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MATHEMATICS FOR ENGINEERS
PART I
The DirectlyUseful Technical Series
See Detailed Prospectus
Mathematics for Engineers
Part II
By W. N. ROSE, B.Sc. Eng. (Lond.)
440 pages. Demy 8vo. 13/6 net.
E two volumes of Mathematics for Engineers form
a most comprehensive and practical treatise on the
subject. Great care has been taken to show the direct
bearing of all principles to engineering practice, and the
complete book will prove a valuable reference work
embracing all the mathematics needed by engineers in
their practice, and by students in all branches of engineering
science.
The second part is on similar lines to the present, and
contains exhaustive chapters on the following
INTRODUCTION TO DIFFERENTIATION
DIFFERENTIATION OF FUNCTIONS
APPLICATIONS OF DIFFERENTIATION
METHODS OF INTEGRATION
MEAN VALUES, R.M.S. VALUES, CENTROIDS,
MOMENT OF INERTIA, ETC.
POLAR COORDINATES
DIFFERENTIAL EQUATIONS
APPLICATIONS OF CALCULUS
HARMONIC ANALYSIS
SPHERICAL TRIGONOMETRY
MATHEMATICAL PROBABILITY, ETC.
Particulars of other books in this Series are given on page 515.
The DirectlyUseful D.U. Technical Series
Founded by the late WILFRID J. LINEHAM, B.Sc., M.Inst.C.E.
Mathematics for Engineers
PART I
INCLUDING
ELEMENTARY AND HIGHER ALGEBRA,
MENSURATION AND GRAPHS, AND
PLANE TRIGONOMETRY
BY
W. N. ROSE
B.Sc. ENG. (LOND.)
Late Lecturer in Engineering Mathematics at the University of
London, Goldsmiths' College ; Teacher of Mathematics
at the Borough Polytechnic Institute
THIRD EDITION
 (p
LONDON
CHAPMAN & HALL, LTD.
II HENRIETTA STREET, W.C* 2
1922
FEINTED IN GRF.AT BRITAIN BY
RICHARD CLAY & SONS, LIMITED,
BUNGAY, SUFFOLK.
EDITORIAL NOTE"
THE DIRECTLY USEFUL TECHNICAL SERIES requires a few words
by way of introduction. Technical books of the past have arranged
themselves largely under two sections : the Theoretical and the
Practical. Theoretical books have been written more for the train
ing of college students than for the supply of information to men
in practice, and have been greatly filled with problems of an academic
character. Practical books have often sought the other extreme,
omitting the scientific basis upon which all good practice is built,
whether discernible or not. The present series is intended to
occupy a midway position. The information, the problems, and
the exercises are to be of a directly useful character, but must at
the same time be wedded to that proper amount of scientific ex
planation which alone will satisfy the inquiring mind. We shall
thus appeal to all technical people throughout the land, either
students or those in actual practice.
. THE EDITOR.
AUTHOR'S PREFACE
AN endeavour has here been made to produce a treatise so
thorough and complete that it shall embrace all the mathematical
work needed by engineers in their practice, and by students in all
branches of engineering science. It is also hoped that it will
prove of special value for private study, and as a work of
reference.
Owing to the vast amount of ground to be covered, it has been
found impossible to include everything in one volume : and accord
ingly the subjectmatter has been divided into two portions, with
the first of which the present volume deals. Stated briefly, Part I
treats fully of the fundamental rules and processes of Algebra,
Plane Trigonometry, Mensuration, and Graphs, the work being
carefully graded from an elementary to an advanced stage ; while
Part II is devoted to the Calculus and its applications, Harmonic
Analysis, Spherical Trigonometry, etc.
It is felt that the majority of books on Practical Mathematics,
in the endeavour to depart from a theoretical treatment of the
subject, neglect many essential algebraic operations, and, in addi
tion, limit the usefulness of the rules given by the omission of the
proofs thereof. Throughout the book great attention has been
paid to the systematic development of the subject, and, wherever
possible, proofs of rules are given. Practical applications are
added in the greater number of instances, the majority of the
exercises, both worked and set, having a direct bearing on engineer
ing practice, thus fulfilling the main purpose of the book : and
strictly academic examples are only introduced to emphasise
mathematical processes needful in the development of the higher
stages.
In order to make the work of the greatest use to the engineer
as a means of reference, many practical features have been intro
viii AUTHOR'S PREFACE
duced, including : Calculation of Weights, Calculation of Earthwork
Volumes, Land Surveying problems, and the Construction of PV
and T diagrams or other general Practical Charts; and great
care has been exercised in order that the best possible use may be
made of mechanical calculators, such as the slide rule and the
planimeter.
Chap. I deals with methods of calculation. The method of
approximating for a numerical result, introduced by Mr. W. J.
Lineham, has been found to be very effective and easily grasped;
and it is felt that the device here described for investigation for
units could be more universally employed, because of its simplicity
and directness.
Simple, simultaneous, quadratic, cubic and all equations solvable
by simple algebraic processes are treated in Chap. II : also fac
torisation by the simple methods, including the use of the Factor
theorem, and the simplification of algebraic fractions. Great stress
is laid on the importance of facility in transposing both terms and
factors from one side of an equation to the other; and in this
respect numerous literal equations are considered.
The various rules of the mensuration of the simple areas and
solids are stated in Chap. Ill ; the conic sections being included in
view of their importance in connection with the theory of structures
and strength of materials. The chapter concludes with the appli
cation of the rules for the calculation of weights ; a variety of types
of machine parts being treated.
All the elementary graph work is included in Chap. IV, in
which attention is specially directed to the derivation of one curve
from another, necessary, for example, in the case of efficiency
curves.
The usefulness of graphical solutions for the problems on arith
metical and geometrical progression is emphasised in Chap. V, in
which also methods of allowing for depreciation of plant are intro
duced as illustrations of the commercial use of series. Here also
are numerous examples on evaluation of formulae containing frac
tional and negative powers; and in these examples the absolute
necessity of analysing compound expressions into their elements is
made clear.
In Chap. VI a departure is made from the old convention of
the measurement of angles from a horizontal line, calling them
positive if measured in a counterclockwise direction. Plane
Trigonometry has its widest application in land surveying, in
which angles are measured by a righthanded rotation from the
AUTHOR'S PREFACE ix
North direction. Hence the north and south line is here taken as
the standard line of reference and all angles are referred to it.
Also by doing this the mathematical work is simplified, since the
trigonometrical ratio of an angle does not alter; the angle of any
magnitude being converted to the " equivalent acute angle," viz.
the acute angle made with the north and south line. The calcula
tion of coordinates in land surveying is introduced as a good
instance of the solution of rightangled triangles. Many rules for
the solution of triangles are stated, but two only, viz. the " sine "
and the " cosine " rules, are recommended for general use. This
chapter contains much of importance to the electrical engineer, in
the way of hyperbolic functions and complex quantities.
The mode of utilising the planimeter for all possible cases is
shown in Chap. VII, including the case in which the area to be
measured is larger than the zero circle area. Graphic integration
is introduced, in addition to the rules usually given for the measure
ment of irregular curved areas.
Chap. VIII should prove of great value to railway engineers
and to surveyors, since in it are collected the various types of
earthwork problems likely to be encountered.
Chap. IX deals with the plotting of difficult curve equations,
and in this chapter it is demonstrated how a curve representing a
rather complex equation may be obtained from a simple curve by
suitable change of scales. Thus all sine curves have the same
form, and accordingly the curve representing the equation
y =* 72 sin(ioo7r/  io6) can be obtained directly from the simple
sine curve y sin x. The work on the construction and use of PV
and T< diagrams should commend itself to students of thermo
dynamics.
In Chap. X emphasis is laid on the advantage of making suit
able substitutions when transforming any equation into the linear
form for the purpose of determining the law correlating two
variables.
Chap. XI provides a novel feature in its presentation of some
methods used in the construction of charts applicable to drawing
office practice. Alignment charts are here explained in the fullest
detail, and it is hoped that the explanation given will further the
more universal employment of these charts.
Chap. XII embraces the more difficult algebra, necessary chiefly
in the study of the Calculus; and in addition, the application of
continued fractions to dividinghead problems.
For extremely valuable advice, helpful criticism and assistance
x AUTHOR'S PREFACE
at all stages of the progress of the book the Author desires to tender
his sincere thanks to MESSRS. W. J. LINEHAM, B.Sc., M.I.C.E.,
J. L. BALE, C. B. CLAPHAM, B.Sc., and G. T. WHITE, B.Sc.
While it is hoped that the book is free from errors, it is possible
that some may have been overlooked; and notification of such
will be esteemed a great favour.
W. N. ROSE.
Goldsmiths' College,
New Cross, S.E.,
January, 19 iS.
NOTE TO SECOND EDITION
THE early demand for a second edition of this volume has
afforded an opportunity for making a number of corrections both
in the text and in the illustrations; whilst a few exercises have
been added.
To those who have contributed to the improvement thus made,
whether by notifying errors or by offering valuable suggestions,
the Author's sincere thanks are proffered.
November,
NOTE TO THIRD EDITION
THE favourable reception accorded the first and second editions
inspires the hope of similar appreciation of the third edition.
In this edition the need for the inclusion of some explanation
of the determinant mode of expression employed in treatises on
aerodynamics has been recognised by the addition of a section
dealing with determinants.
The work has been subjected to thorough revision, corrections
being made where necessary, and further exercises have been
added.
December, 1921.
CONTENTS
PACE
INTRODUCTORY .... i
Previous knowledge Definitions and abbreviations Tables of
weights and measures Useful constants.
CHAPTER I
AIDS TO CALCULATION . 6
Methods of approximation Indices Logarithms: i. Reading from
tables. 2. Determination of characteristic Antilogarithms Ap
plications of logarithms Investigation for units.
CHAPTER II
EQUATIONS 31
Solution of simple equations Solution of simultaneous equations :
i. With two unknowns. 2. With three unknowns Methods of
factorisation The remainder and factor theorems Simplification
of algebraic fractions Solution of quadratic equations : i. By
factorisation. 2. By completion of the square. 3. By use of a
formula Solution of cubic equations Solution of simultaneous
quadratic equations Solution of surd equations.
CHAPTER III
MENSURATION 79
Area of rectangle and triangle Area of parallelogram and rhombus
Areas of irregular quadrilaterals and irregular polygons Areas of
regular polygons Circle: i. Circumference and area. 2. Area of
annulus. 3. Length of chord and maximum height of arc. 4. Length
of arc by true and by approximate rules. 5. Area of sector. 6. Area
of segment by true and by approximate rules Area and perimeter
of the ellipse The parabola The hyperbola Surface area and
volume of prism and cylinder Surface area and volume of pyramid
and cone Frusta of cones and pyramids The sphere : i. Surface
area. 2. Volume. 3. Volume and surface area of a zone. 4.
Volume of a segment. Relations between sides, surface areas, and
volumes of similar figures Guldinus' rules for surface areas and
volumes of solids of revolution Positions of centroids of simple
figures Calculation of weights Tables of areas and circumferences
of plane figures Tables of volumes and surface areas of solids.
xii CONTENTS
CHAPTER IV
PAGE
INTRODUCTION TO GRAPHS 148
Object and use of graphs Rules for plotting graphs Interpolation
The plotting of coordinates Representation of simple equations
by straightline graphs Determination of the equation of a straight
line Plotting of graphs to represent equations of the second
degree Solution of quadratic equations : i. By use of a graph.
2. On the drawing board Plotting of graphs to represent equations
of degree higher than the second Graphs applied to the solution of
maximum and minimum problems.
CHAPTER V
FURTHER ALGEBRA 193
Variation Arithmetic and geometric progression, treated both
algebraically and graphically Practical applications of geometric
progression The laws of series applied to the calculation of allow
ance for depreciation of plant The value of " e " Napierian
logarithms : i. Reading from tables. 2. Calculation from common
logarithms Use of logarithms in the evaluation of formulae con
taining fractional and negative powers Logarithmic equations.
CHAPTER VI
PLANE TRIGONOMETRY 232
Definitions of the trigonometric ratios Reading the values of the
trigonometric ratios from the tables Solution of rightangled
triangles Reading the values of the trigonometric ratios from the
slide rule Calculation of coordinates in land surveying Meaning
of the terms " reduced bearing " and " wholecircle " bearing in
surveying Rules for the determination of the trigonometric ratios
of angles of any magnitude Table of signs of the trigonometric
ratios Rules for the solution of triangles, for any given conditions
The " ambiguous " case in the solution of triangles Proof of the
" s " rule for the area of a triangle Use of tables of the logarithms
of the trigonometric ratios The expansion of sin (A + B),
A
sin (A B), etc. Ratios of 2 A, , 3 A and 4 A in terms of the
ratios of A Rules for the change of a sum or difference of two
trigonometric ratios to a product, and vice versa Solution of trigo
nometric equations Hyperbolic functions Complex quantities
Rule for addition of vector quantities Inverse trigonometrig
functions.
CONTENTS xiii
CHAPTER VII
PACE
AREAS OF IRREGULAR CURVED FIGURES 300
Areas of irregular curved figures by various methods : I. By the
use of the Amsler planimeter and the Coffin averager and plani
meter : the use of the Amsler planimeter for large areas being
fully explained. 2. By averaging boundaries. 3. By counting
squares. 4. By the use of the computing scale. 5. By the trape
zoidal rule. 6. By the midordinate rule. 7. By Simpson's rule.
8. By graphic integration.
CHAPTER VIII
CALCULATION OF EARTHWORK VOLUMES 319
Volumes of prismoidal solids Volume of a wedgeshaped excava
tion Area of section of a cutting or embankment Volume of a
cutting having symmetrical sides Volume of a cutting having
unequal sides Net volume of earth removed in making a road by
both cutting and embankment Volume of a cutting with unequal
sides, in varying ground Surface areas for cuttings and embank
ments Volumes of reservoirs.
CHAPTER IX
THE PLOTTING OF DIFFICULT CURVE EQUATIONS . . . 336
Curves representing equations of the type y = ax" Use of the
loglog scale on the slide rule Expansion curves for gases Special
construction for drawing curves of the type pv n = C Equations
to the ellipse, parabola and hyperbola The ellipse of stress
Curves representing exponential functions The catenary Graphs
of sine functions Use of the sine curve " template " Simple har
monic motion Graph of tan x Compound periodic oscillations
Equation of time Curve of logarithmic decrement Graphic solu
tion of equations insolvable or not easily solvable by other methods
Construction of PV and r<j> diagrams : i. Drawing PV and TQ
diagrams. 2. Drawing primary adiabatics and constantvolume
lines. 3. Drawing secondary adiabatics. 4. Plotting the Rankine
cycle for two drynesses. 5. Plotting the common steamengine
diagram for an engine jacketed and nonjacketed. 6. Plotting
quality curves. 7. Calculating exponents for adiabatic expansions.
8. Plotting constant heat lines PV and T$' diagrams for the
Stirling, Joule and Ericsson engines.
CHAPTER X
THE DETERMINATION OF LAWS 396
b
Laws of the type : i. y = a +; y = a + bx*, etc. 2. y = ax" ;
the usefulness of the slide rule for log plotting being demonstrated.
3. y ^ ae bx . 4. y = a+bx+cx*. 5. y = #+b.x n ; y
y == {? + fo nx ; y = ws m ,
xiv CONTENTS
CHAPTER XI
PAGE
THE CONSTRUCTION OF PRACTICAL CHARTS . . . .419
Correlation charts, including log plotting Ordinary intercept
charts of various types Alignment charts, their principle and
use Alignment charts involving powers of the variable Alignment
chart for four variables.
CHAPTER XII
VARIOUS ALGEBRAIC PROCESSES, MOSTLY INTRODUCTORY TO
PART II 448
Continued fractions Application of continued fractions to dividing
head problems Resolution of a fraction into two or more partial
fractions Determination of limiting values of expressions Per
mutations and combinations The binomial theorem : i. Rule for
the expansion of a binomial expression. 2. Rule for the calculation
of any particular term in the expansion Use of the binomial
theorem for approximations The exponential and logarithmic
series Calculation of natural logs  Determinants.
ANSWERS TO EXERCISES 479
TABLES :
Trigonometric ratios 491
Logarithms 492
Antilogarithms 494
Napierian logarithms . . . . 496
Natural sines 498
Natural cosines 500
Natural tangents 502
Logarithmic sines 504
Logarithmic cosines 506
Logarithmic tangents 508
Exponential and hyperbolic functions . . . . . . .510
INDEX 511
MATHEMATICS FOR ENGINEERS
INTRODUCTORY
Previous Knowledge. While this work is intended to supply
all the mathematical rules and processes used by the engineer,
certain elementary branches of the subject have necessarily been
omitted. It is assumed that the reader has a sound working
knowledge of arithmetic, and also is acquainted with the four
simple rules of algebra, viz. addition, subtraction, multiplication
and division. Thus the meaning of the following algebraic processes
should be known
x 45
a? = axaxa; (* 2 ) 9 = # 18 ; = = x*;
x s
O% _ \)
i2c) = 30*2 426+72^; = '5# i25y;
4
(4076) (0
Again, the use of the loinch slide rule is not explained in detail
as regards multiplying, dividing, involution and evolution ; but the
special application of the slide rule is dealt with as occasion arises.
Definitions and Abbreviations. An expression is any
mathematical statement containing numbers, letters and signs.
Terms of an expression are connected one with another by
+ or signs.
The factors of an expression are those quantities, numerical
or literal, which when multiplied together give the expression.
Thus considering the expression
, 2ga?xb 2 and io8ay 6 are terms; and each of these terms can
be broken up into a number of factors; e. g.
= i$xaxaxb.
Again (ga 46) (5^+76) = 45a 2 +43& 286 2 ; and (9^46) and
+7^) are the/ac/ofs of 45
B
2 MATHEMATICS FOR ENGINEERS
When an expression depends for its value on that given to one
of the quantities occurring in it, the expression is said to be a
function of that quantity. Thus gx 3 7# 2 +5 is a function of x;
and this relation would be written in the shorter form
9 * 3 7*2+5 =/(*).
If a letter or number is raised to a power, the figure which
denotes the magnitude of that power is called the exponent.
An obtuse angle is one which is greater than a right angle.
An acute angle is one which is less than a right angle.
A scalene triangle has three unequal sides.
The locus of a point is the path traced by the point when its
position is ordered according to some law.
The abbreviations detailed below will be adopted throughout.
= stands for " equals " or " is equal to."
+ ,, " plus."
,, ,, " minus."
x ,, " multiplied by."
T ,, " divided by."
.*. ,, ?, " therefore."
,, ,, " plus or minus."
> ,, ,, " greater than."
< ,, " less than."
,, ,, " circle."
Qce ,, ,, " circumference."
oc ,, ,, " varies as."
co ,, ,, " infinity."
/_ ,, ,, " angle."
A ,, ,, " triangle " or " area of triangle."
li or 4! ,, ,, " factorial four "; the value being that of the
product 1.2.3.4 or 2 4
"P, " the number of permutations of n things taken
two at a time."
"C, ,, ,, " the number of combinations of n things taken
two at a time."
n t ,, n (n i) (n 2).
7) ,, ,, " efficiency."
a ,, ,, " angle in degrees."
6 ,, ,, " angle in radians."
I.H.P. ,, ,, " indicated horsepower."
B.H.P. ,, ,, " brake horsepower."
m.p.h. ,, ,, " miles per hour."
r.p.m. ,, ,, " revolutions per minute."
r.p.s. ,, ,, " revolutions per second."
I.V. " independent variable."
INTRODUCTORY 3
F stands for " degrees Fahrenheit."
C ,, " degrees Centigrade."
L.C.D. " lowest common denominator."
E.M.F. ,, ,, " electromotive force."
1 ,, " moment of inertia."
E ,, ,, " Young's modulus of elasticity."
S n ,, " the sum to n terms."
S OT ,, ,, " the sum to infinity (of terms)."
2 ,, ,, " sum of."
B.T.U. " Board of Trade unit."
B.Th.U. " British thermal unit."
T ,, ,, " absolute temperature."
p. ,, ,, " coefficient of friction."
sin 1 x ,, ,, " the angle whose sine is x."
e ,, ,, " the base of Napierian logarithms."
g ,, " the acceleration due to the force of gravity."
cms. ,, ,, " centimetres."
grins. ,, ,, " grammes."
Ly. " limit to which y approaches as x approaches
** a the value a."
Tables of Weights and Measures.
BRITISH TABLE OF LENGTH
12 inches (ins.) = I foot
3 feet (ft.) = i yard
5j yards (yds.) = i pole
40 poles (po.) =i furlong
8 furlongs (fur.) = i mile.
I nautical mile = 6080 feet
i knot = i nautical mile per hour
i fathom = 6 feet.
SQUARE MEASURE
144 square inches (sq. ins.) = i square foot
9 square feet (sq. ft.) = i square yard
30^ square yards (sq. yds.) = i square pole
40 square poles = i rood
4 roods or 4840 sq. yds. i acre
640 acres = i square mile.
4 MATHEMATICS FOR ENGINEERS
CUBIC MEASURE
1728 cubic inches (cu. ins.) = I cubic foot
27 cubic feet (cu. ft.) = i cubic yard.
Weight of i gallon of water = 10 Ibs.
Weight of i cu. ft. of water = 624 Ibs.
i cu. ft. = 624 gallons.
METRIC TABLE OF LENGTH
I kilometre (Km.) = 1000 metres .
I hectometre (Hm.) = 100
I dekametre (Dm.) = 10
metre (m.) = 3937"
i decimetre (dm.) = i metre
i centimetre (cm.) = 01 (254 cms. = i".)
i millimetre (mm.) = ooi
LAND MEASURE
100 links = i chain
i chain = 66 feet
10 chains = i furlong
80 chains = i mile
10 square chains = i acre.
Useful Constants.
<? = 271828 TT = 314159
log<io = 23026 Iog 10 e = 4343
log M N = log.N X 4343 logeN = log 10 N X 2303
g = 3218 ft. per sec. per sec.
i horsepower = 33000 foot Ibs. per min. = 746 watts.
Absolute temperature r = /C.+273 or tF.\^6i.
i radian = 573 degrees.
pressure of one atmosphere = 147 Ibs. per sq. in.
i inch == 254 centimetres. i sq. in. = 645 sq. cms.
i kilometre = 6213 mile. i kilogram = 2205 Ibs.
i Ib. = 4536 grms.
The following are the statements of the propositions in Euclid,
to which reference is made in the text
Euc. I. 47. In any rightangled triangle, the square which is
INTRODUCTORY 5
described on the side subtending the right angle is equal to the
squares described on the sides which contain the right angle.
Euc. III. 35. If two straight lines cut one another within a
circle, the rectangle contained by the segments of one of them shall
be equal to the rectangle contained by the segments of the other.
Euc. III. 36. Corollary. If from any point without a circle
there be drawn two straight lines cutting it, the rectangles contained
by the whole lines and the parts of them without the circle equal
one another.
Euc. VI. 4. The sides about the equal angles of triangles which
are equiangular to one another are proportionals.
Euc. VI. 19. Similar triangles are to one another in the duplicate
ratio of their homologous (i. e., corresponding) sides.
Euc. VI. 20. Similar polygons have to one another the dupli
cate ratio of that which their homologous sides have. [From this
statement it follows that corresponding areas or surfaces are pro
portional to the squares of their linear dimensions.]
CHAPTER I
AIDS TO CALCULATION
Approximation for Products and Quotients. Whatever
may be the calculations in which the engineer is involved, it is always
desirable, and even necessary, to obtain some approximate result
to serve as a check on that obtained by the use of the slide rule or
logarithms ; only in this way is confidence in one's working assured.
Speed in approximation is as important as reasonable accuracy,
and the following method, it is hoped, will greatly assist in such
acceleration, especially in the cases of products and quotients.
The great trouble in the evaluation of such an expression as
4783 x 3142 X 941 X 0076 ..,<.
is the fixing of the position of the
33000
decimal point. The rules usually given in handbooks on the
manipulation of the slide rule may enable this to be done, but they
certainly give no ideas as to the actual figures to be expected.
The method suggested for approximation may be thus stated
Reduce each number to a simple integer, i. e., one of the
whole numbers I, 2, 3, etc., if possible choosing the numbers
so that cancelling may be performed ; this reduction involving
the omission of multiples or submultiples of 10. To allow
for this, for every " multiplying 10 " omitted place one stroke
in the corresponding line of a fraction spoken of as a point
fraction, and for every " dividing 10 " place one stroke in the
other line of this fraction. Thus two fractions are obtained,
the number fraction, giving a rough idea of the actual figures in
the result, and the point fraction from which the position of
the decimal point in the result is fixed. Accordingly, by
combining these two fractions the required approximate result
is obtained.
To illustrate the application of the method consider the
following
AIDS TO CALCULATION 7
*o ^
Example I. Find an approximate value of the quotient .r
The whole fraction may be written approximately as
 (the number fraction) and IT (the point fraction) ;
that is, we state 481 as 5 (working to the nearest integer). By so
doing we are not multiplying or dividing by any power of ten, so that
there would be nothing to write in the point fraction due to this change :
but by writing 5 in place of 05, we are omitting two " dividing tens " ;
therefore, since 5 is in the numerator of the number fraction, two
strokes appear in the denominator of the point fraction. The number
fraction reduces to i ; and the point fraction indicates that the result
of the number fraction is to be divided by 100, since two strokes,
corresponding to two tens multiplied together, appear in the denomi
nator. Hence, a combination of the two fractions gives the approximate
result as i f 100 or 01.
It may be easier to effect the combination cf the two fractions
according to the following plan
The result of the number fraction being !; shift the decimal
point two places to the left, because of the presence of the two strokes
in the denominator of the point fraction, thus
01
Example 2. Determine the approximate value of ^i 
284 x 00074
To apply the method to this example
State 9764 as 10,000, i. e., write i in the numerator of the number
fraction and four strokes in the numerator of the point fraction.
For 0213 write 2 in the numerator of the number fraction and two
strokes in the denominator of the point fraction. The strokes are
placed in the denominator because in substituting 02 for 2 we are
multiplying by 100, and therefore, to preserve the balance, we must
divide the result by 100.
For 284 we should write 3 with i stroke in the denominator, and
for 00074 we should write 7 with 4 strokes in the numerator.
Thus
Number fraction. Point fraction.
1X2 1111 1\\\
3><7 \\ \
. e., i and  by cancelling.
Hence the approximate result is i x io 5 , i. e., 10,000 ; or, alter
natively, the shifting of the decimal point would be effected thus
10000
8 MATHEMATICS FOR ENGINEERS
It will be seen from this method that it is often an advantage
to express a very large or very small number as an equivalent
simpler number multiplied by some power of ten. Not only is a
saving of time obtained, but the method tends to greater accuracy.
Thus 2,000,000 may be written as 2 X io 6 , a very compact form ;
also it is far more likely that an error of a nought may be made in
the extended than in the shorter form. " Young's modulus " for
steel is often written as 29 X io 6 Ibs. per sq. in., rather than
29,000,000 Ibs. per sq. in.
Example 3. Find the approximation for
47'83 x 3142 x 941 x 0076
33000
The method will be understood from the explanation given in the
previous examples ; and for clearness the strokes are separated in the
point fraction.
The approximation is
5x3x1x8 \ \
3 111
which reduces to
40
im
'. e., 40 r io 5 = 0004.
The change in the position of the decimal point would be 00040'
Further examples on approximation will be found on pp. 18
to 21.
Approximations for Squares and Square Roots. An ex
tension of this method can be made to apply to cases of squares
and square roots, cubes and cube roots. As regards squaring and
cubing, these may be considered as cases of multiplication, so that
nothing further need be added. To find, say, a square root approxi
mately, we must remember that the square root of 100 or io 2 is
io, the square root of io 4 is io 2 , and so on ; the approximation,
therefore, must be so arranged that an even number of tens are
omitted or added. Hence the rule for this approximation may be
expressed
Reduce the number whose square root is to be found to
some number between i and 100, multiplied or divided by
some even power of ten ; then the approximate square root of
this number, combined with half the number of strokes in the
point fraction, gives the approximate square root of the number.
AIDS TO CALCULATION 9
In the case of cube roots, the number must be reduced to
some number between i and 100 multiplied or divided by
3, 6 or 9 ... tens; then the approximate cube root of this
number must be combined with onethird of the strokes in the
point fraction.
Example 4. Find an approximation for ^4984.
In place of 4984 write 500, which can be written as 5 x io 2 ,
11
or as 5
Then the approximate square root is 22
or 22.
If the number had been 4984 the number would read
5 o ^
and the square root
7 , 22;
Example 5. Find approximately the cube root of 000182.
If for 182 we write 200, then 000182 is replaced by
111111
so that the cube root of 000182 is that of 200 divided by io a , since
two strokes (viz. ^ of 6) appear in the denominator of the point fraction.
Thus, the cube root is
ss n
or 058.
Example 6. Evaluate approximately I/ 
154 x 2409
Disregarding the square root sign for the moment, the approxima
tion gives
2X I J \_
15x2 i iu
,'.*, 67 n
For the application of the method of this paragraph this result
would be written 67
of which the square root is 82
or the approximate square root is 082.
io MATHEMATICS FOR ENGINEERS
Exercises 1. On Approximations.
Determine the approximate answers for Exercises i to 20.
1. 4957 x 0243 2. 00517 x 1724 3. ?
2 3 4
4. 8965 x 7249 x '094 5. 1167 x 0004 x 981 x 2710
4176 X 25400 _
~ 87235 7 
11540 x 3276 x 3142 x 0078

4176 X 25400 _ 154 X 00905
6> ~ 87235 7  "847 8 
"00346 x 0209
Indices. The approximation being made, the actual figures can
be determined either by logarithms or by the slide rule.
Napier, working in Scotland, and Briggs in England, during the
period 161417 evolved a system which made possible the evalua
tion of expressions previously left severely alone. Without the aid
of their system much of the experimental work of modern times
would lose its application, in that the conclusions to be drawn
could not be put into the most beneficial forms ; and failing loga
rithms, arithmetic, with its cumbersome and exacting rules, would
dull our faculties and prevent any advance.
The great virtue of the system of logarithms is its simplicity :
rules with which we have long been acquainted are put into a
more practical form and a new name given to them. Many are
familiar with the simpler rules of indices, such as a 3 X a 4 = 3+4 = a 7 ;
a8_i_02 = a 8  2 = a 6 ; (a 3 ) 4 = 3X4 = a 12 , etc.
Following along these lines we can find meanings for *, a,
and a~ 3 , i. e., we can establish rules that will apply to all cases of
positive, negative, fractional or integral indices. Thus, to find a
meaning for a fractional power, consider the simplest case, viz.
that in which the index is J.
When multiplying a 3 X a 4 we add the indices ; this can be done
whatever the indices may be, hence
= a = a
i. e., a* is that quantity which multiplied by itself is equal to a, or
in other words, a Ms the square root of a.
AIDS TO CALCULATION n
In like manner, since a& x a$ x a& = a 1 = a, a? may be written
as ty 'a. For example, 27^ = ^27 = 3.
To carry this argument a step further we may consider a numerical
example, e. g., 64$, and from the meaning of this, derive a meaning
for a*
Thus 64^ might be written as 64^ X 64^, which again may be
put into the form ^64 X ^64, i. e., (^64)* or ^S^ 2 .
Hence the actual numerical value = ^64 x 64 = 16.
We see that the denominator of the exponent indicates the
p
root, and the numerator the power; thus a = ^a r t
To find a meaning for a
a m x a = m+0 = m .
Dividing through by a m , a = i, i. g., any number or letter
raised to the zero power equals i : e. g., 465 = i; 2384 = i;
4*0 = 4 x i = 4.
Assuming this result for a, we can show how to deal with
negative powers, for
a m xa~ m = a m ~ m = a = i.
Hence, dividing through by a" 1 ,
m _ T
a" 1
Accordingly, in changing a factor (such as a" 1 ) from the top to
the bottom of a fraction or vice versa, we must change the sign
before its index.
Thus 2
b~ 7
Example 7. Simplify ( 6 6 2 c 8 ) 2 x Va 3 & 4 c 6 .
The expression = a~ 10 &Vxa^&~^ . . Removing brackets.
= o~ 10+i 6 4  2 c 6 +3 . . . Collecting like letters.
17,29 6V 6V
= a yb c =. or 
Vfl 17
Example 8. Simplify (64*"')*
2( 5 AT 2 ) 3
The expression
2X5 3 * 6 2XI25X* 25QX~ r ~ 12
12 MATHEMATICS FOR ENGINEERS
Exercises 2. On Indices.
1. Express with positive indices
2. Find the numerical values of
32!; 6 4 f; (j)~*; 6x512$; fox #6^*} + {i 5 x
3. Simplify (sa 2 6c 3 ) x (^a 9 b 5 c)^.
4. Simplify "343^*^ r 8ix~ l y
5. Simplify na 2 6crf 8 x (a~*b 6 c 3 )^
TJ^ n WI} n ~ ^
6. Find the value of  .^ in terms of v, when = 137.
4 U
7. Find the value of vnCv~ n i in terms of p when n =141
and pv n = C. _
r i / ^Tj 2
8. Simplify j V/ i f , a formula referring to the flow of a
gas through an orifice, a being the ratio of the outlet pressure to the
pressure in the vessel.
9. Simplify 8 (**) x ()** i (e 2 ')* r , and find its value when
e = 2718.
10. Simplify the expression
11. The work done in the adiabatic expansion of a gas from volume
Q
t/i to volume w, may be written W = ^(Vt 1 " v^ 1 "). If p t v t n =
piUi 1 = C, by substituting for C, find a simpler expression for W.
Logarithms. It is necessary to deal with indices at this stage,
because logarithms and indices are intimately connected.
For example, 100 = io 2 , and the logarithm of 100 to the base
10 = 2 (written Iog 10 100 = 2). Here are two different ways of
stating the same fact, for 2 is the index of the power to which the
base io has to be raised to equal the number 100; but it is also
called the logarithm of 100 to the base io, i. e., the index viewed
from a slightly different standpoint is termed the logarithm. Hence
the rules of logs (as they are called) must be the same as those
connecting indices.
In general : The logarithm of a number to a certain base is the
index of the power to which the base must be raised to equal the
number. *
It is not necessary to understand the theory of logs to be able
to use them for ordinary calculations, but the knowledge of the
principles involved is of very great assistance.
AIDS TO CALCULATION 13
Consider the three statements
64 = 2 6 ; 64 = 4 3 ; 64 = 8 2 .
These could be written in the alternative forms
Iog 2 64 = 6; Iog 4 64 = 3; Iog 8 64 = 2;
where the numbers 2, 4 and 8 are called bases.
It will be noticed that the same number has different logs in
the three cases, i. e. t if we alter the " base " or " datum " from
which measurements or calculations are made, we alter the log;
in consequence, as many tables of logs can be constructed as there
are numbers. This shows the need for a standard base, and
accordingly logs are calculated either to the base of 10 (such being
called Common or Briggian Logarithms) or to the base of e, a letter
written to represent a series of vast importance, the approximate
value of which is 2718. Logarithms calculated to the base of e
are called Natural, Napierian or Hyperbolic Logarithms. At present
we shall confine our attention to the Common logs; in the later
parts of the work we shall find the importance and usefulness of
the natural logs.
From the foregoing definition of a logarithm the logs of simple
powers of ten can be readily written down ; thus, Iog 10 iooo = 3,
since 1000 = io 3 , Iog 10 ioooooo = 6, etc. ; Iog 10 iooo = 3 is usually
written in the shorter form log 1000 = 3, the base io being under
stood when the small base figure is omitted.
For a number, such as 5263, lying between 100 and 1000,
*. e., between io 2 and io 3 , the log must lie between 2 and 3, and
must therefore be 2 f some fraction. To determine this fraction
recourse must be made to a table of logs.
To read logs from the tables. The tables appearing at the
end of this book are known as fourfigure tables, and are quite
full enough for ordinary calculations, but for particularly accurate
work, as, for example, in Surveying, five and even sevenfigure
tables are used. One soon becomes familiar with the method of
using these tables, the few difficulties arising being dealt with in
the following pages.
To return to the number 5263 : the fractional or decimal part
of its logarithm is to be found after the following manner : Look
down the first column of the table headed " logarithms " (Table II
at the end of the book) till 52 is reached, then along this line until
under the column headed " 6 " at the top the figure 7210 is found;
this is the decimal part of the log of 526, so that the 3 is at present
unaccounted for. At the end of the line in which 7210 occurs
are what are known as " difference " columns. Under that headed
14 MATHEMATICS FOR ENGINEERS
" 3 " and in the same line as the 7210, the fourth figure of our
number, the figure 2, occurs; this, added to 7210, making 7212,
completes the decimal portion of the log of 5263. The figure from
the tables is thus 7212, and since this is to be the fractional portion
the decimal point is placed immediately before the first figure.
The log of 5263 is therefore 27212, or, in other words, 5263 = 10
raised to the power 27212; similarly the log of 52630 must be
47212, because 52630 is the same proportion of a power of 10
above 10,000 as 5263 is above 100, and also it lies between io 4
and io 5 , so that its logarithm must be 4 + some fraction.
The log thus consists of two distinct parts, the decimal part,
which is always obtained from the tables and is called the mantissa,
and the integral or wholenumber part, settled by the position of
the decimal point in the number, and called the characteristic or
distinguisher. The logs of 5263 and 52630 are alike as regards the
decimal part, but must be distinguished from one another by the
addition of the relative characteristic.
When the number was 5263, i. e., having 3 figures before the
decimal point, the characteristic was 2, i. e., 3 I ; when the
number was 52630, i. e., having 5 figures before the decimal point,
the characteristic was 4, *'. e., 5 i. This method could be applied
for numbers down to i, i. e., 10, but for numbers of less value
we are dealing with negative powers, and accordingly we must
investigate afresh.
So far, then, we can formulate the rule : " When the number is
greater than 1, the characteristic of its log is positive and is one less than
the number of figures before the decimal point."
E.g., if the number is 2507640, the characteristic of its log is 6,
because there are seven figures in the number before the decimal point.
Referring to the figures 5263 already mentioned, place the
decimal point immediately before the first figure, giving 5263.
The number now lies between i and i. Now
i = = io 1 and i = 10
io
so that the log of 5263 lies between i and o, being greater than
i and less than o, and therefore is i + a fraction. The
mantissa is as before, viz. 7212, hence log 5263 = i + 7212,
or, as it is usually written, 17212, the minus sign being placed over
the i to signify the fact that it applies only to the i and not to the
7212, which latter is a positive quantity and must be kept as such.
17212 actually means, then, I + 7212, or, in fact, 2788.
AIDS TO CALCULATION 15
The figures taken from the tables are always positive, and
accordingly the form 17212 is adhered to throughout.
From similar reasoning, log 005263 = 37212 (i)
and log 00005263 = 57212 .... (2)
We can conclude, then, that : When dealing with the log of a
decimal fraction the mantissa is found from the tables in just the
same way as for a number larger than I, or, in other words, no
regard is paid, when using the tables to find the mantissa, to the
position of the decimal point in the number whose log is required.
The characteristic of the log, however, is negative, and one more
than the number of zeros before the first significant figure.
In (i) there are 2 noughts before the first significant figure;
therefore the characteristic is 3. In (2) the characteristic is 5,
because there are 4 noughts before the first significant figure.
For emphasis, the rules for the determination of the charac
teristic of the log of any number are repeated
If the number is greater than 1, the characteristic is positive and one
less than the number of figures before the decimal point : if the number
is less than i, the characteristic is negative and one more than the number
of noughts before the first significant figure.
It will be observed that in the earlier part of the table of loga
rithms at the end of the book there are, for each number in the
first column, two lines in the " difference " column. This arrange
ment (the copyright of Messrs. Macmillan & Co., Ltd.) gives
greater accuracy as regards the fourth figure of the log, since the
differences in this portion of the table are large. The log is looked
out as explained in the previous case, care being taken to read the
" difference " figure in the same line as the third significant figure
of the number whose log is being determined.
Thus the log of 1437 is 31553 + a difference of 21 = 31574,
while the log of 1487 is 31703 + a difference of 20 = 31723.
We are now in a position to write down the value of the log of
any number, and a few examples are given
Number.
40760
2359
7008
0009
500000
Log.
46102
13728
18456
49542
5 ~
(The mantissa for the log of 9,
90, 900, and 9000 is 9542.)
i6
MATHEMATICS FOR ENGINEERS
Values of log 1 and log 0. If a be any number, then, as
proved earlier, a = i ; or in the log form, log a I = o.
Thus the log of i to any base = o.
The log of o to any base is minus infinity ; or if a be any number,
log a o = <x.
For, by writing this statement in the alternative form
a' = o
where x is the required logarithm, we see that x must be an infinitely
small quantity ; in fact, the smallest quantity conceivable.
Antilogarithms. Suppose the question is presented to us in
the reverse way : " Find the number whose logarithm is 29053."
The table of antilogarithms (Table III) will be found more con
venient for this, although the log tables can be used in the reverse
way. Just as the mantissa alone was found from the log tables
when finding the logarithm, so this alone is used to determine the
actual arrangement of the figures in the number. In the case under
consideration the mantissa is "9053, hence look down the first
column until 90 is reached, then along this line until in the column
headed " 5 " 8035 is read off : to this must be added 6, the number
found in the " difference " column headed " 3," so that the actual
figuring of the number is 8035 + 6 = 8041.
The characteristic 2 must now be considered so as to fix the
position of the decimal point. Referring to our rule, we see that
the characteristic is one less than the number of figures before the
decimal point (since 2 is positive), hence, conversely, the number
of figures before the decimal point must be one more than the
characteristic; in this case there must be 3 figures before the
decimal point, i. e., the required number is 8041.
If we had been asked to find the antilog of 20905, the line
through 09 would havebeen followed and not that through 90,
and the antilog is found to be 1231. Many errors occur if this
distinction is not appreciated; and the actual mantissa must be
dealt with in its entirety, no noughts being disregarded wherever
they may occur.
Examples
Log.
No.
81164
130700000 or i
307 x io 8
2549
i 0062
1799
1014
38609
007259
AIDS TO CALCULATION
Failing a table of logs, the log scale on the slide rule can be
used in the following manner. Reverse and invert the slide so that
the S scale is now adjacent to the.D scale: place
the ends of the D and S scales level: then using
the D scale as that of numbers, the corresponding
logarithms are read off on the L scale it being
remembered that although the scale is inverted the
numbers increase towards the right. The mantissa
alone is found in this way, whilst the characteristic
is settled according to the ordinary rules.
Fig. i shows the scales of an ordinary 10" Slide
Rule lettered as they will be referred to throughout
this book. On the front are the scales A, B, C
and D, the B and C scales being on the " slide."
If the slide is taken out and reversed the S, L and
T scales will be noticed (see righthand end of
figure). Any special markings referred to through
out the text are also indicated, and it is to this
sketch that the reader should refer, no other
sketch of the slide rule being inserted. The slide
rule is referred to from time to time, wherever
its use is required, and a word or two is then said
about the method of usage, but no special chapter
is devoted to its use. For a full explanation of
the method of using the slide rule reference should
be made to Arithmetic for Engineers*
Applications of Logarithms. It will be
granted that
2+4 = 6
or
log 100 + log 10,000 = log 1,000,000 from definition.
But 1,000,000 = 100 x 10,000
/.log(ioo x 10,000)= log 100 + log 10,000.
Simple powers of ten have been taken in this
example, for convenience, but the rule demon
strated is perfectly general, holding for all numbers.
In general, log (AxB) = log A+ log B, where A and B are any
numbers. Thus, the log of a product = the sum of the logs of the factors.
This and the succeeding rules hold for bases other than 10; in
fact, they are general in all respects.
* Arithmetic for^Engineers, by Charles B. Clapham, B.Sc. Chapman
and Hall, Ltd., ?s.'6d. net.
C
HEUlJH
to
i8 MATHEMATICS FOR ENGINEERS
In like manner it can be shown that
log( = ) = log A log B
\o/
i. e., the log of a quotient = the difference of the logs of numerator and
denominator.
Again, 3x2 = 6
.'. 3Xlog 100 = log 1,000,000
= log (ioo) 3
or in general, log (A) 1 * = n log A, this holding whatever value be given to n.
E.g., (a) ^4^76 =(42 76)*
6 = J log 4276.
(b) log (05I7) 4 * = 42 x log 0517.
Stated in words this rule becomes : The log of a number raised to
a power is equal to the log of that number multiplied by that power.
Summarising, we see that multiplication and division can be
performed by suitable addition and subtraction, whilst the trouble
some process of finding a power or root resolves itself into a simple
multiplication or division. (The application of logarithms to more
difficult calculations is taken up in Chap. V.)
In any numerical example care should be taken to set the
work out in a reasonable fashion ; especially in questions involving
the use of logs.
Example 9. Find the value of 4821 x 7429.
Actual Working Approximation 50x7 = 350.
Let x = 4821 X 7429
then log x = log 48 2 1 + log 7429= 16831 + 8709
= 25540
= log 3581 from the antilog tables.
/. x = 3581.
V
Example 10. If C=^, a formula relating to electric currents,
find the value of C, a current, when the voltage V is 241 and the
resistance R is 287.
Substituting the values of V and R
.'. log C = log 241 log 287
= 3820 14579
A pproximation.
2J4
3 !
i. e., 8 7 10 or 08
= 29241, since 2 subtracted from =
= log 08397
/. C = 08397
AIDS TO CALCULATION
Example n. If F, the centrifugal force on a rotating body, = ,
find its value when W = 28, v = 475,
g=322, r= 1875.
Substituting the numerical values in place of the letters
F _ 28 x (475) a
322 x 1875
Taking logs throughout
log F = (log 28 + 2 log 475)  (log 322
+ log i 8 75)
= I> 447 2 _ i '579
13534
28006
= 28006
= 10197
= log 1047
' F 1047.
17809
17809
Approximation..
3x5x5 \
3x2 \
i. e., 125.
Explanation.
log 475= '6767
.\2xlog 475 = 13534
j?
Example 12. If/=~r, an equation giving the acceleration pro
duced by a force P acting on a weight W, find / when g = 322, P = 5934,
and m = 487.
Substituting the numerical values
, 322 x 5934
487
Taking logs
log/= (log 32 2 + log 5934) log 4' 8 7
= (i'5079 +7734) 26875
= 22813 26875
= T5938 = log 3924
/. / = 3924.
Approximation.
3x6 \_
5 \1
i. e., 36 ^ 10 or 36.
Example 13. Find the value of *
001872
Let
_ 05229
~~ 001872
then log x = log 05229 log 001872
= 27184 32723
= 14461
= log 2794
.'. ^=2794.
A pproximation.
5 Ul
2 \\
i. e., 25 X 10
or 25.
Note. In the subtraction the minus 3 becomes plus 3 (changing
the bottom sign and adding algebraically) ; and this, combined with
minus 2, gives plus i.
20
MATHEMATICS FOR ENGINEERS
Example 14. Find the value of the expression s =
QII54
4761 x 0000753
Taking logs
log s = log 01154 (log 4761 + log 0000753)
= 2 0622 (l 6777 + 58768)
= 2062235545
= 5077
= log 3219
/. s = 3219.
A pproximation.
48 x 75
\\\
i. e., 033 x 100
or 3.3.
Note. In this subtraction the i borrowed for the 5 from o should
be repaid by subtracting it from the 2, making it 3 : this, combined
with + 3 (the sign being changed for subtraction), gives o as a result.
Alternatively, the 3 must be increased by i to repay the borrowing,
so that it becomes 2 ; and 2 subtracted from 2 gives o.
Example 15. The formula V = X3i42xr 3 gives the volume of
a sphere of radius r. Find the volume when the radius r is 56.
Substituting the numerical values
= X 3142 x
Taking logs
log V = (log 4 + log 3142 + 3 log 56)
log 3
= (6021 + 4972 + 12446)  4771
= '3439 '4771
= 18668
= log 7358
.'. V = 7358.
Approximation.
4x3x6x6x6
3 Hi
i. e., 864 f 1000
or 864.
Explanation.
log 56 = 17482
3 x log 56 = 12446
i. e., there is + 2 to carry
from the multiplication of
the mantissa and this, to
gether with 3 which is ob
tained when T is multiplied
by 3, gives T.
Example 16. Find the fifth root of 009185.
Let x= V 009185 = (009185)^
then log x = \ log 009185
= \ x 3^9630 *
= Jx {5 + 2 9630}
= 15926 = log 3913
.. *= 3913.
* We must not divide 5 into 39630 because the 3 is minus, whilst
the 9630 is plus ; but the addition of 2 to the whole number and of
+ 2 to the mantissa will permit the division of each part separately,
while not affecting the value of the quantity as a whole.
AIDS TO CALCULATION
21
Example 17. Evaluate
(2 1 64)* x
(001762)4 x (49'i8)f
Let the whole fraction = x.
Then log x = {3 log 2164 + \ log 7454} (J log 001762 + f log 4918}.
Explanation.
log 2164 = 13353
3 x log 2164 = 20059
lo 745 '4 = 2 8724
jxlog 7454=14362
log 001762 = 32460
\ x log 001762 = 15410
log 4918 =16918
2 X log 4918 =33836
.. ^=1677. .\fxlog49i8 = 6767.
= (20059+14362)  (15410+ 6767)
= 14421 2177
= 12244
= log 1677
Example 18. If x= \/    5 find the value of x.
v 9004 x 0050
Taking logs
log x = {(log 2917 + log 1245)  (log 9004 + log 0856)}
= & (I4649+ 10951) ~ 039544 + 29325)}
= (5600  28869) = t(3673i) = i(8 + 5673I).
= 17091 = log 5118
.. x = 5118.
The following examples are worked by the slide rule.
Example 19. Find the buckling stress P for a column of length /;
 
from the formula P =
when k z = 575, c = and / = 180.
Substituting these numerical values
48000
30000 575,
The second term of the denominator must be worked apart from
the rest
Thus, to evaluate 4X 180x180 eed as f oUows _
30000 x 5 75
Actual figuring, found from the slide rule, Approximation.
is 751, so that, in accordance with the 4x2x2 \\\\1
approximation the value of this term is 3x6 \\\\
i. e., 88 x 10
or 88.
22 MATHEMATICS FOR ENGINEERS
Example 20. Find the value of E, Young's modulus for steel,
W/ ( I 2 z~\
from the formula E = ~~c \(Tr + jr\ which expresses the result of a
bending test on a girder.
Given that A = 924 / = 60 W = 5000
D = 07 I = 1115.
Substituting values
F 5 000 x 60 / 60 x 60 5 \
8 x 07 1 6 x 1115 924!
= 536000 {5375 + 541}
= 317 x io 8 .
Exercises 3. On the Use of Logarithms and Evaluation of Formulae.
Evaluate, using logarithms or the slide rule, Exs. i to 32; using
approximations wherever possible.
1. 8523x6917 2. 8764 x 1194x2356
3. 7542 xooo2835 4. _ 4 5. 005376 x 1009
6 9543 f] 2896x3472 g i2o8x02H2
08176 8148 01299
0005 1Q 4843^2985 1 154 x 07648
' 007503 75132 " 009914x3642
9867 x 4693 13 3687x257
'* 0863 x 1842 ' 085 x 1377 x 05
.. 2423 x 7529 x 00814 ._ 572 x 0086
3000 xo 115x45 2 7" 4539 x 0037 x 059
16. V9403 17. (0517)3 18.
19. "IX (00 1 769) 3 20. v (1182)3 21.
22 ' ^fx^ 23 ' ( ' 253)3X *^
(0648) 2 xV2 753 2g ^9472x853^9"
('275) s ' (2347) x 5x10*
OR (9i'56) 2 x(3i8 4 )l (472) 3 x V2643
(2 3 ) 2 xV8^
28. V 7 ' 008 ?^;, 0372 29.
423x105 /O5Oo6\ 3 _. V 6463 x (086)
'
(2763) 2 x log 10 3476
* V('4349) B x 5007x1
33. The formula V = nr z l gives the volume of a cylinder. If r =
= 3142, / = 1276 find V,
AIDS TO CALCULATION 23
34. Given that L = . Find L when =117, ^B = 1755.
l z a
35. If R = 7+ find R when / = 51 and a = 087.
oa 2 J '
36. The velocity ratio of a differential pulley block is found from
the formula
2.d
VR = s L (where d lt d t and d a are the diameters of the pulleys)
"a~~ a a
Find VR when ^!= 1457, ^ 2 = 5'72, ^3 = 483.
37. If v = u+ft and 5 = ^+i// 2 , find values of v and 5 when
M = 350, / = 27, and t = 48.
38. Find a velocity, v, from
when g= 322, <Z= 084, A = 30, /= 5000.
39. If p= (7854 x *~) + d, find its value when f s = $, t = oJ5
ft=6, d = i 04 ; p is the pitch of rivets, of diameter d, joining plates
of thickness /.
40. Find the weight of a roof principal from Merriman's formula
W = aliii ) when a=io and /=8o.
41. To compare the cost of lighting by gas and electricity the
b
~~ f
following rule is often used, a = 5
where a = price of i Board of Trade (B.O.T.) unit in pence ; b = price
per 1000 cu. ft. of gas in pence; d= watts per candle power (C.P.) ;
e = candles per cu. ft. of gas per hour; c = cost in pence of lamp
renewals per 1000 candle hours.
Find the equivalent cost per electric unit when lamps take 25
watts per C.P., e = 2, c = i and gas is 25. zd. per 1000 cu. ft.
42. The length / of a trolley wire for a span L when the sag is d
8d 2
is given by the formula /= L+ ^. Find /, when L = 500, d= 12,
3^
43. The input of an electric motor, in H.P., is measured by the
product of the amperes and the volts divided by 746. What is the
input in the case where 872 amps, are supplied at a pressure of 1125
volts ? If the efficiency of the motor at this load is 45 %, what is its
output ? (Output = efficiency x input.)
44. 24 Ibs. of iron are heated from 60 F. to 1200 F. The specific
heat of iron being 13, find the number of British Thermal Units
(B.Th.U.) required for this, given B.Th.U. = weight x rise in temp, x
specific heat.
45. The following rules for the rating of motorcars have been stated
at various times.
(a) By Messrs. Rolls Royce, Ltd.
where d = diameter of cylinder in inches, N = no. of cylinders
S = stroke in inches.
MATHEMATICS FOR ENGINEERS
(6) By the Royal Automobile Club
H.P. = i97d(d i)(r+ 2)N
where N and d have the same meanings as before, and r = ratio of stroke
to diameter. Find the rating of a 4cylinder engine, whose cylinders are
of 4" diameter, and stroke 86" ; by the use of each of the rules.
46. 130 grms. of copper (W) at 95 C. (T) are mixed with 160 grms. (w)
of water at ioC. (t), the final temperature (t^ being iGC. Calculate
the specific heat (s) of copper from
<*<
~
47. The volume t; of a gas at a temperature of oC., or 273 C.
absolute, and at a pressure corresponding to 760 mms. of mercury is
1783 cu. ins. Find its volume at temperature t C. and pressure H
where t = 837 and H = 797 from the formula
V 
48. If P=
find its value when 6 =
/=!. =8, L=I2,
y = 80000.
L = length of a railway spring on each side of the buckle, n =
number of leaves, / = thickness of leaves, /= working stress, P = load
applied and b = width of leaves.
49. The increase in length of a steel girder due to rise of temperature
can be found from the formula, new length = old length (i + at) when
t rise in temperature, and a = coefficient of linear expansion. Find
the increase in length of a girder of 80 ft. span due to change of
temperature of 150 F. when a= 000006.
50. If c = *fo V(i ip + id) (p + id), find its value when p = 3", d = } *.
The meaning of c will be understood by reference to the riveted joint
shown in Fig. 2.
51. Find the thickness (tj of a butt strap
from the B.O.T. rule
pd\.
when p = 4f *, / = !", d = \%'.
52. Find the thickness (t) of a pipe for
pressure p Ibs. per sq. in., when internal
dia. = d, from
d=2, =45
53. Taking p =
which gives the principal (or maximum) stress
p due to a normal stress /and a shearing stress
s : determine p when /= 3800, s= 2600.
54. If P=
Fig. 2. Riveted Joint.
(Gordon's formula for the buckling load
1500
on struts), find P when F = 38, d = 15, T =
AIDS TO CALCULATION 25
55. If p= i ,s 2 (Rankine's formula for the buckling load on
struts), find p when / = 48000, / = 14 \ x 12, c =  , A 2 = 307.
56. The deflection d of a helical spring can be obtained from
, _ 6^wnr 3
'' CD* '
Find the deflection for the case in which w= 48, D = J
n= 1273, r=i5.
C= 12 X IO 6 .
57. If the deflection d of a beam of radius a and length J, due to a
load of W is measured, Young's Modulus for the material of which the
4\V7 3
beam is composed, can be found from E= , .. If in a certain case
37ra 4
the deflection was 42; and W, /, a and IT had the values 148, 1756,
39 and 3142 respectively, find the value of E.
58. For oval furnaces, if
A = difference between the half axes before straining.
8 = . , after
p = pressure in Ibs. per sq. in.
E = Young's Modulus.
D = diameter of furnace.
AX32EI
 
 ==rf 
32EI
Find the value of 8 when A = 5, D = 40, p= 100, E = 30 x io and
I = 0104.
59. The modulus of rigidity C of a wire of length / and diameter d
may be found by attaching weights of m { and w 2 respectively at the
end of the wire and noting the times, t l and t z respectively, taken for
a complete swing. The formula used in the calculation is
p _ I28irl(m 1 W 2 ) 2
Find C when w^gS, m t = !$, t 1 = 2i, /, = 1*6, =32, d= 126,
I = 483, a = 97 and ?r = 3142.
60. The weight W in tons of a flywheel is given by
_
R 3 N 3
Find the weight when R =  , r = 2, n = 30, N = 120, H = 70.
61. The approximate diameter of wire (in inches) to carry a given
current C with 6 rise in temperature can be obtained from
Find D when J = 0935, & = 35. P = ^ c = 55. m = >O02 5
ir = 3142.
62. Find the dimensions for the flanged
castiron pipe shown in Fig. 3 (in each case
to the nearest ^th of an inch), when P = 85,
PD .
d= , = F .
There are n bolts and n = 6D+?,
26 MATHEMATICS FOR ENGINEERS
Investigation for Units. A train covers a distance of 150
miles in 5 hours ; what is its average speed ?
^, . , ... 150 miles . 30 miles
Obviously it is , i. e., ,  or 30 miles per hour.
5 hours ' i hour
It could with equal truth be expressed by ** * ,
J 5 X 60 X 60 sees.
i. e. t ', or 44 ft. per second. The figures in the results differ
' i sec.
because they are measured in terms of different quantities, and it
is essential that the units in which results are expressed should be clearly
stated.
Here we have another form of investigation to be performed
before the actual numerical working is attempted. To find the
units in which the result is to be expressed, these units, with their
proper powers attached, are put down in the form of a fraction,
all figures and constants being disregarded, and are treated for can
celling purposes as though they were pure algebraic symbols.
Suppose a force of 100 Ibs. weight is exerted through a distance
of 15 ft., then the work done by this force is 100 X 15 or 1500 units :
these units will be "foot Ibs." since the result is obtained by multi
plication of Ibs. by feet. This statement might be written in the
form, Work = Ibs. X feet = foot Ibs. If now we are told that the
time taken over the movement was 12 minutes we can determine
the average rate at which the work was done. The work done in
i minute is evidently obtained by dividing the total work done
in 12 minutes by the number of minutes : thus, rate of working
=  = 125. This figure gives the number of foot Ibs. of work
done in one minute, and the result would be expressed as, average
rate of working = 125 foot Ibs. per minute. It will be seen from this
and from the previous illustration that the word per implies division.
To obtain a velocity in miles per hour, the distance covered, in
miles, must be divided by the number of hours taken, or
. number of miles , ,, miles
velocity (miles per hour)  number of hours or, more shortly, j^.
An acceleration = rate of change of speed
= feet per second added every second (say)
feet feet
QJ*
sees. X sees. (sees.) 2
Hence, wherever an acceleration occurs it must be written as
distance . ,, . ,. ,. f .,
T. TJ in the investigation for units.
AIDS TO CALCULATION 27
The " g " so frequently met with in engineering formulae is an
acceleration, being 322 ft. per sec. per sec. or 322 . '^, and
(sees.) 4
therefore must be treated as such wherever it occurs.
Example 21. The steam pressure, as recorded by a gauge, is
65 Ibs. per sq. in. ; the area of the piston on which the steam is acting
is 87 sq. ins. What is the total pressure on the piston ?
Total pressure = area x intensity of pressure
and is Jprgfy. Ibs *. e., is in Ibs.
The true pressure is 65 + 147 Ibs. per sq. in., because the gauge
records the excess over the atmospheric pressure
.*. total pressure = 797 x 87 Ibs.
= 60 Ibs.
= 6930 Ibs.
Example 22. Find the force necessary to accelerate a mass of
10 tons by 12 ft. per sec. in a minute. The formula connecting these
w/
quantities is P =  where W = weight, / = acceleration and g has its
usual meaning.
Dealing merely with the units given, and forming our investigation
for units
_ (sees.) 8 feet
~~ X sees, x mins.
It will be seen that no cancelling can be done until the minutes
are brought to seconds ; then we have
P=Tons x
To find the force, therefore, the minutes must be multiplied by 60;
t. e., the denominator must be multiplied by 60.
Hence P= 10 x x ^ = 0621 ton or 1392 Ibs.
322 oo ^  ^^
Example 23. The modulus of rigidity C of a wire can be found
by noting the time of a complete swing of the pendulum shown in
Fig. 4 and then calculating from the formula, C = * , where / is
O
the length of the wire, d is its diameter, I is the moment of inertia
28
MATHEMATICS FOR ENGINEERS
of the brass rod about the axis of suspension and t is the time of one
swing.
If / and d are measured in inches, / in seconds,
and I in Ibs. x (feet) 2 [I being of the nature of
mass x (distance) 2 ], in what units will C be expressed ?
Investigating for units
I287T / I
C = constant x ins. x Ibs. x feet 2 x
feet ins. 4 sec.
_ constant x Ibs. x feet
ins. 3
If the numerator is multiplied by 12, then
_ constant x Ibs. x ins. _ Ibs.
ins 2 . ~~ ins.*
or the result would be expressed in Ibs. per sq. in.
provided that the numerator was multiplied by 12.
Example 24. The head lost in a pipe due to friction is given by the
I v 2
formula h = 03 . r . . Find its value if the pipe is 3" dia., 56 yards
long, and the velocity of flow is 28 yards per min.
The meanings of the various letters will be better understood by
reference to Fig. 5.
Dealing only with the units given, and disregarding the constants
i yards 2 sees. 8
Head lost = yards x .  X .  9 x 7
ins. rmns. 2 feet
This is not in a form convenient for cancelling; accordingly, bring
all distances to feet and all times to seconds.
i feet 2 sees. 2
feet sees. 2 feet
Then the head
= feet.
Fig. 5. Flow of water through a pipe.
Substituting the numerical values in place of the symbols
Head lost = *= ^3 x 56 x_ 3 x X2 x 28 x 3 x 28 x 3
= 612 foot.
3 x 60 x 60 X 644
AIDS TO CALCULATION
29
Example 25. Find the maximum deflection of a beam 24 ft. long,
simply supported at its ends and loaded with 7 tons at the centre.
The moment of inertia I of the section is 872 ins. 4 units, and Young's
Modulus E for the material is 30 x io 6 Ibs. per sq. in.
W/ 3
The maximum deflection = orT ,
40! >
The investigation for units, as given, reads :
i I
W / 3 T E
Deflection = tons x feet 3 x = . x TC 
ins. 4 IDS.
No cancelling can be attempted until the tons are brought to Ibs. and
the feet to inches or vice versa ; assuming the former, then
i ins ^
Deflection = Ibs. x ins. 3 x . . x ^^^ = ins.
ms. 4 Ibs.
So that, since 7 tons = 7 x 2240 Ibs. and 24 ft. = 288 ins.
7 x 2240 x 288 3
Deflection = 5* ~ ^ ins.
48 x 872 x 30 x io 6
Calculation
log d = (log 7+log 2240 + 3 log 288)
(log 48+ log 872+ log 30,000,000)
= (845 1 + 33502 + 73782)
 (16812 + 19405 + 7'477 J )
= 11.5735 110988 Explanation
'4747 = log 2984 log 288 = 24594
.*. deflection = 2984 ins.
A pproximation
7x2x3x3x3 m
5x9x3 UUVWU
or 28.
3 x log 288 = 73782.
Exercises 4. On the Finding of Units.
6E
1. In what units will / be expressed if / = ^ and 8 is in inches,
E in Ibs. per sq. in. and D in ins. ?
2. If a H.P. = 33000 foot Ibs. of work per minute, find the H.P.
necessary to raise 300 cwts. of water through a vertical height of i6
yards in half an hour.
f2
3. If H=^g: find H in yards when / = 18 tons per sq. in.;
E = 13000 per sq. in. ; p = 480 Ibs. per cu. ft.
4. Determine the stress / in a boiler plate in tons per sq. in. from
bd
f = ~ when / = 63 in., d = 8 feet, p = 160 Ibs. per sq. in.
t is the thickness of plate, p is the pressure inside the boiler, and d is
the diameter of the boiler.
30 MATHEMATICS FOR ENGINEERS
5. The jump H of the wheels of a gun is given by
_
=
Find the jump in inches when A = 40 ins., A=io ft., h=i yd.,
P=475 cwts., R= 115 tons, M= 12 tons, W = g cwts.
6. The tension in a belt due to centrifugal action can be calculated
from T= . If w = wi. per foot run of belt in Ibs., u = veloc. in
8
ft. per sec., and g has its usual value, in what units will T be expressed ?
7. If, in the previous example, w = 43 Ib. per foot length of belt
per sq. in. of surface, find a simple relation between the stress (in
Ibs. per sq. in.) and the velocity (ft./sec.).
8. The I.H.P. of an engine is determined from the formula
IHp=2 PLAN
33000
where P = mean effective pressure in Ibs. per sq. in., L = stroke in
feet, A = area of piston in sq. ins., and N = revolutions per minute.
If / is the stroke in ins. and A = '7854D 2 show that this equation may
be written I.H.P. =  approximately.
1,000,000 J
tw^
9. Given that / =  , where w = weight in Ibs. per cu. in.,
v = veloc. in feet per sec.
(a formula relating to tensile stress in revolving bodies).
Arrange the formula so that / is given in Ibs. per sq. in.
10. Investigate for units answer in the following formula for the
Horse Power transmitted by a shaft.
H.P. =   where R is inches, N is Revolutions per minute,
33000
IT is a constant, and p is in Ibs. per sq. in.
If these are not found to be H.P. units, viz. foot Ibs. per minute,
state what correction should be made.
11. The formula Q = a t a, \f f^ *f{ gives the quantity of water
P\ a i ~ a t i
passing through a Venturi Meter.
In what units will Q be expressed if a t and a, are in sq. ft. ; p^ and
p 2 in Ibs. per sq. ft. ; g in feet per sec. per sec. ; p in Ibs. per cu. ft. ?
12. Given that i Ib. = 454 grms., i*= 254 cms.
i erg. = work done when i dyne acts through i cm.
i grm. weight = 981 dynes.
and i watt = io 7 ergs per sec.;
find the number of watts per H.P.
CHAPTER II
EQUATIONS
Simple Equations. A simple equation consists of a statement
connecting an unknown quantity with others that are known;
and the process of " Solving the equation " is that of finding the
particular value of the unknown that satisfies the statement. To
many, this chapter, on the methods of solving equations and of
transposing formulae, must be as important and useful as any in
the book, for it is impossible to proceed very far without a working
knowledge of the ready manipulation of formulae. The methods
of procedure always followed is the isolation of the unknown,
involving the transposition of the known quantities, which may be
either letters or numbers, from one side of the equation to the
other. The transposition may be of either (a) terms or (b) factors ;
and the rule for each change will now be developed.
To deal first with the transposition of terms :
When turning the spindle shown in Fig. 6
it was necessary to calculate the length of the
" plain turned " portion, or the length marked I
in the diagram. The conditions here are that
the required length, together with the radius
*375" must add to 15". A statement of
conditions may thus be made, "in the form
*+ '375 = i5
The truth of this statement will be unaltered
if the same quantity, viz. 375, is subtracted from each side, so
that
I + '375  '375 = i'5  '375
or / = 15  375 = 1125.
Thus, in changing the 375 from one side of the equation to the
other, the sign before it has been changed ; + 375 on the one side
becoming 375 when transferred to the other side.
Again, suppose the excess of the pressure within a cylinder
over that of the atmosphere (taken as 147 Ibs. per sq. in.) is 862
32 MATHEMATICS FOR ENGINEERS
Ibs. per sq. in., and we require to determine the absolute pressure
in the cylinder.
Let p represent the absolute pressure, i. e., the excess over zero
pressure. Then
p 147 = 862.
To each side add 147; then
P = 862 f I4'7 = iQQ'9 Ibs. per sq. in.
Thus, 147 on the lefthand side becomes + *4'7 when trans
ferred to the righthand side of the equation.
Accordingly, we may say that : When transferring a TERM from
one side of an equation to the other, the sign before the term must be
changed, plus becoming minus, and vice versa.
To deal with the transposition of factors :
Suppose we are told that 3 tons of pig iron are bought for
7 los. : we should say at once that the price per ton was \ of
7 ios., or 2 IDS.
We might, however, use this case to illustrate one of the most
vital rules in connection with transpositions, by expressing the
statement in the form of an equation and then solving the equation.
The unknown in this case is the price per ton, which may be
called p shillings. Our equation then becomes
3x^ = 150 (i)
Divide both sides by 3, which is legitimate, since the equation
is not changed if exactly the same operation is performed on either side.
:. P = ^ = so (2)
or the cost is 505. per ton.
Again, had we been told that  a ton could be bought for 255.
we could express this in the form
\P = ^ (3)
If we multiply both sides by 2 we find that
= 25x2 = 50 . . (4)
which, of course, agrees with the above.
It will be seen that, to isolate p and so find its absolute value,
we transfer the multiplier in equation (i) or the divider in equation
(3) to the other side, when its effect is exactly reversed : thus the
multiplier 3 hi equation (i) becomes a divider when transferred
to the other side of the equation, as in (2) ; and the dividing 2 in
equation (3) becomes the multiplying 2 in equation (4).
EQUATIONS 33
The motion of a swinging pendulum furnishes an illustration
of the transposition of a factor which is preceded by a minus sign.
The acceleration of the pendulum towards the centre of the move
ment increases proportionately with the displacement away from
the centre. Taking a numerical case, suppose that we wish to find
the displacement s when the acceleration / is 46 units and the
relation between/ and s is/ = 255.
Substituting the numerical value for/
46 = 255.
To isolate s we must divide both sides of the equation by 25,
and then
s
"
25
or s = '184 unit.
The rule for the transposition of factors can now be stated, viz.
To change a FACTOR (i. e., a multiplier or a divisor) from one side of an
equation to the other, change also its position regarding the fractional
dividing line, viz., let a denominator become a numerator and conversely ;
and let the sign of the factor be kept unchanged.
We have thus established the elementary rules of term and
factor changing in simple equations. The following examples, as
illustrations of these fundamental laws, should be most carefully
studied, every step being thoroughly grasped before proceeding to
another.
*\X 7
Example i. Solve for x, in the equation, *" r*g
*4. I *O
Transferring the 5 and 4 so that x is by itself, the 5 must change
from the top to the bottom and the 4 from the bottom to the top, since
5 and 4 are factors.
Then * = ? 5 x4 = 3ii.
10 5
'Example 2. Solve for a, in the equation, 40+17 = 2509.
Transposing, to get the unknowns together on one side
4250 = 917.
Here the change is that of terms, hence the change of signs.
Grouping, or collecting the terms
i^a = 26
26
34 MATHEMATICS FOR ENGINEERS
Example 3. The weight of steam required per hour for an engine
was a constant 60 Ibs., together with a variable 25 Ibs. for each H.P.
developed. If, in a certain case, 210 Ibs. of steam were supplied in an
hour, what was the H.P. developed ?
Let h represent the unknown H.P.
Then 25^ represents the amount of steam for this H.P., apart from
the constant, and the equation including the whole of the statement
of conditions is
25/1 + 60 = 210.
Transferring the term + 60 to the other side, where it becomes 60,
25& = 210 60 = 150.
Dividing throughout by 25 h = ^2. = 5
or, the H.P. developed was 6.
Example 4. To convert degrees Fahrenheit to degrees Centigrade
use is made of the following relation
F3, = C.
Find the number of degrees C., corresponding to 457 F.
Substituting for F its numerical value
45732 = C
' 5=
Transposing factors 5 and 9,
2361 = C
i.e., 236 C. correspond to 457 F.
It might happen that in an engine or boiler trial only ther
mometers reading in Centigrade degrees were available, whereas
for purposes of calculation it might be necessary to have the tem
peratures expressed in degrees Fahrenheit. This would mean
that a number of equations would have to be solved ; but the work
involved could be shortened by a suitable transposition of the
formula given above.
F 3 2 = C ........... (i)
F = c+ 3 2 ......... (2)
Equation (2) is far more suitable for our purpose than equation
(i), although the change is so slight.
EQUATIONS
Example 5. Convert 80, 15, 120, and 48 C. to degrees F.
When C = 80, F = (% x 80) + 32 = 176, and so on.
Or, we might tabulate, for the four readings given, thus :
35
C
lC+33
F
80
144 + 32
176
15
27+32
59
I2O
216+ 32
248
48
864 + 32
1184
Example 6. Ohm's law states that the drop in electrical pressure
E when a current C flows through a resistance R, is given by the formula
E = CR. Transpose this for R and C.
To find C,
Transposing the factor R,
In like manner
* f*
= CR
E
R'
E
C'
Brackets occurring in equations must be removed before
applying the rules of transposition, and the same remark applies
to fractions, which may always be regarded as brackets written in
a different form.
Example 7. Solve for w in the equation
Removing brackets
3W4I
Dissociating knowns and unknowns
w^yjo = 9'3
_ i. 3 w = 274
Note that the sign of 13 is kept unchanged.
Example 8. When finding the latent heat L of steam, the following
equation was used
Transpose this for L, i. e., find an expression for L in terms of the
other letters, which must be regarded as representing known quantities.
36 MATHEMATICS FOR ENGINEERS
Here L is the unknown, since the values of all the other letters are
supposed to be known.
Clearing of brackets, wTwt = qL+^T
Transposing terms, wT wt ^pT = qL
r~ wTwtti + T
Transposing the factor q,  = L
Example 9. Solve the equation
4* 7* 81 _ igx , 721
5~2 + T : ~5 3~'
The L.C.M. of 5, 2, 4 and 3 is 60, and multiplication throughout
by this figure will remove the denominators.
(4* x 12) (7* x 30) + (81 x 15) = (19* x 12) + (721 x 20)
48* 2io#+ 1215 = 228^+1442
48* 2IOX 228* = 14421215
or 1848* = 227
Example 10. The electromotive force E of a cell was found on
open circuit, and also the drop in potential V when a resistance of R
was placed in the circuit. The internal resistance of the cell may be
V
calculated from the equation (E V) = ~ x R where R< is the internal
resistance. Find the internal resistance for the case for which E = 134,
V = 8965 and R = 5.
V
It being required to find R< we transpose the ^ and treat the
bracketed letter as one quantity for the time being ; then
*(EV) = R,
which completes the transposition.
Substituting the numerical values
R<= ^siV 1 ' 34 "' 896 ^ =
= 247 ohms.
Example n. Solve the equation
, 5 _
~ 8
4 16 8
Before proceeding to find the L.C.M. it will be found the safest
plan to place brackets round the numerators of the fractions. This
EQUATIONS 37
emphasises the fact that the whole of each numerator is to be treated
as one quantity. Thus
(3y 5) (7^+9) , (Sy+ig) 69 _
rz r Q 1 cf "
4 lo o o
Failing this step, mistakes are almost certain to arise, especially
with signs, e. g., the minus before the second fraction applies equally
to the 9 and to the 7y. This fact would probably be overlooked if the
bracket were not inserted.
Multiplying throughout by 16, the L.C.M. of 4, 16 and 8
4(3y5)(7yl9)+2(8y+i9) + (2X 69) = o
i.e., I2y 20 7y 9+i6y+38+i38 = o
.*. I2y 7y+i6y = 20+938138
2iy = 147
v   T 47
y ~ ~^i
Example 12. If p is the intensity of pressure over an annular plate
of outside diameter D and inside diameter d, then the total pressure
on the plate is given by
Assuming that p, P and D are known, transpose this equation into
a form convenient for the calculation of the value of d.
Treating the 785^ as one quantity, and transposing it
 r
7854^
Transferring D 8 to the righthand side
P
Changing signs throughout
 
