTOR PROBLEMS 43
(app. 32 ft per sec). This would decrease the vertical component.
Hence the equation of the trajectory would be the parabola
X = Vt cos 0
Y = Vt sin 0 gt2/2
e v coso
0 -Ron.3e I
Fig. 21. Parabolic Trajectory
From these equations the time of flight is obtained as
2V sin 6
t and the range (distance from the gun that the pro-
g
jectile strikes the ground) as X = V2 sin 2 0/g. The range is a
function of the initial velocity and angle of elevation, and it is
easily seen that the maximum range for a given velocity is
obtained with an initial elevation of 45. For objects having
relatively small velocities the parabolic trajectory is a good ap-
proximation to the true path. For high speed projectiles air
resistance is such a large factor that the above equation is only
a crude approximation. Air resistance slows the projectile down,
decreases the range and increases the angle of impact, that is,
the angle at which the projectile strikes the ground.
Artillery is usually classified according to the kind of tra-
jectory the projectiles follow; a gun or rifle has a high muzzle
velocity and a comparatively flat trajectory. They are usually
designed to fire at a maximum elevation of 45. Howitzers have
a shorter tube, a lower muzzle velocity and a more curved tra-
jectory, firing at angles of elevation up to approximately 65.
Mortars have still shorter tubes, lower muzzle velocity, more
curved trajectories and are primarily designed for firing at
angles of elevation greater than 45.
EXAMPLE: A bomb is dropped from a plane traveling 200
mph. If the bomb struck the target 25 sec after release, how high
was the plane? How far was the plane from the target at the
instant the bomb was released? Neglect air resistance.
SOLUTION:
In time t an object drops a distance S = gt2/2. t = 25 sec, g =
32.2 ft/sec. Hence S = 16.1(25)2 = 10062 ft = height of plane.
200 mph = 293 ft/sec. In 25 sec the plane will have traveled
44 BACKGROUND MATERIALS FOR WAR MATHEMATICS
25 x 293 = 7325 ft = horizontal distance of plane from target
at the instant of release.
Although air resistance increases with the velocity, the in-
crease does not follow any simple fixed law for all velocities. For
a wide range of velocities, however, air resistance is proportional
to the square of the velocity. If the resistance is measured in
pounds of force and the velocity in miles per hour, the air re-
sistance for ordinary objects is given approximately by 0.01V2.
The increase in air resistance is also proportional to the cross
sectional area, and depends greatly on the shape; for example
air resistance is only one-fifth as great for a long nosed boat-
tailed shell as for a spherical ball.
All objects, regardless of their shape, size, or mass would
fall toward the earth with exactly the same acceleration if it were
not for air resistance. As the object falls, its velocity increases,
hence air resistance increases. The force of resistance will in-
crease until it just balances the downward pull of gravity (i. e.,
the weight of the object). When this point is reached the vector
forces balance, and no further acceleration will result. The ob-
ject is said to have reached its terminal velocity, and it will con-
tinue to fall with this constant velocity.
EXAMPLE: A 144 pound aviator makes a delayed parachute
jump. Assuming the resistance of the air is 0.01V2 lb, when V
is measured in mph, what will the terminal velocity be?
SOLUTION:
For terminal velocity we must have weight equal to air resistance.
Therefore 144 = 0.01V2, V = 120 mph is the terminal velocity.
EXAMPLE: If a 150 lb parachute trooper lands with a terminal
velocity of 10 mph, with what velocity will he land if he is loaded
with 100 lb of equipment, assuming resistance proportional to the
square of the velocity?
SOLUTION:
For terminal velocity we must have w, = kv12, w2 = kV22, whence
w1/w2 =v12/V 2.
wi = 250 Ib, w2 = 150 Ib, v2 = 10 mph whence vi = 13 mph
(app.) is the terminal velocity.
Whenever a projectile is fired, momentum (mass times ve-
locity) is imparted to it. From the principle of conservation of
VECTORS AND VECTOR PROBLEMS
momentum it follows that momentum can be imparted to any
object only if some other object acquires an equal and oppositely
directed momentum. For this reason guns "kick back" when fired.
Momentum problems are an excellent example of inverse pro-
portion.
EXAMPLE: A fighter plane, weighing 4000 lb and travelling
at 200 mph fires a 5 sec burst directly ahead from 8 machine
guns. If the guns fire 600 4 oz bullets per minute at a muzzle
velocity of 2000 ft/sec what will the plane's velocity be after the
burst of fire?
SOLUTION:
Each gun fires 10 bullets/sec. Hence in 5 sec 100 Ib of bullets
are imparted a velocity of 2000 ft/sec. We must have M1 V1 =
M2 V2. 4000 V1 = 100 (2000), V1 = 50 ft/sec = 34 mph (app.)
is the velocity lost by the plane during the 5 sec of fire. The
final velocity will be 216 mph. In this problem we have used
weights in place of determining the masses since only the ratios
are involved.
In these days of tank and airplane combat and rapidly mov-
ing targets, nearly every problem of gun fire is a complex vec-
tor problem. A projectile fired from a moving base will travel
a course which is the resultant of its own velocity and the ve-
locity of the base from which it is fired. Thus to fire a moving
target the gun must "lead" the target, that is fire at the position
the target will occupy when the bullet gets there. To fire from
a moving base at a stationary object the gun must be pointed
enough behind the target to allow for the fact that the bullets
will keep the velocity of the moving base. This is the reason for
the introduction of tracer bullets. A gunner can judge the proper
corrections to be made in angle of fire by watching the smoke
path left by a tracer bullet.
EXAMPLE: A ship is 12,000 yd east of a submarine and travel-
ling a course of 3300 at a speed of 18 knots. In what direction
should a 27 knot torpedo be fired so as to strike the ship?
SOLUTION:
Let A be the position of the submarine, B the position of the
ship, BL the ship's course. To solve the problem graphically, lay
46 BACKGROUND MATERIALS FOR WAR MATHEMATICS
off to scale AB = 12,000 yd = 6 L
mi (nautical). Lay off BL,
course 3300, length of BL = 18
mi, draw LM parallel to AB.
With A as a center and 27 mi
to scale as a radius, strike an c
arc meeting LM at P. The tor- B
pedo should be fired in the di- Fig. 22.
reaction AP. The trigonometric
solution is obvious. If C is the point of collision and x the angle
BAC we have BC/AC = 18/27 = 2/3. Angle ABC = 60. Hence
sin x/sin 60 = BC/AC = 2/3 or sin x = 2/3 sin 600 =
3 x = 350 16'.
VECTOR FORCES ACTING ON AIRPLANES
There are four major force vectors acting on an aeroplane
in flight: gravity, lift, thrust and drag. The force of gravity is
the weight of the plane and acts downward. Lift is the force
that counteracts gravity. Thrust is the forward pull and creates
lift. Drag is applied to the forces such as air resistance which
opposes thrust.
A plane is supported in the air by the lift developed by the
forward motion of its wings through the air. This lifting force
varies primarily according to the speed of the plane, the airfoil,
and the angle the plane makes with the relative wind. The word
airfoil means any surface such as a wing, rudder, etc., which is
designed to obtain a reaction from the air through which it
moves. In practice airfoil is applied only to the curves describing
the contour of the wing. In order to produce lift the air pressure
beneath the wing must be greater than the air pressure above.