7854 p
Taking the square root of both sides
Example 13. If / = 2n /, giving the time in seconds of I swing
(periodic, or to and fro) of a simple pendulum of length / feet ; find^ an
expression for /.
It will be easiest in this case to square both sides (i. e. t to remove
the square root sign which is merely one form of bracket).
Then t z = ^ 2 
S
ef*
or, transposing the factors, 4, n 2 and g,  a = *.
38 MATHEMATICS FOR ENGINEERS
Example 14. Transpose for g, the dryness fraction of steam found
by the Barrus test for superheated steam, in the equation
4 8(T A T B w) = (ij)L+ 4 8(T s T).
T A , TB, Tg and T are temperatures, L is the latent heat of the
steam, and n = loss of temperature of the superheated steam when the
supply of moist steam is cut off.
Treating 48^3 T) as a term, it may be transferred to the other
side with change of sign before it
4 8(T A T B w). 4 8(T s T) = (ig)L
or, since 48 multiplies each bracket, we can take it outside one large
bracket
4 8{T A T B wT 8 + T} = (iff)L.
Dividing both sides by L
(iq) = ^{T A TBWT S +T}
q = I
Example 15. The equation  ^ = W(H+e) refers to the stress
produced in a bar by a weight W falling through a height H on to the
bar. Transpose this equation for / and also for e.
To find/:
2E
Transposing factors, /* = W x ^r (H+ e)
Extracting the square root of both sides of the equation.
/ 2 EW(H+e)
J ** V AL
To find e
2EW
/ 2AL
2EW
Example 16. One hundred electric glow lamps, each of 150 ohms
resistance and each requiring 75 ampere, are connected in parallel.
How many cells, each of 0052 ohm resistance and giving 208 volts,
will be required to light these lamps ? (Cells to be in series.)
Total resistance = Internal resistance + external resistance.
External resistance =  = i<5 ohms (because lamps in parallel
100
offer less resistance, i. e., an easier path is made for the current).
EQUATIONS 39
Suppose x cells are required
Total E.M.F. = #x 208
Total internal resistance = x x 0052
Total resistance = 0052*+ 15
Current = * M *' '
Resistance
and 100 x 75 = 
0052* + 15*
Multiplying across, i. e., multiplying throughout by the common
denominator 0052*+ 15.
75 (0052*+ 15) = zo8x
*39# + 1125 = 208*
1125 = 2o8x '39,v = 169,1;
.. x = 666
Or 67 cells would suffice.
Exercises 5. On Simple Equations and Transpositions.
Solve the equations in Exs. i to 6
1 5*+7(*~2) = 34(*+ 6 )
o I , 2 2fl
2. g+ 3 = 5
3 4'2ft = 958
' 745 469
. .
52
y 3^5 375
=
108 295 "*" 9n
6. 82^ 475(3 2x) + 214(5^ +7) = 17 (i 8^) + 5'43
7. Transpose for c in the equation  =
7 5c
8. If H = ws(T t), find an expression for T.
9. If P = CTAE, find E when A = 1925, C = 000006, T = 442,
P = 1,532,000.
10. Transpose for L, the latent heat of steam, in the equation
w i(ti~ T+L) = w(T t), and hence find its value when Wi = *, / x = 212,
w = i J, T = 145, and t = 70.
11. A formula occurring in connection with Tacheometric Surveying
fS
is D =+/+(?. Determine the value of S to satisfy this when
D = 3600, /= 12, d 6 and S = 36.
12. Using the equation in Exercise n, find the value of /to satisfy
it when = 3107, S = 4'63, 8 =015, and ^=5.
13. If w = ^^V find S when w = 8 ' 15 ' l = 5 ' W ** 83 ' 5> d = 4> and
c = 1400. w is the weight of a girder in tons to carry an external load
40 MATHEMATICS FOR ENGINEERS
W tons, d is the effective depth of the girder in feet, s is the shearing
stress in tons per sq. in., and c is a coefficient depending on the type of
girder.
14. If TM i\ ) = 7^, find E in terms of C for the case when m = 4.
HA m/ L/
In other words, find the relation between Young's modulus and the
Rigidity modulus when " Poisson's ratio " is 4.
V
15. Find the value of R, from E V= ~ xRj when E= 1364, V =
979, R = 5 The letters have the same meanings as in Example 10,
page 36.
16. Given that A=   1^  , transpose for R and hence find
its value when A = 35, D = 65, d= 47, a= 25.
17. The equation  ^ ^ = (i8*+2x6^x^)X5X4& occurred
when finding the thickness of the flange of the section of a girder for
an overhead railway. Find the value of t to satisfy this.
18. Transpose for q in the equation W(A, A x ) = w(qL+h h t ). [q is
the dryness fraction of a sample of steam.]
19. How many electric cells, each having an internal resistance of
i '8 ohms, and each giving 2 volts, must be connected up in series so
that a current of 686 amperes may be passed through an external
resistance of 122 ohms?
SC
20. If D= +K, find n when D = 500, S = 12, C = 950, and K= 15.
21. The tractive pull P that a twocylinder locomotive can exert
is given by
8pd*L
~D~
where p = steam pressure in Ibs. per sq. in., d = diameter of cylinders
in ins., L = stroke in ins., and D = diameter of driving wheels in
inches.
Find the diameter of the cylinders of the engine for which the pull
is 19,000 Ibs., the steam pressure 200 Ibs. per sq. in., the stroke 2 x 3",
and driving wheels are 4'6" in diameter.
22. To determine the diameter of a crank the following rule is used
P/ n
Put this equation in a form convenient for the calculation of the
value of d.
23. Lloyd's rule for the strength of girders supporting the top of
chH
the combustion chamber of a boiler is P = 7vfr T\TVF where P =
(W pjDL,
working pressure in Ibs. per sq. in.; t= thickness of girder at the
centre ; L = width between tube plates ; p = pitch of stays ; h = depth
of girder at the centre ; and D = distance from centre to centre of the
girders.
Find the value of p when c= 825, W= 27, L= 2j, D= 7^, / = i J,
h= 61, and P= 160.
EQUATIONS 41
24. Find the thickness of metal t (ins.) for Morrison's furnace tube
from each of the given formulae
(a) Board of Trade rule
p _ 14000*
D
(6) Lloyd's rule
p _ 1259(1612)
D
where P= pressure in Ibs. per sq. in., and D = diameter (ins.) outside
corrugations. Given that P= 160 and D = 43*.
25. (a) Transpose the given equation for A
where 8 = proof strain of iron, a = area of section of bar of length /
on to which a weight W is dropped from a height h inches ; A being
the extension produced.
(b) Find the value of /, which equals ?, when =30x10*;
a= i'2, 3 = ooi, h 132, and W= 40.
26. If / = \ \ i T ' find the value of L when W = 7000, A = 8000,
and t= 162.
27. Find the pitch p of the rivets in a singleriveted lap joint from
7 8 54 d 2 /, = (p  d)tf t where d = t + fa t= J, /, = 23, and f t = 28.
28. Calculate the value of p to satisfy the equation
B= CVpA. when C= 02, A= 200, B= 253.
29. The stress / in the material of a cylinder for a steamengine
may be found from t~, + \ where p = steam pressure = 80 Ibs.
per sq. in., d = diameter = 14", and t= thickness of metal = ". Find
/for this case.
30. Determine the value of p to satisfy the equation
(pd)tf t = i'57irf 2 /,, relating to riveted joints,
when/, = 23, ft =28, d= i, and /= f.
31. The diameter of shaft to transmit a torque T when the stress
allowable is / is found from T = ^/^ 3  Find the diameter of shaft to
transmit a torque of 22,000 Ibs. ft., if the maximum permissible stress
in the material is 5000 Ibs. per sq. in. (IT = 3142).
32. The formula d V/ , / r~\ occurs in reinforced concrete
v xbc(i\x)
design. Find M (a bending moment) when b g,c= 600, x 36, ^=153.
33. D = dY is Lamp's formula for thick cylinders of outside
JP
diameter D and inside diameter d. Calculate the value of p when
D= 95*, d= 6", and /= 6 tons per sq. in.
M E
34. An important formula in structural work is y = R where M
is the bending moment applied to a beam, I is the moment of inertia
of the section of the beam, E is Young's modulus for the beam, and
42 MATHEMATICS FOR ENGINEERS
R is the radius of curvature of the bent beam. If M= 5600 Ibs. ft.,
I =7854 in.* units, E =28x10* Ibs. per sq. in.; find the value of
R, stating clearly the units in which it is expressed.
35. Compare the deflection d m of a beam due to bending moment
with that d, due to shear, for the following cases
(a) length = 10 x depth, i. e. t I = lod.
b) length = 3 X depth.
You are given that
, W/ 3 i 5 W/ .. d* , _ _
d m = Q ~. , ,, d, = \x, ft , andE = 2'5U.
48EAA 2 4AC 12
oo rr mc k 2tr ,.
36. If =  r and c = r, find an expression for r in terms of m
I I n R
and k : hence find its value when = 36, m = 15.
37. If E = 3K(i) and E = 2C(i+i), find the relation between
K, the bulk modulus, and C, the rigidity modulus.
Find also an expression for E, Young's modulus, in terms of K and
C only.
38. Find an expression for x from the equation
i6W* _ i6W(rx) 2W
ird 3 Trd 3 + ird*
39. Find the internal pressure p for a thick cylinder from Lamp's
formula
where D= 1274*. d = 9* /= 2IO Ibs. /D". State the units in which
P is expressed.
n T T
40. Given that W=  pv ~ *, a formula occurring in Ther
** X J. i
modynamics, and also that \V= 33000, T, is fTj, T!= 2190, v= 124,
and p =2160, find the value of n.
41. A takes 2 hours longer than B to travel 60 miles ; but if he trebles
his pace he takes 2 hours less than B. Find their rates of walking.
42. If H=4=^r and /= 001(1 + ^,), find v when H = 22i, d= i,
2gd I2d'
1= 380, and g= 32.
(H is the head lost when water flows through a length / of pipe of
diameter d, and / is the coefficient of resistance.)
43. If M = moment of a magnet, H = strength of the earth's
field, p = time of a complete oscillation of the magnet, and I = moment
M d 3 T
of inertia of the magnet, then Q = (expressing the result of a
deflection experiment, d being the distance between the centre of the
magnet and that of the needle, and T being a measure of the deflection)
47T 2 I
and also MH = 3L (expressing the result of an oscillation experiment).
Find the values of M and H when ^=20, I = 169, p = 133, TT = 3142,
and T= 325.
44. In finding the swing radius k (ins.) of a connecting rod, the
following measurements were made :
EQUATIONS 43
/ = lime of a complete oscillation = 203 sees.
p = distance of centre of gravity from the centre of suspension =
3I43"
If h = distance of centre of percussion from centre of suspension
t=27r*/; and also k z = ph.
Find k in inches {n = 3142, g= 322 f.p. sec. 2 }.
45. The maximum stress in a connecting rod can be found from the
, D*p , vW
equation /= i 05^ + 00429 ^
If /= 4700, D = diameter of cylinder = 14, d= diameter of rod = 25,
p = steam pressure at mid stroke = 65, v = velocity of crank, r = crank
radius = 8, and / = length of connecting rod = 60, find the value of v.
46. It is required to find the diameter D of one pipe of length L,
equivalent to pipes of length /j and l z and diameters d^ and d t respec
tively, from
I, I, J,
D 5 ~ d? + J?
Put this equation in a form suitable for this calculation.
/' \"
/ i
47. If   = I ^j find an expression for y.
_ y o^
48. Transpose the equation ,  = , 2 , ; occurring in
n y 2/t \~2Cn C
structural design, to give an expression for y.
Simultaneous Equations. So long as only one of the quanti
ties with which we are dealing is unknown, one equation, or one
statement of equality, is sufficient to determine its value.
Cases often present themselves in which two, and in rarer cases
three or even more, quantities are unknown ; then the equations
formed from the conditions are termed simultaneous equations.
Taking the more common case of two unknowns, one equation
would not determine absolutely the value of either, but would
simply connect the two, i. e., would give the value of one in terms
of the other. For two unknowns we must have two sets of con
ditions or two equations. This rule holds throughout, that for
complete solution there must be as many equations as there are unknowns.
The treatment of such equations will be best understood by the
aid of worked examples.
Example 17. What two numbers add up to 54 and differ by 26?
For shortness, take x and y to represent the numbers ; substituting
these to form an equation to satisfy the first condition
5'4 ............ ( J )
44 MATHEMATICS FOR ENGINEERS
Here, by taking various values of y we could calculate corresponding
values of x, and there would be no limit to the number of " solutions."
The first statement in the question is, however, qualified by the second,
from which we form equation (2), viz.
x y=26 (2)
If equations (i) and (2) are added
2X = 80
.'. *=4;
or, in other words, y has been eliminated, i. e., the number of unknowns
has been reduced by one. Our plan must therefore be to " eliminate,''
by some means, one unknown at a time until all become " knowns."
This method will be followed in all cases.
Reverting to our example, x is found, but y is still unknown.
To find y, substitute the value found for x in either equation (i)
or equation (2).
In (i) 4+y = 54
and y = 5'4~4 1*4
x = 40 \
y = i'4J
and we have completely solved our problem.
Example 18. Determine values of a and 6 to satisfy the equations
40 + 36 =43 (i)
3 26 =11 (2)
If equations (i) and (2), as they stand, were either added or sub
tracted, both a and b would remain, so that we should be no nearer a
solution. To eliminate a, say, we must make the coefficients of a the
same in both lines.
E. g., if equation (i) be multiplied by 3
and equation (2) be multiplied by 4, each line would contain I2a,
so that the subtraction of the equations would cause a to vanish.
Thus 120+96 = 129
I2 86 = 44
Subtracting 176 = 85
whence 6=5.
Substituting this value for b in equation (2)
3a 10 = ii
3 = 21
a = 7
Grouping the results
Note. If it were desired to eliminate 6, equation (i) would have to
be multiplied by 2 and equation (2) by 3, and the resulting equations
added, since there would then be +66 in the top line and 66 below,
which on addition would cancel one another.
EQUATIONS 45
Example 19. The effort E, to raise a weight W, by means of a
screw jack, is given by the general formula, E=aW+6. If =25
when W= 5; and if =55 when W= 20, find the values of a and
b, and thence the particular equation connecting E and W.
Substituting the numerical values for E and W
25= 5 a +*> ........... (i)
In this case it is easier to subtract straight away ; thus eliminating b.
Thus 3= i 5 a
or
Substituting in equation (i), 25 = i + b
so that E= 2W+I5.
Example 20. Keeping the length of an electric arc constant and
varying the resistance of the circuit, the values of the volts V and
amperes A were taken. These are connected by the general equation
_ n
Find the value of m and n for the following case
V = 545 when A = 4 ~\
V = 488 when A = ioj
Substituting the numerical values, in the general equation
, n
545 = w+
488 = m+^
Changing the fractions into decimals to simplify the calculation
545 =w+*25n (i)
488 = m+'in (2)
Subtracting 57 = 15^
M = f^ = 38
Substituting this value in equation (2)
488 = w+38
m = 45
or V
Example 21. Karmarsch's rule states that the total strength Pof
a wire in Ibs. is given by P= ad+bd z , where d is the diameter in inches.
For copper (unannealed)
P=42i when d=
P= 55212 when d= i
Find the actual law connecting P and d.
46 MATHEMATICS FOR ENGINEERS
By substitution of the numerical values
55212= (ax 12)+ (6 xi 44) (i)
421= (ax i)+(6x 01) (2)
To eliminate a multiply equation (2) by 12 and subtract.
Thus 55212 = i2a+i'44&
5052 = i2a+ 126.
Subtracting 50160 = 1326
6 = 5i6o
132
Substituting in equation (2)
421 = ia+38o
ia = 41
/. a 410.
P= 410^ + 380000?
i. e., for a diameter of 5", the total strength is
(410 x 5) + (38000 x 25) = 9705 Ibs.
Solution of Equations involving three unknowns. These
may also be solved by the process of elimination, the method being
similar to that employed when there are two unknowns only. Three
equations are necessary and these may be taken together in pairs,
the same quantity being eliminated from each pair, whence the
question resolves itself into a problem having two equations and
two unknowns.
Example 22. Find the values of a, b and c to satisfy the
equations
4~5 6 +7 c = 14 (i)
90+2&+3C = 47 (2)
aby = ii (3)
The unknowns must be eliminated one at a time. Suppose we
decide to commence with the elimination of c. This may be done by
taking equation (i) and equation (2) together, multiplying equation (i)
by 3 and equation (2) by 7, and then subtracting; an equation con
taining a and 6 only being thus obtained. For complete solution one
other equation must be found to combine with this ; if equation (2)
and equation (3) are taken together, equation (2) must be multiplied
by 5 and equation (3) by 3 and the resulting equations then added.
Hence, considering equations (i) and (2), and multiplying according
to our scheme
I2a I5&+2IC = 42
63a+i4&+2ic = 329.
Subtracting 5ia29& = 371 (4)
EQUATIONS
47
Combining equations (2) and (3), multiplying equation (2) by 5
and equation (3) by 3
45a+io& + i5C = 235
3a 3615^ = 33
Adding 480+ jb =268 .......... ( 5 )
Equations (4) and (5) may now be combined and either a or b
eliminated.
To eliminate a, multiply equation (4) by 16 and equation (5) by 17
and add.
Then
Adding
8i6a 4646 = 5936
8i6a+ii9& = 4556
345& = 1380
6= 4
Substitute this value of 6 in equation (5) and the value for a is
found
i. e., 480 + 28 = 268
or 480 = 240
For a write 5, and for b write 4, in equation (2).
Then 45 + 8+35 = 47
or
Collecting the results
Example 23. A law is required, in the form E = a+6T+cT 2 ,
for the calibration of a thermoelectric couple. The corresponding
values of E and T are
T (C.)
IOO
600
1000
E (microvolts)
45
3900
5600
In other words, we wish to find the values of the three unknowns,
a, b, and c.
The three equations formed from the given values are
5600= a+iooo&+ioooooos (i)
3900 = a+ 6oo&+ 3600005 (2)
450 = a+ 1006+ looooc (3)
Grouping equations (i) and (2) and subtracting, a is eliminated;
and similarly for equations (2) and (3).
48 MATHEMATICS FOR ENGINEERS
Thus 5600 = a + 10006 + looooooc
3900 = a+ 6006 f 3600006
.*. 1700= 4006+ 6400006 ........ (4)
Also 3900=0+ 6006+ 3600006
450 = a+ 1006+ 100006
345= 5006+ 3500006 ........ ( 5 )
To eliminate b, multiply equation (4) by 5 and equation (5) by 4,
and subtract.
Then 8500 = 20006 +32000006
13800 =20006+14000006
5300 = 18000006
5300
c = o^  = 00294
1800000
Substituting in equation (4)
1700 = 40061884
or 4006 = 3584
b = 896.
Substituting for b and c in equation (3)
450 = a+8g6 29
a = 417.
Hence the law of calibration is
E = 4I7+896TOQ294T*.
Exercises 6. On Solution of Simultaneous Equations.
Solve the equations in Exercises i to 9.
1. ?x+3V = i 2 * 2a ~ 9& = 3 2
35* 6y = i 3a+io6 = i
3. sm6n = 66 4. 48* 27^ = 48
nn 25 = 2m y$ix = 51
5. y+i37=4* 6  '\x\V = &*+&&
gxijy =4987 19^+27 = 268
4Xyy_ , 8. 2a+ 3 6+ 5C=4 5
' ' 3C 7a+ 156 = 627
9. 2p 55+4^ = 33 10. If E= a+6*+c/ 2 , and also
P ' for
3 PI2* + 2S = 8 9 values
we
have
E
t
46
10
45
find the values of a, b, and c.
11. If P= ad} bd z and P= 17830 when d= 5)
P= 2992 when d= 2)
find the values of a and 6.
(P and d have the same meanings as in Example 21, page 45.)
EQUATIONS
49
12. You are given the following corresponding values of the effort
E necessary to raise a load W on a machine. Find the connection
between E and W in the form E=aW+&, given that E=7 when
W 20 ; and E = 142 when W = 80.
13. Corresponding values of the volts and amperes (obtained in a
test on an electric arc) are
V = 4875 when A = 4 ; and V = 7575 when A = 8.
Find the law connecting V and A in the form V=
14. The I.H.P. (I) of an engine was found to be 319 when the
B.H.P. (B) was 2, and 605 when the B.H.P. was 5. Find the I.H.P.
when the B.H.P. is 37. {I = aB + 6.}
15. The law connecting the extension of a specimen with the gauge
length may be expressed in the form, e = a + feL, where L = length
and e extension on that length.
The extension on 6" was found to be 2062", and that on 8" was
2444*. Find the values of the constants a and b.
16. The electrical resistance R< of a conductor at temperature t
may be found from R t == R (i + at) where R = resistance at o, and
a = temperature coefficient,
If the resistance at 20 is 538 ohms and at 90 is 771 ohms, find
the resistance at o and also the temperature coefficient.
17. Find a simple law connecting the latent heat L with the tem
perature t when you are given that
L
975
800
t
200
450
Find also the latent heat at 212.
18. Unwin's law for the connection between the length, the area,
and the extension of a specimen is
,. c Varea . ,
Percentage elongation e = . T. + 6.
If the area a is 75 and e= 3011 when the length 1=5", and if
e= 256 when /== 8", find the law for this case (Mild steel specimen).
19. Repeat as for No. 18, when a= 212, and /= 3" when e= 592
and 10* when e =245 (Rolled brass specimen).
20. The difference in potential E between the hot and cold junction
of a thermal couple for a difference of temperatures T is given by
E = a + 6T + cT 2 .
Find the law connecting E and T for the values
T
5
IOO
3OO
E
2O22
5701
2058
21. /is the tenacity (in tons per sq. in) of copper at t F. /and t
are connected by an equation of the form/= a b(t 6o) 2 . Find this
equation, given that /= 148 at 60 F. and /= 132 at 400 F.
E
50
MATHEMATICS FOR ENGINEERS
22. Repeat as for No. 21, the values of /and / (for cast phosphor
bronze) being
/
1606 131
1
t
IOO
400
23. Given that W=
Find the law connecting W and p
if W= 2111 when p= 80; and also W= 1656 when p= 126. W is
the weight of water used by a steam engine per H.P. hour, and p
is the absolute pressure.
24. If w = steam per H.P. hour and I = H.P., then
If 12000 Ibs. of steam were used per hour when the H.P. was 1000
and 3554 Ibs. when the H.P. was 180, find the law connecting w and I.
25. 500 cu. ins. of cast iron together with 240 cu. ins. of copper
weigh 2068 Ibs., whilst 13 cu. ins. of copper weigh as much as 16 cu.
ins. of cast iron. Find the number of cubic inches per ton of each
of these metals.
26. Measurements to find the constants of a telescope with stadia
wires resulted in the following. At i chain distance from the instru
ment the difference between the readings on the staff for the top and
bottom wires was 65 ft.; and at 2 chains the difference was 1311 ft.
Find the constants, C and K from CSfK=D where S = difference of
staff readings and D = distance. (i chain = 22 yds.)
27. Three wires A, B, and C are successively looped together and the
resistance of each loop measured. The resistance of A and B is found
to be 260 ohms, of A and C is 280 ohms, and of B and C is 300 ohms.
Determine the individual resistances of A, B, and C.
28. The following equations occurred when finding the fixing couples
of a builtin girder
=7625
200 , 400
*i+ m s = 7186.
o 3
Solve these equations for m^ and m t .
29. The " dead weight " tonnage of a ship is 700 tons, whilst the
cubic capacity of its hold is 42000 cu. ft. To ensure the most profitable
voyage, a mixed cargo of heavy and lighter goods must be carried,
and the complete capacity of the hold must be utilised. Prove the
truth of the following rule : " To obtain the weight of the lighter cargo,
multiply the specific volume (i. e., the number of cu. ft. per ton) of
the heavy cargo by the dead weight tonnage. Subtract this result
from the total cubic capacity and divide the difference by the difference
between the specific volumes of the heavy and light goods."
If, in a certain case, the densities of the heavy and light goods
are 35 cu. ft. per ton (saltpetre), and 80 cu. ft. per ton (ginger in bags)
respectively, determine the weight of saltpetre carried and also the
weight of the ginger.
EQUATIONS 5I
Methods of Factorisation. Reference has already been made
to the word " factor " as denoting a number or symbol that multi
plies or divides some other numbers or symbols in an expression.
Thus 3x5 = 15, and 3 and 5 are called factors of 15, i.e., when
multiplied together their product is 15.
Again 260? = 2Xi3XaXaXa.
Here the quantity has been broken up into 5 factors. The
process of breaking up a number or expression into the simple
quantities, which, when multiplied together, reproduce the original,
is known as factorisation. Little is said about this in works on
Arithmetic, but the process is used none the less for that.
To illustrate by a numerical example
Find the L.C.M. of 18, 24, 15, and 28.
These numbers could be factorised and written as follows
2x3x3, 2x2x2x3, 3x5, 2x2x7.
The L.C.M. must contain each of these; it must, therefore,
contain the first, any factor in the second not already included,
and so on for the four.
2x3x3 2x2 5 7
*. e., L.C.M. = , x ^^ x ^^ X ^^ = 2520.
ist 2nd 3rd 4th
The necessity for the presence of the two 2's in the second
group should be realised. There must be as many 2's as factors
in the result as there are 2's in the number having the greatest
quantity of 2's in its factors : i. e., there must here be three 2's
as factors in the result.
It is, however, in Algebra that this process finds its widest
application. Rather difficult equations can often be put into
simpler forms from which the solution can be readily obtained,
and by its use much arithmetical labour can be saved. Generally
speaking, the factorised form of an expression demonstrates its
nature and properties rather more clearly than does its original
form. For practical purposes the following methods of factorisation
will be found sufficient.
Rule i. Often every term of an expression contains a common
factor : this factor can be taken out beforehand and put outside a bracket.
The multiplication is then done once instead of many times.
35+6055 is, we know = 40
But 35+6055 = (5X7) + (5Xi2)(5Xii)
and the factor 5 is common to each term. If this factor is taken
outside a bracket, the arrangement then becomes 5(7+1211),
52 MATHEMATICS FOR ENGINEERS
or 5 x 8 = 40, which agrees with the previous result. The final
arrangement is to be preferred, because the numbers with which
we have to deal are much simpler. Hence for this numerical
case we see that the common factor must be taken outside a
bracket, ^vhilst the terms inside are the quotients of this factor
derived from the original terms.
Numbers have been taken for clearness of demonstration, but
the method holds equally well for symbols of all kinds.
Example 24. Factorise the expression, 7a 4 fc 2 28a 3 6c 2 + 42a 6 6 3 c 4 .
In this expression, 7 is common to each, term, a 3 is the highest
power of a common to each term, 6 the highest power of b, whilst no
c occurs in the first term, and c is, therefore, not a factor common to
all terms.
Then, the factor to be taken outside a bracket = ja?b.
Hence the expression = ya 3 b(ab~ 4C 2 +6a 3 & 2 c 4 ), or we have broken
it up into two factors.
Example 25. Find the volume of a hollow cylindrical column,
12 ft. long, i ft. external radius, and 9 ins. internal radius, from the
formula
Volume of a cylinder = irr z l (r= 3142)
In this case the net volume will be the difference between the
volumes of the outside and inside cylinders
.*. V= (ffXi 2 xi2) (7rx(J) 2 xi2) ......... working in feet.
= i27r{i 2 () 2 } because i2?r is a factor common to both terms.
= 16*48 cu. ft.
Rule 2. The expression may be of a form similar to one whose
factors are known, and the factors may be written down from inspection.
If (A+B) be multiplied by (A B) the resulting product is
A 2 B 2 .
Conversely, then, the factors of A 2 B 2 are (A B) and (A+B),
or A 2 B 2 = (AB)(A+B),
i.e., to factorise the difference of two squares, multiply the sum of the
quantities by their difference. This rule is of wide application.
Example 26. Write down the value of 9154*
Squaring each and subtracting the results is far longer than making
use of the rule just given
Thus 9i54 2 9i5 l2 = (9154+ 9i5 I )(9i54 9I5 1 )
= 18305 x 3 = 54915
EQUATIONS 53
Example 27. Find the factors of 8i 8 166*.
8ia 8 166* = (9 4 ) 2 (4& 2 ) 2 , which is the difference of two squares,
and therefore = (9 4 4& 2 )(9 4 f 4& 2 )
In this example the rule is applied twice.
Two other standard forms are here added, although their use
is by no means so frequent as that of the above.
A 3 B 3 = (AB)(A 2 +AB+B 2 )
A 3 +B 3 = (A+B)(A 2 AB+B 2 ).
Example 28. Find the factors of 2ja*b 3 + i25 3 c 9 .
Let E denote the expression, then
E= a 3 (27a 3 6 3 + 1256') by Rule i.
Rule 3. In many cases of trinomial, i. e., threeterm expressions,
the factors must be found by trial, at any rate to a very large extent.
There are certain rules applying to the signs, which can best be
followed by first considering the following products :
= * 2 +ii*+3o .......... (i)
(*5)(* 6) = * 2 n*+3o .......... (2)
(x5)(x+6) = x*+x3o ........... (4)
In (i) and (2) there are like signs in the brackets and a plus sign
precedes the third term in the expansion, which must be written in
the order of ascending or descending powers of x or its equivalent.
In (3) and (4) there are unlike signs in the brackets and a mintts
sign comes before the 30. Hence the first rule of signs may be
stated : So arrange the signs that the one before the first term is
plus, an adjustment of signs throughout being made if necessary.
Look to the sign before the third term of the expression ; if this
is a plus then we conclude that the signs in the brackets will be
like, and if this sign is a minus then the signs in the brackets will be
unlike. If they are to be like, they must be either both plus or both
minus, and the sign before the second term in the given expression
indicates which of these is accepted. Thus, a plus sign before the
second term indicates that the signs in the brackets are both plus.
54 MATHEMATICS FOR ENGINEERS
If, however, the signs in the brackets are to be unlike, one product
must be the greater and the sign before the second term indicates
whether it is the product obtained by using the plus or the minus
sign.
E. g., in (3) we have 30 as the third term ; accordingly the
signs in the brackets will be unlike : also the second term is x
so that the minus product is to be the greater; hence the minus
sign in the brackets must be before the 6.
The actual numbers in the brackets must be found by trial.
They must in each of the four instances multiply together to give
30 ; also, in (i) and (2) they must add together to give n, and in
(3) and (4) their difference must be i.
Example 29. Find the factors of # 2 + ijx no.
In the given expression the third term is no, so that there must
be unlike signs in the brackets. Also, the + product must be the
greater, since +17* is the second term.
Since the signs in the brackets are to be unlike, two numbers must
be found which when multiplied together give no, and which differ
by 17.
These numbers are 5 and 22 ; and the signs placed before these
must be so chosen that + ijx results when the brackets are removed.
Thus the plus sign must be placed before the 22, and hence
x z + ijx no = (x f 22) (x 5).
Example 30. Factorise the expression 2X 2 z8x go.
Applying Rule i
The expression = 2(# 2 + 14*+ 45).
(Note the adjustment of signs, to ensure + before the first term.)
Dealing with the part of the expression in brackets : plus signs
throughout denote f in brackets ; hence two numbers are required
that multiplied give 45, and added give 14; these being 9 and 5.
/. The factors = z(x+ 9)(*+ 5).
Example 31. Find the factors of 6m 2 + nm 35.
This expression could be reduced to the form of the previous examples
by dividing by 6, but the fractions so obtained would render the further
working rather involved. It is better, therefore, to proceed as follows :
There will be unlike signs in the brackets, since the sign before the
third term is minus, and the factors of 6 have to be combined with
those of 35 to give a difference of products of +n. The varying of
the factors at either end may result in many arrangements being tried
EQUATIONS 55
before the correct one is found. After a little practice, however, the
student disregards absurd arrangements and so reduces his work.
The correct arrangement in this case is (yn 5)(2nt+j).
The first terms when multiplied together give 6m 2 , the last ones
give 35, the extreme terms give +2im, and the middle terms lorn,
i. e., the last two combine to give +nw.
The arrangement is more clearly shown if written down as
(3xx5)
The end terms are easily settled, but for the middle term the multi
plication must be performed as indicated by the arrows, and the results
must be added or subtracted as the case may demand. When the
correct arrangement of the figures has been found, the letters must
be inserted. Hence, the expression has for its factors (3m 5)(2m+7).
Example 32. Factorise the expression 72a 2 + i8a6 776*.
In the first place disregard the letters; dealing only with the
numbers.
The factors of 72 are to be combined with those of 77 to give a
difference of 18.
72 has many factors, but 77 = 7x11 or 77x1.
The trial arrangements would be of this nature
ox^T For the middle term, the difference = 43.
135
faXX H ., = '8.
The last is the arrangement desired. To allocate the signs: the
net result of the products is to be +18 : 7x12 gives the greater
product, hence the + must be placed before the 7.
.'. The expression = (6a + 7b)(i2a lib).
The Remainder and Factor Theorems. Suppose we
have to deal with an expression such as
If this expression be divided by (xa), the remainder will be
56 MATHEMATICS FOR ENGINEERS
which could have been more simply obtained by substituting a for
x in the original expression.
If (x a) is to be a factor of the original expression then the
remainder after division by (x a] must be zero. Hence we obtain
a rule enabling us to find factors of rather complicated expressions
Find the value of the main quantity (usually the x) which makes the
suggested factor zero ; substitute this value in place of the x in the
expression, and if the result is zero one factor has been found.
E. g~, if it be conjectured that (#(3) is a factor of an expression,
its value would be found when x had the value 3.
Example 33. Find the factors of X s + x z 14* 24.
Let us try if (#4) is a factor; we will substitute, therefore, + 4
for x in the expression, which becomes
( 4 )3 + (4)214(4) 24 = 64+165624 = o
.*. (x 4) is a factor.
Another likely factor would be (#+3), for 3x4 is part of 24, and
there must be a plus sign to combine with the minus in (#4) to
give 24.
Substitute 3 for x, and the expression becomes
(3) s + (3) 2 i4(3)24 = 27+9+4224 = o
.*. ( x + 3) is a factor.
The other factor may be found to be (#+2)
/. x 3 + x z  14* 24 = (x + 2}(x + 3)(x 4).
Multiplication and Division of Algebraic Fractions.
The simplification of algebraic fractions furnishes useful examples
on the appli cation of the rules of indices and of factorisation.
When a number of fractions are to be multiplied together, cancelling
can be performed as in the case of arithmetic fractions, always provided
that the complete factors are cancelled and not portions thereof.
E. ., r^ is in its lowest terms ; we cannot cancel 2X into
' 4*+3
4* or strike out the 3's, because (2^+3) must be treated as one
quantity, as also must (4*43).
.. c . ,.,
Example 34. Simplify ~ .r^. x p , .
* ja*b z c 5 i8a 3 c 4
48a 3 fcc 2 3
The fraction =  Tr*. x 
alb
EQUATIONS 57
Example 3 5 .Simplify *+*+ 15 x * I 4 7_
y 2* 2 + 3* 35 A 20*2 + 28* 96
No cancelling must be made until numerators and denominators
are expressed in terms of their factors. Thus the fraction _
= (*+3)(*+5) x 3 (2* 7) (2 x + 7)
( 2 *7)(*+ 5 ) 4(*+ 3 j ( 5 * 8)
and in this fraction cancels with
cancels with
cancels with
giving the answer j!J^rf j in which no further cancelling can occur.
Example 3 6.Simplify 4* 2 +*~i4 x

 
6xyi 4 y A' 2  4 4* 7 ' 3 #2*i 4
The numerators and denominators are first factorised giving the
fraction in the form
( 4 *7)(*+g) 4*2 (x2) (3* 
which by cancelling reduces to 
2 v3 . * T v __ /t^2 [_ *jf\
Example 37. Simplify the fraction ~ ~^L
^x + nx 20
The factors for the denominator are the more easily found ; they
are (#+5) and (3* 4). The first of these is a possible factor of the
numerator also ; applying the remainder theorem, the value of the
numerator when x 5 is 2( 125) 4i( 5) (5)2 + 70, i.e., o;
hence (#+5) is a factor. In like manner it would be found that
(x 2) was also a factor; and by division of the numerator by the
product of these, viz. by x*+ 3* 10, the remaining factor is found
to be (2X 7).
Hence the fraction^ (*M+5K**7) = <**)(**7)
~4/ \3 X ~4)
Addition and Subtraction of Algebraic Fractions. The
same rules are adopted as for arithmetical fractions.
The L.C.M. of the denominators (L.C.D.) must first be found by factor
ising the separate denominators according to the plan detailed on page 51.
Example 38. Simplify the fraction
_5_ ,_ioo_ 51
407 120 21 20fl 35
5 8 MATHEMATICS FOR ENGINEERS
This becomes (after factorisation of the denominators)
(47) + 3(47) ~~ 5(4 7)
and the L.C.D. = ($a 7) x 3 x 5 = 15(40 7)
whence the expression
Example 39. Simplify ey . ft +
15 12
This becomes (after factorisation of the denominators)
# 15 12
C*+3)(#+ 2 ) (#+7) (#+2) (#+7) (# + 3)
and the L.C.D. is (x + 3) (x + 2) (* + 7) .
Dealing with the first term only and multiplying both numerator
and denominator by this L.C.D.
which after cancelling reduces to
. + / + \(x+ 7)'
In like manner the second and third terms reduce to
i5(*+3) and
" d
7) + I 5( ;r + 3) I2(#+ 2)
Hence the fraction =  g+ 3 &+ 2)(+ 7)
45 "*  24
IPX + 21
(*+3)(*+7)
., a c ., a{b c 4 d
Example 40. Show that if r= j. then ^ = .  and
_
a b ~ c d
From ^ = ^, by adding i to each side
o a
a c ,
b +I= d +1
Taking the L.C.D. of each side
a+ b _ c+ d
EQUATIONS 59
In like manner by subtracting i from each side of the original
e< l uation  ^ = ^r ........... W
Hence, dividing (i) by (2)
q + b _ c+ d
ab ~ T^~d ........... 3
These results are of importance.
Exercises 7. On Factors, and on Multiplication and Addition
of Algebraic Fractions.
Factorise the expressions in Examples i to 20.
1. x 2 + iSx 88 2. x 2  igx + 88
3. x z 26x+ 105 4. 8a 3 i25& 6 5. 24** #44
6. (2fl+6) 2  (3046)2 7. a 2 + 4 a&4 5 & 2
8. I2# 2 73#y + 105^2 9. 88 3*2 13*
lOi 2om+ 2 on 5 8m 11. <^!_^ 3 + 5J^_
4 24 384 16
12. nR 3 $irr 3 , giving the volume of a hollow sphere of outside
radius R, and internal radius r.
13. 94^ 2 +39^963.
wlx 3 wx* wl 3 x
i^EI ~~ 2~EI ~ 2~El' an ex P resslon occurring in connection with
the deflection of beams.
15. 540*6 30oo 2 6c 2 42a 3 6c 16. 4# 2 i6c 2 1206 + 9& 2
17. 64C 5 + _
27
18. V v (giving the volume of the frustum of a cone; R and r
being the radii of the ends of the frustum and h its thickness) where
T , wR 2 /r , .\ Trr z k rh
V = (h + A), v =  and k = ^
3 3 K  y
19. 2# 3 + 7# 2 44*+ 35. {Hint. Try (x + 7) as a factor.]
20. 6p 3 + 2$p z \6p 35. [(p i) is one factor.]
21. Find, by the methods of this chapter, the value of (199 x 46)
+ (398x69) (199x92).
22. Find the value of nR 2 / irr z l, which gives the volume of a
hollow cylinder, when IT = 3142, R= 1272, ^=958, =643.
23. Find the L.C.M. of # 2 #6, 3^21^ + 36, and 4*2  8* 32.
24. Simplify
_ c . ,., 8^ 2 2AX 80
25. Simplify
2ox z + i^xno 4X 2 2xgo
26. Simplify + ^ ^ ^ ^
y X42X+4 x z 2X8
27. Simplify ^ + ^ + ^ + Y&^W^
6o MATHEMATICS FOR ENGINEERS
28. Simplify
_ 5* __ 8*+7* __ 14 8*
18** 100 + 30* 24** 4* 280 fo# 2 + 175 155*
29. Solve the equation _ =
30. Solve the equation I _3__ = __9__
[Hint. Multiply through by the L.C.D.]
31. A unit pole is attracted by a magnetic pole of strength m with
a force ^. t and repelled by a force of
What is the resultant attractive force ? Find the value of this force if
/ is very small compared with d.
32. Find the factors of (a) 3**+ 6x* 189* ; (b) 24 + 37* 72** ;
M (3* + 7X)'  (2*  37)*
33. Find the factors of (** j 7*+ 6)(** + 7*+ 12)  280.
34. M, a bending moment, is given by
M = Pfl ( 2S +3) _ Pa(i 85*4355 + 9)
8(5+2) " 24(65*+ 2 +135)
Find a more simple expression for M.
35. The expression pjt>i  (ptVPivJ p t v t relates to the
7* *~~ X
work done in the expansion of a gas. State this in a more simple form.
36. The depth of the centre of pressure of a rectangular plate, of
width h, immersed vertically in a liquid, the top being a and the bottom
f (&'*')
b units below the level of the surface of the liquid, is given by 4 
^(6* a*)
Express thU in a simpler form.
Quadratic Equations. Any equation in which the square,
but no higher power, of the unknown, occurs, is termed a quadratic
equation. The simplest type, or pure quadratic, is <f* = 25 ; to
solve which, take the square root of both sides. Then d = either
+5 or 5, because (f5) 2 = 25 and also ( 5) 2 = 25. This result
would be written hi the shorter form d = +5.
It is essential that the two solutions should be stated, although
in most practical cases the nature of the problem shows that the
positive solution is the one required.
The solution of the pure quadratic is elementary; but in the
case of an equation of the type Jt 2 +7*+i2 = o (spoken of as
an adfccted quadratic, i. e., one in which both the first and the
second power of the unknown occur) new rules must be developed
or stated. Three rules or methods of procedure are suggested for
the solution of adfected quadratics, viz.
EQUATIONS 61
Method 1. Solution of a Quadratic by Factorisation.
Group all the terms to the lefthand side and factorise the expression
so obtained. Next, let each of these factors in turn = : thus two
solutions are determined.
For all quadratics there must be two solutions or " roots " ; in
some cases they may be equal, and in rare cases " imaginary."
Applying this method to the example under notice :
Example 41. Solve the equation x*+ jx+ 12 = 0.
By factorisation of the lefthand side
Then either * f 3 = o, in which case x = 3,
or x f 4 = o, in which case x = 4,
because, if one factor is zero, the product of the two factors must also
be zero ; e. g., if *= 3, (x+ 3)(x+ 4) = ox i = o.
Hence x= 3 or 4.
Example 42. Solve the equation 24a a f 170 = 20.
Collecting terms, 240* + i ja 20 = o.
Factorising, (8a 5) (30 f 4) = o.
.'. either 83 5 = 0, i.e.,a= 1
or 30 f 4 = o, i. e., a =  J
If no factors can be readily seen we may proceed to
Method 2. Solution of a Quadratic by completion of the
Square.
All the terms containing the unknown must be grouped to one
side of the equation and the knowns or constants to the other
side.
The lefthand side, viz. that on which the unknown is placed,
is next made into a perfect square by a suitable addition, the same
amount being added also to the righthand side, and then the
square root of both sides is taken. The solution of the two simple
equations thus obtained gives the " roots " of the original equation.
Before proceeding further with this method a little preliminary
work is necessary, the principle of which must be grasped if the
reason of the method of solution is to be understood.
Suppose that the first two terms of the righthand side are
62 MATHEMATICS FOR ENGINEERS
given, and it is desired to add the necessary quantity to make it
into a complete square.
/A8\ 2
a 2 +48a+576 might be written as (a) 2 + [2 X (24) X (flVH
\ 2 /
so that if 2 +48a is given, the term to be added is ( ) , *' , is
\ 2 /
the square of half the coefficient of a.
/7\ 2
Similarly, # 2 f 7* could be expressed as a perfect square if (]
(7\
#+' )
2/
Returning to the method; a numerical example will best
illustrate the processes.
Example 43. Solve the equation x 2 + i$x 49 = 0.
Grouping terms x 1 + i$x = 9.
Adding the square of half the coefficient of x, viz. ( ) 'to each side,
or
Extracting the square root of both si
#4^= 688
.*. x= 75 688
=  7'5 + 688 or  75  683
= 62 or I4'38.
The change from X s + 15*4 ^ J to fjr+ ) often presents diffi
culty : the reason for the omission of the i$x does not seem clear.
It must be remembered that it is represented in the second form, for
~ 2 = (i st ) 2 + (2 nd ) 2 +2 (product) =
If the coefficient of x 2 is not unity it must be made so by division
throughout by its coefficient.
Example 44. Find a value of B (the breadth of a flange) to satisfy
the equation 364B 4 5I8B 2 900 = o.
This equation, though not a quadratic, may be treated as a quadratic
and solved first for B 2 ; i. e., if for B 2 we write A the equation becomes
364A 2 5I8A 900 = o.
EQUATIONS 63
Dividing through by 364 (the coefficient of A 2 ) and transferring
the constant term to the righthand side
A 2 I424A = 247.
The coefficient of A is 1424; half of this is 712, hence add 712*,
i.e., 508 to each side
i.e., A 2 I424A+ (7I2) 2 = 247+508 = 2978
or (A 7i2) 2 = 2978.
Extracting the square root throughout
A 712 = db 1726
i.e., A = 712 1726
whence A = 2438 or 1014.
Now A =B a , so that B 2 = 2438 or 1014. Of these values the
former only is taken, since we cannot extract the square root of a
negative quantity.
Thus B 2 = 2438 or B = 494, but evidently the negative solution
has no meaning in this case.
Example 45. If 4a 2 1506+ 2& 2 = o, find the values of a to satisfy
this equation.
Dividing through by 4 and transferring the constant term to the
T  2
righthand side, a 2  ab = 
4 2
The coefficient of a is b; half of this is b : hence we must add
4
(a) to each side.
Thus a*  * +  . + =
4 \ 8 / 2 04 04
Extracting the square root
= 3616 or 146.
Method 3. Solution of a Quadratic by the use of a
Formula.
It will be evident from the foregoing examples that all
quadratics reduce to the general form
Ax 2 + BX + C = 0.
If Method 2 is applied to the solution of this, the result is a
64 MATHEMATICS FOR ENGINEERS
formula giving the roots of any quadratic, provided that the par
ticular values of A, B and C are substituted in it. Thus
A* 2 + Ex + C = o
Dividing through by A and transposing the constant term
x* + B * = 
/ B\ 2
To each side add the square of half the coefficient of x, viz.
. B , /B\ 2 C . /B\ 2
IY& _L v _L I __
+ A + \2\) ' A + Uv
= C+B
A ' 4/
B \ 2 B 2  4AC
QJ I V _J \ _
_C ^ _B 2 4AC
A *1 i \ 2
Extracting the square root of both sides
2A 2A
, B VB 2  4AC
whence x =  ,
2A
Thus the " roots " of the general quadratic A* 2 + BA; + C = o
 B + VB 2  4AC ,  B  B 2  4AC
are  and
2A 2A
Example 46. Solve the equation $x z Sx = 12.
Collecting all the terms to one side, $x z 8x 12 = 0.
Then for this to be identical with the standard form
A = 5, B = 8, C = 12
+ 8 V6 4 + 240
= 8 V3Q4 ^ 8 174
10 10
= 254 or  94.
Great care must be exercised to avoid errors of sign. To obtain
the value of 4AC, first find AC= 5 x ( 12) or 60;
4AC then = 240
and 4AC = + 240.
EQUATIONS 65
Example 47. Solve for y, in ^y z i'$y 32 = o.
It is always advisable to have the first term positive, so change all
the signs before applying the formula.
Then 4>/ 2 +15?+ 32 = o.
Here A =4, 6=15, C=32.
i'5 V225 512
8
 15 ^1738
= 23 or 352.
Example 48. The stresses on the section of a beam due to the
loading are a normal stress / and a shearing stress q. These produce
an entirely normal stress / on a plane known as the plane of principal
stress. Find an expression for / from the equation /(//) = q*.
Removing the bracket and grouping the terms to one side
/ 2 //n<? 2 =0.
Here A=i, B= /, C=q*
f _ +fn V/n 2 +4g 2
The next example is instructive as showing the advantage of
resolving large or small numbers into integers multiplied by powers
of ten.
Example 49. Solve the equation L# 2 + ~Rx + T? = ( an equation
occurring in electrical work) when L= 00:1:5, R = 4OO, K='45XiO" 6 .
Substituting the numerical values for L, R, and K
'  = o.
The last term may be written in the more convenient form 222 x io 8
since ^ = 222 and ^4= io 8 
Thus oo 1 5# 2 4 400^+ (222 x io 6 ) = o
Comparing with the standard form
A = (15 x io 3 ), B= (4 x io 2 ), C= (222 X io)
66 MATHEMATICS FOR ENGINEERS
Hence x = ~ (4 x I{)2 ) V(r6 x *o 4 ) (6 x IP' 3 X 222 x io 6 )
3 x io 3
 (4 x io 2 ) V(i6 x io 4 ) (133 x io*)
3 x io~ 3
the second term under the radical sign being written in this form so
that io* is a factor common to both terms ; and the square root of io 4
is readily found.
[6 X IO 3 X 222 X IO 8 = 1332 X IO 3 = 1332 X IO 4 .]
The square root of io 4 is io 2 ; this may be placed outside the
radical sign, and then
 (4 X io 2 ) io 2 Vi6 133
3 x io 3
_ io 2 (4 VI46;)
3 x io
= (io 6 x 261) or io 5 x 053
= 261000 or 5600.
Example 50. A formula given by Prony (in connection with the
flow of water through channels) connecting the hydraulic gradient i
with the velocity v and the hydraulic mean depth m was of the form
mi = av + bv*. Under certain conditions a = 000044, & = 000094.
Show that this is in close agreement with the formula given by
Chezy, viz. v = 103 Vmi.
mi = av + bv z
or bv* + av mi = o
a Va z
2b
Inserting the numerical values for a and b
4 mib = 4 x oooo94u
Also, a 2 is very small, even in comparison with 000376, and can
therefore be neglected.
000044
Hence v = 
2 x 000094 2 x 000094
= 234 1043 Vmt.
Taking the + sign and neglecting the first term, v = 1043 Vmt,
which agrees well with the v = 103 Vmi given by Chezy.
EQUATIONS 67
Quadratics with "imaginary" Roots. The question may
have presented itself : What is done when (B 2 4AC) in the
formula for the solution of the quadratic becomes negative ? ' How
can the square root of a negative quantity be extracted ?
The square root of a negative quantity is known as an imaginary
quantity, and all imaginaries are reduced to terms of the square
root of i, which is denoted byj. At present no meaning can be
stated for this, but it is referred to again in a later chapter.
Thus ;=V r I, j 2 =i, j 3 = V i, etc.
E.g., V^so = Vsox i = Vsox V^ = 547.7.
Example 51. Solve the equation 2x z 3* + 15 = o, employing
Method 3.
x
4
+ 3 ViTi X V^
Expressions of the type a bj, where a and b may have any values,
occur in Electrical theory and in the theory of Vibrations ; such being
referred to in Chapter VI.
Cubic Equations. Cubic Equations, i. e., equations contain
ing the cube of the unknown as its highest power, may be solved
graphically, in a manner to be demonstrated in a later chapter, or
use may be made of what is known as Cardan's solution.
The three roots of a cubic equation may be either, one real and
two imaginary, or, three real. Cardan's solution applies only to the
former of these cases and gives the real root only.
If x 3 f ax f b = be taken as the standard type of cubic
equation, then the real solution is given by Cardan as
/ b, /^T^Uof b /*V
X== \2 + V 27+TJ + \2~V 27+
The proof of this result is too difficult to be inserted here, but
it is outlined in A Treatise on Algebra, by C. Smith (Macmillan
and Co., Ltd., 75. 6d.).
3 I\2
If + be negative, the three roots are all real, but Cardan's
27 4
solution cannot be applied.
68 MATHEMATICS FOR ENGINEERS
Example 52. Solve the equation x 3 I2X + 65 = o. (Imaginary
roots are not required.)
Here a= 12, b = 65, in comparison with the standard form.
= {325 + 315}* + {325 315}*
= (!) + (4)= 5
If the equation is not of the form, x 3 f ax + 6 = o, it can be
made so in the following manner.
Example 53. Find a solution of the equation
t; 2 + 1441; 1944 = ' v being a velocity.
For this to be reduced to the standard form, the term containing
v* must be eliminated.
By writing (V+a) for v and suitably choosing a, this can be done,
for
(V + a) 8 + 2 4 (V + a) + I44(V+ a)  1944 = o
'.., V'+3V 2 a+3a 2 V+a 3 +24V 2 +24a 2 +48aV + I44V+ 1440
 1944 =o ............ (i)
Equating the coefficients of V 2 to zero (since the term containing V 2
is to be made to vanish), 30 + 24 = o, *. e., a = 8 ;
so that v = V8.
Equation (i) can now be written ( 8 being substituted for a)
V 3 24V 2 +I92V 512 + 24V 2 +1536 384V +I44V 1152 1944 = o
or V 8 48V 2072 = o
Comparing with our standard formula
a= 48, 6= 2072.
Therefore, by Cardan
V
= (2070)*+ (2)* = I275+I26 = 14.
Hence v = V 8 = 6.
Equations of degree higher than the third (if not reducible to
any of the forms already given) are best solved graphically. (Com
pare with Chapter IX.)
EQUATIONS 69
Exercises 8. On Quadratic and Cubic Equations.
Solve the equations in Exercises i to 10.
1. x z + $x + 4 = o 2. 2X Z 7* + 15 = 4*
3. 3* 2 +9*+i4 = o 4. 8p*7p = 4p* + 5 p+i6
5. ooia 2 234a 764 = 417(1 3250? 6. 9^ + 5^+2 =
7 *+4 ^295* g 2* 17 _5*_
~ ' * *+2 #+3
z 10. 1700 + oi26F = 000003F 2
r , V a 2X230X/ J 2 fi , , , V=33"O
11. If =   z  , find the values of I when J I
2g 2x36 #=322)
02 i ^2
12. If r= T , find h when f = 15, a = 55. This equation gives
the radius of a circle when the height of arc A and length of chord 2a
are known.
13. Solve for F the equation
3F 2 3F
 S    = 100.
30 x io 6 7200
14. We are told that W( + A) = FA (a formula relating to the
jr
strength of bodies under impact), and also A =  , W=45, and
h = 24. Find values of F to satisfy these conditions.
ph z
15. The equation 6 a ab a? =  relates to masonry dams,
where 6 = width in feet of base of a dam a feet wide at the top, and
h feet deep; w being the weight of i cu. ft. of masonry, and p being
the weight of i cu. ft. of water. Find b for the case when a = 5,
h = 30, w = 144, and p = 624.
16. Find expressions for /, from
17. If L solve (a) for u and (6) for v.
2U* + 3U 2
18. To find n (the depth from the compression edge to the neutral
axis of a reinforced concrete beam of breadth 6) it was necessary to
solve the equation bn z + 2A T ww 2wA T d= o. Determine the value
of n to satisfy the conditions when ^=15, A T = i56, 6 = 5, and
d= io.
19. Solve for C the equation 75 x ioC 2  io 10 C + 12 x io 10 = o.
/P _ /wx _ /)
20. Find the values of t when L = s p Ap ' X oogC, and
P=i20, C=3375,T=67, 1^=1765.
21. Find the ratio, length of arc of approach a
pitch p
wheels) from '^ = + , where n = number of teeth in the follower
a np 4
wheel == 24.
70 MATHEMATICS FOR ENGINEERS
22. The equation mi = av + bv z relates to the flow of water in
channels. If a =000024 an( i &= 000014, P ut this in the form
v = cVmi, making any justifiable approximation. (Compare Example
50, p. 66.)
23. Solve the equation 2# a 53 + $x
*54
24. Find values of / to satisfy the equation W=$al(i\ J
(Merriman's formula for the weight of roof principals), given that
W = 5400 and a = 10.
25. x is the distance of the point of contraflexure of a fixed beam
of length / from one end. If x and / are connected by the equation
I T H 72" = fi 11 ^ the positions of the points of contraflexure.
i if
26. To find the position of a mechanism so that the angular velocities
of two links should be the same it was necessary to solve the equation
/ 3 I9'5/ 2 + 42 5/ + 546 = o. Find values of / to satisfy this.
27. To find d, the depth of flow through a channel under certain
conditions of slope, etc., it was necessary to solve the equation
d 3 i '305*2 1305 = o. Find the value of d to satisfy this.
28. The values of the maximum and minimum stresses in the metal
of a rivet due to a shearing stress q and a tensile stress / due to con
traction in cooling are given by the roots of the equation
/(//n)=<? 2 
If q = 4^ tons per sq. in. and / = 3 tons per sq. in., find the two
values of /.
29. The length L of a wire or cable hanging in the form of a parabola
is given by
L  S
3
where S = span and D = droop or sag.
Find the span if the sag is 3'9* and the length of cable is 1004023 ft.
Simultaneous Quadratics. Consider the two equations
= 18 (i)
y
and 5'6(5*6 y) = x 2 (2)
Values of x and y are to be found to satisfy both equations at
the same time (hence the term " simultaneous ") : also the second
equation is of the second degree as regards x, and is therefore a
quadratic.
In most practical examples (the above being part of the investi
gation dealing with compound stresses) one equation is somewhat
more complicated than the other, and therefore, for purposes of
elimination, we substitute from the simpler form into the more
difficult.
EQUATIONS 7I
In this example equation (i) is the simpler, and from it, by
transposition
i8y
X =  2 = ay.
2 ^
Substitute for x, wherever it occurs in equation (2), its value in
terms of y. Then
56(56 y) = (.gy) = 8iy2
3136 5 6y = 8iya
or '8iy 2 +56y 3136 = o.
= ~ 5 ' 6 V ^'3^ + 4 x 81 x 3136
56 i
Hence, y =
162
*62
To find *
* = gy and, substituting in turn the two values found for y,
x= 9 x 1056 or 9 x 367
= 951 or 330
x= 951 or 330 \
and y= 1056 or 367 /
Example 54. In a workshop calculation for the thickness * of a
packing strip or distance piece in a lathe the following equations
occurred
(99)2 = (i7 5 + y) + * ........... (2)
Solve these equations for x and y. The packing strip was required
for a check for a gauge, and great accuracy was necessary in the
calculation.
By removal of the brackets the equations become
9801 = 225 + y 2 +3y+64 + * 2 +i6* .... (3)
9801 = 3o625 + y 2 +35y + * 2 ........ (4)
By subtracting equation (4) from equation (3) an equation is
obtained giving y in terms of x, thus
o= 8125 5y+ 64+ i6x
or *5y = i'6x '1725
whence y = 32^ 345.
Substituting for y in equation (4)
9801 = 30625 + (32* 345) 2 + 35(32* 345)+**
= 30625+ 1024**+ 12002208*+ 1 1 2X I 2075 + x*.
72 MATHEMATICS FOR ENGINEERS
Collecting terms, n 24^+ 8992.* 96035 = o
, 8992 V / (8992) 2 + (4 X 96035 x 1124)
whence #= ^ .a
8992 663219
2248
_ 573299 nr 753I39
' 2248 2248
= 25503 or 33503
i.e., the thickness of the strip was 25503 (inches); the negative
solution being disregarded.
The two values of y would be obtained by substituting the two
values found for x in the equation y = 32* 345.
Thus y = (32 x 25503)  345 or y = (32 x  3'353)  '345
y = 7816 or 11066.
The positive solutions alone have meaning in this example, so that
x = 25503 and y = 7816.
Example 55. Solve the equations
5# 2 + y 2 +2# 77 4 = c* (i)
7*+ 3? =9 (2)
From equation (2) yy = 97*
or y = 9 ~ 7 
Substituting in equation (i)
.. 5*' +  +  7 ^ = 95
Multiplying through by 9
45# 2 + 81 + 49# 2 126*+ iSx 189+ 147* = 855.
Collecting terms
94* 2 + 39* 963 =
Factorising
(9 4 #+32i)(*3) = o
321
i.e., x=  or 3.
94
Now,
EQUATIONS 73
Substituting the two values for x
64
orv 9
21
3
3
1031
or =
4
94
* = 3
or
1
94
y = 4
1031
or
94
]
The method of substitution indicated in the previous three
examples is to be recommended in preference to the " symmetrical "
methods usually given, but which only apply to special cases.
Occasionally one meets with an equation or pair of equations
to which this method is not applicable. Thus if the equations are
homogeneous and of the second degree, i, e., all the terms contain
ing the unknowns are of the second degree in those unknowns,
proceed as in the following example ; the method being in reality
an extension of the preceding.
Example 56. Solve the equations
Divide equation (i) by equation (2).
Then ~ =
Multiplying across
2i6y a .
Collecting terms
115** + zgixy 2i6y z = o.
Factorising
72 3
whence x = y or x = =y,
Substitute each of these values for x in turn in equation (2).
72
Thus, taking x = *^y
74 MATHEMATICS FOR ENGINEERS
..2^*15*23
y ' 20
and # = y = 
.e., y = 36 when # = f
and y = + 36 when x = ~
Taking x = y y z + ^y z = 1 15
23
and y = 5
and x = 3
Grouping results .'. ^ = 3 or ^ 36
y = 5 or
Surds and Surd Equations. One often meets with such
quantities as y/3, 3/7 or $ a : such are known as surds or irrational
quantities, since their exact values cannot be found.
The value of V$ can be found to as many places of decimals
as one pleases, but for ordinary calculations two, or at the most
three, figures after the decimal point are quite sufficient : thus
\/3 = 173 approximately, or 1732 more nearly.
It is both easier and more accurate to multiply by a surd than
to divide by it, and therefore, if at all possible, one must rid the
denominator of the surds by suitable multiplication.
The process is known as rationalising the denominator.
Example 57. Rationalise the denominator of ~~
To do this, multiply both numerator and denominator by VJl
since V^x ^3 =3.
Then 5* = ^ ^ =  , or if the result is required in
v 3 V 3 x V 3 _3_
decimals it is 289.
Example 58. Find the value of ' ._
4 V5
Multiply numerator and denominator by 4+ V^.
EQUATIONS 75
Then the fraction  7(4 + V5) __ = 7(4+ ~
= 7(4 + 2236) = 7 x 6236
l6 5 H
= 3968.
Surds occurring in equations must be eliminated as early as
possible by squaring or cubing as the case may demand.
Example 59. Solve the equation 3/p 2 = 7.
Cubing both sides p 2 = 343
P = 345.
Example 60. Find x from the equation
4  2#+3 = 5.
Transposing so that the surd is on one side by itself
Vzx +3 = 1.
Squaring both sides
2X + 3 = i
whence x = i.
Note. The solutions of all surd equations should be tested.
Reverting to the original equation and substituting i for x
4V2#+3 = 4V2 + 3 = 41 = 3
and not 5, so that i is not a solution of this equation, but it would
be of the somewhat similar equation
4+ \2#+3 = 5.
When squaring, either ( \/2#+3) 2 or ( + V2# +3)* = 2^+3,
so that the solution obtained may be that of either the one equation
or the other.
In this case, then, there is no solution to the equation as given.
Example 61. Wohler's law for repeated stresses can be expressed
where f l = original breaking stress, S = stress variation in terms of / if
and f t = new breaking stress.
Find f.j, when S = 537/1, /i = 52, x = 2.
Substituting the numerical values
y 2== ;53_7/2 + \/52 2 (2 x 537/2X52)
*. e., f t = 2685/ t + A/2704 
76 MATHEMATICS FOR ENGINEERS
Arranging so that the surd is isolated
/,  2685/3 = V270 4  55 8/,
or 7315/2 = V270 4  55 8/,
Squaring '535/2 2 = 2704  558/5,
or '535/ 2 +558/  2704 = o.
Solving for /, by means of the formula
f  55' 8 V3"Q + 579Q
/a  ^o7~~
 55'8 94'5 = 387 Qr _ 50^3
107 107 107
= 362 or 1405.
The negative solution has no meaning in the cases to which this
formula is applied, hence the positive solution alone is taken.
Example 62. Find the value or values of x to satisfy the equation
V^x 7 + 3 Vzx + 17 = 18.
It is best to separate the surds thus
3 Vzx +17 = 18 V^x 7.
Square both sides and then
9(2* + 17) = (18)* +( V^x 7)* 2 x 18 x V4# 7 *
i. e. t i8x + 153 = 324 + 4* 7 36 V<ix 7
or, isolating the surd
36 V4* 7 = 164 14*
or i8V4# 7 = 82 7*.
Squaring again
324(4*  7) = 6724 + 4Q* 2  1148*
whence 49# 2 2444* + 8992 = o.
Factorising (49* 2248) (# 4) = o.
2248
* = W r4 '
To test whether these values satisfy the original equation
When x = 4, lefthand side = Vi6 7 + 3 V8 +17
= 3+15 = 1 8, which is the value of the
righthand side x = 4 satisfies.
* Always remember that when squaring a "twoterm" expression
three terms result ; of the character ; (ist squared) + (and squared)
(twice product of ist and 2nd).
. e., (A + B) 2 '= A 2 + B 2 + 2AB
(A  B) 2 = A 2 + B 2  2AB.
EQUATIONS 7?
When * = , lefthand side = >  7 + 3
49
_ 93 , 2ig
7 ' 7
312
= r = 44$ which does not = 18.
Hence # =  is not a solution of the given equation ; it however
satisfies the equation 3 Vzx + 17 V^x 7 =a 18.
Exercises 9. On the Solution of Simultaneous Quadratic Equations
and Surd Equations.
1. Find values of x and y to satisfy the equations
*3V= 16
* 2 + 3V Z 2*+ 4y = 50
2. Solve the equations a 2 = 8 + 4y 2
2a + 2y = 7
3. Solve for p and <y the equations
3p*pq7i 2 = 5
4. Solve the equations 5# 2 9^+ i2Ary zy z = 132
7^+8y = 54
5. Determine the values of a and b to satisfy the equations
a 2 2ab +3 =
2a + 6 = 4
6. Solve for m the equation \/3w a qm = 5.
7. The following formula is used to calculate the length of hob
required to cut a worm wheel, for throat radius r and depth of tooth d
f= 2 Vd(zr d)
Find the depth of tooth when the hob is 3* long and the throat
radius is 2".
8. Find a from the equation V8a+ g 3 = 4.
9. Solve for H the equation 25*6H 346 VH= 10000.
g _
10. The formula /, = h V/i 2 xSfi is that given by Unwin, and
refers to variation of stress. Bauschinger's experiments in a certain
case gave S= '4 1 /*' /i = 228, and #=15. Find the value of/,.
11. Using the same formula as in the previous example (No. 10)
find / 2 when S = / 2 , / x = 30, and x = 2.
12. Solve for a the equation Va +2+ Va = 77=
V Ge ~\~ 2
13. Find a value or values of x to satisfy the equation
Vi + gx = Vx+ i V^x + i
78 MATHEMATICS FOR ENGINEERS
14. The length x of a strut in a roof truss was required from the
equation Vx z 36 + Vx* 4 = 16. Find this length.
15. Find the value of k to satisfy the equation referring to the
discharge of water from a tank
12 " ^7
Given that /,= 45x60, ^=26x60, A=i56, a =  ^, =322,
144
A 1= =36, and A, = 25.
16. The following equations occurred when calculating a slope of
a form gauge^
w 2 = (*o6 + n}n
vn_ _ 0175 n
03 ~ 035
Find the values of m and n to satisfy these conditions.
17. If / = span of an arch d = rise
A = height of roadway above the lowest point of the arch
c = highest
l z
then c+d = h and also c = J^+o
Find the values of d when h = 23 ft., and / = 24 ft.
18. The dimensions for cast iron pipes for waterworks are related by
the equation
where H = head of water in feet
t = thickness of metal in inches
d = internal dia. of pipe in inches
If H = 300 and t = .5 find d.
CHAPTER III
MENSURATION
Introduction. Mensuration is that part of practical mathe
matics which deals with the measurement of lengths, areas, and
volumes. A sound knowledge of it is necessary in all branches of
practical work, for the draughtsman in his design, the works'
manager in his preparation of estimates, and the surveyor in his
plans, all make use of its rules.
Our first ideas of mensuration, apart from the tables of weights
and measures, are usually connected with the areas of rectangles.
How much floor space will be required for a planer 4 ft. wide and
12 ft. long ? Here we have the simplest of the rules of mensuration,
viz. the multiplication together of the two dimensions. Thus, in
this case, the actual space covered is 4 X 12 =48 sq. ft.
Area of Rectangle and Triangle. If the rectangle is
bisected diagonally, two equal triangles result, the area of each
being onehalf that of the original rectangle, or we might state it,
\ (length X breadth), or as it is more generally expressed, \ base X
height or \ height X base. (Note that the \ is used but once;
thus we do not multiply \ base by \ height.) This rule for the area
of the triangle will always hold, viz. that the area of the triangle
is onehalf that of the corresponding rectangle, i, e., the rectangle on
the same base and of the same height. Thus in Fig. 7, the
triangles ABC, AB'C, and AB"C are all equal in area, this area
being onehalf of the rectangle ACB'D, i. e., %bh. It is the most
widely used of the rules for the area of the triangle, because if
sufficient data are supplied to enable one to construct the triangle,
one side can be considered as the base, and the height (i. e., the
perpendicular from the opposite angular point on to this side or
this side produced) can be readily measured, whence onehalf the
product of these two is obtained.
A special case occurs when one of the angles is a right angle;
then the rule for the area becomes : Area (to be denoted by A)
equals onehalf the product of the sides including the right angle.
8o
MATHEMATICS FOR ENGINEERS
One further point in connection with the rightangled triangle
must be noted, viz. the relation between the sides.
The square on the hypotenuse (the side opposite the right angle,
*'. e., the longest side) is equal to the sum of the squares on the other
sides. (Euclid, I. 47.)
D B B (
A
Fig. 7
Fig. 8.
In Fig. 8, AB is the hypotenuse, because the right angle is at
C, and
.(AB) 2 = (AC) 2 f(BC)
or c 2 = b z +a*.
A word about the lettering of triangles will not be out of place
here. It is the convention to place the large letters A, B and C
at the angular points of the triangle, to keep these letters to repre
sent the angles, e. g., the angle ABC is denoted by B, and to letter
the sides opposite to the angles by the corresponding small letters.
Thus the side BC is denoted by a, because it is the side opposite the
angle A.
Rule for Area of Triangle when the three sides are
given. As previously indicated the rule %bh can here be applied
if the triangle is drawn to scale and a height measured. (The
triangle can be constructed so long as any two sides are together
greater than the third.) If, however, instruments are not handy
proceed along the following lines :
Add together the three sides a, b, and c, and call half their sum s ;
s=
Then the area is given by
A = Vs(s  a)(s  b)(s  c)
This rule will be referred to as the " s" rule, and the proof of it
will be found in Chapter VI.
Logarithms or the slide rule can be employed directly when
using this formula, since products and a root alone are concerned.
MENSURATION
81
Example i. One end of a lock gate, 7 ft. broad, is 2 ft. further
along the stream than the other when the gates are shut: find the
width of the stream.
Let 2x = width of stream. (See Fig. 9.)
Then f = x z +2* ~ JC 
x * = 72  2 *= (72)(7 + 2)
or x = 67
so that the width of the stream = 2x = 134 ft. Fig. 9.
Example 2. Find the area of the triangle ABC, Fig. 10.
Area = base x height
= $x 146 x n 4
= 832 sq. ins.
C.
(Note that 114 is the perpendicular
on to AB produced.)
Fig 13
Example 3. The pressure on a triangular plate immersed in a
liquid is 45 Ibs. per sq. ft. The sides of the plate measure 181", 253",
and 17*4" respectively : find the total pressure on the plate.
Let a = i8'i, b = 253, c = 174.
Using these figures, the area will be in sq. ins.
s =
181
253+174 608
" =   = 304
2 2
Then A = V3O 4 (3O 4  i8i)(3O 4  253X304 174)
= A/304 x 123 x 51 x 13
Taking logs throughout
log A = i{log 304 + log 123 + log 51 + log 13}
'I4829 1
10899
7076
11139
= 21972
U'3943'
= log 1575
A = 1575 sq. ins.
Then total pressure = ^^ X 45 Ibs. [feet" x
144 L.
Ibs.
feet 1
Ibs
i
82
MATHEMATICS FOR ENGINEERS
A further rule for the area of a triangle will be found in
Chapter VI.
A rule likely to prove of service is
Area of an equilateral triangle = '433 x (side) 2 .
Thus if the sides of a triangle are each 8 units long its area is
'433 X 8 2 , i. e., 277 sq. units.
Exercises 10. On Triangles and Rectangles.
1. A boat sails due E. for 4 hours at 137 knots and then due N.
for 7 hours at 104 knots. How far is it at the end of the n hours from
its starlingpoint ?
2. Find the diagonal pitch of 4 boiler stays placed at the corners
of a square, the horizontal and vertical pitch being 16*.
3. If a rightangled triangle be drawn with sides about the right
angle to represent the electrical resistance (R), and reactance (2nfL),
respectively, then the hypotenuse represents the impedance. Find the
impedance when / = 50, L,= 'i^g, R = 5O, and ?r = 3142.
4. It is required to set out a right angle on the field, a chain or
tape measure only being available. Indicate how this might be done,
giving figures to illustrate your answer.
5. A floor is 29'$" long and n'io" broad. What is the distance
from one corner to that opposite ?
6. At a certain point on a mountain railway track the level is
215 ft.; 500 yds. further along the track the level is 227 ft. Express
the gradient as
(a) i in x (x being measured along the track).
(b) i in x (x being measured along the horizontal).
Fig. Tl.
7. For the Warren Girder shown at (a), Fig. n, find the length of
the member AB.
8. A roof truss is shown at (fc), Fig. n. Find the lengths of the
members AB, BC and AC.
MENSURATION 83
9. A field is 24 J chains long and 650 yds. wide. What is its area
in acres ? (Surveyors' Measure is given on p. 87.)
10. Find how many " pieces " of paper are required for the walls
of a room 15 ft. long, i2'6" wide and 8 ft. high, allowing 8 % of
the space for window and fireplace (a " piece " of paper being 21"
wide and 9 ft. long).
11. A courtyard 15 yds. by 12 yds. is to be paved with grey stones
measuring 2 ft. x 2 ft. each, and a border is to be formed, 2 ft. wide,
of red stones measuring i ft. x i ft. How many stones of each kind
are required ?
12. A room 15 ft. by 12 ft. is to be floored with boards 4^* wide.
How many foot run will be required ?
13. Before fracture the width of a mild steel specimen was 2014"
and its thickness '387". At fracture the corresponding dimensions
were i'524" and 250". Find the percentage reduction in area.
14. A rectangular plot of land J mile long and 400 ft. wide is to be
cut up into building plots each having 40 ft. frontage and 200 ft. depth.
How many such plots can be obtained ?
15. The top of a tallboy is in the form of a cone ; the diameter
of the base is 4", and the vertical height is ij". Find the slant
height.
16. A bar of iron is at the same time subjected to a direct pull of
5000 Ibs., and a pull of 3500 Ibs. at right angles to the first. Find the
resultant force due to these.
17. At a certain speed the balls of a governor are 5" distant
from the governor shaft ; the length of the arms is 10". Find the
" height " of the governor h and hence the number of revs, per sec.
. 816
n from h = .
n*
18. A load on a bearing causes a stress of 520 Ibs. /a*. If the stress
is reckoned on the " projected area " of the bearing, the diameter of
which is 4" and the length 5%', find the load applied.
19. The sides of a triangle are 174", 84" and 157" respectively.
Find its area, by
(a) Drawing to scale and use of i base x
height rule.
(6) Use of " s " rule. (See p. 80.)
20. Find the rent of a field in the form of
a triangle having sides 720, 484 and 654 links
respectively, at the rate of 2 los. per acre.
(See note to Ex. 9.)
21. Find the area of the joist section shown
in Fig. ua. (Thickness of flange is 02*.)
22. Neglecting the radii at the corners, calculate the areas of the
8 4
MATHEMATICS FOR ENGINEERS
sections in Fig. 12 : viz. (6) channel section, (c) unequal angle, and (d)
tee section.
h 4 "H
Ay/'y/jT A
Fig. 12. Mild Steel Rolled Sections.
Area of Parallelogram and Rhombus. From the three
sided figure one progresses to that having four sides, such being
spoken of generally as a quadrilateral.
Of the regular quadrilaterals reference has already been made
to the simplest, viz. the rectangle (the square being a particular
example), for which the area = length X breadth.
Since the area of a triangle is given by the product, \ base X
height, it follows that
(a) Triangles on the same base and having the same height are
equal in area, and
(b) Triangles on equal bases and having the same height are
equal in area.
MENSURATION
Thus, if in Fig. 13 the length FC is made equal to the length
ED, the triangles AED and BFC will be equal in area, since the
bases are equal and the height is the same. Also it will be seen
that the sides AD and BC are parallel, so that the figure ABCD is
a parallelogram. Then
The area of the parallelogram
ABCD = area of figure ABFD + area of triangle BFC
= area of figure ABFD + area of triangle AED
= area of rectangle ABFE
= AB x BF.
This result could be expressed in the general rule, " Area of a
parallelogram = length of one side x the perpendicular distance from
that side to the side parallel to it."
In the case of the rhombus (a quadrilateral
having its sides equal but its angles not right
angles) one other rule can be added.
Its diagonals intersect at right angles, and
hence its area can be expressed as equal to
onehalf the product of its diagonals; i. e.,
Area = X BD X AC, the reference being to
the rhombus in Fig. 14.
This rule should be proved as an example
on the J base X height rule for the triangle.
Area of Trapezoid. A trapezoid is a quadrilateral having
one pair of sides parallel.
Its area = mean width X perpendicular height.
= (sum of parallel sides) X perpendicular distance
between them.
Fig. 15
* 30' H
Fig. 1 6 Cross Section of a
Cutting.
In Fig. 15, AB and CD are the parallel sides, and the area
86
MATHEMATICS FOR ENGINEERS
Example 4. Calculate the area of the cross section of a cutting,
having dimensions as shown in Fig. 16.
Area = {70 + 30} x 16 sq. ft.
= 800 sq. ft.
Example 5. The kathode, or deposit plate, of a copper voltameter
has the form shown in Fig. 17. Calculate, approximately, the area
and hence the current density (i. e., amperes per sq. in. of surface) if
142 amperes are passing.
Fig. 17.
Fig. 1 8.
We may divide the surface of the plate into three parts, A, B, and C.
Area of A = 26 x 265 = 69 sq. ins.
Area of B = ('7+ 2 ' 6 5\ x .3. = x
\ 2 /
Area of C = 6 x 7 = 42
Total area of one side = 874
This is the area of one side ; but the deposit would be on both sides
total area = 2 X 874 = 1748 sq. ins.
and current density = Jj g = '0812 amp, per sq. in.
or i amp. for 123 sq. ins. of surface.
Example 6. Find the area of the rhombus, one side of which
measures 5" and one diagonal 8".
Let 2X= length of other diagonal in inches (Fig. 18).
Then, by the rightangled triangle rule,
** = 5 2 4 2 = 9
x = 3 and 2X = 6.
Area = i (product of diagonals) = J x 8 X 6 = 24 sq. ins.
MENSURATION 87
This example could also have been worked as an exercise on the
" s" rule, the sides of the triangle being 5, 5 and 8 respectively.
Areas of Irregular Quadrilateral and Irregular Poly
gons. Having dealt with the regular and the " semi " regular
quadrilaterals, attention must now be directed to the irregular
ones. No simple rule can be given that will apply to all cases of
irregular quadrilaterals : the figure must be divided up into two
triangles and the areas of these triangles found in the ordinary way.
This method applies also to irregular polygons (manysided
figures) having more than four sides; but these figures split into
more than two triangles.
Example 7. Find the area of the quadrilateral ABCD, Fig. 19,
in which AD = 17 ft., DC= 15 ft., BC= 19 ft., ^
AC= 26 ft., and the angle ABC is a right angle.
It will be necessary to find the length of AB.
By the rule for the rightangled triangle,
(AB) 2 = 26 2 ig 2 = 7 x 45 = 3'5
AB = 1776 ft.
The quadrilateral ABCD = Triangle ADC +
Triangle ABC.
Dealing first with the triangle ADC, its area must be found by the
" s" rule.
s =
2
A = A/29 X 12X14X3
= 1209 sq. ft.
Area of triangle ABC = x 1776 x 19 = 169 sq. ft.
/. Area of quadrilateral ABCD =121+169
= 290 sq. ft.
Example 8. Find the area of the plot of land represented in Fig. 20
(being the result of a chain survey).
Some of the dimensions are given in chains : it is worth while to
remind ourselves of the magnitude of a chain.
SURVEYORS' MEASURE
i chain = 22 yards = 66 feet.
i chain = 100 links (i link = 792 .)
10 chains = i furlong.
80 chains = i mile.
i sq. chain  22 2 = 44 sq. yards = T V of an acre.
or 10 sq. chains = i acre.
10 sq. chains = 100,000 sq. links.
i acre 100,000 sq. links,
88
MATHEMATICS FOR ENGINEERS
The given figure is divided by the " offsets " into triangles and
trapezoids ; the offsets being at right angles to the main chain lines.
It will be convenient to work in feet.
Dealing with the separate portions.
Area AC J = \ xig8 X2&4 = 26136 sq. ft.
ACB =ixig8x24 = 2376
192
H5 2 M
II2O
4 80
650
28OO
225
CDB =
CDEF =
FEGH = (20+8)x8o =
HGJ =x8xi2o =
JKL =ix 100x13 =
LKMN= ^(13 + 15) x 200 =
NMA =
35131 sq.ft.
.'. Total area = 35131 sq. ft.
= 807 sq. chns.
= 807 acre.
Fig. 20.
Areas of Regular Polygons. Regular polygons should be
divided up into equal isosceles triangles ; and there will be as many
of these as the figure has sides. The areas of the triangles are best
found (at this stage) by drawing to scale, and as an aid to this the
following rule should be borne in mind.
The angle of a regular polygon of n sides
X 90 degrees.
Thus, for a regular pentagon (a fivesided figure) n = 5 and the
angle = [ (2X5) ~ 4 1x9o = 108.
Alternatively, the following construction may be used. Suppose
that the area of a regular heptagon, i. e., a sevensided figure, is
required, the length of side being i"; and we wish to find its
area by drawing to scale. Set out on any base line (Fig. 21) a
semicircle with A as centre and radius ij" (the length of side).
Divide the semicircumference into seven (the number of sides)
equal parts, giving the points a, B, c, d, e, f, G (this division to be
done by trial). Through the second of these divisions, viz. B,
draw the line AB; drawing also lines Ac, Ad, etc., radiating from
A. With centre B and radius i" strike an arc cutting Ac in
C ; then BC is a side of the heptagon. This process can be repeated
until the figure ABCDEFG is completed,
MENSURATION
89
Bisect AG and GF at right
D
To find the area of ABCDEFG.
angles and note the point of
intersection O of these bisectors ;
this being the geometrical centre
of the figure. Measure OH (it
is found to be 156").
Then area of AOG
= \ AG X OH = i x 15 X 156
= 1*17 sq. ins.
/. Area of ABCDEFG
= 7XAAOG = 7x117
= 819 sq. ins.
Fig. 21. Area of Polygon.
Exercises 11. On Areas of Quadrilaterals and Polygons.
1. The central horizontal section of a hook is in the form of a
trapezoid 2$" deep, the inner width being 2" and the outer width $*.
Find the area of the section.
2. The diagonals of a rhombus are I9'74" and 528" respectively.
Find the length of side and the area.
3. Find the area of the quadrilateral ABCD shown at (a), Fig. 22.
What is the height of a triangle of area equal to that of ABCD, the
base being 5* long ?
4. A field in the form of a quadrilateral ABCD has the following
dimensions in yards : CD = 38, DA = 29, AC = 54, BE (perpendicular
from B on to AC) = 23. Find its area in acres.
5. Reproduce (6), Fig. 22, to scale, and hence calculate the area of
ABCDEF.
6. Find the area, in acres, of the field shown at (c), Fig. 22.
Fig. 22.
90 MATHEMATICS FOR ENGINEERS
7. A retaining wall has a width of 4 ft. at base and 2'6* at top.
The face of the wall has a batter of i in 12, and the back of wall is
vertical. Find the area of section and also the length along the face.
8. The side slopes of a canal (for ordinary soil) are ij horizontal
to i vertical. If the width of the base is 20 ft. and the depth of water
is 5 ft. find the " area of flow " when the canal is full.
9. Find the hydraulic mean depth ( i. e.,  Area of flow } for the
\ Wetted perimeter/
canal section for which the dimensions are given in Question 8.
10. The end of a bunker is in the form of a trapezoid. Find its
area if the parallel sides are 9'$", and 15'! i" respectively, the slant
side being 248", while the other side is perpendicular to the parallel
sides.
11. A regular octagon circumscribes a circle of 2* radius. Find
its area.
12. Find the area of a regular hexagon whose side is 428".
13. The " end fixing moment " for the end A of the builtin girder,
Fig. 220, is found by making the area ABEF equal to the area ABCD.
Find this moment, i. e., find the length AF.
14. A plate having the shape of a regular hexagon of side 2.\" is
to be plated with a layer of copper on each of its faces. Find the
current required for this, allowing 16 amperes per 100 sq. ins.
15. An irregular pentagon
of area 5908 sq. ins. is made
up of an equilateral triangle
with a square on one of its
sides. Find the length of
side.
16. Neglecting the radii at
the corners, find the approxi
mate area of the rail section
shown at (a), Fig. 12.
22a
Circumference and Area of Circle. When n, the number
of sides of a polygon, is increased without limit, the sides merge
into one outline and the polygon becomes a circle.
A circle is a plane figure bounded by one line, called the circum
ference and is such that all lines, called radii, drawn to meet the
circumference from a fixed point within it, termed the centre, are
equal to one another.
The meanings of the terms applied to parts of the circle will
best be understood by reference to Fig. 23 and Fig. 24.
If a piece of thread be wrapped tightly round a cylinder for,
say, five turns and the length then measured and divided by 5,
the length of the circumference may be found. Comparing this
with the diameter, as measured by callipers, it would be found to
be about 3, times as long.
Repeating for cylinders of various sizes, the same ratio of these
lengths would be found. The Greek symbol TT (pi) always denotes
MENSURATION
this ratio of circumference to diameter, which is invariable ; but
its exact value cannot be found. It has been calculated to a large
number of decimal places, of which only the first four are of use
to the practical man. For considerable exactness TT can be taken
as 3'i4i6 : however, \ 2  or 31428 is quite good enough for general
use, the error only being about 12 in 30,000 or about '04%. Even
 2 . 12  need not be remembered if a slide rule be handy, for a marking
will be found to represent TT (see Fig. i, p. 17).
Fig. 24.
Then circumference = TT x diameter = ird or 2irr
where d = diam. and r = radius.
Also Area = nr 2 or ^d 2
4
 = 7854 : a marking on the slide rule indicating this.
4
(The mark M on the slide rule is  J
It is sometimes necessary to convert from the circumference to
the area; thus
ATT ATT ATT
[ stands for circle and 0ce for circumference.]
Example 9. Find the diameter of the driving wheel of a locomotive
which in a distance of 3 miles makes 1010 revolutions (assuming no
slipping).
In one revolution, the distance covered = 0ce.
total distance 3 x 5280 f .
i* )CC ; ~~ " ' ""^ " A*
number of revolutions 1010
and Aiim W =
92 MATHEMATICS FOR ENGINEERS
Example 10. Find the area of the cross section of shafting, 3^"
diam.
Area =  x 35* = 962 sq. ins.
4
Notice that  or '7854 is in the neighbourhood of '75 or f ; therefore,
4
for approximation purposes, square the diameter (to the nearest round
figure) and take f of the number so obtained.
In this case, (3'5) 2 = 12 approximately,
and f of 12 = 9.
Areas of circles can most readily be obtained by the use of the
slide rule, the method being as follows
Set one of the C's (marked on the C scale) (refer Fig. I, p. 17)
level with the diameter on the D scale, place the cursor over i on
the B scale, then the area is read off on the A scale ; the approxima
tion being as before. This method is of the greatest utility, and
several examples should be worked by means of it for the sake
of practice.
Examples
Dia.
Area
Approximation
48
181
fx 25=18
79'5
4965
 x 6400 = 4800
65
332
 x 50 = 36
If the C's are not indicated on the C scale of the slide rule,
markings should be made for them at 1128 and 3569 respectively.
The reason for these markings may be explained as follows
The area of a circle = ^ 2 , or, as it might be written, d 2 ^*
4 TT
Now 2 = 1286, of which the square root is 1128. A marking is
7T
thus placed at 1128, so that when this mark is set level with the
diameter on the D scale, the reading on the D scale opposite
the index of the C scale gives the value of d i */4. By reading
the figure on the A scale level with the index on the B scale, the
square of d 4 A /3 or d 2 f  is found; this being the area of a
V 7T
circle of diameter d.
For convenience in handling the rule a marking is made at
MENSURATION 93
3569 on the C scale also ; this figure being obtained by extracting
the square root of 1286 instead of that of 1286.
Some slide rules are supplied with a threeline cursor. If the
centre line is placed over the dia. on the D scale then the left hand
line is over the area on the A scale.
Example n. Find the connection between circumferential pitch
and diametral pitch (as applied to toothed wheels).
The circumferential pitch
. _ Qce of pitch circle _ ird
number of teeth N
i 4 v, .A number of teeth N
The diametral pitch p d =    =
diam. of pitch circle d
I 7T
Hence, p c ir X =
N "P*
~d
circumferential pitch = diametr " al pitch
E. g., if pc is ~ , then = 375 or p d = ~ = 837".
To find the area of an Annulus, i.e., the area between two
concentric circles.
It is evident that the area will be :
Area of outer area of inner circle =
TrR 2  T' 2  (Fig 25.) r///y .
This can be put into a form rather \///\ ^^..
more convenient for computation, thus
Area of annulus = 7r(R 2 f 2 ) or^(D 2 rf 2 ) VAmTuTus^
X^VVX///' '
or, in a form more easily applied Fig. 25.
Area of annulus = 7r(RrXR+r) or (Dd)(D+d).
This rule can be written in a form useful when dealing with
tubes, thus
. iR 4 zr r\
Area = n(R  r}(R + r) = 2^(R  r)( )
= 2v X t X average radius
= TT X average diameter X /
where t is the thickness of the metal of the tube.
94 MATHEMATICS FOR ENGINEERS
Example 12. What is the area of a piece of packing in the form of
a circular ring, of outside diameter gj" and width 3 J" ?
Here, R = 475", r = 4'75 ~ 3'25 = i'5"
Hence the area = (R r)(R + r) = ^(4*75 + i'5)(4'75 ~ 1 '5)
= x 625 x 325
= 637 sq. ins.
Example 13. A hollow shaft, 5* internal diam., is to have the
same sectional area as a solid shaft of n* diam. Find its external
diam.
Area of solid shaft =xn a = x 121.
4 4
Let D = outside diam. required.
Area of hollow shaft = (D 2  25)
4
The two areas are to be the same ; equating the expressions found
for these
(D 2 25) = x 121
4 4
whence D 2 25 = 121
and D 2 = 146
.'. D = 1207".
Products, etc., of TT. Certain relations occurring frequently
are here stated for reference purposes.
TT = 3142  = 318 =
ff * = 987 4^2 = 39'4 8 sa Y 39'5 4* 1 = 4 I 9
o
TT TT 47T , (often taken
i = .^236 = 7854  = 1256 v
6 J 4 10 as f).
log TT = 4972
Exercises 12. On Circumference and Area of Circles.
1. Find the circumference of a circle whose diam. is 713*.
2. The semicircumference of a circle is 914 ft. What is its
radius ?
3. Find the area of a circle of diam. i^'^".
4. The following figures give the girth of a tree at various points
along its length. Find the corresponding areas of cross sections :
428, 519, 647, 210, 87.
{Suggestion : area =  ; first find value of constant multiplier
4T
(approx. 08). Keep the index of the sliderule B scale fixed at this ;
4*
MENSURATION 95
place cursor over Qce on the C scale and read off result on A scale ;
the squaring and the multiplying are thus performed automatically.}
5. If the circumferential pitch of a wheel is i J*. find the diametral
pitch. (See Example n, p. 93.)
6. A packing ring, for a cylinder 12* diam., before being cut is
12 5" diam. How much must be taken out of its circumference so that
it will just fit the cylinder ?
7. A circular grate burning 10 Ibs. of coal per sq. ft. of grate per
hour burns 66 Ibs. of coal in an hour. Find the diameter of the grate.
8. Assuming that castiron pulleys should never run at a greater
circumferential speed than i mile per min., what will be the largest
diam. of pulley to run at 1120 revs, per min. ?
9. The wheel of a turbine is 30* diam. and runs at 10600 R.P.M.
What is the velocity of a point on its circumference ?
Note. The rule used in questions such as this is v = 271fN, where
v = velocity in feet per min., r = radius of wheel in feet, and N =
number of revolutions per minute.
10. A piece, 4" long, is cut out of an elastic packing ring, fitted to
a cylinder of 30" diam., so that the ends are now J* apart. Find the
diam. of the ring before being cut.
11. Find the diameter of an armature punching, round the circum
ference of which are 40 slots and the same number of teeth. The width
of the teeth and also of the slots (at circumference) is '35*.
12. While the load on a screw jack was raised a distance equal to
the pitch of the screw ("), the effort was exerted through an amount
corresponding to i turn of a wheel 1051" in diam. Find the VelocRy
, . f .... distance moved by effort"!
Ratio of the machine \ V.R. = ,rr
L distance moved by load )
13. The stress / in a flywheel rim due to centrifugal action is given
by / = , where w = weight of rim in Ibs. per cu. in., v = circum
ferential speed of rim in ft. per sec., and g = 322. Find the revs, per
min. if /= 12 x 2240 when w = 28 and diam. = 10 ft.
14. Find the bending stress in a locomotive connecting rod revolving
at revs, per sec. from the equation
, p ATr 2 n z yrl z . 480
stress = x * ^ where y = J, p = ^g
r = 12, / = 120, k 2 = 3, and g = 322.
The driving wheels are 6 ft. in diam., and the locomotive travels
at 40 miles per hour.
15. Find the area of the section of a column, the circumference
of which is 1847".
16. Calculate the diameter of a circular plate whose weight would
be the same as that of a rectangular plate measuring 2'6* by 3'2*, both
plates being of the same thickness and material.
17. Find the area of the section of a rod of 498" diam.
18. If there is a stress of 48000 Ibs. per sq. in. on a rod of 566*
diam., what is the load causing it ?
19. Find the total pressure on a piston 9* diam., when the other
9 6
MATHEMATICS FOR ENGINEERS
side of the piston is under a back pressure of 3 Ibs. per sq. in. above
a vacuum.
Gauge pressure (pressure above atmosphere) = 64 Ibs. per sq. in.
i atmosphere = 147 Ibs. per sq. in.
20. The driving wheel of a locomotive, 5 ft. in diameter, made
10000 revolutions in a journey of 26 miles. What distance was lost
owing to slipping on the rails ?
21. The total pressure on a piston was 8462 Ibs. If the gauge
registered 51 Ibs. per sq. in. (i. e., pressure above atmosphere) and
there was no back pressure, what was the diameter of the piston ?
22. Find the area of section of a hollow shaft of external diam. 5^*
and internal diam. 3".
23. A circular plot of land is to be surrounded by a fencing, the
distance between the edge of the plot and the fencing being the same
all round, viz. 6 ft. The length of the fencing required is 187 ft. Find
the area of the space between the plot and the fencing.
24. Find the resistance of 605 cms. length of copper wire of diam.
068 cm. from
R*
a
where a = area in sq. cms., / = length in cms., and k = resistivity =
0000018 ohm per centimetre cube.
25. The buckling load P on a circular rod is given by
Af c where A = area of section
diameter
P =
(_L\
VK/
and K =
Find the diameter when
P = 188500, f c = 67000, c
26. A pair of spur wheels with
pitch of teeth ij" is to be used to
transmit power from a shaft running
at 120 R.P.M. to a counter shaft run
ning at 220 R.P.M. The distance
between the centres of the shafts is to
be 24" as nearly as possible.
If the diameters of the pitch circles
are inversely as the R.P.M., find the
true distance between the centres and
the number of teeth on each wheel.
27. Calculate the area of the zero
circle (the circle of no registration of
the wheel), the radius of which is BD,
for the planimeter shown in outline in
Fig. 26.
28. The resistance of i mile of
copper wire is found from
5600
, and L = 50.
Tracioq
Po'iot
Fig. 26. Amsler Planimeter.
R
04232
area in sq. ins.
Find the resistance of I mile of wire of No. 22 B.W.G. (diam.
03*).
MENSURATION ^
Length of Chord and Maximum Height of Arc In
Fig. 27 let h = maximum height of the arc, 2a = length of the
chord, and r = radius.
Then, by the rightangled triangle rule, applied to the triangle
AGO
Transposing terms
a* = 2rhh 2
whence
a = V2rhh?
or length of chord
= 2\/2r/i/i 2
a rule giving the length
of chord when the radius
and maximum height of
arc are given.
B
If h is expressed as a fraction of the radius, say h = fr, the rule
for the length of chord becomes
length of chord = 2r\/2/f 2 .
From equation (i) 2rh = a 2 }/; 2
r =
2/1
a rule giving the radius when the chord and the maximum height
of arc are known.
From (i) also, h 2 2rh\a z = o, and from this, by solution of
the quadratic
, 2r v 2 2
or
Vr*i
giving two values for h (i. e., for the arc less than, and the arc
greater than, the semicircumference) when the radius and length
of the chord are known.
If two chords intersect, either within or without the circle,
the rectangles formed with their segments as sides are equal
in area, Euc. Ill, 35 and 36. Thus, in Fig. 28, at both (a)
and (&)
AP.PB = CP.PD
MATHEMATICS FOR ENGINEERS
If C and D coincide as at (c), Fig. 28, then
(PI) 2 = AP ._PB
Fig. 28.
Example 14. The hardness number of a specimen, according to
load
Brinell's test, is given by curved area ofdepression'
Express this as a formula.
The curved area (of segment of sphere) = 2nrh (see p. 120).
r is radius of the ball making the indentation.
D = diameter of depression.
Then corresponds to a in the foregoing formulae,
I * D 2
= r./r*
V 4
so that h
and hardness number =
Length of Arc. Exact Rule. The length of the arc depends
on the angle it subtends at the centre of the circle : the total angle
at the centre is 360, this being subtended by the circumference.
An angle of 36 would be opposite an arc equal to onetenth of
the circumference, whilst if the arc was = of Oce, the angle
at the centre would be 120.
arc angle in degrees
In general  = ^
Oce 360
2trr X angle in degrees __ angle in degrees X radius
or, cure 7=
360 573
If the arc is exactly equal in length to the radius, the angle
then subtended ought to serve as a useful unit of measurement,
MENSURATION 9g
for one always expresses the circumference in terms of the radius.
This angle is known as a radian.
If the chord were equal to the radius, the central angle would
then be 60, so that when the arc is involved in the same way the
angle must be slightly less than 60.
Actually, the radius is contained 2ir times in the Oce, hence
2?r radians = 360, i. e., i radian =  = 573.
2ir
Therefore, to convert from degrees to radians divide by 57'3.
Thus 273 = 21 = 4.76 radians.
Radian or circular measure is the most natural system of angular
measurement ; ah 1 angles being expressed in radians in the higher
branches of the subject.
A simple rule for the length of an arc can now be established.
,, , 2nrx angle (degrees)
Length of arc =  & ^
2irrx angle (radian) /since 360 \
27r \ = 27r radians/
= rx angle subtended by the arc, expressed in
radians.
Now it is usual to represent the measurement of an angle in
radians by 6, and when in degrees by a. Thus the angle AOB in
Fig. 27, subtended at the centre of the circle by the arc ADB would
be expressed as 0, if in radians ; or a, if in degrees.
Hence, length of arc = g or re
Example 15. A belt passing over a pulley 10" diam. has an angle
of lap of 115 : find the length of belt in contact with the pulley.
In this case r = 5 and a = 115
/. length in contact = length of arc =
Example 16. What angle is subtended at the centre of a circle of
148 ft. diam. by an arc of 374 ft. ?
Arc = rd
. a _ arc _ 37'4 v<2 _ .. .
6  r^8 x
Thus the angle required is 505 radians or 290 degrees.
TOO
MATHEMATICS FOR ENGINEERS
It may be found of advantage to scratch a mark on the C scale
of the slide rule at 573, so that the conversions from degrees to
radians can be performed without any further tax on the memory.
Example 17. It is required to find the diameter of the broken
eccentric strap shown in the sketch (Fig. 29).
Here a = 2* and h = iz".
Then r =
a z + h z
2h
4+ i44
24
" 24
= 2265.
diam. = 453* (probably
Fig. 29.
An approximate rule for the length of arc is that known as
Huyghens' ; viz.
Length of arc = *~ *
o
where c 2 and c x represent the chord of half the arc and the chord of
the arc respectively (*'. e., c^ = 2a}. (See Fig. 27.)
To find the Height of an Arc from any Point in the Chord.
It is required to find the height EF (see Fig. 27) of the arc
ADB, the length of chord AB, the maximum height CD of the arc
ADB and the distance CF being given.
If O is the centre of the circle, OE is a radius and its length
can be found from r =
2h
Then
(OE) 2 = (EG) 2 + (GO) 2
= (EG) 2 + (CF) 2
and of these lengths, OE and CF are known ; therefore EG is found.
But FG is known, since FG = OC = r h.
:. the height EF, which = EG FG, is known also.
A numerical example
will demonstrate this
more clearly.
Example 18. A circular
arc of radius 15* stands on
a base of 24". Find its
maximum height, and also
its height at a point along
the base 5* from its ex
tremity. (Deal only with
the arc less than a semi
circle.) (See Fig. 30.) Fig. 30.
MENSURATION I0 i
To find h. We know that r 15*, and a = 12*
hence h = r Vr z a*
= 15 V225I 44
= 15 9 = 6* or 24*.
According to the condition stated in brackets h must be taken as
6*, i. e., the maximum height of the arc is 6*.
Then I5 2 = EG 2 +7
EG 2 = is 2 ; 8 = 22 x 8 = 176
or EG = 1326*
CO = rh = 156 = 9*
.*. EF = 13269 = 426*
or the height of the arc at the 5* mark is 426*.
Area of Sector. A sector is a portion of a circle bounded by
two radii and the arc joining their extremities; it is thus a form
of triangle, with a curved base (see Fig. 24). Its area is given by
a rule similar to that for the area of a triangle, viz.,  base* X height,
but in this case the base is the arc and the height is the radius (the
radius being always perpendicular to the circumference).
Hence Area of sector = J arc x radius,
or, in terms of the radius, and angle at the centre (in radians).
Area = ^r 2 ^, since for the arc we may write rd.
The area of the sector bears the same relation to the area of
the circle as the length of arc does to the o ce, i. e.
Area of sector angle (in degrees)
area of O 360
.*. Area of sector = ^7; x Tir 2
BQU
Area of Segment. The area of the segment, being the area
between the chord and the arc (see Fig. 24), can be obtained by
subtracting the area of the triangle from that of the sector. Thus,
in Fig. 24
Area of segment ADB = area of sector OADB area of triangle
OAB.
In place of this exact rule, we may use an approximate one,
viz.
2h f 1 chord + 3 arc\
Area of segment = y { ^ t
where h = maximum height of segment.
102
MATHEMATICS FOR ENGINEERS
When the arc is very flat the chord and arc become sensibly
the same, so that
2hfio chordl
Area of segment = ! 
2
=  X h x chord
2
= X maximum height X width.
The area of a segment may also be found from the approximate
rule
4 Id
Area of segment = ^h z *J r '608
o v a
where d = diam. of circle, and h = maximum height of segment.
Example 19. The Hydraulic Mean Depth (H.M.D.) a factor of
great importance in connection with the flow of liquids through pipes
or channels is equal to the section of flow divided by the wetted
perimeter.
Find this for the case illustrated in Fig. 31.
Here, section of flow = area of segment ACB
= area of sector OACB area of tri
angle OAB
= ~ X7rx6 2 4x6x6
360
= 977 18
= 103 sq. ins.
Wetted perimeter
_ arc ACB = ^ x 2 X n X 6 = ^JT
= 942"
.*. H.M.D. (usually denoted by tri)
e
IQ'3
= 1094
Note that for a pipe running full bore the H.M.D.
MENSURATION
103
Exercises 13. On Arcs, Chords, Sectors and Segments of Circles.
In Exercises i to 5, the letters have the following meanings as in
Fig. 27, v = radius, c^ = chord of arc, c 2 = chord of half arc, h = maxi
mum height of arc, and a = angle subtended at the centre of the circle
by the arc.
1. r = 8*, c t = 24"; find c l and h.
2. c t = 80", r = 50"; find c a and h.
3. Ci = 49*, c s = 25*; find r and h.
4. Ct = 6", r = 9*; find arc and area of segment.
5. c t = io*, h = 134*; find area of segment and a.
6. A circular arc is of 10 ft. base and 2 ft. maximum height Find
the height at a point on the base i'6* distant from the end, and also
the distance of the point on the base from the centre at which the
height is i ft.
7. A circular arc has a base of 3* and maximum height 4*. Find
(a) radius, (&) length of arc, (c) area of segment, (d) height of arc at a
point on the base i* distant from its end.
8. A crank is revolving at 125 R.P.M. Find its angular velocity
(t. e., number of radians per sec.).
9. If the angular velocity of a flywheel of i2'6* diam. is 45, find
the speed of a point on the rim in feet per minute.
10. Find the area of a sector of a circle of 97* diam., the arc of the
sector being 123" long.
11. One nautical mile subtends an angle of i minute at the centre
of the earth; assuming a mean radius of 20,890,000 feet, find the
number of feet in i nautical mile.
12. Find the difference between the apparent and true levels (i e
CE), if AC = 1500 yards and R = 3958 miles. [See (a), Fig. 32.]
Fig. 32.
13. (6), Fig. 32 (which is not drawn to scale), shows a portion of a
curve on a tramway track. If R = radius of quickest curve allowable
(in feet), T = width of groove in rail (in inches), and B = greatest
permissible wheel base (in feet) for this curve, find an expression for
B in terms of R and T.
14. A circle of 24" diam. rolls without slipping on the circum
ference of another circle of 614" diam. What angle at the centre is
swept out in i complete revolution of the rolling circle ?
104
MATHEMATICS FOR ENGINEERS
15. A railway curve of J mile radius is to be set out by " i chain "
steps. Find the " deflection angle " for this, i. e., the angle to which
the theodolite must be set to fix the
position of the end of the chain.
The deflection angle is the angle
between the tangent and the chord.
16. Fig. 33 shows a hob used for
cutting serrations on a gauge. It
was found that the depth of tooth
cut when the cutting edge was along
AB was not sufficiently great. Find
how far back the cutter must be
ground so that the depth of serration
is increased from 025* to '027*,
i. e., find x when AB = 025" and
CD = 027".
C25
Fig. 33. Gauge Hob.
The Ellipse. The ellipse is the locus of a point which moves
in such a way that the sum of its distances from two fixed points,
called the foci is constant. This constant length is the length of
the longer or major axis.
In Fig. 34, if P is any point on the ellipse, PF f PF 1 =
constant = AA 1 , F and F 1 being the foci.
Fig. 34. The Ellipse.
Let major axis = 2, and minor axis = 2&.
Then from the definition, FB = F X B = a.
; In the triangle FOB, (FB) 2 = (FO) 2 + (OB) 2
or a 2 = (FO) 2 + (b) z
FO = Vo 2 ^ 2
so that if the lengths of the axes are given the foci are located.
Area = nab. (Compare with the circle, where area = nrr.)
The perimeter of the ellipse can only be found very approximately
as the expression for its absolute value involves the sum of an
MENSURATION
105
infinite series. Various approximate rules have been given, and of
these the most common are, perimeter = 7r(a+6), or
the second of which might be written in the more convenient form
2 + b 2 . These rules, however, do not give good results
when the ellipse is flat. A rule which appears to give uniformly
good results is that of Boussinesq, viz.
perimeter = 7r{l5(a f b) Vab} .
The height of the arc above the major axis at any point can
most easily be found by multiplying the corresponding height of
the semicircle described on the major axis as diameter by ,
a
e. g., referring to Fig. 34, QN = XMN.
Example 20. The axes of an ellipse are 48" and 74". Find its
perimeter and its area.
According to our notation, viz. as in Fig. 34, ia = 74, a = 37
26 = 48, 6 = 24.
Then the perimeter = ir(a + b) = 77(61) = 1915"
77 A/2(a 2 +& 2 ~) = TT V2(i9 45 ) = 1958"
7r{i5(a + 6) Vab} = 7r{i5(6i) V^y x 24}
or
or
Area
Fig. 35. The Parabola.
106 MATHEMATICS FOR ENGINEERS
The Parabola. The parabola is the locus of a point which
moves in such a way that its distance from a fixed straight line,
called the directrix, is always equal to its distance from a fixed
point called the focus.
Referring to Fig. 35, PZ = PF, where F is the focus, and P
is any point on the curve.
The distance BA, which is equal to AF, is always denoted by a.
The chord LL 1 through the focus, perpendicular to the axis,
is called the latus rectum, and from the definition it will be seen to
be equal to 40. The latus rectum, in fact, determines the propor
tions of the parabola just as the diameter does the size of the circle.
If PQ = y and AQ = x, then FQ = AQ AF = (x a) and
PF = PZ = BQ = x+a.
Then in the triangle FPQ,
(FP) 2 = (PQ) 2 +(FQ) 2
or x z +a z \2ax = y 2 +# 2 +0 2 2ax
whence y 2 = 4ax
or ( width) 2 = latus rectum X distance
along axis from vertex, Fig. 36.
e. g., (MR) 2 = 4 x AR in Fig. 36.
If a semicircle be drawn with F as centre and with FP as
radius, to cut the axis of the parabola in T and N, PT is the tangent
at P and PN is the normal. (Fig. 35.)
The distance along the axis, under the normal, *. e., QN in Fig. 35,
is spoken of as the subnormal. For the parabola, the length of
the subnormal is constant, being equal to 2a, i. e., % latus rectum.
Use is made of this property in the design of governors. If
the balls are guided into a parabolic path, the speed will be the
same for all heights, for it is found that the speed depends on the
subnormal of the parabola, and as this is constant so also must
the speed be constant.
2
The area of a parabolic segment = = of surrounding rectangle, . e.,
B
area of P X AP (Fig. 36) = xPP x xAQ. Length of parabolic
8 D 2
arc = Sfs g approximately, where S = span and D = droop
or sag, as indicated in Fig. 36.
Circular and other arcs are often treated as parabolic when
the question of the areas of segments arises; and if the arcs are
very flat no serious error is made by so doing. The rule for the
MENSURATION
107
area of a parabolic segment is so simple and so easily remembered
that one is tempted to use it in place of the more accurate but more
complicated ones which may be more applicable.
Take, for example, the case of the ordinary stressstrain diagram,
as in Fig. 37. To find the work done on the specimen up to frac
ture it is necessary to measure the area ABCDE. Replacing the
irregular curve (that obtained during the plastic stage) by a portion
of a parabola BF, and neglecting the small area ABG, we can say
that area ABCDE = rectangle AGHE f parabolic segment BFH
A Extension E
Fig 37 Stressstrain Diagram.
L .
Fig. 38. The Hyperbola.
If the ratio ^ is denoted by r, then the result may be written
eMY . L\ Me, . .
area ABCDE = (2+^) = y ( 2 +')>
which is Kennedy's rule.
So, also, in questions on calculations of weights, circular seg
ments are often treated as parabolic.
Example 21. The bending moment diagram for a beam 28 feet
long, simply supported at its ends, is in
the form of a parabola, the maximum
bending moment, that at the centre being
49 tons feet. Find the area of the
bending moment diagram, and find also
the bending moment at 6 feet from one
end (this being given by the height of
the arc at D, Fig. 39).
Area of parabolic segment ACB
=  x 49 X 28 = 915 units.
These units are tons feet X feet or tons feet*.
Fig. 39.
io8 MATHEMATICS FOR ENGINEERS
Now it can be shown that the moment of onehalf the bending
moment area (viz. AMC), taken round AG determines the deflection
at A and also at B.
Actually, the maximum deflection (at A or B) = x area of
AMC x L where E = Young's modulus for the material of the beam
and I = second moment of its section. Since E would be expressed
in tons per sq. foot and I in (feet) 1 the deflection would be expressed in
feet 2 x feet 2 tons x feet . . , 4
? ri t. e., in teet.
tons x feet*
To find the height ED
(MB) 2 = 40 x MC from definition
(MB) 2 14'
jl / _* f ' ~ A
MC 49
EF* = 4 a x CF
4 4
DE = MC  CF = 49  16 = 33.
.*. Bending moment 6 feet from end = 33 foot tons.
Example 22. Find the length of the subnormal of the parabola
y* 6y i6x 23 = o.
The equation might be written
(y* 6y + 9) i6x 32 = o
or (y s) 2 = i6(# + 2).
This is of the form Y 2 = 4X
where Y = y 3, X = # + 2, a = 4
.*. Length of subnormal = 2 a = 8 units ; and is a constant.
The Hyperbola. The hyperbola is the locus of a point which
moves in such a way that the difference of its distances from two
fixed points, called the foci, is constant. There are two branches
to this curve, which is drawn in Fig. 38. If P 1 is any point on the
curve, then PT I^F 1 = AA 1 = za.
AA 1 = transverse axis, and BB 1 = conjugate axis = 26.
DOD 1 and EOE 1 are called asymptotes, i. e., the curve approaches
these, but does not meet them produced : they are, as it were,
its boundaries.
PM and PQ are parallel to EOE 1 and DOD 1 respectively : then
a most important property of this curve is that the product
PM X PQ is constant for all positions of P.
If BB 1 = AA 1 , the asymptotes are at right angles and the
MENSURATION 109
hyperbola is rectangular : e.g., the curve representing Boyle's
law (the case of isothermal expansion) is a rectangular hyperbola,
the constant product being denoted by C in the formula, PV = C.
Exercises 14. On the Ellipse and the Parabola.
1. A parabolic arc (as in Fig. 36) stands on a base of 12*. The
latus rectum of the parabola being 8", find
(a) Maximum height of arc; (6) area of segment; (c) width
at point midway between the base and the vertex.
2. A parabola of latus rectum 5* stands on a base of 6", find
(a) Maximum height of arc ; (6) height at a point on the base
2* from the centre of the base ; (c) area of segment ; (d) position
of focus.
3. A parabolic segment of area 24 sq. ins. stands on a base of 12*.
Find the height of the arc at a point 2j* from the centre of the base
and also the latus rectum.
4. The axes of an ellipse are 10* and 6* respectively. Find
(a) The area ; (6) distance between foci ; (c) height of arc at a point
on the major axis 4" from the centre; (d) perimeter by the 3 rules.
5. The lengths of the axes of an ellipse can be found from a* = 30,
b 2 = 15, where a and 6 have their usual meanings (see Fig. 34). Find
(a) Area of ellipse; (6) distance of foci from centre; (c) peri
meter by the three given rules.
6. A manhole is in the form of an ellipse, 21* by 13*. Find, approxi
mately, the area of plate required to cover it, allowing a margin of 2*
all the way round and assuming that the outer curve is an ellipse.
7. A cantilever is loaded with a uniform load of 15 cwts. per foot
run. The bendingmoment diagram is a parabola having its vertex at
wl*
the free end, and its maximum ordinate (at the fixed end) is , where
' 2
w = load per foot run, and / is the span which is 18 ft. Find the
bending moment at the centre, and at a point 3 ft. from the free end.
8. It is required to lay out a plot of land in the form of an ellipse.
The area is to be 6 acres and the ratio of the axes 3 ; 2. Find the
amount of fencing required for this plot.
9. There are 60 teeth in an elliptical gear wheel, for which the
pitch is 235*. If the major axis of the pitch periphery is twice its
minor axis, find the lengths of these axes.
10. Find the number of feet per ton of oval electrical conduit
tubing, the internal dimensions being f* x f* and the thickness
being 042* (No. 19 B.W.G.). Weight of material = 296 Ib. per
The Prism and Cylinder. A straight line moving parallel
to itself, its extremities travelling round the outlines of plane
figures generates the solid known as the prism. If the line is always
at right angles to the plane figures at its extremities the prism is
known as a right prism. If the plane figures are circles the prism
becomes a cylinder.
no
MATHEMATICS FOR ENGINEERS
A particular case of the prism is the cuboid, in which all the
faces are rectangular, i. e., the plane figures at the extremities of
the revolving line are rectangles.
For all prisms, right or oblique
Volume = area of base x perpendicular height.
The lateral or side surface of a right prism
= perimeter of base x height.
Total surface = lateral surface + areas of ends.
Applying to the Cuboid.
Volume = area of base X height
= acxb = axbxc. (Fig. 40.)
Lateral surface = 2ab{2bc
Total surface = zab+zbc+zac
= 2(ab\bc+ca).
Fig. 40.
If a = b = c, the cuboid becomes a cube,
and then vol. = a 3
and total surface = 2( 2 + 2 + 2 ) = 6a 2
If the diagonal of a cuboid is required it can be found from,
diagonal = V 2 +6 2 +c 2 ; whilst for the cube, diagonal = aVs.
Example 23. An open tank, made of material y thick, is 2'6*
long, 10" wide and 15* deep (these being the outside dimensions). Find
the amount of sheet metal required in its construction if the plates
are prepared for acetylene welding, and find also the capacity of the
tank.
If the plates are to be joined by acetylene welding no allowance
must be made for lap ; the plates would be left as shown in the sketch
at A, Fig. 41.
MENSURATION
in
Total Surface = 2 x (15  J)[io (2 x J)]
H 2 x Us  i)[3o  (2 x J)] + [30  (2 x i)][io  (2 x i)]
= 280 + 870 + 280 sq. ins. = 1430 sq. ins. = 994 sq. ft.
Volume = (30 J) x (10 \) x (15 J)
= 2 95 x 95 x 1475 f .
1728
x 95 x 1475 x 625
Capacity =
1728
149 gallons.
ions
or jd 2 h, where r = radius
If the weight of water contained is required
Weight = 149 x 10 = 149 Ibs.
Note. I cu. ft. of fresh water weighs 624 Ibs.
I cu. ft. of salt water weighs 64 Ibs.
i gallon of fresh water weighs 10 Ibs.
6 gallons occupy i cu. ft.
i cu. cm. of water weighs i grm.
Applying the foregoing rules to the cylinder.
Vol. = area of base X height
i. e., Volume of cylinder = irr z xh = ir
of base, d = diam. of base, h = height or length.
Lateral surface = 2rh.
Total surface = 2jrrh\2Trr z
= 2irr(h+r).
Volumes of cylinders can most readily be obtained by the use
of the slide rule, adopting an extension of the rule mentioned on
p. 92.
It is repeated here with the necessary extension :
Place one of the C's, on the C scale of the rule, opposite the
diameter on the D scale : then place the cursor over the length
on the B scale, and the volume is read off on the A scale.
Rough approximation, Vol. = f d 2 /!.*
E. g., Diam. = 463"
Length = 1875".
Vol. (by approximation) = fx 20x20 = 300 cu. ins.
Vol. (by slide rule) = 316 cu. ins.
Exercises
Dia.
Length.
Vol.
23
47'3
300
28
1245
4945
ii2 MATHEMATICS FOR ENGINEERS
Example 24. Find the weight, in Ibs. and grms., per metre of
copper wire of diam. 045 cm. (Copper weighs 32 Ib. per cu. in.)
Note 254 cms. = 1 in.
453'6 grms. = 1 Ib.
Then I cu. cm. = , I >, cu. in.
1 2 '54J
.'. Weight of i cu. cm. of copper = . t 32 . 3 Ib.
Vol. of i metre of wire = x (O45) 2 xioo cu. cms.
4
= 159 cu. cm.
.*. Weight = >:[ 59X 32 = OQ
(254)3 2
or weight = 00311 x 4536
= 1409 grms.
Example 25. A boiler contains 480 tubes, each 6 ft. long and
2 ins. external diameter. Find the heating surface due to these.
The heating surface will be the surface in contact with the water,
i. e. t the outside surface of the tubes.
Lateral surface = ird x length x no. of tubes
= jr x Q x 6 x 480
4 s
= 2070 sq. ft.
Exercises 15. On Prisms and Cylinders.
Prisms
1. A room 22 ft. long by IS'IQ" wide is g'^" high. Find the
volume of oxygen in it, if air contains 21 % of oxygen and 79 % of
nitrogen by volume.
2. A block of wrought iron is'xg'xf weighs 142 Ibs. Find the
density of W.I. (Ibs. per cu. in.) and also its specific gravity if i cu. ft.
of water weighs 624 Ibs.
3. The weight of a brass plate of uniform thickness, of length
6'5" and breadth n" was found to be 794 Ibs. If brass weighs 3 Ib.
per cu. in., find the thickness of this plate.
4. The sectional area of a ship at its waterline is 5040 sq. ft. ; how
many tons of coal would be needed to sink her i f t ? (35 cu. ft. of sea
water weigh i ton.)
5. The coefficient of displacement of a ship
volume of immersed hull of ship
volume of rectangular block of same dimensions
If the displacement is 4000 tons and the hull can be considered to
have the dimensions 32o'x35'xi5' find the coefficient of displace
ment.
MENSURATION II3
6. The ends of a right prism 8'4" long are triangles having sides
19", 272" and 114" respectively. Find the volume of this prism.
7. Water is flowing along a channel at the rate of 65 ft. per sec
The depth of the channel is 9", the width at base 14", and the side
slopes are i horizontal to 3 vertical. Find the discharge
(a) In cu. ft. per sec. ; (b) in Ibs. per min.
8. A tightlystretched telephone cable, 76 ft. long, connects up two
buildings on opposite sides of the road. The points of attachment of
the ends are 38 and 64 ft. above the ground respectively, one being
37 ft. further along the road than the other, and the buildings each
standing 10 ft. back from the roadway. Find the width of the road.
9. The section of an underground airway is as shown in Fig. 42.
Air is passing along the airway at 105 ft. per sec. ; find the number of
cu. ft. of air passing per minute.
Fig. 42.
Fig. 420.
10. Find the volume of stone in a pillar 20 ft. high, the crosssection
being based on an equilateral triangle of i foot side, having three
circular segments described from the angular points as centres, and
meeting at the mid points of the sides. Find also its weight at 140 Ibs.
per cu. foot. (Fig. 42.)
Cylinders
11. The diameter of a cylinder is 387", and its length is 28'3".
Find its curved surface, its total surface and its volume.
12. Find the ratio of total heating surface to grate area in the case
of a Caledonian Railway locomotive. The heating surface in the
firebox is 119 sq. ft., the grate area is 2063 sc l ft and there are 275
tubes, of if external diameter, the length between the tube plates
being io'j".
13. A current of 6 ampere at 100 volts was passed through the
two field coils of a motor. If the diam. of the coils was 4* and the
length 4^', find the number of watts per sq. in. of surface. (Curved
surface only is required.)
14. Find the weight of 5 miles of copper wire of 02* diam., when
copper weighs 32 Ib. per cu. in.
15. Find the weight of a hollow steel pillar, 10 ft. long, whose
external diam. is 5* and internal diam. is 4 (i cu. ft. of steel weighs
499 Ibs.). (See Area of Annulus, p. 93.)
16. Water flows at the rate of 288 lb.s. per min. through a pipe of
ij" diam. Find the velocity of flow in feet per sec.
17. Find the heating surface of a locomotive due to 177 tubes of
1 1" diam., the length between the tube plates being io'6*,
I
114 MATHEMATICS FOR ENGINEERS
18. A piston is moving under the action of a mean effective pressure
of 382 Ibs. per sq. in. at a speed of 400 ft. per min. If the horsepower
developed is 70, find the diam. of the piston.
r_j p _ Feet per min. x total pressure in lbs.~
33000
19. In a tencoupled locomotive there were 404 tubes of 2* diam.
and the heating surface due to these was 3280 sq. ft. Find the length
of each tube.
20. The diameter of a hydraulic accumulator is 12" and the stroke
is 6 ft. Find the work stored per stroke if a constant pressure of
750 Ibs. per sq. in. be assumed.
21. In calculating the indicated horsepower of an engine at various
loads it was found that a saving of time was effected if an " engine
constant " was found.
Vol. of cylinder
If the engine constant =
12 x 33000
find this, if diam. = 55* and stroke = 10*.
22. The weight of a casting is to be made up from 414 Ibs. to
416 Ibs. by drilling a ^" diam. hole and plugging with lead. To
what depth must the hole be drilled if the weights of lead and cast
iron are 41 and 26 Ib. per cu. in. respectively ?
23. The conductivity of copper wire can be expressed by its resist
ance per gramme metre. Find the " conductivity " of a wire 5 metres
long and of 762 cm. diam. (No. i S.W.G.) if the Resistance is given
by 00000017 x 6  ; the units being cms. and the weight of i
cu. cm. of copper being 891 grms.
24. Find the weight, in Ibs. per 100 feet, of electrical conduit tubing
of external diam. 2" and internal diam. I872", the weight of the
material being 296 Ib. per cu. in.
25. A 10* length of i" diam. steel rod is to be forged to give a bar
ij" wide and " thick. Assuming no loss in the forging, find the
length of this bar.
Pyramid and Cone. If a straight line of variable length
moves in such a way that one extremity is always on the boundary
of a plane figure, called the base, whilst the other is at a fixed
point, called the vertex, the solid generated is termed a pyramid.
If the line joining the vertex to the geometrical centre of the base
is at right angles to the base, then the pyramid is spoken of as a
right pyramid.
When the base is circular the figure is termed a cone ; right
circular cones being those most frequently met with. These are
cones for which all sections at right angles to the axis are circles.
The lateral surface of a right pyramid will evidently be the
sum of the areas of the triangular faces.
Consider the case of a " square " pyramid, i. e., where the base
is square [see (), Fig. 43]. The triangular faces are equal in area.
MENSURATION II5
Area of each = base X height
= i X AB x VL [see (a), Fig. 43]
where VL is spoken of as the slant height of the pyramid ; its value
being found from
VL = WO 2 + OL 2 [see (b), Fig. 43]
LO being \ side of base.
Total lateral surface = 4 x AB x VL
= 2 . AB x VL
or lateral surface of pyramid = J perimeter of base x slant height.
This rule will hold for all cases in which the base is regular.
[Note that if the base is rectangular, there will be two distinct
slant heights.]
V
B L
o
Fig. 43. Square Pyramid.
Length of edge of pyramid = VB = VVO 2 f OB 2 [see (b),
Fig. 43], where OB = diagonal of base.
The three lengths or heights should be clearly distinguished.
VO = perpendicular height, or more shortly, the height
VL = slant height, and VB = length of edge.
Volume of pyramid is onethird of that of the corresponding prism
(i. e., the prism on the same base and of same vertical height).
.'. Vol. of pyramid = J x area of base x perpendicular height.
Example 26. A flagstaff, 15 ft. high, is kept in position by four
equal ropes, one end of each being attached to the top of the staff,
whilst the other ends are fastened to the corners of a square of 6 ft.
side. Find the length of each rope.
n6
MATHEMATICS FOR ENGINEERS
Diagonal of base = 6\/2 (the diagonal of a square always =
\/2~ x side). The length required is the length of the edges, viz. VB
[see (6) Fig. 43].
Now VO = 15, OB = 3 Vz, hence VB = \/ (3 V
(i 5)* = ^18+225
= ^243
VB = length of each rope = 156 ft.
Applying to the Cone. If the lateral surface of the cone is
developed, i. e., laid out into one plane, a sector of a circle results,
the radius being the slant height /, and the arc being the circum
ference of the base of the cone or 2irr (see Fig. 44).
Now area of sector of circle = arc X radius =
=
i. e., area of curved surface of a cone = vrl.
Notice that this agrees
with the result obtained
from the rule for the pyra
mid, viz. perimeter of base
X slant height.
If the development of the
cone were actually required
it would be necessary to find
the angle a (Fig. 44).
Now
Janldeufit
I
a
360
arc
Oce
360?
I
2irl
Lateral surface, then, =
Total surface =
Fig. 44.
= irr(l+r).
As the cone is a special form of pyramid its volume will be
onethird that of the cylinder on the same base and of the same
height.
1 T
Vol. of cone
or ~d 2 h or 2618</ 2 /i
\Lt
d being the diameter of the base and h the perpendicular height.
The approximation for the volume is x(diam.) 2 x height.
Example 27. A projectile is cylindrical with a conical point (see
Fig. 45). Find its volume.
MENSURATION
117
As the cone is on the same base as the cylinder its volume can be
accounted for by adding J of its length to that of the cylinder, and
treating the whole as one cylinder.
Fig 45
Hence, net length = 4 / "+(Jxi8") = 46"
.*. total vol. = x(i6) 2 X46
4
= 926 cu. ins.
Frusta. If the pyramid or cone be cut by a plane parallel to
its base the portion of the solid between that plane and the base is
known as a frustum of the pyramid or cone.
The lateral surface and the volume can be found by subtracting
that of the top cone from that of the whole cone or by the following
rules, which give the results of this procedure in a more advanced
form.
Lateral surface of frustum of pyramid or cone
= {sum of perimeters of ends} X slant thickness.
Vol. of frustum of pyramid or cone =
where A and B are the areas of the ends, and h is the perpendicular
height or thickness of the frustum. (The proofs of these rules are
given on p. 123.)
For the frustum of a cone these rules may be expressed in rather
simpler fashion
Lateral surface of frustum of cone = 7r/(R+r)
/ being the slant height of the frustum.
Volume of frustum of cone = ^~{R 2 } r 2 + Rr}
where R and r are radii of ends, and h is the thickness of the frustum.
Example 28. A friction clutch is in the form of the frustum of a
cone, the diameters of the end being 6$" and 4!*, and length 3$".
Find its bearing surface and its volume (see Fig. 46).
n8 MATHEMATICS FOR ENGINEERS
The slant height must first be found
.. / = 368*.
Now R = 3*25, and r 213.
.*. Lateral surface
= 77 X 368(325 + 213)
= 7T x 368 x 538 = 622 sq. ins.
Also
Volume = {R 2 + y 2 + Rr}
= 3l 5 {1054 + 453 + 692}
TT X 35 X 2199 o
2? 805 cu. ms.
3   
Fig. 46. Friction Clutch.
Exercises 16. On Pyramids, Cones and Frusta.
1. The sides of the base of a square pyramid are each 137" and
the height of the pyramid is 95*. Find (a) the volume, (b) the lateral
surface, (c) the length of the slant edge.
2. The volume of a pyramid, whose base is an equilateral triangle
of 52" side, is 796 cu. ins. Find its height.
3. Find the total area of slating on the roof shown at (a) Fig. 47.
Plan
27^
Fig 47
4. Find the volume of a hexagonal pyramid, of height 512*, the
base being a regular hexagon of 174* side.
5. A square pyramid of height 5 ft., the sides of the base being
each 2 ft., is immersed in a tank in such a way that the base of the
MENSURATION 119
pyramid is along the surface of the water. Find the total pressure on
the faces of the pyramid if the average intensity of pressure is the
intensity at a depth of i'3* below the surface ; the weight of i cu. ft.
of water being 624 Ibs.
6. A turret is in the form of a hexagonal pyramid, the height being
25 ft. and the distance across the corners of the hexagon being 15 ft.
Find the true length of the hip (i. e., the length of a slant edge), and
also the lateral surface.
Cones.
7. The curved surface of a right circular cone when developed was
the sector of a circle of 1142" radius, the angle of the sector being
127. Find the radius of the base of the cone, and also its height.
(Refer p. 116.)
8. A piece in the form of a sector (angle at centre 66) is cut away
from a circular sheet of metal of 9" diam., and the remainder is made
into a funnel. Find the capacity of this funnel.
9. A right circular cone is generated by the revolution of a right
angled triangle about one of its sides. If the length of this side is
324 ft. and that of the hypotenuse is 559 ft., find the total surface
and the volume of the cone.
10. A vessel is in the form of a right circular cone, the circum
ference of the top being 1974 ft anc ^ the ^ uu depth of the vessel being
12 ft. Find the capacity in gallons. Find also the weight of water
contained when the vessel is filled to onehalf its height.
11. A conical cap is to be fitted to the top of a chimney. The cap
is to be of 7" height and the diam. of the base is 12". Find the amount
of sheet metal required for this.
If this surface be developed, forming a sector of a circle, what will
be the angle of the sector ?
Frusta of Pyramids and Cones.
12. A pier is in the form of a frustum of a square pyramid. Its
ends are squares, of side 3 ft. and 8'6" respectively, and its height is
6 ft. Find its volume and its weight at 140 Ibs. per cu. ft.
13. A circular brick chimney is too ft. high and has an internal
diam. of 5 ft. throughout. The external diam. at base is n ft. and
at the top 7 ft., the thickness being
uniformly reduced from bottom to top.
Find its weight at 120 Ibs. per cu. ft.
14. Find the lift h of the valve
shown in Fig. 48, given that BC = i f *
and AD = if*. It is necessary that
the area of the lateral surface of ABCD
should be 13750".
15. One of a set of weights had
the form of a frustum of a cone, the
thickness being 4^", the diam. at the Fig. 48.
too being 10", and the diam. at the
tStom bling 2*'. Find its volume and its weight at 26 Ib. per cu m
16. A square pyramid of height g" and side of base 15 is ' cut into
two parts by a plane parallel to the base and distant 4* from it. Find
the volume of the frustum so formed, and also its lateral surface.
I2O
MATHEMATICS FOR ENGINEERS
17. A cone 12" high is cut at 8" from the vertex to form a frustum of
a cone of volume 190 cu. ins. Find the radius of the base of the cone.
18. The parallel faces of a frustum of a pyramid are squares on sides of
3* and 5" respectively, and its volume is 32! cu. ins. Find its altitude
and the height and lateral edge of the pyramid from which it is cut.
19. A conical lampshade is 2 \" diam. at the top and 8J" diam. at
the bottom. The shortest distance between these ends is 5*. Find
the area of material required for this, allowing 4 % extra for lapping.
By drawing to scale, find the area of the rectangular piece from
which the shade would be cut.
20. A pyramid, having a square base of side 18*, and a height of
34*, is cut by a plane distant n" from the base and parallel to it. Find
the total surface of the frustum so formed, and also its volume.
The Sphere. If a semicircle revolves about its diameter as
axis it sweeps out the solid known as the sphere.
The portion of the sphere
between two parallel cutting
planes is . known as a zone :
thus CDFE in Fig. 49 is a
zone.
F The portion included be
B tween two planes meeting
along a diameter is known as
a lune.
A plane section through the
centre is called a great circle :
any other planes will cut the
sphere in small circles.
Thus, the section on AB
(Fig. 49) would be a great
circle, and the sections on
The portion CMD is a segment.
Fig. 49 .
CD or EF would be small circles.
Let the radius of the sphere = r, and diam. = d.
Then the surface of the sphere = 4 X area of a great
= 4 X Try 2 = 4rrr 2 OF Trd 2
Vol. of sphere = ^nr 3 =  . d 3 = or '5236d 3
Surface of a zone = curved surface of circumscribing cylinder
(h being the distance between the parallel planes).
Vol. of zone = ^{3 (/y 5 + r 2 2 ) f h 2 }
[The proof of these two rules will be found in Vol. II of Mathematics
for Engineers.]
MENSURATION
121
The zone may be regarded as a form of frustum, r v and r z being
the radii of the ends and h being the thickness.
If r 1 = o, the zone becomes a segment, and then
Vol. of segment = ^{Srj 2 + A 2 }
k being the height of the segment.
A relation that exists between the volumes of the cone, sphere
and cylinder should be noted. Consider a sphere, of radius r; its
circumscribing cylinder (i. e., a cylinder with diam. of base = zr and
height = 2r], and the cone on the same base and of the same height.
Then. Vol. of the sphere = Trr 3 = 7ry 3 X2
3 3
2
Vol. of the cylinder = Try 2 x zr =  Try 3 x 3
o
TtY^ 2
Vol. of the cone = X 2r =  nr 3 x i.
3 3
Hence the respective volumes of the cone, sphere and cylinder
of equal heights and diameters are in the proportion 1:2:3.
Example 29. A disc of lead 14* diam. and 8" thick is melted down
and cast into shot of (a) " diam., (6) J* diam. How many shot can
be made in each case, supposing no loss ?
Case (a). VoL Qf digc = ^ x I4 2 x . 8 cu ^
4
= 39'27r cu. ins.
Vol. of i shot = I x () 3 =  r ^
3927T x 6 x 512
.* No. of shot = ^
7T
= 120,300.
Case (6). The diam. is twice that of Case (a); therefore the vol.
of i shot is 2 3 , i. e., 8 times as great.
A No. of shot gggg 15.038.
Example 30. Find an expression for the weight in Ibs. of a sphere
of any material, having given that the weight of a cu. in. of copper is
318 Ib. (approx.).
Weight of a copper sphere of diam. D
= volume x density
= ^D 3 x 318
122
Hence the weight of a sphere of any material, its diameter
being D
_ D 3 x specific gravity of solid
"~ 6 x specific gravity of copper
Example 31. Find the total surface of a hemispherical dome, of
inside diam. 5^" and outside diam. 74*.
Outside surface = J X 4?r x (3'7) 2 = 856 sq. ins.
Inside surface = \ x 477 x (2'75) 2 = 475 ,,
Area of base = n(3'7 2 275 2 ) = 192
/. Total surface area = 1523 sq. ins.
Similar Figures. Similar figures are those^ having the same
shape : thus a field and its representation on a drawingboard are
similar figures. Triangles, whose angles are equal, each to each,
are similar figures.
On every hand one comes across instances of the application of
similar figures ; and in connection with these, three rules should be
remembered.
(1) Corresponding lines or sides of similar figures are proportional.
(Euclid, VI. 4.)
(2) Corresponding areas or surfaces are proportional to the squares
of their linear dimensions. (Euclid, VI. 20.)
(3) Volumes or weights of similar solids are proportional to the cubes
of their linear dimensions.
E. g., consider two exactly similar cones, the height of one being
three times that of the other.
Then (i) the radius and hence the circumference of the base of
the first are three times the radius and circumference of the second
respectively.
(2) The curved surface of the first = 3 2 X that of the second.
(3) The volume or weight of the first = 3 3 X volume or weight
of the second.
To generalise, using the symbols L, S, and V for side, surfaces
and volumes respectively
If the ratio of the linear dimensions of two similar figures is
. , , L! Sj /LA 2
represented by j, then ^ = ( ~ ] (i)
and
^Hri)
MENSURATION
If it is desired to connect up volumes with surfaces
/ C ^ 3 / T \H
By cubing equation (i) [*) = ( tl)
W2/ VLg
By squaring equation (2) (*] = /~i
123
Hence
or
(3)
Example 32. A conical lampshade has the dimensions shown in
Fig. 50. Find the height of the cone of which it is a part.
Let x inches be the height of the
top triangle, viz. ABC.
Then ABC and ADE are similar
triangles, hence the ratio v  is
OclSG
the same for both.
x
i. e., ^ for the small triangle must =
x + 4
for the large triangle.
Then, by multiplying across
iox = 6x + 24
4* = 24
x = 6"
Total height of cone = 6 + 4 = 10*.
Fig. 50.
It is convenient at this stage to insert the proofs of the rules
for the lateral surface and the volume of a frustum, given on p. 117.
In Fig. 50 let the height or thickness FG of the frustum BCED
be denoted by h; let A be the area of the end DE and let B be
the area of the end BC. [Note. The figure is taken in these proofs
to be the elevation of a Pyramid, so that the proofs may be perfectly
general.]
Then, from the similarity of the triangles ABC and ADE
perimeter of end DE AD AB + BD =
:_^_i _ _ J ~Df~* A "D A "D '
BD
perimeter of end BC AB
, p. ofDE BD
whence  , p/ ^ i = 7^5
p. of BC AB
p. of DE p. of BC _ BD
AB ' AB
(p. being written to denote
perimeter)
or
p. ofBC
AB
124 MATHEMATICS FOR ENGINEERS
Lateral surface of frustum BCED = lateral surface of pyramid ADE
lateral surface of pyramid ABC
= J(p. of DExAD)(p. of
BCxAB)
. = i[(p. of DExAB) + (p. of DE
^ xBD)(p. of BCxAB)]
= [AB(p. of DEp. of BC)
+ (p. of DExBD)]
Substituting from equation (i) = [(p. of BCxBD) + (p. of
DExBD)]
=  X BD X sum of perimeters of
ends
=  sum of perimeters of ends
X slant thickness.
Again, since ABC and ADE are similar solids, the areas of their
respective bases are proportional to the squares of their respective
heights
_
~~
B ~~ (AF)
By transposition B = A X / AG ( 2 ......... (2)
(3)
Also by extraction of the square root
VA_AG
VB~AF
Volume of frustum BCED
= vol. of pyramid ADE vol. of pyramid ABC
= JxAxAGJxBxAF
By substitution from equation (2)
(AF^ 2
jfzzJL v A TT
/ A ,/ \ A /\ Xi.1/
_ r(AG)3(AF)31
 aiil / A ^\o
Factorising the numerator (see p. 53)
_ JA[(AG AF)] [(AG) 2 + (AG X AF) + (AF) 2 ]
(AC) 2
FAG AFM i^rAxjAG) 2 AxAGxAF , Ax(AF) 2 ]
[AGAFA] = ^LlAGF + (AG) + @ST.
MENSURATION
Substituting from equation (3)
125
= p[A+ VAB+B].
Example 33. A surveyor's chain line is to be continued across a
river. Describe a method by which the line may be prolonged and
show how the required distance may be deduced.
Suppose C is a point on the line : select some station A on the
opposite bank (Fig. 51) and put A, B and C in line. Set off BD and
CE as offsets at right angles, so that E, D and A are in a straight line.
AB _ DF BC
BD
Then
*'. e., AB =
FE ~ CE  BD
BC x BD
CE  BD
or AB is found.
Fig. 51
Example 34. The actual area of a field is 5 acres : on the plan it
is represented by an area of 50 sq. ins. To what scale is the plan
drawn ?
We are told that 50 sq. ins. represent 5 acres or 50 sq. chains.
Hence i sq. in. represents i sq. chain
or i* represents i chain.
So that the scale is i* to a chain, or the representative fraction
_ __
22 X 36 ~ 7Q2
126 MATHEMATICS FOR ENGINEERS
Example 35. The heating surfaces of two exactly similar boilers
are 850 and 996 sq. ft. respectively. The' capacity of the second being
750 gallons, what is the capacity of the first ?
It is not necessary to determine the ratio of the linear dimensions,
for statement (3) on p. 123 can be used, since the capacities are pro
portional to the volumes.
Now S x = 850, S a = 996, V 2 = 750, and V x is required.
V, \S,
or V, 7S .x()
log Vj = log 750 + i 5 (log 850  log 996)
= 28751 + 15(29294  29983)
= 28751 15 x 0689
= 27717
.*. V\ = 591 gallons.
An application of similar figures is found in the engraving
machine and in the reducing gear used in connection with indicators.
In Fig. 52 such a gear is represented. The movement of the cross
head is reduced, the ratio of reduction being
movement of crosshead OC DC
__ ..i. _ OT* _
movement of pencil OP AP
The performance of large ships can be investigated by comparing
with that of small models. Here, again, the laws of similarity are
of great importance.
Suppose the model is built to a scale of , i. e., any length on
the ship is fifty times the corresponding length on the model.
Then its wetted surface is  of that of the ship; while its
2500
displacement is   U. e., gj of the ship's displacement. Also
the resistance to motion of the ship would be 5o 3 times that of the
model.
An instance of the use of the rules for similar figures is seen in
the following :
If the circumference of a circle of 3" diam. is 9426" and its
area is 7069 sq. ins., then the circumference of a circle of 30" diam.
will be 9426 X 10, i. e., 9426", and its area = 7069 X I0 2 = 7069
sq. ins.
MENSURATION
127
Hence one can form a most useful table, to be used for all sizes
of circles.
Diana.
Circumference.
Area.
I
3142
785
2
6283
3142
3
9426
7069
4
I2 5 66
12566
5
15708
I9635
6
18850
28274
7
2I99I
38485
8
25I33
50265
9
28274
63617
Suppose the circumference of a circle of 375" is required,
ce of circle of 3" diam. = ^ of ce of Q of 3" diam. = 9426
ce of circle of 07" diam. = ^ of ce of of 7" diam. = 2199
ce of circle of 005" diam. = y^ of ce of of 5" diam. = 0157
.'. ce (375" diam.) = 11782
Again, the area of a circle of 8" diam.
= g X area of circle of 8" diam.
= 503 sq. in.
Exercises 17. On Spheres.
1. Find the surface and volume of a sphere of 714* diam.
2. A sphere of 8" diam. is weighed in air and its weight is found
to be 80 Ibs. Its weight in water is 7035 Ibs. If Specific Gravity
weight of solid . , , ,
= ^rr r  7  and loss of weight = weight of water
weight of equal vol. of water
displaced, find the specific gravity of the material of which this sphere
is composed and the weight of i cu. ft. of it.
3. Find the volume of a spherical shell whose external diam. is
492", the thickness of the metal being ".
4. A storage tank, in the form of a cylinder with hemispherical
ends, is 23^ ft. long over all and 4 ft. in diam. (these being the internal
measurements). Calculate the weight of water contained when the
tank is half full.
5. A sphere of diameter 22 cms. is charged with 157 coulombs of
electricity. Find the surface density (coulombs per sq. cm.), which is
, quantity in coulombs
given by . .
area in sq. cms.
6. The volume of a sphere is 842 cu. cms. : find its diam.
7. Find the surface and volume of the zone of a sphere of radius
8* if the thickness of the zone is 2" and the radius of its larger end
is 6*.
128 MATHEMATICS FOR ENGINEERS
8. The weight of a hollow sphere of gunmetal of external diam. 6*
was found to be 223 Ibs. Find the internal diam., if the gunmetal
weighs 3 Ib. per cu. in.
9. In a Brinell hardness test a steel ball of diam. 10 mm. was
pressed on to a plate, and the diam. of the impression was measured
to be 315 mm. Find the hardness number for the material of the
plate if the load applied was 5000 kgrms. and hardness number
= 5 TJ = . (Compare Example 14, p. 98.)
curved area of depression
On Similar Figures.
10. Find the area of section of the masonry dam shown at (b),
Fig. 47
11. The symmetrical template shown at (c), Fig. 47, was cut too
short along the bottom edge ; the length dimensioned as 206" should
be 222". Find the amount x to be cut off in order to bring the edge
to the required length.
12. A plan is drawn to a scale of $$. The area on the paper is
4280*. What is the actual area of the plot represented ?
13. Find the diam. of the small
end of the conical roller for a
bearing shown in Fig. 53.
14. The wetted surface of a
ship of 6500 tons displacement
is 260003'. What will be the
wetted surface of a similar vessel
whose displacement is 3000 tons ?
15. One side of a triangle is 12*. Where must a point be taken in
it so that a parallel to the base through it will be cut off a triangle
whose area is that of the original triangle ?
16. The parallel sides of a trapezoid are 10* and 16*, and the other
sides are 5" and 7*. Find the area of the total triangle obtained by
producing the nonparallel sides.
17. The surface of one sphere is 6 times that of another. What is
the ratio of their volumes ? Find also the ratio of their diameters.
18. The area of a field was calculated, from actual measurements
taken, to be 527 acres. The .chain with which the lines were measured
was tested immediately after the survey and found to be 1008 links
long. Find the true area of the field (i chain = 100 links and 10
sq. chains = i acre).
19. A plank of uniform thickness is in the form of a trapezoid
where one end is perpendicular to the parallel sides and is 12 ft. long.
The parallel sides are 12" and g" respectively. At what distance from
the narrower end must the plank be cut (the cut being parallel to the
12" and 9* sides) so that the weights of the two portions shall be the
same ?
20. A trapezoid has its parallel sides 24* and 14* and the other
sides each 8". Find the areas of the 4 triangles formed by the diagonals.
21. The length of a model of a ship was 1075 ft., whilst that of the
ship itself was 430 ft. If the displacement of the ship was 11600 tons,
what was the displacement of the model ?
22. To ascertain the height of a tower a post is fixed upright 27 ft.
from the base of the tower, with its top 12 ft. above the ground. The
MENSURATION 129
observer's eye is $'4* above the ground and at 3 ft. from the post
when the tops of the tower and post are in line with the eye. Find the
height of the tower.
23. What should be the diameter of a pipe to receive the discharge
of three pipes each J* diam. ?
The Rules of Guldinus. These deal with surfaces and
volumes of solids of revolution.
A solid of revolution is a solid generated by the revolution of a
plane figure about some axis ; e. g., a rightangled triangle revolving
about one of its perpendicular sides traces out a right circular
cone ; and a hyperbola rotating about either of its axes generates
a hyperboloid of revolution.
For the cases with which we deal here the axis must not cut
the revolving section, and all sections perpendicular to the axis
of revolution must be circular.
The rules are
Surface of solid of revolution = perimeter of revolving figure
X path of its centroid.
Volume of solid of revolution = Area of revolving figure
x path of its centroid.
The centroid of a plane figure is the centre of gravity of an extremely
thin plate of the same shape as the figure. The motion of the
centroid may be taken to be the
mean of the motions of all the
little elements of the curve or
area.
These rules are of great value
in dealing with awkward solids ;
e. g., suppose the volume of the
nose of a projectile is required,
it being generated by the re
volution of a curved area round
the axis of the projectile (see Fig. 54).
The area of ABCD and the position of its centroid G can be
found by rules to be detailed later, and then
Vol. of nose = area of revolving figure X path of its centroid
= (ABCD)x(2irXOG)
A simpler example is that of a flywheel rim.
Example 36. Find the weight of the rim of a castiron flywheel
of 5 ft. outside diam. ; the rim being rectangular, 8' across the face
and 4' thick radially. (C.I. weighs 26 Ib. per cu. in.)
K
130 MATHEMATICS FOR ENGINEERS
Here, area of revolving figure =8x4
also the mean diam. = 56*
whence path of centroid = TT x 56
and vol. of rim = x 56 x 32 cu. ins.
.*. Weight of rim = x 56 x 32 X 26 Ib.
= 1460 Ibs.
The positions of the centroids (G) for a few of the simple figures
is here given (Fig. 55).
Triangular area (i) ..... OG = \h /BD is the median,"^
GD = BD I i. e., AD = DC j
2f
Semicircular arc (2) ..... OG = = 637 r
TT
Semicircular area (2) . OG, = = 424 r
3*
2r
Semicircular perimeter (2) . . OG 2 = = 389 r
(i. e., arc + diameter).
Parabolic segment (3) .... OG = f A
Semiparabolic segment (4) . . OQ = f h, QG = f b
Area over parabolic curve (5) . OG = '^h; GP =
4
Area of quadrant of circle (6) . OG = GP = 424 r
Area over circular arc (quadrant) or Fillet (7). OG = GP = 223 r
Trapezoid (8). Bisect AB at E and DC at F. Join EF. Set
off BM = DC and DN = AB. Intersection of MN and EF is at G,
or, by calculation, OG =  (
Quadrilateral (9). Bisect AC at F and BD at E.
Make OP = OE and OQ = OF
Through Q draw a parallel to BD and through P, a parallel to AC.
The intersection of these gives G, the centroid of ABCD.
Exercises 18. On Guldinus* Rules.
1. An isosceles triangle, each of whose equal sides is 4 ft. and
whose altitude is 3 ft., revolves about an axis through its vertex parallel
to its base. Find the surface and volume of the solid generated.
2. Find the surface and volume of the anchorring described by a
circle of 3* diam. revolving round a line 4* from the nearest point on
the circle.
3. Find the surface and volume described by the revolution of a
semicircle of 4" diam. about an axis parallel to its base and 5* distant
from it.
4. An equilateral triangle of 5* side revolves about its base as axis.
Find the surface and volume of the double cone thus generated.
MENSURATION
Fig. 55 ._Positions of Centroids (G) for Simple Figures.
132
MATHEMATICS FOR ENGINEERS
5. A parabola revolves about its axis. Compare the volume of the
paraboloid thus generated with that of the circumscribing cylinder.
6. At (a). Fig. 56,
T
T
10
L
Fig. 56.
is shown in section jy~J"
the winding of the
secondary wire of an
induction coil. Find
the volume of the
winding.
7. Calculate the
weight, in mild steel
weighing 287 Ib.
per cu. in., of the
spindle weight for a
spring compressor
shown at (b), Fig. 56.
[Hints. Area of
a fillet, as at A,
= '215r 2 where r is the radius of the circular arc.
For the position of the centroid of a fillet refer to (7), in Fig. 55,
and also to p. 130.]
Application to Calculation of Weights. When calculating
weights two rules should be borne in mind in addition to the
foregoing.
(a) The solid should be broken up into simple parts, i. e., those
whose volumes can be found by the rules already given; and
(b) suitable approximations should be made wherever possible.
Circular segments may be replaced by parabolic segments if the
rules for the latter are easier, the rounding of corners may be
neglected, unless very large, mean widths may be estimated, etc.
For purposes of reference the table of weights of materials and
other useful data are inserted here ; but the values given must be
considered as average values.
WEIGHTS AND DENSITIES OF METALS.
METAL.
Weight in
Ibs. per cu. in.
Weight in
Ibs. per cu. ft.
Specific Gravity
(gnus, per cu. cm.).
Cast iron ...
Wrought iron
Steel .....
26
28
2Q
450
485
<?OO
721
776
804
Brass or Gunmetal
Copper (Cu) ...
Lead (Pb) ...
Tin (Sn)
3
.32
HI
27
518
553
710
/i6^
831
887
n34
7M&
Aluminium (Al)
Zinc (Zn)
0932
26
161
ACQ
2 5 8
7*21
MENSURATION
WEIGHTS AND DENSITIES OF EARTH, SOIL, ETC.
133
MATERIAL.
Slate.
Granite.
Sandstone.
Chalk.
Clay.
Gravel.
Mud.
WEIGHT \
(cwt. per cu. yd.) J '
43
42
39
36
31
30
2 5
Useful Data. Wroughtiron plate weighs about 10 Ibs., and
steel 104 Ibs. per sq. ft. of area per J" of thickness, i. e., 8 sq. ft.
of W.I. plate " thick would weigh 10 X 8 x 3 = 240 Ibs.
Wroughtiron bar or rod weighs about 10 Ibs., and steel 104 Ibs.
per yard for every sq. in. of section.
Wroughtiron bar or rod, i"diam., weighs 8 Ibs. and steel
82 Ibs. per yard : also the weight is proportional to the diameter
squared; thus, a yard of steel bar 2" in diam. would weigh
2 2 X 82 or 328 Ibs.
Four hundred cu. ins. of wrought iron, 430 cu. ins. of cast iron,
390 cu. ins. of steel, each weigh about i cwt.
A few examples are here worked out to give some idea of the
method of treatment.
Example 37. Calculate the weight, in cast iron, of the D slide
valve shown in Fig. 57.
e*
IT ]]
.11
Fig. 57. D Slide Valve.
In many cases where the solid is partially hollowed it is best to treat
first as a solid and then subtract the volume cut away,
134
MATHEMATICS FOR ENGINEERS
First, considering as a solid
Vol. above AB = 155 x 85 x 3 cu. ins. = 395 cu. ins.
Vol. below AB = 1625 x J 3 x I>2 5 = 26 4 "
.*. Total vol. (as a solid) = 659
To be subtracted
Vol. of cavity = 14 X 7 x 35 = 343
/. Net vol. = 3i6
and weight = 316 X 26 = 821 Ibs.
Example 38. Find the weight of a plate for a castiron tank. The
plate (see Fig. 58) is 24* square and f thick; there are 20 ribs, each
\" x ij" x ij", and 24 boltholes in the flanges, each f* square; also
the flanges are 23!" x J* x ij".
r jj
i"
Z
S
(.
T ; !
5"^
JJ
::
i
a
1 :::
~li
8^
a
! :.;
*;;j i
IV*

;:
*
a
..
*"""!
A
1
1
?
.c
LZ
.0 ^" *"'1
B^
i
U II U U D >]f I;
a
El
Fig. 58. Plate for Tank.
J
Dealing with the separate portions :
Flat Plate (A).
Vol. = 24 x 24 x  . = 216 cu. ins.
Flanges (D).
Length = (2 x 22") + (2 X 23!") = 93"
.*. Vol. =93*fx = 5 8 ' 1
Ribs (B).
Area of face of one = \ x X f
/. Vol. of 20 each \" thick = \ x f X f X \ X 20 = 78
Gross vol. = 2819 >,
Subtract for 24 boltholes (C) ; 24 x f x f x J = 47
Net vol. = 2772
.*. Weight = 2772 X 26 = 72 Ibs.
MENSURATION I35
Example 39 Find the weight of the wroughtiron stampings for a
dynamo armature as shown in Fig. 59) 14* diam. and IO * long io/
of the length being taken off by ventilation and insulation. There
are 3 ventilating ducts, each
6* internal diam. and i*
thick, the gaps between
these being i%" long; and
also 60 slots, each " by f*.
The shaft is 3* diam.
Note. The stampings
are only thin and are separ
ated one from the other by
some insulator; also there
would be a small gap for
ventilation purposes, and
hence the actual length of the stampings is less than 10
is to be taken as 90 % of 10", i. e., g".
Area of face of stamping =  x i4 2 = *54 sq. ins.
To be subtracted
Area of 60 slots = 60 x J x f
Mean length of ventilating ducts = (IT x 7) (3 x
= 175*
/. Area = 175 x i
Fi S 59 Stamping for Dynamo Armature,
in this case it
= 197
Area of hole for shaft =  x 3 2 ..
4
Thus the total area to be subtracted
= 71
= 44'3 ..
or the net area of the stamping = 1097
Then the volume = 1097 x 9 cu  i ns 
and the weight = 1097 x 9 x 28 Ibs.
= 277 Ibs.
Example 40. Find the weight of 150 yards of steel chain, the links
of which have the form shown in
Fig. 60.
The effective length A of a link
is the inside length, provided that
a number of yards of chain are being
considered. (For small lengths this
is not quite correct.)
In this case the effective length
of a link = i^*, so that in i yard
36
of the chain there are
i. e., 24
Fig. 60. Chain Link.
&
links, or in 150 yards of the chain there are 3600 links.
The mean length of i link = 0ce of circle of i J* diam. + (2 x f")
~~" 3 V3 v "^ *) TV
136
MATHEMATICS FOR ENGINEERS
Now i* diam. steel rod weighs 82 Ibs. per yard (see p. 133) ; therefore
\" diam. steel rod weighs ^, i. e., 205 Ibs. per yard.
Hence, weight of i link = 5_^> x 205 Ibs.
and weight of 3600 links = 5 '43 * 205 x 3600 lbs = III5lbs
Example 41. Two straight castiron pipes, making an angle of
135 with one another, have the centres of their ends 2 ft. apart (in a
straight line). They are to be joined by a curved pipe (as in Fig. 6r),
4* external and 3* internal diam., with flanges 8" diam. and \" thick.
Find the weight of the curved pipe if the flanges each have five bolt
holes, of \" diam.
Fig. 61. Curved Castiron Pipe.
This is a useful example on the application of Guldinus' rule.
Path of centroid = arc of circle, which is ^ or 5 of the circum
360 8
ference.
By drawing to scale (or by Trigonometry), the radius is found to
be 26 ft.
.*. Path of centroid = Jx 77X52 = 204 ft.
and length of the path of the centroid between the flanges^
% =20 4 ft.(2X")
5= ;9oft,  235*,
MENSURATION
Area of revolving section = f X4 2 J ( X3 2 ) = 55 sq. ins.
hence the volume of the solid between the flanges = 235x55 cu. ins.
= 129 cu. ins.
Vol. of 2 flanges, each J* thick, 8* external and 3* internal diam.
= 432 cu. ins.
.*. Gross vol. of bend = 1722 cu. ins.
To be subtracted
Vol. of ten " diam. holes :
Diam.
Length.
Vol.
625
5"
i'5
/. Net vol. of bend = 1707 cu. ins.
and weight = I7O7X26 = 444 Ibs.
Example 42. Find the weight of the wroughtiron crank shown in
Fig. 62, allowing for the horns at the junctions of the web and bosses.
Dealing with the three parts
separately :
Vol. of the upper boss is the
difference of the volumes of two
cylinders
Diam.
Length.
Vol.
12*
8*
908
6*
8*
227
net volume = 681 cu. ins.
Similarly, vol. of the lower
boss
Diam.
Length.
Vol.
*5',
9
725
7 2 5
1282
462
 8
i
04
net volume = 820 cu. ins.
Fig. 62. Wroughtiron Crank.
The horns can be allowed for by adding J of the height of each
to the length of the web (i. e., we replace the circular segment by a
parabolic segment, because the rule for the area is simpler).
To find the height h^ of the top horn A, {a x = 5, r z = 6}.
hi = fi VrfaJ = 6 ^3625 = 6332 = 268*.
Hence add J of 268", i, f ,, >f tq the length of the web,
138 MATHEMATICS FOR ENGINEERS
For the lower horn B, a 2 = 6", r t = 75*
3"
Hence add on i* to the length of the web.
Thus the effective length of the web
its mean width
so that its vol.
= 184*
= n*
= i84X 11X45 = 9 IQ
Total vol. of crank = 2411 cu. ins.
Weight = 241 1 x 28 = 675 Ibs.
Example 43. Determine the number of i* diam. rivets, as at (a)
Fig. 63 (i. e., with snap or spherical heads) to weigh i cwt. (Given that
d = /+ t V and length = 2t.)
1 17
Fig. 63.
If d = i* then t = &" and length = i *.
For the heads, a rough approximation is that the two together
are onehalf the volume of a sphere of diameter i8d, this being the
diameter of the sphere of which the heads are segments ; but the
result will be somewhat more accurate if 52 is taken in place of 5.
(This figure is arrived at by the use of the rule given on p. 121 for the
segment of a sphere.)
Then vol. of heads = ^x^nxg 8 = 158 cu. ins.
vol. of body {Diam. = i", length = 1125"}= 88
or
vol. of i rivet = 246
Number of i* rivets per cwt. =
2 46 x 29
Example 4^. Find the weight of the castiron hanger bearing
shown in Fig. 64.
This example illustrates well the method of breaking a solid up into
its component parts ; the different parts being dealt with according
to the letters on the diagram.
139
cub. ins.
MENSURATION
Treating first as a solid throughout
A. Cuboid, length = 12", breadth = 675", thickness = 75".
Volume = 12x675x75 = 6075
B. 4 cylinders, of diam. 1625* an d total length = 5*
Volume (obtained from the slide rule) = 1035
C. Area of section = semicircle + rectangle
(55x25)
= 1086 + 1375 = 2561
Volume = 2561x275
D. Cylinder, diam. = 4*, length = 4*
Volume
E. Cylinder, diam. = 45*, length = 75
Volume
F. 4 cylinders, diam. = 2", total length = i"
Volume ,
= 7048
= 5030
= 1192
= 314
Gross Volume = 20694
Fig. 64. Castiron Hanger Bearing.
To be subtracted
cub. ins.
= 2820
G. Cylinder, diam. = 3*, length = 4*
Volume
H. Cylinder, diam. = 2%", length = 3^*
Volume = I7 >:1 5
J. 4 cylinders, diam. = 75", total length = 9"
Volume = 3 '97
Total volume to be subtracted = 49'3 2
Net volume = I57' 62
Hence, weight = 1576 x 26
= 41 Ibs.
140
MATHEMATICS FOR ENGINEERS
Exercises 19. On Calculation of Weights.
1. Find the weight of the castiron Veeblock shown at (6), Fig. 63.
2. Find the weight in steel of the crank axle shown in Fig. 65.
Centre Lir\e
of Engine.
'
P369"
Fig. 65. Steel Crank Axle.
3. Find the weight of sheet iron in a rectangular measuring tank ;
the metal being i" thick. Inside dimensions of the tank are 4'6*
by 3 '6* by 7 / o"deep. Cut from the sides are openings to accommodate
fittings as follows : One rectangular hole 4 / o* by 2", two elliptical
holes 4" X 2", two circular holes 4" diam. and eight fdiam. bolt holes.
4. Determine the weight of a wroughtiron boiler end plate, 8 ft.
diameter and ^" thick. There are two flue holes, each 2 ft. diam. and
an elliptical manhole i8*xi2*.
5. Find the weight of 22 yards of iron chain. The links are
elliptical and are made of elliptical metal i"x%", the greatest width
of section being at rightangles to the plane of the link. The mean
lengths of the axes of the link are 4* and 2^".
6. How many fdiam. snapheaded rivets weigh i cwt. ? (Compare
with Example 43, p. 138.)
7. Find the weight in cast iron of the flywheel of a steam engine
having a rectangular rim, 7" wide by 4* radial thickness ; six straight
arms of elliptical section, the axes of the ellipse being 4^* and 2" ;
a boss 7** wide, 9* diam. and 4^" bore. The outer diameter of the
wheel is 79*.
8. Required the weight of the castiron anchor plate shown in Fig. 66.
L,
ajfef
ii_
\
/
/
"<
8'
M
t
eb
i
M 16"
Fig. 66. Anchor Plate,
T
._. ?af^
L IK
Fig. 67. Planer Tool Holder.
MENSURATION
141
9. Calculate the weight in cast iron of the tool holder for a planer
shown in Fig. 67.
10. Find the weight of the castiron roll for a rubber mill as in Fig. 68.
(Use the slide rule throughout.)
Fig. 68. Roll for Rubber Mill.
11. A mild steel sleeve coupling for 3* shaft is shown at (a), Fig. 69.
Find its weight.
12. The steelwork for Hobson's flooring has the sectional form
shown at (b), Fig. 69. There are 20 such plates for each span of the
bridge, each ^" thick and 22 ft. long. Find the total weight of the
steelwork, neglecting the angle and Tbar.
Mild Steel Sleeve Coupling. Section of Hobson's Flooring.
Fig. 69.
13. Find the weight in cast iron of the simple plummer block
shown in Fig. 70.
Fig. 70. Plummer Block.
142
MATHEMATICS FOR ENGINEERS
14. Fig. 71 shows the worm shaft for a motorcar rear axle. It
is made of nickel steel, weighing 291 Ib. per cu. in. Find its weight.
L3T
Fig. 71. Worm Shaft.
15. Calculate the weight in cast iron of the half coupling shown in
Fig. 72.
Fig. 72. Wroughtiron Coupling.
16. Find the weight in cast iron of the cylinder cover shown in
Fig. 73
 73 C.I. Cylinder Cover.
MENSURATION
143
17. Fig. 74 shows the brasses for the crankshaft of a 61"x6 T
launch engine. Find the weight of one of these in gun metal.
!
r
Hf
r*