This difference in pressure is obtained by shaping the airfoil so
that the air stream from the propeller travels a greater distance
and at a higher velocity above the wing than it does below. The
wing is also tilted at an angle with the horizontal'so that the air
stream creates an impact pressure on the lower surface of the
wing. The pressure exerted by any fluid (liquid or gas) on a
surface over which it moves varies inversely as the velocity. This
physical law is known as Bernoulli's theorem. Since the airstream
over the top surface of a wing moves faster than it does below
VECTORS AND VECTOR PROBLEMS
the wing, the pressure above is less than the pressure below.
This difference in pressure results in an upward lift.
The relative wind is the direction of airflow with respect to
the wing. If the wing is moving horizontally forward, the relative
wind is horizontally backward. The angle of attack is the angle
between the relative wind and the lower surface of the wing. If
we consider the resultant of the forces acting on a plane, lift is
the component of the resultant force perpendicular to the relative
wind. Drag is the component of the resultant force parallel to
the relative wind. In level flight at constant speed the resultant
of all the forces acting must be zero, lift equals weight, thrust
equals drag. Both lift and drag depend on the airfoil, the density
of the air, the angle of attack, wing area and velocity.
The equation, giving lift in lb is
(P)
L = C AV2,
2
where p is air density in slugs per cu ft. (p = 0.002378 slugs/cu
ft under standard atmospheric conditions), A is wing area in
square feet. V is velocity in ft/sec, C is the lift coefficient which
is determined experimentally for each airfoil at different angles
of attack. The formula for drag is exactly similar to that for lift
except that the drag coefficient must be used instead of the lift
coefficient.
Since in level flight lift equals weight we have at once
W = C () AV2, or V= -2w
2 CpA
The ratio W/A, weight divided by wing area, is the wing loading.
If this is constant, velocity will vary inversely as the square root
of the lift coefficient. For the same airfoil and angle of attack,
assuming air density constant, velocity is directly proportional
to the wing loading, or
V, I(W )/(Wi)
V2= A, A2
EXAMPLE: An airplane weighing 3000 lb lands at 45 mph;
with 500 lb added load with what speed should it land?
48 BACKGROUND MATERIALS FOR WAR MATHEMATICS
SOLUTION:
Here Ai = A2 and therefore
V1 =V2 VW7WT= 45 300j = 48.6.
The plane should land at a velocity of 49 mph.
When a plane makes a turn centrifugal force will act upon
it in addition to the other forces. The value of centrifugal force
depends upon the speed and the radius of turn. In order to main-
tain stable flight in a turn without losing altitude, slipping or
skidding, the plane must be banked at an angle to the horizontal,
with the inner wing lower than the outer. This angle is called
the angle of bank. The resultant of the weight and centrifugal
force will be a force directed outward and downward. In a cor-
rect turn this resultant force should balance with the lift, which
is always perpendicular to the wings. The resultant of the weight
and centrifugal force must be perpendicular to the line of the
wing, and equal to the lift.
L'
\ \ ^ L\ \13
SCF CF\ \CF
W W
,ASE CASE Z CASE
Fig. 23. Vector Forces in a Horizontal Turn
Case 1. Angle of Bank too Small. Forces result In Outward Skid
Case 2. Angle of Bank too Large. Forces result in Downward Slip.
Case 3. Correct Angle of Bank. Forces In Equilibrium
EXAMIPLE: Find the correct angle of bank for an airplane
making a horizontal turn. Show that the lift in a turn must be
greater than that in level flight if the plane is not to lose altitude.
SOLUTION:
Let B be the angle of bank. It is the angle between the line of
the wings and the centrifugal force vector. Since the weight (W)
is perpendicular to centrifugal force (CF) and for a correct turn
we must have the resultant R perpendicular to the line of wings,
VECTORS AND VECTOR PROBLEMS
the angle between R and W must also be B. Hence we must have
CF,
tan B = C
W
or the angle of bank should be the angle whose tangent is the
ratio of the centrifugal force to the weight. Since CF = MV2/R,
W = mg, we have tan B = V2/Rg, where V = speed, R = radius
of turn, g = acceleration of gravity.
Since lift must equal the resultant of W and CF and R =
VW ~ + (CF)2 we must have L = VW2 + (CF2 which is greater
than W. Hence if altitude is not to be lost the lift must be in-
creased in a turn; greater power will be required in a turn than
in level flight.
The bank indicator of an airplane is a simple instrument
utilizing the principle of balance of vector forces. It consists
usually of a hollow glass tube slightly curved, containing a steel
ball. In level flight gravity keeps the ball centered at the lowest
point. In a turn gravity tends to pull the ball to the inside end
of the tube, centrifugal force to throw it to the outside, at the
proper angle of bank these forces balance and the ball remains
centered.
EXAMPLE: An airplane of 2000 lb is turning at 175 mph with
an angle of bank of 500. What should be the radius of turn?
SOLUTION:
V = 175 mph = 770/3 ft/sec = 256.7 ft/sec.
tan B = V2/Rg, whence R = V2/g tan B
R (256.7)2 = 1717 ft is the radius of the turn.
32.2 tan 500
SECTION IV
Navigation
The basic problem of all forms of navigation is that of de-
termining direction and position on the earth's surface. A navi-
gator must constantly be concerned with direction in order to
steer his craft properly and must determine his geographic
position frequently in order to make certain he is making good
his intended track. Commonly navigation is divided into three
main subdivisions.
1. Pilotage is the art of directing a ship or plane over a
desired course by reference to visible landmarks, the location of
which are known or can be determined by means of maps or
charts.
2. Dead reckoning (D.R.), or more accurately deduced
reckoning, is the method of keeping track of position by means
of known courses steered and distances run from a known point
of departure. Distances are obtained from the known speeds and
elapsed time of travel. In practice most dead reckoning and
pilotage problems are solved by plotting courses and positions
directly on maps or charts.
3. Celestial navigation is the method of determine geo-
graphic position by means of altitude observations of celestial
bodies such as the sun, stars or planets, together with the exact
time of observation. Celestial observations are used mainly to
check and correct dead reckoning positions and not as an inde-
pendent form of navigation.
The course of a ship or plane is the direction of travel over
the earth. It is the angle measured clockwise from north to the
course line or track.
The heading of a ship or plane is the angle between north
and the longitudinal axis of a plane or the keel of a ship. Heading
and course will only coincide when there is no wind or current. In
both naval and aircraft work courses and headings are given in
three figures from 0000 at north to 360.
DETERMINATION OF DIRECTION THE COMPASS
In practice, true directions of travel are determined by
plotting the desired course line on a map or chart and measuring
NAVIGATION
the angle between the course line and the geographic meridians.
In order to determine whether the desired course is being fol-
lowed or not some form of compass instrument is essential. The
oldest and still the basic instrument is the magnetic compass,
which depends for its directional effect upon the earth's magnetic
field. The earth is similar to a great magnet, having like any
magnet, magnetic poles. These magnetic poles are not located at
the geographic poles. The north magnetic pole is located near
Hudson's Bay at approximately 71 N latitude, 960 S latitude,
156 E longitude. Since the earth is not composed of homogene-
ous magnetic material, the lines in which the earth's magnetic
forces act are not regular curves and in places differ greatly
from the great circles passing through the poles. These irregular
curves are called the magnetic meridians. At any point on the
earth's surface a magnetic compass points in a direction parallel
to the magnetic meridian if it is not influenced by any other
magnetic forces.
Variation is the difference in direction between the geo-
graphic and magnetic meridians. It is the angle measured in
degrees from true north to magnetic north and is marked east
or west according to whether the magnetic meridian is east or
west or true north. The value of magnetic variation depends upon
the geographic location. At some places magnetic and true direc-
tions coincide. The magnetic meridian of 0 variation just misses
the east coast of Florida and passes near Savannah and Chicago.
At Pensacola the variation is 50 E, at San Diego it is 15 E, at
New York it is 10' W, in Maine it is 200 W.
Any iron or steel placed in a magnetic field takes on mag-
netic properties. This magnetism tends to become permanent if
the iron or steel is subjected to any strains and hence any ship
or plane containing steel parts becomes, in effect, a magnet dur-
ing the process of building. This magnetism will affect the com-
pass of any ship or plane and cause it to deviate from magnetic
north.
Deviation is the angle between magnetic north and the
north indication of the compass. It is measured in degrees east
or west from magnetic north to compass north. Variation is fixed
by geographic position but deviation will differ for each ship
and compass. The effect of the ship's magnetism on a compass
changes with the direction the ship is heading, and hence devia-
52 BACKGROUND MATERIALS FOR WAR MATHEMATICS
tion varies with the heading. The effects of a ship's or plane's
magnetism can be partially compensated for but cannot be elimi-
nated entirely. In practice the deviations of a ship's compass are
determined for every 150 change in magnetic heading; for nor-
mal aircraft operations it is sufficiently accurate to determine
the deviations for every 30'. Intermediate values are found by
interpolation. A typical airplane compass deviation card would
be as shown below
given.
except that the deviation row would not be
Magnetic heading 0 30 60 90 120 150
II
Compass heading 2 33 64 93 122 150
Deviation 2W 3Wi 4W 3WI 2W 0
I I I 1
Magnetic heading 180 210 240 270 300 330
Compass heading 179 208 237 267 299 330
Deviation 1El 2EI 3E1 3E 1E 0
In any navigation problem there will always be three ways
of expressing any given course or heading:
1. The true course (T.C.) or the angle measured clockwise
from geographic north to the desired track over the earth's sur-
face
2. The magnetic course (M.C.), for which the origin of
measurement is magnetic north
3. The compass course (C.C.), for which the origin of
measurement is compass north.
Any given direction may be expressed in all three of these
ways and the problem of changing from one to the other is con-
NAVIGATION 53
stantly before a pilot or a navigator. The relation between the
three courses is shown in figure 24.
Sn C
Angle NAB = True course
Angle MAB = magnetic course
Angle CAB = compass course
Angle NAM = variation
Angle MAC = deviation
Angle NAC = compass error
Fig. 24. True, Magnetic and
A Compass Course
EXAMPLE: To find the magnetic and compass courses from
the true course, knowing variation and deviation.
SOLUTION:
If the variation is east, magnetic north is east of true north
and the magnetic course is less than the true course by the
amount of the variation; if variation is west, the magnetic course
is greater than the true course by the variation,
M. C. = T. C. + Var. west,
M. C. = T. C. Var. east.
Similarly, if deviation is east the compass course is less than the
magnetic course; if west the compass course is greater than the
magnetic course by the amount of the deviation,
C. C. = M. C. Dev. east
C. C. = M. C. + Dev. west
The process of applying variation and deviation to a true
course is called uncorrecting. The reverse process of finding a
true course from a compass course is called correcting. The ex-
ample above is sufficient to show that,
When correcting, easterly errors are additive,
westerly errors are subtractive;
When uncorrecting, easterly errors are subtractive,
westerly errors are additive.
54 BACKGROUND MATERIALS FOR WAR MATHEMATICS
In following this rule it must be born in mind that the true course
is the most correct, the magnetic course is incorrect by the
amount of the variation, and the compass course is the least cor-
rect since it differs from the true by the compass error which is
the algebraic sum of variation and deviation.
EXAMPLE: A pilot wishes to fly a true course of 2300, the
variation of 170 W, assuming deviations as shown in the sample
card, what compass course should he steer?
SOLUTION:
Here we are uncorrecting, westerly errors are additive, hence
M. C. = 230 + 170 = 247.
From the card the deviation is 30 E. hence
C. C. = 247 3 = 2440
is the compass course that he should steer.
EXAMPLE: The compass course is 0400, deviation is 50 W,
variation is 90 E. What is the true course?
SOLUTION:
We are correcting, hence,
T. C. = 0400 + 9 50 = 044 is the true course
EXAMPLE: A ship is on a compass course of 0120. The navi-
gator sights two distant peaks in a line bearing 0670 by compass.
From the chart the true bearing is 0490. The variation of the
locality is 120 W. Find the true course and the compass deviation.
SOLUTION:
The true bearing is smaller than the compass bearing, hence
compass error = 0670 049' = 180 W.
Therefore deviation = 18 W 12 W = 60 W.
To find the true course we are correcting, westerly errors are
subtractive. Since 18 from 12 is negative we first add 3600 to
the compass course.
T. C. = 12 + 360 18 = 354' is the true course.
A ship or plane compass differs from the ordinary boy
scout or woodsman's compass in that the magnetic needles are
rigidly attached to the compass card and the whole card rotates.
Directions are read by means of a lubber line marked on the
compass bowl. The card of a ship's compass is flat while that of
an aeromagnetic compass is a cylindrical section. Both are grad-
NAVIGATION
uated in degrees. In the aerocompass intervals of 100 are num-
bered but the final zero is omitted. Thus a reading of 4 means
400, while 34 would indicate 340.
In addition to the magnetic compass, large ships and prac-
tically all naval vessels are equipped with gyroscopic compasses.
The gyro compass has one great advantage over the magnetic
in that within small limits of error gyro bearings are true. It
can be placed anywhere in a ship and be made to operate various
devices like course recorders. Since it is a complex instrument
liable to mechanical failure it cannot entirely replace the mag-
netic compass. A gyro compass suitable for aircraft has not yet
been developed.
PILOTAGE PROBLEMS
Position in pilotage is determined by means of observations
of visible objects. Once a course line has been laid down on the
chart from a fixed point it may be regarded as a statement of
intentions. It is a locus of D. R. positions and is sometimes called
the D. R. track. A course line on a chart is always labelled with
the course above the line and the speed below.
A line of position is a straight line or circle on which a ship
is located at some instant. The simplest line of position is the
range, which is the line of sight connecting two objects. The
next and commonest is the bearing which is the direction of the
line of sight joining a ship and an object.
A fix is an accurate determination of position and is found
by means of the intersection of two or more lines of position. A
line of position should be labelled with the time it is obtained
and also as to type, if a bearing, the direction should be shown,
etc. The use of bearings and ranges to fix position and the proper
labelling is shown in figure 25. Note that the 1950 fix is ob-
tained by advancing the 1930 bearing parallel to itself in the
direction and distance the ship moved between 1930 and 1950.
A fix obtained in this way is called a running fix. The position of
a ship can also be fixed by means of radio bearings. In principle
this method does not differ from that of taking bearings on
visible objects. The direction from which a radio signal is com-
ing can be determined by radio direction finders. If mercator
charts are being used a correction must be applied to a radio
direction to convert it into a true direction since radio waves
56 BACKGROUND MATERIALS FOR WAR MATHEMATICS
travel great circle paths and not rhumb line paths. Consequently
to draw a radio bearing on the chart, the great circle bearing
must first be converted into the mercator bearing.