^
\
1
1
1
1
1
T

ft
t
MI
!
1
't
1
i
^i
r ' "^
Fig. 74. Crank Shaft Brasses.
18. The brasses for a thrust block are shown in Fig. 75. Calculate
the weight of one of these in gun metal.
Fig. 75. Brasses for a Thrust Block.
19. An air vessel is shown in Fig. 76. Find its weight in cast iron.
Oval 4^x3'
ty,
3*3
T.
Vj.2 VKO
..i?....S?.
li'
6'
Fig. 76. Air Vessel for Pump.
144 MATHEMATICS FOR ENGINEERS
TABLE OF AREAS AND CIRCUMFERENCES OF PLANE FIGURES.
145
TABLE OF AREAS AND CIRCUMFERENCES OF PLANE FIGURES (continued).
TUle.
Hollow circle
(annulus) .
Hollow circle
(eccentric)
Sector of
circle . .
Sector of hol
low circle
Fillet . .
Segment of
circle . .
Figure.
Ellipse
Irregular
figures
ircumference or
Perimeter.
57'3
Area.
or ir x mean dia. x thick
ness
or 7r(R*  r 2 )
approx
irnr* Ir
~?6o~ or 
360
2I5K 2 or approx.
Area = sector triangle
Various approx. for
mulae on p. 102.
Vab]
more nearly
Step round
curved por
tions in small
steps, with
dividers ; add
in any straight
pieces.
Divide into narrow
strips ; measure thei
midordinates. Then
Area = aver, midordi
nate x length /
146 MATHEMATICS FOR ENGINEERS
TABLE OF VOLUMES AND SURFACE AREAS OF SOLIDS.
Title.
Figure.
Volume.
Surface Area.
Any prism .
Rectangular
prism or
cuboid . .
Cube . . .
Square prism
Hexagonal
prism .
Octagona!
prism .
Cylinder . .
Hollow cylin
der .
Elliptical
prism ,
Sphere . .
Hollow
sphere . .
Area of base
X height
llh
S 3
S 2 /
26S 2 /
or 866/ 12 /
Circumference of base
X height
or Sag/ 2 /
or 7854^
r(R 2  r z )h
rabh
or 523&D 3
Whole
area/
= 2(lb+Ui+bh)
Whole area = 6S*
Lateral surface = 4S/
Ends = 2S a
Whole 1 ^c/^7 i o\
surfacej= 2S ( 2/+S )
Lateral = 6S/ or y^bfl
(For ends see Table on
p. 1 44.)
Lateral = 8S/ or 332/1
Lateral = 2irrh
Two ends = 2irr*
Whole area = 2jrr(& + r)
Outer lateral)
surface /
Inner lateral)
surface
Lateral
/
= 2irrh
or Tr(a+b)h
(less accurate)
MENSURATION 147
TABLES OF VOLUMES AND SURFACE AREAS OF SOLIDS (continued).
Title.
Figure.
Volume.
Surface Area.
Segment of
sphere .
Zone of
sphere .
Any pyramid
Square pyra
mid . . .
Cone . . .
Frustum o:
any pyra
mid . .
Frustum o
square
pyramid
Frustum o:
cone .
Anchor ring
urved surface = 2irRA
or 5236/i(3' 2 +A 2 )
area of base
X height
^=height of frus
tum
A=area of large end
B=area of small end
(A + B + VAB)
here R=rad. of sphere
ateral = \ circum. of
base X slant height
Lateral = 2S/
Lateral = *y/
Lateral= mean circum
X slant height
Lateral = 2/(S +'s)
(I = slant height)
Lateral = ir/(R + f
(1 = slant height)
Round section
Square section
DS
rDS
These four tables are reproduced from Arithmetic for Engineers by kind
permission of the author, Mr. Charles B. Clapham.
CHAPTER IV
INTRODUCTION TO GRAPHS
Object and Use of Graphs. A graph is a pictorial statement
of a series of values all drawn to scale. Such a diagram will often
greatly facilitate the understanding of a problem ; for the meaning
is more readily transmitted to the brain by the eye than by descrip
tion or formulae. When reading a description, one has often to
form a mental picture of the scenes before one can grasp and fully
appreciate the ideas or facts involved. If, however, the scenes
are presented vividly to us, much strain is removed from the brain.
A few pages of statistics would have to be studied carefully before
their meaning could be seen in all its bearings, whereas if a " graph "
or picture were drawn to represent these figures, the variations of
their values could be read off at a glance.
To take another example : a set of experiments are carried
out with pulley blocks; the results will not be perfect, some
readings may be too high, others too low : and to average them
from the tabulated list of values would be extremely laborious;
whereas the drawing of a graph is itself in the nature of an
averaging.
Or, again, a graph shows not only a change in a quantity, but
the rate at which that change is taking place, this latter being often
the more important. On a boiler trial a graph is often drawn to
denote the consumption of coal : from which is shown during what
period the consumption is uniform, or when the demand has been
greater or less than the average, and so on.
A graph, then, is a picture representing some happenings, and
is so designed as to bring out all points of significance in connection
with those happenings. The full importance and usefulness of
graphs can only be appreciated after many applications have been
considered.
To commence the study of this branch of our work let us consider
an example based on some laboratory experiments.
INTRODUCTION TO GRAPHS T49
Example i. In some experiments on the flow of water over notches
the following figures were actually obtained.
RIGHTANGLED VNoxcH
Head (ft.) H . .
1888
2365
2617
2878
3065
336i
Quantity flowing
(Ibs. per min.) Q
J4I5
2498
3235
4114
4836
608
The flow, in subsequent experiments, was to be gauged by the
" head " of water at the notch, so that a good " calibration " curve was
desired.
The figures were plotted as shown in Fig. 77, H along a horizontal
axis and Q parallel to a vertical axis.
In such plotting as this the following points of detail should
be observed.
Select two lines at right angles for the main axes and thicken
them in : these lines should be as far over to the left and as low
down, respectively, as will permit of the scales being written to
the outside of each.
Look to the values to be plotted, noting the " range " in either
direction, the scales for the plotting being selected so that the whole of
the available space is utilised : but care must be taken to select a
sensible scale. Generally a decimal scale is to be preferred, e. g.,
in the present case we take \" to represent 02 ft. of head,
horizontally and " to represent 100 Ibs. per min. vertically.
Write figures along the axes to indicate the scales adopted, and also
indicate clearly which quantity is plotted along the horizontal axis and
which along the vertical axis ; for attention to such details greatly enhances
the value of the graph.
To plot : We wish to illustrate the fact that for each value of H
there is a value of Q ; which we can do by selecting some value of H,
running up the vertical through the marking denoting that value until
we meet the horizontal through the given corresponding value of Q,
and then making a small mark, e. g., the point denoting that H = 2878
when Q = 4114, as shown on the diagram by the point P.
The use of paper ruled in squares will ease matters, although
in a good many instances a series of horizontal and vertical lines
through points specified in a table of given values will suffice.
When all the points have been plotted, the best average or
150
MATHEMATICS FOR ENGINEERS
smooth curve must be drawn through them : the points above the
line should about balance those below it, and any obviously in
accurate values must be disregarded. For good results the curve
should be drawn with the aid of either a spline or a French
curve.
The curve is now what is called a calibration curve for the notch,
i. e., for any head within the range for which experiments were
carried out, the quantity flowing can be read off.
600
100
18
20
24 26 28 30
Values of H(feel)
 77 Calibration Curve for Vnotch. (Full size.)
32
34
This process of reading off intermediate values is spoken of as
"interpolation." Without the graph, for any values not given
in the table one would have either to estimate or to repeat the
experiment if intermediate values were required. Also one further
point should be noticed : even the figures in the table may not be
quite the best, and better approximations can be obtained from
the curve.
Ex. To find the quantity when the head is 24 ft. : erect the
perpendicular SQ through 24 on the scale of head, meeting the curve
at Q. Draw QR horizontally to cut the axis of quantity at Q = 260.
Then for a head of 24 feet, 260 Ibs. per min. are flowing.
Ex. Find the head when Q = 480. From the diagram, H = 30.5 ft.
INTRODUCTION TO GRAPHS 151
Example 2. The following figures were obtained in some trials
on a gas engine. Draw the efficiency curve, i. e., the curve in which
the efficiency is plotted against the output.
I.H.P. (Input)
i'54
309
458
567
650
B.H.P. (Output)
o
162
3'33
471
58!
The efficiency (to be denoted throughout this book by rj, the Greek
, . output B.H.P.
letter eta) = ^ t or  and could be calculated by taking
corresponding values of B and I from the table.
Fig. 78. Test on Gas Engine.
It is better, however, to first plot B.H.P. against I.H.P. and average
these points by a straight line, which can be drawn with more certainty
than a curve (see Fig. 78). The efficiencies at various loads can now
T>
be calculated from this " curve " by taking the ratios of y for con
venient values of B; e. g., when B = i, I = 243 and / = '4 12 
Plotting the values of the efficiency so obtained to a base of output,
a welldefined smooth curve is obtained, as in Fig. 78.
The efficiencies worked from the experimental figures are
B.H.P. . .
O
162
3'33
471
581
T) ...
525
726
831
895
If now these values are plotted to a base of B.H.P. the points lie
fairly equally about the efficiency curve already drawn.
152
MATHEMATICS FOR ENGINEERS
The efficiencyoutput and the inputoutput curves now agree,
whereas they would not do so in all probability if plotted quite
separately.
This derivation of one curve from another is of wide application.
To illustrate by another example :
Example 3. A test on a MorrisBastert pulley block gave the
following results :
Load lited\
(Ibs.) W. j
275
47'5
675
875
1075
1275
1475
1675
1875
Effort re]
quired (Ibs.)
207
2'5
315
405
452
5'2
585
64
71
P. j
The velocity ratio (V.R.) of the machine was 48.
Draw the efficiency curve to a base of loads.
W
Theoretical effort to raise a weight W = P = T?~
Actual effort = P! and efficiency = ^5
Fig. 79. Test on Pulley Block.
First plot the given values, W horizontally and P vertically, and
draw the straight line which best fits the points (see Fig. 79).
To calculate values of P corresponding to the values of W set 48
on the C scale of the slide rule level with i on the D scale. Then the
readings on the C scale will correspond to values of W and those on the
D scale level with these to values of P; e. g., place the cursor over
275 on the C scale and 572 is read off on the D scale, so that the value
of P when W = 275, is '572. All values of P can thus be read off with
one, or, at the most, two settings of the rule.
INTRODUCTION TO GRAPHS
153
The values of P are 572, 99, 141, 182, 224, 266, 307, 349, 39.
Plotting these to the same scale as chosen for P l the lower line in
Fig. 79 is obtained.
By division of corresponding ordinates of these lines the efficiency
can be calculated for any load, e. g., when W = 80, P = 163, P! = 365
and r) =  = 447. A scale must now be chosen for efficiencies,
and the curve can then be put in ; this will be a smooth curve, because
it is obtained from two straight lines..
Example 4. In some experimental work, only gramme weights
were available, whilst for calculation purposes the weights were re
quired in pounds. To save the constant division by 4536 (the number
of grms. equivalent to i Ib.) a straight line could be drawn from which
the required interpolations could be made. To construct such a chart :
6
4 JT
3 
577
r
~r
/
/
J
7
46
/
jr
 co
o
/
'
o
/
/
/
U?232
210
Y\
iOi
>
/
\
/
i
207*
44 i^~ "
y/2QO
*7<5
(O5O
985 '
Sec
lie
of
2t
iOO
400 600 12OO iQOO OOO E4OO 28OO
Fig. 80. Chart to convert grammes to Ibs.
Suppose that the readings in grms. were
200, 476, 985, 1050, 2072, 2600.
Plotting grms. along the horizontal as in Fig. 80, a scale must be
chosen to admit of 2600 being shown. Draw a vertical through 453 6
to meet a horizontal through i on the " Ib. " scale. The line joining
this to the origin (i. e., the zero point for both scales) is the conversion
154
MATHEMATICS FOR ENGINEERS
line. The required values can now be quickly read off as in the
following table :
grms.
200
476
985
1050
2072
2600
Ibs.
'44
105
216
232
46
577
One axis might take the place of the two in the above diagram.
Along this on one side the graduation would be in Ibs. and on the
other side, in grms. ; thus amounting to putting a scale of Ibs.
alongside one of grms. Tables of logarithms might be, and in fact
are (in Farmer's Log Tables), replaced by a number of lines,
graduated in numbers and also in logarithms. For great accuracy
a great number of lines are required so that two pages do not suffice
as in the case of the tables, this being rather a disadvantage : never
theless there is much to be said for this method of table construc
tion. There are no differences to add, nor is it necessary to remem
ber when differences have to be subtracted, since for any definite
value in the one set of units the corresponding value in the other
is read off directly.
Exercises 20. On Simple Plotting.
1. In a test on Hobson's flooring the following figures were
obtained.
Total load (tons) 35
4
5 I 6o
70
80
90
IOO
no
Deflection (ins.)
*
A
A 1
i*
ii
i A
2
Plot a graph to give the deflection for any load between 35 and no
tons ; and read off the deflection for a load of 55 tons and also the
load causing a deflection of i*.
2. Plot a curve to show the decrease in the tenacity of copper
with increase of heat, from the following table :
Temperature
F. . . .
212
350
380
400
500
530
580
620
720
Tenacity (Ibs.
per sq. in.)
32000
30000
29500
29000
26500
255
23500
21500
2OOOO
Read off from your graph : (a) the tenacity at 302 F. ; (b) the
temperature at v.hich the tenacity is 21000 Ibs. per sq. in. ; (c) the
tenacity at 545 F.
INTRODUCTION TO GRAPHS 155
3. Draw the calibration curve for a rectangular notch, given
Head (foot) ....
0871
1115
1588
1838
2124
Quantity (Ibs. per min.)
I39H
199
323'3
4062
5028
Find the discharge when the head is 19 ft.
4. The following figures are given for the working stress allowable
on studs and bolts :
Diam. of stud (ins.) .
I
3
I
I*
ii
1
2
Stress (Ibs.persq. in.)
2OOO
3OOO
3900
4700
5500
6300
7OOO
Find the stress allowable on a stud of " diam. and also the stud
to be used if the stress is 5100 Ibs. per sq. in.
5. Castiron pulleys should never run at a greater circumferential
speed than i mile per minute. In the table the maximum revolutions
per minute (R.P.M.) allowable are given for various diameters. Find
the R.P.M. for a pulley of 14". diam. Check this figure by the
ordinary rule of mensuration.
Diam. (ins.)
5
6
8
10
12
15
18
2O
25
30
R.P.M. . .
4034
336i
2524
2017
1681
1345
1120
I008 807
673.
6. Plot a curve to give the diameter of a shaft for any twisting
moment from 7 ton per sq. in. to 360 tons per sq. in.
Equivalent twisting ~
moment (tons per >
sq. in. }
701
2367
5611
1095
1894
3007
44'9
63'9
877
152
359
Diam. of shaft (ins.)
i
i'5
2
2'5
3
3'5
4
4'5
. 5
6
8
7. The table gives the " time constant " of the coils of an electro
magnet for gaps of various lengths. Represent this variation by a
graph.
Distance apart (cms.) .
125
5
75
i
i5
2
25
3
Time constant (sees.) .
25
17
i4
i4
ii
II
9
9
8. The relation between pressure p and temperature t of steam shown
in the table was found experimentally. Plot a curve to represent this,
finding the value of t when p is 105, and the value of p when t is 300.
p Ibs. per sq. in. . .
5
10
15
205
27
3i
36
44
50
60
70
80
90
100 IIO
120
<(F.)
235
243
251
260
270
276
282
290
296
306
3'4
322
329
336 342
34
9. and 10. Plot curves of Magnetic Induction for (i) Iron, and
(2) Cobalt, from the figures given in the tables following.
156 MATHEMATICS FOR ENGINEERS
(9. Iron.)
H (magnetising!
force) t
o
5
10
i?
25
30
38
45
52
Co
65
B (mag. induci
tion density) /
2400
4500
6000
7100
7800
8300
8500
8600
8600
8700
(10. Cobalt.)
H.
o
155
3'io 465
6.2O
775
1240
I55
2325
3i
3875
465
B.
99
268 j 642
1128
1298
2405
2995
4070
4860
5390
5810
11. Plot a curve to show the variation in the ratio Q
/weight of armament and protection)
\ load displacement /
as given for a speed of 21 knots, from the following table :
Load displace\
mentP(tons)/
18000
22OOO
24000
26000
30000
34000
38000
4000O
Ratio Q . .
383
401
409
416
428
438
446
45
Find the weight of armament and protection when the displacement
is 28000 tons.
12. Plot a curve, as for Question n, but the figures belonging to a
speed of 27 knots.
p . .
18000
2OOOO
24OOO
26000
28000
30000
32OOO
36000
40OOO
Q
236
252
275
286
295
303
3 IO
324
336
Find the value of Q when P = 34000.
13. The temperature of the field coils of a motor was measured at
various times during the passage of a strong current, with the following
results :
Time (mins.) ....
2
5
10
15
20
25
30
35
40
45
50
55
60
Temperature (C.) . .
14 16
23
324
39
43'4
47
505
525
55
568
586
593
59'4
Find the time that elapses before radiation losses, etc., balance the
heating effect of the current, viz. when there is no further sensible rise
of temperature ; and find also the maximum rise of temperature.
14. Repeat as for Question 13, taking the following results :
Time (mins.) ....
o
5
10
15
20
25
30
35
40
45
50
55
60
65
Temperature (C.) . . .
20
26
325
4i
46
49
525
54'5
565
58
595
61
617
62
15. The following figures were obtained by reading spring balances
at the ends of a beam on which a weight of 7 Ibs. was hung. Plot
INTRODUCTION TO GRAPHS
157
curves to give the values of the reactions for any position of the weight.
Note their point of intersection.
Distance (ins.) of weight \
from R.H. end /
2
4
6
8
10
12
14
16
18
24
28
30
32
Lefthand reaction (Ibs.)
'4
8
125
17
2*1
2'55
3
3'45
3'9
5'2
605
65
7
R.H. reaction (Ibs.) . .
7
65
6*05
5'6
5'2
48
4'3
45
3H5
3
18
8
*4
o
In Questions 16 to 19 draw to a base of loads (W) curves whose
ordinates gives
(a) Actual effort P x ; (6) theoretical effort P ; (c) efficiency 17.
16. Test on a 6 to i pulley block, i. e., V.R. = 6.
W
28
48
68
88
108
128
148
1 68
188
208
P!
975
1475
2025
2 575
3075
3575
4025
4525
4925
5525
17. Test on a Single Purchase Crab (V.R. = 27).
W . .
501
921
137
1 80
224
266
310
354
394
PI . .
36
535
7'9
99
n8
139
147
169
195
18. Test on a Screw Jack (V.R. = 605).
W .
34
54
74
94
114
134
T 54
174
194
214
234
PI .
i73
285
3'93
517
619
770
895
10
n3
12
I2'9
19. Test on a Weston Pulley Block, when raising (V.R. = 24).
W . . .
25
45
65
85
105
125
145
165
185
205
P, . . .
4
675
875
875
10
135
15
1875
21
225
20. The table gives the current absorbed by a carbon brush at
various pressures. Plot, to a base of amperes of current, curves giving
resistance and voltage. ! Resistance = '
I amperesj
The resistance curve should be obtained from that for voltage.
Volts. . . .
'35
65
9
88
i
1875
13
215
145
I'S
165
i'75
37'5
i77
18
1825
185
Amps. . . .
4
135
24'5
275
325
405
42
45'5
47'5
21. To a base of frequency plot curves giving (a) voltage, (b) current
taking the following figures :
Frequency
40
435
47
50
52
54
56
60
64
75
80
88
Current
539
875
I4'35
1867
1473
1166
933
683
519
305
264
214
Voltage .
52
32
195
15
19
24
30
41
54
93
1 06
131
158
MATHEMATICS FOR ENGINEERS
22. The following figures were obtained in a tensile test on a sample
of 25% nickel steel.
Stress (Ibs. per sq. in.) .
4000
12000
20000
28000
36000
48000 52000
56000
60000
Extension (inches per j
inch length)
00015
00047
0007
001 1 1
00145
00195 00213
00235
00264
64000
68000
72000
76OOO
80000
84000
88000
92000
96000
IOOOOO
104000
00365
0065
02 1
035
052
068
0853
1025
134
171
201
Plot the " stressstrain " diagram, the stresses being vertical and
extensions along the horizontal; also determine the stress at the
" yield point," where the sudden change occurs.
23. The voltage supplied to a 4volt lamp was varied, and the
candlepower (C.P.) then measured for various values of the voltage,
the results being as follows :
C.P. .
o
5
IO
!'5
2
2'5
3
Volts .
303
3'57
396
4 2 5
444
4'75
Amps. .
116
129
136
148
163
171
If watts = volts x amps, plot to a base of C.P. curves whose
ordinates represent
(a) volts; (6) amps, and by a combination of corresponding
ordinates of these (c) watts per C.P.
24. The drop in potential due to a standard resistance of 3 ohm
was measured by a potentiometer, for various currents. The current
was also measured on an ammeter.
volts
If current = , calculate the true currents flowing. Also
resistance
plot a curve of true current against registered current, and hence
find the percentage error of the ammeter.
Ammeter reading)
(Registered current)/
I
I< 5
2
2'5
3
3'5
4
4'5
475
Volts .... ^093
458
6149
7629
92
10487
1204
I37 1
1*437
25. From the following figures (taken from a test on a 10 H.P.
Diesel engine) plot curves, to a base of B.H.P., to show
(a) I.H.P., from which deduce (b) mechanical efficiency : (c) oil
per hour, and hence (d) oil per B.H.P. hour.
B.H.P. . . .
o
O.TO
671
8i;
9Q4.
I.H.P. . . .
4'5
7^7
1066
1169
1295
Oil per hour (Ibs.)
i'5
237
363
4'35
5'45
INTRODUCTION TO GRAPHS
159
26. From the given figures plot to a base of I.H.P., curves with
ordinates to represent (a) steam per hour and thence (b) steam per
I.H.P. hour.
Steam per hour (Ibs.)
513
452
436
403
37
327
182
I.H.P
1312
iQ'54
983
885
815
657
184
27. Results of an efficiency test on a small motor gave the follow
ing :r
Output (watts) . . .
646
242
338
37'5
40
553
138
615  649
77'i
9*
"7
Input (watts) . . .
576
824
IO2
1042
1072
1422
1411
1624
1875
228
To a base of output plot curves giving (a) input and thence (b)
efficiency. (Efficiency = ^J
28. The voltage of an accumulator, when discharging, fell according
to the following : At 2 o'clock voltage = 215, at 2.30 o'clock and also
at 3.30 voltage = 206, at 6.30 voltage = 187 and at 9 o'clock voltage
= 172. Another cell was charged at a uniform rate from 2 o'clock to
7 o'clock, the voltage rising from 175 to 238. Assuming that the
discharge was uniform, find the time at which the cells had the same
voltage.
Coordinates. So far, in these graph problems, we have been
concerned with positive quantities only; the question now is, How
to deal with negative quantities ? If the plotting " movement "
has been in a certain direction for the positive, then clearly for a
negative the motion must be reversed. The convention adopted
is that to the right and upwards are positive directions for the
horizontal and vertical axes respectively; and therefore to the
left and downwards will be the corresponding negative directions.
These are indicated in the diagram (Fig. 81). To admit of all
arrangements of signs the paper must be divided into four parts
or quadrants as shown, the point of intersection of the axes being
termed the origin, viz. the point O.
The points A! A 2 A 3 and A 4 are all distant 4 units from the
vertical axis and 3 units from the horizontal, so that to distinguish
between them we must make some mention of the quadrant in which
each is placed by affixing the correct signs.
The distances from the axes together are spoken of as coordinates,
that along the horizontal being usually called the abscissa, while
vertical distances are called ordinates. In representing a point by
its coordinates the abscissa is always stated first.
i Go
MATHEMATICS FOR ENGINEERS
Point A! is thus + 4 and + 3 or more shortly (4, 3)
A 2 is 4 and + 3 or more shortly ( 4, 3)
A 3 is 4 and 3 or more shortly (4, 3)
A 4 is +4 and 3 or more shortly (4,  3).
Note that ( 4,  3) does not imply 7, but a movement of
4 units to the left of the vertical axis and then 3 units down from the
horizontal axis.
E. g., Point B is (15, i)
Point C is ( 32, o)
Point Dis ( 14, 23).
D....
{14, 23)
I
^
4 3
2
1
(I5.I)
2
oAa
(4,:
3
(4,3:
Fig. 81. Coordinates of Points.
To fix the position of a point in space it would be necessary
to state the three coordinates, viz. the distances from three axes
mutually at right angles. For example, a gas light in a room would
be referred to two walls and the floor to give its position in the air.
Representation of an Equation by a Graph. If two
quantities x and y depend in a perfectly definite way, the one upon
the other, the relation between them may be illustrated by a graph
which will take the form of a straight line or a smooth curve. From
this curve much information can be gleaned to assist in the study
of the function as it is called. [Explanation. If y = 2X + 5,
INTRODUCTION TO GRAPHS
161
y is said to be a function of x, for y depends for its value on that
given to x; if y = 4z 2 +jz 3 8 log z, y is a function of z or, as it
would be expressed more shortly, y = /(), meaning that y has a
definite value for every value ascribed to z : e.g., in the case first
considered, y = /(*) = 2*f 5, then /(a) would indicate the value of
y when 3 was written in place of x, i. e.,/(3) = (2x3) +5 = u.]
Dealing first with the simplest type of graph, viz. the straight
line, whenever the equation giving the connection between the
variables is of the first degree as regards the variables, . e., it
contains the first power only of the variables, a straight line will
result when the equation is plotted.
Example 5. Plot a graph to represent the equation y = 5* 9.
In all cases of calculation for plotting purposes it is best to
tabulate in the first instance; for any error can thus be readily
detected, and in any case some system must be adopted to reduce
the mental labour and the time involved.
The general plan in these plotting questions is to select various
values for one of the variables, which we can speak of as the " in
dependent variable " (I.V.), and then to calculate the corresponding
values of the other, which may
be spoken of as the " dependent
variable" In questions where x
and y are involved it is customary
to make x the I.V., and to plot its
values along the horizontal axis.
We may take whatever values
for x we please, since nothing is said
in the question about the range.
Let us suppose that x varies from
4 to +4. The table, showing
values of y corresponding to values
of x would be as follows :
X
5*
 9
y
*4
20
 9
 29
3
 15
 9
 24
2
10
 9
 19
I
 5
 9
 14
o
 9
 9
I
5
 9
 4
2
10
 9
i
3
15
 9
6
4
20
 9
II
* i.e., 5* = 5* (4) =  20 '
M
Fig. 82. Curve of y = 52 9
162 MATHEMATICS FOR ENGINEERS
When we come to the plotting we see that it is advisable to select
different scales for x and y, since the range of x is 8 and that of y is 40.
On plotting the above values a straight line passes through them all
(Fig. 82).
A straight line would be definitely fixed if one knew its slope
or inclination and some point through which it passes. As regards
the slope, a line sloping upwards towards the right has a positive
slope, because the increase in the value of x is accompanied by an
increase in the value of y, and the slope is measured by r
change of x
In measuring the slope of a line, the denominator is first decided
upon, a round number of units, say 2 or 10, being chosen, and the
numerator corresponding to this change is read off in terms of the
vertical units from the diagram.
In the case of the line representing y = 5^9 the slope is
2 1 )
seen to be = 5, i. e., the slope is the coefficient of x in the original
*J
equation.
The fixed point, a knowledge of which is necessary before the
line can be located, is taken on the y axis through x = o, *'. e., the
point of intersection of the line with the vertical axis through x = o
must be known. In the case shown in Fig. 82 the line intersects
at the point for which x = o, y = 9 : also 9 is noted to be
the value of the constant term in the equation from which the graph
is plotted.
In general, if the equation to a straight line is written, y = ax+b ;
a is the slope of the line and b is the intercept on the vertical axis through
the zero of the horizontal scale.
All equations of the first degree can be put into this standard
form, and hence will all be represented by straight lines.
Example 6. Consider the three equations
= 8 ........... (i)
= o ........... (2)
A similarity is at once noticed between the equations ; a short
investigation will show the full interpretation of that similarity when
regarded from the graphical standpoint.
Whenever an equation is to be plotted it is always the best plan to
find an expression for one variable in terms of the other; and it is
usual to find y in terms of x in these simpler forms.
INTRODUCTION TO GRAPHS
Q ,*.
From (i) 5)/ = 84*
= 124*,
From (2)
From (3)
y 5  16 8*
y = 24 8x
163
 (4)
 (5)
(6)
Fig. 83. Straight Lines and their Equations.
Evidently all three equations, viz. (4), (5) and (6), are of the form
y = ax + b, the value of a being constant throughout, viz. 8, whilst
the value of 6 varies. From our previous work, then, we conclude that
the three lines representing these equations have the same slope and
are therefore parallel, being separated a distance vertically represented
by the different values of b.
To plot, first calculate from the equations
(i) y = 168*. (2) y = 8*. (3) y = 248*.
and tabulate the numerical values :
(i)
*
i6 8*
y
4
16+32
48
2
16+ 16
32
O
i6 o
16
2
i6 16 i o
4
1632
16
*
24 8x
y
4
24+32
8
2
24 + 16
 8
o
24
24
2
24 i6
40
4
2432
56
These lines are parallel (see Fig. 83) and cross the y axis, (i) at 16,
(2) at o, and (3) at 24, or the values of b in the three cases are 16,
o and 2 4 respectively.
164
MATHEMATICS FOR ENGINEERS
Solution of Simultaneous Equations by a Graphic
Method. Knowing that a firstdegree equation can be represented
by a straight line, our attention must now be directed to some useful
application of this property. One of the greatest advantages of
graphs is that they can be utilised to solve equations of practically
every description. As a first illustration we shall solve a pair of
simultaneous equations by the graphic method.
Example 7. To solve, by the graphic method, the equations
5*+3y =19 (i)
gx zy =12 (2)
Each of these equations can be represented by a straight line ; and
these lines will either be parallel or meet at a point, and at that point
only. Such a point represents by its coordinates a value of x and a
value of y ; and since this point is common to the two lines, these values
must be the solutions of the given equations.
Fig. 84. Solution of Simultaneous Equations.
[If the given equations were 5#+3J> = 19 and 5#+3y = 9 it would
be found on plotting that the lines were parallel ; there could thus be
no values of x and y satisfying the two equations at the same time, or,
in other words, the equations are not consistent.]
For the example given, the lines are not parallel.
Two points are sufficient to determine a line, and therefore two
values only of y need be calculated, but for certainty three are
here taken, because if two only were taken, and an error made in
one, the line would be entirely wrong.
INTRODUCTION TO GRAPHS
Equation (i) 5* + yy = 19 from which $y = 19 $x
or y = 633 167*.
Table of values reads :
165
X
633
i6jx
y
 3
633 +
5
n33
o
633
633
4
633
668
 '35
Equation (2)
whence
gx 2y = i2
2y = 12 9*
Table of values reads :
2y = 9^ 12
y = 4'5# 6.
X
4*5^ ~~"
y
2
 9
 6
 15
o
 6
 6
4
18
 6
12
These two lines must be plotted (see Fig. 84) to the same scales and
on the same diagram and their point of intersection noted, viz. (2, 3).
x = 2, y = 3 are the solutions of the given equations.
[The scales chosen must be such that the point of intersection
will be shown ; to ensure that this shall be the case a rough mental
picture of the diagram should be formed. This is not a difficult
matter, as one soon becomes accustomed to reading a table from
its graphical aspect. E. g., one can see at a glance in which direc
tion the line is sloping, and a little further consideration decides
the rate of its rising or falling.]
Exercises 21. On plotting Coordinates, and plotting of Straight Lines
representing Linear Equations.
1. On the same diagram plot the points (2, 5) ; (3, 4) ; (9, 3) .'
(o, n); and (12, o). Indicate each point clearly.
2. Join up the four points (10, 10) ; (5, 10) ; (15, 25) ; and
(10, 25) in the order given, and find the area in sq. units of the
figure so formed.
3. On the same diagram plot the points (14, 2500); (75, 374) '
(182, 1140); (32, 4816). Indicate clearly the scales chosen.
4. Plot the straight line $x8y .= 19 from x = 4 to x = +5.
What is the slope of this line, and what is its intercept on the vertical
axis through o on the horizontal ?
5. Plot a straight line to show the change of x consequent on
change of y between 10 and +15; the connection between y and x
being 'i6y = 428 406*.
166 MATHEMATICS FOR ENGINEERS
6. The illumination I (foot candles) of a single arc lamp placed
22 ft. above the ground, at d feet from the foot of the lamp is given by
I = i 4 oid.
Plot a graph to show the illumination for distances o to 12 ft. from
the foot of the lamp.
7. Unwin's law states that the velocity of water in ft. per sec. in
town supply pipes is v = 145^+ 2, where d is the diam. of pipe in
ft. Plot a graph to give the diam. of pipe for any velocity from o to
13 ft. per sec.
8. The law connecting the ratio (A i e, ^^  of a journal with
the speed (N, R.P.M.) is 3 = oo^N + i.
Plot a graph to show values of this ratio for values of N from 20
to 180. If the diam. is 45* what should the length be at 95 R.P.M. ?
9. Plot a conversion chart to give the number of radians correspond
ing to angles between o and 360. (i radian = 573.)
10. The law connecting the latent heat L with the absolute tem
perature T, for steam is
L = 1437  "Jr.
Plot a graph to give the latent heat at any temperature between
460 and 1000 F. absolute.
11. Plot a graph giving the resistance R of an incandescent lamp
at any voltage V between 40 and no. You are given that
R = 25 V + 75.
What is the slope of the resulting graph ?
Solve graphically the equations in Exs. 12 to 16 :
12. yn 6n = 66 13. 48* 27^ = 48
nn 25 = am. y$ix = 51.
14. y+i37=4* 15. 7#+3y=io
gxijy =4987. 35*6y = i.
16. y = 14* 3
26# y = 13.
17. The coordinates of two points A and B are :
A. Latitude (vertically) N 400 links ; Departure W (horizontally)
700 links
B. Latitude S 160 links; Departure W 1500 links.
Plot the points A and B and find the acute angle which the line
AB makes with the N and S line.
Determination of Laws. The straight line as the representa
tion of an equation finds its most direct and important application
in the determination of laws embodying the results of experiments.
An experiment has been made with some machine and a number of
readings of the variable quantities taken; and it is desirable to
express the connection between these quantities in a simple yet
conclusive manner. If this is done the law of the machine is
known for the range dealt with.
INTRODUCTION TO GRAPHS
167
Example 8. A test is carried out on a steam engine, and trials are
made with the engine running at various loads. The amount of steam
used per hour (W) and the Indicated Horse Power (I.H.P.) are calculated
from the readings taken at each load, and the corresponding values
are as follows :
I (I.H.P.)
4
5
7
IO
12
W (Ibs. of steam per hour)
7i
103
121
153
197
234
Find a simple relation connecting W and I.
It is reasonable to assume that to just start the engine a certain
amount of steam would be required, which would in a sense be wasted,
and that after once starting, the steam used would be practically
240
200
teo
120
QO
o
W
 3
<x
5
J
z
Values c
O2. 46 8 10 /
Fig. 85. Test on Steam Engine.
proportional to the power developed : accordingly we should expect
a formula of the type W = b + al where a and b are constants to
be determined. This we see is of the standard type y = ax + b, or
putting it in a more general form (Vertical) = a (Horizontal) + b, where
(Vertical) stands for the quantity plotted along the vertical ; therefore,
a straight line should result when W is plotted against I.
On plotting (see Fig. 85) we see that a straight line fits the points
very nearly, being above some and below others, i. e., averaging the
results.
The values of a and b may be found by either of two methods.
The first is that used in the laboratory and is to be recommended
when the slope of the line is more important than the intercept :
it can be used on all occasions when the quantities given admit
i68
MATHEMATICS FOR ENGINEERS
of the vertical axis through the zero of the horizontal being drawn
without diminishing the scale. This method is very quick, measure
ments on the paper being scaled off and a quotient easily found.
The second method is the more general, but involves rather more
calculation; both methods should, however, be studied.
First Method W = a I + b
where a the slope of the line and b = intercept on the vertical axis.
To find the slope, select some convenient startingpoint, say, where
the line passes through the corner of a square, and measure a round
number of units along the horizontal, in this case (Fig. 85) 5 being taken.
(Note. Distances are measured in terms of units, and not in inches.)
The vertical from the end of the 5 to meet the sloping line measures
79 units ;
, increase in W 79
hence slope =  ; ^ = = 158, .'. a = 158.
increase in I 5
Intercept on axis of W through I = o is 40 units, .'. b = 40.
Thus the equation is W = 1581+40
Second Method, or Simultaneous Equation Method
Select two convenient points on the line, not too close together
e. g., W = 1675 \ and w = 8 7'5 \
when I = 8 / when I = 3 J
Substituting these corresponding values in the equation W = a I + b
two equations are formed, the solutions of which are the required
values of a and b.
Thus 1675 = 8a + b (i)
875 = 3 + b (2)
Subtracting 80 = $a
whence a = 16.
Substituting in equation (2) b = 875 48 = 395
.*. as by first method (very closely) W _= 1 6 1 + 3 9 ' 5
This particular line connecting the weight of steam per hour with
the indicated horsepower is known as a Willans' line (named after
Mr. Willans, who first put the results of steamengine tests into this
form).
To take a further example
Example g. In a test on a crane the following values were found
for the effort P! required to raise a weight W. Find the law of the
crane.
W (Ibs.) .
IO
20
30
40
50
60
70
80
90
IOO
P! (Ibs.) .
I
i6 3
213
263
325
375
425
5
5'5
6
INTRODUCTION TO GRAPHS
169
To find the equation in the form P t = aW + 6 plot W along the
horizontal (Fig. 86).
First Method Slope = = 0564, .*. a = 0564
u
Also the intercept on the axis through o of W = 41, /. b = 41
PI= Q564W+ 41.
Second Method
when
Subtracting
Substituting in (2)
and Pj
when W
38 = 6oa + b
7= 5+&
(i)
a = 0564
6 = 7  282 = 418
! = 0564 W + 418.
or
282
5O
Values of W,
20 50 40 50 GO 70
Fig. 86. Test on a Crane.
SO
30 100
This result suggests that 41 Ib. is required to just start the machine,
i. e., to overcome the initial friction, and that after that point for every
pound lifted only 0564 Ib. of effort is required.
If we are told, in addition, that the velocity ratio of the machine is
39, we can calculate the efficiency of the machine for any load.
distance moved by effort
Velocity ratio = b>Twei g ht
and work done by effort = work done by weight ; hence, theoretically,
i Ib. of effort should just lift 39 Ibs. of weight ;
i. e., the connection between P and W (theoretically) is P = W.
170
MATHEMATICS FOR ENGINEERS
Theoretical effort P
Then the efficiency at any load =
Actual effort
LW
39 _ .
PI
0256W
_ ____
+ 418 0564 W + 418
22
1635
W
e. g., if W = 50, efficiency = rj
22 +
396.
1635
5
Example 10. The following are the results of a test on a 6ton
Hydraulic Jack (V.R. = 106).
Load (Ibs.) 1 600
IO2O
1445
I88 5
2320
2740
3210
3625
4010
Effort (Ibs.) ii
17
22
279
327
37'5
43'4
45'2
49
It is required to find an expression for the efficiency at any load,
and also the maximum efficiency.
O . 5OO /OOO 2OOO 3OOO
Fig. 87. Test on Hydraulic Jack.
4000
To a base of W (load) we plot the values of PI (practical effort)
and average the results by a straight line, as in Fig. 87.
Theoretically, each pound of effort applied should lift 106 Ibs.
of load, hence a straight line can be drawn giving the theoretical effort
(P) for all loads within the range dealt with,
p
Now, efficiency ? = p ; and therefore for any load find the quotient
INTRODUCTION TO GRAPHS
171
p, which will be the efficiency at that load. A new scale must be
chosen for efficiency, and the curve, a smooth one, because obtained
from two straight lines, is plotted.
[e.g., If W = 2000, P = 189, Pj = 28, , = ^ = 675}
To find the maximum efficiency, i. e., the efficiency at 6 tons load.
Also
P = V
PI = 5'6+,^4W 
P
P,
2oo
56+OH2W . . (See Fig. 87)
i
5'6 OII2W
"  7 + oo9 44 W
or efficiency at any load =
593
W
+ 119
Then for the efficiency at 6 tons load we must write 6 x 2240 for
W, hence
maximum efficiency = *
^3_ + ii 9 I>23
= 814.
13440
Exercises 22. On the Determination of Laws.
P
1. Find the average value of ^ (coefficient of traction) from the
following figures (i. e., find the slope of the resulting straight line).
W (Ibs.) .
3
5'5
7'5
95
"5
135
!5'5
175
P (Ibs.) .
125
225
2'75
375
4'25
525
625
725
This was for the case of wood on wood.
2. Recalculate but for cast iron on cast iron (dry).
W (Ibs.) .
33
533
632
729
93'2
"3
P (Ibs.) .
n3
19
22
2 5
28
37'5
In Exs. 3 and 4 the slope of the line gives the value of the Young's
Modulus E for the material. Find E in each case, stating the units.
(Note that the stress is to be plotted vertically.)
3. For i" round, crucible cast steel.
Stress (Ibs. per sq. in.)
2000
4000
6000
8000
IOOOO
I2OOO
14000
16000
Extension (inch per)
inch length) . . . /
00008
oooiS
OOO2I
00028
00034
00041
00048
00053
i;2 MATHEMATICS FOR ENGINEERS
4. For i" round, hardrolled phosphorbronze.
Stress (Ibs. per sq. in.)
2000
4000
6000
8000
IOOOO
12000
Extension (inch per\
inch length) . . J
oooi
00022
00034
00044
00055
00067
5. Find the simple law connecting the Indicated Horse Power I
with the Brake Horse Power B, given the following values of I and
B;
B
o
333
671
835
9'94
I
4'5
727
1066
1169
1295
{I = aB + b}
6. The diameter under the thread for various diameters of bolts is
given in the table for the Whitworth standard thread. Find the law
connecting the smaller diameter, d lt with the larger, d.
d
0625
09375
125
15625
1875
25
375
'5
625
'75
d,
0411
067
0929
1162
1641
1859
2949
3932
5085
6219
7. Recalculate as for Ex. 6, but taking the figures for the British
standard fine thread.
d
i
i
4
f
i
t
.1
ii
d,
199
3 ii
420
534
643
759
872
1108
Find the law connecting T and 6 in the following cases (Exs. 8
and 9). T = ad + 6. (T = twisting moment and = angle of twist.)
T
O
30,60
90
1 2O
150
1 80
2IO
240
270
300 330
360
o
49
156
21
27
3'4
4
4'5
5'i
53
625
682
9.
T
o
1 200
2400
3600
4800
6OOO
7200
e
o
34
67
I'O2
136
I7I
2O6
Express the results of the tests on incandescent lamps given in
Exs. 10, ii and 12 in the form R = aV + b. (R = resistance and
V = voltage.)
10. Test on a metallic filament lamp.
V
75
78  80
82
84
86
88
90
92
94
96
98
100
102
R
144
147 148
149
151
153
155
157
158
159
1 60
161
1625
1645
INTRODUCTION TO GRAPHS
11. Test on two metallic filament lamps in parallel.
V
54
60
65
70
75
80
8 5
90
95
I0 5
A
5
55
'57
59
61
63
65
67
69
72
(Values of resistance R must first be calculated from R = , where
A = ampere.)
12. Test on a metallic filament lamp.
V
86
80
70
60
5
40
30
R
277
267
259
231
208
174
I5<>
The following two examples refer to tests on the variation of the
resistance of a conductor with variation of temperature. Find the
values of R,, (resistance at o) and a (temperature coefficient) in each
case. [R, is intercept on the vertical axis through o of the temperature
Slope of line T
scale, and a ~ J
tv<>
13. Equation is of form R, = R u (i + at) where R t = resistance at
temperature t.
Temperature (/) .
10
2 5
35 50
80
90
IOO
Resistance (R t )
1039
ii
1141
1198
132
I 357
1402
14.
t
171
25'4
303
362
41
49'4
613
67
743
8o'i
i'532
938
R,
1214
1259
1285
1317
I34I
1369
144
1473
1505
1622
15. The following results were obtained from the testing plant of
the Pennsylvania Railroad Co. :
x (B.Th.U. across heat \
ing surface per min.) /
207200
24/500
295900
331000
367500
393500
44 3000
448500
481300
I (I.H.P.)
3657
4547
5876
650
7793
8033
95l'4
9751
1036
Find the law connecting I and x in the form I = ax + b.
16. From the following figures find the value of g. (g = 329 x slope
of line : obtained by plotting t 2 horizontally and / vertically.)
/
34'5
30
28
25
21
16
12
t
187
176
167
16
149
126
I'll
[t = periodic time in seconds of a pendulum swing, / = length of
simple pendulum in inches, and g = acceleration due to gravity, in feet
per sec. per sec.]
i 7 4
MATHEMATICS FOR ENGINEERS
17. Find the value of Young's Modulus E for the material of a
beam, from the following :
Load (W Ibs.) .
365
565
965
1365
1765
2165
2565
2965
3I65
Deflection (d in.)
12
198
34
5i
63
'79
925
107
117
W/ 3
Also d = O FT . I = 40", and I = 0127.
( Hint. Slope of line = ^.J
Graphs representing Expressions of th'e Second Degree.
Consideration must now be paid to the graphs of such equations
as y = 5* 2 + 7* 9, or x = ay z f by + c. As mentioned be
fore, the curves representing these equations will be smooth and of
standard forms. The preliminary calculation must be performed
in a manner similar to that already employed for the straightline
graphs. The only trouble likely to be experienced is with the
signs : it must be remembered that 3 or + 3 squared each gives
9, so that if x = 3 and x 2 is required, the value is (9), i. e.,
9 ; also 6x z would be 6 X ( 3) 2 = 6 x 9 = 54.
Since we are no longer dealing with straight lines, two points
are not sufficient to determine the curve, so a number of values
must be taken.
Example n. Plot, from x=$ to # = +4, the graph repre
senting the equation
y = 5 X * + 7* ~ 9
Arranging the calculation in tabular form :
X
x*
5 * 2 + 7*  9
y
 5
25
I 2 5  35  9
81
 4
16
80 28 9
43
 3
9
45219
15
2
4
20 14 9
 3
I
i
579
ii
O
o
o +09
 9
I
i
5+79
3
2
4
20+14 9
25
3
9
45 + 219
57
4
16
80 + 28 9
99
The scale for x must admit of a range of 9 units, whilst that for y
requires a range of no units : and as the greater part of the curve is
to be on the positive side of the x axis, this axis should be drawn fairly
INTRODUCTION TO GRAPHS
175
.90
80
.60
5o
low down on the paper and not in the centre (see Fig. 88). After
plotting the points from the table of values, a smooth curve should be
sketched in, passing
through all the points;
and if any one point is
not well on the curve,
the portion of the table
in which the calculation
for that point occurs
must be referred to.
The curve is a form of
parabola, whose axis is
vertical, and whose ver
tex is at the bottom of
the curve : indeed, in
all equations of the type
y = ax 2 + bx + c the
curve will be of the
form shown if a is posi
tive ; while if a is
negative the axis will
still be vertical, but the
vertex will be at the
top of the curve.
As an illustration
of the latter type
Example 12. Plot the curve 4 y =  3* 2 ~ **+ 2 '44 from x =  6
to x = +3
2 1
20
10
Fig. 88. Curve of y = 5* 2 + 7*  9
Division by 4 gives, y =
Table of values :
= _ 75AT 2 2X+'6l.
X
x*
'75* 2 2#+6i
y
 6
~ 5
36
25
16
27 + 12 + 61
 1875 + 10 + 61
12 + 8 + 61
 1439
 814
 339
4
 3
2
I
9
4
I
o
 675 + 6 + 61
_ 3 + 4 + 61
 75 + 2 + ' 6l
o + o + 61
 14
+ 161
+ 186
+ 61
I
2
3
i
4
9
 '75  2 + ' 6l
3 4+ 6i
675 6 + 61
214
 039
1214
Here the greater part of the curve is negative; hence the axis o
176
MATHEMATICS FOR ENGINEERS
x must be higher than the centre of the paper. The plotting is shown
in Fig. 89.
i 1 1 1 ,^^ 2 
Solution of Quadratic
Equations. The equations
5* 2 +7*9 = o
and 75* 2 2*+6i = o,
or, in fact, any quadratic equa
tion, can be solved by the
aid of graphs. For the equa
tions y = 5* 2 +7# 9 and
5* 2 +7# 9 = o to be alike,
y must equal o. Now y is = o
anywhere along the x axis : if,
then, we wish to arrange that the
y value or ordinate of the curve
is to be o, we must select the
value or values of x that make it
so; or, in other words, we must
find those values of x at the points
where the curve crosses the x axis.
These values of x are the solu
tions or roots of the equation
5^2+7^9 = o. From the dia
gram (Fig. 88) we see that the
curve crosses the x axis when
x = 82 and also when x = 222 :
Fig. 89. Curve of
4? =  3* 2  8 * + 2 '
therefore x = 82 or 222 gives the two solutions of 5* 2 +7# 9 = o.
In like manner the roots of 75** 2#+6i are 295 and 28.
(See Fig. 89.)
Solution of Quadratic Equations on the Drawing Board.
Whilst on the question of the graphical solution of quadratic
equations, mention may be made of a method that is simple and
requires the use of set squares and compasses, but not squared paper
The general quadratic equation is ax 2 f bx f c = o.
To solve this equation by the method of this paragraph : Set
off a length OA (see a, Fig. 90) along a horizontal line, working
from left to right, to represent a units to some scale. Through A
draw AB perpendicular to OA ; if & is positive a length to represent
& must be measured, giving AB, so that the arrows continue in a
righthand direction. If c is positive draw BC perpendicular to
AB, making BC to represent c units to the same scale as before, the
INTRODUCTION TO GRAPHS
arrows still continuing to indicate righthand movement about O.
(If c were negative BC would be measured to the other side of AB.)
Join OC. On OC as diameter describe a circle to meet AB in the
points D and E. Then the roots of the equation are 7^7 and W?.
OA OA
Proof of the construction. Let F be the centre of the circle ODC
(a, Fig. 90). Draw FG parallel to OA to cut AB in G and join
C to H, the point at which the circle cuts OA.
Then, from the property of intersecting chords
OA X AH = EA X AD
Fig. 90. Solution of Quadratic Equations.
Dividing both sides by (OA) 2
OA AH
OA X OA
or
AH
OA
EA
OA
EA
AD
X OA
AD
OA X OA
Now the angle OHC is a right angle since it is the angle in a
semicircle and since angle OAB is a right angle also, CH and AB
are parallel and AH = BC = c.
Also, since FG and OA are parallel and F bisects the line OC,
then GA = GB.
Then
EA+DA = ED+DAfDA = 2GD+2DA = 2GA = BA = b
EA DA_ _6_ _& /
OA + OA ~ OA a
DA
or
Let
^ OA =
178
MATHEMATICS FOR ENGINEERS
AH or BC
= ap
b
OA
= a/3 or aft = 
a
Then from equation (i)
and from equation (2)
Vtr
The original equation ax z + bx + c = o might be written
Or X* (a f ft)x + a/3 = O
which after factorisation becomes (x a)(x ft) = o
whence % = a or (3.
D\ EA
In other w r ords, a and /3 or ^ and ^K are the ro ts of the
original equation.
Example 13. Solve the equation ^x* + yx 9 == o by this method.
Starting from the point O (&, Fig. 90) set out OA to represent 5 units to
some scale. Draw AB downwards from A, since 7, the coefficient of x, is
positive, and make it 7 units long. From B draw BC 9 units long, to the
left of the positive direction of AB (since the constant term is negative).
Join OC and on it describe the circle cutting AB in D and also in E.
Then DA = + 404 units, EA = ni units and OA = 5 units
DA
or the roots are
and
EA
OA'
t. e.,
i. e.
5
iii
or 81
or 222.
Example 14. Solve by the
same means the equation
1'5X* + 4* f 122 = O.
First, change the signs
throughout to make the co
efficient of x z positive, i. e., the
equation becomes
I'5# 2 4* 122 = O.
Set out, in Fig. 91, OA = 15
units, AB (upwards, for b is
negative) = 4 units, and BC (to
the right, to reverse the direction
of movement about O, for c is
negative) = i 22 units. The circle
on OC as diameter cuts AB in D
and E.
DA = 42 (for this would
give lefthand rotation about O) ;
EA = + 4 45 , OA =15.
Fig. 91. Solution ot Quadratic
Equation.
INTRODUCTION TO GRAPHS
= 28
179
Then the roots are 7=rr =
OA
and
EA
15
Graphs representing Equations of Higher Degree than
the Second. This work will best be understood by some
examples.
Example 15. Plot a curve to show the cubes of all numbers between
o and 6. Use this curve to find the cube roots of 30 and 200.
_
Fig. 92. Curve of y = x a .
If x represents the numbers and y the cubes then the equation of
the curve will be y = # s .
i8o
MATHEMATICS FOR ENGINEERS
A few values of x may be taken, and the corresponding values of y
calculated, the curve being plotted to pass through these points. All
intermediate values can be interpolated from the curve.
The table of values reads :
*
o
2
3
4
5
6
y = x*
o
I
8
27
64
I2 5
216
The points all lie on a smooth curve (see Fig. 92), which is known
as a " cubic " parabola.
To read cubes, we must work from the horizontal scale to the
curve and thence to the vertical scale; thus the cube of 48 = in
while for the determination of cube roots the process is reversed ;
thus v'so = 31 and ^200 = 585.
Example 16. Represent the equation y = x 3 8x z + 3* + 15, by
a graph (x to range from 4 to +4).
Fig. 93. Curve of y = x 3 8x 2 + $x + 15.
The table is arranged thus :
X
x*
x 3 8x* + 3* + 15
y
 4
16
64 128 12 + 15
 189
 3
9
 27  72 9+15
 93
2
4
 8  32 6 + 15
3i
I
I
 i 83 + 15
+ 3
O
o
o 0+0 + 15
+ 15
I
i
i 8+3 + 15
+ ii
2
4
8  32 +6+15
 3
3
9
27  72 +9+15
21
4
16
64 128 +12+15
37
INTRODUCTION TO GRAPHS
181
The greater part of this curve is negative, hence the axis of x is
taken well up to the top of the paper (Fig. 93).
A warning is again given concerning the evaluation of 8x 2 ;
e. g., when x = 4. First find x 2 , i. e., ( 4)* or + 16, then find 8# 2 ,
j. e>> + 128, and finally Sx 2 = 128.
Example 17. Solve, graphically, the equation
2# 3 gx 2 2X + 24 = o.
Fig. 94. Curve of y = 2x 3 gx* 2x + 24.
We shall first plot the curve y = 2* 3  gx 2 2^+24 and then deter
mine the values for x at the intersections of the curve with the x axis.
Let x range from 3 to +5 ; and arrange the table as indicated :
X
x 2
X 3
2# 3 gx 2  2* + 24
y
3
9
2 7
54 81 +6+24
 105
2
4
 8
16 36 +4 + 24
24
_. _ T
i
i
2 9+2 + 24
15
O
o
o o + 24
24
I
i
i
2 92 + 24
15
2
4
8
16  36  4 + 24
o
3
4
9
16
27
64
5481 6+ 24
128  144 8 + 24
9
o
5
25
125
250 225 10 + 24
39
On plotting the values of y against those of x the curve in Fig. 94
is obtained.
182
MATHEMATICS FOR ENGINEERS
We observe that the curve is of a different character from
the " square " parabola, in that it bends twice whereas the latter
bends but once ; there is thus one bend for a seconddegree equation,
two bends for a thirddegree equation and so on. One can form
some idea of the form of the curve from the equation by bearing
in mind this fact.
The curve crosses the x axis at three points and three points
only; and the three values of x satisfying the given equation are
found from these points of intersection. Thus in Fig. 94
x = 15, 2, and 4.
A cubic equation has three roots, although in some cases only
one may be evident, the others being imaginary : if the curve
were drawn to represent an equation, two of the roots of which were
imaginary, it would cross the x axis at one point only, the bends
being either both above or both below it.
Example 18. A cantilever, 30 ft. long, carries a uniformlydis
tributed load of w tons per foot run. The deflection y at distance x
from the fixed end is given by the formula
where I = moment of inertia of section of cantilever
E = Young's Modulus of material.
/ = span.
If w = 5, I = 200, and E = 12500, show by a graph the deflected
form of the cantilever.
Span.
10 12 14 16 18
2O 22 24 26 28 30
20
Fig. 95. Deflection of Cantilever.
Substituting values
(5400**
+ x*)
24 X 12500 X 200 v
= '833 X IQ 7 (54OO* 2 I2O# S + X 4 )
= 833 x 10 7 x Y
(Y is substituted in place of the expression 54OO* 2 i2O# 3 + x*.)
Since the powers of x combined with their respective coefficients
give large numbers, it is found to be better to express all these large
INTRODUCTION TO GRAPHS
183
numbers as simple numbers multiplied by a power of ten Thus the
product 5400*2 when x = 5, which has the value 135000, is written
135 x io 5 , and similarly the other products are written in this abbrevi
ated form. One has thus to deal with the addition and subtraction
of small numbers, performing the multiplication or division by io
at the end once instead of three times. To find values of y from those
of Y we must divide by io' and multiply by 833, and according to our
scheme we find it convenient to note the values of Y x io 5 (shown in
the sixth column) and then multiply these by 833, dividing by io 2 .
By arranging the work in columns one setting of the slide rule suffices
for the multiplication by each particular constant, i. e., in evaluating
the values of 5400*2, 54 on the D scale would be set level with i on the
C scale ; and the figures in the second column would be taken on the
C scale, while the figures on the D scale level with these would be the
products of 5400 and # 2 .
Tabulation :
X
X*
X 3
X*
5400# a I20X 3 + X*
Yfio 5
y
o
O
o
+
o
o
5
25
125
625
i35Xio 5  i5xio s + o63Xio 5
126
0105
10
IOO
1000
I0
54 X io 5 12 Xio 5 f i xio 5
4'3
0358
15
225
3375
50600
I2I4XI0 5 4'05Xio 5 + 5o6xio 5
86
0716
20
400
8000
160000
216 xio 5 96 xio 5 +i6 xio 5
136
'"33
25
625
15630
390000
337 Xio 5 i8'75Xio 5 +39 xio 5
1885
I 57
30
900
27000
810000
486 xio 5 324 xio s +8i xio 5
24'3
2025
The deflected form is shown in Fig. 95, the scale for deflections
being magnified in comparison with the linear scale.
Turningpoints of Curves : Maximum and Minimum
Values. A quadratic curve has one bend, and a cubic has two :
there must therefore be some one point on each of these bends
which is either higher or lower than all other points in its immediate
neighbourhood, for the curves are perfectly smooth and continuous.
Such points are known as turningpoints of the curve, and it is
with these that we must now deal. If the curve is an ordinary
parabola, let us say that representing the equation y = $x 2 + jx q
(see Fig. 88), there can only be one turningpoint, and that is
lower than all points on the curve round about it. Referring now
to the ordinate at that particular point we note that it is less,
algebraically (i. e., taking account of sign), than any other ordinate
near to it; it is therefore spoken of as a minimum value of the
function. What is usually required is the value of the " inde
pendent variable " that makes the function a maximum or mini
mum : hence the highest or lowest point on the curve must be
184
MATHEMATICS FOR ENGINEERS
found, by sliding a straight edge parallel to the x axis until it just
touches the curve, the abscissa of this point being noted. Thus
the function 5# 2 f 7* 9 has its minimum value when x = 7.
The curve y = *75# 2 2x \ 61 would have no minimum
value (" minimum ", being understood to imply " less than any
other value in the immediate vicinity "), but would have its ordinate
a maximum when x = 133 (see Fig. 89). It is possible for
a minimum value of an ordinate to be greater than a maximum.
Many instances occur in practice in which greatest or least
values have to be found, or, more generally, values of some variables
which cause some function to have maximum or minimum values.
Questions of economy of material or time, best dimensions for
certain conditions, etc., all arise, and may be classed under the
heading of " maximum and minimum " problems. Before dealing
with any of these, an ordinary theoretical example will be treated
as a clear demonstration of the principles involved.
Example 19. Find the value or values of x that make the function
x 9 + 2X Z 4* + 7 a maximum or minimum. State clearly the nature
of the turningpoints.
First plot the curve y = x 3 + 2X* 4^+7.
is as follows :
For this, the tabulation
X
x z
x 3 + 2x z 4* + 7
y
 4
16
64 + 32 + 16+7
 9
 3
9
27+18+12 + 7
10
2
4
8 +8 +8+7
15
I
i
i +2 +4+7
12
O
o
o +o o + 7
7
I
i
i +2 4+7
6
2
4
8 +8 8 + 7
15
3
9
27+1812 + 7
40
4
16
64+3216+7
87
A rough plotting is made (in Fig. 96) from the figures in this table ;
and for greater accuracy the portion between x = 3 and x = i
and that between o and 1*5 are drawn to a larger scale and more values
of x are taken. One should always adopt such refinements as this ;
and especially does this apply when solving equations, viz. disregard
the portion of the curve that is of no immediate use and deal with the
useful portion in greater detail.
Apparently one turningpoint is in the neighbourhood of 2 and
another in the neighbourhood of i, therefore take as additional
INTRODUCTION TO GRAPHS
185
values for x, 25, 15, 5 and 15. Thus the subsidiary table
reads :
X
x z
x 3 + 2X 2 4* + 7
y
25
625
 1563 + 125 + 10 + 7
1387
 i'5
225
 338 +4'5 +6+7
1412
'5
25
13 + '5 2 + 7
563
i'5
225
338 +45 6+7
888
Y
2.
40
30.
Jc
2.
/
JJ
Fig. 96.
Drawing only these portions of the curve (see (a) and (6), Fig. 97)
we find that the trend is horizontal when x = 2 and also when x = 67.
Therefore, y = x 3 + 2# 2 4* + 7 is a maximum when * = 2, and a
minimum when x = 67.
Example 20. We require to find for what external resistance R
the power supplied from a battery of internal resistance r and electro
motive force E is a maximum. We are told that E = 84 and r = 57.
The power = (Current) 8 X external resistance
E.M.F. ) x external resistance
Uotal resistance/
RE 2 R X (8 4 ) 8
 (R + r) a (R + 57) 2
Since (8 4 ) 2 is a constant it can be disregarded throughout as it
i86
MATHEMATICS FOR ENGINEERS
does not affect the resistance for which the power is a maximum, but
only the magnitude of the power.
Let W = 7^ r ; then we require a value of R that makes W
(R + 57) 2
a maximum, and R must be treated as the I.V., i. e. t R is plotted along
the horizontal.
No negative values need be taken for R, but otherwise we have no
idea as to its magnitude ; a preliminary tabulation, and if necessary
a preliminary graph, must consequently be first made
The table reads :
R
(R + 57)
(R + 57) 2
R  w
(R+57) 2
oo
57
325
oooo
5
107
114
438
I'O
i'57
246
406
i'5
207
427
352
2'O
2'57
66
303
Apparently the curve rises fairly rapidly from R = o to R = 5
and then falls again : hence we conclude that the maximum value of
W will be obtained when R = 5 or thereabouts. (If this reasoning
cannot be followed from mere inspection of the table, a rough graph
should be drawn to represent it.)
Accordingly, let us take values between R = 2 and i o.
R
R+57
(R + 57) 2
R W
(R + 57) 2
2
77
592
338
4
97
94 1
425
5
107
II45
4367
6
117
137
4383
8
i'37
188
4263
io
i'57
246
406
INTRODUCTION TO GRAPHS
187
Plotting the portion between R = 4 and 8 (as in Fig. 98) we
find that W has its maximum value when R = 57, i. e ., the external
resistance is equal to the internal resistance.
44
43
42
"T*^
j
/
(
s
\
/
\
S
/
\
/
\
I
i
u
Cp 5 '
7
4 5 '6 'J ^8
Fig. 98. Curve of Power from an Electric Battery.
Example 21. The horse power transmitted by a belt passing round
a pulley and running at v feet per sec. is given by
H.P. =
g
where T = maximum stress permissible in belt = 350 Ibs./D*
w mass of i foot length of belt = 4 Ib.
g = 322. {The belt is 4* wide and i* thick.}
Find the speed at which the greatest horsepower is transmitted
under these conditions : find also the maximum horsepower trans
mitted.
Substituting the values of T, w and g, the equation becomes
/ 3 N
H.P. = Usoy^
IIOO\ 32*2>
IIOO
(35 oy 
The factor
may be disregarded in the curve plotting, as it
is simply a constant, and does not affect the value of v without similarly
affecting H.P.
i88
MATHEMATICS FOR ENGINEERS
Hence we plot the curve, Hj_ = 3501; oi24t; 3 ; and taking values
of v from o to 160 we obtain the following table :
V
V 3
3501* OI24?; 3
H,
o
o
20
8000
7000 99
6100
40
64000
14000 794
13206
60
216000
21000 2680
18320
80
512000
28000 6250
21750
IOO
I0 6
35000 12400
22600
I2O
1728 x io 6
42OOO 2 1 IOO
20900
140
2352 x io 6
49OOO 28700
20300
1 60
4096 x io 6
56000 50OOO
6000
H! is evidently a maximum somewhere in the neighbourhood of
v = 100 ; accordingly, taking some intermediate values, the subsidiary
table reads :
90
729000
31500  9040
22460
95
855000
33200 10600
22600
105
116 x io 6
36800 14400
22400
97
913000
33930  11310
22620
224OO
9Q
1976
100
Values af V
Fig. 99 Curve of H.P. transmitted by Belt.
Plotting the portion of the graph from v = 90 to v = 105 (Fig. 99),
we find that H x is a maximum when v 976. Maximum value of
H! is 22615, i.e., the maximum H.P. = y $ = 206. Hence
I IOO
we con
clude that the greatest H.P. is transmitted at a speed of 976 ft. per
sec. and that the greatest H.P. transmitted is 206.
INTRODUCTION TO GRAPHS 189
Exercises 23. On the plotting of Graphs of Quadratic and Cubic
Expressions : and on Maximum and Minimum Values.
1. Plot from x = 5 to x +3 the curve y = $x z 5* + 13.
2. Plot from x = 3 to x = +6 the curve y = 415* 23*2 + 1.94.
3. The centrifugal force on a pulley rim running at v ft. per sec. is
found from T  . If w = 336 and g = 322, plot a curve to give
o
values of T for values of v ranging from 70 to 200.
4. Plot a curve giving the H.P. transmitted by a belt running at
velocity v from H.P. = 4<  when t = 400 and v is to range
from o to 165.
5. Indicate by a graph the changes in B consequent on the variation
of T from 10 to 50 when
6. If w = Ibs. of water evaporated per Ib. of fuel, and / = Ibs. of
fuel stoked per hour per sq. ft. of grate
w = ^ + 85.
Plot a curve to give values of w as / ranges from 12 to 40.
7. The weight per foot W of certain railroad bridges for electrical
traffic can be calculated from W = 50 + 5/, where I = span in feet.
Plot a graph to give the total weight of bridges, the span varying from
12 to 90 ft.
8. Johnson's parabolic formula for the buckling stress (Ibs. per sq. in.)
of struts is (for W.I. columns having pin ends)
p = 34000  67 Jj
Plot a curve to give values of p for values of rj from o to 150.
9. Plot as for Ex. 8, but for C.I. columns, for which the relation
25 / I \ z
is expressed by the formula p = 60000   ( r ) ; the range of the ratio
/ 4 \/
v being from o to 55.
K
10. For Yorke's notched weir or orifice for the measurement of the
flow of water, the quantity flowing being proportional to the head,
Shape op
Opeaiaq
iaa Plcfte
_^__,.
h
^_^^x
Lirae of No Head
i
Fig. 100. The Yorke Weir.
igo
MATHEMATICS FOR ENGINEERS
the half width w (see Fig. 100) at head h is given by w = ^jj* Show
the complete weir for a depth of 6*, taking the range of h from 095*
to 6095*.
11. The length of hob /to cut a worm wheel with teeth of i* circular
pitch, N being the number of teeth, is found from
/" = .8 74 2N  1373.
Plot a curve to show values of /for N ranging from 10 to 120.
12. The resistance R, in Ibs. per ton for the case of electric traction,
o(V + 12) V 2
at a speed V miles per hour is given by R =  v  \
If V ranges from o to 40, show the variation of R by a graph.
13. The following equation occurs in connection with the reinforce
ment of rectangular beams k Vzrm + r z m z rm.
Plot a curve to give values of k for values of r ranging from 005 to
02, taking the value of w as 15.
Solve, graphically, the equations in Exs. 14 to 17.
14. x* 5* 6 = o. 15. 6x* 56 = $x.
16. 14** + 87*  154 = o.
17. (3 x io 6 * 2 ) + (28 x io*x) + (31 x io 4 ) = o.
18. Find a value of x which makes M maximum or minimum, it
being given that M = 342* ix 2 .
19. The following values were given for the B.H.P. and I.H.P. for
different values of the valve cutoff. Find the cutoff when the engine
uses least steam, (a) per I.H.P. hour; (b) per B.H.P. hour.
Cutoff ....
9f
6*
44*
3"
B.H.P
in
H5
"5
no
I.H P
118
I2S
127
114
Steam per hour
2160
2116
2080
2020
20. If 40 sq. ft. of metal are to be used in the construction of an
open tank with square base, the dimensions being chosen in such a
way that the capacity of the tank is to be a maximum for the metal
used : Let x ft. be the length of the side of the base ; then the volume
By taking values of x from o to 7, find that value
Hence find also the height
, ., , / Volume \
of the tank I 1 ).
21. The table gives figures dealing with gasengine tests.
is IQX cu. ft.
4
which gives the greatest volume of water.
PvHn nf ait >
117
1043
913
774
538
440
360
314
gas J
Gas per I.H.P.)
hour J
3i'9
22
208
19
216
248
298
34'5
INTRODUCTION TO GRAPHS lgi
What are the best proportions of the mixture for least consumption of
gas per I.H.P. hour?
22. In a noncondensing engine running at 400 revs, per min. the
following results were obtained ;
Ratio of expansion r
4
4'4
48
5'2
56
60
8
Ibs. of steam per 1
I.H.P. hour /
2075
2048
2035
2016
20
2032
2314
Find the most economic ratio of expansion.
23. The work done by a series electric motor in time / is given by
w = g(E  e)t
where e = back E.M.F., E = supply pressure, R = resistance of
armature.
The electrical efficiency is =.. Find the efficiency when the motor so
runs that the greatest rate of doing useful work is reached. {R = 035,
E = no, / = 20.}
24. The total cost C, in pounds sterling, of a ship's voyage of 3000
nautical miles is given by
P _ 3Ooo/ v 3 \
V \ 22OO/
where v is the speed in knots. Find the speed at which the cost has
its minimum value and state the cost at this speed.
25. To find the best angle of thread for a worm gear with steel
worm and brass worm gear, calculations were made with the following
results :
Angle (degrees)
o
10
20
30
45
60
75
Efficiency .
o
466
61
671
68
605
28
Find the best angle and the maximum efficiency.
26. If W = 4C a + ^, find a value of C between o and 5 that
Vx
makes W a maximum or minimum.
27. The efficiency t; of a Pelton wheel is given by ? = ^, '
Find the value of u in terms of v which makes 17 a maximum. Find
also the maximum efficiency.
28. If r, = 2U ( V ~ U ^ and v = 25, find the value of u for the maximum
i/ 8
value of 77 ; find also the maximum value of 17.
29. The ratio of horsepower to weight of a petrol motor is j^j
where D = diam. of cylinder in inches. Find the value of D which
makes this ratio a maximum.
30. Sixteen electric cells are connected up, in  rows of x cells
per row. The current from them is
16
+ 4
Find the arrangement
for maximum current.
31. Find a value of V between o and 10 that makes R a maximum,
when
_ 3 (Vi2) t
V+I2
54
32. Plot from x = 4 to #= + 4 the curve
33. Plot from x = 2to x = + 6 the curve
zy = 56*2 107* 148*3 + 88.
Solve, graphically, the equations in Exs. 34 to 37.
34. 2* 3  x 2  7* + 6 = o. 35. 2o# 3 + n# 2 + 27 = 138*.
36. x 3 + 5# 2 o8x 882 = o. 37. $op 3 + 4 = 2$p $p*.
38. Find the turningpoints of the function 2x 3 + 3# 2 36* + 15,
stating their nature.
39. If x is the distance of the point of contraflexure from the end
of a builtin girder whose length is /, find x in terms of / by the solution
fax fax^
of the equation i j + , a  = o.
40. To find d, the depth of flow through a channel under certain
conditions of slope, etc., it was necessary to solve the equation
d 3  i 305*2 1305 = o.
Find the value of d to satisfy this equation.
41. From tests with model planes Thurston calculated the following
figures :
Inclination of plane to
horizontal (degrees).
2
i
i
2
3
4
5
6
8
10
it
Weight supported
perH.P.
168
3i'i
SI'S
9<>'5
157
203
230
2565
2592
233
196
1285
Plot these values, to a base of angles, and find for what inclination
the greatest weight is lifted per H.P. developed.
CHAPTER V
FURTHER ALGEBRA
Variation. If speed is constant during a journey, the time
taken is proportional to the distance, i. e., the bigger the distance
the longer is the time taken, or, to extend this statement, twice the
time would be required for twice the distance.
This is expressed by saying that the time varies as the distance,
or more shortly t <x d where the sign oc stands for varies as. We
cannot say that t = d, but the statement of the variation is well
expressed by the equation  =J or r=^ 2 , where ^and d : are
1 2 2 <*1 **2
the values of the time and distance in one instance, and t 2 and d 2
are corresponding values in some other.
If, in the second arrangement, k is the number to which each
fraction is equal, it will be seen that
 2 = k ' t = kd
or, in general, t = kd.
Hence the sign of variation may be replaced by the sign of
equality together with a constant factor.
e. g., suppose the time for a journey of 300 miles is 15 hours,
then 15 = k X 30
or k = ^
i. e., the constant factor is ^ so long as the units are miles and
hours, and the speed is uniform.
Variation such as this is known as direct variation, since / varies
directly as d. Suppose now that the length of journey is fixed,
then the bigger the speed the less will be the time taken ; halve
o
i 9 4 MATHEMATICS FOR ENGINEERS
the speed and the journey takes double the time. Here the time
varies inversely as the speed when the distance is constant;
or t cc 
v
i. e., t = lx = 
v v
where / is some constant.
If both speed and distance vary, the time will vary directly as
the distance and inversely as the speed ;
or t cc d and also t cc 
v
. d
i. e., t cc 
v
md
or / =  (i)
v
This variation is known as joint variation.
A proof of statement (i) is here given, as the reason for it is
not selfevident.
Suppose the original values of time, distance, and speed are
/!, d v and v r respectively.
Change the distance to d z , keeping the speed constant : the time
will now be t, the value of which is determined from the equation
t d z , \
fj flj
Now make another change; keep the distance constant at d 2
but let the speed become v z , then the time will change to t 2 and
' (3)
'2
Multiplying equations (2) and (3) together
or = T^
or M = ~ constant = m, say.
md md,
= ~ or ^ =
. md
or, in general, t=
v
Questions on variation should be worked in the manner out
lined in the following examples.
FURTHER ALGEBRA I95
Example i The loss of head of water flowing through a pipe is
proportional to the length and inversely proportional to the diameter.
If in a length of 10 ft. of \ H diam. pipe the head lost is 46 ft., what
will it be for 52 ft. of 3^" diam. pipe ?
Taking the first letters to represent the words
h oc / when d is constant ; and h oc i when / is constant.
Then, when both I and d vary
h oc ^ or h = ^, where k is a constant.
We must first find the value of k. In the first case
,. k x 10
46 
Substituting this value in the second case
A='23X^ = '^f =368 ft.
Example 2. The weight of shafting varies directly as its length
and also as its cross section. If i yard of wroughtiron shafting of
i" diam. weighs 8 Ibs., what is the weight of 50 ft. of W.I. shafting
of J* diam. ?
If for weight, length and area, W, / and a respectively are written,
then W oc / and also W oc a; and when both I and a vary W oc la.
Also we know that the area of a circle depends on the diameter
squared; hence
a oc d 2
and W oc Id* or W = kid*
In the first case 8 = & x 3 X i 2
k= and W = ld*
3 3
Substituting this value in the second case
Example 3. The diam. d of a shaft necessary to transmit a certain
horsepower H is proportional to the cube root of the horsepower.
If a shaft of 15* diam. transmits 5 H.P., what H.P. will a 4" diam.
shaft transmit ?
Here d oc vH or
d = AH*
196 MATHEMATICS FOR ENGINEERS
Substituting the first set of values
i 5 = k x 5*
,s
When d = 4
4 "^
Transposing H* = 4 5 '
_ , . T 3
Cubing
An application of this branch of the subject occurs in con
nection with the whirling of shafts. It is known that the deflection
d of a shaft, as for a beam, is proportional to the cube of its length /,
and also that the critical speed of rotation c is inversely propor
tional to the square root of the deflection.
In mathematical language
d oc I s ............ (i)
i
and c oc ~ ............ (2)
\ d
We desire to connect c with /.
From equation (i) d = kP
Substituting in the modified form of equation (2), viz. c = 
vd
c==
where /> is some constant, *. e., the critical speed is inversely pro
portional to the f power of the length.
Thus if the equivalent lengths of the shaft under different
modes of vibration (i. e., for the higher critical speeds) are I, , ,
2 3
etc., the critical speeds are in the ratio i, 282, 52, etc.; for
comparing the first and third
/! = I / 2 = \
Cj = I C 2 = ?
but tl =
1 a
*. e., P = c l /^
also p = c 2 1
FURTHER ALGEBRA Ig7
Thus c 2 if = Cl if
= 52
Hence c * = $2
Ci i
Example 4. The energy E stored in a flywheel varies as the fifth
power of the diameter d and also as the square of the speed n.
Find the energy stored in a flywheel of 6 ft. diam., whilst it changes
its speed from 160 to 164 revs, per min., if the energy stored at 100
R.P.M. is 25000 ft. Ibs.
E <x d*n*
E = kd s n*.
When n = 100, d = 6, E = 25000,
so that 25000 = k x 6 s x ioo 8
k = f 5000^
6 s x io 4
Thus E at n = 164 = k x 6 5 x 164*
and E at n = 160 = k x 6 5 x 160*
Difference = k x 6 5 (i64 2 i6o 2 )
= 25000 x 6 S ( 4 )( 3 24)
6 6 x 10*
= 3240 ft. Ibs.
Example 5. A directacting pump having a ram of io* diam. is
supplied from an accumulator working under a pressure p of 750 Ibs.
per sq. in. When no load is on, the ram moves through a distance
of 80 ft. in i min. at a uniform speed v. Estimate the value of the
coefficient of hydraulic resistance or the coefficient of friction, viz. the
friction force when the ram moves at a velocity of i ft. per sec. ; the
total friction force varying as the square of the speed.
Find also the time the ram would take to move through 80 ft.
when under a load of 15 tons.
If the whole system is running light, the full pressure is used to
overcome the friction, i. e., p oc v 2 , since total friction force varies as
(velocity) 2 .
Thus p = kv z where k is the coefficient of hydraulic resistance ;
also v = g = i 33 ft. per sec., and p = 750
then 750 = k x (i33) a
or *ri^~ 4M
. e., the coefficient of hydraulic resistance is 422 if the units of pressure
and velocity are Ibs. per sq. in. and ft. per sec. respectively.
I 9 8 MATHEMATICS FOR ENGINEERS
The intensity of pressure due to a load of 15 tons
i"5 x 2240
= = 428 IDS. per sq. m.
If a
 X I0 2
4
Then, to find the velocity in the second case
Total pressure = pressure to overcome the friction + pressure
to move the load,
i. e., p = kv^ + pi
where v l is the new velocity, and p^ = 428.
Then 750 = kvf + 428 = 422 1^ 2 + 428
or v *  75 ~ 42S  76^2
Vi ~ 422
and P! = 8736.
Hence the time required for 80 ft. of the motion
80 i
x j~ = 1526 mms.
8736
Example 6. The linear dimensions of a ship are X times those of
a model. If the velocity of the ship = V, find the speed of the model
at which the resistance is p times that of the ship, given that the fluid
resistance varies as the area of surface S and also as the square of the
velocity.
Let R = resistance of ship ; then from hypothesis R oc S, and also
R oc V 2 .
Then R = KSV 2
and r = resistance of model = Ksv*.
Now
s i /for surfaces of similar solids are proportional to the\
*{
S X 2 \ squares of corresponding linear dimensions J
and we are told that
Hence
i.e.,
or
*'. e.,
v and V (which is v V\) are spoken of as " corresponding speeds."
FURTHER ALGEBRA IQ9
Exercises 24. On Variation.
1. The weight of a sphere is proportional to the cube of the radius
A sphere of radius 34" weighs 478 Ibs. ; what will be the weight of a
sphere of the same material, of which the radius is 417* ?
2. The candlepower (C.P.) of a lamp is proportional to the square
of its distance from a photometer. A lamp of 16 C.P. placed at 58
cms. from a screen produced the same effect as a second lamp placed
94 cms. from this screen. If this second lamp was absorbing 100
watts, find its efficiency, where 17 = watts per C.P.
3. The velocity of sound in air is proportional to the square root
of the temperature r (centigrade absolute, i.e., t C. + 273). If the
velocity is 1132 ft. per sec. at temperature 18 C., find the law con
necting v and T ; find also the velocity at 52 C.
4. The force of the earth's attraction varies inversely as the square
of the distance of the body from the earth's centre. Assuming that
the diameter of the earth is 8000 miles, find the weight a mass of 12 tons
would have if it could be placed 200 miles above the earth's surface.
5. The total pressure on the horizontal end of a cylindrical drum
immersed in a liquid is proportional to the depth of the end below the
surface and to the square of the radius of the end.
If the pressure is 1200 Ibs. when the depth is 14 ft. and the radius
is i yard, find the pressure at a depth of 6 yards when the radius is
8ft.
6. The loss of head due to pipe friction is directly proportional to
the length, to the square of the velocity and inversely proportional to
the diameter. If 2235 ft of head are lost in 50 ft. of 2* pipe, the
velocity of flow being 4 ft. per sec., find the diameter of pipe along
which 447 ft. of head are lost, the length of the pipe being i mile and
the velocity of flow 87 ft./sec.
7. The electrical resistance of a piece of wire depends directly on
its length and inversely on its diam. squared. The resistance of 85
cms. of wire of diam. 045 cm. was found to be 214 ohms. Find the
diam. of the wire of which 128 cms. had a resistance of 833 ohms.
8. The power in an electric circuit depends on the square of the
current and also on the resistance. The power is 1534 kilowatts
when 23 amps, are flowing through a resistance of 29 ohms. If a
current of 9 amps, flows through a resistance of 17 ohms for 50 mins.,
what would be the charge at 2d. per unit ?
(i unit = i kilowatthour.)
9. The electrical resistance of a conductor varies directly as the
length and inversely as the area of cross section. The resistance of
70 cms. of platinoid wire of diam. 046 cm. was found to be 1845 ohms.
Find the resistance of 194 metres of platinoid wire of diam. 028 cm.
10. The number of teeth T necessary for strength in a castiron
wheel varies directly as the H.P. transmitted, inversely as the speed
and inversely as the cube of the pitch p of the teeth.
If T = 10 when p = 2" and ratio of H.P. to speed (in R.P.M.) = 101,
find the H.P. transmitted when there are 30 teeth, the pitch of the
teeth being 6", and the speed being 30 revs, per min.
11. The coefficient of friction between the bearing and shaft varies
directly as the square root of the speed of the shaft and inversely as
the pressure. The coefficient was 0205 when the speed was 10 and
200
the pressure was 30 ; find the pressure when the coefficient is 0163
and the speed is 45.
12. The I.H.P. of a ship varies as the displacement D, as the cube
of the speed v, and inversely as the length L. If I.H.P. = 2880 when
D = 8000 tons, v = 12 knots, and L = 400 ft., find the speed for which
I.H.P. = 30600, the displacement being 20000 tons and the length
being 580 ft.
13. The pressure of a gas varies inversely as the volume and directly
as the absolute temperature r (see proof in Question 18). The pressure
is i kgrm. per sq. in. when the volume is 690 and the absolute tem
perature is 468; find the absolute temperature when the pressure is
892 kgrms. per sq. in. and the volume is 139.
14. In some experiments on antirolling tank models, the number
of oscillations per min. of a model of length 1075 ft was 2 7 I* the
number of oscillations per min. is inversely proportional to the square
root of the ratio of the linear dimensions, find the number of oscillations
of a similar ship 430 ft. long.
15. Assuming the same relations between volume, pressure and
absolute temperature as in Question 13; if the pressure is 108 Ibs. per
sq. in. when the volume is 1304 cu. ins. and the absolute temperature
is 641, find the absolute temperature when the pressure is 413 Ibs. per
sq. in. and the volume is 283 cu. ins.
16. The time of vibration of a loaded beam is inversely propor
tional to the square root of the deflection caused by the loading. When
the deflection was '0424* the time was 228 sec. ; find the deflection
when the time was 45 sec.
17. If the cost per foot of a beam of rectangular section of breadth
b and depth h varies as the area of section, and the moment of resist
ance of the beam is proportional to the breadth and also to the square
of the depth, find the connection between the cost per foot and the
moment of resistance.
18. Boyle's law states that the pressure of a gas varies inversely
as its volume, the temperature being constant ; Charles's law states
that the pressure is proportional to the absolute temperature, the
PV
volume being kept constant. Prove rigidly that = constant.
Series. A succession of numbers or letters the terms of which
are formed according to some definite law is called a series.
Thus 6, 9, 12 is a series for which the law is that
each term is greater by 3 than that immediately preceding it.
Again, 40,, i6ab, 6^ab z is a series in which any term
is obtained by multiplying the next before it by 46. In these
particular series, taken as illustrations, the terms are said to be in
progression, the former in Arithmetical Progression, written A. P.,
and the latter in Geometrical Progression, written G.P.
Other series with which the engineer has to deal are those
known as the Exponential and the Logarithmic ; and in the expan
sion or working out of certain binomial or multinomial functions
or expressions a "series" results.
FURTHER ALGEBRA 201
Arithmetical Progression. Consider the series of numbers
2, 9, 16, 23 ... etc.
The 2nd term is obtained from the ist by adding 7.
3rd 2nd 7.
4 ln 3rd 7.
i. e., each term differs by the same amount from that imme
diately preceding it. The numbers in such a series are said to be
in Arithmetical Progression ; and since the terms increase, this is
an increasing series.
Again, I, 4, 9, 14 ...... is an A.P., the common
difference in this case being 5. This is a decreasing series.
In general, an A.P. can be denoted by
a, (a + d), (a + zd)
where a is the ist term and d is the common difference.
Now the 2nd term = a+d = a+(2 i)d
and the 3rd term = a+2d = a+(3i)d
So that the 20th term = a+igd
i. e., the general term, or the /1 th term = a + (n l)d.
Thus the I5th term is obtained by adding 14 differences to the
ist term, or I5th term = a+md.
If three numbers are in A.P., the second is said to be the
arithmetic mean between the other two ; e.g., 95, 85, 75 are three
numbers in A.P., where 85 is the A.M. between 95 and 75 and
^ = 95 + 75 . or t k e arithmetic mean of two numbers is onehalf
3 2
their sum.
To find the sum of n terms of an A. P., which is denoted by S n
S n = a + (a + d) + (a + 2d) + ...... [a + (ni}d}
Also, by writing the terms in the reverse order
S n = {a + (n  i)d} + {+( 2}d} + {a + (n 3)^} + ..... a
Adding the two lines
2 S n = {2a +(n i)d} + (2a + (n  i)rf} +
{za f (n L)d} ...... to n terms
or 2S n = n {za +(n i)d}
If we call the last term /, then / = +( i)d, and the
formula for the sum can be written
s, = ?(,+/} or
202
MATHEMATICS FOR ENGINEERS
i. e., the sum can most easily be found by multiplying the average
term, i. e.,  , by the number of terms n.
Many problems on A. P. can be worked by means of a graph.
If ordinates represent terms,
and abscissae the numbers
of the terms, an A. P. will
be represented by a sloping
straight line for which the
" slope " is the common
difference d and the ordinate
on the axis through i of the
horizontal scale is the first
term.
The sum will be the area
under the line, with onehalf
the sum of the first and last
terms added.
Term VW
Fig. 101. Arithmetical Progression.
For the area under the line, viz. ABCD (Fig. 101)
IS
but
J(AD f BC) X AB =
(ni)
^areaunderline
Example 7. Find the sum of 12 terms of the series, 4, 2, o
find also its loth term.
18
Fig. 102 Sum of a Series.
2
12
I
14
16
18
FURTHER ALGEBRA
203
case, n = 12, a = 4 , d = 4 subtracted from 2 =  2 .
Si,  ^{(2X 4 ) + (12  I) X  2}
= 6 {8  22} = 84.
Also the lothterm =a + gd = 4 ~g X2
or graphically, area under the line""
= i (4 x 2} $ {9 x  18}. (Fig. 102 )
= 4  81 =  77
and J(a + l) = l^LlI = _ 7
S =  77  7 =  84
and ordinate AB represents the loth term and = 14.
Example 8. Insert 4 arithmetic means between z6 and 04
i.e., insert 4 terms between 16 and 94 equally spaced so that together
with the terms given they form an A.P.
94
2345
Te.rmJV
Fig. 103. Arithmetic Means.
6
The total number of terms must be 6 (two end terms together with
the 4 intermediate), so that
ist term = 16
and 6th term = 94
but the 6th term = a + 5^ and a = 16
16 + sd = 94
and $d = 78 or d = i 56.
Hence the means are 316, 472, 628, and 784.
The graphical construction would be quicker in this instance.
Referring to Fig. 103, draw a vertical through i on the horizontal
scale to represent 16, and a vertical through 6 to represent 94; join
204 MATHEMATICS FOR ENGINEERS
the tops of the ordinates by a straight line and read off the ordinates
through 2, 3, 4 and 5.
Example 9. In calculating the deflection of a Warren girder due
to the strain in the members of the lower flange, if U = the force in a
member caused by a unit load at the centre of the girder, F = the
force in the bar due to the external loads, a = area of section of member,
d = length of one bay, h = height of the girder, and n == number of
P
bays, then deflection = ^ x sum of all the separate values of the
product U x length of member.
p
If d = 20 ft., h = i2'6", n = 8, = 4 tons per sq. in., and E = 12500
tons per sq. in., find the deflection.
I
Fig. 104. Deflection of a Warren Girder.
Dealing with the first bay (see Fig. 104) and taking moments round
the point A
UiXAD or Jx^UjX*
d
u 1 =  A
i.d
For the second bay, by taking moments round E, U a = , ', while
cd 4"
for the third bay U 8 = ^r, and so on.
4* j _ j
Hence the sum of the separate values of U = ? + ~ + . . . .to
4* 4* d
n terms, i. e., it is the sum of an A. P. of which the first term is * and
d * h
the common difference is =
2*
Hence S. = i {(a x .*) + (.  ,)
The sum of the products of U x length of member is this total x d,
since all the members have the same length, viz. d.
Then the deflection
F
~ X
and for this particular case, by substituting the numerical values,
4 x 64 x 400 ,,
deflection =   ft.
12500 x 4 x 125
FURTHER ALGEBRA
20 5
7 Q . 8
"'
and the
Geometrical Progression.The numbers
are part of a series in which each term is
one by the use of a common S^
as a Geometrical Progression, or a G.P.
conLn muldp^er or ratio is  l^' ^ ISt term
Generally a G.P. may be expressed by
a ' ar>ar * ...... (' being the common ratio).
The 2nd term = ar 1 = ar z ~^
the 3rd term = ar z = a^ai
the n lh term =
e. g., the 5ist term =
If three terms are in G.P., the middle one is said to be the
geometric mean of the other
two : it is equal to the square
root of their product, for
if a, m and b be in G.P.
 = and m 2 = ab 2
or m =
Fig. 105 Geometrical Progression.
[If the true weight of a body
is required, but the weighing _ (
balance has unequal arms,
weigh in each pan, and call 2
the balancing weights W x and
W 2 respectively : then the
true weight W is the geometrical mean between W x and W 2 , or
W= VW x W a .]
To find the sum to n terms, written S n
S = a f ar + ar 2 + ar n ~ 2 f ar n ~ l
and r S,, = ar + ar 2 + ar n ~ z + ar"~ l f ar"
then S,,(i r] = a ar n (Subtracting)
(lrJ or (rl)
1 r r 1
and
the first form being used when the ratio is less than i.
Referring once again to the series 4, 2, i the
numerical value of the terms, plus or minus, soon becomes so small
that the sum, say, of 60 terms is practically the same as that of
50, and the series is said to be rapidly converging. This fact is
well illustrated by the graph of term values plotted to a base of
206 MATHEMATICS FOR ENGINEERS
term numbers as in Fig. 105 ; the area between the curve and the
horizontal axis being extremely small after even the fifth term of
the series has been reached.
Hence the sum of the entire series, called the sum to infinity
(of terms) and written S^, can be expressed definitely.
j. c a(i r")
r rom the rule S B =
i r
if r is very small and n is very great, r" will be very small indeed
compared with I, and may be neglected.
a
O
. 3 on
1r
Example 10. Find the sum of 5 terms of the series 2, 002,
000002 and compare with the sum to infinity.
In this case a = 2 and r = ooi.
Then S  a   ' 5
r i ooi
= {i (ix io~ 15 )}
999 '
d 22
whereas Soo = = = , and therefore the
i r i ooi 999
two are essentially the same.
Example n. The 5th term of a G.P. is 243 and the 2nd term is 9 ;
find the law of the series, viz. find the values of r and a.
5th term = ar* = 243 (i)
2nd term = ar l = 9 (2)
Dividing equation (i) by equation (2)
ar* 243
= 3* = 27
ar g
r 3 = 27
and r = 3.
Substituting in equation (2), a X 3 = 9 and = 3.
Hence the series is 3, 9, 27, etc.
It is of interest to note that the logarithms of numbers in G.P.
will themselves be in A.P.
Thus, if the numbers are 284, 284, 2840
(i. e., in a G.P. having the common ratio = io),
then their logs are i'4533, 24533, 34533
(i. e., are in A.P. with common difference i).
FURTHER ALGEBRA 207
Use may be made of this property when a number of geometric
means are required to be inserted between two numbers.
Suppose that five geometric means are required between 2 and
89. Mark off on a strip of paper a length to represent the distance
between 2 and 89 on the A or B scale of the slide rule. Divide
this distance into 5 + i, *. e., six equal divisions : place the paper
alongside the scale with its ends level with 2 and 89 respectively :
then the readings opposite the intermediate markings will be the
required means to as great a degree of accuracy as is required in
practice.
The means are 376. 71, 133, 251, and 473.
To check this by calculation
a = 2, and ar* = 89.
Hence, by division r 6 = = 445.
Taking logs 6 log r = i 6484
.*. log r = 2747
Now log av = log 2 + log r = 3010 + 2747 = 5757
.'. ar = 3763.
Also log ar z = log ar + log r = 5757 + 2747 = 8504
/. ar 2 = 7084.
Similarly, the other means are found to be I3'34, 2 5' Il > an d 4726.
It has already been demonstrated that the plotting of the
values of the terms in an A.P. to a base of " term numbers " gives a
straight line. Consequently it will be seen that if the logs of the
values of the terms in a G.P. are plotted to a base of " term
numbers," a straight line will pass through the points so obtained,
since the logs of numbers in G.P. are themselves in A.P. Conse
quently many problems on G.P. can be solved by means of a straight
line plotting.
Example 12. The values of the resistances of an electric motor
starter should be in G.P. Thus if r t = resistance of armature and
rheostat on the first step, and r 2 , r 3 , r t , etc., are the corresponding
values on the subsequent steps, then = =  8 , etc., and the value
*2 *3 M
of this ratio is ^, where C, = starting current and C = full load working
^
current.
Find the separate resistances of the 9 steps in a motor starting
switch for a 220 volt motor, if the maximum (i.e. starting) current
208
MATHEMATICS FOR ENGINEERS
must not exceed the full load working current of 80 amps, by more
than 40%, and the armature resistance is '133 ohm.
Q
Here we are told that ^ = 14 or the common ratio of the G.P.
\^
is : while the value of r can be calculated by Ohm's law, viz.
'
voltage 2;
ft.
= 1964 ohms.
starting current 14 x 80
The problem now is to insert 7 geometric means between 1964
and 133 ; and this can be done in the following simple manner. Along
a horizontal line indicate term numbers as in Fig. 106, and erect verticals
through the points i, 2 9.
Set the index of the A scale of the slide rule level with the point i,
and mark the point P at 1964 (at the righthand end of the rule) :
similarly the point Q should be indicated, the distance gQ representing
133 (at the lefthand end of the rule). Join PQ.
I 23456789
Term N?
Fig. 106. Resistances of an Electric Motor Starter.
Then the ordinates to this line through the points 2, 3 .... 8,
read off according to the log scale (i. e., by the use of the A scale of the
slide rule, the index being placed at the horizontal in each case), give the
required means, which are 1403, 1002, 716, 511, 365, '261, and 186.
Compound Interest furnishes an example of geometrical
progression.
If the original principle be P and the rate of interest be r
Then the interest at the end of ist year
= Pr and the amount = A x = Pf Pr
interest at the end of 2nd year
= rA,
= r(P+Pr) and amount == A^Ia
FURTHER ALGEBRA
209
i.e., I 2 = Pf(i+r) and A 2 =
I 3 = Pr(i+r) 2 and A 3 =
/. I n = Pr(i+r)i and A =
The consecutive interests are thus
Pr, Pr(ifr), Pr(i+r) z ......
. e., they are in a G.P. of ist term Pr and common ratio (i + r).
Hence total interest for n years
or the amount at the end of n years = P+Interest
Further Applications of G.P. If an electric condenser be
discharged through a ballistic galvanometer, and the lengths of
the consecutive swings of the needle are measured, it will be found
that they form a G.P. ; the ratio, of course, being less than I, because