EXAMPLE: A ship leaves a fix at point L at 1400 on course
0700 true with a speed of 12 knots. At 1520 the ship is on the
range of points A and B and point C bears 1300. Determine the
1520 D. R. point and fix. At 1700 the ship changes course to 120
true. At 1930 point C bears 2050, at 1950 point D bears 600.
Advance the 1930 bearing of C and determine the 1950 running
fix.
SOLUTION:
The solution of this problem is illustrated in figure 25.
L in o P iti i i a F i
0
I I l
Fig. 25. Diagram Illustrating the Plotting and Labelling of Course Lines,
Lines of Position, Deadreckoning Points and Fixes
EXAMPLE: To determine the position of a ship by means of
two bearings on the object, knowing the direction of the course
and the distance travelled between the bearings. Find also the
predicted distance at which the object should be passed abeam.
SOLUTION:
Let the line A B D be the course line, angle A the first bearing
of object C, angle 13 the second bearing as shown in figure 26.
From C drop a perpendicular C D to the course line. Then B D is
the distance of the ship from C at the time of the second bear-
ing. CD will be the predicted distance at which C will be
NAVIGATION 57
passed abeam. A B is the
known run between bearings. /
From elementary trigonometry ty|
it follows that / "i-
BC = AB '
sin A sinA N
sin C AB _w couUsE
sin C sin (B-A) A UN aB o -N-
AB Fig. 26. Distance and Predicted
CD Distance Abeam from Two Bear-
cot A cot B ings on one Object
In practice the navigator solves these triangles for BC and CD
by the use of tables which tabulate values of BC and CD for a
run of 1 mile and relative bearings for even numbered degrees
from 200 to 70. By multiplying the known run by these tabular
values the distance BC and CD are obtained.
Certain special cases of the above problem are frequently
used in practice. In particular if the first bearing is 450 the
second 900. Then BC = CD = AB, or the distance abeam equals
the known run.
EXAMPLE: What should the second bearing be in order that
BC equal the known run AB?
SOLUTION:
If BC = AB we have an isosceles triangle and obviously angle B
must be equal to twice angle A. This case is known as Doubling
the angle on the bow.
If A = 220, B = 450 it is easily shown that the predicted
distance abeam is approximately 0.7 of the known run. If A =
300, B = 60, the predicted distance BD is V312 = 7/8 (app.)
of the known run AB.
EXAMPLE: For what pairs of relative bearings A and B will
the predicted distance abeam equal to the known run AB?
SOLUTION:
Since CD = AB/ (cot A cot B) it is clear that the predicted
distance CD = AB when
cot A Cot B = 1
An infinite number of such pairs of angles can be found. In
practice pairs fairly widely separated and close to whole degrees
58 BACKGROUND MATERIALS FOR WAR MATHEMATICS
are used. Several examples are shown
First Bearing A 22 2
I 29
Second bearing B 34 41 46 51 59
A frequent problem in pilotage is that of steering a ship to avoid
dangerous obstacles. A single example of this type of problem is
given.
EXAMPLE: A navigator wishes to steer his ship to clear a
shoal S near a coast by a given distance. Two prominent land-
marks A and B are visible and identified on the chart.
SOLUTION:
With S as center and the distance it is wished to clear the shoal
as radius describe a circle. Construct a circle through A and B
tangent externally to the first circle, (figure 27). Measure the
angle subtended at the second circle by the chord AB. Set this
angle on the sextant and continuously observe the angle sub-
tended at the ship by AB. As long as this angle is less than that
set on the sextant the ship will be outside the danger circle.
Fig. 27. Danger Angles
NAVIGATION 59
A slight modification of the above problem will enable a ship
to steer between two danger areas. Construct a circle around
each danger area S and R, pass a circle through AB tangent
externally to the danger circle S closest to the coast as before,
pass a second circle through AB tangent to the second danger
circle R on the coast side. Measure the two angles subtended on
these circles by AB. The ship should steer so that the angle
subtended by AB at the ship lies always between the angles
subtended by AB on the two circles through AB.
The problem of locating a ship when three known objects
are visible can be solved by the methods treated under the prob-
lem of map resection.
VECTOR PROBLEMS IN AERIAL NAVIGATION
There are always two factors involved in the motion of an
airplane with respect to the ground; the movement of the air-
plane with respect to the air and the movement of the air with
respect to the ground. The track of the airplane with respect to
the ground will be the resultant of these two motions. The power
plant of an airplane drives it along its heading, that is, the direc-
tion of motion with respect to the air. The air in which the plane
is supported carries the plane in the direction the wind is moving
and at.wind speed. In order to make good a desired course or
track over the ground a pilot or navigator must constantly make
allowances for wind effects. Only in the rare case when there is
no wind will the heading and the track of an airplane coincide,
and the speed made good with respect to the air equal the speed
made good over the ground. Moreover it must be stressed that
wind is not constant in direction or velocity; it varies with time,
distance and altitude. It is a variable factor that will always be
present in every problem of aerial navigation. Winds are de-
scribed by their direction and force. The direction of a wind is
always that from which the wind is blowing, the force is the
speed of the wind in miles per hour or knots.
In practice, wind heading course, or wind airspeed -
groundspeed problems are solved graphically by vector diagrams,
in which all the quantities involved are drawn to scale in the
proper direction. In the problems that follow E will always rep-
resent the point of departure with respect to the earth, W repre-
sents the wind and EW the wind vector. Lines through E except
60 BACKGROUND MATERIALS FOR WAR MATHEMATICS
EW will represent track or courses to be made good over the
earth; lines through W will represent headings, or direction of
the plane with respect to the air. P will indicate the position of
a plane at the end of one hour flying time, it will always be at
the intersection of track and heading then WP or wind-plane
will be the true airspeed vector, EP or earth-plane the ground
speed vector. The course and heading lines always bear in the
same general directions. True north should always be indicated
in these diagrams.
In the solution of these problems it is important to note that
the true air speed must be used and not the indicated air speed
given by an airspeed meter. Since the functioning of airspeed
meters is based on air pressure, its accuracy is limited by condi-
tions affecting atmosphere density primarily altitude and tem-
perature. Roughly the correction to be applied is approximately
2 per cent per 1000 ft of altitude, the true air speed being greater
than the indicated air speed. Also in drawing the vectors true
directions should be used, and then compass directions deter-
mined by applying compass corrections for variation and devia-
tion. In the solution of these problems it is sufficiently accurate
to read distances to the nearest mile. Since an airplane cannot
be steered with an accuracy greater than one degree, and this
only under exceptionally stable conditions, it is sufficient to read
all angles to the nearest degree. From distance speed relations,
time of flight should be computed to the nearest minute.
EXAMPLE: To determine the true heading and groundspeed,
knowing the course to make good, airspeed, and the wind direc-
tion and velocity.
SOLUTION:
From E lay off the true course line EC. Using any convenient
scale lay off EW to represent the wind. With W as center and
radius equal to the airspeed on the scale chosen strike an arc
meeting EC at P. Then the direction of WP is the heading to be
flown in order to make good the track EC. The length of EP to
scale represents the speed made good over the ground. That
this is the correct solution is obvious if one visualizes the forces
acting on the plane. If the plane left E and flew in the direction
EA parallel to WP, at the end of one hour it would be at point A,
a distance EA equals WP equals airspeed from E, if there were
NAVIGATION
Fig. 28. Vector Triangle of Velocities to Determine Heading and Groundspeed
no wind. But during the hour the wind has carried the plane a
distance EW equals AP in the direction of the wind. Therefore
if the wind is as shown at the end of one hour the plane will be
at P. Since EP is the ground distance covered in one hour it
represents the speed over the ground.
The angle EPW equals PEA is the wind drift correction and
must be applied to the true course to find the true heading. The
wind drift correction must always be applied into the wind. If
one visualizes the plane as headed outward from the center of a
clockwise compass rose it is clear that if the wind is from the left
the action of the wind is to drift the plane clockwise and thus
increase the compass reading, if from the right to drift the plane
to the left and decrease the compass reading. Hence to find the
heading from the course, if the wind is from the right of the
course add the wind drift angle, if from the left subtract. To find
the course from the heading reverse this rule.
To find the time of flight it is only necessary to divide the
distance to be flown by the ground speed. Adding this to the time
of departure will determine the estimated time of arrival. In
connection with distance-rate-time problems it is important
for any pilot to be able to change decimal parts of an hour into
minutes, and conversely. While this is a simple procedure, it
is exactly in these simple operations that prospective pilots are
found most deficient. To change tenth and hundredths of an
hour into minutes it is only necessary to note that 0.1 hr equals
6 min, 0.05 hr equals 3 min, or 0.01 hr equals 36 sec equals I min
62 BACKGROUND MATERIALS FOR WAR MATHEMATICS
plus 6 sec and any operation of this kind becomes a very simple
mental problem. Thus 0.43 hr is 4 x 6 plus 3/2 min plus 18 sec
equals 26 minutes to the nearest minute.
EXAMPLE: To determine the heading out, the heading back,
and ground speed out and back, knowing the course, airspeed
and wind conditions in a return flight.
SOLUTION:
Since the course from a point B to a point A differs from the
course A to B by 1800 it is unnecessary to draw separate dia-
grams. The problem can be solved using one course line. From E
lay off the course EC on the outward leg. Extend EC backward
from E to B. Then EC represents the outward, EB the return
course. From E lay off the wind as before on any convenient
scale. From W strike an arc with true airspeed as radius cutting
EC in P1, EB in P2. Then WP, is the heading out, WP2 is the
heading back, EP1 is groundspeed out, EP2 is groundspeed back.
P2 Awdleh hack W
Se
60c IC
E
Fig. 29. Velocity Triangle for the Out and In Problem
If the wind is as shown in the diagram it is from the right on
the outward course and the true heading will be the course NEC
plus the drift angle EPiW. On the return trip the wind will be
from the left (assuming of course that it does not change during
the course of the flight) and the true heading will be NEC plus
1800-angle EP2W. To find the compass heading to be flown, the
true heading is corrceted for the magnetic variation of the
locality, and the compass deviation of the headings flown.
NAVIGATION 63
The above problem is closely connected to the so-called radius
of action problems. By radius of action is meant the distance a
plane can fly on a given course, with a given amount of fuel
(which determines the flying time) and given wind conditions
and return to its starting point, before its fuel supply is ex-
hausted or at the end of a given time interval. The solution of
the problem involves the out and in headings and ground speeds,
the time to turn back and distance flown before turning back.
In this connection we first solve the old rowboat on the river
problem dressed up in modern language.
EXAMPLE: Suppose the speed of an airplane in still air is S,
the change in speed due to the wind along the line of flight is W,
the plane flies a distance R and returns to its starting point.
What is its average speed?
SOLUTION:
The wind will increase the speed in one direction and decrease
it in the other. Hence the plane will fly a distance R at a speed
of S + W and a distance R at speed S-W, using time equals
distance/rate we have the time of flight T,
R R 2RS
S- S + S S W2'
Since average speed is total distance flown divided by time taken,
the average speed V will be
2R S2 W2 W2
2RS/(S2 W2)- S S S
Hence the average speed in going first with and then against
the wind will always be less than the speed in still air by an
amount W 2/S, where W is the speed of the wind along the line
of flight. A constant wind is always a drawback in a return flight.
Since the effective component of the wind's velocity is the wind
force times the cosine of the angle between the wind direction
and the course flown it follows, as is otherwise obvious, that the
maximum effect of the wind on the plane's speed is obtained
when the wind is parallel to the flight path, the minimum effect
when it is perpendicular.
This simple problem is included here because of the com-
mon misconception that the effect of the wind would cancel out
on a return flight, that is, that the average speed would be the
64 BACKGROUND MATERIALS FOR WAR MATHEMATICS
numerical average of the out and in speeds. That this will never
be true is obvious as soon as it is pointed out that the time taken
to cover the two equal distances is different, and that average
speed means the average distance covered per unit of time.
EXAMPLE: To derive the radius of action formula.
SOLUTION:
Let R equal radius of action, T the total flying time (fuel hours),
T1 the time taken on the outward flight, T2 the time taken on the
return flight, S1 the ground speed out (EP1), S2 the ground
speed back (EP2). Then using time equals distance divided by
rate we have
R R
T = T + T2 + whence solving for R,
SSS2
R= T
S1 + S2
or, the radius of action of an aircraft is the number of fuel hours
times ground speed out times ground speed back divided by the
sum of ground speed out and back. Since R equals S1T1 (distance
equals rate times time) we have substituting for R and cancelling
the common factor Si.
S T S2
the time in hours after the departure to turn back. The clock
time of turning is obtained at once by changing T1 into hours and
minutes and adding to the time of departure.
A frequent problem in air navigation is that of planning a
flight so that in the event of adverse weather conditions making
it impossible to land at the desired destination, the plane can be
turned from its original course and arrive at an alternate airport
within an allotted flying time. The distance that can be flown
before changing course for the alternate airport and the head-
ings to be flown can be determined from a vector diagram by the
use of a line of relative motion. A line of relative motion is the line
of constant bearing or direction joining two objects moving in dif-
ferent but fixed directions at constant speeds. For example if a
plane leaves a ship and travels a course EP while the ship travels
a course ES, the line SP is the line of relative motion, the bearing
of iSP will remain constant as long as the plane and ship maintain
their courses and speeds. If EP represents the velocity of the
NAVIGATION 65
plane, ES the velocity of the ship, then SP will represent in direc-
tion and magnitude the velocity of the relative motion.
N
d
4-spe-
Fig. 30. Line of Relative Motion
Before solving the problem stated above we first solve the
simpler problem of finding the course and heading to be followed
in order to intercept a moving object.
EXAMPLE: An airplane is at E. Given the speed and course
of a ship at point B, airspeed of plane and the wind direction
and velocity to find the heading and course to be flown by the
plane in order to intercept the ship.
SOLUTION:
In order to intercept the ship along the shortest path the plane
must fly in such a direction that the line of relative motion join-
ing ship and plane maintains a constant bearing. If the plane
heads directly toward the moving ship its direction of motion
must constantly change and it will follow a curved path and take
longer to overtake the ship. Locate E and B, from B lay off the
ship's course and locate S to represent the ship's position at the
end of one hour. From S draw a line SA parallel to EB. Lay off
the wind EW. With W as center and airspeed as radius locate P
on SA. Then EP is the course and WP the heading to be flown
to intercept the ship. Clearly if the plane follows the course EP
it will intercept the ship at C. The time of interception may be
found by dividing EC by the ground speed or BC by the ship's
speed. In case the ship should change course or speed, it is
necessary to rework the problem from the positions at the time
of change. Whether one plane has speed enough to intercept
66 BACKGROUND MATERIALS FOR WAR MATHEMATICS
Fig. 31. Interception of a Moving Object
another can always be determined by the relative bearing. If
the pursuer cannot keep a constant relative bearing on the pur-
sued he cannot intercept him. An interceptor plane trying to
overtake another plane can quickly determine whether it is
possible to overtake it by observing the angular bearing between
the direction he is flying and the direction toward the other
plane. If this bearing increases he can turn toward the other
plane until the bearing remains constant and intercept the plane
by maintaining that heading. If the bearing decreases he cannot
overtake the other aircraft.