the amplitude of the swing decreases.
If ! = ist swing and a % = 2nd swing,
then a = ka^
and On = k n ~ l a r
The logarithm of the ratio (
i. e., log (^ J according to our notation,
is called the logarithmic decrement of the
galvanometer.
Thus if the respective swings were,
in divisions on a scale, 36, 31*4, 2175,
etc., the ratio k = 3 ^ and the logar
ithmic decrement of the galvanometer
^6
= log ^ = 1594
To find the practical mechanical
advantage (^j for the pulleyblock
aw
shown in Fig. 107. The pull P on one Fig I07 ._puii e y Block,
side of the pulley becomes cP after
passing round the pulley (due to friction, and bending of the rope) :
after passing round the second pulley, the pull is now c 2 P, and so on.
2io MATHEMATICS FOR ENGINEERS
Hence W = cP(i+c+c 2 +c 3 +c 4 +c 5 )
if there are 6 strings
i c
for the case of 6 strings from the lower block.
This result may be put into a more general form by writing n
in place of 6 ; n being the velocityratio of the blocks.
W c
Thus J = ^(i' n )
In an actual experiment with a i : I block, the value of c was
found to be 837. Taking this value, the result given above may
then be written
By the use of this formula the maximum efficiency of any pulley
block can be determined. Thus for a 4 : i block n = 4
W
and p = 5'i3[i('837) 4 ] = 5'i3(i 497) = 2613.
W
Theoretically, p = 4, and hence the maximum efficiency
= = 653
Series may occur which, whilst not actually in arithmetical or
geometrical progression, may be so arranged that the rules of the
respective series may be applied.
Example 13. Find the sum of n terms of the series, the rth term
of which is
(i) 3' + 1 1 (2) 5 x 3 r 
(i) rth term = 37+1
Hence the ist term = (3 x i) + I
the 2nd term = (3x2) + !
and S n = (3X i) + (3X 2) + (3 x 3)+ .... +(1 + 1+1 . . . . to n terms)
= 3 (sum of natural numbers to n terms) + n
= 3 x "{2 +(ni)i}+
FURTHER ALGEBRA 211
(2) rth term = 5 x 3 r
Hence the ist term = 5 x 3 1
2nd term = 5 x 3*
and the nth term = 5x3"
also S = (5 x 3) + (5 x 3 2 ) + (5 x 3 3 ) +
or 7.5(3' i).
Methods of allocating Allowance for Depreciation. The
principles of the previous paragraph may be applied to deal with
the various systems of allowance for the depreciation of machinery,
etc., which may be calculated by one of three methods.
First Method, involving arithmetical progression, and sometimes
spoken of as the " straightline method."
According to this scheme, the annual contribution to the
depreciation fund is constant, and no interest is reckoned.
Let P = the original price of the machine, R = its residual
value at the end of its life, n years, and let D = the annual con
tribution to the depreciation fund.
E.g., if a machine costs 500, has a scrap value of 80, and its
life is 21 years, the annual contribution = *^ r = 20.
21
Then
Value at end of ist year = P R D
(i. e., neglecting its value as scrap).
Value at end of 2nd year = P R 2D
and Value at end of wth year = P R D
whilst the contributions to the depreciation fund would total nD.
Its value as a working machine would be o at the end of the
period, i. e., P R nD = o, or nD = P R; hence its value as
p p>
scrap would be taken into account in fixing D, for D = ,
which is not so great as 
n
Taking the figures suggested above
Value of the machine at end of ist year = 5008020
= 400, . e., its value as a working machine, and the depreciation
fund would then stand at 20.
At end of 2nd year, value = 5008040 = 380 and
depreciation fund = 40.
Thus the value + depreciation fund always = 420 = " work
ing " value of machine, which is as it should be.
212 MATHEMATICS FOR ENGINEERS
Second Method. According to this method of reckoning, the
same amount is added to the depreciation fund yearly, but interest
is reckoned thereon.
Let the rate of interest = r per annum per i.
At end of ist year, depreciation fund = D
2nd t , = DffD+D (since rD is
the interest on the
first contribution).
3rd = (2D+rD)+f( 2 D+fD)+D
We wish to find a general expression giving the magnitude of
the depreciation fund at the end of any year; to do this, the ex
pression last obtained must be slightly transposed.
3D+yD+rD = D ( a ^ ^multiplying and\
r ^ ' \ dividing by r )
This is the value of the fund at the end of the 3rd year.
In like fashion, the value of the fund at the end of the 4th year
so that at the end of the wth year, the depreciation fund stands at
This must be equal to the working value or P R,
. e., ?{( I+r )._i} = PR
p
E. g., if the original value = 500
the scrap value = 80
no. of years = 21
and rate of interest =3%, *' , r = 03
FURTHER ALGEBRA 2I3
Then
 D = /"3(Soo8o)
(i03) 21 i
x 420
Explanation.
Let x = (io 3 ) 2 i
log X =21 log 103
73^ = 21 x 0128
= 2688
x = 1857
There is one disadvantage in connection with the second method :
the depreciation fund does not grow rapidly enough in the early years.
Keeping to the same figures as before,
depreciation fund at end of ist year = 147
2nd = (2 X 147) + (03 X 147)
= 2984
3rd = (3 x 147) f (09 X 147)
+ (0009 x 147)
= 4544.
If the value of the machine decreases each year by 20, the
depreciation fund would not be sufficiently large to ensure no loss
in the event of the loss of machine in the first few years of its life :
on the other hand, provided nothing untoward happens, only about
threequarters of the depreciation has to be allowed for yearly,
i. e., 147 as against 20.
Third Method. The disadvantage of the second method may be
eliminated by setting aside each year a constant percentage of the
value of the preceding year.
Let this constant percentage be K : then at the end of the
first year KP will be assigned to the depreciation fund.
At the end of the 2nd year the fund will stand at KP + per
centage of value at end of ist year
= KP+(PKP) X K
= P(2KK 2 )
= P{i(i2K+K 2 )}
At the end of the 3rd year
depreciation fund = KP+K(PKP)+K(PKP)(iK)
= P{K+KK 2 fK2K 2 +K 8 }
= P{ 3 K 3 K 2 +K 8 }
= P{i(i3K+3K 2 K)}
214
MATHEMATICS FOR ENGINEERS
Hence at the end of nth year
depreciation fund = P{i (i K) n }
This must = P R
so that PR = PP(iK)
or P(iK)* = R
(!*) = ?
Taking the nth root of each side
or
R
To compare with the results by the other method, take the
figures as before, viz. P = 500, R = 80, and n = 21.
Explanation.
= i 9164
= 0836
Then
depreciation fund at end of ist year
Ditto end of 2nd year = 80
Ditto end of 3rd year = 116
/. e., the yearly allowance is greater at
commencement .
Let
21/8~
= Vo
={90311699}
= 7959
21
= " '0379
= 19621
x = 9164
Exercises 25. On Arithmetic and Geometric Progressions.
1. Find the 7th term and also the 2gth term of the series 16, 18,
20 ....
2. Which term of the series 81, 75, 69 ... is equal to 33?
3. The 3rd term of an A. P. is 34 and the i7th term is 8; find
the sum of the first 30 terms.
4. Insert 8 arithmetic means between 28 and 109.
5. Three numbers are in A. P. ; the product of the first and last is
216, and 4 times the second together with twice the first is 84. Find
the numbers.
6. How many terms of the series 18, 14, i . . . must be taken
so that the sum of them is 672 ?
7. In boring a well 400 ft. deep the cost is 2S. 3^. for the first foot
and an additional penny for each subsequent foot ; what is the cost
of boring the last foot and also of boring the entire well ?
FURTHER ALGEBRA 215
8. A manufacturer finds that his expenses, which in a certain year
are 4000, are increasing at the rate of 28 per annum. He, however,
sells 4 more machines each year than during the preceding, and after
1 6 years his total profit amounts to 14240. Find the selling price of
each machine and the total number sold over this period if his profit
the first year was 800.
9. A tank is being filled at the rate of 2 tons the first hour, 3 tons
the second hour, 4 tons the third hour, and so on. It is completely
filled with water in 10 hours. If the base measures 10 ft. by 15 ft.
find the depth of the tank.
10. A body falls 16 ft. in the ist second of its motion, 48 ft. in the
2nd, 80 ft. in the 3rd, and so on. How far does it fall during the igth
second and how long will it take to fall 4096 ft. ?
11. A slow train starts at 12 o'clock and travels for the first hour
at an average speed of 15 m.p.h., increasing its speed during the second
hour to one of 17 m.p.h. for the hour, and during the third hour to
19 m.p.h., and so on. A fast train, starting at 130 from the same place
travels in the same direction at a constant speed of 32 m.p.h. At what
time does this train overtake the first ?
12. Find the 5th term of the series i, 12, 144 ...
13. Find the sum to infinity of the series, 40, 10, 25 ...
14. Insert 3 geometric means between ij and 6.
15. Calculate the sum of 15 terms of the series 5, 65, 845 . . .
16. In levelling with the barometer it is found that as the heights
increase in A. P., the readings decrease in G.P. At a height of 100 ft.
the reading was 100; at a height of 300 ft. the reading was 80; at
500 ft. the reading was 64. What was the reading at a height of
2700 ft. ?
17. Find the sum of the series 15, 12, 96 ... to 7 terms and the
sum to infinity of the series 8, 02, 005 . . .
18. When a belt passes round a pulley it i? known that the tensions
at equal angular intervals form a G.P. If the tension for a lap of 15
is 2108 Ibs. and that for 90 is 2738 Ibs., find the least tension in the
belt, i. e., at o (the angular intervals are each 15).
19. The sum of the first 6 terms of a G.P. is 1020 and the common
ratio is 24; find the series.
20. Find the 2oth term of the series 3, 12, 33, 72, 135 . . . [the
nth term is of the form n(a + bn + en 2 )].
21. A contractor agrees to do a piece of work in a certain time
and puts 150 men on to the work. After the first day four men drop
off daily and the work in consequence takes 8 days longer than was
anticipated. Find the total number of days which the work actually
takes.
22. Find the deflection of the Warren girder shown in Fig. 104
due to the strain in the members of the upper flange. [H int. By
taking moments about the point B, the value of U AB = ~ v etc.]
23 A lathe has a constant countershaft speed, four steps on the
cone and one double backgear. There are thus 12 possible speeds
for the spindle; the greatest being 150 ^ vs. per sec. and the least
being 3 revs, per sec. If the spindle speeds are in 6.P, find the respective
speeds.
216 MATHEMATICS FOR ENGINEERS
Napierian Logarithms. Suppose that i is lent out at 2%
compound interest per annum.
Then the amount at the end of ist year = (i + 02) ; and
this is the principal for 2nd year.
I 2 = (i + 02) X 02
and A 2 = (i + '02) 2
If, however, the interest is to be reckoned and added on each
month the amount at the end of the first year will be greater, for
*O2
the interest = (*'. e., per month),
12
and, amount at end of ist month = ( i \  )
\ 12 /
(O2\ 2
I + j
/ 'O2\ 12
ist year = (* + ^) ....... (i)
Assume now that the interest is added day by day, i. e., practi
cally continuously, then at the end of ist year
(02 \ 365
i + ffi) ........ (2)
If the interest is calculated and added on each second, that
being as near continuity as we need approach
Amount at end of year
>? \31536000
1 +
3i536ooo/ ........
By means of laborious calculation the actual values of these
amounts could be found, and it would be observed that the amount
in (2) was greater than that in (i), and the amount in (3) was greater
than that in (2) ; the difference in the values being very slight,
and not perceptible unless a great number of decimal places were
taken.
It would appear at first sight that by increasing the number
of additions of interest to the earlier amounts, the final amount
could be made indefinitely large : this, however, is not the case,
for the amount approaches a figure beyond which it does not rise,
but to which it approximates more nearly the larger the value
of the exponent (*. e., 12, 365, etc.). This final amount is 2718
for a principal of i; in other words, when the interest, added
continuously, is proportional to the previous amount, the final
amount will reach a limiting value, being 2718. The symbol
FURTHER ALGEBRA 2I7
" e " is given to the figure 2718 . . . from Euler, the discoverer
of the series.
Later work will show that e can be expressed as a sum of a
series, viz.
1
^
_ ___
1x2 1x2x31x2x3x4
and from the foregoing reasoning it will be seen that it is a natural
number: it occurs as a vital factor in the statement of many
natural phenomena.
E.g., a chain hanging freely due to its own weight lies in a
curve whose equation may be written
or more simply Y = \ {e* + e~*}
i. e., it hangs in its natural curve (known as the " catenary "),
and this curve, depending for its form entirely on e, can only have
this one form if e is a constant, and, further, a particular constant.
Again, if an electric condenser discharges through a large resist
ance, the rate at which the voltage (i. e. t the difference in potential
between the coatings of the condenser) is diminishing is proportional
to the voltage. The equation which gives the voltage at any time
t
t, is v = ae where K, R and a are constants settled from the
given conditions. Then e is a constant, but one determined quite
apart from any particular set of conditions.
Actually the most natural way to calculate logarithms is to
work from e as base, such logs being called natural, Napierian, or
hyperbolic logs ; the common logs, *. e., those to base 10, which
are far more convenient for ordinary use, being obtained from
the Napierian logs. In higher branches of mathematics all the logs
are those to base e, for if natural laws are being followed, then any
logs that may be necessary must, of course, be natural logs.
It is, therefore, desirable to understand how to change from
logs of one base to logs of another. The rule can be expressed in
this form
log e N = log 10 Nxlog e io ........ (i)
To remember this, omit the logs and write the law in the
fractional form
N_ N 10
~i~ io e
which is equally correct, as proved by cancelling.
218 MATHEMATICS FOR ENGINEERS
Again log 10 N = log e N X Iog 10 5
N N e
or  =  x 
10 e 10
Proof of statement (i)
Let log e N = x, log 10 N = y and log e io = z
then N = e*, N= 10" and 10 = e*
and e* = io y = (f?) y = e y *
or x = yxz
i. e., log e N = log 10 Nxlog e io
Taking e as 2718 (its actual value, like that of TT, is not com
mensurate) Iog 10 = Iog 10 2718 = 4343 and logeio is the reciprocal
of this, viz. 23026 or 2303 approximately.
Hence, from the rules given above
log e N = 2303 log 10 N
logio N = '4343 logN.
To avoid confusion with these multipliers it should be borne
in mind that e is a smaller base than 10, and therefore it must be
raised to a higher power to equal the same number.
Hence the log to base e of any number must be greater than
its log to base 10.
If tables of Napierian logs are to hand, the foregoing rules
become unnecessary ; but a few hints as to the use of such tables
will not be out of place, for reading from tables of Napierian logs
is somewhat more involved than that from tables of common logs.
Examples are here added to demonstrate the determination of
natural logs by the two processes.
Example 14. Using the tables of natural logs (Table IV at the end
of the book), find log* 4872, log,. and log e 00234.
log, 4872 = log, (4872 x 10) = log e 4872 + log* 10
= 1 '5 8 35 + 23026
= 38861
log ' 4Ti = log ' Je? = loge 5 ' 7 ~ loge 4 ' 61
= 16233 ~~ 15282
= 0951
FURTHER ALGEBRA 2I9
log, 00234 = log, ^24 = log, 234  log, 1000
= log, 234  3 log, 10
= 8502 3 x 23026
= ^8502 69078
= 9424
It will be observed that for each power of ten in the number 23026
has to be added or subtracted as the case may be.
Example 15. Without tables of natural logs, find values of
log, 963, log, ?, and log, 2357.
log, 963 = Iogj 963 x 2303
= 9836 x 2303
= 2266
x 2303 . : '
= 2303 {Iog 10 i7i 7  Iog lo453 }
= 2303{32347  26561}
= 2303 x 5786
log,23 5 7 = logi 2357 x 2303
= 13724 x 2303
Separating the two distinct parts
= (f x 2303) + (3724 x 2303)
=  2303 + 8576
= 2:5546
{the subtraction being performed so that the mantissa is kept positive}.
Application of Logarithms to harder Computations.
In the first chapter the method of applying logs for purposes of
evaluation of simple expressions was shown. Such values were
found as (2I25) 5 , ^03, etc., *. e. t numbers raised to positive
powers only. The rules there used are applicable to all cases,
whatever the powers may be. A negative power may be made
into a positive power by changing the whole expression from top
to bottom of the fraction or vice versa (for a~ n = ) so that the
\ a n /
evaluation is obtained on the lines already detailed ; or it may be
obtained directly as here indicated.
N.B. Great care must be observed in connection with the signs : when
ever distinct parts (e.g., a positive and a negative) occur in a logarithm,
these should be treated separately.
220 MATHEMATICS FOR ENGINEERS
Example 16. Evaluate (
Let the expression = x
then x = (005 134)' 184 and by taking logs,
log x = 134 x log 005134
= 134 x 37 I0 4
= ('134 X 3) + (134 x 7104)
= 402 + 0952 = 16932 = log 4934
x = 4934
Notice that 402 is subtracted from 0952 although the former is
the greater; this being done so that the mantissa of the log shall be
positive.
Example 17. Find the value of (1473) ~ l1
Let *=
Then log x = 2*1 x log 1473 = 21 x 11682
= ( 21 x I) + ( 21 x 1682)
= + 21  3532 = 17468
= log 5582
x = 5582
Example 18. Evaluate {^3187} **
Let x = {log* 3 i87} 2
_ y'ost where y = log 3187.
The value of y must first be found.
From the tables log* 3187 = 11591
Hence y = 11591 and x = (11591) OM
Now log x = 024 x log 1159
= 024 x 0641
= 001538 = 1998462 or 19985
= log 9965
x = 9965
Example 19. Evaluate 
Let x = this fraction.
Then
log x ={J log 4217  2 log 0145} {2 log 891  116 log 5827}
=ii log 4217 + 116 log 5827} {2 log 891 + 2 log 0145}
FURTHER ALGEBRA 22 j
= (5417 + 2046)  (18998 + 43228)
B 7463 2'2226
= 25237 = log 3339
= 3339
Explanation.
log 4217 = 16250
X log 4217= 5417
log 5827 =17654
116 x log 5827= 2046
log 891= 9499
2 x log 891 = 18998
log 0145 =2 1614
2 X log 0145 =43228
When substituting figures for the letters in formulae and thence
evaluating the formulae, the importance of the preceding rules
will be recognised. Empirical formulae and also the direct results
of rigid proofs are of no value at all if one cannot use them efficiently.
It is necessary for this purpose that one or two rules, in addition
to, or in extension of, those already given should be rigidly observed,
viz.
Work one step at a time : keep all terms quite distinct until their separate
values have been found : and remember that statements including +
and cannot be directly changed into log forms.
e. g., x = 45 + (29) 1 ' 2
would not read, when logs were taken throughout,
log x = log 45 + 12 log 29 which is wrong.
To evaluate this equation, (29) 1>a would be found separately
and its value afterwards added to 45.
In cases in which a number of separate terms have to be evaluated
it is advisable to keep the separate workings for these to one side
of the paper and quite distinct from the body of the sum.
Example 20. A gas is expanding according to the law ptf* = C.
Find the value of the constant C when p = 85, v = 293 and n = 13.
Substituting values C = 85 x (293) 1 *
In the log form log C = log 85+13 log 293
= 19294+ (13 x 4669)
= 19294 + 607 = 25364
C = 3439
Example 21. The insulation resistance of a length / inches of
fibrecovered wire, of outside radius r,, and inside radius r l ; the specific
resistance of the insulator being S, is given by the formula
Find the resistance of the insulation of 50 ft. of wire, of outside
diam. 25 cm. and inside diam. I2 cm., when S = 3000 megohms.
222
R = 366 x
MATHEMATICS FOR ENGINEERS
Explanation.
J ** le. o
3000
50 X 12
125
^06"
366 x 5 x 734
= 1343 megohms.
is a ratio, and .*. r t and
?! may be in any units so
long as both are in the
same
log*
06
125
= logi25 log, 6
= 2231 14892
 7339
Example 22. For the flow of water over a rectangular notch, the
quantity
Q = aLH 1  6 + 6H 2 ' 5
Find Q when a = 27, L = 115, 6 = 28, and H = 517.
Making substitutions for the separate parts
Let Q = x + y.
Then the values of x and y must be first found quite separately and
then added. It is preferable in this example to treat the determination
of the values of x and y as the main portion, . ., to work these in the
centre of the page.
x = aLH 1  6
= 27 x 115 X (5I7) 1 ' 5
Then log x = log 27 + log 115 + 15 log 517
= 14314 + 10607 + (15 X I7I35)
= 14314 + 10607 15 + 10703
= 0624
x = 1154
Also y = bW'* = 28 x (5I7) 2 ' 5
Then log y = log 28 + 25 log '5 1 ?
= 14472 + (25 x 17135)
= 14472  25 + 17838
Alternative Method
of Setting Out.
V = 5383
Q = x + y
+ 5'383 = 6537
No
Log
517
ii'5
27
i'7 I 35
i'5
35675
7135
10703 +
i'5 
i 5703
10607
I43I4
II54
0624
FURTHER ALGEBRA
223
Example 23. The dryness fraction q of a sample of steam, ex
panding adiabatically, viz. without loss or gain of heat, can be found
from
where r lt q^ and LJ are the original conditions of absolute temperature,
dryness and latent heat respectively; and T, q and L are the final
conditions of absolute temperature, dryness and latent heat.
One Ib. of dry steam at 1151 Ibs. per sq. in. absolute pressure,
expands adiabatically to a pressure of 208 Ibs. per sq. in. absolute :
find its final dryness.
From the steam tables, r t (corresponding to pressure 1151 Ibs. per
sq. in.) = 799 F. absolute temperature, and r (for pressure = 208
Ibs. per sq. in.) = 691 F. absolute temperature.
Also the respective latent heats are L t = 879 and L = 954.
Then if qi = i, since the steam is originally dry
725{ii + 145}
725 x 1245
903
Explanation.
2303(29025  28395)
2303 x 0630
145
Example 24. For an airlift pump for slimes (a mixture of water
and very fine portions of crushed ore, of specific gravity = 11013)
the formula for the horsepower per cu. ft. of free air can be reduced to
H.P. = 015042 {P*(p) " p i}
Find H.P. when P! = 125, P, = 15
Substituting values
H.P. = .oi50 4 2{i 5 (^ 5 ) ?1  125}
= 015042 {15 x 8786  125}
= 015042 (1318 125)
= 01023
Explanation.
/I25Y 71
Let AT = (^P)
log x = 71 (log 125  log 15)
= 71(10969  11761)
= 71 x 0792
=  0562
= I9438
... x = 8786.
224 MATHEMATICS FOR ENGINEERS
Logarithmic Equations. Whenever it is required to solve
equations containing awkward powers it is nearly always the best
plan, and in many cases the only one, to use logarithms. Little
explanation should be necessary after the previous work, and a few
examples will suffice.
Example 25. Find v from the equation, pv n = C, when C = 146,
n = 137, and p = 22.
Substituting values 22 x v 1  37 =146
Then log 22 + 137 log v = log 146
x '37 log v log J 46 log 22
log 146 log 22
log v =  * *
_ 21644 13424 _ 822 _
137 ~ i'37~~
v = 3981
Example 26. If h = ^ , giving the head h lost in length /
of pipe of diam. d, the velocity of flow of the water being v, find d
when h = 87, v = 47 and I = 12.
Transposing for d
OOO4U 1 ' 87 /
* = j
n
Taking logs of both sides
14 \ogd = log 0004 + 187 log v + log / log h
= log 0004 + 187 log 47 + log 12 log 87
f46o2i
J 12568 19395 Explanation.
10792
log 47 = 6721
129381
29986
187 x log 47 = 12568
Then logrf = ^6 _  2 + 9986 =  10014
14 14 14
=  7 1 53 =
d =
Example 27. It is required to express the clearance in the cylinder
of a gas engine as a fraction of the stroke. We are told that the tem
perature at the end of compression is 1061 F. abs. and at the end of
expansion is 661 F. abs. ; and that expansion is according to the law
pv 1 '* = C. Also = K. (This example is important and should be
carefully studied.)
FURTHER ALGEBRA 225
Let p c , r e and v e be the pressure, absolute temperature and volume
respectively at the end of compression; and let p e , r t and v t be the
corresponding quantities at the end of expansion.
Then we can say p c v c 1 ' 3 = C = p e v e 1 ' 3
(Pc\ _ V 8
'' w'v 7 " 3
and also = K = tS.
Hence from equations (i) and (2)
or, dividing through by '
2 . (3)
T t
What is required is   ; and this can be found if is known.
v e v e v c
For.simplicity let x = 
Then from (3)
g
In the log form '3 log * = log 1061 log 661
log 1061 log 661 30257 28202
or log x = 
^=685
.. x = 4842 which is thus the value of ^
Hence v e = 48421;,.
and v e v e = 4842^ iv e
= 3 '842^
v c v e
3 '
260^
Exercises 26. On Evaluation of Difficult Formulae and on Logarithmic
Equations.
1. Find the natural logs of 2142 ; 318; 164.
i 1871
2. Find the values of log, 007254; log* 7254; log,. ^
3. Tabulate the values of log ^p when r = 461, 500, 560, 613,
800, and 1000 respectively.
4. Evaluate [log,
226 MATHEMATICS FOR ENGINEERS
5. Find the value of pv log e r when p = 120, v = 471 and r = 513.
Evaluate Exs. 6 to 14.
6. (2491) ' 7l 7. (1183)" 8.
(0054)' 1 '
9. (3418)" x (4006) 34 10 . S*2Lj3*5J>:*
11. (04I05) 2 ' 3 12. (3724) 2 ' 43
(lo ge i62)3 X (log 10 3256) 247
14. 1163 x (0005) 7 ' 76 i V(log 10 2i67) 1
15. The heat (B.Th.U) generated per hour in a bearing = dlv 1 ' 33
where d = diameter of bearing in inches, / = length of bearing in
inches, v = surface velocity of shaft in feet per sec. Find the number
of B.Th.U. generated per hour by a shaft of 5" diam., rotating in a
bearing 2 ft. long with surface velocity of 50 ft. per sec.
16. Find the value of a velocity v from
when c = 97, K = 63, g = 322 and h = 495.
17. The collapsing pressure P Ibs. per sq. in. for furnace tubes with
longitudinal lapjoints may be calculated from Fairbairn's formula
p= 7363 x i
where t = thickness in inches, / = length in inches and d = diameter
in inches. Find P when t = 043", / = 38", and d = 4".
18. Similarly for tubes with longitudinal and cross joints.
Calculate P if / = 12", I = 60", and d = 5^* from
fU
P = 15547000^^^,
19. The theoretical mean effective pressure (m.e.p.) in a cylinder is
calculated from
ft. , Eiii'og.") _ Ps
where P = boiler pressure, Pj = back pressure, and r = ratio of
expansion. The actual m.e.p. = p m x diagram factor.
Find the actual m.e.p. in the case when P = 95, P& = 15, cut off
is at 3 of stroke (. e., r = J and diagram factor = 8.
20. The H.P. required to compress adiabatically a given volume of
free air, to a pressure of R atmospheres, is given by
H.P. = oi5P(R' 29 i) when the compression is accomplished
in one stage and H.P. = O3P(R' 145 i) when the compression is
accomplished in two stages.
Find H.P. in each case if P = 147 and R = 46.
21. Find H, a hardness number, from
i6PD*
' n(2d)
Given that D = 24, d = 5, P = 58, n = 235.
FURTHER ALGEBRA 227
22. Mallard and Le Chatelier give the following rule for the deter
ruination of the specific heat at constant volume (K,,) of CO, (carbon
dioxide)
/ } V367
44K,, = 433 ^ j where t = C.
Find K v when t = 326.
23. Find H.P. from H.P. = oi 5 o 4 p 2 (? 1 )' n P 1 
when Pj = 125, and P 2 = 22; the letters having the same meanings
as in worked Example 24.
24. Find the efficiency i; of a gas engine from
/j\n~l
T; = I ~V~J when n = i4 and r 5
25. The H.P. lost in friction when a disc of diameter D ft. revolves
at N revs, per min. in an atmosphere of steam of pressure p Ibs. per sq.
in. abs., is given by
H.P. = io 13 D 5 N 3
Find the H.P. lost when the diameter is 5 ft., N = 500, and p = i.
26. If p = P(  }* =l and n= 141 find
\i + nj
27. Calculate the entropy of water <j> w , and that of steam <t>, at
absolute temperature T from
and ^log
The value of r is 682.
28. In the case of curved beams, as for a crane hook
where R = radius of inside of crane hook in ins. = 15, D = diam. of
cross section in ins. = 21, P = safe load of hook in Ibs., and 5 = maxi
mum allowable tensile stress = 17000 Ibs. per sq. in. Find the value
of P.
29. A sample of steam of dryness 83 at 380 F. expands adiabatic
ally to 58 F. ; calculate its dryness at the latter temperature from
is the initial temperature and L = 1115 '\
r = F. abrol. V
i.e., t+ 46iJ
30. Steam 20% wet at 90 Ibs. per sq. in. ^S^f
adiabatically to 25 Ibs. per sq. in. absolute. Find its
second pressure. Note that :
p = go ,t= 3 20F.; p = 2 5 ,t = wF.;
[Note also the difference between examples 29 and 30 as
given data.]
228 MATHEMATICS FOR ENGINEERS
31. The efficiency rj of a perfectlyjacketed engine is given by
+ 6(r 1 r 1 )
e l + a + &T!
TJ
where a = 1437, b = 7; r t and r, being the extreme temperatures
(F. abs.).
Find the efficiency of a jacketed engine working between 66 F.
and 363 F.
32. Calculate the efficiency of an engine working on the Rankine
cycle between 60 F. and 363 F., using the formula
LI + T! T,
T X and T Z are absolute temperatures and L= 1437 "jr.
33. Calculate the flow Q over a triangular notch from the formula
Q = 1 tan  V2i H*
A
where g = 322, H = 28, tan  = 577.
34. Find the number of heat units Hf supplied for the jacket to an
engine working between 60 F. and 363 F. from the formula
H, = 1437 !og 7^  ( r i  TI)
where r t and T X are absolute temperatures, initial and final respectively.
35. Francis' formula for the discharge of water over a rectangular
notch is
Q (cu. ft. per sec.) = 333 (L
If the breadth L = 54, the head H = 4, and n = 2, find Q.
36. If * = ^ , find * when m = 216, v = 165.
37. The volume of i Ib. of steam may be calculated from Callendar's
equation
where o> = 017, c = 1*2, R = 154
V = vol. in cu. ft., p = pressure in Ibs. per sq. foot, T = temperature
in centigrade degrees absolute (i. e., t C. + 273).
Find V when p = 10 Ibs. per sq. in. and t = 896 C.
38. Recalculate, when P = 7200 Ibs. per sq. foot and t = 1382 C.
39. Similarly, when p = 100 Ibs. per sq. in. and temperature is
437 C. absolute.
40. In calculating the tensions of ropes on grooved pulleys we have
the formula ~.
**
where 6 is the angle of lap in radians, /x is the coefficient of friction,
r is a coefficient depending on the angle of the groove, and T and t
are the greatest and least tensions respectively. Calculate the value
of T if the angle of lap is 66, /x = 22, t = 45 and r => 184.
FURTHER ALGEBRA 22g
41. The efficiency of an ideal or perfect engine (working on the
Diesel principle) is given by
where d = volume at cut ff = maximum volume
volume of clearance' volume of clearance
Find the efficiency when d = 156, r = 143 and n = 14.
42. Find the tensions T and * in a belt transmitting 20 H P the
belt lapping 120 round the pulley, which is of 3 ft. diam. and runs at
1 80 R.P.M. The coefficient of friction /x between the belt and pulley is 3.
Given that  = (P 9 and 6 = angle of lap in radians ; and
7rND(Tfl XT
""33060" ' = revs P 61 min ' and D ft  = diam  of pulley.
43. The pressure of a gas is 165 Ibs. per sq. in. when its volume is
2257 cu. ft. and the pressure is 98 Ib. per sq. in. when the volume
is 286 cu. ft. If the law connecting pressure and volume has the form
pv n = constant, find the values of n and this constant.
44. Find y from 4 2y = 587.
45. Solve for x in the equation x l ' n = i^x' n
46. When e Sc = 41 aS 2 ' 9 , find the value of c.
47. If (# 2 )*' 8 = gx : solve this equation for x.
48. Given that /^pt'J =/,*p,t, and also that Pl = 283, /i = 28,
and /, = 195 : find p a .
49. In the law connecting pressures and temperatures of a perfect
gas, find p, from the equation
having given that = 137, p^ = 2160, T, = 1460 and T^ =2190.
50. For a gas engine, P i; 1 ' 83 = p(v + s) 1 ' 83
where P = compression pressure, p suction pressure, v = clearance
volume and s = total volume swept out by the piston.
If P = 891 p and s = 138, find v.
51. If v = aH n , and H = 3 when v = 387 : and H = 80 when v =
2000 ; find the values of a and n.
52. If the pressure be removed from an inductive electric circuit,
the current dies away according to the law
where C is the current at any time / sees, after removal of the voltage,
R and L are the resistance and selfinductance of the circuit respec
tively, and V is the voltage. If R = 350, L = 55 and V = 40000, find
the time that elapses before the current has the value 80 amperes.
53. 120 was lent out at r% per annum compound interest, the
interest being added yearly; and in 5 years the amount became ^150.
Find the rate per cent. [Amount = Principal (i + j I
54. If P,VV = P<jV 33 ; = 206; and P<, = 44000; find P,.
230 MATHEMATICS FOR ENGINEERS
55. The insulation resistance R of a piece of submarine cable is
being measured ; it has been charged, and the voltage is diminishing
according to the law
_ t
v = be KR
where b is some constant, and t = time in sees, and K = 8 x io~ 6 .
If v = 30, and at 15 sees, after it is noted to be 2643, find the value of R.
56. Calculate the efficiency of a Diesel Engine from the formula
^CH
where n = 141, r c = compression ratio = 138 and r e = expansion
ratio = 74.
57. Determine the ratio of the maximum tension to the minimum
tension in a belt lapping an angle 6 radians round a pulley, the co
efficient of friction being \L, from
T ma *.
~
A min. 3
The coefficient of friction is 18 and the angle of lap is 154.
58. The work done in the expansion of a gas from volume i/ t to
volume v x is given by
i n
Find this work when v^ = 10, v t = i, and n = 113.
59. If T = te* 9 (the letters having the same meanings as in Example
40) : 6 = 288 radians, p. = 15 and t = 40, find the value of T.
T
60. Similarly if 6 = 165, and j = 178, find /i.
61. In the expansion of a gas it is given that pv n = c, and that
p = 1073 when v 3 : and p = 405 when v = 6 : find the law connecting
p and v in this case.
62. In a " repeated load " test on a rotating beam of fa" rolled
Bessemer steel, the connection between the stress F in Ibs. per sq. in.
and the number of revolutions N to fracture was found to be
F _ 214300
 N '147
Find the value of N when F = 40700.
63. In a similar test on a specimen of \" bright drawn mild steel
p _ 733QO
 N '04
Determine the value of F which makes N = 48300.
64. The total magnetic force at a point in a magnetic field
~ (r* + * 2 )i
Find this force when C = 4, n = 10, r = 4 and x = 59.
65. From the results on a test on the measurement of the flow
of water over a rectangular notch, complete the following table; it
FURTHER ALGEBRA
23I
being given that coeff. of discharge = j^^L$!^Lge_ . .
theoretical discharge'
retical discharge = 4015 bh? (Ibs. per min.).
ft
k
Actual Discharge
(Ibs. per minute).
Theoretical
Discharge.
c.
i75
829
35
i'75
I 4 I
79
i'75
181
1126
66. Also calculate as in the preceding Example, but for a submerged
rectangular orifice, for which the theoretical discharge
*2
*i
6
Actual
Discharge.
Theoretical
Discharge.
c.
2325
1075
I2 5
888
3'34
209
125
1096
4'4I5
3165
I2 5
133
61 1
486
I2 5
1566
67. The skin resistance per sq. ft. of a ship model is proportional
to some power of the speed. If the resistance is 0821 at velocity 5,
and 612 at velocity 14, find the law connecting resistance and velocity.
68. The loss of head due to pipe friction is proportional to some
power of the velocity. If loss of head was 1413 when velocity was 1023,
and loss was 631 when velocity was 676, find the law connecting loss
of head h, and velocity v.
69. Relating to the flow of water through pipes it is required to find
a value of d (the diameter of the pipe) to satisfy the two equations
OOO45V 1 ' 95 , it ,,
=  and *vX
If * (hydraulic gradient) = ^ , find this value.
2040
70. When a disc revolves in air the H.P. lost in air friction varies
as the 55 power of the diameter of the disc and the 35 power of the
revolutions. If H.P. lost is i when diameter is 4 and disc makes
500 R.P.M. find diameter when 10 H.P. is lost, the disc revolving at
580 R.P.M.
71. When a disc revolves in a fluid it is found that the friction F
per sq. foot of surface is proportional to some power of the velocity V.
For a brass surface
F per sq ft. .
22
126
V ft./sec. . .
10
25
Find a formula connecting F and V.
CHAPTER VI
PLANE TRIGONOMETRY
Trigonometric Ratios. If the ordinary 30 : 60 setsquare
be examined it will be found that for all sizes the ratios of corre
sponding sides are equal. If one of the angles is selected and the
sides named according to their position with regard to that angle,
the ratios of pairs of sides may be termed the trigonometrical ratios
of the angle considered. The word trigonometry implies measure
ment of angles ; the measurement of the angles being made in terms
of lengths of lines.
For example, let the sides of the setsquare be as shown in
Fig. 108 : then the angle 30 can be
denned as that angle in a right
angled triangle for which the side
opposite to it is 2", whilst the
hypotenuse is 4", i. e., the ratio of
, opposite side 2
hypotenuse 4
Again, the side 346" long is that
" lying next " or adjacent to the
angle 30, so that the angle 30 could
thus be alternatively denned by the ratio of its adjacent side to
the hypotenuse, or by the ratio of the adjacent side to the opposite
side.
To these ratios special names are given.
The ratio g pposite side is called the " sine " of the angle considered,
hypotenuse
Tne ra tio ad]acent Slde is called the "cosine" of the angle considered,
hypotenuse
The ratio ffl oate s "* e is called the "tangent "of the angle considered,
adjacent side
These three are the most important ; if they are inverted
PLANE TRIGONOMETRY
three other ratios are obtained, viz. the cosecant or ( J'
and
233
secant
'
or
As a general rule these ratios,
which, as denned, only apply to
rightangled triangles, are written
in the abbreviated form : sin, cos,
tan, cosec, sec and cot.
In the triangle ABC, Fig. 109
sin A = PP site to A _ a
hypotenuse c
whilst
sin B = PP site to B _ b
hypotenuse c
cos A
tan A
cosec A
sec A
cot A
_ adjacent to A
b
cos B
tan B
cosec B
sec B
cot B
a
hypotenuse
_ opposite to A
c'
a
c
b
adjacent to A
hypotenuse
~ b'
c
~ a'
c
a
c
~ b
c
~ a
opposite
hypotenuse
adjacent
adjacent
~ b'
b
opposite
~ a'
~b
The angles A and B together add up to 90 ; each being called
the complement of the other ; and it may be noticed that any ratio
of one of the angles is equal to the coratio of its complement.
Hence the syllable " co " in cosine, cosec and cotan, indicates
the complement of the sine, sec and tan respectively.
Thus sine A = cosine of its complement B.
tan B = cotan of its complement A.
For any angle the ratios could be found by careful drawing
to scale and measurement of sides; this is not very accurate,
however, and is certainly very tedious, and therefore tables are
provided, in which the ratios of all angles from o to 90 are ex
pressed. The changes in the values of the sine and cosine as the
angle increases from o to 90 are illustrated by Fig. no, in which
the quadrant is that of a circle of unit radius
"i, e., OA =s QC  op~ i'.
234
MATHEMATICS FOR ENGINEERS
Now sin i
L. EOC = EC
*** * JJ.il. j j . _ . _
nT = ^n = BA, and in like
manner sin
OA i
and sin L FOD = FD. Also cos L BOA = OB,
cos L EOC = OE and cos L FOD = OF. Thus the sine of the
angle depends on the horizontal distance from the line ON of the
end of the revolving line, while the cosine depends on the vertical
distance from OP.
When the angle is very small, A is very near to ON and conse
quently the sine is small ; and as O*A approaches ON more closely,
the value of the sine decreases until, when the angle is o the sine
is o, because the revolving line lies along ON. When the angle is
90, the revolving line
'^^ Curve
N %. /
\f
Curve of Cosines
the
lies along OP and the
horizontal distance of its
end from ON has its
greatest value, viz. i.
Thus the value of the sine
increases from o to i as
the angle increases from
o to 90.
Along OA, produced,
set off ABj = AB =
sin L BOA; and in like
manner obtain the points
Ej, Fj and O v Draw a
smooth curve through the
points M, B 1( Ej, 1 and
O 1 : then this is a curve
of sine values, since the
intercepts between the quadrant perimeter and this curve give the
values of the sine, thus sin L MOR = RR r
Similarly the curve of cosine values can be drawn, and it is seen
that it is of the same form as the curve of sine values, but it is
reversed in direction.
To read Table I at the end of the book. In this table one
page suffices for the various ratios, these being stated for each
degree only from o to 90. This table is compact and has educa
tional advantages, for it demonstrates clearly that as the angle
increases the sine increases whilst the cosine decreases; and that
a ratio of an angle is equal to the coratio of its complement, and
so on.
Down the first column and up the last are the angles expressed
PLANE TRIGONOMETRY 235
in degrees, whilst in the adjacent columns the corresponding values
in circular measure (radians) are given. Thus 31 = 5411 radian,
and 73 = 12741 radians.
The values of the sines appear in the 4th column from the
beginning and the 4th from the end, as do also the cosine values;
but for cosines the tables must be read in the reverse direction.
No difficulty should be experienced in this connection if it be
remembered that one must always work away from the title ol the
column. Thus for cosines read down the 7th column and up the
4th column.
Values of tangents and cotangents appear in the 5th and 6th
columns; again working away from the title
E.g., sine 17= 2924, sine 61 = 8746
cos 23 = 9205, cos 49 = 6561
tan 42 = 9004, tan 88 = 286363
cot 5 = 114301, cot 59 = 6009.
To read Table V at the end of the book, which should be used
when greater subdivision of angles is required. Suppose that sin
43 22' is required : if Table I is followed, sin 43 must be found,
22
viz. 6820, and sin 44, viz. 6947, and g of their difference must
be added to 6820.
'  ^
22
Thus sin 43 22' = 6820+ (6947 6820)
= 6867
This process is rather tedious : accordingly, referring to Table V,
look down the ist column until 43 is reached, then along the line
until under 18', the figure is 6858 ; 4' have now to be accounted
for; for this, use the difference columns, in which under 4', 8 is
found
.'. sine 43 22' = 685840008 = 6866.
The tangent tables, Table VII, would be applied in the same
manner, but here the value of the ratio gets very large when in the
neighbourhood of 90 so that the difference columns cannot be
given with accuracy. When the angle = 45, the tangent = I
and the tangent continues to increase as the angle increases, there
fore it happens occasionally that the integral part of the value has
to be altered in the middle of a line. To signify this a bar ( )
is written over the _first figure: e.g., tan 63 = 19626, whilst
tan 63 30' is written 0057, and this means 20057, the bar indicating
that the integer at the commencement of the line must be increased
by i.
236 MATHEMATICS FOR ENGINEERS
When using the cosine table, viz. Table VI, it must be remem
bered that an increase of the angle coincides with a decrease of the
cosine, so that differences must be subtracted: e.g., if the value
of cos 52 55' is required.
cos 52 54' = 6032 ; diff . for i' = 2
/. cos 52 55' = 6032 0002 = 6030.
Values of cosecants and secants can be found by inverting the
values of sines and cosines respectively.
Example i. The angle of advance 6 of an eccentric in a steam
engine mechanism can be found from sin 6 = j^  = . Find 6 when
the lap is 72", the lead is 12" and the travel is 36*.
Substituting the numerical values, sin 6 =  =
I "O I*O
= 4667.
We have now to find the angle whose sine is 4667.
Turning to the table of natural sines we find 4664 (the sine of
27 48') to be the nearest figure under 4667; this leaves 0003 to be
accounted for. In the difference columns in the same line we see that
a difference of 3 in the sine corresponds to a difference of I min. in the
angle; hence i' must be added to 27 48' to give the angle whose sine
is 4667. Hence 6 = 2 7 49'.
 a
Example 2. If cos A =  T  ' a = 42, 6 = 78 and c = 6 ; find A.
Substituting the numerical values
_ 7 8'+6 2 42 2 _ 6084+361764
'2x78x6' 936
= .3 6I .
93
From the table of natural cosines we find that the angle having
the ratio the nearest above 8461 is 32 12'; for this the cosine is 8462,
and therefore the difference of oooi has to be allowed for. In the
difference columns we see that a difference of 0002 corresponds to i';
and thus oooi corresponds to 30*. Hence A = 32 12' 30*.
Exercises 27. On the Use of the Tables of Trigonometric Ratios.
1. Read from the tables the values of: sin 61; tan 19; cos 87;
tan 2269 radian.
2. Find the values of sin 77$; cos 15 24'; tan 58 13'; cos 1283
radians.
3. Evaluate '
PLANE TRIGONOMETRY 237
4. In a magnetic field, if H = horizontal component and T = the
total force due to the earth, then H = T cos d. Find T when H = 18
and d = 63.
5. The tangent of the angle of lag of an electric current = reactance
resistance
and reactance = 2n x frequency x inductance. If frequency is 40,
inductance 0021 and resistance 17, find the angle of lag.
6. The mean rate of working in watts = amperes x volts x
cos (angle of lag). Find the mean rate when A = 243, V = no and
lag = I9i. What is the mean rate of working if the current lags 90
behind the voltage ?
rise
7. The pitch of a roof =  = 4 tan A, where A is the angle of the
span
roof. Find the angle of the roof for which the span is 36 ft. and the
rise is 12 ft.
8. If an axle of radius r runs on a pair of antifriction wheels of radius
R, and 6 is the angle between the lines joining the respective centres,
then
2
where F = force required to overcome the friction on a plane axle
and F! = force required to overcome the friction when using the
antifriction wheels. Find F if 6 = 47^, r = 3", 'R = 10* and F x = 47.
9. If D = pitch diameter of spiral toothed gear, N = number of
teeth in gear, P = normal diametral pitch, and a = tooth angle of gear,
then
PCOS a
If D = 5108, N = 24 and P = 5, find a.
10. In calculating principal or maximum stresses, if tan 26 = j t
s = 2852 and /= 3819, find d.
11. The number of teeth in the cutter for spiral gears
no. of teeth in the gear
cos 8 (angle of spiral)
Find the number of teeth in the cutter when the angle of the spiral
is 50 and there are 48 teeth in the gear. (#.B. cos 3 A means the cube
of the cosine of A ; but cos A 8 is the cos of A 8 .)
12. In connection with the design of water turbines the equation
" = tan 6 occurs, where w = tangential velocity of the water at
inlet, = radial velocity of water at inlet, V = velocity of the blade
at inlet, and & is the inclination of the blade at inlet,
per sec., V = 477 ft. per sec. and 6 = 6oJ, find w.
13. In the formula giving the value of the horizontal p ress ure
on a retaining wall of height h, the earth surface being ^***"
. , _ _ *. ~. earth and 4, i s the angle of repose c .he e artn,
Find the value of p when w = 130, * = 2 3 J and h  24 ft.
238 MATHEMATICS FOR ENGINEERS
14. Calculate the value of M, the moment of friction of a collar
bearing, from
__
asinaCRi 2  R, 2 )
when R! = 45, R z = 375, W = 2000, p. = 17 and a = 12.
15. The total pressure P on the rudder of a ship is given by
P= 4 6KAV 2  * in .
"39 + '61 sin a
where V = speed of ship in knots, A = area of rudder, K = 7, and
a = angle of rudder with fore and aft plane. Calculate P, given that
V = 16, A = 8 and a = 15%.
16. The force Pj applied horizontally to move a weight W up a
rough plane inclined at an angle a to the horizontal, is given by
p = WQi + tan a)
i p. tan a
Find P a if W = 3000, a = 8, and /*, the coefficient of friction, = 12.
17. The total extension d of a helical spring is given by
Wa 2 /.
d = FQ(I 2 sin 2 a)
If a radius of coil = 4", G = 12 x io 6 Ibs. per D*, J = 15,
/ = length = 29*, W = 12 Ibs. and a = 14, find the total extension.
V 2 sin 2A
18. The range of a projectile is given by   , where V =
o
velocity of projection, A = elevation of gun and g = 322. Find the
range, if the projectile is fired at an elevation of 29 15' with a velocity
of 1520 ft. per sec.
19. p n = intensity of the normal pressure of wind on a surface
inclined at 6 to the direction of wind, and p = intensity of pressure on
the surface perpendicular to its direction
2 sin 6
**~*'i + *a*e
If P = 35 and 6 = 22 J, find p n .
TT "p i
20. The maximum powerfactor of a motor = cos <f> = ' ' ~]
If H.P. is 478 find <f>, the angle of lag of the current.
21. If P = effort on crosshead of a steam engine, T = crankpin
connecting rod
effort, 6 crank angle, n =   ; and if P = 450 Ibs., = u
Sm 2 *
and 6 = 15 radians ; find T, from T = P/sin 6 + .
I Vw 2 sin 2 6}
{Hint. Sin 1719 = sin 81}
22. Calculate the value of y = Re~ K ' sin (wt + d) when R = 35,
K = 4, t = 02, w = 5, 6 = 16 ; the angle being expressed in radians.
23. The electrical induction B in an air gap is given by
Csin(i+)Rxio 9
B=  2
An x io 7
Find B when A = 3515, n = 20, X = 0867, C = 4205, R = 10382
and tan 20 = 1052.
PLANE TRIGONOMETRY 239
24. Find a value of 6 to satisfy the equation
4. an A $d(l 2x)
~!*~
where d = 5, / = 30 and x = 45. This equation refers to stiffened
suspension badges, where 6 is the angle of inclination of the cable to the
horizontal at a horizontal distance x from one end of the bridge, / is
the span of the bridge and d is the sag.
Application of Trigonometric Ratios. We will first deal
with a very simple case.
Example 3. The angle of elevation of the top of a chimney at a
point on the ground 120 ft. from the foot of the chimney is 25. Find
the height of the chimney.
Before proceeding to the actual working of the example, the
term angle of elevation must be explained. The zero of the
theodolite (an angle measuring instrument) would be observed
when the telescope was directed along the horizontal : the telescope
would then be moved in a vertical plane until the top of the chimney
was seen and the angle then noted. This angle is called the angle
of elevation and is the angle between the horizontal and the line joining
the eye to the objeet.
If the instrument be placed on the chimney top, the same angle
would be read, but it would now be called the angle of depression
because the object (the earth) is below the level of the eye.
In the example before us, let h ft. = height of chimney (Fig. in)
TT " _ PP!./< rt r or\
"T
and therefore = tan 25.
Now, from the tables
tan 25 = 4 66 3
.'. = 4663
120
and h = 120 x 4663 = 55 '96, say 56 ft.
Example 4. Two coils are connected in series over a 220 volt
alternatingcurrent main, and the drop across each coil is 126 volts.
If the diagram illustrating the relation between the voltage drops j
as in Fig. 112, find the difference in phase between the voltages in tl
two coils, i. e., find the angle a.
Since the sides AB and BC are equal, the perpendicular from B on
to AC bisects AC. or DC = no; and also /.DBC =
240
MATHEMATICS FOR ENGINEERS
Then
and
whence
sin  = ^ == 8730
2 126 /3
= sin 60 49'
=*>
a = I2I (
49
Example 5. In a test on the Halpin thermal
storage system, as fitted to a Babcock and
Wilcox boiler, the volume of water taken from
the storage tank to the boiler is to be deter
mined by the difference in water level between
start and finish. The tank being a cylinder of 57*81* diam. and 2^1*
length, with its axis horizontal, see Fig. 113, the water level is 5296"
from the tank bottom at the start and 1486* at the finish. Find the
volume of water abstracted in cu. ft.
We have to find the area of ABCD, viz. the difference between the
area AEBCD (at the start) and the area AEB (at the finish), and then
multiply by the length of the tank.
To find the area of the segment AEB
and
OF = OE  EF
OF
:OS a = prf = Q
OA 2891
a = 60 56' or
H
2891 1486 = 1405
= 4859 = cos 60 56'
6093
57'3
1063 radians.
Fig. 113. Halpin Thermal Storage System.
We can now use the rule previously given for the area of a segment,
9*
viz. area = (8 sin S) where 6 is the central angle in radians, for
r = 2891, 6 = u AOB = 2a = 2126, and sin 6 = sin 121 52'
= sin(i8o 121 52^ = sin 58 8' = 8493.
PLANE TRIGONOMETRY 241
proof of the rule sin A = sin (180 A) is given later
in the book.]
Thus area of AEB = Mi
( 2 . I26 _
= 533*8 sq. ins.
To find the area of the segment DHC
OG = EGOE = 5296  2891 = 2405
and
ft = 33 43' or 588 radian
sin 2,3 = sin 67 26' = 9234.
'83.8 cos 33' 43"
Hence area of DHC =
 923)
= 1059 sq. ins.
Area of the whole circle =  x 578i 2 = 2625 sq. ins.
area of ABCD = 262553381059 = 1985 sq. ins.
and volume = Jlj cu ft> = 2 88 4 cu. ft.
Outcrop
Example 6. A seam dips at an
angle of 62 to the horizontal for
a distance of 900 ft. measured
along the seam and then continues
dipping at an angle of 40 to the
horizontal. A shaft is started to
cut the seam at a distance of 1200
ft. horizontally from the outcrop;
at what depth will it cut the
seam ?
This example introduces the
solution of two rightangled tri
angles : the lengths AD and AC
(Fig. 114) are given and we require
to find DF.
AB
Fig. 114. Problem on a Coal Seam.
In the triangle ABC, = sin 28 = 4695
Also
Hence
In the triangle CEF,
AB = 900x^4695 = 4226'
= sin 62 = 8829
900
BC = 900 x 8829 = 79^4'
CE = BD = I200AB = 7774'
^g = tan 40 = 8391
EF = 7774 x 8391 = 6524
DF = DE+EF = BC+EF = 794'4+ 6 5 2 '4 = I 44P'8ft.
242 MATHEMATICS FOR ENGINEERS
Trigonometric Ratios from the Slide Rule. The sine
and tangent scales of the slide rule may be usefully employed in
trigonometry questions; the multiplication of the side of the
triangle by the trigonometric ratio being performed without the
actual value of the ratio being read off.
To read values of trigonometric ratios : Reverse the slide so
that the S scale is adjacent to the A scale and the T scale to the
D scale. The sines of angles on the S scale will then be read off
directly on the A scale. If the number is on the lefthand end of
the rule, then o must be prefixed to the reading, but if on the right
hand end of the rule, then a decimal point only.
e. g., to find sin 4 : place the cursor over 4 on S scale, and on
A scale read off 698; this being on the lefthand end of the rule
sin 4 = 0698.
Again, sin 67 = 921 for 921 is read off on the A scale above 67
on the S scale and is on the righthand end of the rule.
As the angle approaches 90 the sine does not increase very
rapidly and therefore the markings for the angles on the S scale
in this neighbourhood are very close together. From 70 the
usual markings are for 72, 74, 76, 78, 80, 85 and 90, the
longer mark being at 80.
To use the S scale for a scale of cosines, first subtract the angle
from 90, *. e., find its complement, and then find the sine of this.
e. g., cos 37 = sin 53 = 799.
To combine multiplication with the reading of ratios, use the
S scale just as the ordinary slide or B scale, multiplying, as it were,
by the angles instead of by mere numbers.
e.g., suppose the value of the product 185 x sin 72 is required.
The right hand of the S scale is set level with 185 on the A scale, the
cursor is placed over 72 on the S scale, and the product 176 is read off
on the A scale.
The tangents of angles from o to 45 will be read in a similar
fashion, the T and D scales being used. Tan 45 = i, and after
this the tangent increases rapidly, being infinitely large at 90.
For an angle greater than 45 : subtract the angle from 90 and
divide unity by the tangent of the resulting angle.
e. g., suppose tan 58 is required.
i
Actually tan 58 = ,
tan 32
Hence : set 32 on the T scale level with i on the D scale ; then
PLANE TRIGONOMETRY 243
the reading on the D scale opposite 45, ,'. ., the end of the T scale
is the required value and is 16.
A further example. Find the value of  7
tan 64
talV = 87Xtan 26 = 42 '4'
[The setting being : 45 on the T scale against 87 on the D scale
the cursor over 26 on the T scale; then 424 on the D scale.]
Example 7. A boat towed along a canal is 12 ft. from the near
bank and the length of rope is 64 ft. The horse pulls with a force of
500 Ibs. : find the effective pull on the boat, and that tending to pull
the boat to the side of the canal.
Fig. 115. Forces on Boat towed along a Canal.
The " space " diagram is first set out and from this x is calculated,
viz. x = v/64 2 i2 2 = 628 (a, Fig. 115).
If a triangle be drawn (see b, Fig. 115) with sides parallel to those
of the triangle ABC so that EF represents 500 Ibs. to some scale, then
EG and GF represent the pulls required to the same scales.
Or by calculation
= cos E = cos A = 2, *' GE = 500 cos E
500 64
i. e., the pull towards the bank = 938 Ibs.
Also = sin E = sin A = ^, *. e., GF = 500 sin E
500 64
, 49I
i. e., the effective pull in the direction of the boat's motion = 49* Ibs.
In general the components of a force R in two directions at
right angles to one another (see Fig. 116) are R cos a, and R s'.n a
244
MATHEMATICS FOR ENGINEERS
where a is the angle between R and the first of the components.
As a further example of resolution into components, if T (Fig. 117)
is the total magnetic force on a unit pole at some place and d
is the angle of dip, H the horizontal component of the force
is = T cos d, and V the vertical component = T sin d.
H
Fig. 1 1 6.
Components of Forces.
Fig. 117.
Calculation of Coordinates in Land Surveying. When
plotting the notes of a traverse survey, in which the sides of a
polygon and the " included " or internal angles are measured in
the field, it is necessary to first transform the dimensions of the
lines and angles so as to give the coordinates of the comers as
measured from the north and south line (or meridian) on the one
hand, and from some chosen east and west line on the other hand.
The survey is then plotted from the coordinates, with the object
of introducing an accuracy of drawing which is impossible if the
fieldbook dimensions are directly set out. In the latter case the
angular error is cumulative, and, further, the plotting of angles
at all times is more productive of error than the plotting of lines
(e. g., coordinates).
Quadrant bearings. The coordinate axes being chosen as just
stated, viz. NorthSouth and EastWest, every line of the traverse
is referred to the meridian in terms of the smallest angle between
it and the meridian, with the further statement of the " quadrant "
(N.E., S.E., S.W., or N.W.) in which it is placed. Such angles are
termed quadrant or reduced bearings.
Thus in Fig. 118
The reduced bearing of the line A is 27 N.E., that of the line
B is 36 S.E., that of the line C is 66 S.W., and that of the line D
is 11 N.W.
PLANE TRIGONOMETRY
245
handed direction. This is better than the quadranf method
requiring but one simple numerical statement
N.W. QUADRANT* N N.E. QUADRANT
S.W. QUADRANT c. S.E. QUADRANT
Fig. 118. Reduced Bearings.
Fig. 119. Wholecircle Bearings.
For example, in Fig. 119, the wholecircle bearings of the lines
A, B, C and D are respectively 27, 144, 246 and 349, all measured
from the north line ON.
Example 8. Measurements on a triangular plot of land ABC,
Fig. 120, resulted in the following : AB = 7073 links, BC = 7736 links,
CA = 5462 links, A = 75, B = 43 and C = 62. The reduced bearing
(R.B.) of AB is 9 N.E. and the point A is taken as the origin for the
coordinates. Find the reduced bearings of BC and CA, the coordinates
of the points B and C, and also the area of ABC.
Righthand order should be adhered to throughout, as indicated
by the letters ABC.
To find the R.B. of BC. [It should be grasped that the bearing
of C to B is not the same as the bearing B to C.] Mark on the diagram
all the known angles, and then by combination with 90 or 180 all the
required bearings can be found. Thus R.B. of BC = 43 9 = 34 S.E. ,
since 34 is the acute angle made by BC with the N. and S. line : the
quadrant must also be stated, to definitely fix the direction of movement.
Similarly, the R.B. of CA = i8o62 3 4 = 84 S.W.
To calculate the coordinates of B
also
= sin 9 and therefore BD = AB sin 9
t5
AD = AB cos 9.
246
MATHEMATICS FOR ENGINEERS
Thus the departure of B (i. e., its distance E. or W. from A)
= AB x sin (R.B. of AB)
and the latitude of B (i. e., its distance N. or S. from A)
= AB x cos (R.B. of AB)
Then
In the log form
BD = 7073 x sin 9
log BD = log 7073 + log sin 9
= 38496 + II943 = 30439
BD = 1106 links, which is the departure of
B east of A
JSL
Fig. 1 20. Plot of Land.
Again AD = AB cos 9 = 7073 x cos 9
In the log form log AD = log 7073 + log cos 9
= 38496 + T994 6 = 3844 2
.". AD = 6985 links, which is the latitude
B north of A
Hence the coordinates of the point B are 1106, 6985.
For the point C
In the log form
BE = 7736 sin 34
log BE = log 7736 + log sin 34
= 3888 5 + 17476 = 36361
BE = 4326 links, which is the departure of
C east of B.
PLANE TRIGONOMETRY 2 4?
Again CE = 7736 cos 34
In the log form log CE = log 7736 + log cos 34 = 38885 + 19186
= 38071
= 6413 links, which is the difference of
latitude between B and C.
Thus the coordinates of Care (1106 + 4326) and (6985  6413)
or (5432, ,572).
The figure may now be accurately plotted by means of the co
ordinates.
To calculate the area
A ABC = ADEFABDBECACF
= (5432 x 6985) (x 6985x1106)
= (3795 x io)(387X io)(i388x io 6 )(i5 53 x io
= 18647000 sq. links
Dividing by ioo 2 , = 18647 S( l chns.
Dividing by 10, = 18647 acres.
For greater precision tables of log sines and log cosines (viz.
Tables VIII and IX at the end of the book) have been utilised in
the working of this example. For general work the accuracy of
the slide rule is sufficient, but in all cases these tables, which are
used in the same way as the tables of natural sines and natural
cosines, are convenient.
As shown earlier in the chapter the value of the sine or cosine
of an angle varies between o and I, and accordingly the values of
the logs of these ratios vary between oo (;'. e., the smallest quantity
possible) and o, since log o = oo (refer Chapter I) and log i = o.
Except for small angles, therefore, the log sine will be of the nature
of 7 ..... or ~2' ..... whilst the value of the log cosine will
be 7 ..... or 2 ..... unless the angle is large.
e. g., sin 27 = 454 and Io g sin 2 7 = lo g '454 =
sin o33' = 0096 and log sin o33' = log 0096 = 39823
cos 87 = 0523 and log cos 87 = log 0523 = 27185
Example g. From the following coordinates compute the true
length, the bearing, and the angle with the horizontal of the line AB.
Station.
Feet.
Feet.
Feet above Sea Level.
A
B
Northing 45012
Southing 201
Westing 561
Easting 47881
Reduced level 2492
Reduced level 3292
248
MATHEMATICS FOR ENGINEERS
By plotting the points A and B from the coordinates given, their
actual positions are represented. Complete the triangle ABC by
drawing AC vertically and BC horizontally (see Fig. 121).
Then AC = 45012 + 201 = 45213
and BC = 47881 + 561 = 48442
561
To express the results with the
same accuracy as that with which
the figures have been measured we
must use five figure log tables.
To find the angle CAB
tan CAB =484412
45213
i.e., log tan CAB = log 48442
 log 452 1 3
= 368523  3655 2 7
= 01996
= log tan 46 19'.
CAB = 46 19'
The whole circle bearing of AB
is thus 180 46 19' = 133 41^.
To find the length (on the plan) of AB
pjs = sin 46 19'
AB
' Bearing angle
reauired
I
inclined length
CB
In the log form
sin 46 19'
= 48442
sin 46 19'
log AB = log 48442 log sin 46 19'
= 368523  185924 = 382599
= log 66987
AB= 66987
This length found is that of AB on the plan ; the true length will be
slightly greater than this, since it is the hypotenuse of the triangle
of which the base is 66987; and the height is 80.
In the triangle ABB 1 :
tan B'AB = 
66987
and log tan B 1 AB = log 80 log 66987 = 190309 382599
= 207710
= log tan 41' 4*
.*. the inclination of AB to the horizontal is 41 '4*
and the true length of AB (or AB 1 ) = V (66987) ~^8o*
= 66993
PLANE TRIGONOMETRY
249
Exercises 28. On the Solution of RightAngled Triangbs, and the
Calculation of Coordinates.
In the following Examples i to 7, ABC is a triangle rightangled
at C. (In each case the figure should be drawn to scale.)
1. c = 45", A = 15, find a and b.
2. a = 12", B = 36, find 6 and c.
3. c = 65", A = 48, find a and b.
4. b = 34", B = 27, find a and c.
5. c = 2737", A = 54, find a and 6.
6. 6 = 72 5", A  3 8i, find a and c.
7. c = 234", B = 27i, find a and b.
8. A bomb dropped from an aeroplane strikes a building which is
known to be one mile away from an observing station, at which the
elevation of the aeroplane is seen to be 29. Find the " range," i. e.,
the distance of the aeroplane from the observer, and also its height.
9. A mountain railway at its steepest rise has a gradient of i in 7.
What is the inclination to the horizontal of this gradient ? [Note that
the gradient is always the *;
hypotenuse J
10. From the top of a house, 37 ft. high, a bench mark (Government
height above sealevel) is sighted, and the angle of depression is 48.
Find the horizontal distance from the house of the B.M., which is
placed at a point 3 ft. above the ground.
11. The crank and connecting rod of a reciprocating engine are at
right angles to one another. If the value of the ratio
connecting rod length
length of crank
is 47, find the angle which the crank makes with the line of stroke.
12. The rise of a roof is n ft. and the span is 84 ft. : find the angle
of the roof.
13. The tangent of the angle of a screw is given by the pitch divided
by the circumference of the screw. If the diameter is 5* and the
pitch angle is 7 15', find the pitch.
14. If the screw in Ex. 13 becomes (a) double or (b) treble
threaded, what are now the angles of the thread ?
665 
FT'
Fig. 122. " Setover " of Lathe
Tailstock.
Fig. 123. Brown and Sharpe
Worm thread.
15. Calculate the " setover " of the tailstock of a lathe for turning
a taper (the angle being 9 and the length of job 152). See Fig. 122.
16. Find the angle of thread 6 for the Brown and Sharpe worm
thread shown in Fig. 123.
250
MATHEMATICS FOR ENGINEERS
17. When using the Weldon Range Finder, one determines a length
AB
AB by comparison with a base AD. Find ratio of ^~. for the case
illustrated (Fig. 124).
B B
90
Base
Fig. 124.
D
Fig. 125.
18. Determine the coordinates of the points A, B, C and D (Fig. 125)
with references to the axes marked. Find the area of ABCD; and
state also the " reduced bearings " of BC, CD and DA. The bearing
of ABis 505 N.E.
19. In Fig. 126 calculate the coordinates of the points B and C,
the reduced bearings of BC and CA, and the area of ABC, if the bearing
of AB is 60 S.E.
N
Fig. 126.
20. In finding the length of a line CB, a line CA was set out by means
of the optical square at right angles to CB and the distance CA was
chained and found to be 114 chains. The angle CAB was then ob
served by a box sextant and found to be 71 54'. Calculate the length
ofCB.
21. The coordinates of two stations A and B are
A. Latitude N 400 links ; Departure W 700 links
B. Latitude S 160 links; Departure W 1500 links
Find the whole circlebearing of AB.
PLANE TRIGONOMETRY
251
22. You are 220 ft. horizontally away from the headgear of s
mine. From a point on the same level as its base you finf that the
headgear subtends a vertical angle of !8 3 o'. Find the height
23 A ball fitting down to the taper sides was used to test the
correctness of the cupshaped check shown in Fig. I2 6a. The test ra
made by measurement of the distance AB. Calculate' this
 to 10000
Toper of Sides 5 IP 8
A measured on diarn.
Hole to be I
drilled
Fig. I26a. Test for Gauge.
95"dia "
Fig. 1266. Block for Jig.
24. Determine the diameter of the largest drill that could be used
for the hole in the jig block shown in Fig. 1266, when you are told that
the drilled hole, which is made first to clear away part of the metal,
must cut the taper hole at the level AA.
Angles of any magnitude. Up to this point our work has
been confined to angles of 90 and under, whose trigonometrical
ratios can easily be found from tables or by the use of the slide rule.
Angles greater than 90 must be reduced to those less than 90 by com
bination with 180 or 360, i. e., they must be reduced to the equivalent
acute angle made with some standard line, which in all this work will be
taken as the N and S line.
N
Fig. 127.
Fig. 128.
Fig. 129.
If the N. and S. line and the E. and W. line be drawn, they
divide the space into four " quadrants," and the position of an angle
can always be stated by reference to the quadrant in which it lies.
Angles are measured in a righthand direction from the N. and S.
line, and the quadrants are numbered as shown in Fig. 127. A
minus sign before an angle indicates a movement from the north
in a lefthand direction.
252
MATHEMATICS FOR ENGINEERS
e. g., referring to Figs. 128 and 129
154 is in the 2nd quadrant ; and its equivalent acute angle
= 180 154 = 26
258 is in the 3rd quadrant ; and its equivalent acute angle
= 258i8o = 78
76 is in the 4th quadrant ; and its equivalent acute angle
= 76
472 is in the 3rd quadrant ; and its equivalent acute angle
= 68
To sum up, it will be seen that the equivalent acute angle
(written e.a. angle) is always the angle made with the N. and S. line ;
*'. e., it is obtained by compounding with 180 or 360.
It is now necessary to find the algebraic signs to be prefixed to
the trigonometric ratios of any angle. Thus although the sine
of 472 is numerically equal to the sine of +68, since 68 is the
e.a. angle for 472 (see Fig. 129), it would not necessarily be
correct to state that sin 472 = sin 68, because we have not
yet examined for the algebraic sign. As a matter of fact,
sin 472 = sin 68.
Suppose that a line of unit length rotates in a righthand direc
tion, starting from the north, thus sweeping out the various angles.
Its " sense " will always be considered positive, whilst the usual
convention will fix the signs for horizontal and vertical distances.
[Note. In all that follows, be sure to measure every angle from
the north point : thus in Fig. 130, the
angle (180 A) is the angle aod, and the
angle (360 A) is the angle aoh measured
in a righthand direction.]
Let Laoc (Fig. 130) represent the
magnitude of the e.a. angle in all the four
quadrants : *. e., L aoc = L eod = L eof =
t_ aoh = A, say.
In the ist quadrant
+ ac + ac
sin A =
cos A =
tan A =
+ oc
+ oa
+ oc
f ac
~4 oa
i
oa
= ac
= + oa
ac
oa
In the 2nd quadrant
sin (180  A) = ^ =
ed : but ed = ac
PLANE TRIGONOMETRY
253
so that sm (i8oA) = sin A. Hence the reason for compounding
with 180 to find the e.a. angle is seen.
Again cos (180 A) = ~ oe
= oe = oa
 oe indicating that oe is a negative length, because measured
downwards
tan (i8oA) = + ed e ac
oe
In the 3rd quadrant
sin (i8o+A) = 
oe
oa
= cf = ac
cos (i8o+A) =
oe
= oe = oa
tan (i8o+A) = e J =
oe
In the 4th quadrant
/ r A \ Qh
sin (360 A) =
cos (360 A) = 
oe
oa
/ r A \
tan (360 A) =
oa
oa
ah
oa
oa
i. e., summarising for the equivalent acute angles in all four
quadrants, the algebraic signs vary as follows
Quadrant.
ISt
2nd
3rd.
4 th.
sine and cosec .
cos and sec .
tan and cot .
+
+
+
+
+
+
i
OS
i
joLL
I i
+ +
1
i
+
on
sin
'+ 
 +
@
SINE AND COSINEAND TANGENTAND
COSECANT. SECANT. COTANGENT.
Fig. 131. Variation in Sign of Ratios.
This variation in sign may be better or more plainly denoted
by the diagrams (a), (b), (c) and (<*), Fig. 131. Fig. (a) 131 may
need an additional word of explanation. In each quadrant is written
254 MATHEMATICS FOR ENGINEERS
the word to indicate which ratio or ratios is or are positive in that
quadrant. Thus in the 3rd quadrant, the tangent alone is positive,
and in the 4th quadrant the cosine alone. Fig. (b), (c) and (d) 131
are merely a representation of the table just given.
Hence, to find the trigonometric ratio of an angle of any magni
tude : find first its e.a. angle and the quadrant in which the angle
occurs, and then apply the sign of the quadrant for the ratio re
quired. (Numerically, the ratio of any angle is that of its e.a.
angle.) In all cases it will be found that a diagram simplifies matters.
Example 10. Find the value of sin 172.
sin 172 = sin (180172) = sin 8, for 8 is the e.a. angle
= +I392
since 172 is in the 2nd quadrant, and the sine there is +.
Example n. Find cos 994.
994 =[(2 X 360) + 2 74]
2 x 360 brings us back to the starting line, and so we deal only with
the 274. Now 274 is in the 4th quadrant, and thus its cos is + ;
also the e.a. angle = 360 274 86.
cos 994 = + cos 86 = +0698.
Example 12. Find tan 327.
The angle 327 is in the ist quadrant, and hence its tan is + ;
also the e.a. angle = 33.
tan 327 = + tan 33 = +6494.
Example 13. Find the sin, cos and tan of 115. What connection
is there between them ?
sine is +1
;os is J
:an is }
The angle is in the 2nd quadrant, hence sine
cos
tan is .
also the e.a. angle 180 115 = 65
sin 115 = +sin 65 = + "9063
cos 115 cos 65 = 4226
tan 115 = tan 65 = 21445
sin 115
Now
= tan 115.
PLANE TRIGONOMETRY 255
This most important relation always holds, viz. that
tanA = sin  A
cos A
The "reduced bearing/' in surveying, may be regarded as
identical with the " equivalent acute angle " here used.
In the general solution of triangles only angles up to 180 occur
e we are concerned mainly with the 1st and 2nd quadrants.
Exercises 29. On the Trigonometric Ratios of Angles of any Magnitude.
Find from the tables, the values of the sin, cos and tan of the
following angles (Exs. i to 5).
1. 116; 322; 218. 2. 82; 398; 1562.
3.1992; 34i5'; 9842 3 '. 4. 4 ; u62; 85; 116 radians.
5. 1194; 245 radians; 787 n'.
6. Find values of cot 126; COSCCTT; sec (52). [Note. The
angle TT radians is that subtended at the centre by the halfcircumference
and is thus 180.]
7. Find a value of A between o and 180 if
cos A  b * + cZ ~ a * and b = 98'
\*\Jj X*. _ _ _
zbc c = 64*
a = 1445*
2 V2 __ 7/2 zoTT
8. The equation cot 6 =  ^y  relates to the design of
water turbines. If V = 534, u = 10, H = 100, e = 322, find 6 (between
o and i So ).
9. As for the preceding question, but taking V = , n = V8o,
g = 322 and H = 80.
10. If a = angle of the crank of a steam engine from the dead
centre, m = ratio of connecting rod length to length of crank and
/ = 833 : find values of a to satisfy the equation
cos a = m Vm 2 +i 2m/ when m = 4.
Solution of Triangles. The " solution " of a triangle consists
in the determination of the magnitudes of the six parts, viz. the
three sides and the three angles. In many cases sufficiently accurate
results can be obtained by careful drawing to scale, but for great
precision the values of the parts of the triangle must be calculated.
In such calculation extremely exact tables, giving the relations
between the sides and angles, are employed, and the results obtained
are superior to those given by even skilled draughtsmanship.
Again, it sometimes happens that the triangle is difficult to construct:
thus if in Fig. 136 the base AC was very small compared with
the sides AB and BC, the intersection of AB and CB would not easily
256 MATHEMATICS FOR ENGINEERS
be detei mined, and, therefore, the lengths of the sides as measured
would only be approximate. The angle at B would under these
circumstances be termed " badly conditioned."
There are a number of rules developed for the general solution
of triangles, but of these the following will be found to be of the
greatest service, while even this list may be reduced to the first
two rules.
Adopting the usual notation for the triangle, viz. A, B and C
for the angles, and a, b and c for the sides opposite these angles
respectively, the rules for the solution of all triangles are
ft h C
(1) 7 r = =r =  ^r, usually referred to as the sine rule.
' sin A sin B sin C
(2) a 2 = b 2 + c 2 2 be cos A, usually referred to as the cosine rule.
/ \ / \ A / (s b)(s c)
(3) () sin =</* g
/IA a)
(b) 
A /(s  b)(s  c)
(c) tan  = A / v  j^ ^ 
2v $(s a)
B C fb c\ A A
(4) tan   = ( j.  } cot 
2 \b + c) 2
These may be employed under the following conditions
I. Given two sides and included angle: use either rule (2) to
find the third side and then either rule (i) or rule (3) to find
another angle ; or use rule (4) to find the remaining angles together
with rule (2) for the third side.
e. g., suppose b, c and A are given.
Then from rule (2) the value of a can be found,
, sin B sin A r , , , .,
also  r = [from rule (i)l
b a L
B is found
and C = i8o (AfB), since A+B+C = 180;
or alternatively
BC fb c\ . A
tan   = ( j  ) cot 
2 \b + cj 2
T> _ />
.*. the angle  is found, and hence also (B C).
But (B+C), i. e., 180 A is known,
and therefore B and C are found by solving the simultaneous
equations. Also a can be found from rule (2).
PLANE TRIGONOMETRY 257
II. Given two angles and a side, say a, A and B
Then C = i8o(A+B)
From rule (i)
c a b a
and
sin C sin A' sin B sin A
and therefore all the sides are found.
III. Given two sides and an angle not included by them, say
b, c and B
, , , sinC sinB
From rule (i) = j
C
/. C is found, and also A. {For A = 180 (B+C)}
a b . , ,
and since  T =  a is found,
sin A sm B
IV. Given the three sides : it is more convenient in this case to
use rule (3) to find one of the angles ; because logarithms can be
applied.
From Rule (3) c
tan 
tan 
s(s a)
Then use Rule (i) to find B.
Otherwise a? = & 2 + c 2 2bc cos A
.*. cos A = L i" e > A is found
and thence by the sine rule B may be found.
Thus if rules (i) and (2) are remembered, any triangle may be
solved.
Proof of the "Sine" Rule.
Consider Figs. 132 and 133.
In both figures j = sin B
t p = c sin B
also in Fig. 132 = & sin C
and in Fig. 133 p = bsm (i8oC) =
Hence c sin B = & sin C
c &
" sinC ~ sin B
s
258 MATHEMATICS FOR ENGINEERS
Similarly it could be proved that
b c a
B
D
a
Fig. 132.
c B a c
Fig. 133
Or, the sides of a triangle are proportional to the sines of the opposite
angles.
Proof of the Cosine Rule.
In Figs. 135 and 136 let BD be perpendicular to AC. In Fig. 135
in the triangle ADB
= p 2 +b*+n*2bn ....... (i)
and in the triangle BDC
/>+n 2 = a 2 .............. (2)
B
6
Fig. 134 Fig 135.
Hence, by substitution from (2) into (i)
Fig. 136.
Again in Fig. 136, in the triangle ADB
C 2 =^) 2
= 2
and in the triangle BDC
(3)
(4)
(5)
PLANE TRIGONOMETRY 259
Hence, by substitution from (5) into (4)
c 2 = a 2 +b 2 +2bn (6)
Now in Fig. 135  = cos C or = a ccs C
a
so that, writing a cos C in place of n in (3)
Also in Fig. 136
Tii
 = cos L BCD = cos (180 C) = cos C or n = a cos C.
Substituting this value for n in (6)
c 2 = a z +b z 2ab cosC.
We have thus proved that the rule holds for the case in which
C is an acute angle, and also for the case in which C is obtuse.
When C is a right angle, as in Fig. 134, its cosine is zero and accord
ingly it is correct to write
_ = a 2 \b 2 2ab cosC.
Hence the rule is perfectly general.
The two other forms of the cosine rule can be written down by
writing the letters one on in the sequence a, b, c, a.
i, e , t a 2 = b z +c z 2bc cos A
and 6 2 = c 2 + 2 2ca cos B.
By transposition
c sC=  : S :f
cosB =
2ca
J,2_
cos A =
2bc
the forms in which the rule must be used if the three sides are
given and the angles are required.
In every case of a solution of a triangle the figure should be drawn
to scale, for this serves as the best check on the results obtained by
calculation.
The following examples should be carefully studied
Examples on the use of the Sine Rule.
Example 14. Solve the A ABC completely when c  1916 ft..
6 = 1748 ft., and C = 59. [This triangle is drawn to scale in Fig. 137.]
260
MATHEMATICS FOR ENGINEERS
sin B sin C
To find B
u c
b sinC 1748 x sin 59
and hence sm B =  =  J 1016
Taking logs throughout
log sin B log 1748 + log sin 59 log 1916
= 32425 + 19331  32823
= ^'8933 = log sin 51 28'
B = 5 i28'
Then
A = 180  (59 + 51 280
Fig. 137
Fig. 138.
To find
sin A sin B
b sin A
a = . =
sinB
In the log form
log a = log 1748 + log sin 69 32' log sin 51 28'
= 32425 + 19717 ~ 18933
= 3'3 2 9
a = 2093*
Example 15. Solve the A ABC completely when a = 126*.
b = 178", A = 40. (This is similar to the last Example up to a certain
point.)
To draw this to scale (see Fig. 138). Make the angle 40 with a
horizontal line and along AC mark off a length to represent 178*;
this is the side b. With centre C and radius = 126* (to scale) strike
an arc to cut the horizontal ; and two points of section being found,
call them B and B'. Both the A ABC and the A AB'C satisfy the
given conditions, because AC = b = 178, CB = CB' = a = 126 and
A = 40, so that in this case there are too solutions. This case is known
as the " ambiguous " case in the solution of triangles.
Since CB = CB', L CBB' = L CB'B
L CB'A = 1 80 L CBA
or B'=i8oB
PLANE TRIGONOMETRY 261
and the two values of the angle B, which are indicated on the figure,
are supplementary, i. e., together they add to 180. AB and AB'
are the two different lengths for c for the different cases, while ACB
and ACB' give the two values for the angle at C.
To solve by calculation. Two sides and one opposite angle are
given, and therefore the sine rule is to be used. Taking the same
diagram
To find B
sin B = &sinA = 1 7'8 sin 40
a 126
In the log form
log sin B = log 178 + log sin 40 log sin 126
= 12504 + i 8081 11004
= 19581 = log sin 65 13'
B = 6 5 i3'
The value of B', which is alternative to B must be 180 B
= 114 47'. The mode of calculation would be unchanged, for
sin 114 47' = sin 65 13'.
To find C
In the first case C = i8o(4O+ 65 130
= 74 47'
In the second case C = i8o(4O+ 114 47')
= 25 13'
To find c. This is the base, which is AB' or AB. Either the sine
or the cosine rule can be here used, but the sine rule is more adapted
for logarithmic computation.
_ a sin C
sin A
In the first case
log c = log 126 +_log sin 74 47'  log sin 40
= 11004 + 19845  i 8081
= 12768
c = 1891*
In the second case
log c  log 126 + log sin 25 13'  log sin 40
= 11004 + 16295 i 8081
= 9218
c = 8352
Grouping the results
B = 6.si3' or 114 47'. C=7447' or 2SiV. c=i8gi" or 8352"
all respectively.
The sine scale on the slide rule could be used with advantage in
this example. To multiply or divide by sines of angles, multiply
262 MATHEMATICS FOR ENGINEERS
or divide by the angles, as marked on the scale, in the ordinary
way. E. g.
_ 126 x sin 74 4 7*
sin 40
Set the cursor over 126 on the A scale, move the sine scale until
40 is level with the cursor; then place the cursor over 74 47' on the
S scale. The value of c is read off on the A scale, and = 189.
A little confusion may arise regarding the graduations on the
S and T scales. The markings usually shown are not for decimals
of a degree, but for minutes. As regards the S scale : up to 10,
a line is shown at every 5', *. e., there are 12 divisions for each
degree. From 10 to 20 every 10' is shown, from 20 to 40 every
30', from 40 to 70 each degree, and thence 70, 72, 74, 76, 78,
Fig. 139. Solutions of Triangles.
80, 85 and 90. On the T scale, up to 20, markings are at each
5' and then at every 10'.
Whenever two sides and an opposite angle are given, we must
consider the possibility of the two solutions.
Ths drawing to scale is an excellent test, for the arc B'B in
Fig. 138 must either cut or touch the base if the triangle is to be
possible.
The various cases that arise are illustrated in Fig. 139 : in
which the sides a and b, and the angle A are given. Drawing a
horizontal line of unlimited length to serve as a base, the angle A
can be set out and the point C fixed, since the length of AC is given.
Then an arc of radius equal to b is described from the centre C.
If b is very small, the arc does not cut the base and case (i) arises ;
there being no triangle to satisfy the conditions. If the radius of
the circle, i. e., the length of the side b, is increased, we arrive at
case (2), in which the arc just touches the base and so gives one
PLANE TRIGONOMETRY
263
triangle only, viz. the rightangled triangle ACB 2 . By further
increasing the length of b cases (3), (4) and (5) are found, in which
there are two, one, and one, solutions respectively.
It will thus be seen that there may be two solutions if two
sides of a triangle and an angle opposite the shorter of these is
given. In all cases, however, the triangle should be drawn to scale
before any trigonometrical rules are applied.
Example 16. A mill chimney stands on the even slope of a hill,
which has a gradient of 4 (Fig. 140). Two points are chosen on the
same side of the hill and in
the same vertical plane as that
including the chimney. These
points are 75 ft. apart measured //3/i
up the slope, and, viewed from
the points, the chimney subtends
angles of 48 and 59 from the
horizontal. Find the height of
the chimney above the ground
on which it stands.
It should be noted that the
angles of elevation are measured
from the horizontal, since the
scale of the theodolite vertical
circle reads zero when the tele
scope is horizontal.
Hence L ABC = 4 84 = 44
and L ACD 55
Thus /.ACB=i8o55 = i25 , ABAC=n
/.ADC =94, A CAD = 31
Here we have two triangles, viz. ACB and ACD, one containing
the known length and one containing the unknown length ; and these
must be connected up through a side common to b >th. viz. AU
Let the required height AD = h
Then, in the A ACB
sn 44 sum
In the A ACD
_ 75 sm 44_
sin 11
C AC
55
sin 94 sin 86
AC x sin 5j
since sin 86 = sin 94
264
MATHEMATICS FOR ENGINEERS
Substituting for AC its value
= 75 x sin 44 x sin 55^
sin 11 x sin 86
= 225 ft. (from the slide rule).
Example 17. The elevation of the top P of a mountain (see Fig. 141)
at a point A on the ground is 32. The surveying instrument is directed
to another station B, also on the ground, and 4600 ft. distant from A,
the angle PAB being found to be 48 ; also L PBA is 77. Find the
height of the mountain.
The sloping triangle PAB is shown laid flat on the ground in Fig. 142.
From this ground plan
PA 4600
sin 77 ~ sin 55
pA = 4600 x sin 77
sin 55
P<z/?spec//W vieuf
"^ t i
or survey lints.
A
In the rightangled A PAQ
and PQ = AP x sin 32
Substituting for AP
_ 4600 x sin 77 x sin 32
sin 55
Height of mountain = 2900 ft.
Example 18. It is required to lay out a circular arc to connect the
two straight roads AB and CD (Fig. 143) : the radius r of the arc is
known, but the meeting point E of AB and CD is inaccessible.
Select two convenient stations F and G, and by directing a theodolite
first along Fe and then along FG the angle EFG is measured. Similarly
measure
265
Let the sum of L EFG and L EFG
Then /_AED = i8o
and L AEO =
L. EOS = a
za
= goa
Now
ES
Qg = tan a, since L ESO is a right angle
ES = OS tana = r tana ( x )
and also ET = r tan a, since ET = ES . . . . (i)
To find FS and GT, EF and EG must first be found.
In the A EFG
_EF
sin
FG
or EF = FG
sin EGF
sin FEG ~ sin FEG
EF is known (2)
Also, in the same way
EG = FG i in _ EFG .
sirTFEG
EG is known (3)
Finally, FS is found from (i) and (2) since FS = ES EF, and
GT is also found from (i) and (3) since GT = EG  ET. Thus the
points F and G having been taken at random, S and T can now be
plotted therefrom, which show the startingpoints of the curved road.
Examples on the use of Cosine Rule.
Example 19. In the triangle ABC (Fig. 144) find L C when a = 445',
b = 785', and c = 1194'.
C= 1194
Fig. 144.
The longest side is always opposite the largest angle ; and therefore
C is the largest angle.
266
MATHEMATICS FOR ENGINEERS
C 
V^
Now
I fe 2 ~ c 2 = (445) 2 +(785) 2 (n94) 2
2ab 2 X 445 X 785
198 + 615 1425
698
612
^ = "o'y^
OQ'o
= cos 28 43'
= cos (180 28 43') = cos 151 17'
/. C = 151 i7 / .
It will be seen that a negative value for the cosine implies that
the angle is obtuse.
To avoid remembering too many rules the reader is advised
to work entirely with the sine or cosine rules : this example, how
ever, is worked out, in addition to the above, by another rule, to
demonstrate its usefulness and ease of application.
a = 445, 6 = 785, c = 1 1 94
S = 1212
C _ /(sb)(s~a) _ / 4 2 7 x 767
^ 2 ~ V S (S  C) ~ V 1212 X 18
.. log tan  = J
^{(6304 +8848)
427 + log 767)  (log 1212 + log 18)}
(10835 + 12553)}
= 5882
= log tan 75 32'
and C = 151 4'
=75 32'
i. e., an error of 13' was made when using the slide rule.
[Note that if this rule is used and the angle is required correct to
the nearest minute, we must work throughout correct to a halfminute
since the rule gives as the direct result the value of a halfangle.]
Example 20. If in the triangle ABC
find the side b (Fig. 145).
Using the cosine rule
b 2 = a 2 +c 2 2ac cos B
= (5'93) 2 +(294) 2
(2 x 5 93 x 2 94 x cos65)
= 351+862
(2X593X294X4226)
= 539'
= 593". c = 294", B = 65,
a. = 5 93"
Fig. MS
PLANE TRIGONOMETRY
267
Example 21. Two forces, of 472 Ibs. and 984 Ibs. respectively,
making an angle of 63 with one another, act on a small body at A'.
Find the magnitude of their resultant, or single equivalent force.
If AB and AD in Fig. 146 represent the given forces, AC represents
their resultant, as shown in Mechanics.
Then
XI 7
Fig. 146.
CD = 472, AD = 984, tADC = i8o6 3 <
(AC) 2 = (AD) 2 +(DC) 2 (2xADxDCxcos 117)
= (984)*+ (47'2) a  (2 x 984 x 47'2 x 4540)
= 9670+2230+4220
= 16120
AC = 127 Ibs. and this is the resultant.
Area of a Triangle. The following rule gives the area when
two sides and the included angle are given ; it is simply an extension
of the \ base X height rule, for
AD = AB sin B or AC sin C (Fig. 147)
= c sin B or b sin C
.*. Area = \ X base X height
= XXcsinB or xaX&sinC
= \ac sin B or \ ab sin C
or, generally, area of triangle = J product of two sides x sine of
included angle.
AA
Fig. 147
Fig. 148.
This gives a rule for the area of a parallelogram
Area of ABCD
= 2 {area AOB + area AOD} (Fig. 148)
= 2{ AO.OB sin L AOB + \ AO.OD sin L AOD}
268 MATHEMATICS FOR ENGINEERS
= sin L AOB {AO.OB + AO.OD} since sin L AOB
= sin (180 AOB)
= sin L AOD
= sin L AOB x AO {OB + OD}
= AO.BD sin L AOB
= \ AC.BD sin L AOB
= \ product of diagonals X sine of angle included between
them.
Example 22. Find the area of AABC in which a = 593*, c = 294"
and B = 65.
Area = ac sin B = \ x 593 x 294 x sin 65
= i x 593 x 294 x 9063
= 7*91 sq. ins.
This result should agree with that found by the " s" rule given
in Chapter III ; it being possible to apply this rule since the three
sides are known and are 593, 539 and 294 respectively (compare
Example 20).
Thus 5= 593 + 539 + 294 = ^
and area = ^713 x 120 x 174 x 419 = 791 sq. ins.
Proof of the "s" Rule for the Area of a Triangle.
It has been demonstrated in the previous paragraph that
Area of triangle = ab sin C.
Now for any angle it is true that (sine) 2 + (cosine) 2 = i : hence
sin 2 C + cos 2 C = i, or sin C = i cos 2 C
a 2 +6 2 c 2
Also cos C =  j
2ab
2r (a 2
then cos 2 C:
zr
and i cos 2 C = J '
[Factorising difference of two squares]
_ (2ab a 2 6 2 +c 2 )(2a6+a 2 +ft 2 c 2 )
PLANE TRIGONOMETRY 269
[Factorising difference of two squares]
= (c<*+b)(c+ab)(a+bc)(a+b+c)
= 2(s a) x 2(s b) x 2(sc) x 2s
i. e., sin C = ~y' S (s a)(sb)(s c)
:. area of triangle ABC = $ab x , X Vs(s a)(s b)(s c)
= Vs(sa)(sb)(sc)
In all of these worked examples the results have been given to
as great a degree of accuracy as fourfigure log tables or the slide
rule allow.
When extremely careful observations have been made it is
advisable to employ five or even sevenfigure log tables in any
necessary calculations ; but it should be remembered that the results
must not be given to a greater degree of accuracy than the observa
tions or measurements warrant. Thus it would be useless to
express a length " correct " to eight figures when the least possible
error in measurement was %.
The rules used in such cases are those stated in this chapter,
except that the cosine rule is put into a form more adapted for
logarithmic computation by means of the following artifice
a z = 6 2 +c 2 2bc cos A
In place of this rule we may write
a = (b+c} cose ......... (i)
provided that is found from
zVbc A , v
sm * = T+^ COS 2 ......... (
Both (i) and (2) can be solved by the aid of logs ; and the angle
6 thus introduced is known as a subsidiary angle.
Let us illustrate this by taking the figures of Example 20.
Given a = 593, c 294, B = 65.
To find b
From the above
b = (c+a) cos 6
zVca B
and sin 6 =  cos 
j. e., sin
2^294 X 5 '93 cos ,2 1
=  . 5 s 3 2 i
270
MATHEMATICS FOR ENGINEERS
In the log form
log sin 6 = log 2 + {log 294 + log 593} + log cos 32 \ log 887
= 18998 = log sin 52 33''
/. = 52 33'
Then b = 887 x cos 52 33'
In the log form
log b = log 887 + log cos 52 33' = 7318
Exercises 30. On the Solution of Triangles.
In Exs. i to 14 solve the triangle ABC completely, being given
that
1. a=3*,6 = 52", B= 7 8i. 2. a=795",C = 5 i 32',B = 4 7 3 6'.
3. C = 26 50', 6 = 886", c = 568". 4. 6 = 597", C = 6 4 18', A = 75.
5. c = 92, a = 1031, = 46. 6. 6 = 6ift,,c = 93ft.,A = 73i6'.
7. a =1244, 6 = 937, c = 993. 8. a = 137*. 6= 105*, 0=130.
9. a = 427", A = 29, b = 586*. 10. c = 688o, B = 30, b = 5141.
\D
crosssection o/'u/'V'e
Fig. 149. Solution of Triangles.
11. A = 50 50', 6 = 9224, c = 10038.
12, B = 353o', 6 = 386, c= 4 357. 13. = 218,6= 157,0 = 47 32'.
14. c = 327, 6 = 394, B = 55 30'. Find also the area.
15. The area of a triangle is 120 sq. ft. and the angles are 75, 60
and 45. Find the longest side.
16. The connecting rod of an engine is 8 ft. in length and the crank
i '6*. Find the inclination of the connecting rod to the line of stroke
when the crank has moved 52 from its inner dead centre position.
17. The sides of a " triangle of forces " represent the forces 37 tons,
2275 tons an d 3' 02 5 tons respectively. Find the angles of this triangle.
18. Forces of 216 and 197 Ibs., making an angle of 126 with one
another act at a point. Find the magnitude of their resultant and its
inclination to the larger force.
19. In setting out a railway curve to connect the lines AO and OD
((a) Fig. 149), a line CB was measured and found to be 1*4 74 chains.
PLANE TRIGONOMETRY 2?I
points are E and F find the lengths of BE and CF
20. The diagram (6) Fig. 149, is necessary for the calculation of
lag by the 3voltmeter method. If <p is the angle of lag, find Ss value
for the case illustrated. {V 8 = 107, V x = 90, V, = 48 }
21. The jib of a crane is inclined at 57 to the horizontal the
kes '
23. The tangents to a curve meet at 120. On the bisector of this
angle is a point 100 ft. distant from the point of meeting of the tangents
and through which the curve must pass. Find the radius of the'
required curve and also the tangent distances.
24. It is required to find the height of a house on the opposite
bank of a river. The elevation of the top of the house is read at a
certain point as 17; approaching 86 ft. nearer to the bank, towards
the house, the elevation is found to be 31. Find the height of the
house.
25. A theodolite is set up at two stations A and B at the water's
edge of a lake which is 1240 ft. above sealevel. A staff on a hill at C
is sighted from each station. From A the elevation of C is 15 14' and
the horizontal angles CAB and CBA are 59 10' and 71 48' respectively.
If AB = 820 yds., find the height of C above sealevel.
26. From a station C on a hill, two stations A and B, on opposite
sides of the hill are observed. The horizontal projection of L ACB is
43 23', the horizontal projection of CA is 3633 links and of CB is
4275 links. The angle of elevation of C at A is 44 37' and at B is 33 24'.
Determine the horizontal distance between A and B and the difference
of level between them.
27. It is required to set out a curve of mile radius between two
straight portions of a railway, AB and DC, which intersect in an in
accessible point E. Rods are set up at points B and C on the two
straight portions and the angles ABC and BCD are measured and
found to be 110 20' and 120 30' respectively.
If BC = 830 links, determine the distances of the tangent points
G and H from B and C respectively.
28. In a theodolite survey to find the positions of two visible but
inaccessible points B and C, the following measurements were made
AD = 51775 links, L BAC = 70 44' 10", L BAD = 108 9', L ADB =
36 i8'3o", and /_ADC= ioiiS'3o". Find the lengths of AB, DC,
AC and BC in order.
29. When setting out the centre line for a tunnel between the two
ends A and B, an observatory station C is chosen on the top of a hill
from which both A and B are visible, but it is not on the centre line of
the tunnel. Let D be a point on a vertical through C. The horizontal
projection of L ACB = 45 58', the vertical angle ACD = 4945' and
the vertical angle BCD = 5 7 42'. The horizontal projection of CA
is 750 yards, and of CB is 800 yards. Find the horizontal distance
between A and B and the difference of level.
30. A light railway is to be carried round the shoulder of a hill, and
272
MATHEMATICS FOR ENGINEERS
its centre line is to be tangential to each of the three lines AB, BC and
CD as follows
Line.
Bearing.
Length.
AB
E. 30 N.
BC
E
600 feet
CD
S