EXAMPLE: A plane leaves point E for a destination D, an
alternate airport is at A. Knowing the airspeed, wind force and
direction and the courses and distances from E to D and E to A,
how far along the course ED can the plane fly before turning
and arriving at A in a specified time (T) ? Find also the headings
to be flown.
SOLUTION: From E lay off the course lines ED and EA. Draw
in the wind vector EW. As before draw the airspeed vector from
W meeting ED in P1. EWP1 is the triangle of velocities for the
flight toward D.
w
PP
Fig. 32. Radius of Action in a Turn to an Alternate Airport
Next we assume that if it is necessary to turn back, we return
to the starting point E, but that meanwhile E is moving toward
NAVIGATION
A at a speed equal to the distance EA divided by the total hours
of flight. Locate L so that EL equals EA divided by time of
flight in hours. EL is the relative velocity of the point of de-
parture. Draw P1L and extend it beyond L. P1L is the line of
relative motion. With W as center and airspeed as radius locate
P2 on P1L extended. EWP2 is the triangle of velocities for the
flight to A. WP2 being the true heading and airspeed to A after
turning, EP2 being the course and ground speed to A. P1L is the
rate of departure, V1, P2L is the rate of return, V2. The time to
turn can now be found from the radius of action formula using
the rates of departure and return instead of the ground speeds.
T V2
T_
V1 + V2
The distance along ED to the point of turning can be found by
multiplying t by the ground speed EPI, or the point of turning on
ED can be located on the diagram by drawing the line AR from
A parallel to the ground speed EP2. The point R where this line
meets the course ED is the point to turn. Note that it is the
ground speed line, which always represents the course, that is
to be drawn from A, not the airspeed.
The above problem is identical with that of operations from
a moving base. If a plane left an aircraft carrier at E and scouted
on track ED with orders to return to the carrier in T hours, the
ship meanwhile travelling on course EA at a speed EL, the prob-
lem of finding the headings out and back to intercept the ship at
a specified time is identically that solved above. A typical prob-
lem of this kind is given.
EXAMPLE: At 0800 a plane, airspeed 100 knots, takes de-
parture from a carrier to scout on track 270' with orders to
return at 1200. The carrier, speed 20 knots, steams north for
one hour and then is to change course to 0450. The wind at the
flight altitude is from 3200, velocity 25 knots. Find the head-
ings out and back and the time of turning.
SOLUTION:
From E lay off the given courses and the wind vector. Locate
the point A where the ship will be at the end of 4 hours. Draw
EA and locate the point S one-fourth of the way from E to A.
EA represents the virtual motion, ES the virtual speed of the
ship during the time of flight. Locate P1 as before, with WP,
equal to airspeed of 100 knots.
68 BACKGROUND MATERIALS FOR WAR MATHEMATICS
AI0
"1200
Fig. 33. Radius of Action and Return to a Moving Base
Draw the line of relative motion P1S and locate P2 on this line
extended with WP2 equal to airspeed.
Then WP1, the true heading out, equals 2700 plus angle EPi W,
or 279.
WP2 the true heading back is 0600
SPI the rate of departure is 95 knots
SP2 the rate of return is 93 knots.
Time of flight before turning is
TVs 4 x 93
T = V. 4 x 93 1 hr 59 min
V1 + V2,- 93 + 95
The plane should turn back at 0959.
In connection with these problems it should be noted that
no allowance has been made for the time taken for the plane to
get into the air and gain the required altitude. Departures should
always be taken after the plane is in the air at the altitude at
which the flight is to be made. Also in computing flying time on
the basis of fuel supply, pilots are required in all civilian flights,
to figure flight times on a basis of approximately a 25 per cent
reserve supply of fuel. In case of turning to an alternate airport
the time of turning should be figured on a basis of allowing a
minimum of 45 minutes fuel supply remaining on arrival.
In the above examples it has been assumed that the wind
was constant. Since the wind is rarely constant it will frequently
NAVIGATION
be necessary to determine wind direction and force while in
flight. It is frequently possible to estimate the wind by observa-
tions of water surfaces, smoke, etc., but at best these observa-
tions are doubtful, particularly at high altitudes. From the tri-
angle of velocities we can find the wind if we know the course-
groundspeed, and heading-airspeed legs. The true heading can
always be found by correcting the compass heading for variation
and deviation and the true airspeed by correcting the airspeed
meter. When known landmarks are visible, the true course can
be determined from charts and the groundspeed calculated by
measuring the time taken to fly over known landmarks.
EXAMPLE: 15 minutes after leaving town A on a true head-
ing of 0850 and true airspeed of 110 miles per hour a pilot passes
over town B, 32 miles away and bearing 0700 from A. Find the
wind force and velocity.
SOLUTION:
Since it took 15 min to travel 32 mi the groundspeed is 128 mph.
Knowing the true heading 085 and airspeed 110 mph plot WP.
Since the true course is 0700 and groundspeed is 128 mph plot
EP. Then EW represents the wind. The wind is from 189,
force 35 mph.
05
E
Fig. 34. The Determination of Wind from Known Course and Heading
Wind direction and force can also be determined by flying
two or more different headings at a constant air speed and ob-
serving the drift angles and then constructing velocity triangles
closing in the wind line.
EXAMPLE: Flying a true heading of 0400 a pilot observes a
drift angle of 200 to the right. Heading 2950 a drift angle of 100
70 BACKGROUND MATERIALS FOR WAR MATHEMATICS
to the left is observed. If the true air speed is 120 mph find the
wind direction and force.
SOLUTION:
From W lay off the true headings WP, at 0400 and WP2 at 295,
with WPI and WP2 equal to airspeed to scale. At P1 lay off an
angle of 20 to the right of P1W, at P2 an angle 100 to the left
of P2W. These two lines will intersect at E. EW represents
the wind.
Fig. 35. Double Drift Method of Determining Wind Force and Direction
The wind is from 3150, force 42 mph.
NAVIGATION 71
TIME AND CELESTIAL NAVIGATION
Geographic position can be fixed by observations of the
position of any of the heavenly bodies. As seen from the earth
celestial objects appear to be located on the surface of a sphere,
the celestial sphere, with the earth at the center. Position on this
sphere is determined by the intersection of certain circles just
as position on the earth is fixed by latitude and longitude. If we
project the earth's axis outward it meets the celestial sphere in
the celestial poles. The projection of the earth's equator onto
this sphere is called the celestial equator or the equinoctial. The
point directly above any observer on the earth is called the zenith.
The great circle 900 from the zenith is the celestial horizon. The
great circles through the celestial poles, corresponding to the
earth's meridians are called hour circles. The small circles paral-
lel to the equinoctial, corresponding to parallels of latitude, are
called declination circles. Great circles perpendicular to the hori-
zon and passing through the zenith are called vertical circles. An
observer's celestial meridian is the hour circle which is the pro-
jection of his geographic meridian. Since this circle passes
through the zenith it is also a vertical circle.
The position of a heavenly body is determined by its hour
angle and declination, or by its altitude and azimuth.
The declination of a body is its angular distance from the
equator measured along the hour circle of the body, in degrees
and named north or south according as the body is north or south
of the equator.
The hour angle of a body is the angular distance measured
along the equator westward from the upper meridian to the foot
of the hour circle through the object; or it is the angle at the
pole between the meridian and the hour circle of the body. Hour
angle is measured from 0 to 3600 or from 0 to 24 hours.
The altitude of a body is the angular distance of the body
above the horizon, measured along the vertical circle through
the body.
The azimuth of a body is the arc of the horizon intercepted
between the north point of the horizon and the foot of the vertical
circle through the body. It is measured and marked east or west
according as the object is east or west of the meridian.
72 BACKGROUND MATERIALS FOR WAR MATHEMATICS
These definitions are illustrated in figure 36, in which P is
the elevated pole, Z the zenith, Q the intersection of the upper
Fig. 36. The Celestial Sphere and the Astronomical Triangle
meridian and the equator. The pole P, the zenith Z and the posi-
tion of the sun, star or planet such as S forms an astronomical
triangle.
It is an elementary and well known exercise in geometry
to show that the latitude of an observer equals the altitude of
the celestial pole or the declination of the zenith. Hence the arc
PZ in the astronomical triangle is 900 latitude; the arc PS
is 900 declination; the angle ZPS = t is the meridian angle. If
S is west of the meridian t = hour angle, if S is east of the merid-
ian t = 360' the hour angle. The declination of celestial ob-
jects is tabulated in nautical almanacs. If t, d, and latitude are
knowns, three parts of the triangle are known and the remain-
ing parts can be solved for by any of the well known methods
of solving spherical triangles. In particular the navigator desires
NAVIGATION 73
the sides ZS which is the complement of the altitude and the
angle PZS which is the azimuth or bearing of the object.
The point directly below a star is known as its geographic
position. As one moves away from the geographical position the
altitude decreases. From any point on a circle with center at the
geographic position the observed altitude of a star would be
the same. It is an elementary geometric exercise to prove that
the angular distance of an observer from the geographic posi-
tion of a star is the complement of the observed altitude.
EXAMPLE: Prove that the altitude of a celestial object at
any instant is the complement of an observer's distance from
the geographic position measured in arc.
SOLUTION:
If the observer is directly at the geographic position, the object
is in his zenith and the altitude is 900. If the observer was 1
nautical mile away his zenith would be displaced 1' of arc and
the zenith distance of the object would be 1' of arc. Hence the
altitude, which is the complement of the zenith distance would
be 89 59'. The relation would obviously hold for any other dis-
tance expressed in nautical miles. Hence if we can measure the
altitude of a star and find the point on the earth directly beneath
the star we know we are located on a circle. We have a celestial
line of position.
In practice an assumed latitude and longitude is taken, which
may be the actual dead reckoning position or some convenient
point near the D.R. position. The altitude of the body is observed
with a sextant or octant. This altitude is corrected for instru-
mental errors, refraction of light, height of the observer above
the earth's surface (dip) etc., to obtain the observed altitude
Ho. From the time of observation and the assumed longitude the
meridian angle t is computed. The triangle is then solved for
the computed altitude He (complement of SZ) and the azimuth.
If the observer were at the assumed position H, = Ho.
EXAMPLE: To prove that the distance from the assumed po-
sition to the circle of position is equal to the difference of the
computed and observed altitudes.
SOLUTION:
Draw a circle to represent the earth cut by the plane of the
74 BACKGROUND MATERIALS FOR WAR MATHEMATICS
azimuth of the body. Let A be the
assumed position, B the actual po-
sition of the observer. The hori- M-A -
zon at A is MN, at B it is PQ.
Then OA is perpendicular to MN
at A, OB is perpendicular to PQ
at B. Let SA and S'B be rays of
light from the body. Since ce-
lestial bodies may be regarded as
at an infinite distance these lines
are parallel. Then angle SAN is Fig. 37. Geometric Proof that
He, angle S'BQ is Ho. Let a be the a H H
angle NLQ where L is the intersection of MN and PQ. Let G be
the intersection of PQ and SA.
Since SA is parallel to S'B, angle SGB = Ho.
But SGB =He+a,
Therefore He + a =Ho, or a = Ho He.
Since OAL = OBL = 90,
angle C + angle ALB = 180.
But a + angle ALB = 180
Therefore C = a = H, Ho.
The azimuth will give the direction in which the geographic
position of the star will lie. If Ho is greater than the computed
altitude He then we must be toward, or closer to the star than
the assumed position by a distance in nautical miles equal to
Ho -He in minutes of arc. If He is greater than Ho then we must
be farther away from the geographic position by a distance
equal to He Ho in minutes of arc. From the assumed position
the computed azimuth is laid off on the chart. The distance
Ho He or He Ho is laid off on this line either toward or away
from the assumed position as the case may be. From the point
thus determined a perpendicular is constructed to the azimuth
line. This perpendicular represents a portion of the celestial circle
of position. Since the radius of these circles is usually quite large
-the radius being the complement of the altitude-no serious
error is introduced by replacing a portion of the arc by a straight
line. At least two lines of position are necessary to determine a
fix. If it is only possible to obtain one line of position it is as-
sumed that the ship or plane is on the line of position at the foot
of the perpendicular dropped from the D.R. position. Two celes-
tial circles of position of course intersect in 2 points but these
NAVIGATION 75
are usually widely separated and only one will be anywhere near
the D.R. or assumed position.
C
c.
Fig. 38. Celestial Lines of Position Away and Toward
In practical navigation the astronomical triangle is solved
by the use of navigational tables. These are tables of preconiputed
altitudes and azimuths tabulated so that the tables can be en-
tered with known values of latitude, declination and meridian
angle t as arguments and the computed altitude He and azimuth
can be taken out directly or by interpolation. The average navi-
gator will rarely have occasion or time to use a formula from
spherical trigonometry.
The most difficult part of the practical computation is in-
volved in the finding of the hour angle or meridian angle t of the
object observed. This involves a thorough knowledge of the vari-
ous kinds of time and the relation between time and longitude,
and time and hour angle.
The position of any celestial object changes constantly in
relation to any fixed point on the earth's surface, due to the rota-
tion of the earth and the body's own motion in the celestial
sphere. Time is measured by the rotation of the earth. The east-
ward rotation of the earth causes all heavenly bodies to appar-
ently rotate westward along their declination circles and their
hour angle increases constantly. Since the sun changes in declina-
tion during the course of a year from 230 south to 23o north,
and also during the year the distance of the earth from the sun
varies, the apparent motion of the sun varies; the actual length
of the solar day is not constant. The rate of change of the hour
76 BACKGROUND MATERIALS FOR WAR MATHEMATICS
angle of the sun is not uniform, consequently time for general
use is measured by the rotation of a fictitious body called the
mean sun. Mean time is simply averaging sun time. A mean
solar day is the average length of time it takes the sun to make
one revolution about the earth.
Time is essentially hour angle. Since a local hour angle is
measured from the instant the object crosses the upper or noon
meridian, and the local civil day begins at midnight, when the
hour angle of the mean sun is 12h, it follows that civil time =
hour angle of the mean sun + 12h. Of course if the sum is
greater than 24h, the 24h is dropped.