Calculate the radius of the curve and the lengths required for setting
out the tangent points. [Note. E. 30 N. means 30 north of east.]
31. In taking soundings from a boat the position is fixed by
observations taken to three stations A, B and C on the shore. The
lines AB and BC have been measured by the following traverse : A
to B, 542 ft., bearing 70 14'; B to C, 714 ft., bearing iio33'. From
the boat in a certain position P, the angles APB and BPC were read
as 32 16' and 44 21' respectively. Calculate the distances AP, BP
and CP.
32. The speed of the blades of a turbine is 600 ft. per sec., the
velocity of the steam at entrance to the wheel is 1780 ft. per sec., and
the nozzle is inclined at 20 to the blades. Find the relative velocity
of the steam at discharge, and the inclination of the direction of this
velocity to the line of motion of the blades.
33. Find the diameter of the wire, whose section is shown in (c)
Fig. 149, in terms of the pitch p of the Vthreaded screw. This wire
is used as a gauge to test the accuracy of the form of the thread.
Further Mensuration Examples.
34. A circular arch has a rise of 20 ft. and a span of 80 ft. Find
the angle at the centre of the circle which is subtended by this arc,
and also the length of the curved portion of the arch.
35. A wooden core, having as section an equilateral triangle, is
placed in the tubes (internal diameter f *) of a surface condenser. Find
the ratio of the tube surface to the watercarrying section.
36. A roof is in the form of the surface of a segment of a sphere
of 6 ft. radius. The tangents at the eaves make 48 with the hori
zontal. Find the area of the roof surface, and the weight of sheet
lead required to cover it at 7 Ibs. per sq. ft.
37. Find the diagonals of a rhombus in which one side is 65* and
one angle is 70.
38. A quadrilateral has two adjacent sides equal and containing
a right angle. The other pair of sides are equal and contain 60. The
area is i sq. ft. Find the lengths of the sides.
39. A quadrilateral has two adjacent angles each 120. The side
between them is 24 ft., and the perpendiculars on this side from the
other angular points are 7 ft. and 10 ft. respectively. Find the area
of the quadrilateral.
40. A trapezoid has its parallel sides 82" and 38* and two of its
angles each 60. Find its area and the area of the triangle obtained
by producing the non parallel sides.
41. A quadrilateral with two opposite angles right angles and one
of the remaining angles 60 is described about a circle of 2" radius.
Find its area.
1 tan A tan B
tan A tan B
PLANE TRIGONOMETRY 273
The Addition Formulae. It is sometimes necessary, more
particularly in electrical work, to express the ratio of a compound
angle in terms of the ratios of the simpler angles, or vice versa;
e. g., it might be easier to state tan (A+B) in terms of tan A and
tan B, and then evaluate, than to evaluate directly. The following
rules must be committed to memory for this purpose
sin (A + B) = sin A cos B + cos A sin B
sin (A B) = sin A cos B cos A sin B
cos (A + B) = cos A cos B sin A sin B
cos (A B) = cos A cos B + sin A sin B
tan (A + B)
tan (A B) .
1 + tan A tan B
Considering sin (A+B) one might be tempted at a first glance
to apply the ordinary rules of brackets, and write the expansion
as sin A+sin B. That this is not correct may be readily seen
by referring to any angles.
e. g., suppose A = 46, and B = 15, then (A+B) = 61
*. e., sin (A+B) = sin 61 = 8746
whereas sin A+sin B = sin 46+sin 15
= 7193 +2588
= 9781
and 9781 does not equal 8746.
It will be observed, however, that the above rule holds, at any
rate for these particular values of A and B.
Thus
sin 46 cos 15 + cos 46 sin 15 = (7193 X 9659) + ( >6 947 X 2588)
= 8745
= sin 61
= sin (46+i5).
A more general proof is necessary to establish the truth of these
rules for all angles ; and the proofs are here given for the simplest
cases only.
To prove that sin (A+B) = sin A cosB + cos A sinB.
Taking the simplest case, when A and B are both acute
In Fig. 150 let L. PQR = A, and L RQM =
L PQM = (A + B) ; also let QR be perpendicular to PM.
Then APQM = APQR + AQRM
.'. iPQ QM sin (A + B) = iPQ . QR . sin A + *QR . QM . sin B.
T
274 MATHEMATICS FOR ENGINEERS
Dividing through by PQ . QM
sin (A+B) = ^ sin A + ^ sin B
= cos B sin A + cos A sin B
or sin A cos B f cos A sin B.
Fig. 150. Fig. 151.
To prove that cos (A + B) cos A cos B sin A sin B.
In Fig. 151 let L MOQ = A, and L QOP = B
_ PQO = right angle.
Drop QN perpendicular to RP, RP being perpendicular to OM.
Then L OQN = A, L NQP = 90 A, and therefore L QPN = A.
TVT /A 1 TN TD^T, OR OM MR
Now cos (A+B) = cos L ROP
OP
OP
\JXi IX \J r . XT .~ uTTn
= T&r [ smce N Q = MR 1
OM_NQ
OP OP
OM OQ_NQ QP
= OQ'OP QP'OP
= cos A cos B sin A sin B.
To prove that tan
ton ^ + ton B

i tan A tanB
Assume the rules for sin (A+B) and cos (A+B).
Then tan(A+B)= sin i A +g
1 cos (A+B)
[Dividing both numerator and
denominator by cos A cos B.]
sin A cos B + cos A sin B
cos A cos B sin A sin B
sin A , sinB
cos A cos B
sin A sin B
_
cos A ' cos B
_ tan A + tan B
~ i tan A tanB
PLANE TRIGONOMETRY 275
23 .Verify the rules for sin (A  B), cos (A + B) and
tan (A  B) for the case when A = 164 and B = 29.
sin (A  B) = sin (164  29) = sin 135
= sin 45
= 7<>7
Also sin A cos B  cos A sin B = sin 164 cos 29  cos 164 sin 29
[cos 164 =  cos 16] = sin 16 cos 29 + cos 16 sin 29
= (2756 x 8746) + (9613 x 4848)
= 241 + 465
= 706.
For brevity we shall denote the side containing (A + B) by L.H.S.
(lefthand side) ; the other by R.H.S. (righthand side).
/. L.H.S. = R.H.S.
For cos (A + B)
L.H.S. = cos (A + B) = cos (164 + 29) = cos 193 =  cos 13 =  9744
R.H.S. = cos A cos B sin A sin B = cos 164 cos 29 sin 164 sin 29
= cos 16 cos 29 sin 16 sin 29
= ( 9613 x 8746)  (2756 x 4848)
= 841133
= 974
.*. L.H.S. = R.H.S.
For tan (AB)
L.H.S. = tan (164 29) = tan 135 = tan 45 = i
R TT s tan 164 tan 29 _  tan 16 tan 29
i + tan 164 tan 29 ~ i  tan 16 tan 29
2867 5543
" i (2867 x 5543)
= ^4L = 841 = t
1159 841
/. L.H.S. = R.H.S.
Example 24. Find the value of cos (A + B) when sin A = 5,
cos B = 23. (Tables are not to be used.)
Before proceeding with this example, a little preliminary
investigation is necessary.
In the rightangled triangle ABC (Fig. 152)
b* , fl 2
Cos 2 A + sin 2 A = 1, since cos A =  and sin A = 
276
MATHEMATICS FOR ENGINEERS
This is a most important relation between these ratios ; and it
holds for every value given to the angle A.
Two other rules obtained by similar methods are
sec 2 A = 1 + tan 2 A
cosec 2 A = 1 + cot 2 A.
JCL
Fig. 152.
Fig 153.
Returning to Example 24 :
cos (A + B) = cos A cos B sin A sin B.
Values must first be found for
cos A and sin B.
Now cos* A + sin 2 A = i
from which cos 2 A = i sin 2 A, or sin 2 A = i cos 2 A
or cos A = Vi sin 2 A, sin A = Vi cos 2 A
Then cos A = Vi  (s) 2 = V^j$ = 866
and sin B = Vi (23) 2 = ^947 = 973
.*. cos (A + B) = cos A cos B sin A sin B
= (866 x 23) (5 x 973)
= 1991 4865
2874.
It is often necessary to change the binomial or twoterm expres
sion a sin qt\b cos qt into an expression of the form M sin (qt + c) ,
where c is an angle. We must therefore find the values of M and c
in terms of a and b, so that
M sin (qt\c) = a sin qt + b cos qt (i)
Take the addition formula, viz.
sin (A+B) = sin A cos B + cos A sin B.
Replacing A by qt and B by c, this statement becomes
sin (qt\c) = sin qt cos c + cos qt sin c.
Multiplying through by M
M sin (qt+c) = M sin qt cos c + M cos qt sin c . . . . (2)
PLANE TRIGONOMETRY 2?7
Since the righthand sides of (i) and (2) are equal in total value,
they can be made equal term for term by choosing suitable values
for the coefficients.
Thus M sin qt cos c = a sin qt
and M cos qt sin c = b cos qt
.'. M cos c = a and M sin c = b
a, b
i.e., cosc = ^ andsinc = v?
If, now, a triangle be drawn (Fig. 153) with sides a, b and
hypotenuse M, it will be seen that the angle opposite the side b
is the angle c, for its adjacent side is a, and therefore its cosine = ^
M
Hence c is found, for tan c = 
a
Knowing the values of b and a, the value of c is read off from
the table of tangents and is usually expressed in radians.
Also M 2 = 2 f6 a
M = V a + b*, so that M is found.
This investigation is valuable in cases of harmonic motion.
Example 25. The voltage necessary to produce an alternating
current C, after any particular period of time /, and in a circuit of
resistance 2 ohms in which the current varies, being given by
C = 100 sin 6oot, can be expressed as
V = 200 sin 6oot + 300 cos foot
Find a simpler expression for V.
Let 200 sin 6oo/ + 300 cos 6oo/ = M sin (6oo/ + c). Then by the
previous work
M = V2OO 2 + 300* = 3606
and tan c = ^ = 15 = tan 563
c 563 = ^ radians = 983 radian
/ J
V = 3606 sin (6ool + '983)
or, as it might be written, V = 3606 sin 600 (/ + 00164).
Note. If the current were continuous, then
voltage = current x resistance
= 200 sin 6oo/.
When the current is interrupted, "inertia" or "induction" effects
set up another current to oppose that due to the impressed voltage,
and therefore the amperes are not a maximum when the voltage is,
278 MATHEMATICS FOR ENGINEERS
i. e., the current lags behind the E.M.F. ; in this case to the extent
of 00164 second. The coefficient 600 in the formulae = zirf where
/= frequency: thus in this case the frequency = =95*6 cycles
per second.
As a further example of transformation consider the following
case :
Example 26
Let S w = displacement of the main steam valve of an engine from"\
its central position
S = displacement of the expansion plate from its central J
position J
for the case of an engine with Meyer valve gear.
Then Sm = r m cos (6 + a x ), and S e = r e cos (6 + a,).
To find a simple expression for the displacement of the expansion
plate relative to the main valve.
This relative displacement = S m S e = r m cos (d + a t ) r e cos (6 + a,).
Then
SmS e = r m cos (6 + a x )  r e cos (6 + a s )
= r TO cos tfcosaj r m sin Osma^ r e cos 6 cos a t + r e sin tfsina,
= cos 6(r m cos G! r e cos a,) + sin 6(r e sin a, r m sin aj)
= A cos 6 + B sin 6
= VA* + B 2 sin (6 + c) as before proved
= VA + B* cos ((0 + c)\
VA + B* cos (6 + p)
where
A = r m cos Oi r e cos a,, J5 = r e sin a, r m sin aj
and #> = c~ = tan 1 g  {tan 1 ^ is the awg/e whose tan is ^ j
We have thus reduced the expression for the relative displacement
to a form of a simple character which shows that this displacement
is equivalent to that caused by an imaginary eccentric of radius
VA 2 + B 8 and of angular advance p.
Exercises 31. On the Addition Formulae in Trigonometry.
1. If sin A = 45, find cos A and tan A (without reference to the
tables).
2. If sin B = *i6, cos A = 29, find the value of
sin A cos B cos A sin B.
3. Find the values of cos (A + B), and sin (A B), when
sin A = 65, sin B = 394.
PLANE TRIGONOMETRY 279
4. Tan A == 162, tan B = 58; find the values of tan (A + B) and
tan (A B).
5. The horizontal force P necessary to just move a weight W down
a rough plane inclined at a to the horizontal, the coefficient of friction
between the plane and the weight being /*, can be obtained from the
formula
P
^ = tan (*  a)
P
If tan <(> = ft, find an expression for ^ in terms of p. and tan a. Hence
find the value of P when W = 48, /i = 21, and a = 8.
6. The effort P required to raise a load W by means of a screw,
of pitch p and radius r, is given by
P = W tan (<t> + a)
where a =. angle of screw and tan <f> p = coefficient of friction. Find
an expression for P in terms of W, p, r and /*.
7. If ^Tr = , and tan A = u, find a simple expression for P.
W cos <t>
8. Given that tan (A B) = 537 and tan B = 388, find tan A.
9. Express 4*2 cos 5^ + 27 sin 5^ in the form M sin (5* + c).
10. Express 200 sin 50* 130 cos 50^ in the form M sin (50* + c).
11. If a bullet be projected from a point on ground sloping at an
angle A to the horizontal, the elevation being 6 to the incline, the
range R is given by the formula R = ut \gt z sin A. Find a simpler
2V sin Q
expression for R, if u = V cos 6 and t = g A
12. The efficiency of a screw jack = j^^ + * where 6 is the
angle of the screw and <f> is the angle of friction. In a certain experi
ment the efficiency was found to be 3, and by measurement of the
pitch and the mean circumference of the screw tan 6 was calculated
as 083. Find tan<f>, which is the coefficient of friction between the
screw and nut, and thence find $>.
13. If the E.M.F. in an inductive circuit is given by
E = RI Sin 2irft + 2r/LI COS 2irft
find a simpler expression for E, i. e., one having the form M sin (a*/* + c),
when R = 46, / = 60, L = 02 and I = 138.
Formula for the Ratios of the Multiple and Submultiple
Angles. In the addition formulae let B be replaced by A; by so
doing, expressions may be found for the ratios of 2A.
Thus sin (A+B) = sin A cos B + cos A sin B
sin (A+ A) = sin A cos A + cos A sin A
or sin 2A = 2 sin A cos A.
Also cos (A+B) = cos A cos B  sin A sin B
cos (A+ A) = cos A cos A  sin A sm A
or cos 2A = cos 2 A  sin 2 A.
280 MATHEMATICS FOR ENGINEERS
If for cos 2 A we write i sin 2 A, which is permissible since
cos 2 A f sin 2 A = i,
then cos 2 A = i sin 2 A sin 2 A
= i 2 sin 2 A.
Also cos 2A = cos 2 A (i cos 2 A)
= 2 cos 2 A i.
,. tan A + tanB
Again tan (A+B) == I _
. . . . tan A + tan A
tan (A+A) == ^
2 tan A
i tan 2 A
Grouping the results
sin 2 A = 2 sin A cos A
cos 2A = cos 2 A sin 2 Al
= 2cos 2 A 1
= l2sin 2 A J
2 tan A
tan 2A =
1  tan 2 A
If the ratios of the halfangles are required they can be obtained
" from the foregoing by dividing all the angles by 2.
E. g., cos A = cos 2 sin 2
2 A
= 2 COS 2 1
2
2 A
= i 2 sin 2 
2
A  A A
sin A = 2 sin cos
2 2
A
2 tan
tan A =
itan 2 
2
Similarly, by multiplying all the angles by 2, expressions can
be found for the ratios of the angle 4A
e. g., sin 4A = 2 sin 2A cos 2A
and this expansion can be further developed if necessary.
Formulae for ratios of 3A can be obtained by writing 2A in
place of B in the (A+B) formulae, and using the rules for the
ratios of 2A
PLANE TRIGONOMETRY 28l
E. g., sin 3A = sin (aA+A)
= sin 2A cos A + cos 2 A sin A
= 2 sin A cos 2 A + (i_ 2 sin 2 A) sin A
= 2 sin A cos 2 A + sin A  2 sin 3 A
= 2 sin A (i S i n 2 A) + sin A  2 sin* A
= 3 sm A 4 sin 8 A.
In like manner cos 3A = 4 cos 3 A  3 cos A
tan 3 A = 3tanAtanA
i 3 tan 2 A
for
For cos 2A
L.H.S. = cos 2A = cos 48 = 669
R.H.S. = cos a A  sin 2 A = cos* 24 sin* 24
= (9I35) 2 ~ (4067)'
= '835 ~ 165
= 670
.'. L.H.S. = R.H.S.
For tan 2A
L.H.S. = tan 2A = tan 48 = 11106
R H S = 2tan A  2 tan 24 2 x 4452
i tan 2 A i tan 2 24 i (4452) 2
89
= .855  ri08
/. L.H.S. = R.H.S.
(the small differences being due to sliderule working).
For sin 3A
L.H.S. = sin 3 A = sin 72 = '9511.
R.H.S. = 3 sin A 4 sin 3 A = 3 sin 24 4 sin* 24
= 3 X 4067 
= i 220 269
A A A
Example 28. If sin A = 85, find sin , cos and tan , without
22 2
the use of the tables. (Examples are set in this manner so that the
reader may become familiar with the formulae and the method of
using them ; but in practice the tables would be used.)
282 MATHEMATICS FOR ENGINEERS
A A
To find sin . The formula that contains sin , only, of the ratios
2 2
of the halfangles is
. A
cos A = i 2 sin 2
2
and, therefore, to use this, cos A must first be found.
cos A = Vi sin 2 A = Vi (85) 2 = 526.
Then 526 = 12 sin 2
2sin 2  = 1526 = 474
or sin 8  = 237
.% sin = '487 {the positive root only being taken}.
To find cos
2
. f or alternatively ~\
cos A = 2 cos 2 i
2
/. 526 = 2 cos 2 i
2 COS 2 = 1526
cos 2 = 763
2
cos  = 875.
To find tan 
2
. A
.sin
A 2 48'
tan
T
2 A
cos
2
Example 29. Find the value of tan 2A if cos A = *g6.
sin A = Vi cos 2 A
= Vi  (96) a
= 2795
. sin A 2795
Then tan A = r = ^f 2 = 292
cos A '96
2 tan A 2 x 292 584
and tan 2A = r 5* = ~ 7 ^rs = ^  = '638.
i tan 2 A i (292) 2 915  >
PLANE TRIGONOMETRY
283
Example 30. It was required to find, to an accuracy of oooi',
the dimension marked c in Fig. 154; the figure representing part of
a gauge for the shape of a boring tool. There is a radius of 5* at the
top of the sloping side, which is tangential to an arc of 34* radius at
the bottom ; and other dimensions are as shown.
B
329KB
Fig. 154. Gauge for Boring Tool.
Introduce the three unknowns x, y and z as indicated on the figure ;
y being the distance along the slant side from the point of contact
with the arc to the base.
Let the angle ACB = o, then L APT = 
CB = 1704 + x  1375 = '329 +
In the triangle ACB tan a = ^j
O * *
In the triangle APT tan  = ^ = *z
(i)
(2)
3.4 A From the properties of
tana = ....... (3)! the circle.
y 2 = (68 + *)* . . . . (4) ' (* EDC * a ri g ht an 6 le )
also
and
Connecting (i) with (3)
329 + x
y
i. e., y*
136 (329 + *)
18496 (329 +
Hence from (4)
whence
(68 + x]x = 18496 (329 + *)
8496*'  5583* + '20021  O
284 MATHEMATICS FOR ENGINEERS
+ 5*583 V3II699 6804
so that x = J J  T?
16992
or the required value of x =  = 0361*.
16992
Now
y 329 + x
so that y = 4965
also tan a = ^> = 68482.
4965
It would be unwise to use the tables to find a from the previous
equation, for in the neighbourhood of the required value the change
in the value of the tangent is extremely rapid ; hence it is a good plan
to make use of the rule for tan aA or its modification.
2 tan 
2
Thus tan a = 
itan z 
2
i. e., 68482 = i for tan  = 2* from (2).
i 42* 2
This is a quadratic in terms of z, and the solution applicable to
this case is z = '4323.
.*. C = 275 2 X '4323 = 18854*.
Further Transpositions of the Addition Formulas
sin (A+B) = sin A cosB+cosA sinB
sin (A B) = sin A cos B cos A sin B
Hence, by addition
sin (A+B)+sin (A B) = 2 sin A cos B
and by subtraction
sin (A+B) sin (A B) = 2 cos A sinB
Also cos (A+B) = cos A cos B sin A sin B
cos (A B) = cos A cos B+sin A sin B
.'. cos (A+B)+cos (A B) = 2 cos A cosB
and cos (A B) cos (A+B) = 2 sin A sin B.
[Note the change in the order on the lefthand side in this last
formula.]
Now
A = ' ' '*\ L i.e.,= % sum of the two angles,
and B = ( A +B)(AB) { e . > = $ difference of the two angles.
2
PLANE TRIGONOMETRY 28s
Hence, the first of these formulae could be written
Sine (one angle) + sine (another angle)
= 2 sine (i their sum) x cos ( J their difference)
A substitution is very often of great service ; thus,
let (A+B) = C and (AB) = D '
Then
sin C + sin D = 2 sin S cos =? etc
2 2
and we have the summary
If the change is to be made from a sum or difference to a product,
use the (C+D) formulae
sin C+sin D = 2 sin cos
sin Csin D = 2 cos sin . . ( 2 )
a 2
cos C+cos D = 2 cos ~ cos ^^ ...... (3)
C+D C D
cos D cos C = 2 sin ~ sin ^ ...... (4)
If, however, the change to be made is from a product to a sum
or difference, use the A and B formulae, which follow
sinA cosB = {sin(A+B)+sin(AB)} ..... (5)
eosA sinB = J{sin(A+B)sin(AB)j ..... (6)
cosA cosB = {cos(A+B)+eos(AB)} ..... (7)
sinA sinB = {cos (AB) cos (A+B)} ..... (8)
In later work it will be found that certain operations can be
performed on a sum or difference of two trigonometric ratios that
cannot be done with products ; hence the great importance of this
last set of formulae.
It may appear to the reader that his memory will be severely
taxed by the above long list of formulae, but a second thought
will convince him that all are derived from the original (A + B)
and (A B) formulae, which must be committed to memory to
serve as the first principles from which all the later formulae are
developed.
Example 31. Express 17 sin 56 sin 148 as a sum or difference.
sin 56 sin 148 = i {cos (148  56)  cos (148 + 56)} . . from (8)
{A = 148, B = 56}
/. 1 7 sin 56 sin 148 = 8j, {cos 92  cos 204).
286 MATHEMATICS FOR ENGINEERS
To check by the use of tables
L.H.S. = 1 7 sin 56 sin 1 48 = 17 sin 56 sin 32
= 17 x 8290 x '5299
= 747
R.H.S. = 85 {cos 92 cos 204}
= 85 {  cos 88 + cos 24}
= 85 { 0349 + 9135} = 85 x 8786 = 747.
/. L.H.S. = R.H.S.
Example 32. The voltage V in an A.C. circuit, after a time /, is
given by V = 200 sin 360*. and the current by C = 35 sin(36o/ + c).
Find an expression for the watts at any time, expressing it as a sum
or difference.
Watts = amps x volts
= 3'5 sin (360* + c) x 200 sin 360*
= 700 sin (360^ + c) sin 360*
= 222 {cos c cos (720* + c)} from (8)
= 350 {cos c cos (7201 + c)}.
Example 33. Express (4 sin 5*) (5 cos 3*) as a sum or difference.
(4 sin 5/) (5 cos 3<) = 20 sin 5* cos 3*
= 10 {sin 8t + sin 2t} . . . from (5)
Exercises 32. On Transpositions of the Addition Formulae.
1. If sin aA = 824, find cos 2A and tan 2A.
2. If sin A = $, find sin 2A and cos 2A.
3. Express cos 2 14 in terms of cos 28.
4. Find an expression for sin 2B in terms of cos B alone. Hence
find the value of sin 2B when cos B = 918.
A A
5. If sin A = '317, find sin , cos and sin 3 A.
* " 22 3
^
6. If sin 2 A = 438, find cos 4 A and tan
7. Change 5 sin 2 2t into a form containing the first power only of
the trigonometric function.
8. Express 157 cos 160 sin 29 as a sum or difference.
9. Simplify sin 15* + sin 3* + cos lit cos yt.
/k
10. Sin 2 A = "504. Find sin A, tan A and cos
j i 2
11. A rise of level is given by 100 sin a cos a x s where 5 = difference
between the readings of the top and bottom hairs of a tacheometric
telescope. Express this statement in a more convenient form.
If the angle of elevation a is 11 37' 30*, and the staff readings are
572 and 841, find the rise.
PLANE TRIGONOMETRY 287
12. Express as products, and in forms convenient for computation
(a) sin 48  sin 17 ; (b) cos 99 + cos 176 ; (c) 12 cos 365  12 cos 985'.
13. When using a tacheometer and a staff it is found that if C
and K are the constants of the instrument, 6 is the angle of depres
sion, s the difference of the staff readings, then depth of point below
level of station = sin 20 + K sin 6  E + Q, and distance of point
from station = CS cos 2 6 + K cos 6. Find the depth and the distance
when C = 9887, sin 6 = 2753, K = 75, S = 69, E = 488 and Q = 955.
14. If tan a =   and tan ^ = 22, find values of z to satisfy
1 o75 ~ * 2
the equations. [Refer to Fig. 154 and the worked Example 30.]
15. If V = 94 sin27r/* and A = 2 sin (zirft '117), express the
product AV as a sum or difference.
Trigonometric Equations. Occasionally one meets with an
equation involving some trigonometric ratios; if only these ratios
occur, *. e., if no algebraic terms are present in addition, the
equations may be solved by the methods here to be detailed.
The relations between the ratios themselves, already given,
must be borne in mind, so that the whole expression can be put
into terms of one unknown quantity, and the equation solved in
terms of that quantity.
For emphasis, the relations between the ratios are here
repeated
tan A =^, cotA = r, sec A = ,  T r
cos A tan A cos A sin A
sin 2 A + cos 2 A = i, whence sin 2 A = i cos 2 A
or cos 2 A = i sin 2 A
sec 2 A = i+tan 2 A, cosec 2 A = i+cot 2 A.
The idea in the solution of these trigonometric equations is to
eliminate all the unknowns except one, by the use of the above
relations, and then to apply the ordinary rules of equations to
determine the value of that unknown.
Example 34. Solve the equation 4 sin Q = 3*5.
4 sin 6 = 35
and sin 6 = 5 = 875.
4
Hence one value of 6, viz. the simplest, is 61 3',
since sin 61 3' = 875
but sin (180  61 30, *' . sin n857', also  875.
so that a possible solution is n857'.
288 MATHEMATICS FOR ENGINEERS
Again, 360 + 61 3' or 360 f 118 57' would also satisfy, and so
an infinite number of solutions could be found ; but whilst these could
all be included in one formula, it is not at all necessary from the
engineer's standpoint that they should be, for, at the most, the angles
of a circle, viz. o to 360, are all that occur in his problems.
Hence, throughout this part of the work the range of angles will be
understood to be o to 360.
.'. The solutions in this example are 61 3" and 118 57'.
Example 35. If tan 6 = 5 sin 6, determine values of 6 to satisfy
the equation.
Apparently, in this one equation two unknowns occur, or the data
are insufficient, but in reality two equations are given, for tan 6
Thus ' * = 5 sin 6.
cos 6 J
Dividing through by sin 6 [and in doing this we must put sin Q = o
as a possible solution, since * = o and 5x0 = 0]
COS v
Then ^ = 5
cos 6 J
COS = '2 = COS 78 28'
/. 6 = o; or 180; or 78 28'; or 360  78 28', i.e., 281 32'.
Example 36. Solve the equation sin 6 + tan 6 = 3 cos 6 sin 6.
sin 6 + tan 6 = 3 cos & sin 6
By substituting for tan 6 its value
sin 6 A A
sin 6 H 2i = 3 cos ^ sin
COS C7
or '^^ {cos 6 + 1} = 3 cos 6 sin 0.
COS t/
Dividing through by sin 6 {sin 6 = o thus being one solution} and
multiplying through by cos 6
cos 6 + i = 3 cos 2
or 3 cos 2 6 cos 6 i = o.
It may appear easier to solve this equation if X is written for cos 6
. e., 3 X 2  X  1=0
+ I Vl + 12
whence X =
3 '606
6
4606 2 '606
" 6 f 6
= 7677 or  4343
cos 6 = 7677 or cos 6 = '4343
PLANE TRIGONOMETRY 2 8 9
Now for the cosine to be positive, the angle lies in the first and
fourth quadrants ; and, since the smallest angle having its cosine =7677
is 39 51'. the values of 6 are 39 51' or 36o  3 9 5I ', i. e ., 320 9'. '
For the cosine to be negative, the angle lies in the second and
third quadrants. Now cos 64 15' = 4343, and therefore the values
of 6 are i8o64i5', i.e., H545', or i8o+6 4 i 5 ', i.e., 24 4 i 5 '.
Hence the solutions are
= o; 180; 395i'; 320 g'\ 11.5 4.S' or 2 4 4is'.
Example 37. The velocity of the piston of a reciprocating engine
is given by the formula
/ . n . Y sin 20\
v = 2nnr sin 6 \  = )
\ 21 I
where r = crank radius, n = R.P.M., 6 crank angle from dead centre
position, and / = length of connectingrod.
Y
The velocity is a maximum when cos0+jcos20 = o; find the
crank angles for the maximum velocity when / = 8r.
We require to solve the equation
., . cos 20
cos 6 H  o = 
o
To change into terms of cos 6 write 2 cos 2 6 i in place of cos ^6.
A , 2 COS 2 6
Theii cos 6 \  ~
8 cos 6 + 2 cos 2 6i = o
or 2X 2 + 8X 1=0 where X = cos 6.
The solutions of this equation are given by
x _  8 V6 4 + 8
4
_ 8 8485
4
= ~ 16485 or 485
4 4
= 41212 or 12 12, which are the values of cos 6.
But cos 6 cannot =  41212, since cos 6 is never greater than i,
hence the first root is disregarded.
cos & = 12 12, which gives the required solutions,
i.'e.. = 83 2' or 276 58'.
If a skeleton diagram is drawn it will be observed that when 6 has
these values the crank and connectingrod are very nearly at nghi
angles to one another.
U
290
MATHEMATICS FOR ENGINEERS
Exercises 33. On the Solution of Trigonometric Equations.
Solve the equations (for angles between o and 360).
1. Sin 2 A + 2 sin A = 2 cos 2 A.
3. Cos0
6cos 2  =
5. 2 sin* 6 5 cos 6 = 4.
7. Tan x tan 2X = i.
9. Cos 2 A + 2 sin 2 A 25 sin A = o.
2. 2 sin 2 6 + 4 cos 2 6 = 3.
4. Cot 14 tan 6 = 5.
6. 15 cos 2 $ + 9 sin = 126.
8. Tan A + 3 cot A = 4.
11. 3 tan 1 B 2 tan B i
13. 3 tan 2 Q + i = 4 tan 6.
15. Cos 2X + sin 2* = i.
rj
17. Cos x sin # = ^
16
10. Cos x tan # = 5842. 11. 3 tan 1 13 2 tan 13 i = o.
12. Tan # + cot x = 2.
14. Cos* # = 3 sin 2 x.
16. Cos # + \/3 sin # = i.
18. 235 sin # 172 cos x = 64.
19. The velocity of a valve actuated by a particular Joy valve
gear is maximum when
i'2p* cos pt + i8p* sin pt = o
where p = angular velocity of the crank shaft.
Find the values of the angle pt for maximum velocity.
20. To find the maximum bending moment on a circular arch it is
necessary to solve the equation
;R 2 sin 6 cos 6 + 9340^* sin 6 = o.
Find values of 6 to satisfy this equation.
21. The following equation occurred when taking soundings from
a boat, the position of the boat being fixed by reference to three points
on the shore. [Compare Exercise 31, p. 272.]
sin (70 14" + *) = 1195 sin (4 8 5 6/ + *)
Find the value of x to satisfy this equation, x being an acute angle.
Hyperbolic Functions. Consider the circle of unit radius
(Fig. 155) and the rectangular hyperbola whose halfaxes are also
unity (Fig. 156), '. e., OA in either case = i.
Fig. 155
Fig. 156.
Draw any angle EOA in each diagram, and let the " circle
angle " EOA = Q, and let the " hyperbola angle " EOA = ft.
PLANE TRIGONOMETRY 291
radkns T * ****** ***** " circular " or "hyperbolic"
length of circular or hyperbolic arc
""mean length of radius vector ' or by 2 x area of sector OAE
EX
Now in Fig. 155 _ =E X=sin0, and the corresponding
ratio, viz. EX, of the hyperbolic angle is termed sinh p.*
Similarly
OX = cos in Fig. 155 and OX = cosh ft in Fig. 156
At = tan in Fig. 155 and A* = tanh p in Fig. 156.
In Fig. 155 (EX)* + (OX) 2 = i
i.e., sin 2 + cos 2 = i ......... (!)
In Fig. 156 (OX) 2 (EX) 2 = i, since the equation of the
rectangular hyperbola is x*y* = i if the semiaxes are each
equal to unity and the centre is taken as the origin.
Hence cosh 2 p sinh 2 p = i
or cosh 2 p + ( i x sinh 2 )8) = i
*. e., cosh 2 p + (V~i X sinh ) 2 = i
or cosh 2 p + (j sinh /3) 2 = i
.where / is written to indicate V^i.
Comparing the last equation with equation (i), we see that we
may change from circular to hyperbolic functions if we write
/ sinh ft for sin 0, and cosh p for cos 6, and hence / tanh P for tan 6,
If these substitutions are made, the ordinary rules for circular
functions follow.
E. g., sin (x f y) = sin % cos y f cos x sin y
and the corresponding expansion with hyperbolic functions is
/ sinh (X+ Y) = / sinh X cosh Y + cosh X . ; sinh Y
or sinh (X+ Y) = sinh X cosh Y + cosh X sinh Y
or again, cos 2x = sin 2 x f cos 2 x ........ see p. 280
t. e., cosh 2X =  (/ sinh X) 2 + (cosh X) 2
= sinh 2 X + cosh 2 X, since ; 2 = (V 11 !) 8 = i.
It can be shown that these hyperbolic functions can be expressed
in terms of the exponentials in the forms
e  x = cosh x sinh x
? = cosh x f sinh x
e e
i.e., cosh*= *T5 + i. 2.3.4"*
and
 , .2.3.4.5 .
* To avoid confusing with the circular functions, sinh is usually
pronounced " shine," and tanh " tank."
292 MATHEMATICS FOR ENGINEERS
The corresponding relations for the circular functions are
e ix = cos x f j sin x
e~ 5x = cos x ; sin x
y*
cos* =  _ = j
1.2 1.2.3.4
/ O .A .
sm A; = = = x
2; 1.2.31.2.3.4.5
Hyperbolic functions occur frequently in engineering theory;
e. g., in connection with the whirling of shafts the equation
y = A cos mx + B sin mx + C cosh mx + D sinh mx
plays a most important part : the equation of the catenary is
y = cosh x ; and so on.
It is in electrical work that these functions occur most fre
quently; thus, for a long telegraph line having a uniform linear
leakage to earth by way of the poles the diminishing of the voltage
is represented by a curve of the form y = cosh x, whilst the curve
y = sinh x represents the current.
Example 38. A cable weighing 3 Ibs. per foot hangs from two points
on the same level and 60 feet apart ; and it is strained by a horizontal
pull of 300 Ibs. The form taken by the cable is a catenary. Find the
length of the cable from the formula
Length = 20 sinh
zc
, , horizontal tension
where L = span and c = JTT 7 7 r 7 TT
weight of i foot of cable
Here we have L = 60 and c =  = 100 ; hence = = 3
3 2C 200
Thus length of cable = 2X 100 sinh 3
Table XI at the end of the book may be utilised to find the value
of sinh 3, in the following manner : Look down the first column until
3 is seen as the value for x : follow the line in which this value occurs
until the column headed sinh x is reached. The value there shown is
that of sinh 3 and is 3045.
Hence length of cable = 200 x '3045 = 60*9 ft.
This rule gives the exact length of the cable, but in practice the
form of the cable is assumed to be parabolic, and the approximate
length is given by
Length = span + 8 (sag) '
3 span
PLANE TRIGONOMETRY 293
the sag also being calculated on the assumption of the parabolic form
of the cable. In this case the sag is found to be 45 ft. and hence _
length of cable = 60 + = 6o . g
In this instance the result obtained by the true and approximate
methods agree exactly: and in the majority of cases met with in
practice the approximate rule gives results sufficiently accurate.
Example 39. The resistance of the conductor of a certain telegraph
line is 83 ohms per kilometre and the insulation resistance is 600 meg
ohms per km. The difference in potential E between the line and earth
at distance L kms. from the sending end is found from the formula
E = A cosh Vrl. L + B sinh Vrl. L
where A and B are constants, r = resistance of unit length of the
conductor and I = conductance of unit length of the path between the
line and earth.
If the total length of the line is 100 kms., the voltage at the
sending end is no, and at the receiving end is 85, find the values of
A and B.
We have two unknowns and we must therefore form two equations.
At the sending end L = o
and then 1 10 = A cosh Vrl . o + B sinh Vrl . o
= A cosh o + B sinh o = Ax i = A
Hence A= no.
NnwrZ =  = iiSix io~ 8 and Vrl= 0001176; also at a
600 x io 6
distance of 100 kms. from the sending end the value of E is to be 85.
Substituting these numerical values in the original equation
85 = no cosh (0001176 x 100) + Bsinh (0001176 x 100)
= no cosh 01176+ Bsinh 01176 ......... (i)
In order to solve this equation for B the values of cosh 01176 and
sinh 01176 must first be found; and as the given tables of values of
cosh and sinh are not convenient for this purpose we proceed according
to the following plan
and sinh, = ^; . = ."";
and to evaluate we must take logs.
Let y = e' 01176 and then
log y = 01176 x log e
= 01176 x 4343 = 0051
so that y =1012.
Thus * 1176 = 1012 and e' 01 "' which is the reciprocal of ' 01
is 9883.
294
Then
and
MATHEMATICS FOR ENGINEERS
2 + 988
cosh 01176 =
sinh 01176 =
2
1012
= I
012
Substituting these values in equation (i)
85 = (no x i) + (B x 012)
whence 012 B = 25
or B = 2083
Hence E = no cosh 0001176 L 2083 sinh 0001176 L.
Complex Quantities. Algebraic quantities generally may be
divided into two classes, real and imaginary, and the former of these
Fig. 157. Complex and Vector Quantities.
may be further subdivided into rational and irrational or surd
quantities. Thus, Vs and also 7 are real, whilst V 15 is
imaginary; indeed, all quantities involving the square root of a
negative quantity are classed as imaginary. An expression that is
partly real and partly imaginary is spoken of as a complex quantity ;
thus 4 + 7^9, and Vzx}i6V zy are complex quantities.
The first expression might be written as 4 f (jxVq X V i),
. e., 4+21; where / stands for V i. The general form for
PLANE TRIGONOMETRY 395
these complex expressions is usually taken as a f jb, where a and
b may have any real values.
According to the ordinary convention of signs, if OA ((a) Fig. 157)
represents + a units, then OA' would stand for a if the length
of OA' were made equal to that of OA ; in other words, to multiply
by I, revolution has been made through two right angles. Now
X V I X Vi = a, so that the multiplication by V^i
must involve a revolution onehalf of that required for the multiplica
tion by i; or OA* must represent ja. Accordingly a meaning
has been found for the imaginary quantity /, and that is : If fa
is measured to the right and a is measured to the left, from a
given origin, then ja must be measured upward, and differs from
the other quantities only in direction, which is 90 from either
fa or a.
To represent a f jb on a diagram, therefore, we must set out
a distance OA to represent a, erect a perpendicular AB making
AB equal to b, choosing the same scale as that used for the hori
zontal measurement, and then join OB ; then OB = a f jb. For
OB' would represent a + b and AB = / . AB', so that OB must
be the result of the addition of a to jb. The addition is not
the simple addition with which we have been familiar, but is
spoken of as vector addition, i. e., addition in which attention is paid
to the direction in which the quantity is measured as well as to its
magnitude.
It is often necessary to change from the form a + jb to the
form r (cos 6 f / sin 0} ; and this can be done in the following
way
If r(cos0 + / sin0) is to be identically the same as a+/6
then the real parts of each must be equal, and also the imaginary.
i. e. t r cos 6 = a (i)
rj sin = jb
whence r sin = b (2)
By division of (2) by (i) tan 6 = 
and by squaring both (i) and (2) and adding
f 2 cos 2 + f* sin 2 = a 2 +&*
or r z a 2f &a since cos 2 f sin 2 = i
Thus, at (b) Fig. 157
OB = r
and L BOA =
Example 40. To change 3  yjj into the form r (cos 6 + / sin 6).
296 MATHEMATICS FOR ENGINEERS
From the above, since a = 3, and b = 57
r* = 3 2 + (~5'7) 2 = 9 + 324 = 4iH
r = 644
and tan = ^2 =* ig or 6 = 62 14$'.
This case is illustrated at (6) Fig. 157, in which OB represents r.
and the angle BOA is the angle 6.
Vector quantities, such as forces, velocities, electrical currents
and pressures, may be combined by either graphic or algebraic
methods; in the algebraic addition, for example, the components
along two directions at right angles are added to give the com
ponents along these axes of the resultant. Thus if the vector
2+15; were added to the vector 4 + 6;', the resultant vector
would be 2 4+i"5/+6;, i.e., 2+21;. The addition is really
simpler to perform by the graphic method, thus : OB at (c)
Fig. 157 represents the vector quantity 2+15; and OD represents
4+6/. Through B draw BE parallel and equal to OD and
join OE; then OE is the resultant of OB and OD. It will be
seen that OE is the vector 2+21; since OF = 2 units measured
in a negative direction and FE = 21 units.
To multiply complex quantities. Let OA ((d) Fig. 157) represent
a + jb, i. e., r (cos 6 + / sin 0)
and let OB represent a 1 + jb lt i. e., ^(cos 0j + / sin X )
Then : OA x OB = (r cos 6 + rj sin 0)(r 1 cos X + rjj sin X )
= TTj cos cos 6j + rr^j sin Oj cos +
rrrf sin cos X + rr^' 2 sin sin X
= rrL (cos cos 0j sin sin X ) +
rrtf (sin cos 0! + cos sin 0J
= rr^cos (0 + ej + / sin (0 + X ) j
^ ,. ., .... T , a + jb r (cos +;sin0)
To divide complex quantities, Let  f v = 4 '. . .. \
a i + 1i fi(cos X + ; sin 0j)
Rationalise the denominator by multiplying by (cos 0j ;' sin X )
~^~ r cos 9 ~^" sm ecos e ~ sm 9
Th
(cos 2 X + sin 2
which can be expressed in the form A + ;'B if desired.
These results might have been arrived at by expressing
r (cos + ;' sin 0) as rei and r (cos X + ;' sin X ) as r^ l .
Thus (a + jb)(cii + jb^ = rei* x r^ = rr^'+W
= rr^cos (0 + Oj) + / sin (0 + 0j)}
PLANE TRIGONOMETRY 297
Example 41. The electric current C in a starconnected lighting
system was measured by the product Potential P x admittance y.
If P = (068  0015;) (28 + 30;) and y = 9 + 18; find C.
P = (068  0015;) (28 + 30;) = 1949 + 1998; (by actual multi
plication).
= 2791 (cos 45 43' + ; sin 45 43')
y = 9 f i8j = 9179 (cos 11 19' + j sin 11 19')
.'. C = Py = 2563 (cos 57 2'4y sin 57 2') or, alternatively,
t395 + 21497'.
Inverse Trigonometric Functions. If sin * = y, then x is
the angle whose sine is y, and this statement may be expressed
in the abbreviated form x = sin~ 1 y. (Note that sin~ 1 y does not
mean ^ ^ but the I indicates a converse statement, y being the
value of the sine and not the angle.)
sin" 1 y is called an inverse circular function.
Similarly sinh 1 y is called an inverse hyperbolic function
Angles are sometimes expressed in this way instead of in degrees ;
e. g., when referring to the angle of friction for two surfaces : if
the coefficient of friction between the surfaces is given, that is
the value of the tangent of the angle of friction, and the angle of
friction = tan~V*, where /t is the coefficient of friction.
Example 42. Given sin" 1 * = y, find the values of cos y and tan y.
sin" 1 x = y
i. e., sin y = x
and tan y =
cos y = Vi sin'y = Vi X*
sin y x
cosy Vi  x*
Example 43. The transformation from the hyperbolic to the
logarithmic form occurs when concerned with a certain integration.
If cosh y = x, show that cosh" 1 * = log ,(* + V* 2  i).
cosh y =
or
whence
/ :
(Solving the quadratic)
* + Vx*i
MATHEMATICS FOR ENGINEERS
since x Vx 2 i =   ._
x + Vx*  i
as is seen if we multiply across
log, (x + Vx*  i) = y, or log*
i. e., y = log* (x + Vx z i)
or if only the positive root is taken
y = loge (x + V* 2 i)
cosh" 1 * = loge (x + Vx* i)
In like manner it can be proved that
. , , x , x + Vx z + a 2
smh. 1  = log,
a a
, x . x f Vx z a 3
 1  = log e 
Exercises 34. On Hyperbolic and Inverse Trigonometric functions.
1. Read from the tables the values of cosh 7 and sinhi'5.
2. Evaluate 5 cosh 015 + ! sinhoi5.
3. Find the true length of a cable weighing 18 Ibs. per foot, the
ends being 120 ft. apart horizontally and the straining force being
90 Ibs. weight. [Refer to worked Example 38, p, 292.]
4. Calculate the sag of the cable in Question 3, from the rule
sag = c ( cosh  i )
\ 2C I
straining force in Ibs. wt.
where c =  . ^ 77 T TT 
wt. of i ft. of cable
Hence find the approximate length of the cable, from the rule
length = span +**J&
3 X span
5. The E.M.F. required at the transmission end of a track circuit
used for signalling can be found from
E, =
Put this expression into a simpler form, viz. one involving hyper
bolic functions.
6. If the " angle of friction " for iron on iron is tan~ 1 i9, find this
angle.
7. The lag in time between the pressure and the current in an
...... , period , 2nL ,
alternating current circuit is given by * ^ x tan 1 p where
n = number of cycles per second, L = selfinduction of circuit and
R = resistance of circuit, the angle being expressed in degrees. If
the frequency is 60 cycles per second, L = 025 and R = 12, find the.
lag in seconds.
PLANE TRIGONOMETRY 299
8. If cosh y = 14645, find the positive value of y.
9. A block is subjected to principal stresses of 255 Ibs. and 171 Ibs.,
both tension. The inclination of the resultant stress on a plane in
clined at 27 to the plane of the greater stress is tan^K? tan A where
Vi /
/! and / 2 are the greater and lesser stresses respectively and 6 is the
inclination of the plane. Find the inclination of the resultant stress
for this case.
10. The solution of a certain equation by two different methods
gave as results
s = sin ( 7* + tan~ x ) and s = sin ( 7* 2 tan" 1 2
53 V/ 45 7 53 V/ 2
respectively. By finding the 'numerical values of the angles tan" 1
and tan 1 , show that the two results agree.
11. The following equation occurred in connection with alternator
regulation
a = 6 + <(>
lid = sin" 1 ^4r anc * cos" 1 55, find sin a.
OIoO
12. The speed V knots of waves over the bottom in shallow water
is calculated from
V 2 = i8
.L,
where d = depth in feet
L = wave length in feet.
If d = 40 ft., and L = 315 ft., calculate the value of V.
13. By calculating the values of the angles (in radians) prove the
truth of the following relations :
tan 1 \ + tan 1 =
4
4 tan 1 tan 1 ?fa = 
4 tan 1 tan 1 ^ + tan 1 ^ = ~
Illustrate the first of these by a diagram.
14. An equation occurring in the calculation of the arrival current
in a telegraph cable contained the following :
N _ 9* _ 3_ cos 2a gink 2,b cos za cosh 2b.
10 2
Find N when a = 45 and b = 2.
15 If C = 54 (cos 62 + j sin 62) and y = x8 (cos 12 + ; sin 12 )
find P (in the form a + /&). The letters have the same meanings
Example 41, p. 297.
CHAPTER VII
AREAS OF IRREGULAR CURVED FIGURES
Areas of Irregular Curved Figures. Rules have already
been given (see Chapter III) for finding the areas of irregular figures
bounded by straight sides; if, however, the boundaries are not
straight lines, such rules only apply to a limited extent.
The mean pressure of a fluid such as steam or gas on a piston
is found from the area of the " indicator diagram," the figure
automatically drawn by an engine " indicator," correlating pressure
and volume. By far the quickest and most accurate method of
determining the area of this diagram is (a) to use an instrument
called the planimeter or integrator. Other methods are (b) the
averaging of boundaries, (c) the counting of squares, (d) the use
of the computing scale, (e) the trapezoidal rule, (/) the midordinate
rule, (g) Simpson's rule and (h) graphic integration.
To deal with these methods in turn :
(a) The Planimeter. The Amsler planimeter is the instrument
most frequently employed, on account of its combination of sim
plicity and accuracy. It consists essentially of two arms, at the
end of one of which is a pivot (see Fig. 158), whilst at the end of
the other is the tracingpoint P. By unclamping the screw B the
length of the arm AP can be varied, fine adjustment being made by
the adjusting screw C : and this length AP determines the scale
to which the area is read. The rim of the wheel W rotates or
partially glides over the paper as the point P is guided round the
outline of the figure whose area is being measured; the pivot O
being kept stationary by means of a weight. The motion of the
wheel W is measured on the wheel N in integers, and on the wheel
D in decimals, further accuracy being ensured by the use of the
vernier V.
To use in the ordinary manner, the pivot O being outside the
figure. By rough trial find a position for the pivot so that the
figure can be completely traversed in a comfortable manner. Mark
some convenient startingpoint on the boundary of the figure and
s: AS: : 8 r A ^^ ^  .
case the area would be 844 sq. units.
Along the arm AP are marks for adjustment to different scales
f A is set at one of these marks the area will be in sq. ins., at
Fig. 158. The Amsler Planimeter.
another, in sq. cms., etc.; but if there be any doubt about the
scale, a rectangle, say 3* X 2* should be drawn, and the tracer
guided round its boundary. Whatever reading is thus obtained
must represent 6 sq. ins. so that the reading for i sq. in. can be
calculated therefrom.
If, in the tracing for which the figures are given above, the
zero mark A had been set at the line at which " 01 sq. in." is
found on the long bar, then the area would be 844 sq. ins., since
the divisions on the vernier scale represent 01 sq. in. each.
For large areas it may be found necessary to place the pivot
inside the area ; and in such cases the difference between the first
and last readings will at first occasion surprise, for it may give an
302
MATHEMATICS FOR ENGINEERS
area obviously much less than the true one. This is accounted for
by the fact that under certain conditions, illustrated in Fig. 159,
the tracer P traces out a circle, called the zero circle, for which the
area as registered by the instrument is zero, since the wheel does
not revolve at all. For a large area, then, the reading of the instru
ment may be either the excess of the required area over that of the
zero circle, or the amount by which it falls short of the zero circle
area. These areas are shown respectively at EEE and III in
Fig. 159, while the zero circle is shown dotted. [Noie. For the
ordinary Amsler planimeter the area of the zero circle is about
220 sq. ins., but it is indicated for other units by figures stamped
on the bar AP.]
Fig. 159. Zero Circle of Planimeter.
To use in the special manner. By means of a set square, adjust
the instrument so that the axis of the tracing arm is perpendicular
to a line joining the fixed centre O in Fig. 159 to the point of con
tact of the wheel and the paper. Measure the radius from the fixed
centre O to the tracingpoint P, and draw a circle with this radius
on a sheet of tracingpaper. Place this over the plot whose area
is being measured, and endeavour to estimate whether the figure
is larger or smaller than the zero circle. If this is at once apparent
trace round the figure in the ordinary way and add the area of the
zero circle to the reading, or subtract the reading from the zero
303
circle area as the case may demand. If not apparent, proceed
thus
Set the planimeter to some convenient reading, say 2000 and
trace the area in a righthanded direction. Then, if the final
reading is greater than 2000, the area is greater than that of the
zero circle and vice versa. Then to obtain the area
(1) If the area is greater than the zero circle, trace in a right
handed direction and add the excess of the last reading over the
first to the area of the zero circle; i. e., if x is the excess of the
final reading over the initial reading, the true area = x f zero
circle area.
(2) If the area is less than the area of the zero circle, trace in a
lefthanded direction and subtract the difference of the first and last
readings from the area of the zero circle. For x = excess of
the last reading over the first, if the tracing is in a righthanded
direction, and this becomes f x if the tracing is in a lefthanded
direction. The true area = x + zero circle area, and the tracing
is performed in a lefthanded direction in order to get a positive
value for x.
If the instrument is to be used as an averager, as would be the
case if the mean height of an indicator diagram was required, LL
in Fig. 158 must be set to the width of the diagram and the outline
must be traced as before. Then the difference of the readings
gives the mean height of the diagram. Further reference to the
planimeter is made in Volume II of Mathematics for Engineers.
The Coffin Averager and Planimeter (Fig. 160) is somewhat
simpler in construction as regards the instrument itself, but there
are in addition some attachments. It is, in fact, the Amsler instru
ment with the arm AO (Fig. 158) made infinitely long so that A,
or its equivalent, moves in a straight line and not along an arc of
a circle.
Referring to Fig. 160 it will be seen that the pointer is con
strained to move along the slot GH.
To use the instrument to find the mean height of a diagram :
Trace the diagram on paper and draw a horizontal line and two
perpendiculars to this base line to touch the extreme points on the
boundary of the figure. Place the paper in such a way that the
base line is parallel to the edge of the clip B and set the clips AE
and CD along the perpendiculars already drawn. Then start from
F, the reading of the instrument being noted, and trace the outline
of the figure until F is again reached. Next move the tracing
point along the vertical through F, *'. e., keep the tracer P against
304
MATHEMATICS FOR ENGINEERS
the clip, until it arrives at M at which stage the instrument records
the initial reading. FM is then the mean height of the diagram.
Fig. 1 60 The Coffin Averager.
If the area of the figure is required, the reading of the instru
ment must be made when the tracingpoint is at F. The outline
305
of the figure is then traced until F is again reached and the reading
is again noted. Then the difference between the two readings is
the area of the figure.
(b) Method of averaging Boundaries. The area of a figure of
the shape bounded by the wavy
line in Fig. 161 being required,
proceed as follows : Draw the
polygon ABCRED so that it shall
occupy the same area as the
original figure, viz. the portions
added are to be equal to those
subtracted, as nearly as can be
estimated.
Then, by joining BD, BE, CE,
etc., the polygon is divided into a
number of triangles and the area
of each is : X base X height.
Therefore draw the necessary per
pendiculars, scale off the lengths of the bases and the heights, and
tabulate as follows :
Fig. 161. Area by Averaging
Boundaries.
Triangle.
Base.
Height.
Sura of Heights.
Area of the two
triangles = J base X
sum of heights.
ABD
BED
BD
BD
AG
EF
 AG + EF
BD(AG+ EF)
CBE
CRE
CE
CE
BM
RN
 BM+RN
iCE (BM+RN)
The triangles are thus grouped in pairs and the area of the
figure is the sum of the quantities shown in the last column.
(c) Method of counting Squares. Draw the figure, whose area
is required, on squared paper, choosing some convenient scales.
Then count the squares, taking all portions of a square greater than
onehalf as one, and neglecting all portions smaller than a half,
square.
If i linear inch represents x units horizontally and y units
vertically and the paper is divided into n squares to the linear
inch : each square is ^ sq. ins., and I sq. in. on the paper represents
xy
xy sq. units of area, so that each square represents ^ sq. units.
x
306 MATHEMATICS FOR ENGINEERS
If the total number of squares = N
Area of figure = : a  sq. units
(d) The Computing Scale is often employed in the drawing
office to find the areas of plots of land. It consists of two main
parts, viz. a slider A, Fig. 162, and a fixed scale C. The slider can
Tracing
Fig. 162. Area by the Computing Scale.
be moved along the slot B by means of the handles D; and it
carries a vertical wire, which is kept tightly in position by means
of screws.
Along the fixed scales are graduations for acres and roods
according to a linear scale of 4 chains to i", and a scale of square
poles, 40 of which make up i rood, is indicated on the slider.
To use the instrument. Rule a sheet of tracingpaper with
a number of parallel lines exactly J" apart, i. e., i chain apart
according to the particular scale chosen. Place the tracingpaper
AREAS OF IRREGULAR CURVED FIGURES 307
over the plot in such a way that the whole width in any one directio
is contained between two of these parallel lines.
Place the slider at the zero mark, and move the whole instrument
bodily until the wire at a cuts off as much from the area as it adds
to it. Next move the slider from left to right until b is reached
Remove the instrument and without altering the position of
the wire, place the scale so that the wire is in the position c : then
run the slider along the slot until the wire arrives at d, and so on.
Take the final reading of the instrument, and this is the total area
of the plot. If the slider reaches the end of the top scale before the
area has been completed, the movement can be reversed, *'. e., it
becomes from right to left and the lower scale must be used'.
It will be observed that by the movement of the slider the
mean widths of the various strips are added. Now the strips are
each I chain deep, so that if the mean lengths of the strips measured
in chains are multiplied by i chain, the total area of the plot is
found in square chains. But 10 sq. chains = i acre, and the
scale to which the plan is drawn is i" = 4 chains. Hence 2j* =
10 chains, and the scale must be so divided that 2\" = i acre,
since the strip depth is i chain.
If the plot is drawn to a scale other than the one for which
the scale is graduated the method of procedure is not altered in
any way, but a certain calculation must be introduced. Thus if the
figure is drawn to a scale of 3 chains to the inch and the computing
scale is graduated according to the scale i" = 4 chains, then the
true area = (f) 2 or T \ of the registered area.
(e) The Trapezoidal Rule. When using this rule divide the
base of the figure into a number of equal parts and erect ordinates
through the points of division. The strips into which the figure
is thus divided are approximately trapezoids. For a figure with a
very irregular outline the ordinates should be drawn much closer
than for one with a smooth outline. Then the area of the figure
is the sum of the areas of the trapezoids, *. e., in Fig. 163,
Area = %!+y 2 ) + % 2 +y 3 ) + ....... i%io+yii)
yn) + y 2 + y 3 +
Or, the area is equal to the length of one division multiplied by
the sum of half the first and last ordinates, together with all the
remaining ordinates.
Example i. Find the area of the figure ABCD in Fig. 163. which is
drawn to the scale of half full size.
308
MATHEMATICS FOR ENGINEERS
The base is divided into 10 equal parts and the ordinates are measured.
Then the calculation for the area is set down thus
y\ = 25
y n = 20
sum of first and last = 225
y 2 = 440
ys = 5 >I0
y\ = 534
y& = 413
y^ = 383
y s = 380
Vg = 3'63
i(yi + yu) = 225
Sum = 3988
Width of one division of the base = h = i"
Area = i x 3988
sq. ins.
At
Fig. 163. Area by Trapezoidal Rule.
(/) The Mid  ordinate rule is very frequently used and is
similar to the trapezoidal rule. The base of the figure is divided
into a number of equal parts or strips, and ordinates are erected
at the middle points of these strips; such ordinates being called
midordinates as distinct from the extreme ordinates through the
actual points of section. The average of the midordinates multi"
plied by the length of the base is the area of the figure.
Example 2. Find the area of the figure ABCD in Fig. 164, which is
an exact copy of Fig. 163, and is drawn to the scale of half full size.
The lengths of the midordinates are 366, 490, 524, 524, 4*4, 392,
3'8, 3'73, 3'3 and 213 ins. respectively, and the average =  = 4032.
AREAS OF IRREGULAR CURVED FIGURES 309
Hence the area = 4032 sq. ins., as against the previous result of
3988 sq. ins., showing a difference of i%.
The midordinate rule is much in vogue on account of its sim
plicity. As a modification of this method we may ascertain the
total area by the addition of the separate strip areas. It is not
necessary to divide the base into equal portions : but the divisions
may be chosen according to the nature of the bounding curve. If
3s
HO I
Al _
Fig. 164. Area by Midordinate Rule.
the latter is pretty regular for a large width of base, the division
may be correspondingly wide; but sudden changes in curvature
demand narrower widths. Assuming that the area has been
divided into strips in the manner suggested, find the lengths of the
midordinates and the widths of the separate strips and tabulate
as in the following example.
Example 3. Calculate the area of the figure ABMP (Fig. 165).
Strip.
Width
(inches).
Length of
Midordinate (ins.).
Area of Strip
(sq. ins.)
Sum of Areas of Strip
(sq. ins.)
AB
18
3'3
5'94
5'94
BC
4
502
2OI
7'95
CD
'3
4'35
I3I
926
DE
5
189
1115
EF
4
385
i'54
1269
FG
GH
i'3
6
3'54
2'53
460
152
1729
1881
HJ
J K
KL
i'3
6
8
36
4'54
3'4
468
272
272
23H9
2621
2893
LM
IO
231
231
3124
Thus the total area = 3124 sq.
MATHEMATICS FOR ENGINEERS
(g) Simpson's Rule is the most accurate of the strip methods
and is scarcely more difficult to remember or more complicated in
its application than the trapezoidal rule.
In this rule, the base must be divided into an even number of
equal divisions ; the ordinates through the points of section being
added in a particular way, viz.
first ordinate + last ordinate +
4 (sum of even ordinates) +
2 (sum of odd ordinates,
excluding the first and last). J
If the portions of the curve joining pairs of ordinates are
straight or parabolic, i. e., if the equations to these portions are of
the form y = a}bx\cx 2 , the ordinates being vertical, the rule gives
rn
Area =
width of one division of base
3
i I i 4
Scale of In
heights and widfhs. 1
Fig. 165. Modification of Midordinate Rule.
3 ins.
perfectly correct results ; and the strip width should be chosen to
approximately satisfy these conditions.
Taking an example
Example 4. Find the area of the indicator diagram shown in
Fig. 1 66.
A convenient horizontal line is selected to serve as a base and, in
this instance, is divided into 10 equal parts. The ordinates are
numbered y\, y z> y t , etc., and their heights are measured, being
those between the boundaries of the figure, and not down to the
base line.
AREAS OF IRREGULAR CURVED FIGURES
The working is set out thus
Width of i division = i foot.
Ordinates.
Even. Odd.
y, = 80 y 3 = 66
?4 = 55'5 y&= 48
y 6 = 43'5 Vi= 385
Vs = 34'5 y= 305
Vw= 24
ist = y l = o
last = y u = o
Sum =o
Sum = 2375
and (Even) 4Xsum = 950 2xsum = 366 (Odd)
.". Area = f[o + 950 + 366} = 439 sq. units, which in this case
would represent to some scale the work done per stroke on the piston.
KO
ho
Fig. 1 66. Area by Simpson's Rule.
Notice that this rule agrees with our notion of
" average height X base " for
Number of ordinates considered = i+i+4(5)+ 2 (4) = 3<>
and if A = sum of ordinates according to the particular scheme
A h
Area = base X average ordinate = ioAx = XA
If the area is of such a character that two divisions of the base
are sufficient
Area = {ist + last + (4Xmid.)}
O
= k^ ( Ist + last + (4Xmid)} since length = 2*
312
MATHEMATICS FOR ENGINEERS
(h) Graphic Integration is a means of summing an area with
the aid of tee and setsquare, by a combination of the principles
of the " addition of strips " and " similar figures." An area in
Fig. 167 is bounded by a curve a'b'z' ', a base line az and two vertical
ordinates aa' and ztf . The base is first divided as in method (/),
where the widths of the strips are taken to suit the changes of
curvature between a' and z' , and are therefore not necessarily equal ;
and midordinates (shown dotted) are erected for every division.
Next the tops of the midordinates are projected horizontally on
Fig. 167. Graphic Integration.
to a vertical line, as BB'. A pole P is now chosen to the left of that
vertical; its distance from it, called the polar distance p being a
round number of horizontal units. The pole is next joined to
each of the projections in turn and parallels are drawn across the
corresponding strips so that a continuous curve results, known as
the Sum Curve. Thus am parallel to PB' is drawn from a, across
the first strip; mn parallel to PC' is drawn from m across the
second strip, and so on.
The ordinate to the sum curve through any point in the base
gives the area under the original or primitive curve from a up to
the point considered.
AREAS OF IRREGULAR CURVED FIGURES 313
Referring to Fig. 167
Area of strip abb' a' = ab x AB
but, by similar figures
B'a or BA _ bm
P ~ ab
whence AB x ab = p x bm
area of strip
i" , . or area of strip = pxbm
i. e., bm measures the area of the first strip to a particular scale,
which depends entirely on the value of p.
T , v / area of second strip
In the same way nm' =
P
and by the construction nm' and bm are added, so that
cn _ area of ist and 2nd strips
P
or area of ist and 2nd strips = p x cn
Thus, summing for the whole area
Area of aa'z*z = pXzL
Thus the scale of area is the old vertical scale multiplied by the polar
distance ; and accordingly the polar distance should be selected in
terms of a number convenient for multiplication.
e. g., if the original scales are
i" = 40 units vertically
and i" = 25 units horizontally
and the polar distance is taken as 2", i.e., 50 horizontal units;
then the new vertical scale
= old vertical scale X polar distance
= 40 X 50 = 2000 units per inch.
If the original scales are given and a particular scale is desired
for the sum curve, then the polar distance must be calculated as
follows
. . ., new vertical scale
Polar distance in horizontal units = old yerticaHclOe"
e. g., if the primitive curve is a " velocitytime " curve plotted
to the scales, i" = 5 ft. per sec. (vertically) and i" = = i sec. (hori
zontally) and the scale of the sum curve, which is a " displacement
time " curve, is required to be i" = 25 ft., then
2*5
Polar distance (in horizontal units) = e = 5
314
MATHEMATICS FOR ENGINEERS
= ! unit along the horizontal, the polar distance
and since I* =
must be made 5".
The great advantages of graphic integration are
(a) Its ease of application and its accuracy.
(b) The whole or part of the area is determined without separate
calculation ; the growth being indicated by the change in the sum
curve. Thus, if the load curve on a beam is known, the sum curve
indicates the shear values, because the shear at any section is the
sum of the loads to the right or left of that section.
Example 5. Draw the sum curve for the curve of acceleration
given in Fig. 168. Find the velocity gained in 20 seconds from rest,
and also in 35 seconds : find also the average acceleration.
Polo,
Scale .of time (sees)
Fig. 1 68. Construction of Sum Curve.
Method of procedure. Project be horizontally to meet the vertical
AB in c. Draw AM parallel to PC to cut the second ordinate in M.
Project de horizontally and draw MN parallel to Pe. Continue the
construction till Z is reached on the last ordinate.
The polar distance was chosen as 3", or 15 horizontal units, so
that, whilst the old vertical scale was i" = 2 units of acceleration, the
sum curve vertical scale (in this case a scale of velocity) will be
i" = 2 X 15 = 30 units. This new scale is indicated on the extreme
right by the title " scale of velocity." Note that Z is at the point
* 3
tt^tttt^ m ^2**~
this being the average height of the
Graphic integration can only be immediately applied when the
base is a straight Hne. If it is otherwise, the figure m
duced to one wrth a straight base by stepping oQ the
re
jjualion :u oxVkx+ocH J
Fig. 169. Comparison of Sum Curves.
with the dividers. Therefore, if the full area is required, as in
the case of an indicator diagram, the additional complication would
neutralize any other gain ; but if separate portions of the area are
wanted the method is the most efficient.
It is of interest to note that if the original curve is a horizontal
line
The first sum curve is a sloping straight line,
The second sum curve is a parabola of the second degree, or a
" square " parabola.
The third sum curve is a parabola of the third degree, or a
" cubic " parabola.
316 MATHEMATICS FOR ENGINEERS
These cases are illustrated in Fig. 169, the poles being chosen
to bring the curves to about equal scales for comparison.
Graphic Integration will be again referred to when dealing with
the Calculus generally.
Calculation of Volumes. All the rules for finding areas can
be extended to the calculation of volumes. The area of the figure
should then represent the volume : e. g.,if the crosssection at various
distances through an irregular solid be noted or estimated, and
ordinates be erected to represent these crosssections at the proper
distances along the base of the diagram ; the area of the figure on
the paper will represent the volume of the solid. Thus
If i" represents x feet of length
and i" represents y sq. ft. of crosssection, then
i sq. in. of area represents xy cu. ft. of volume.
Example 6. Find the capacity of a conical tub of oval cross section,
the axes of the upper oval being 28" and 20", those of the base being
21" and 15", and the height being 12".
In this case the rule for the three sections may be applied; the
axes of the midsection are 24^" and 17^" and the areas of the three
sections are
A = TT X 14 X 10 = 14077 sq. ins.
B = 77 x 105 x 75 = 7 8 '75 7r ..
M = 77 x 1225 x 8*75 = 107277 ,,
Volume = {A + B + 4 M} = ^{140 + 7875 + 4288}
= 277 x 6476 = 4070 cu. ins.
.*. Capacity = 4 ' gallons = 147 gallons.
Other worked Examples on the calculation of volumes will be
found in Chapter VIII.
Exercises 35. On the Areas of Irregular Curved Figures.
1. A gas expands according to the law pv = 150, from volume 3
to volume 25. Find the work done in this expansion.
2. An indicator card for a steam cylinder is divided into 10 equal
parts by 9 vertical ordinates which have the respective values of 100,
84, 63, 50, 42, 36, 32, 28 and 26 Ibs. per sq. in. ; and the extreme
ordinates are 100 and 25 Ibs. per sq. in. respectively. Find the mean
pressure of the steam.
3. The end areas of a prismoid are 628 and 205 sq. ft., the section
midway between is 367 sq. ft. and the length of the prismoid is 15 ft.
Find the average crosssection and the volume.
AREAS OF IRREGULAR CURVED FIGURES 317
4. The midship section of a vessel is given, the height from keel
to deck being 19$ ft. ; and the horizontal widths, at intervals of 321 ft
are respectively 468, 462, 454, 43, 362, 262 and 144 ft., the first
being measured at deck level and the last at the keel. Calculate the
total area of the section.
5. To measure the area of the horizontal water plane, at load line
of a ship, the axial length of the ship was divided into nine abscissa
whose halfordinates from bow to stern were 6, 285, 91, 1554, 18,
187, 1845, I 7' 6 I 5'i3 an d 67 ft. respectively; while the length of
the ship at load line was 270 ft. Find the area of the water plane.
6. The velocity of a moving body at various times is as given in the
table
Time (sees.) .
o
i'5
28
3'6
5
62
77
89
103
12
Velocity (ft."l
per sec.) . J
37'3
315
275
25H
22 4
20'3
182
169
158
15
Find the total distance covered in the period of 12 seconds ('. e.,
find the area under the velocity curve plotted to a time base.)
7. To find the crosssection of a river 90 ft. in breadth, the following
depths, marked y, in feet, were taken across the river ; x, in feet, being
the respective horizontal distances from one bank.
X
o
10
20
30
40
50
60
70
80
90
y
3
4'5
56
6
5'7
48
4'7
4'5
4
3
Find the area of the crosssection. If the average velocity of the
water normal to the crosssection is 51 ft. per sec., find the flow in
cu. ft. per sec.
8. A series of offsets was measured from a straight line to a river
bank. Find, by Simpson's rule, the area between the line and the
river bank.
Offsets (ft.) .
7
9
8
5
2
3
7
9
ii
'.5
20
13
3
Dist. along \
line (ft.) /
o
IOO
200
300
400
450
500
600
700
725
750
775
800
900
1000
9. The mean spherical candlepower (M.S.C.P.) of a lamp can be
determined by calculating the mean height of a Rousseau diagram
(candle power plotted to any linear base). Find the M.S.C.P. for the
arc lamp for which the Rousseau diagram is constructed from the
following figures :
Dist. from one end\
of base (ins.) J
i
V
18
2'5
3'5
42
4'9
5'4
58
6
Candle power .
on5
i
350
650
I IOO
1350
1500
1 200
400
10. Reproduce (a) Fig. 12 to scale and then determine its area.
MATHEMATICS FOR ENGINEERS
11. Fig. 170 is a reproduction of an indicator card taken during a
test on a 10 H.P. Diesel engine. Calculate the mean pressure for this
case, i.e., find the mean height of the diagram.
O
O
O
O
N
O
I I I
CHAPTER VIII
CALCULATION OF EARTHWORK VOLUMES
IN this chapter a series of examples will be worked out to illus
trate the method of calculating volumes of earthwork, such as
railway cuttings, embankments, and other excavation work, mostly
for the purpose of estimating the cost of earth removal.
Definitions of Terms introduced in these Examples.
The formation surface is the surface at the top of an embank
ment or at the bottom of a cutting, and in all the cases here con
sidered it will be regarded as horizontal. The line in which the
formation surface intersects the transverse section of the cutting
or embankment is spoken of as the formation width.
The natural surface of the ground is the surface existing before
the cutting or embankment is commenced.
The sides of a cutting or an embankment slope at an angle which
is less than that of sliding for the particular earth ; and the slope
is usually expressed as x horizontal to one vertical.
A few typical values of the slope are given for various soils :
SOIL.
Compact
Earth.
Gravel.
Dry
Sand
Vege
table
Earth.
Damp
Sand.
Wet
Clay.
ANGLE WITH HORIZONTAL
50
40
38
28
22
16
SLOPE (i. e., x horizontal
to i vertical)
8 39 I
IIQ2
128
188
2475
3487
The unit of volume usually adopted in questions of earth removal
is one cubic yard, and accordingly the weights in the following
table are expressed in terms of that unit :
MATERIAL.
Slate.
Granite.
Sand
stone.
Chalk.
Clay.
Gravel.
Mud.
WEIGHT (cwts. per cu. yd.)
43
42
39
36
31
30
25
Volumes of Prismoidal Solids. To find the volume of any
irregular solid having two parallel faces or ends, find the average
320 MATHEMATICS FOR ENGINEERS
crosssection parallel to these faces and multiply by the axial
distance between them.
+ 4M}
Then
volume = {A
and average section =  {A + B + 4M}
where L = axial length; and A, B, and M are the end sections
and the middle section respectively.
Example i. A solid with vertical sides.
Let the base be horizontal and all the sides be vertical as in
excavating foundations for a house. Referring to Fig. 171
A = (I2 + IO)
X 20 = 220 sq. ft.
_ (4 + 8)
_ (7 + I0 ) x 20 = 170 sq. ft.
L= 50
/. Volume = ^{220 + 120 + ( 4 x 170)} = 8500 cu. ft. = 3148 cu. yds.
Fig. 171. Fig. 172.
Example 2. Calculate the weight of clay removed in making the
simple wedgeshaped excavation shown in Fig. 172.
In this case A = \ x 12 x 84 = 504 sq. ft.
B = ^ x 20 x 140 = 1400 sq. ft.
M = x i6x 112 = 896 sq. ft.
and L = 60 ft.
60
then volume = ^{504 + 1400 + (4 x 896)} = 54880 cu. ft.
= 20326 cu. yds.
and weight of clay removed = 2 32 ' 6 X 3I tons
20
= 3151 tons.
CALCULATION OF EARTHWORK VOLUMES 321
Example 3. A more difficult wedgeshaped excavation, which is
shown in Fig. 173. To calculate the volume of earth removed.
The earth removed is
represented by a wedge
figure ADEF and a tri
angular pyramid AFBC.
The volume of the pyra
mid can be found if the
area of the base is first
obtained.
Axis of
Sloping Ground
whence AD = 6265
sin L BAD = .
6265
but sin L BAC =
sin ( 1 80 BAD)
= sin L BAD
and hence
sin ,1 BAC =7^
6265
Also
AC = 250 6265
= I8 7'35
Fig. 173. Wedgeshaped Excavation.
Then area of triangle ABC = BA . AC sin L BAC
= \ x 78 x 1874 x
60
= 7000 sq. ft.
6265
Height of pyramid = 30 ft.
/. Volume of pyramid = \ x 30 x 7000 = 70000 cu. ft.
= 2592 cu. yds.
For the volume of the prismoidal solid ADEF, using the general
rule
A = \ x 20 x 60 = 600 sq. ft.
B = \ x 30 x 78 1170 sq. ft.
M = \ x 25 x 69 = 8625 an d L = 60 ft.
.*. Volume = ^{600 + 1170 + 3450} = 52200 cu. ft.
= 1935 cu  y ds 
.'. Total volume removed = 4527 cu. yds.
Sections of Cuttings. It will be convenient at this stage to
demonstrate the mode of calculation of the areas of simple sections.
In Fig. 174 we have the first case, of a cutting whose sides are
sloped and whose natural surface of ground DC is horizontal.
Let AB be the base or " formation width " and let its value
be 2.
Y
322
MATHEMATICS FOR ENGINEERS
GH = height from centre of base to the natural surface = h.
6 = inclination to the horizontal of the sloping sides.
EC = horizontal projection of slope.
Then cot is usually denoted by s; or, in other words, the
slope of the sides is s horizontal to i vertical.
EC
FB
= cot 6 = s and FC = FB x s = hs
GC = half width of surface = a\hs.
Area ABCD (/. e., the area of the section of the cutting)
= (DC+AB) xh = (za+2a+2hs}
= h(2a\hs)
)M JO. _ _
Fig. 174.
Fig 175
Fig. 175 shows the cutting section when the natural surface of
the ground takes a slope DC.
Let a. = inclination to the horizontal of the natural surface,
and let cot a = r.
CM and DN, though not equal, are called the " halfwidths "
of the section ; let these be represented by m and n respectively.
To find m and n
Also
From (2)
MG i , .. m
 = tan a = , so that MG = 
m r r
= tan = , so that HK = 
s s
MK i m
 = tan e = , whence MK = 
GK = GH+HK = h+
S
(i)
(2)
(3)
From (i) and (3)
GK = MKMG =   
CALCULATION OF EARTHWORK VOLUMES 323
Hence fcf.5'.. 2_2
s s r
and
Similarly n = ^(a+hs).
To find the extreme heights CE and DF
BE = HEHB = ma
BE
= cot = s, whence CE x s = BE
.*. CExs = ma or CE =
s
72 n
and similarly DF =
To find the area of the section
Area ABCD = CDFE DFACBE
FF /CE+FD\ ! .IBECE
\ 2 J 2 2
if(m\n)(ma\na) (na}(na)
21 S S
_(m a)(m a)}
s
2s\ \2an m 2 a? \2amj
nina z
fm' 2 \n 2 \2mn2am2ann 2 a z
Example 4. A cutting is to be made through ground having a
transverse slope of 5 horizontal to i vertical, and the sides are to
slope at ij horizontal to i vertical. If the formation width is Co ft.
and the height of the cutting (at centre) is 12 ft., find the halfwidths,
the extreme heights and the area of the section.
Adopting the notation as applied to Fig. 175
2a = 60, h = 12, s =' ij, and r = 5
Then m = (a+hs) = ^[30 + (12 x ij)]
= 60 ft.
: ^  (30 + 15)
36 ft.
324
MATHEMATICS FOR ENGINEERS
CE =.
=. ^^ = 6 " 3 =
= 24 ft.
Area =
5 I25
na _ 36 30 = g f
5 125 S
mna 2 (60 x 36)
125
= looSsq. ft.
Example 5. Volume of a cutting having symmetrical sides, the
dimensions being as in Fig. 176.
Calculate the volume of earth removed, if the cutting enters a hill
normally to the slope of the latter and emerges at a vertical wall or
cliff.
Fig. 176. Cutting on a Hill.
The volume is found by application of the general rule.
Volume = ^{A+B+ 4 M}
To find A h = 24, s = i, 2a = 40
/. Area = h(za + hs) = 24(40 + 36) = 1824 sq. ft.
In the case of the other end section, h = o and thus B = o.
For M h = 12, s = i, 20 = 40
Area = 12(40 + 18) = 696 sq. ft.
also L = 170 ft.
170
Hence Volume = ^{1824 + o + 2784} = 130600 cu. ft.
or 4837 cu. yds.
Example 6. To find the volume of a cutting having unequal sides.
In this case, shown at (a), Fig. 177, the cutting enters the hill in
an oblique direction, although the outcrop is vertical as before. The
CALCULATION OF EARTHWORK VOLUMES 325
sides of the cutting slope at i horizontal to i vertical, while the natural
surface of the ground slopes upward at 4^ horizontal to i vertical.
The solid can be split up into a prismoidal solid SRFE, together
with the two pyramids SE and RF.
To deal first with the prismoidal solid SRFE : its volume can be
found from the general rule Volume = ^{A + B + 4M}, and in order
to find the values of A and B the lengths of BS and AR must first
be found.
Fig. 177. Cutting with Unequal Sides.
Referring to (b), Fig. 177
CQ = m a, and m = (a+hs)
Y 5
Also
so that
Hence
Now
Also
Again
and
Hence
r =
s =
h = 30, and 2a = 40
m = ^(20+45) = 97'5 ft 
CQ = 975 20 = 77'5 f t
S
= tan a =
4'5
SO = = 5
4'5 4'5
= I7 . 22 f t .
BQ = 22^5 = 51 . 66 ft<
* i5 3
BS = BQ  SQ = 5166  1722 = 3444 ft.
v di
n = ~(a+hs) = ^(20+45) = 4875 ft.
f~T~S D
PD = 4875  20 = 2875 ft 
^ = , whence PR = = 639
PD 45 4*5
AP = PD tan 6 = *^p 1917 ft 
AR = AP + PR = 1917 + 6 '39 = 2556 ft
326 MATHEMATICS FOR ENGINEERS
We can now proceed to find the volume of the solid SRFE.
A = xBSxBE = 1x3444x150 = 2583 sq. ft.
B = xARxAF = 1x2556x100 = 1278 sq. ft.
M = X3OXI25 = 1875 sq. ft.
and L = 40
.'. Volume = ^{2583+1278+7500}
= 75750 cu. ft. = 2805 cu. yds.
To find the volume of the pyramid SE
The base is the triangle CSB, of which the area = JxBSxCQ
= *x 3444x775
= 1335 sq. ft.
The height = 150 ft., and hence
Volume =  x *~ x 1335 cu. yds. = 2471 cu. yds.
To find the volume of the pyramid RF
The base is the triangle ARD; and its area = xARxPD
= 2X2556x2875
= 3675 s q ft 
The height = 100 ft., and hence
Volume = x X 3675 cu. yds. = 4536 cu. yds.
3 /
.*. Total volume = 2805 + 2471 + 4536 = 5730 cu. yds.
Cutting and Embankment continuously combined ; the
Sides being Symmetrical. If a road or a railway track has
to be constructed through undulating ground, both cuttings and
embankments may be necessary. The cost of the roadmaking
depends to a large extent on the " net " weight of earth removed,
seeing that the earth may be transferred from the cutting to the
embankment. The calculation of the net volume removed will be
dealt with according to two methods :
First Method.
Example 7. A cutting is to be made through the hill AC (Fig. 1 78)
24^' and an embankment in the
j^6QL94OH i050^772V584'f. <* valley BC so as to give
^^~ JL^^^ a straight horizontal road
t\S ? g *""JC ic el JR fromAtoB. The formation
* "V. ~ ^ 7
^ V '"* s jb__JJ<f width is to be 40 ft., and the
>i i* sides of the cutting and the
Fig. 178. embankment slope if hori
zontal to i vertical. Calculate the net weight of vegetable earth
removed (25 cwt. per cu. yd.).
The volume of the cutting will be found by considering it made up
of three prismoidal solids, and the volume of the embankment will be
CALCULATION OF EARTHWORK VOLUMES 327
found in the same way. Then the net volume is the difference of
these separate volumes.
Dealing with the portion between A and C, i. e., with the cutting :
For the portion Aae
A = o
B = i8[40 + (18 x 1 1)] = 1287, since h = 18
M = 9[40 + (9 X 1 1)] = 50175, since h = 9
L = 560
/. Volume = =^{o + 1287 + (4 x 50175)} = 307400 cu. ft.
For the portion abfe
A = 1287
B = 15^0 + (15 X if)] = 994. since h = X 5
M= i65[40+ (165 X if)] = 1136, since h = 165
L = 940
.'. Volume = ^{1287 + 994 + (4 X 1136)}
= 1,069,000 cu. ft.
For the portion fbC
A = 994
B = o
M= 7  5 [ 4 o+ (75 x 1 1)] = 398
L = 1050
/. Volume = 6 5 {994 + o + (4 X 398)} = 453oo cu. ft.
Thus the total volume removed to make the cutting
= 307400 + 1,069,000 + 453000 = 1,829,400 cu. ft.
Dealing with the embankment portion, viz. that from B to C : .
For the solid Cch
A = o
B = 17^0 + (17 x i)l = Il8 5
M = 8 5 [ 4 o + (85 x if)] = 466
L = 772
.. Volume = 2?{o + 1185 + (4 x 466)} = 392000 cu. ft.
For the solid chid
A =1185
6=1185 L = 584
M=n85
.. Volume = 1185 x 584 = 692000 cu. ft.
For the solid dl~B
A =1185
B = o
M=85[40 + (85 X if)] =466
L = 246
/. Volume = ^{1185 + o + (4 x 466)} = 125000 cu. ft.
328
MATHEMATICS FOR ENGINEERS
Hence the total volume required for the embankment
_ ( 3g2 __ 692 + 125) X I0 3 CU. ft. = 1,209,000 CU. ft.
Then net volume removed = (1829 1209) x io 6
= 620000 cu. ft. or 22960 cu. yds.
22960 x 25 ,
and the net weight removed =  tons
= 28700 tons
Second Method.
Example 8. Fig. 1 79 shows the longitudinal section of some rough
ground through which the road AC is to be cut. The sides of the
cutting and of the embankment slope at ij horizontal to i vertical,
Lonqiludinol Section
Fig. 179. Volume of Earth removed in making Road.
and the road is to be 50 ft. wide. Calculate the net volume of earth
removed in the making of the road.
Divide the length AC into ten equal distances and erect mid
ordinates as shown. Scale off the lengths of these, which are the
heights of the various sections.
The areas of the sections at a, b, c, d, etc., can be found by cal
culation as before, or, if very great accuracy is not desired, the various
sections may be drawn to scale and the areas thus determined. To
illustrate the latter method : Draw DE = 50 ft., and also the lines
DF and EG, having the required slope, viz. ij to i. Through R,
the middle point of DE, erect a perpendicular RS, and along it mark
distances like RM, RN, etc., to represent the respective heights of the
sections : thus RM = 24, and RN = 48. Then to find the area of
the section at a, which is really the figure DPQE, add the length of
PQ to that of DE and multiply half the sum by RM. The area of
the section at 6 is % (TV + DE) x RN, and so on. The areas of the
CALCULATION OF EARTHWORK VOLUMES 329
respective sections are 2064, 5856, 4746, 1826, and 650 sq. ft., these
being reckoned as positive ; and 650, 3744, 3444, 1386, and 496 sq. ft.,
these being regarded as negative. The average of all these sections,
added according to sign, is 5422 sq. ft. or 6024 sq. yds. Then the net
volume of earth removed = 6024 x 1000 = 602400 cu. yds.
Cutting with Unequal Sides, in Varying Ground.
First Method.
To find the average crosssection of ground with twisted surface
(Fig. 180) ; an end view being shown in Fig. 181.
The surface slopes downwards to the left at A and to the right
at B.
Fig. 180.
End View.
Fig. 181.
Let WL n lt h lt and r t be the halfwidths, etc., for A; and
w 2 , 2 , h. z , and r 2 the corresponding values for B.
Then Area of A =
For the midsection M
m =
,
and 3 =
33 
and the area of M =  "
/. Average crosssection
A+B+4M
6
_ m 1 n l
6s
6a 2
6s
6s
330
MATHEMATICS FOR ENGINEERS
Example 9. Find the average crosssection of ground with twisted
surface, when the formation width is 20 ft. and the sideslopes are
i horizontal to i vertical. At the one end of the embankment the
height is 12 ft. and the natural surface of the ground slopes at 20
horizontal to i vertical downwards to the right; while at the other
end the height is 6 ft. and the slope of the ground is 10 to i down
wards to the left.
Adhering to the notation employed in the general description ;
For the section A
{10 + (12 X 15)} = 303 ft.
{10 + 18} = 2605 ft.
20 15
2O
For the section B
10
+ (6 x 15)} = 165 ft.
 10  i
* = , I o+ I . 5 ( I + 9) = 224 ft.
Hence the average crosssection
_ (303 x 26) + (165 x 224) + (468 x 484) 600
9
= 3136 sq. yds.
Second Method.
Example 10. Calculate the volume of earth removed in making a
cutting of which AE is a longitudinal centre section (Fig. 182). The
formation width is 20 ft., the length of the cutting is 4 chains, the
Fig. 182.
2O 1
Section oT B.
Fig. 183.
sections are equally spaced, and the slope of the sides is 2 horizontal
to i vertical. All the sections slope downwards to the left, as indicated
in Fig. 183.
The heights of the sections, in feet above datum level, are :
Section.
Left.
Centre.
Right.
A
O
O
O
B
26
46
66
C
4'i
64
87
D
40
62
84
E
o
o
o
The areas of the sections may be found by drawing to scale and
CALCULATION OF EARTHWORK VOLUMES 331
then using the planimeter; and Simpson's rule can afterwards be
employed, since there are an odd number of sections.
The results in this case are as follows :
Section.
M
Area
A
10
10
O
B
32
137
1695
C
422
157
280
D
399
1555
260
E
10
10
O
Then the volume = {o + o + 4(1695 + 260) + 2(280)}
= 50100 cu. ft. or 1855 cu. yds.
Surface Areas for Cuttings and Embankments.
The area of land required for a cutting or an embankment can
be determined when the halfwidths of the various transverse
sections are known ; the method of procedure being detailed in
the following example :
Example n. Fig. 184 represents the horizontal projection of the
cutting dealt with in
Example 10. Find the
area of land required
for this cutting if a
space of 5 ft. between
the outcrops and a
fence be allowed.
The width RM is
the extreme width of
the section, i. e., its
value = m + n; accord
ingly, allowing 5 ft.
on each side, the widths
to be considered are of
the form m + n + 10.
Taking the values of
m and n as in the previous example, the widths are as in the
table :
Fig. 184. Surface Area for Cutting
Section.
A
B
c
D
E
m + n + 10
30
557
678
65H5
30
332
MATHEMATICS FOR ENGINEERS
Applying Simpson's rule
Area of land required = {30 + 3 + 4 (55 '7 + 6 5'45) + (2 x 678)}
= 14960 sq. ft. or 1663 sq. yds.
Volumes of Reservoirs.
Example 12. Find the volume of water in the reservoir formed
as shown in Fig. 185, when the water stands at a level of 45 ft. above
datum level, the bottom of the reservoir being at the level 22 ft.
Fig. 185. Volume of Reservoir.
In the diagram the land is shown contoured, i. e., the line marked
40, for example, joins all points having the level 40 ft. above datum.
The problem, then, is to find the volume of an irregular solid, and
this may be done in either of two ways, viz.
(a) By taking vertical sections. According to this method, we should
find the extreme length of the reservoir, which is about 320 ft., and
then draw the crosssections at intervals of, say, 40 ft. The area of
each crosssection would then be found, preferably by the planimeter,
and the volume calculated by adding the areas according to Simpson's
rule.
This process is somewhat tedious, as each section must be plotted
separately; and consequently it is better to proceed as in method (6).
(b) By taking horizontal sections, i. e., sections at heights of 45, 40,
35, etc., ft. respectively.
To find the area of the section at the height 45 ft., determine the
area of the figure ABCD by means of the planimeter. This area is
found to be 5083 sq. ins. Now the linear scale is i" = 80 ft., and
therefore each square inch of area on the paper represents 80 x 80 or
CALCULATION OF EARTHWORK VOLUMES 333
6400 sq. ft. Thus the area of the section at the level of 45 ft.
= 5' 8 3 x 6400 = 32500 sq. ft. ; and in the same way the areas at
the levels 40, 35, 30, 25, and 22 ft. are 21550, 10560, 3780, 577, and
o sq. ft. respectively. The length of the irregular solid is 23 ft.,
i. e., 45 22, and we may plot the various areas to a base of length'.
24000 
I6ooo 
8000
as indicated in Fig. 186. The area of the figure EFG, which is found
to be 1633, gives the volume of water in the reservoir, to some scale.
In the actual drawing i" = 10 ft. (horizontally), and i" = 16000 sq. ft.
(vertically), so that i sq. in. on the paper represents 10 x 16000
or 160000 cu. ft.
Hence volume of the reservoir = 160000 x 1633 = 261300 cu. ft.
or its capacity = 1630000 gallons.
Exercises 36. On the Calculation of Volumes and Weights of
Earthwork.
1. Calculate the volume of the solid with vertical sides shown in
Fig. 187.
Fig. 187.
334
MATHEMATICS FOR ENGINEERS
2. Fig. 188 shows the plan of a wedgeshaped excavation, where
the encircled figures indicate heights. Calculate the weight of clay
removed in making the excavation.
3. Fig. 189 is the longitudinal section of some rough ground through
which a straight horizontal road is to be cut, the width of the road
being 64 ft. The soil is vegetable earth (25 cwts. per cu. yd.), and
Af
ROAD
 ZOOOyds.
Fig. 189.
the sides of the cutting and embankment slope at 2 horizontal to I
vertical. Calculate the weight of earth removed in making the road,
if the natural surface of the ground is horizontal.
4. Determine the area of land required for making the cutting
from A to B in Fig. 189. The sideslopes are 2 horizontal to I vertical,
the formation width is 64 ft., and a fence is to be built round the
working at a distance of 6 ft. from the outcrops.
5. Calculate the capacity of a reservoir for which the horizontal
sections at various heights have the values in the following table :
Height above sea level (ft.)
180
170
1 60
155
IS
147
Area of section in sq. ft.
47200
31000
21700
19000
11300
o
6. The depth of a cutting at a point on the centre line is 20 ft.,
the width of the base being 30 ft. The slope of the bank is i hori
zontal to i vertical, and the sidelong slope of the ground is 12 horizontal
to i vertical. Find the horizontal distances from the vertical centre
plane to the top of each slope.
7. Find the volume of earth removed from a cutting, if the forma
tion width is 20 ft., the sideslopes are i to i, and the slope of the
surface is 10 to i. The depth of the cutting at the first point is 25 ft. ;
at the end of the cutting (200 ft. long) it is 30 ft. ; and halfway between
it is 26 ft.
8. The base of a railway cutting is 32 ft. in width, the depth of the
formation is 34 ft. below the centre line of the railway, the sideslopes
are ij to i, and the surface of the ground falls i in 8. Calculate the
halfbreadths for the cutting.
At a distance of i chain along the centre line the depth of forma
tion level is 28 ft., and at a distance of 2 chains it is 20 ft. Find the
volume of earth to be removed.
9. On the centre line of a railway running due N. the difference in
level between the natural ground and the formation level of the em
bankment is 56 ft., 84 ft. and 6 ft. at the 23rd, 24th, and 25th chain
pegs respectively. The width of the formation level is 20 ft., and the
sides of the embankment slope at 2 to i.
CALCULATION OF EARTHWORK VOLUMES 335
The natural ground slopes down across the railway from E. to W.
at i in 10. Determine at each chain peg the distances of the toes of
the embankment from the centre line and the area of the crosssection ;
determine also the volume of the embankment between the 23rd and
25th chain pegs.
10. A cutting runs due E. and W. through ground sloping N. and S.
The formation level is 15 ft. below the surface centre line and is 20 ft.
wide. The ground slopes upwards on the north side of the centre
line i vertical to 6 horizontal, and on the south side the ground slopes
downwards i vertical to 10 horizontal. The sides of the cutting slope
I vertical to i horizontal. Calculate the positions of the outcrops.
CHAPTER IX
THE PLOTTING OF DIFFICULT CURVE
EQUATIONS
Plotting of Curves of the Type y = ax". The plotting in
Chapter IV was of a rather elementary character in that integral
powers only of the quantities concerned were introduced. All
calculations could there be performed on the ordinary slide rule;
e. g., such curves as that representing y = 5# 2 + 7^
possible. If, now, a formula occurs in
were
which one, say, of the
quantities is raised to
a fractional or negative
power, and a curve is
required to represent the
connection between the
two quantities for all
values within a given
range, the necessary cal
culations must be made by
the aid of logs. Suitable
substitutions will in some
cases make these calcu
lations simpler, but unless
great care is exercised
over the arrangement
of the calculations and
the selection of suitable
values for the quantities,
Fig. 190. Curve of y = 22* 1  78 . .  1
a great deal of time will
be wasted. In fact, the method of tabulating values is of more
importance than is the actual plotting.
Example i. To plot the curve y = 22A 1 ' 75 , values of x ranging
from o to 4.
y 2*2#^* 7 ^
/. log y = log 22 + 175 log x.
Arrange a table according to the following plan : In the first
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 337
column write the selected values of x; in the second column write
the values of log x. With one setting of the slide rule the values of
I 75 log x can b 6 read off; and these must be written in the third
column. In the fourth column we must write the values of log y.
which are obtained by addition ; then the antilogs of the figures in
column 4 will be the values of y in column 5.
The advantage of working with columns rather than with lines is
seen ; thus we write down all the values of log x before any figure is
written in the third column, and this saves needless turning over of
pages, etc.
Table :
X
log AT
i75 log * + log 22
logy
y
00
00 + 3424
00
5
1699 = '3 01
 527 + 3424
18154
654
io
o
o + 3424
3424
22
1 '5
1761
308 + 3424
6504
447
2O
3010
527 + 3424
8694
740
2'5
3979
696 + 3424
10384
1092
30
4771
835 + 3424
11774
1504
3'5
'544 1
952 + 3424
12944
197
40
0021
1054 + "3424
13964
2491
The plotting is shown in Fig. 190.
Use of the Log Log Scale on the Slide Rule. The use of
the log log scale now placed on some slide rules would obviate a
great amount of the calculation in this and similar examples.
e. g., taking y = 22* 1 ' 75 and disregarding the factor 22 until the
end
logy = i 75 log x
log (log y) = log 175 + log (log x)
i.e., log Y = log i 75 + log X
where Y = log y, and X = log x.
Therefore if a length on the ordinary log scale, say the C scale, be
added to a length on the log log scale, which is usually the extreme
scale, the result on the log log scale will be that required.
jf y = ^175 ; an d supposing the value of y is required when x = 25.
Set the index of the C scale level with 25 on the log log scale ;
move the cursor until over the power, 175 on the C scale : then the
reading on the log log scale (496) is the value of 25"". Multip icatwn
by 22 for y = 22*, can be done with one setting of the rule afte
338
MATHEMATICS FOR ENGINEERS
X
% vn
y = 22* 1  75
2'5
496
1092
all the powers have been found. The tabulation would in this case
reduce to
as an example.
The log log scale is most useful for finding roots.
E. g., to find >/432. Set 5 on the C scale level with 432 on the
log log scale ; then the reading on the log log scale opposite the index
of the C scale is 337, i. e., the 5th root of 432.
Expansion Curves for Gases. The formula PV" = C, for
the expansion or compression curves of gases, is of the same type
as that in the last example.
600
10., , p ~ r 2O
Values of u.
Fig. 191. Expansion Curves for Gases.
25
30
In this formula p is the pressure in Ibs. per sq. in. or per sq. ft.
and v is the " specific volume," i. e., volume in cu. ft. of i Ib.
whilst n and C are constants varying with the conditions
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 339
Thus for air expanding adiabatically, . <?., without loss or gain
of heat, n = 141 : for the gas in the cylinder of a gas engine
n = J> 37 : f r isothermal expansion, i. e., expansion at constant
temperature, n = i. It is instructive to plot two or three expan
sion curves on the same diagram, n alone varying, and thus to
note the effect of this change.
Example 2. Plot, on the same diagram and to the same scales,
from v = 4 to v = 30, the curves representing the equations : (a)
PV*" = 2500, (6) pv = 2500, (c) pv = 2500. The plotting is shown
in Fig. 191.
Each equation is of the form pv n = C
log p + n log v = log C
or log p = log C n log v.
Dealing with the separate cases
(a) Adiabatic expansion of air; n = 141
log p = log 2500 141 log v.
The arrangement of the table is as follows :
9
logy
log 2500 141 log v
log/>
P
4
602
3.398 847
255 1
356
7
845
1191
2207
161
10
o
141
1988
973
14
146
1615
1783
607
18
255
1770
1628
425
20
301
1834
1564
366
24
380
1945
1453
284
27
43i
2 O2
1378
239
30
477
2O82
1316
207
(b) Expansion of superheated steam; n = 13.
Values of v and log v are as above ; and the table is completed as
shown :
log 2500
13 logu
log/)
P
3398 
783
2615
412
1097
2301
200
3
2098
125
49
1908
809
632
1766
583
691
1707
5 '9
794
1604
402
860
1538
34'5
920
1478
301
340 MATHEMATICS FOR ENGINEERS
(c) Isothermal expansion ; n = i .
V
4
7
10
M
18
20
24
27
30
 2 5
625
357
250
179
139
125
IO4
926
833
P V
It will be seen that the bigger the value of n, the steeper is the
curve, or, in other words, the slope of the curve depends on n.
All these curves are hyperbolas, that for (c) being the special case
of the rectangular hyperbola.
A Construction for drawing Curves of the Type pv" = C.
In this construction the position of one point on the curve must
be known.
LetP, (piVj, be the
given point. Fig. 192.
Choose any angle
o, say 30 : set it off
as shown, and also
an angle ft, calculated
from the equation
tan ft = i (i tan a)r.
Draw the horizon
tal PM to meet OA in
M, and the vertical PR
to meet OR in R.
Draw MN making 45
with ON and RS
making 45 with OR.
A horizontal through
N meets a vertical
through S in Q ; then
Q is another point on
Fig. 192. Construction for Curves of the
pv n = C type.
the curve : also the construction for the point L on the other side
of P is indicated.
ProoJ oj the Method
MT TN ^ ^ ^
tan a = OT = OT =
tan 
SW
RW
Pi
v 9 v
but
OW ' " OW "
tan/3 = I (i tan a)
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 341
=*
*>2 W Pi
or p^f = p z vf, i. e., pv* = Constant.
Example 3. If n = 9 and a = 30, calculate the value of /3.
tan /3 = i (i  tan 30)^ = i(i .5774)1111
= I(4226) l ' ul
Let x = (4226) 1 ' 111
then log* = i i 1 1 x log 4226 = iin x 16259
= i i 1 1 + 696
= 1585
whence x = 3846
Then tan j3 = i  3846 = 6154 = tan 31 36'
or = 31 36'.
If n = i, then tan ft = tan a
i. e., ft = a.
Note. 30 is rather a large angle for a if the range to be covered
is small. Accordingly, the value of /? is stated here, for a = 10
and n = 137.
tan/3 = i (i tan io)^7 = i (i 1763)' 73
= i 871 = 129
.'. ft = 7 8 ai'.
Example 4. A tube 3* internal and 8* external diameter is sub
jected to a collapsing pressure of 5 tons per sq. in. : show by curves
the radial and circular stresses everywhere, it being given that at a
point r ins. from the axis of the cylinder
T? T^
The radial stress p = A + , and the circular stress q = A ,
Note that p = 5 tons per sq. in. when r = 4"; and p = o when
r = i 5* ; and the object is to first find the values of the constants
A and B from the data given.
From the given conditions
A+ B
225
Subtracting 5 = B (^5
= B (0625 4444)
5 = 382B
or B ='3x.
342
MATHEMATICS FOR ENGINEERS
Also
Hence
5 = A + (0625 x  131) = A 818
A = 5818.
p = ">8i8 5. a = 5818 f i
f */ y ' 3 J yZ
[Note that (p + q) = 11636 = constant. The material is sub
jected to crushing stresses p and q in two directions at right angles to
one another and in the plane of the paper : therefore dimensions at
z:
15
35
Fig 193 Curves of Radial and Hoop Stresses.
right angles to the paper must elongate by an amount proportional to
(P + q) If the crosssection is to remain plane this elongation must
be constant ; hence (p + q) must also be constant.]
To calculate values of p and q the table would be set out as
follows :
r
"
131
I3 .!
Sifi I3>1 *
r*
J r* ~ g
5 OI r i P
15
225
5818
11636
2O
4
3275
9093
2'543
25
625
2095
79I3
3'O
9
1455
7273
4363
3'5
40
1225
16
1068
818
6886
6636
4750
5
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 343
It is customary, how
Circular
v/ on
Hoop Stress
The curves are shown plotted in Fig. 193.
ever, to plot the curves of
radial and hoop stress in the
manner shown in Fig. 194,
where curve (i) gives the
radial stress at any point
between a and b, and curve
(2) gives the circular or hoop
stress at any point between
a l and b t .
Example 5. According
to a certain scheme (refer
to p. 212), the depreciation
fund in connection with a
machine can be expressed
by
Fig. 194. Curves of Radial and Koop
where Stresses.
D = amount contributed yearly to the sinking fund, and
ioor = percentage rate of interest allowed on same.
For a machine whose initial value is 500 and scrap value is 80,
D is found to be 14 145., if 3% interest per annum be allowed. If
the life of the machine is 21 years, plot a curve to show the state of
the sinking fund at any time, i. e., plot the curve
A = 2 {io3 i}, n varying from o to 21.
wo
A
Si
350
/
/
280
<>
y
<&
<
><
&

^210
&]
<n
u
A
S
A
r
~JC\
/
~
A
~
O
^
.
.
i
01 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 '9 2O 21
Values of n
Fig. 195. Curve of Depreciation Fund for Machine.
344
MATHEMATICS FOR ENGINEERS
It will be advisable to work out i 03" separately.
Let 103" = x; then log* = wlog 103 = oi28
also
147
^ = 490.
03
Taking a few values only for n, between o and 21, the tabulation
will be as follows :
n
oi28n = log x
X
X I
!*(,_,).,. A
03
o
I
o
4
0512
1126
126
617
8
1024
1266
266
130
JO
128
1343
343
1 68
12
1536
1424
424
2075
16
2048
1603
603
296
21
2688
1857
857
420
and the plotting is shown in Fig. 195.
Equations to the Conic Sections. A knowledge of the form of
the curve that represents some particular type of equation may ensure
a great saving of time and thought. Values for the variables need
not then be chosen at random and beyond the range of the curves.
The equations to the conic sections are here given because
many of the curves occurring in practice are of one of these forms.
The Ellipse. If the origin be taken at the centre of the
ellipse, the equation is
where a and 6 are the halfmajor and halfminor axes respectively,
or the maximum values of x and y.
If the equation is given in a slightly different form, it should
be put into the standard form before any values are selected.
Example 6. Plot the curve representing the equation 3# 2 + $y z = 60
3* 2 + 5y* = 60 is the equation of an ellipse, and can be written
60 60 "~
i. e., the equation is divided throughout by 60, so that the righthand
side becomes unity.
Thus
2O
so that
12
and
and
a z = 20, and a = 4472
fc 2 = 12, and b = 3464.
Hence the range of x is from 4472 to +4472, and no lower
or higher values respectively should be taken.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 345
If the values of y are to be calculated, we have
5y 2 = 60 3# 2
y* = 12 6x*
y= Vi2 6# a
Dealing only with onehalf of the ellipse, the table of values reads
X
X 2
12 6# 2
y2
y
o
12 O
12
3464
I
I
12 '6
n4
338
2
4
12 24
96
310
3
9
12  5'4
66
255
4
16
12 96
24
J 547
4H72
20
12 12
o
o .
The other half can be obtained by projection, and Fig. 196 is
plotted. If the graphic method of drawing an ellipse is known this
Fig. 196. Curve of Equation to Ellipse.
calculation is unnecessary : all that is required from the equation
being the lengths of the axes.
An application of the ellipse is found in the Ellipse of Stress,
in the subject of Strengths of Materials. It is required to deter
mine the magnitude and direction of the resultant stress on a
346
MATHEMATICS FOR ENGINEERS
plane BD, due to the stresses /j and / 2 acting as indicated in
Fig 197
It is found that the resultant stress / = \//! 2 cos 2 + / 2 2 sin 2 0,
and if a is the angle made with/!
tan a = Y tan #
/i
If an ellipse be constructed with axes to represent the original
stresses, the resultant stress can very easily be read from it.
Along OQ in Fig. 198 and perpendicular to BD, mark off a
length OQ to represent /j, and a length OR to represent / 2 . Draw
Fig. 197
Ellipse of Stress.
a horizontal QM to meet a vertical PR in P; then OP represents
/and ^.MOP = a.
To show that P lies on an ellipse, we must prove that the equa
yv2 /y2
tion governing P's position is of the nature ^ f j z =i.
OM = OQ cos =f 1 cos 6
MP = RN = OR sin 6 =/ 2 sin
:. (PO) 2 = (OM) 2 + (MP) 2 = /! 2 cos 2 +/ 2 sin 2 = f
i. e., OP =/
If the origin is at O, and x and y are the coordinates of P
then x = MP =/ 2 sin 0, and y = OM = / x cos e
X V
y = sin 6, fss cos 0.
/2 /I
but
sin 2 6 + cos 2 = i for all values of
x 2 v 2
4 < T
f 2 ~ / 2 *
h i
or P lies on an ellipse the lengths of whose axes are 2/ 2 and 2/ x .
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 347
The circle may be regarded as a special case of the ellipse,
x z v 2
where a = b, i. e., ^ + ^ = * or x z + y 2 = a 2 , a being the radius
of the circle.
e. g., 5*.+ 5 y 2 = 45
can be written x* f y z = g,
which represents a circle of radius 3 units.
The Parabola. If the axis is horizontal, and the vertex at
the origin, then the equation is y 2 = 40*.
If the axis is vertical, the equation is x z = qay, where 40 =
length of the " latus rectum," the chord through the focus
perpendicular to the axis.
To make the investigation more general, let x be changed to
x\ c = say, x\j: and y to y+c x = say, y+n45; also let
4 = 2.
Then the case will be that of the parabola having a latus rectum
of 2, and the axis will be vertical, with the vertex at the point
7, 1145
The equation is
(*+7) 2 =
5* 2 +7*+245H'45 = y
or y = 5* 2 +7*~ 9
(This curve is shown plotted in Fig. 88.)
Conversely, the equation y = $x z + jx 9 might be put into
the standard form, thus
y = 5(*+i4*i8)
= 5(*+7) 2 '45
= (*+7) 2
which equation is of the form 4#Y = X a
where 4* = T> Y = yHMS. and X = *+7
This analysis is useful if the position of the vertex, say, is
desired and the curve itself is not needed. (Compare maximum
and minimum values.)
For the parabolas occurring in practical problems the simpler
forms are sufficient.
MATHEMATICS FOR ENGINEERS
The Hyperbola. If the centre of the hyperbola is at the
origin, the equation is
^ 2 _y 2 _
a 2 b 2 ~
where 20, = the length of the transverse axis (along the x axis)
2b = the length of the conjugate axis (along the y axis).
No values should be taken for x between a and \a, for
there is no part of the curve there.
Example 7. Plot the curve representing the equation
2# 2  5V 2 = 4 8  (Fig I99)
By dividing throughout by 48 the equation may be written
so that a = ^24 = 49
and & = Vg6 = 31
If a rectangle be constructed by verticals through x = 49 and
+ 49, and horizontals through y = 31 and + 31, the diagonals of
this rectangle will be the " asymptotes " of the hyperbola, i. e., the
boundaries of the curves are known.
To calculate values
 5 y 2 = 482*2
5 y z = 2*248
y 2 = 4# 2 96
y = V'4# 2 96
i. e., an expression is found for y in terms of x,
The table of values reads :
X
x z
4*2 96
y z
y
4'9
24
96 96
o
o
5'0
25
10 96
4
632
5'5
303
I2'I 96
25
158
60
36
144  96
48
219
This is the calculation for one branch of the curve only ; and the
other branch may be obtained by writing x for + x throughout,
e. g., when x = 5, y = 632 ; therefore project across.
_. _ T
a 2 a 2
If a = b, then
r a 2
or * 2 y 2 = a 2 .
For this case the asymptotes are at right angles, and the hyperbola
is rectangular.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 349
To find the equation oj the hyperbola when rejerred to the
asymptotes as axes (see Fig. 199.) From P a point on the curve,
draw PN parallel to OF and PM parallel to OE. Let PM = p,
/
\
\
\
y
\
i
X 1 
<* A
\
Fig. 199. The Hyperbola.
and PN = q ; then the coordinates, when the asymptotes are axes,
are (p, q}. Note that PN and PM are parallel to the asymptotes,
and not perpendicular to them.
Let L EOA = a ; then tan a = 
i.e.,
also
COS a =
and sin a =
OM = NP = q.
QM = OM = ML = q (From equality of angles.)
PL = PMML = pq
PQ = PM+MQ = p+q
PR h . V &
PL
= =
also
P9 = \
or
QS .
i.e.,
Va*+b*
(2)
350 MATHEMATICS FOR ENGINEERS
By subtracting (i) from (2)
= a?+b z ..... since =  2 = I
a* o*
But a and b are constants, therefore the product of the co
ordinates p and q is constant : this is a most important relation.
If the hyperbola is rectangular, b = a (the asymptotes being at
right angles)
a 8
and pa =
2
Compare the equation PV = C, for the isothermal expansion of
a gas.
Example 8. Find the equation of the hyperbola x 3 $y* = 3,
referred to its asymptotes. Answer : Pq = i.
Exercises 37. On the plotting of Equations of the Type y=ax" + b.
1. Plot, for values of x ranging from i to 9, the curve y = 5j6x 1 ' 29 .
2. Plot the curve zy = od^x"" from x = o to x = 2.
3. Plot on the same axes the curves y 1 =^2x 1 ' 63 and y t = ^i* 3 ' 47
and by adding corresponding ordinates obtain the curve
y = 42* 1 ' 83 + '3i.* 3 ' 47 . {x to range from 2 to 35.}
4. Plot, from y = 5 to y = +'5, a curve to give values of C,
when C = 169 (log, 3  j )
5. Formulae given for High Dams are as follows :
where x = depth in feet of a given point from the top
y horizontal distance in feet from such point to flank of dam
z = horizontal distance in feet from such a point to face of dam
P = safe pressure in tons per sq. ft. on the masonry
Draw the section of a dam 30 ft. deep, allowing P = 45.
6. For a steam engine, if x = mean pressure (absolute) expressed
as a percentage of the initial pressure (absolute), and y = cutoff
expressed as a percentage of the stroke, then
* = y(56o5iogy)
Plot a curve giving values of x for values of y between o and 70.
7. If a number of observations have been made, say, for a length
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 351
of a chain line in a survey, then the probable error e of the mean of
the observations can be calculated from
where r difference between any observation and the mean observa
tion and n = number of observations. If
2 f 2 _. j. 2> p^t a curve to give values of e
for values of n between 2 and 30. ///^7>v^A Worm
8. Plot on the same axes the curves
(a) pv 1 ' 13 = 4000 and (6) pv' 8 = 2540,
v ranging from 4 to 32.
9. In Fig. 200, T^TrA Worm Wheel.
D = the outside diameter of a worm wheel
= 2A( i cos j + d.
If d = 4 and A = 75, show by a graph the Fig 200.
variation in D due to a variation in a from 20 to 60.
10. The calculated efficiency jj of worm gearing is found from
_ tan a(i n tan a)
p + tan a
where /x = coefficient of friction and a = angle of the worm.
If p. = 15, plot a curve to show efficiencies for angles from o to 50.
11. The ideal efficiency 17 of a gas engine is given by 77 = i fJ
If n = 141, and r = compression ratio, plot a curve giving the efficiency
for any compression ratio between 3 and 18.
12. A machine costs 500 ; its value as scrap is 80.
Plot curves to show the state of the depreciation fund as reckoned
by the two methods
(a) Equal amounts put away each year.
(b) A constant percentage of the value of the preceding year set
aside each year.
The fund at the end of n years = 500 [i(i 0836)**], and the
life of the machine is 21 years.
13. The capacity K per foot of a single telegraph wire far removed
from the earth is K = ^4^ microfarads. Plot a curve to
2log6i8
give the capacity for wires for which the ratio  increases from 500 to
20000.
14. Mutton's formula for wind pressure on a plane inclined to the
actual direction of the wind is
where P = pressure on a plane at right angles to the direction of the
wind,
p = pressure on a surface inclined at 6 to the direction of the wind.
If P = 20 Ibs. per sq. ft., plot a curve giving values of p for any
angle up to 90.
352
MATHEMATICS FOR ENGINEERS
15. Plot a curve showing the H.P. transmitted by a belt lapping
1 80 round a pulley for values of the velocity v from o to 140, the
coefficient of friction p being 2.
T = 350, w = 4, g = 322, 6 = angle of lap in radians.
16. Aspinall gives as a rule for determining the resistance to motion
of trains
T 6582
where R = resistance in Ibs. per ton, V = velocity in miles per hour.
Plot a curve to give values of R for all velocities up to 55 m.p.h.
17. Find the value of r (the ratio of expansion), which makes W
(brake energy per Ib. of steam) a maximum.
r
27
00833 ,
^+000903
18. The efficiency j of a threestage air compressor with spray
injection is given by
where n = 12 and r = ratio of compression.
Plot a curve giving the efficiency for any compression ratio between
2 and 12.
19. Determine the length of the latus rectum and also the co
ordinates of the vertex of the parabola $y = 2X z nx2y.
20. A rectangular block is subjected to a tensile stress of 5 tons
per sq. in. and a compressive stress of 3 tons per sq. in. Draw the
ellipse of stress and read off the magnitude and direction of the resultant
stress on the plane whose normal is inclined at 40 to the first stress.
[Hint. Refer to p. 346.]
Curves representing Exponential Functions. To plot the
curve y = e x , where e has its usual value, one may work directly
from the tables, or a preliminary transformation of the formulae
may be necessary. If tables of powers of e are to hand, the values
of y corresponding to certain values of x are read off at a glance ;
and in such a case the values of x selected are those appearing in
these tables.
Example 9. Plot the curves y = e x and y = e* from x 4 to +4.
From Table XI at the end of the book the figures are found thus :
X
 4
 3
2
I
O
I
2
3
4
y = e x
0183
0498
1353
3679
I
27183
73891
2008
54'6
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 353
When x =  4,  is required, and this is found in the 3 rd column.
x = 3, e 3 is required, and this is found in the 2nd column.
The plotting for these figures is shown in Fig. 201, by the curve (i).
If tables of powers of e are not available, proceed as follows :
y = e*, and therefore log y = x log e = 4343*
and the table is arranged thus
X
4343* = log x
y
2
8686
7389
Having drawn the curve y e?, draw the tangent to it at some
point and measure its slope; and it will be found that the value of
the slope is also the value of the ordinate to the point of contact of the
tangent and the curve. Thus, the tangent is drawn to touch the curve
at the point for which x = 35 : its slope is measured and found to
be 33, and this is seen to be the value of y when x = 35.
If for x we write x, i. e., we plot the curve y = e~*, we find that
this gives a curve exactly similar to the last, but on the other side of
the y axis : such would be expected, since x must now be measured
as positive towards the left instead of to the right. The curve y = e~*
is shown plotted in Fig. 201, and is curve (2).
All equations of this type will be represented by exactly the
same form of curve, drawn to different scales.
A A
354
MATHEMATICS FOR ENGINEERS
Example 10. To plot the curve y = e 3x .
Write this as Y = e*, where Y = y and X = 3*.
Plot the curve Y = e x exactly as before, and then alter the hori
zontal scale in such a way that i on it now reads J, and so on.
For X = 3*
i. e., construction scale = 3 x required scale
construction scale
or required scale =
Example u. Plot the curve 5^ = 4K
This can be written
i. e.,
where Y =
4
Y =
, X = x.
4 7
Hence, plot Y = e x from the tables, and alter both scales in such a
way that the
New scale for y =  x construction scale
> ** 7 ^ > >
so that where the vertical construction scale reads 5, 4 must be written ;
and 7 must be written in place of i along the horizontal.
Example 12. If the E.M.F. is suddenly removed from a circuit
containing resistance R, and selfinduction (coefficient of self inductance
L), the current C at any time t after removal of the E.M.F. is given by
the equation B<
C = C e~i
Plot a curve to show the dying away of the current for the case
when C = 50 amps, R = 32 ohm, and L = 004 henry.
Substituting the numerical values
^32*
C = 50<T<x> 4
= so* 80 '
It will be sufficient to plot values of C for values of / between
t = o and 05 sec.
C = so* 80 '
C = e" 1 [ C is spoken of as C bar]
f*
where C =  and T = 8o/
50
If the maximum value of t is 05, the maximum value of T must
be 80 x 05, i. e., 4.
Hence from the tables :
T . . .
5
i
2
3
4
Ci.e.. e~ T
i
6065
3679
1353
0498
0183
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 355
These values are shown plotted in Fig. 202, and then the scales
are altered so that i on the vertical becomes 50, and i on the hori
zontal becomes Q .
oo
5CL
45
40
^
c
I
$
Q
C*"
So
15
*>
5
tt
v
A
A
i\
* \
^ \
CSOe" 60 *
?
V
\
>
3
V
&
X
\
X
^_
/OI25
^
[ l^
Conslr? Seal* *
O 77uo Sco/e 025 O375 "OS
Fig. 202. " Dyingaway " of Current m an Electric Circuit.
The saving of time and thought in the calculation of values
more than compensates for the somewhat awkward scales that
may result, and even this difficulty may be avoided by choosing
the original or construction scales suitably.
If it is found that the necessary values of the x or t cannot
readily be used, i. e., if values are necessary for x for which no
values of <?, etc., are given in the table, recourse must be made to
calculation.
In this case the work would be arranged thus :
log C = log 50  Sot log e
= 1699 80 y. 4343'
= 1699 34 74/
t
1699 3474'
logC
C
0025
005
etc.
05
i 699 o
1699 1737
1699
1962
50
9162
356
MATHEMATICS FOR ENGINEERS
Example 13. If a pull t is applied at one end of a belt passing
over a pulley and lapping an angle 6 (radians), the pull T at the other
end is greatly increased owing to the friction between the belt and
the pulley.
If ft, = coefficient of friction between belt and pulley
T = lw*
Plot a curve to show values of T as 6 increases from o to 180,
taking t = 40, and /i = 3.
The angle 6 ranges from o to 314. (n radians = 180.)
100
4o
5 1 iS ~ 5^ 3
Values of ft
Fig. 203. Pull on a Belt.
It will be rather more convenient in this case to calculate
Substituting values T = qoe' 39
Then log T = log 40 + 3^ log e
= 16021 + 3 x 43430
= 16021 + 13030
6
16021 + 1303$
logT
T
o
16021 + o
16021
40
5
+ 0652
I 6673
465
io
+ 1303
17324
54
i'5
+ '*955
17976
628
2O
+ 2606
18627
729
2'5
+ 3258
19279
847
30
+ 3909
19930
984
314
+ 4180
2O2OI
1047
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 357
i. e., when the belt is in contact for half the circumference of the pulley
the tension is increased in the proportion of 26 to i. In practice a
ratio of 2 to i is very often adopted. The plotting for this example
is shown in Fig. 203.
Example 14. If an electric condenser of capacity K has its coats
connected by a wire of resistance R, the relation between the charge q
at any time t sees, and the initial charge q at zero sec. is given by
3. = e~Es.
ft
Find the time that elapses before the charge falls to a value
=  X initial charge, and indicate the form of the curve which repre
6
sents the discharge.
If * = RK, then q = q^ = & = ^q
i. e., the charge falls to  5 f its initial value in time RK sees.
2*710
This time is termed the " time constant " of the condenser circuit.
The curve representing this discharge would be similar to that
plotted for Example 12, viz. in Fig. 202.
The Catenary. Referring to the curves y = e* and y = e~ x
if the " mean " curve of these is drawn it will represent the equation
e* + e*
y =   i. e.. y = cosh x.
' 2
This curve is known as the " catenary " ; and it is the curve
taken by a cable or wire hanging freely under its own weight.
The catenary when inverted is the theoretically correct shape for
an arch carrying a uniform load per foot curve of the arch.
If the cable is strained to a horizontal tension of H Ibs., and
the weight per foot run of the cable is w Ibs., then the equation
becomes
* _
y __fl_+_l
c 2
H
where c = 
The proof of this rule is rather difficult, and is given in Volume
II of Mathematics for Engineers.
From what has already been mentioned it should be seen that
the catenary is the curve y = cosh x with the scales in both direc
tions multiplied by c, since its equation can be written
358
MATHEMATICS FOR ENGINEERS
Provided
Y = y  and
c
X= then U =
Therefore, to plot any catenary one can select values of x,
read off corresponding values of cosh x from the tables and plot
one against the other, afterwards multiplying both scales by c.
If a definite span is suggested, the range of values for X must
be selected in the manner indicated in the following example.
Cor\st"rvr Scale.
Fig. 204. The Catenary.
Example 15. A cable weighing 3*5 Ibs. per ft. has a span of 50 ft.,
and is strained to a tension of 40 Ibs. Draw the curve representing
the form of the cable. Find the sag, and the tension at 10 ft. from
the centre.
Here
C = = H42.
3'5
Also the span is to be 50 ft., i. e., on the " new " or " final " scale
25 ft. must be represented on either side of the centre line.
But, new scale = c x construction scale
.*. 25 ft. on new scale 1142 x X on construction scale
or X =  = 219
1142
so that no values of X need be taken beyond, say, 22.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 359
Taking values of X from o to 22, the values of cosh X are found
from Table XI, thus :
X
'5
io
i'5
2O
22
cosh X = Y
i
1128
1543
2'352
3762
4568
The curve is now plotted, as in Fig. 204, and then for unity on the
construction scales 1142 must be written, and the 25 ft. is marked off
on either side of the centre line. The sag is read off as 393 ft., using
the final scale.
The tension in the cable at any point is measured by the ordinate
to the curve multiplied by w; e.g., the tension at 10 ft. from the
centre = 35 x 159 = 556 Ibs.
Exercises 38. On the plotting of Curves representing Exponential
Functions.
1. Plot, for values of x from 8 to 29, the curve y = 2e~*. Find
its slope when x = 16.
2. Plot the curve y = 250$* from x = o to x = 15.
3. Plot, from # = 5 to #=+3, the curve y = 021 x 162*.
4. If C = C e~ at , C = 146, a = 410, and t ranges from o to 023,
represent by a graph the change in C (the dyingaway of a current).
5. Plot a curve to give the tension T at one end of a belt for various
coefficients of friction p,; the angle of lap (6 radians) being constant.
Given that
T = lev*, 6 = 165, and / = 50 ; p. ranges from i to 35.
6. A cable weighing 218 Ibs. per ft. and strained to a tension of
56 Ibs. hangs freely. Depict the form taken by the cable when the
span is 30 ft., and find the tension in it 12 ft. from the centre.
Rt
7. C = 487(1 e~i*). If R = 56, L = 008, plot a currenttime
(C /) curve for values of t from o to 062.
8. Trace a graph to show the drop in electric potential down a
uniform conductor, if the potential at the receiving end is 200 volts,
the resistance per kilometre r of the conductor is 10 ohms, the leakage
g of the insulation is 5 x io~ 6 megohms per kilometre, and the dis
tance from the " home " end to the receiving end is 500 kilometres.
If e = the potential at distance x from the receiving end
e = 200 cosh Vgr . x.
Graphs of Sine Functions.
Consider the equation y = sin x. We have already seen in
Chapter VI that as the angle x increases from o to 90, the sine y
increases from o to i; and as x increases from 90 to 180, y
decreases from i to o. Continuing into the 3rd and 4th quadrants :
for x increasing from 180 to 270, y decreases from o to i ; and
for x increasing further to 360, y increases from i to o.
360
MATHEMATICS FOR ENGINEERS
After 360 has been reached the cycle of changes is repeated,
i. e., 360 is what is called the period, for the function y = sin x.
Because y and x are connected by a law, we conclude that the
changes will not be abrupt or disjointed, or in other words, the
curve representing y = sin x will be a smooth one.
The sine curve is perhaps the most familiar of ah 1 curves, there
being so many instances of periodic variation in nature.
Thus, if a curve be plotted showing the variation in the magnetic
declination of a place over a number of years, its form will be
that of a sine curve : so also for a curve showing the mean tem
perature, considered over a number of years, for each week of the
year.
2
10
l/=SLM.JG
Radionsi
Angle
300"
360"
Fig. 205. Sine Curve.
Sine curves occur frequently in engineering theory and practice ;
in fact, a sine curve results whenever uniform circular motion is
represented to a straight line base.
All sine curves are of the same nature, and therefore it is neces
sary to carefully study one case, and that the simplest, to serve as
a basis.
To plot y = sin x : select values of x between o and 90, thus :
x degs.
o
25
45
60
80
90
y
o
423
707
866
985
i
Choose suitable scales so as to admit the full period to be plotted
and plot for these values, as in Fig. 205.
No further recourse to the tables is necessary, this portion of
the curve being simply drawn out three times.
For sin 100 = sin (180 100) = sin 80
and therefore for 10 to the right of 90 the value of y is the same
as that for 10 to the left of it, i. e., the curve already drawn can
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 361
be traced and pricked through to give the portion of the curve
between x = 90 and x = 180.
Again, sin 205 = sin (180 + 25) = sin 25
and sin 240 = sin (180 + 60) = sin 60
*. e., the 3rd portion of the curve is the ist portion " folded over "
the horizontal axis. Similarly, the 4th will correspond to the 2nd
" folded over " ; and accordingly we need only concern ourselves
with calculations for the ist quarter of the curve.
The maximum value of y, viz. i, is spoken of as the amplitude
of the function. Thus in the case of a swinging pendulum, the
greatest distance on either side of its centre position is the amplitude
of its motion.
If y = 5 sin x, then the amplitude is 5, and the curve could be
obtained from y = sin x by multiplying the vertical scale by 5.
Example 16. Plot the curve y = 5 sin 4*.
v
Writing this as f = sin 4*
o
or Y = sin X
[where Y =  = 2y, and X = 4*]
we see that the simple sine function is obtained.
Accordingly we plot the curve Y = sin X (making use of the table
on p. 360), and then alter both scales so that x = and y =
360
Dealing with the last example, we see that the period is
or 90 ; i. e.,ii x is multiplied by 4, the period must be divided by 4.
Similarly for the curve representing y = sin \x, the period
would be 360 r r ~ 1800. We thus obtain the important rule :
" To obtain the period for a ' sine ' function, divide 360 by the coefficient
of x or t (whichever letter is adopted for the base or ' independent
variable ' ") or briefly
360
Period in degrees = eoeffieient ot ]^T,
Since 27r radians = 360, wherever we have written 360 above
we should write 2ir, if the angle is to be expressed in radians, >'. e.,
the period in radians or seconds (of time)
27T
= coefficient of the x or t
Thus if y = 4 sin 6x
Period = > = 
6 3
and amplitude = 4
Period = = 1 or 6o<
362
MATHEMATICS FOR ENGINEERS
Example 17. The current in an electric circuit at any time / sees.
is given by the expression C = 45 sin loowt.
Plot a curve to show the change in the current for a complete
period.
The general formula is C = C sin 2nft, where / = number of cycles
per second = frequency. In this case 2irf = IOOTT, .'. / = 50.
If / = 50, the time for one cycle, or the period, must = ^ = 02 sec.
Thus the periodic time = 02 sec.
Notice that the period is given in terms of seconds (of time) in this
case, and not in degrees.
The same periodic time would have been obtained if our previous
rule had been applied, for
,. . , 2ir 2ir
Period = ~ T. =  = 02 sec.
~ T
coeff. of
IOOTT
Fig. 206. Change in Current in Circuit.
Either of two methods can be used for the calculation of values
(a) Plotting from the simple sine function.
According to this scheme write the equation in the form
C
= sin looirt
4 '5
C = sin T
, and T = ioont
4'5
or
where
Hence to plot the curve (Fig. 206) C = sin T, select values of T
between o and  (o and 1571), and thus read off values for C so that
the first quarter of the curve can be plotted, remembering always that
the base must be numbered in radians.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 363
Values for this portion would be of this character :
T
o
2
4
7854
96
ii
i'4
I57I
C
o
198
385
707
819
891
985
I
To obtain the scales so that the given equation is represented,
multiply the vertical scale by 45, i. e., i on the original scale now
reads 45 ; and divide the horizontal scale by IOOTT, so that  now
reads 005
The curve is shown in Fig. 206.
\2OO7T
(6) According to the second method, the simple sine curve is not
used and no alterations are necessary. Having found the period,
02 sec., it is known that values of t need only be taken for one quarter
of this, i. e., between o and 005.
The tabulation would be arranged as follows :
t
lOOirt O
(radians)
r iS.ooot
(degrees)
sin IOOTT*
C = 45 sin looirt
o
O
o
o
OOI
3H
18
309
139
OO2
628
36
588
2645
003
9 4 2
5<
809
364
004
1256
**l
951
4275
005
I57 1
90
i
4'5
Note that the 2nd column is not really necessary; it is only in
serted here to make clear the reason for the 3rd column.
Example 18. A crank i'6* long rotates uniformly in a righthand
direction, starting from the inner dead centre position, and making
30 revs, per minute. Construct a curve to show the height of the end
of the crank above the line of stroke at any time, assuming pure
harmonic motion.
Time for i revolution =
60
30
2 sees.
or, in 2 sees., 2ir radians is the angular distance travelled.
' In i sec. TT radians is the angular distance travelled, or the " angular
velocity," usually denoted by a, = it radians per sec.
Construction. Draw, to some scale, a circle of radius i'6*, to the
left of the paper. (Fig. 207.)
Divide its circumference into a number of equal parts, say 10 or
12 ; in this case 10 is chosen, lines making 36 with one another being
drawn. These division lines will correspond to the positions of the
crank at time divisions of a tenth of a period, . e., 2 sec. apart.
Number these divisions of the circumference o, i, 2, 3 . . .
364
MATHEMATICS FOR ENGINEERS
Produce the horizontal through O and along it mark off to some
scale a distance to represent 2 sees., and divide this into 10 equal
parts.
When the crank is in the position Oi, i. e., at time 2 sec. after start,
its projection on the vertical axis is OA : hence produce lA to meet
Fig. 207.
the vertical through 2 at i x ; and this will be a point on the curve
required. Proceeding similarly for the other positions of the crank,
the full curve is obtained, and from its form we conclude that it is a
sine curve.
To prove that it is a sine curve
Suppose that in time / sees, the crank moves to the position OC
(Fig. 208).
Fig. 208.
Fig. 209.
In I sec. the angle moved = ir (in this case)
.'. In t sec. the angle moved = itt (in this case)
where a>t is the angle in radians.
or a> (in general)
or <ot (in general)
OA = CB = CO sin L COB = r sin f
where r = radius of crank circle.
Therefore the curve obtained by the construction is that repre
senting the equation y = r sin at.
Hence a graphic means of drawing sine curves can be employed in
place of that by calculation. Great care must, however, be taken in
connection with the magnitudes involved.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 365
e. g., to plot C = 45 sin iooirt by this means.
Radius of circle = 45, the amplitude of the function
and tat = looirt or <> = ioor
. e., 100 TT radians must be swept out per sec.
27T radians are swept out in 02 sec.
Therefore, if the circle were divided into 10 equal parts, the dis
tances along the time base corresponding to the angular displacements
would be 002 sec. each.
Simple Harmonic Motion. If the crank in Fig. 209, which
is supposed to revolve uniformly, were viewed from the right or left,
it would appear to oscillate up and down the line OA. Such motion
is known as simple harmonic motion, or more shortly S.H.M.
Looking, also, from the top, the motion as observed would be
an oscillation along OB, and this again would be S.H.M. ; there
fore, if the connectingrod were extremely long compared with the
crank the motion of the piston would be approximately S.H. In
the case of the valve rod it would be more nearly true that the
movement of the valve was S.H., for the valve rod would be very
long compared with the valve travel.
At a later stage of the work it will be shown that the accelera
tion along OB, say, is proportional to the displacement from O;
and this is often taken as a basis for a definition of S.H.M.
S.H.M., then, is the simplest form of oscillatory motion, and
can be illustrated by a sine curve.
Suppose that the crank does not start from the inner dead
' centre position, but from some position below the horizontal, what
modification of the equation and of the curve results?
If at time t sees, after starting, the crank is at OC (Fig. 209)
(Oo is the initial position of crank}
then L COo = <>*
and L COB = ut c
where c = L. BOo
y = r sin L COB = r sin (<at c).
Similarly, if the crank is inclined at an angle c above the hori
zontal at the start, y = r sin (wt+c).
A moment's thought will show that the curve will be shifted
along the horizontal axis one way or the other, but that its shape
will be unaltered.
Example 19. Plot a curve to represent the equation
C = 45 sin (loont ii) for a complete period.
366
MATHEMATICS FOR ENGINEERS
Let us reduce the equation to a form with which we have already
dealt; thus
C = 45 sin (looirt i i)
C I, ii \
= sm loon! t
45 \ I007T/
= sin ioo?r(/ 0035)
j. e., C = sin loonTi = sin 1
C
where T l = t 0035, T = loouT, and C = 
4 5
*$ r/vWw?
36
, I
^ AX ' 5
^
\,
^0
Mi
s >
/
\
18 &
5.1
L /
\
Vu C=45 sin dooTfl 11}
O o

i/
\
?2 '/
00*
36
O8 .Ol <
12 \<
14 O
16
10 <
2 Scale for I
OO35 V
9 /,
2 ^
/O
157 '
1
>co/e jQ
T' \
\
.
/
/
OO5
'
\
/
27 /
/
1
g
\
/
?'/
1
I
\
/
45
11
V
^
'
Fig. 210.
We have already seen, viz. in Example 17, how to plot this curve,
the period being 02 sec. Then, having altered the two scales accord
ing to previous instructions, the vertical axis must be shifted a dis
tance of 0035 sec. to the left (see Fig. 210), because < = Tj+oo35.
Hence the scale for t, which is the final scale, must be measured from
an axis 0035 unit to the left of that used in the construction, i. e., the
horizontal scale must again be altered, not in magnitude but in position.
The changes in the scales may appear rather confusing, but on the
other hand calculations have only to be made for the one fundamental
curve (the table for this being given on p. 360), and all the others are
derived from it. Therefore, when once the table of values for the
simple sine curve has been set out, it will serve for all sine and cosine
curves, i. e., it is a " template."
It serves for the cosine curve because this curve is merely the
sine curve shifted along the axis a distance equal to one quarter of
the period.
Thus y = sin t
and y = cos t = sin (90 /)
will be the same curve measured from different vertical axes.
Graph of tan x. The graph representing y = tan x is not
of the same type as the sine and cosine curves. As x increases
from o to 45, y increases from o to i, but after x has the value 45
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 367
y increases very much more rapidly; while at 90 the value of y
is infinitely large. After 90 the tangent is negative, for the angle
is in the 2nd quadrant. Supposing some form of continuity in
3 Degrees 90
Radians
Period /ao*
tT rae/iano
Fig. 211. Graph of tan x.
the curve, it must now approach from infinity from the negative
side and come up to cross the axis at 180. After this the curve
is repeated, so that the period for the simple tangent function is
180 or TT.
368
MATHEMATICS FOR ENGINEERS
Selecting values for x, those for y can be read off from the
tables :
x
y = tan x
o
o
25
466
45
io
60
1732
80
5671
85
n43
90
+ 00
X
y = tan x
90 +
00
95
 n43
100
 567I
120
 1732
135
I
155
 466
180
o
Note that 90 indicates that a value of x is supposed to be
taken just less than 90, but practically differing nothing from 90 ;
thus 90 would be of the nature 8999. Similarly 90+ would
indicate 9001, say.
The curve on either side is asymptotic to the vertical through
90, as will be seen from the curve plotted in Fig. 211.
All other simple tangent curves can be obtained from this
fundamental curve by suitable change of scales.
Example 20. Plot the curve representing the equation
y = 8 tan 40O/.
Rewrite the equation as Y = tan T
where Y = ^ and T = 400*.
Then plot Y = tan T from o to 180, t. e., for a complete period,
and afterwards alter the scales so that i on the vertical scale becomes
8, and i on the horizontal scale becomes
400
sin x
cos*'
and therefore, if we had drawn the curves
y x = sin x (i) and y a = cos x (2), we should
obtain the value of the ordinate of the curve y = tan x by dividing
any ordinate of curve (i) by the corresponding ordinate of curve (2).
Example 21. The efficiency of a screwjack is given by
tan 6
' "~ tan (0 + <t>)
where 6 is the angle of the developed screw, and <f> is the angle of fric
tion. If 6 varies from o to 12, plot a curve to give the value of the
efficiency; p, the coefficient of friction, being 1465.
The angle of friction is such that its tangent is equal to the co
efficient of friction, i. e., <f> = tan 1 u.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS
tan 6
369
Thus tan <j> = 1465 and <f> = 8 20'; also ; =
The tabulation of values is as follows :
tan (6 + 8 20')
6
6+ <*>
tan 6
tan (6 + <t>)
n
o
8 20'
o
H^S
2
10 20'
0349
1823
191
4
12 2O'
0699
2186
320
6
14 20'
1051
2555
4i3
8
l620'
1405
2931
480
10
l820'
1763
33H
533
12
20 20'
2126
3706
574
and the plotting is shown in Fig. 212.
Efficiency 7J
O .1 ro (>i A ch <i
r*. .. _ E

S
^

X
?