The relation of time, longitude and hour angle is most easily
seen by means of a time dia- EST
gram. Draw a circle to repre- ,
sent the equator. The center of
the circle is then the pole, di-
ameters represent meridians or G
hour circles. Let Gg be the
Greenwich meridian, Mim, the 5
meridian of an observer, M2 m2,
the meridian of a second place,
S the position of the mean sun.
Capital letters indicate the up-
per meridian, small letters the
lower. If we let T1 and T2 rep-
resent the local times, t, and t2 m,
the hour angles of the sun it is Fig. 39. Time Diagram Illustrating
the Relation of Time, Hour Angle
apparent from figure 39 that and Longitude
T1 T2 = tl t2 = arc M1M2,
or local civil times of any two places differ as the hour angles
of the mean sun as measured from those places and this differ-
ence is equal to the difference of longitude of those places. This
relation will hold for any other kind of time as well as mean time.
Since the sun travels from east to west in its rotation the
local times of any two meridians are different. If it is noon for
any observer, the sun is past the meridian of any place east of
the observer and hence the local time at such a place is later;
for any place west the sun has not yet reached the meridian and
the local time is earlier. To find the time at any other place whose
NAVIGATION 77
longitude is known from a local time it is only necessary to apply
the difference in longitude converted into time units, adding if
the place is east, subtracting if west.
In order to avoid the inconvenience of keeping local times,
the earth is divided into time zones or belts extending approxi-
mately 150 = 1h in longitude. In each of these zones the local
civil time of a standard meridian is kept. The standard meridians
used are the multiples of 150 of longitude. The time kept in these
zones is known as standard or zone time. For convenience in
navigational work the zones are numbered, 0 for the zone 71
each side of Greenwich, thence zones west are numbered con-
secutively from 1 to 12, zones east, 1 to 12. The zones 12
are each only 70 wide and are on either side of the international
date line. To find Greenwich time from zone time it is only neces-
sary to add the zone description. Eastern standard time is the
local civil time of the 750 W meridian. It is zone + 5. Eastern
war time or daylight saving time is that of zone + 4. The dif-
ference between zone time and local time is equal to the differ-
ence in longitude between the local meridian and the central
meridian of the zone, expressed as time.
While zone time is used for convenience it does not lend
itself readily to direct use in navigational problems, since the
navigator is concerned with local hour angles and hence local
times. In particular when the sun is the object observed, the
navigator observes the true sun and not the fictitious mean sun.
The hour angle of the true sun is determined from apparent time,
which is measured by the apparent motion of the true sun. This
is "sun-dial" time. The difference between apparent and mean
times is called the equation of time. The equation of time varies
between + 16 and 16 minutes during a year. Its value is tabu-
lated in nautical almanacs for every hour of Greenwich civil
time (G.C.T.).
The hour angle of the navigational bodies is also tabulated
in nautical almanacs for Greenwich civil time.
To determine the local hour angle of any body a navigator
must first determine the Greenwich civil time and date for the
instant of observation in order to use his tables. The actual steps
involved in the determination of local hour angle of a body at
the instant of observation will be as follows:
78 BACKGROUND MATERIALS FOR WAR MATHEMATICS
To the watch time of observation apply any watch error on
zone time to obtain the zone time of observation. If a chronome-
ter is available keeping G.C.T., the difference between the chro-
nometer and watch readings may be noted and hence the
chronometer time of observation can be found. G.C.T. will be
chronometer time corrected for any error. With the G.C.T. as
argument, the values of the Greenwich hour angle (GHA) and
the declination of the body observed are found in the tables.
From the GHA and the assumed longitude the local hour angle
is found.
A ship's chronometer is simply a well regulated watch or
clock set to keep Greenwich civil time. There is one difficulty
that must constantly be kept in mind in finding G.C.T. The
chronometer, like all other clocks and watches, reads only to 12
hours and does not indicate whether the time is PM or AM, nor
does it indicate the date at Greenwich. This must be determined
from the longitude. This is most easily done by use of a time
diagram.
EXAMPLE: The Greenwich hour angle of the sun is 62. What
is the local hour angle and meridian angle at a place in longitude
160 W? in longitude 160 E?
SOLUTION:
For a place in west longitude we must have
LHA = GHA Long. = 62 160 + 360' = 262
360 is added to the difference to avoid negative angles, since
hour angle is positive by definition. The hour angle is greater
than 180 or 12" hence
t = 3600 2620 = 98 E.
For east longitude
LHA = GHA + Long. = 62' + 1600 = 2220
t = 360 222 = 1380 E.
There is only one instant of the day when the date is the
same over the whole earth and that is noon, Greenwich Civil
Time. At any other instant not only the time but the date may
be different at two places separated in longitude. The date is
changed at midnight when the mean sun crosses the lower stand-
ard meridian of the time zone. Since the relative motion of the
NAVIGATION
sun is westward the date is changed progressively to the west,
starting at the 180th meridian or international date line. At any
place east of an observer in longitude the time will be later and
the date will be the same or one day greater; for places west the
time is earlier and the date the same or one day earlier. The
easiest way to determine the date is by use of a time diagram.
EXAMPLE: Determine the GCT and date at Greenwich when
the observer's time, date and longitude are given as
(a) 4" June 16, 1942 long. = 900 W
(b) 4h June 16, 1942 long. = 900 E
(c) 21h June 16, 1942 long. = 900 W
(d) 21h June 16, 1942 long. = 900 E
SOLUTION:
(a) (b) (c) U)
M M M AM
m Ifl P fl
Fig. 40. Time Diagram to Determine Greenwich Civil Time and Date
The time diagrams are shown in figure 40. Westward is
counterclockwise, eastward clockwise. Once the local meridian
Mm and the Greenwich meridian Gg are placed by means of the
known longitude and the sun S placed from the given time, a
glance at the diagram will show whether it is AM or PM at
Greenwich and whether the date is the same at M and G. If the
sun is east of G, it is AM, if west of G, it is PM at Greenwich.
If the sun lies on the arc mg the date at Greenwich is one greater
or one less than at M depending on whether M is in west or east
longitude. If the sun is on arc gMGm the date at M and G is the
same. For the cases given above we have, since 90 difference in
longitude is 6 hours
(a) GCT = 10h June 16, 1942
(b) GCT = 22h June 15, 1942
(c) GCT = 3h June 17, 1942
(d) GCT = 15" June 16, 1942.
EXAMPLE: A navigator in longitude 1250 W observes a star
on August 15, 1942. The time is 8h 30m 40s PM by watch, which
80 BACKGROUND MATERIALS FOR WAR MATHEMATICS
is keeping zone time. The watch error is 12 sec (fast). Determine
the GCT and date.
SOLUTION:
Longitude 1250 W is in the time zone of
Hence the zone description is + 8.
Watch time = 20h 30m 40s
Watch error (F) = 12s
Zone time 20" 30m 288
Zone description = +8
G CT = 28h 30m 28
= 4 30m 28s
the 120th meridian W.
August 15, 1942
August 16, 1942.
The above discussion of time is intended to present only the
simplest elements. No attention has been paid to sidereal or
star time. Most modern navigational tables can be entered with
GCT, and GHA and declination for the instant in question ob-
tained directly from the tables, so that sidereal time is no longer
of primary importance to the navigator.
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