A
/

T
f 1
1
1
i
1
\
32 A 6 8 IO 15
0, Aragle of Thread
Fig. 212. Efficiency of Screwjack.
Compound Periodic Oscillations. In engineering practice
one often meets with curves which are quite periodic, but are not
of the sine or tangent type. Many of these can be broken up or
" analysed " into a number of sine curves. The process is spoken
of as harmonic analysis, and reference to this is made in Volume
of Mathematics for Engineers. At this stage, however, it is well
to consider the work from the reverse or the synthetic point of view,
in which the resultant curve is constructed from its components by
the addition of ordinates.
B B
370
MATHEMATICS FOR ENGINEERS
An example of importance to surveyors concerns the " equation
of time," which is the difference between the " apparent " and the
" mean " time of day. The apparent time is the actual time as
recorded by a sundial, whilst the mean time is calculated from
its average over a year. Two causes contribute to the difference
between the two times, viz.
(a) The earth in its journey round the sun moves in an ellipse
, , . ., /distance between focf\ . i
having an eccentricity ( ~
sequence of the laws of gravity its speed is greater when nearer to
the sun than when more remote.
(b) The earth's orbit is inclined to the plane of the equator.
i\ , 
1 of ^,and in Con
Set Watch FAST over J
this Period by Amount"!
given bv Ond"
Set I Watch
Period b
Curve giving
Variation duel ho
J ^'Curve giving
. Variation due to
IE EarftA Orbital Spee
Curve Of
Equation of Time
Fig. 213. Curves for " Equation of Time."
The corrections due to these two causes are found separately,
and are represented by the respective curves (a) and (b) in Fig. 213.
For curve (a) the period is one year, and the period of (b) is half a
year.
These, when combined by adding corresponding ordinates, due
attention being paid to the algebraic sign, give curve (c), for which
the period is one year. By the use of this curve the correction to
be added to or subtracted from the observed " sun time " can be
obtained. Thus to determine the longitude, i. e., the distance in
degrees east or west of Greenwich, of, say, a village in Ireland, it
would be first necessary to find the meridian of the place by
observation of the pole star. Next the time of the crossing of the
meridian by the sun i. e., the local time, would be noted, and this
would be corrected by adding or subtracting the equation of time
for the particular day. Then the difference between the corrected
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 371
local time and Greenwich mean time as given by a chronometer
would give the longitude, since one hour corresponds to fifteen
degrees.
Example 22. Plot the curve y = 4 sin / + 5 sin 2t sufficiently far
to show a complete period.
Let y\ = 4 sin * (i), and y, = 5 sin it ( 2 ) ;
then the curve required is y = y^ + y t , i. e., it is the sum of two curves
of different periods.
The period of y = 4 sin / is 2ir, while the period of y = 5 sin 2t is
27T
V_J
X
= 4sint
\^
^T\
\
Fig. 214. Complete period of curve y = 4 sin t + '5 sin 2 t.
Therefore the curves must be plotted between / = o and / = lit to
give the full period of the resultant curve, so that there will be one
period of curve (i) and two of curve (2).
The curves are now dealt with separately, because, being of different
periods, values suitable for the one would not be so for the other.
For curve (i) period = 2n, and amplitude = 4.
The two curves must be plotted to the same scales. The simple
sine curve " template " already mentioned would serve for curve (i),
but curve (2) must be previously adjusted in scale to make it possible
to apply the " template."
It may be sometimes easier to set out the work as follows instead
of using a template :
Curve (i). Values of / need only be taken between o and ^
372
MATHEMATICS FOR ENGINEERS
Curve (2). Values of t need only be taken between o and ; there
fore take values onehalf of those in the previous case, so that the
calculation is simplified.
Curve (i)
Curve (2)
t
sin t
y x = 4 sin t
o
o
o
2
198
792
4
385
i54
7854
707
2828
960
819
3276
ii
891
3564
i'4
985
3*94
I57I
I'D
40
t
zt
sin 2t
y a = 5 sin 2/
o
o
o
I
2
198
099
2
4
385
193
3927
7854
707
354
48
96
819
4i
55
ii
891
446
7
i'4
985
493
7854
I57I
IO
5
These curves are plotted as shown in Fig. 214, and the resultant
curve is obtained by adding corresponding ordinates, paying careful
attention to the signs.
One further example of this compounding of curves will be
given.
Example 23. The current in an electric circuit is given by
C = 50 sin 628*, whilst the voltage is given by V = 148 sin (628^ + 559).
Plot curves to represent the variations in the current, voltage and
power at any time.
i
Fig. 215. Variations in Current, Voltage and Power in Electric Circuit.
Dealing with the three curves in turn : (See Fig. 215.)
Curve (i). This is the curve of current.
27T
C = 50 sin 628*, and the periodic time = ^x = 01 sec
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 373
Plot the curve = sin T from o to 2ir, and then multiply the
vertical scale by 50 and divide the horizontal by 628.
Curve (2), the curve of voltage
= 148 sin 628Tj = 148 sin T
provided that T = 628"]?! and Tj = t + 00089.
This is the first curve with its axis moved to the right a distance
of 00089 sec. and with all ordinates multiplied by 4 or 296. Thus
AB = 296 x ab.
Curve (3), the curve of power, is obtained by multiplying correspond
ing ordinates of curves (i) and (2).
Confusion is avoided by plotting curve (2) along a different hori
zontal axis from that used for (i).
The reader will find it convenient to draw out the simple sine
curve on tracing paper to a scale convenient for his book or paper,
and to use that as a template ; much time and labour being saved
by this means.
Curves for Equations of the Type y = e~* x sin(6x+c).
In plotting such a curve it is not wise to select values of x
and then calculate values of y directly : it is easier to split the
function up into y 1 = e~ ax and y 2 = sin(&#fc), and plot the
curves representing these equations separately, obtaining the final
curve y = y"iXy 2 by multiplication of ordinates.
The forms of the two component curves are already known.
They must, however, be plotted to the same horizontal scale,
which should always be a scale of radians (if an angle is measured
along the horizontal) or one of seconds (if time is measured along
the horizontal).
Example 24. Plot the curve y = e~** sin (5*+ 24), showing two
complete waves.
Let y = yixy, where y t = e~* x and y, = sin (5*424).
To avoid any trouble with the scales, this example is worked in
full, i. e., templates are not used.
It will be slightly more convenient to deal first with curve (2).
Curve (2) y, = sin (5* + 24) = sin 5(#+'4 8 ) = sin 5X
where X = x + 48
Hence the vertical axis through the zero of x in Fig. 216 will be
48 unit to the right of that for X ; hence, since the second scale has
to be used again in the plotting, the construction vertical axis must
be chosen 48 unit to the left of some convenient startingline.
y, = sin 5X, the period being ^ = 1256 radians
374
MATHEMATICS FOR ENGINEERS
Hence values of x need only be taken between o and or to 314.
4
X . .
o
04
08
I57 1
192
22
28
314
5 X . .
2
4
7854
96
I I
i'4
I57I
sin 5 X .
o
198
385
707
819
891
985
I
This curve can now be plotted, reckoning the horizontal scale
from the construction vertical axis; and then the zero is shifted to
its correct position, 48 unit to the right, as shown in Fig. 216.
Fig. 216. Curve of y = e  J* sin (5* f 24).
Curve (i) yi = "". For one complete wave of curve (2)
v = 1256, and therefore for two waves it will be more than sufficient
if values of x are taken up to 3.
The table of values is as follows :
X .
4
8
12
16
2O
24
28
30
\x or X
o
2
4
6
8
10
12
14
i'5
~ x = yi
I
819
67
'549
449
3 68
301
247
223
A word of explanation regarding this table is necessary. Consider
the value x = 12 ; then to find y, the value of e~* x1 ' 2 or e~' 6 must be
found. Therefore X = 6 is read in the first column of Table XI at
th.7 end of the book, and e~' 6 is read off in the third column.
This curve can now be plotted, always to the same horizontal
scale as that chosen for curve (2), but not necessarily to the same
vertical scale. In this example, however, the same scale is convenient
for both.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 375
Curve (3), or y = y x x y, can next be obtained by selecting corre
sponding ordinates of the two curves and multiplying them together.
When x = 106, y^ = 58 and y t = i ; hence in this case the par
ticular product of y>i and y a has the same value as y lt and accordingly
the vertical scale chosen for curve (3) is advisedly that for (i), so
that the curve when plotted touches the curve (i) at its highest points.
Glancing at the curve (3) we observe that the amplitude is now
diminished in a constant ratio, although the period remains the
same, i. e., there is some damping action represented.
If a condenser discharges through a ballistic galvanometer and
deflections left and right are taken, then by plotting the readings
a curve is obtained (naturally of a very small period) of the char
acter of curve (3). The logarithm of the ratio of the ampli
tudes of successive swings is called the logarithmic decrement of the
galvanometer.
For the case considered, the ratio of consecutive amplitudes is
,}*+l256 f \x v fl'256
'~ =  ~  = * 1 ' 256 = 35 (approx.)
.'. logarithmic decrement = log 35 = 1253.
Again, imagine a horizontal metal disc within a fluid, hung by
a vertical wire. If the wire is twisted and then released, the disc
oscillates from the one side to the other. Measurements of the
amplitudes of the respective swings demonstrate the facts that
(a) the ratio of the amplitude of one swing to the amplitude of the
preceding swing is constant for any fluid, and (b) this ratio is less
for the more viscous fluids.
Thus if the disc osciUated in air, the successive swings would
be very nearly alike as regards amplitude ; or, in other words, the
motion is practically simple harmonic, and its representation in the
usual manner gives a sine curve. If the medium is water or thick
oil, the motion is represented by a curve like No. (3) in Fig. 216,
but the damping effect would be much more marked in the case of
the oil.
Exercises 39. On the Plotting of Graphs Representing Trigonometric
Functions.
1 Write down the amplitudes and periods of the following func
tions : i "cos 4 * : * sin (3* 4): 51* sin 314' (< 1S in seconds);
"*XA?^*1*
the amplitude and also the period of this function
376 MATHEMATICS FOR ENGINEERS
3. The range of a projectile fired with velocity V at elevation A is
"y"2 gjn 2A
given by ' . Plot a curve to show the range for angles of
elevation up to 45, the velocity of projection being 1410 ft. per sec.
4. On the same diagram and to the same scales plot the curves
y t = 2 sin x and y t = 5 sin \x, and also, by addition of ordinates, the
curve y = 2 sin x + 5 sin \x.
5. A crank rotates in a righthand direction with angular velocity
10, starting from the inner dead centre position. To a time base
draw a curve whose ordinates give the displacement of a valve, the
connectingrod (or valverod) being many times as long as the crank.
The travel of the valve is to be i*.
6. Plot the curve s = 283 sin(4< 016) for one complete period,
the angle being in radians.
7. Plot the curve y 81 cos 3$ for a complete period.
8. Plot the curve 5^ = 472 tan 4$ for a complete period.
9. The current from an alternator is given by C = 15 sin iirft, and
the voltage by E = 100 sin (zirft n). If the frequency / is 40 and
n (the lag) = 611, draw curves of current and E.M.F., and by multi
plication of corresponding ordinates plot the curve of power.
10. The acceleration A of the piston of a reciprocating engine is
given by
A 2 1 f /I , COS 20}
A = 47r 2 n 2 ri cos \
Plot a curve to give values of the acceleration for one complete revo
, ,. ,, connectingrod length
lution when r = crank radius = i ft., m = ^ = 10,
crank length
n = R.P.S. = 2.
11. The displacement y of a certain slide valve is given by
y = 2 6 sin (0+ 32) + 2 sin (20+ 105).
Plot a curve to give the displacement for any angle between o
and 360.
12. Plot the curve y = e'^sin $x, showing two complete waves.
13. Plot a curve to give the displacement x of a valve from its
centre position when x = 12 cos pt 18 sin pt and p = angular
velocity of the crank, which revolves at 300 R.P.M.
14. Plot the curve y = 5 cosec 0, showing a complete period.
15. What is the period of the curve 7^ = 28 sec3#? Plot this
curve.
16. An E.M.F. wave is given by the equation
E = 150 sin 3i4/ + 50 sin 942^.
Draw a curve to show the variation in the E.M.F. for a complete
period.
17. The " range " of an object from a point of observation is found
by multiplying the tangent of the observed angle by the length of the
base. Draw a curve to give ranges for angles varying from 45 to 70,
the measured base being two chains long.
Graphic Solution of Equations. The application of purely
algebraic rules will enable us to solve simple or quadratic equa
tions. Equations of higher degree, or those not entirely algebraic,
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 377
can best be solved by graphs ; and in some cases no other method
is possible.
The general plan is to first obtain some approximate idea of the
expected result, either by rough plotting or by calculation, and to
then narrow the range, finally plotting to a large scale the portion
of the curve in the neighbourhood of the result.
Occasionally the work is simplified by plotting two easy curves
instead of the more complex one.
Example 25. Solve the equation e 3 * $x z 17 = 0.
The equation may be written e 3 * = $x z +17.
Then if the two curves y x = e sx and y t = $x z + 17 are plotted, their
point or points of intersection will give the value or values required.
Tabulating :
For Curve (i) y l = e 3x . For Curve (2) y t = $x z +17.
X
3# or X
e* = y t
o
o
i
5
i'5
448
io
3
2009
15
4'5
90
2
6
404
X
5* 2 + I 7
y
o
0+17
17
5
125+ 17
1825
i
5 +17
22
i'5
1125 + J 7
2825
2
20 + 17
37
3
45 + J 7
62
We conclude from an examination of these tables that y t and y t
are alike when x has some value between io and 15.
Values of x are next taken between io and 15 ; thus :
X
3* = X
Vi
5 *2+I7
y\
io
II
30
3'3
2OI
271
5 + J 7
605 + 17
22
2305
Therefore, the solution is evi
dently between io and ii; hence
plot the two curves for these values
and note the point of intersection
(see Fig. 217).
This is found to be at the point
for which x = 1032, and therefore
x = 1032 is one solution, and as the
curves intersect at one point only,
it is the only solution.
Numerous examples of this,
method of solution of equations
102 104. 106 108
Fig. 217.
Solution of e* x 5** 17 = o.
MATHEMATICS FOR ENGINEERS
occur in connection with problems in hydraulics. As an example
take the following :
Example 26. Water flows at 745 cu. ft. per sec. through a pipe
of diameter d ft., and the loss of head in 10 miles is 350 ft. The co
efficient of resistance is /= 005(1 \ A Find the diameter of the
pipe, given that
Head lost = i where
2gm
d
m 
4
Area of pipe = ~d z
Then the velocity = Z&. = 7J45 x
area ird z
_
and
Substituting for /
d x 644 x d*
_ 40 x 5280 x 948 x 948 ,
350 x 644
= 8 3 8/.
o.
from which d 6 419^ 35
To solve this equation, we know that no negative values need be
taken ; hence as a first approximation
Let y = d 6 419^ 35
Then
d d
 4igd 35
y
o o
I I
2 64
 o  35
 419  '35
 838  35
 '35
454
5527
Since
d = i makes y negative
and d = 2 makes y positive
the value of d that makes y = o must lie between i and 2 and nearer
to i.
For d = 15, y = (i'5) 6 (4'i9Xi5)35 = 4'76
Thus the required value of d is between i and 15.
If d = 13, y = 98, and we see that the required value is between
i 3 and 15. Plot the values of y for the values of d 13 and 15, as
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 379
in Fig. 218, allowing a fairly open scale for d, and join the two points
by a straight line. The inter
section of this line with the
axis of d gives the value of d
required, which is seen to be
Example 27. The length of
an arc is 267*, and the length
of the chord on which it stands
is 2 5*. Find the angle sub
tended at the centre of the
circle. [This question has refer
ence to the length of sheet
metal in a corrugated sheet.]
Arc = radius x 6
radians.
Now
Also
where is in
rd = 267 and r = ^ L
. 6 125 . i'25
sin = *. t. e., r = .,
2 r ' sini
2
267 125
15
Fig. 218. Solution of Equation
giving Diameter of Pipe.
267
or
sin  = ^d = 480.
2 267
Making our first approxi
mation, taking 6 from o to
6
d_
6
sin 
4680
2
2
O
o
5
25
247
234
io
5
479
468
15
75
682
702
2O
io
842
936
25
125
945
117
30
I 5
998
1403
314
i57
io
147
0T*
Fig. 219. Length of Metal in
Corrugated Sheet.
we see that the solution must lie between 6 = io and d = 15.
When 6 = 12, sin  = 565, and 468$
562.
380 MATHEMATICS FOR ENGINEERS
Q
Plotting the two curves in Fig. 219, y x = sin , and y a = 468$ for
values of 6 from 6 = io to 15; we note the point of intersection to
be at 6 = 119. .*. 6 = 119 radians or 682.
Exercises 40. On the Graphic Solution of Equations.
1. Find a value of x in terms of / to satisfy the equation
3# 3 3l z * + I 3 = o
x being a distance from one end of a beam of length /.
2. Solve for z the equation I 3 $lz z z 3 = o when / = 10.
3. In order that a hollow shaft may have the same strength as a
solid one the following equation must be satisfied
*f D*d* *f
16 * D 16 *'
Writing x for r this equation reduces to x* x i = o. Find
the ratio of the diameters so that the given condition may be satisfied.
4. Find a value of d (a diameter) to satisfy the equation
where r = 32, /= 6, P = 15.
5. Solve the equation e x = 4*.
6. Find values of x between 4 and +3 to satisfy the equation
to
los = 16 + 4* x z
7. Find a value of x between i and 5 to satisfy the equation
x* log* x = 8
8. Solve for positive values of x the equation 50 ^ sin 4*= 18.
(Note that the value of x must be in radians.)
9. Determine a value of x between o and it to satisfy the equation
x 1 ' 5 3 sin x = 3
10. To find the height of the water in a cylindrical pipe so that
the flow shall be a maximum it is necessary to solve the equation
6(2 3 cos 6) + sin 6 = o
Find the value of 6 (radians) to satisfy this equation.
11. Solve for / in terms of L the equation
56/ 3  uiL/ 2 + 72/L 2  i 4 L 3 = o
which occurs when finding the most economical arrangement of the
three spans of a continuous beam ; I being the length of each of the
end spans and L being the total span.
12. In finding the ratio of expansion r for a direct acting single
cylinder steam engine of 14" diameter and 22" stroke, the equation
i+log e r 389^ = o was obtained.
Find the value of r to satisfy this equation.
13. The maximum velocity of flow through a circular pipe is
reached when the angle 6 at the centre of the circular section sub
tended by the wetted perimeter has the value given by the equation
Find this value of 0.
sin 6
; cos 6 =
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 381
14. Solve, for positive values of/ (the length of a link of a certain
mechanism), the equation
/ 3  I 9'5/ 2 + 425/ + 546 = o.
15. Forty cu. ft. per sec. are to pass through a pipe laid at a slope
of i in 1500, the pipe to run half full. The velocity is given by^
. where m = 1530 and the quantity = D 2 u
I 4 4
Simplifying and collecting these equations we arrive at the simpler
form
Find the value of D to satisfy this equation.
16. The bottom of a trapezoidal channel (the slope of the sides
being 2 vertical to i horizontal) is 4 ft. wide. Find the depth of flow d,
if the discharge is 12000 gallons per min., the slope is i in 500, and
the coefficient of resistance is 006.
/Equations reduce to 2 '3 2 (&*+*)* =
V8 + 447^
17. Find a value of r, the ratio of expansion, to satisfy the equation
 1083 log e r 225 = o
18. A hollow steel shaft has its inside diameter 3*. What must be
the outside diameter so that the shaft may safely stand a torque of
200 tons ins., the allowable stress / being 5 tons per sq. in. ? Given
that
Torque 2/
(D*  3 )
32 v
19. Find a value of 6 (the angle of the crank from line of stroke) to
satisfy the equation
sin 6 6 n z sin 4 6 n* sin 2 + n* = o when n = 5.
[Hint. Let X = sin 2 6 and then solve for X.]
20. An equation occurring in connection with the whirling of
shafts is j
cosh x H = o
cos x
Find a value of x between o and it to satisfy this equation.
[Note that the values of cosh x should be taken from Table XI at
the end of the book.]
21. Find the height above the bottom of a cylindrical tank of
diameter 10 ft. at which a pipe must be placed so that the water will
overflow when the tank is twothirds full.
Construction of PV (pressurevolume) and r<f> (temper
atureentropy) Diagrams. It is impossible to proceed far in
the study of thermodynamics without a sound working knowledge
of the indicator and entropy diagrams of heat engines ; and to assist
in the acquisition of this knowledge these paragraphs are addressed
mainly to students of the theory of heat engines. Although we
are not concerned in this volume with the full meaning of these
382
MATHEMATICS FOR ENGINEERS
curves, we can deal with them as practical examples of graph
plotting. More can be learned about the advantages and useful
ness of an entropy diagram by actual construction and use than
by absorbing the remarks of some one else, and taking for granted
all that he says. Careful attention should, therefore, be directed
to the following exercises, which should be worked out step by
step by the reader.
Example 28. Draw a PV diagram (Fig. 220) and also a T$ diagram
(Fig. 221) for i Ib. of steam expanding from a pressure of 100 Ibs. per
sq. in. absolute, to atmospheric pressure, the steam being dry and
E B
O E
4 6 12,  16 "I/" 2O
Fig. 220. Pressure Volume or PV Diagram.
saturated throughout. [Note. Since these diagrams are to be used
for subsequent examples, they must be so constructed that the lowest
pressure indicated is 5 Ibs. per sq. in. absolute.]
To calculate for points on the expansion line BD in Fig. 220 steam
tables must be used; the volumes (V) of i Ib. weight of steam for
various pressures (P) between TOO Ibs. per sq. in. and 147 Ibs. per
sq. in. absolute being read off from the tables and tabulated thus :
P
IOO
80
60
40
20
I 4 7
V
4 '4
5'48
716
1050
20
268
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 383
Horizontals through 100 and 147 on the pressure scale complete
the diagram in Fig. 220. BD is the saturation or 100 % dryness curve.
For the T$ diagram (Fig. 221) rather more calculation is necessary.
The entropy of water at any absolute temperature T Fahrenheit
= log* , if the entropy is considered zero at 32 F., i. e., at 461 + 32
or 493 F. absolute.
For our example we require the " water " line from about 160 F.
to 320 F., since these temperatures correspond approximately to
pressures 5 and 100. Hence the range of T = 621 to 781 F. absolute,
or, say, 620 to 780. The tabulation is next arranged as follows, it
being noticed that
log, =
6 493
log, 493  2303(log 10 rlog lo4 93)
T
Iog 10 rlog, 493
2303 x column (2) = log, 
493
62O
660
7OO
750
780
27924 26928
28195 26928
28451 26928
28751 26928
28921 26928
0996 x 2303 = 229
1267 x 2303 = 292
1523 x 2303 = 351
1823 x 2303 = 42
1993 x 2303 = 459
It is unwise to plot this line until the calculations for the " steam "
line have been made.
The width of the diagram, . e., from the water line to the steam
line (a to b, r to /, etc., in Fig. 221), is always , where L is the latent
heat at the temperature r considered. The values of the latent heat
are read from the steam tables and are set down thus :
[Taking 460 instead of 461.]
tF.
r F. absol.
L
L
T
160
620
IOO2
I6I5
200
660
974
1475
240
700
947
1353
290
750
912
1215
320
780
891
1142
Hence the scale for entropy must be chosen so that the largest
value may be shown, viz. 1844, which is obtained by adding 1615
to 229.
Plotting the values of T, taken from the last two tables, to a hori
zontal base of Q, we obtain the water and steam lines, which are
straight lines over short distances.
The vertical scale may also be numbered to read pressures, which
384 MATHEMATICS FOR ENGINEERS
may be obtained for the temperatures required from steam tables.
Thus :
T
788
753
710
672
622
P
IOO
60
30
147
5
A horizontal through 100 on the scale of P gives the line ab (corre
sponding to AB on the PV diagram), and the intersection of the hori
zontal through 147 Ibs. per sq. in. with the steam or saturation line
gives the point d.
Example 29. On the T$ diagram (Fig. 221) draw the adiabatic
line be, and also the constant volume line dcf, the latter on the assump
tion that qV is constant throughout the curve ; q being the dryness
fraction and V the volume of i Ib. of dry steam. Draw also the corre
sponding line BC in Fig. 220, and the constant volume line DCF.
The line DCF (Fig. 220) is a vertical through D, which meets the
horizontal through 5 on the pressure scale in F; but certain calcula
tions are necessary before the line dcf (Fig. 221) can be drawn.
As the pressure decreases, the volume increases. Thus at 147 Ibs.
pressure the volume of i Ib. weight of steam = 268 cu. ft., while at
10 Ibs. pressure the volume of i Ib. weight is 384 cu. ft. Consequently
if only 268 cu. ft. of steam are present at the lower pressure instead
of the 384 cu. ft., the dryness of the steam must be  r , i. e., 698 ;
3'4
and accordingly the latent heat is only 698 of its true value. Hence
if we make, on the horizontal through 10 Ibs. pressure, kx t = 'bq&kx
(Fig. 221), the point x^ lies on the line of constant volume, viz. 268 cu. ft.
At 5 Ibs. pressure, volume of i Ib. weight = 724 cu. ft.; hence
, 268
wf =  ws = 'ZJiws
724
A number of points can be found in this manner, and the smooth
curve through them, viz. dcf, is obtained.
All adiabatics on the r<f> chart are vertical lines, so that be may
be drawn. To draw the line BC in Fig. 220, proceed as follows :
Select any convenient pressure, say 60, and calculate the value of the
ft
ratio y in Fig. 221. Referring to Fig. 220, determine the position of
T! on the horizontal through 60,
RT T vti
so that 1 = 1
and other points on the curve BC may be found in like manner.
It should be noted that the adiabatic BC lies under the saturation
curve BD, since the steam is not dry throughout the expansion ; and
T?HT
the dryness fraction at any pressure is the value of a ratio like W
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 385
Example 30. Draw the adiabatics through / and F, the final
pressure being 5 Ibs. per sq. in. absolute. (Figs. 22l and 220!)
Dryncss
Fraction
Enfropy
Fig. 221. Temperatureentropy or r$ Diagram.
The point /, on the constant volume line dcf, has already been fixed ;
and a vertical through / gives the adiabatic ef.
EF is obtained from BD in just the same way as BC was derived ;
4 r t t
i. e.. ^r^ = 7 etc.
lii rt
C C
386 MATHEMATICS FOR ENGINEERS
Example. 31. Draw the Rankine cycle for the case in which the
steam is initially dry; and also for the case in which the steam at the
commencement of the expansion has its dryness fraction = ae j ab.
The initial and back pressures are 100 and 30 Ibs. per sq. in. absolute
respectively.
The Rankine cycle is made up of (i) expansion at constant pressure,
(ii) adiabatic expansion, (iii) exhaust at constant pressure, and (iv) com
pression at constant volume.
Thus the horizontals PL and pi (Figs. 220 and 221) must be drawn,
and the Rankine cycle is given by the figures ABPL and abpl for the
one dryness, and AEHL and aehl for the other.
Example 32. Draw the common steam engine diagram with a
toe drop from 30 Ibs. to 5 Ibs. per sq. in. absolute ; showing the case
when the engine is jacketed and also that when there is no jacket.
(See Figs. 220 and 221.)
If the engine is jacketed, the steam expansion line lies along the
Saturation curve, so that the diagram is ABMNX on the PV diagram
(Fig. 220) and abmnw on the r^> chart (Fig. 221) ; the line tnn being a
line of constant volume obtained in the same way as cf.
If there is no jacket, the diagram is ABPQX in Fig. 220, and abpqw
in Fig. 221; pq being a line of constant volume.
Example 33 Calculate the dryness fraction from the entropy
diagram for various temperatures, and thence plot on this diagram
the " quality " curve for the adiabatic be (Fig. 221).
At 100 Ibs. pressure the dryness fraction is i, whilst at 60 Ibs.
ft
pressure the dryness fraction = y ; and at 30 Ibs. pressure the dryness
Ip
fraction = fr. Selecting some vertical line as the base set off hori
zontals to represent these various dryness fractions, taking 9 as the
base of the curve : thus the position of y represents the dryness at 60 Ibs.
pressure. A curve through the points so obtained is the quality curve.
 Example 34. Calculate the values of the exponent in pv n = C for
the expansions represented by BC and EF, Fig. 220.
For the line BC p = 100 when v = 444
p = 13 when v = 268
also log p + n log v = log C
Thus log 100 + n log 444 = log C
log 13 + n log 268 = log C
or 2 + 6474% = log C
and 11139 + 14281^ = log C
whence by subtraction 8861 = 780771
8861
In like manner the exponent for the expansion EF is i 06.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 387
We may compare these values with those given by Zeuner's
rule; viz.
= 1035 + 'I? where q is the initial dryness.
For BC q = i and therefore n = 1035 + ** = i'i35
For EF q = 332 and therefore n = 1035 + 0332 = 1068.
Constant heat lines may be plotted on the r<f> diagram ; but
before showing how this may be done, we must indicate what is
meant by the term " constant heat line." If steam is throttled
by being passed through an orifice its dryness is greater than it
would be if the expansion were free. Thus in Fig. 223, at the
temperature r t the dryness fraction = ~? and not =J~ as for
DC* DC,
adiabatic expansion ; and the line BCj is known as a line of constant
heat.
hh Line, of Constant" Heat
Fig. 222. Constant Heat Lines.
Four cases of the drying effect of expansion without doing
external work, known as " throttling," are possible, these being
represented by (a), (b), (c) and (d) in Fig. 222.
Case (a) illustrates the expansion of a mixture of water and
steam from temperature ij to temperature r 2 . At the commence
ment of the expansion the dryness fraction of the mixture is q lt
its latent heat is L x and its sensible heat h lt while q z , L 2 and h z
are the corresponding quantities at the. temperature T Z . Then,
since the heat content is unchanged
in which equation q lt L 1( A lt L 2 and h 2 would be known, and thus
q 2 could be calculated.
Case (b) is that of water being dried, thus becoming a mixture
of steam and water. The equation here is
A! = ? 8 L 2 + h r
388
MATHEMATICS FOR ENGINEERS
Case (c) is that of dry saturated steam becoming superheated,
and for this change
^i + L! = h z + L 2 + 5(1, T 2 )
T, being the temperature to which the steam is raised by the
throttling ; r s r 2 thus being the degrees of superheat (only
obtained internally).
In Case (d) steam of a certain wetness is completely dried by
expansion under constant heat. (Any further throttling would
naturally superheat.)
For the change shown in the diagram
?iLi + hi = L 2 + &2
= 1115 7*2 + 2 2 60
= 1055 + 3/! 2
from which equation t z , the temperature at which the steam is
just dry, can be found.
From a consideration of the foregoing cases it will be seen that
lines of constant heat appear in either the " saturated area," viz.
the area between the water and steam lines, or the " superheated
area," viz. the area beyond the steam line; and these two cases
will be dealt with in the following examples :
Example 35. Steam 3 dry at 400 F. expands to 150 F., being dried
by throttling. Draw the constant heat line representing this expansion.
If T! and r a are the absolute temperatures, and h t and h t are the
sensible heats
T 1 T* r * g fr4
To draw the line of constant heat it is necessary to calculate
the dryness fraction at various temperatures. From the equation
=
2
In this equation q lt L x , and / 4 are known, whilst values of t t may
be assumed and values of L 2 calculated therefrom, or taken from
steam tables.
Now /! = 400, L x = 835, and q { = 3
Then, taking convenient drops of temperature, say 50 or 100, a
table may be arranged as follows :
t,
L 2
<i<i + fcLi
?l
400
835
o + 251
3
300
905
100 + 251
388
20O
975
2OO + 251
462
15
IOIO
250 + 251
495
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 389
[X = 835 and ?1 L t = 3 x 835 = 251 ; also ^ = 388, Jll = 462
^*^O y I j
and = 
The line of constant heat (Fig. 223) may be drawn after points such
LK
as K have been determined ; K being so placed that ^^r =* 388.
Example 36. On the r<f> chart (Fig. 223) plot the line of constant
heat for superheated steam, which is dry at 350 F.
Lines of
CorisfanTHeaF
. I I I 1 1 1 I 1 1 1 1
O 2 A 6 8 10 12 I 4 16 18
Fig. 223. T<)> Diagram showing Constant Heat Lines.
This example is a numerical illustration of Case (c), Fig. 222, and
hence we must use the equation
i = h t + L a + '5( r * ~ r i)
^
By transposition
_ ^i + LI h t L,
r, = 2(A X h 2 + L! L 2 ) + r, (absolute temp.)
or /, =2(< 1 /,+ L 1 L t ) + / I (F.temp.)
We know that * x = 350, L x = 870 ; and it is convenient to take
drops of temperature of 50 F.
390
MATHEMATICS FOR ENGINEERS
Then the table for the calculation is arranged in the following
manner :
<
L 2
Li
fjf.+LjJt,
2 x column (4)
t
350
870
870
+
o
35
300
905
870
50 35
30
330
250
940
870
100 70
60
310
2OO
975
870
150  105
90
290
Horizontals through these temperatures meet the constant pressure
lines (drawn on all charts, the equation being $ = Kp log e , i. e., the
T O
curve is of the same character as the " water" line) through 350, 300,
etc. (on the " steam " line), at points on the line required ; join these
and the line db is obtained (Fig. 223).
Example 37. Steam of 2 dryness at 266 F. is dried further by
the addition of heat and then allowed to expand through an orifice
down to 200 F., where it is 69 % wet. Find the number of heat
units added at 266 F.
This may be worked by calculation, or by use of the chart.
(a) By calculation. L at 266 F. = 929. Let x heat units be added,
and then the dryness at the end of the addition of heat
x
=  h '2
929
Let this dryness = q t
q t at 200 F., = 931% = 931
L a = 975
Then
Also
But
20
(^9 + ' 2 ) 929 + 266 = (* 931
i e., x = 907 66 186 = 655.
(b) By use of chart. Draw the constant heat line MN (Fig. 223),
,, fRM 1
starting from M. j jrg = '931 f
Then QN = 9 rank, or the heat units added = 9 x (460 + 266),
i. e., x = 655 heat units.
Construction of PV and r$ Charts for Engines other than
Steam ; e.g., The Stirling, Joule and Ericsson Engines.
Example 38. Trace the PV and T< diagrams for the Stirling
engine working between 62 F. and 1000 F., the ratio of expansion
being 3 to i. (Work with i Ib. weight of gas.)
The PV diagram consists of two constant volume lines together
with two isothermals. See Fig. 224.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 391
Starting from the point A, the pressure = 147, r = 461 f 62 = 523,
and the volume (read off from the steam tables) = 1314 cu. ft. To
find the position of the point B : It is true for all values of p, v and T
that = constant. At B the temperature is 1000 F., or 1461 F.
absolute : also the volume is 1314, hence
*. e.,
so that the point B is fixed.
147x1314x1461.,
314 x 5 2 3
40 _
B
Stirling.
15
K>
10 15 80 5 y 30 55
Fig. 224. PV Diagram for Stirling Engine.
For the isothermal BC, pv = constant, and since p B = 411 and
Va _ 1314, the value of the constant is 411 x 1314 = 540.
Using the equation pv = 540, points on BC may be found thus :
If p = 30 v = 18; p = 20, v = 27, etc. ; and the isothermal must be
continued' until C is reached, the volume at C being three times that
at B, . e., f = 3 x 1314 = 39'42
CD is vertical ; and also
T D
540 x 523 _
"f 1461 x 3942
so that the position of D is fixed.
39 2
MATHEMATICS FOR ENGINEERS
The constant for the isothermal DA is 491 X 3942 = 193 '. and
accordingly the points on the line may be obtained.
To draw the T<> diagram (Fig. 225) Suppose the entropy is zero
at the start. Then points on the line ab are calculated from the
equation <t> = K c lo&. , where K = specific heat at constant
523
volume = 1691.
4> = 1691 log e ^ = 1691 x 2303 (Iog 10 r Iog 10 523)
= 39(log 10 rlog 10 523)
J500L
/300.
1100
90O .
5oo <
O 05 / 15 '2
Fig. 225. r<p Diagram for Stirling Engine,
and the table of values reads as follows :
T
log M r10g lo5 2 3
39 x column
(2)=*
7OO
28451 
27185
049
IOOO
3*0 
27185
1096
I2OO
30792 
27185
1405
1461
31647 
27185
174
The position of c is fixed, since the work done = r ^
and thus the distance be =
log,, r
774
532 x log* 3
_ JJ _toj _ .
774
The lines be and ad are parallel, and cd is the curve ab shifted
to the right a horizontal distance be; and thus the diagram can be
completed.
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 393
Example 39. Plot PV and r$ diagrams for the Joule engine, when
the compression pressure is 60 Ibs. per sq. in. and the lower temper
ature is 62 F. Work with i Ib. weight of the gas, and take for the
adiabatics pv 1 ' 11 = C.
Dealing with the PV diagram (Fig. 226) : At C the pressure = 147,
the volume = 1314 cu. ft., and r = 523 : hence p v v = 147 x 1314 = 193.
The point A, at pressure 60, is on the isothermal through C; then
/>Af A = P<,v = 193
whence V A = ^ = 322
40
20 
10
Fig. 226. PV Diagram for Joule Engine.
For the adiabatic AD pv 1 '* 1 = K (say)
so that K = 60 x 322 1 ' 41
log K = log 60 + 141 log 322 = 17782 + (141 x 5079
= 24943
K = 3121.
Hence points on the line AD may be found from pv l ' tl = 3121.
The pressure at D = 147, and the volume = V 14^7" 8 '73 2 
Also t^ = P^
322 x 60
= 347 . 7 F. absolute.
394
MATHEMATICS FOR ENGINEERS
For the adiabatic CB, the constant=/> u 1 " 41 =i47 x i3i4 l ' 41 = 5551 ;
and thus this line may be drawn.
Substituting the
values of p, U A , T A , p B ,
1^(4845) in the equa
tion
P*v p t v A 7OO
t 5  2 = , T B is found
TB TA .
to be 7877 F. abso /
lute.
600
For the r$ diagram
(Fig. 227) : Starting
from the point c, draw
the horizontal through
it; this being the iso
thermal for 523 F.
absolute.
The distance
50O
400
i
2375 log.
3477
7877.
or
the 30
1
O
Fig. 227. T<(> Diagram for Joule Engine.
540
ratios of the tempera
tures being the same.
Points on the line ab
are obtained from the
equation
<t> = 2375 log, ~,
as also are those on cd ;
the latter values of $
being measured back
wards, i. e., towards the
left of the diagram.
The tabulation for this
calculation would be
arranged as in the
previous example, so
that there is no need
for a detailed list of
values here : and the
diagram is completed
by the verticals cb and 1o Q d ^ 18 24 y 3o 36
Fig. 228. PV Diagram for Ericsson Engine.
Example 40. Plot PV and r<p diagrams for the Ericsson engine,
when working between 62 F. and 1000 F., the compression pressure
being 60 Ibs. per sq. in. absolute. (Work with i Ib. weight of the gas.)
^o _
THE PLOTTING OF DIFFICULT CURVE EQUATIONS 395
The calculation is left as an exercise for the reader; but his results
may be checked from Figs. 228 and 229.
In Fig. 228 AB and CD are isothermals, the equations to which are
P v = 193 and pv = 540 respectively
/5oo
soo
/
Fig. 229.
O /
T(J> Diagram for Ericsson Engine.
Exercises 41. On the Construction and Use of the PV and r<j> Diagrams.
1. Construct a r<t> chart, the temperature range being 120 F. to
380 F. ; and by the use of this chart solve the problems in Exercises
2 to 6.
2. Steam 42 dry at 350 F. expands adiabatically to 140 F. What
is now its dryness fraction ?
3. Three hundred heat units are added to a sample of steam dry at
310 F. Find the dryness after the addition of the heat.
The steam is now allowed to expand by throttling to 185 F. ; find
the number of heat units that must be added so that the steam becomes
dry saturated at this lower temperature.
4. Draw the Carnot cycle, the upper pressure being 150 Ibs. per
sq. in. absolute, and the lower being 147 Ibs. per sq. in. absolute.
5. Show on the chart constant volume lines for volumes 5, 10, 15
and 20 cu. ft. respectively.
6. Draw constant heat lines in the superheat area for steam dry
saturated at 250 F. and 65 F. respectively.
7. Draw on a PV diagram the adiabatic line mentioned in Exercise 2,
working with i Ib. of steam. The equation of this expansion line
being PV* = C, find the value of n
(a) Directly from the diagram.
(b) Using Zeuner's rule, viz. n= 1035 + ' J <7> <7 being the initial
dryness.
8. Draw constantdry ness lines for dryness fractions of 2 and 3
respectively.
9. Calculate the dryness fraction for which the constantdryness
line is straight; assuming that L = 1437 "jr and <f> r = log,:
CHAPTER X
THE DETERMINATION OF LAWS
IT is often necessary to embody the results of experiments or
observation in concise forms, with the object of simplifying the
future use of these results. Thus the draughtsman concerned with
the design of steam engines might collect the results of research
concerning the connection between the weight of an engine and its
horsepower, and then express the relation between these variable
quantities in the form of a law. He might, however, prefer to plot
a chart, from which values other than those already known might
be read off. The object of this chapter is to show how to fit the
best law to correlate sets of quantities : and before proceeding
with this chapter the reader should refer back to Chapter IV,
where a method of finding a law connecting two quantities was
demonstrated. The results of the experiments there considered
gave straight lines as the result of directly plotting the one quantity
against the other, and from the straight line the law was readily
determined.
The values of the quantities obtained in experiments, except
in special cases, do not give straight lines when plotted directly
the one against the other, but, by slight changes in the form of one
or both, straight lines may be obtained as the result of plotting.
The general scheme then is to first reduce the results to a " linear " or
" straightline " equation, to plot the straight line and then to calculate
the values of the constants.
The general equation of the straight line may be stated as
Y = aX + b
or (Vertical) = a (Horizontal) + 6
where a is the slope of the line. It is the only " curve " for which
the slope is constant ; hence the reason for our method of procedure.
e. g., suppose we know that two quantities P and Q are con
nected by an equation of the form
THE DETERMINATION OF LAWS 397
We can rewrite this as
I = aQ + b
where P = P 3 and Q = Q 2
and this equation is then of the straightline form. Therefore by
plotting P against Q a straight line must result.
Conversely, if the plotting of P 3 against Q 2 gives a straight line
the equation must be of the form
, P 3 = aQ 2 + b.
In dealing with the results of any original work there will
probably be no guide as to the form of equation, and much time
will therefore be spent in experimenting with the different methods
of plotting until a straightline form is found. Sometimes the shape
of the curve plotted from the actual values themselves will give
some idea of the form of the equation, but a great deal of experi
ence is needed before the various curves can be distinguished with
certainty.
It will be found of great value to work according to the scheme
of substitutions here suggested, for by the judicious use of the
method much of the difficulty will be removed. Thus small
or large letters stand for the original quantities, and large or
" bar " letters respectively stand for the corresponding " plotting "
quantities.
e. g., we are told that given values of x and y are connected by
an equation of the type
y = bx 2 + c.
If we write Y for y and X for x z the equation becomes
Y = 6X + c
which is of the straightline form required. The change here made
is extremely simple but very effective.
Again, suppose the equation H = aD n is given as the type.
Seeing that a power occurs we must take logs : thus
log H = log a + n log D.
As this equation stands, it is not apparent that it is of the
straightline form; but by rewriting Has
H = A + nU
where H (H bar) = log H, A = log a and D = log D,
it is seen to be of the standard linear form.
We shall deal in turn with the various types of equation that
occur most frequently.
MATHEMATICS FOR ENGINEERS
Laws of the Type y = a + ; y = a + bx z , etc.
Example I. The following quantities are connected by a law of
the form y ax 3 + b
X
o
2
5
9
IO
y
8
5
3i
212
291
Test the truth of this statement and find the values of a and 6.
If we write the equation y = ax 3 + 6 as Y = aX + b, which is
permissible provided that Y = y and X = x 3 ; then if the law is true,
a straight line should result when Y is plotted against X.
300
250
100
50
810
100 ZOO MO 4OQ 5OO 6OO ^OO Boo 9OO tOOO
Fig. 230. Determination of Law for Equation of y = ax 3 f b type.
Hence the table for the plotting reads ;
X = x 3
o
8
125
729
IOOO
y = y
8
5
3i
212
291
Plotting these values, as shown in Fig. 230, we find that a straight
line passes well through the points ; and therefore the statement as
to the form of equation is correct.
Selecting two convenient points on the curve
X = 80 when Y = 15!
and X = 890 when Y = 26oJ
Inserting values 260 = 8900+6 (i)
15= 8oa+fc (2)
THE DETERMINATION OF LAWS
Subtracting
Substituting in (2)
399
245 = 8ioa
a = 302
15 = 242 + b
b = 92
Y=. 3 o2X+( 9 . 2 )
i. e., y = '$O2X 3 92.
Alternatively, a and b might be found from the graph; since
a = slope = g^ = 302 ; and b = intercept on vertical axis through
o of X =  92. .*. Y = 302X4 ( 92)
and y yzx* 92.
65
60
CP
55
50
I 234567
Amperes  A
Fig. 231. Law connecting Volts and Amperes of Electric Arc.
Example 2. An electric arc was connected up in series with an
adjustable resistance. The following readings of the volts V and the
amperes A were taken, the length of arc being kept constant and the
resistance in the circuit being varied :
V
67
63
597
58
56
538
522
Si'4
A
i95
246
3
3'44
396
4 '99
5'95
7
Find the law connecting V and A.
400
MATHEMATICS FOR ENGINEERS
By plotting V against A, as in Fig. 231, a curve is obtained which
shows clearly that the connection between V and A must be of an
inverse rather than a direct character, since A increases as V decreases.
Hence a good suggestion is to plot ^ against V, or, in other words,
to assume an equation of the form
Rewriting this equation as V = b +
equation for a straight line.
we see that this is the
Fig. 232. Law connecting Volts and Amperes of Electric Arc.
The plotting table will then be as follows :
V
67
63
59'7
58
56
538
522
5i4
A
513
407
333
291
253
2
168
143
The values of A, i. e., reciprocals of values of A, are obtained from
the slide rule. To do this, invert the slide so that the B scale is now
adjacent to the D scale. Then the product of any number on the B
scale with the number level with it on the A scale equals unity, i. e., if
the numbers are read on the B scale, their reciprocals are read on the
A scale.
The plotting of V against A gives a straight line (see Fig. 232).
THE DETERMINATION OF LAWS
401
Selecting two sets of values
and Y = 645 when
Inserting values 645 = b + 456 .
V = 52 when X = 15"!
5 = 45/
(i)
Subtracting
125 = y
c = 417
and by substitution in (i)
645 = 6+1875
whence b = 4575
=
V = 4575 +
Notice that this problem could have been attacked in a slightly
different way.
A
Multiplying through by A
AV = 6A + c
but the product of amps and volts gives watts (W).
/. W = 6A + c.
Therefore a straight line results if the power (watts) is plotted
against the current (amperes).
The table for the plotting would then read :
A
i'95
246
3
344
396
4*99
595
7
W = AV
1305
155
1791
1994
2215
269
3ii
3598
and thence the procedure is as before.
Laws of the Type y = ax". If there is no guide to the form
of equation, it is most usual to assume it to be y = ax*, or, in
more special cases, y = ax n + b ; this latter form embracing those
already discussed. To avoid the quite unnecessary expenditure of
time in searching for the form, this will be indicated before each
example or set of like examples.
If y = ax n , then, by taking logs
log y = log a + n log x
or Y = A + nX
the large letters being written for the logs of the corresponding
small ones.
This last form is the equation of a straight line, the coordinates
of the points thereon being X and Y, i. e., log x and log y. Accord
ingly, if corresponding values of two quantities are given, and it is
D D
4 oz MATHEMATICS FOR ENGINEERS
thought that they are connected by an equation of the type with
which we are now dealing, a new or " plotting " table must be
made, in which the given values are replaced by their logarithms.
These must next be plotted, and if a straight line passes through
or near the points, the form of equation is the correct one.
The values of the constants n and a may be found, as before,
by either of two methods : (a) by simultaneous equations, or
(b) by working directly from the graph.
4 60
478
476
474
E
472
470
4ee
466
08
62
44 46 48 5O 52 54
Fig. 233. Endurance Tests on Mild Steel Rods.
56
To illustrate by an example :
Example 3. In some endurance tests on mild steel rod the following
results were obtained :
Maximum skin ^
stress F in Ibs. \
per sq. in. . . J
45200
47500
48700
49000
52100
54000
56750
58700
60150 64800
1
Revolutions to \
tracture R . . /
420000
223300
207300
186200
128600
85400
69000
45000
40000 '23200
4 Find the connection between F and R in the form F = aR n .
In the log form
or
where
log F = log a + n log R
F = A + nR
F = log F, A = log a and R = log R
THE DETERMINATION OF LAWS
The table of values reads :
403
F = logF . .
4655I
46767
46875
46902
47168
47324
4  7540
47686
47793
48116
R = log R . .
56232
53489
53166
52700
51093
493I5
48388
4'6532
46021
43655
Plotting from this table, we see from Fig. 233 that a straight line
passes well through the points.
To find the values of n and a :
By method (a). Select two convenient points on the line, giving
the values
F = 468 when K = 536
F = 476 when R = 474
4'76 = A+47 4 n ' . . . . (i)
468 = A+536 (2)
08 = 62
and
Inserting values
Subtracting
whence
n =
08
62
= 129
Substituting 129 in place of n in equation (i)
476 = A + ( 129 x 474)
A = 537
but A = log a and therefore a = antilog of 537 = 234400
/. F = 234400R' 129
By method (6). F = A + nR
Hence if R be plotted horizontally n is the slope of the resulting
line. In measuring the slope, ordinary scales must be used, since the
question of logs does not arise at all; and from the equation it is
observed that n is a small letter, and therefore represents a number
and not a log.
A is the intercept on the vertical axis through the zero of the R
scale, and since the zero of any log scale is the reading corresponding
to i, A is the intercept on the vertical axis through i on the scale of R.
Obviously in the example under notice, it would be impossible to show
this axis on the diagram, at the same time choosing a reasonable scale
for R ; and consequently method (a) is the better.
The slope of the line = ^ = 129.
.'. n = 129
In many practical examples it is only the value of the exponent
that is of importance, so that only the slope of the line is required.
The slide rule may be used to great advantage in this connection,
since its scales are scales of logarithms : and therefore there is no
need to consult the log tables, for the logs of the given quantities
404
MATHEMATICS FOR ENGINEERS
are plotted directly from the rule. After plotting, the slope is
calculated, both horizontal and vertical distances being measured
in centimetres or in inches, the scales on the rule being used : this
slope is the value n.
Note. If the B scale of the rule is used for both horizontal and
,, , difference of vertical
vertical measurements, then the slope = 77^ ^, the
difference of horizontal
same units being employed for both lengths.
If, however, a more open scale is required, say, for the vertical,
i. e., the B scale is used for the horizontal and the C scale for the
vertical, then the vertical difference must be divided by 2 before
comparing with the horizontal difference.
4OG5 477
12
Fig. 234. Hardness Tests of Mild Steel.
Example 4. As a result of some tests for hardness, on mild steel,
the following figures were obtained ;
Pressure (tons per\
inch width) . . /
12
209
250
2925
3i8
4065
446
477
Indentation (ins.) .
0045
0065
0085
0105
on
0145
0155
0165
If i = indentation in inches, p = tons per inch width, and c is a
constant for the material, ci = p*.
Find the value of n.
THE DETERMINATION OF LAWS 405
For the actual plotting, shown in Fig. 234, the C scale of the slide
rule was used along both axes, and therefore n = slope = ^ = i.
For d = p n
In the log form log c + log i = n log p
or log i = n log p log c
1 = P_C
, vertical difference ., T .
= horizontal difference' lf Z 1S P lotted vertically and P hori
zontally.
Laws of the Type y = ae bx , where e = 2718, the base of
natural logs. We have already seen that many natural phenomena
may be expressed mathematically by an equation of the type
y = ae bx ; so also is it possible that an equation of this type may
best fit a series of observations so as to correlate them.
If y = ae 1 "
then log y = log a + bx log e
and, since log e is a constant and equal to 4343,
logy = log a + 43436*
or Y = A + Cx
where Y = log y, A = log a, and C = 43436.
Y = A + Cx is the equation of a straight line of slope C, and
whose intercept on the vertical axis through the zero of the hori
zontal scale is A; provided that Y, *'. e., log y, is plotted against x.
In the cases in which this law applies we have to employ both
direct and log values in the same plotting, and hence there is little
advantage in using the slide rule ; in fact, it seems better to take
the logs required from the tables only. Also, in finding the con
stants, simultaneous equations must be formed and solved.
Example 5. The following are the results of Beauchamp Tower's
experiments on friction of bearings. The speed was kept constant,
corresponding values of the coefficient of friction and the temperature
being shown in the table :
t
1 20
no
IOO
90
80
70
60
f
0051
0059
0071
0085
OIO2
0124
0148
Find values of a and 6 in the equation p = ae bt for the set of results
given.
M = ae bt
In the log form log /* = log a + bt log e = log a + 43436*
or M = A + Ct
where M  log /*, A = log a, and C = 43436.
406 MATHEMATICS FOR ENGINEERS
Hence the plotting table reads :
t
1 20
no
IOO
90
80
70
60
M = log fj.
37076
37709
38513
3.9294
20086
20934
21703
In plotting the values of M it should be remembered that 37076
is 3 + '7076, and that therefore the marking for 37076 on the
vertical scale is above that for 3, to the extent of 7076 unit.
22 _
58 .
37
6o 70 80 90 100 HO /2O
Fig. 235. Experiments on Friction of Bearings.
Plotting these values, as in Fig. 235, we find the straight line that
best fits the points. Selecting two sets of values of M and I
viz., M = 213 when t = 65 \
and M = 373 when t= 115)
we substitute these values in the equation M = A + Ct.
Thus 213 = A+ 656 (i)
373 = A+ii 5 C (2)
Subtracting 40 = 506
but
Substituting for C in (i)
whence
and
50
C = 43436
C
b =
4343
.008
'4343
= 0184.
213 = A+( 1195)
A = 1325
a = antilog of A = 2113
M = zuie' 018 * 1
THE DETERMINATION OF LAWS
407
Laws of the Type y = a + bx f ex 2 . Suppose that given
values of x are plotted against those of y and instead of the straight
line a fairly welldefined curve suits them best. The curve is
most likely to be a portion of some parabola, if not of the types
of the two previous paragraphs. Its equation may then be of
the form y = a + bx + ex 2 + dx? , any terms of which
may be absent. This case thus includes types already discussed
(e. g., y = a + bx 2 , and y = a + dx 3 ). If nothing is stated to the
contrary, and it is thought that the curve is some form of parabola,
it is usually sufficiently accurate to assume as its equation
y = a\bx\ ex 2 .
In this equation there are three constants a, b and c\ and to
determine them in any case three equations must be stated.
If, then, the equation is to be of this type, plot the given values,
sketch in the best smooth curve to pass well amongst the points,
and select three convenient points on this curve : the three equa
tions can now be formed and solved in the manner indicated in
Chapter II. If possible, one point should be on the y axis, for
then x = o and y = a f o + o ; or the value of y is such that
the value of the unknown a is found directly.
Example 6. Readings were taken as follows in a calibration of a
thermoelectric couple :
Temperature C. (T) . .
o
490
840
1003
E.M.F. (microvolts) (E) .
o
3152
5036
5773
Find (a) a formula connecting E and T in the form
E = a + 6T + cT 2
and hence (6) an expression, enabling values of T to be calculated
from any value of E.
The plotting of the values from the tq,ble is shown in Fig. 236.
Selecting three sets of values
E = 150 when T = o
E = 2600 when T = 400
and E = 5800 when T = 1000.
a = 150 {for 150 = a + o + o}
5800 = 150+ 10006+ io e c (i)
2600 = 150 + 4006 + 16 x io*c (2)
Multiplying (i) by 4 and (2) by 10 and subtracting
23200 = 600 + 40006 + 4,ooo,oooc.
36000 ~ 1500 f 40006 + 1,600,000?
408
MATHEMATICS FOR ENGINEERS
Subtracting 2800 = 900 + 2,400,0005
370 =c
2,400,000
c = 00154
Substituting in (i)
5800 + 150 = iooo& 1540
whence b 749
E = 150 + 7'49T OOI54T 2 .
6ooo_
000
/ooo _
Zoo 4oo 600 800 /ooo
Fig. 236. Calibration of a ThermoElectric Couple.
To find an expression for T, solve the quadratic
ooi 54 T 2  7'49T+ (150 E) = o.
_ _ 749 V$6 '00616(150 E)
Thus
00308
= 2430 325^/5508 + oo6i6E.
Equations of Types other than the Foregoing. Very occa
sionally one meets with laws in the form y = a + bx n , y = b(x \ a) n ,
y = a f be, or y = ax"z m . These may be dealt with in the
following manner :
(a) Type y = a + bx*.
This may be written : y a = bx or Y = bx
and is of the type already discussed ; but for the change from the
pne form to the other to be effective, the value of a, roust be known,
THE DETERMINATION OF LAWS 409
a is the value of y when % = o, so that if possible the curve
with y plotted against x should be prolonged to give this value ;
and it is worth while to sacrifice the scale to a certain extent to
allow of this being done.
Otherwise select two points on the curve, draw the tangents
there, and measure their slopes. Let the slopes be s x and s 2 when
x has the values # x and x z respectively.
Then n, b and a can be calculated from
logs, logs, , >
= ,  + i
log *! log x t
a = y l
(b) Type y = b(x + ).
If X = x f a, then y = 6X n , a standard type already discussed.
When y = o, x\ a = o or # = a, so that the value of x
where the curve crosses the x axis is a. Values of b and n can
then be found in the ordinary way.
An alternative, but rather tedious, method is as follows :
Select three sets of values of x and y, viz. * lf x z , x 3 , and y lf y 2
and y s .
A _ i
~ log (*!+*) log (* 3 + )
Then Y = A, because log y l = log b + n log (*,. + a)
log y 2 = log 6 + n log (* 2 + a)
log y 3 = log b + n log (*, + a)
Whence by subtraction
log y x log y 2 = {log (*,. + a) log (* a + a)}
and log yj  log y 3 = n{log (*! + a)  log (x 3 + a)}
{By division w is eliminated.}
For various values of a plot values of (Y A) until this equals
o ; thus the required value of a is found : and values of n and b
can now be obtained by logarithmic plotting.
(c) Type y = a +
Plot y against x; select two points on the curve and draw the
tangents there ; call the slopes of these s l and s,.
410
MATHEMATICS FOR ENGINEERS
Then
= log 5j  log S 2
4343(*i  * 2 )
b  *i
" ^^ M'T
(d) Type y = ax n z m .
The method of dealing with this form of equation will be demon
strated in the following examples :
Example 7. Assuming that the loss of head A in a unit length of
pipe in which water is flowing with a mean velocity v can be expressed
in the form
h cv 3 ~d n
find the numerical values of c and expressed in feet and second units
for a pipe of 4* diameter and 28 ft. long, using the experimental data
of the annexed table :
Loss of head in feet
58
1064
1635
Discharge in Ibs./min.
I55<>
2138
2690
The corresponding values of h, i. e., loss per foot, will be found by
dividing the first line in the table by 28, and are 0207; 0381; 0584
respectively.
To find the velocity
1550 Ibs. per min. = ^ cu. ft./sec.
60 x 624
= 415 cu. ft./sec.
Area of 4* diam. pipe = 0873 sq. ft.
Velocity = _ = 4 . 75 ft./sec.
Similarly, when Q = 2138, v = 654, and when Q = 2690, v 822.
Now h = cv 3 ~ n d n
log A = logc+ (3 n) \ogvnlog_d f d = _'3333\
= log c + (3 ri) log v n x i 5228 \log d Y5228J
= logc+ (3 n)\ogv+ 477W ......... (i)
Selecting two convenient points on the curve shown in Fig. 237,
which is obtained by plotting h against v
h = 03 when v = 58
h = 047 when v = 73
and substituting in (i) we have the equations
26721 = log c+ (3 ) x 8633 +  4 7 7 n ...... ( 2 )
2477i=logc+(3) X 7634+477W < , , , , , (3)
THE DETERMINATION OF LAWS
Subtracting 195 = (3 w) 0999
= 3 in
In = 3  195 = 105
n = 105.
Substituting in (2)
26721 = log c + (195 x 8633) + (477 x 105)
= log c + 1682 + 501
log c = 4489
.'. c = 0003083
Hence h = 0003083
411
rfl05
OE
45678
Fig. 237. Experiment on Loss of Head in Pipe.
Alternatively, we might have proceeded from (i) in the following
manner : Plot log h against log v ; find the slope of the resulting
straight line, this being the value of 3 n; find also the intercept on
the vertical axis through o of the horizontal scale which gives the
value of log c n log d, in which everything is known except c, and
then calculate the value of c.
Example 8. During experiments on the loss of head in a 6* diam.
pipe on a measured length of 10 ft. the following observations were
made :
Experiment.
Quantity
(Gals, per min.).
Loss of head
(ins.).
I
294
I 7 2
2
441
366
3
588
614
4
735
918
412
MATHEMATICS FOR ENGINEERS
Assuming that the loss of head in feet per foot run =
m + n = 3, find values of n, p. and m.
m = 3 n
and that
02
01
A 5 6 789 10
Fig. 238. Experiments on Loss of Head in 6"diameter Pipe.
d = 6", area = 196 sq. ft.
294 gals, per min. = ^ . cu. ft. per sec.
624 x 60
= 785 cu. ft. per sec.
Hence
Similarly
Q
294
441
588
735
V
4
6
8
10
Each value of loss of head (in ins.) must be divided by 10 x 12 to
bring it to feet per foot, so that our final table reads :
v (ft. per sec.)
4
6
8
10
h (ft. per foot)
0143
0305
0512
0765
THE DETERMINATION OF LAWS
413
Plot h against v (Fig. 238) and select two convenient points on the
grapn, viz.
V = 5, h= 022
v = 85, h = 057/
XT v n f d = '5 \
Now A = te^r I log d = 1699 [
I =. 3 oiJ
log h = log p + n log v + (n  3) log d.
Substituting the above values
2 7559 = log /i + 9294 + (3  n) x 301 (i)
2 '34 2 4 = log/i + 699 +(3 w) X 301 (2)
Subtracting
2304
Substituting in (i)
27559 = log/* + 1672 + 361
log p, = 4723
p. = 0005284
j,l8
Hence h = 000528^
Exercises 42. On the Determination of Laws.
[In the following exercises it should be understood that " finding
the law " means finding the constants in the equation.]
1. Find the law to express the following results of a test on an
arc lamp, in the form
W = m + A
where W = watts = volts x amps.
V (volts) .
65
72
62
68
64
66
684
A (amps)
85
5
92
8
90
105
65
2. The law connecting p, and v, for the following figures, has the
form
Find this law, which connects /x the coefficient of friction between
belt and pulley, with v the velocity of the belt in feet per minute.
V
500
IOOO
2OOO
4OOO
6000
r
29
33
38
45
51
MATHEMATICS FOR ENGINEERS
3. The working loads for crane chains of various diameters are
given in the table. Find a law connecting W and d of the form
W = a + bd*.
i
3
*
f
J
I
i
8
Load on chain W (tons)
20
45
Si
127
183
249
325
4. Bazin gives the following results on the discharge over a weir;
H being the head and m being a coefficient :
H
164
328
656
984
1312
164
1968
m
448
432
421
417
414
412
409
If m = a + u , find the law connecting m and H.
rl
5. The table of allowance for the difference / between the hypo
tenusal and horizontal measurements per 66 ft. chain in land surveying
is given for various angles of slope a :
a . .
5
6
7
8
9
IO
15
20
25
30
35
40
/ (links) .
4
6
7
i
12
i'5
3'4
60
9'4
134
181
23H
The connection between / and a can be expressed by a law of the
form / = 6o a . Find this law.
6. The following table gives the weight W of castiron pedestals
for various diameter of shaft d :
d(it.) .
i
i
\
1
i
2
W (Ibs.)
18005
18017
18138
18464
191
26
Find a law of the form W = ad s + b to connect W and d.
7. The results of experiments at Northampton Institute with model
aeroplanes were as follows :
Space (ft.) .
i
24
44
6
76
1 1 2
I 5 6
204
Time (sees.)
2
4
6
7
8
io
12
i'4
Find the law connecting S and t in the form S = Kt n .
8. Find a law connecting horse power H with speed v in the form
H = av n , the following values being given :
V
2O'I
249
302
H
I54
2135
3850
THE DETERMINATION OF LAWS
415
9. Given the following values of torque T, and angle of twist 6,
find a law connecting these quantities in the form T = a6 n .
T (Ibs.in.)
800
850
900
95
IOOO
1050
IIOO
1150
I2OO
6 (degrees)
104
1253
1541
192
2367
2928
3558
4249
5 I2
10. If d = diam. of rivet, / = thickness of plate, and d = al n , find
values of a and to agree with the figures :
d ft I
I ift
'ft
11. The following are results of a test on a Marcet Boiler :
*F
320
315
3ii
3075
303
300
297
293
287
281
277
Gauge pressure
88
80
75
70
6 4
60
55
50
45
40
35
271
265
258
251
244
240
30
25
20
15
125
10
Find a law connecting the absolute temperature T (t + 460) and
the absolute pressure p (gauge + 15) in the form r = ap n .
12. h and v are connected by a law of the form h = av n . Find this
law if corresponding values of h and v are as in the table :
V
804
1167
1443
1741
I990
h
303
. 6n
907
1221
1562
13. As a result of Odell's experiments on the torque required to
keep a paper disc of diam. 22* rotating at various speeds we have the
following :
TorqueT(lbs.ins.)
33
56
875
129
176
24
R.P.M.(n) . . .
400
500
600
700
800
900
Assuming that T = an m , find the values of a and tn.
14. The following figures were obtained in a calibration test of the
discharge of water through an orifice :
Head H .
22
18
i'4
ii
8
6
Quantity Q
89
803
723
64
5'5
485
The law connecting H and Q has the form Q = aH. Find this law.
416 MATHEMATICS FOR ENGINEERS
15. Find a law, of the form v = aH n , connecting the values :
H
25
40
60
IOO
150
250
350
V
1119
1414
1732
2238
2740
3535
4180
16. / and t are connected by a law of the form I = at z + b. Find
this law when corresponding values of / and / are :
t
187
176
167
161
149
127
iii
79
I
34'5
30
28
25
21
16
12
6
17. The resistance R of a carbon filament lamp was measured at
various voltages V, with the following Jesuits :
V (volts)
62
64,
66
68
70
72
74
76
78
R (ohms)
73
72'7
721
717
707
704
705
697
692
80
82
84
86
88
90
92
94
678
684
677
672
672
666
663
662
Find the values of a and b in the equation R = ^ + 6.
18. The following are results of a test on a loovolt carbon filament
lamp. Find values for a and b as for Ex. 17 above.
V (volts) .
54
60
65
70
75
80
85
90
95
IOO
A (amps)
7
79
86
94
104
i'ii
12
i'3
i'4
i'5
(Values of R must first be calculated from R = r )
19. The difference between the apparent and the true levels owing
to the curvature of the earth are given by
Distance in\
feet d j
300
600
900
1 200
1500
1800
2400
3000
3900
Difference of\
level h (ins.)J
026
103
231
411
643
925
1645
2'57
4'344
i
Find a law for this having the form h = Kd n .
20. If pv n = C, find n and C from the given values :
V
I
2
3
4
5
p
205
II 4
80
63
5 2
THE DETERMINATION OF LAWS
417
21. y and x are connected by a law of the form y = ax* + bx + c.
Given that
X
o
4
10
y
15
168
1875
find values of a, b and c.
22. Coker and Scoble give the following results of a test on a thermo
electric couple :
Hot junction temp. T (C.)
o
327
419
657
E.M.F. E (millivolts) .
015
384
4'5
632
Find the coefficients in the equation E = a+bT+cT z .
23. Find a law connecting E and T, in the form E = a+6T+cT 2 ,
for the case in which
T
o
490
840
1003
1283
E
o
3152
5036
5*773
6382
24. The results of some experiments by Edge with a Napier car
were
Area of wind re }
sisting surface >
42
38
34
32
28
24
22
18
16
12
A (sq. ft.) J
Speed V in m.p.h.
479
529
54
55'5
576
625
642
703
75
79
The law fitting these results has the form A = a+bV+cV 2 ; find
this law.
25. Given the equation R = a+&VfcV 2 , and a table of the
corresponding values of R and V, find the values of a, b and c.
R
o
93
21
35
V
16
14
12
10
26. The velocity of the Mississippi river was measured at various
depths with the results :
Proportional depth"!
D below surface /
i
2
3
4
5
6
Velocity (ft. per sec.)
3195
323
3253
3261
3252
3228
3181
7
8
9
3127
3059
2976
If v and D are connected by a law of the form v = a+bD+cD*.
find this law.
E E
4i8
MATHEMATICS FOR ENGINEERS
27. Find values of a and b in the equation y = ae bx for the following
case :
X
i
i*5
2
25
3
35
4
4'5
y
1328
1504
^SS
I980
2311
26
305
34'4
In Exercises 28 to 30 the law is T = 20^.
28.
29.
SO.
T
222
2466
2886 3556
524
1047
1833 2880
T
234
2738
3466
47H4
524
1047
1833
2880
T
2466
3042
4164
6326
6
524
I0 4 7
1833
2880
Find
Find p.
Find
31. Find values of a and b in the equation y = ae bx when values of
y and x are as in the table :
X
230
310
4
492
591
72
y
33
39i
503
672
856
125
32. The following particulars were obtained from an experiment on
the flow through a V notch. Determine a formula connecting the
quantity Q with the head H for the notch (Q = aH n )
Quantity (cu. ft. per sec.)
II2
88
72
17
Head (ft.)
900
815
7 = 7
"422
CHAPTER XI
THE CONSTRUCTION OF PRACTICAL CHARTS
IT has been seen that the correlation of two variables consti
tutes a graph. If two or more interdependent variables are plotted
on the same axes so as to solve by intercepts problems of all con
ditions of related variability, the result is a chart. Charts may be
classified as (a) correlation charts or graphs, (b) ordinary intercept
charts, or (c) alignment charts.
X Y
30
27 ,L
152 3A 4 5 6
5. . units H
8 IO 12 IS 20 3O AO 50
Number
Fig. 239. Chart giving Fifth Roots.
Correlation Charts may be regarded as forms of the graphs
already treated, but specially adapted for particular circumstances.
The modification in the construction of the graph frequently con
sists of the substitution of a straight line in place of a curve, the
former being far the easier to draw, and when powers occur, this
necessitates logarithmic plotting.
Example I. Construct a chart to read the fifth roots of all numbers
up to 100.
Along OX and OY in Fig. 239 mark out log scales, using the B scale
of the slide rule for both directions. The scale of numbers being along
420
MATHEMATICS FOR ENGINEERS
OX, extend this axis to show 100 at its highest reading. Set off
OA = 5 units, say 2$", and set off AB = I unit, i. e., y. Join OB
and produce to C.
Then to find the fifth root of 38, erect a perpendicular through 38
on the horizontal scale to meet OC and project horizontally to meet
OY in D, i. e., at the reading 207 : then '^38 = 207.
2 3^56 8/0 14 20
Fig. 240. Chart to show Values of t> )41 .
The value of the exponent is thus the slope of the line, and
hence this method can be used to great advantage when the power
is somewhat awkward to handle otherwise.
Example 2. In calculating points on an expansion curve, it was
required to find values of v 1 ' 41 , v ranging from i to 30. Construct a
chart by means of which the value of i; 1 ' 41 for any value of v within
the given range can be determined.
THE CONSTRUCTION OF PRACTICAL CHARTS 421
In Fig. 240 draw the axes OX and OY at right angles, and starting
from i at the point O set out log scales along both axes ; the same
scale of the slide rule being used throughout.
Make OM = i unit of length and MN = 141 units of length
(i. e., actual distances) : join ON and produce to cover the given
range. Then for v = 5, w 1 ' 41 = 97, the method of obtaining this value
being indicated on the diagram.
If it be desired to have a more open scale along one axis, allow
ance must be made in the following way :
Referring to Example i, suppose that the B scale of the slide
rule is used for the scale of numbers and the C scale of the rule
for the scale of roots. Then the slope of the line O^ (Fig. 239)
must be made = and not . The scale for roots for this case is
shown to the left of the diagram, viz., along O^j.
Ordinary Intercept Charts. A combination of two or
more graphs is often of far greater usefulness than the separate
graphs, since intercepts can then be read directly and from the one
chart.
Intercept charts may take various forms, and the following
examples illustrate some of the types :
Example 3. Construct a chart to give the horsepower transmitted
by castiron wheels for various pitches and at various speeds. The
speeds vary from 100 to 1500 ft. per min. and the pitch from in.
to 4 ins.
Working with the units as stated, and allowing for the whole
pressure to be carried by any one tooth at a time, the formula reduces to
110
V i) 2
This formula might be written as H = p z x or H = ^xV, sc
that if p is constant H oc V
or if V is constant H oc p 2 .
We may thus draw on one diagram (see Fig. 241) a number of
graphs : for on the assumption that p = 2, say, H = 4_ = 0364 V,
and this relation may be represented by a straight line. By varying
* other lines may be obtained, and as they are all straight lines passing
through the origin (for H = o when V = o) only one point on each
need be calculated, though as a check it is safer to make the calculation
for a second point.
E. g., when p = * and V = 440, H = i, giving a point on the 1
Plot values of V vertically and H horizontally. Join the origin to the
422
MATHEMATICS FOR ENGINEERS
point for which H = i and V = 440 and produce this line to cover the
given range. Indicate that this is the line for pitch = \" .
For p = 2" and V = 440, H = 16.
Hence join the origin to the point (16,440) ; produce this line and
mark it for p= 2*. By two simple calculations in each case a number
of such lines may be drawn, say for each \" difference of pitch.
To use the chart. To find the H.P. transmitted when the pitch is
3j" and the velocity is 560 ft. per min. : Draw a horizontal through
560 on the V scale to meet the sloping line marked p = si", and project
from the point so obtained to the scale of H, where the required value,
viz., 54, is read off.
1200
1KX)
>IOOO
0900
800
Jf 700
600
5oo
4OO
3OO
aoo
loo
/v
I <
if
/I
Values of H
to 50 50 7O 9O 12O I4Q 16Q /8O
Fig. 241. Chart giving H.P. transmitted by Castiron Wheels.
200
Again, if the pitch is ij*, what speed is necessary if 3^ H.P. is to
be transmitted ? Draw a vertical through 35 on the H scale to meet
the line marked p = if* and project to the vertical scale, meeting it
in V = 125.
In an exactly similar fashion a most useful chart might be con
structed to give values of the rectangular moments of inertia for
rectangular sections of various sizes. Since I (moment of inertia
of a rectangular section) = ^bh 3 , then I oc b if h is constant. Then
for each value of h a straight line can be drawn, and the chart can
be used in the same way as before.
Example 4. Construct a chart to give the diameters of crank shaft
necessary, when subjected to both bending and twisting actions, the
THE CONSTRUCTION OF PRACTICAL CHARTS 423
greatest stress allowable in the material being 6000 Ibs. per sq. in.
Given that
Equivalent twisting moment
and also X.
T, = M+ VM 2 +T 2
^/D 3
i6 7
where/ = 6000 and D = diam. of shaft in inches.
Although there are three variables, viz., M, T and D, one simple
chart suffices ; it being constructed in the following manner :
Referring to Fig. 242, select an axis OY near the centre of the page,
and along this axis set out the scale of torque in Ibs. ins. Along the
.300,000
z
7
7
A
Diam
X
100,660 sopoo 1 2 3 4 5 6
Fig. 242. Chart to give Diameters of Crankshaft subjected to Stresses.
horizontal axis OX indicate a scale for diameters, taking the maximum
value as 6^". Two of the three variables may be combined by the
following device : Suppose T = 75000 Ibs. ins. and M = 125000 Ibs.
ins. ; then set off along OP a distance to represent T, using the same
scale as along OY; make OL to represent M. With centre L and
radius LP strike an arc to cut OY in R. Then OR = T,, since
OR = OL+LR = OL+LP
= OL+ V(LO)+(OP)
= M+ VM 2 +f 2 = T e .
Now T and D are connected by an equation which can be repre
sented by a curve, and
irX6oooD 8
16
= H76D 1 .
424 MATHEMATICS FOR ENGINEERS
For this curve the plotting table is
D
i
2
3
4
5
6
D 3
i
8
27
64
I 2 5
216
T e = iiyGD 3
1176
9420
31800
75300
147000
254000
By plotting T e against D complete the chart.
For use : let it be asked what diameter of shaft is required which
is to be subjected to a bending moment of 125000 Ibs. ins. and a twisting
moment of 75000 Ibs. ins.
Set off OP = 75000 and OL = 125000 : with centre L and radius
LP strike the arc PR. Draw the horizontal through R to meet the
curve and thence project vertically to the scale of D, where the diameter
is read as 612 ins. Again, if M = 20000, and T = 30000, then D = 357,
the method of obtaining this value being as before.
A chart representing an equation similar to that in Example i
might be constructed in a slightly different and better manner;
thus :
Example 5. Construct a chart to show the quantity of water flowing
through pipes of various diameters, the velocity of flow also varying.
Let Q! = quantity in cu. ft. per sec. = area in sq. ft. x velocity of
flow in ft. per sec.
then Q t quantity in cu. ft. per min. = 6oQi
and Q = quantity in Ibs. per min. = 60 x 624 x area in sq.ft.
X velocity in. ft. per sec.
so if the diameter is given in inches and the rate of flow in ft. per sec.
area X velocity 60 x 624 x ird 2 v
^ 4XM4 =
where d = diam. of pipe in inches, and v = velocity of flow in ft. per sec.
We will assume a maximum diameter of 6 ins., and a maximum
velocity of 10 ft. per sec.
Draw two axes at right angles in Fig. 243, the vertical axis being
in the middle of the horizontal. Along OX! indicate a scale of diameters,
the range being o to 6, and along OX indicate a scale of quantities, the
range being o to 7500, to include the maximum value of Q, viz. 7350,
the value of the product 2O'4x6 2 xio. Along OY set out values of
204^2, the maximum value being 204x36= 735; and draw the
curve having the equation y = 20 4^, a table for which is :
60 x 624
d
o
i
2
3
4
5
6
204^2
o
20 4
816
1836
326
5io
735
thus obtaining the curve OA.
THE CONSTRUCTION OF PRACTICAL CHARTS 425
In the righthand division of the diagram lines must be drawn of
various inclinations, the slopes depending on the values given to v.
E. g., if v = 2, when the value of y (i. e., zo'^d*) is 500, the value
of Q is 1000, therefore join the origin to the point for which Q = 1000,
y = 500, and mark this as the line for v = 2. The diagram is com
pleted by the lines for v = i, 3, 4 . . . . . 10.
Use of the chart. To find the discharge when the pipe is 2 \" diam.
and the velocity of flow is 5 ft. per sec. : Erect a perpendicular from
2j on the d scale to meet the curve OA; then move across on the
horizontal till the line for v = 5 is met ; and a vertical from this point
on to the scale of Q gives the required value, viz. 637 Ibs. per min.
pC'
lily
sec
Quantity. Q
i . i . i . i i . ! , i
Inches
OOO SOOO 400O SOOO 600O 7OOOV
Lbsper* minute
Fig. 243. Chart to give Quantity of Water flowing through Pipes.
Again, if the quantity is 3000 Ibs. per min. and the velocity is
9 ft. per sec., to find the diameter : Erect a perpendicular through
3000 on the Q scale to meet the line marked v = 9 : draw a horizontal
through this point to cut the curve, and finally drop a perpendicular
on to the scale of diameters. The diameter required is seen to
be 4*.
If desired, the scale of Q may be modified to show values of Q
(cu. ft. per sec.) or Q, (cu. ft. per min.).
Example 6. The weight in Ibs. of a cylindrical pressure tank with
flat heads (allowing for manhole, nozzles, and rivetheads) may be
expressed, approximately, by W = ioDT(L+ D), where L = length in
feet, D = diam. in feet, and T = thickness of shell in sixteenths of an
inch. Construct a chart to show weights for tanks of any diameter
up to 5 ft. and lengths up to 30 ft. ; the maximum thickness of metal
to be *.
426
MATHEMATICS FOR ENGINEERS
Let W = ioDT(L+D) = W/T, . e., Wj = ioD(L+D).
On the left of the diagram (see Fig. 244) no notice is taken of the
thickness, i. e., W x is plotted against (L+D) for various values of D.
A number of straight lines result, since WiW (L+D).
Along OXi indicate the scale from o to 35 for (L+D), and along
OY the scale for W x from o to 1750. The scale along OX will be that
for W, the maximum value required being 8x1750, i. e., 14000 Ibs.
For the lefthand portion. Suppose D = 2, then for L = 30
Wi = loxax (30+2) = 640.
Join the origin to the point for which (L+D) = 32 and W\ = 640,
and mark this as the line for D = 2. Proceed similarly for other
values of D.
, &? I , . iff
4<X>0 6060, 8OOO 10OOO
Lbs.
Fig. 244. Chart to give Weights of Pressure Tanks.
For the righthand portion. Suppose T = ", i. e., $".
When W x = looo, W = W\T = 1000 X 8 = 8000.
Join the point for which W = 8000, Wt = 1000 to the origin, and
mark this as the line for T = $*. Draw lines for T =
similar manner.
*, etc., in a
To use the chart. Let it be required to find the weight of a tank of
length 18 ft. and of diameter 4 ft., with thickness of shell f.
Here (L+D) = 18+4 = 22. Hence erect an ordinate through 22
on the scale of (L+D) to meet the line for = 4; draw a horizontal
to meet the line for which T = f * ; then project to OX, and the value
of W is read off as 5250 Ibs.
THE CONSTRUCTION OF PRACTICAL CHARTS 427
Again, what will be the length of the tank, of diameter 4 J ft., the
thickness of shell being J*. and the weight 7000 Ibs. ?
Erect a perpendicular through 7000 on the scale of W to meet the
sloping line for which T = J", and draw a horizontal to meet the line
for which D = 45. A perpendicular through this point cuts OX t in
the point for which L+D = 385, but as D = 45, then L must = 34 ft.
Example 7. The next chart involves a considerable amount of
calculation, which, however, once done serves
for all cases. We wish to find the volume of
water in a cylindrical tank for various depths
and various lengths.
Preliminary calculation. Let the depth of
the water be h (Fig. 245).
Then OC = r h, or, taking the radius as
i ft., i  h.
Let L DOC = , then cos = L
2 21
245.
E. g., for h =
i/*
cos = i i = 9 = cos 25 50'
2 = 25 50', .., = 5 i4o'
Now, the area of the crosssection of the water = area of segment
2
6 sin Q
=
2
where & is expressed in radians
A
i. e., 6 (radians) = (degrees)
Hence our table, giving areas of crosssection for different heights,
may be arranged as follows ; h being expressed as a fraction of the
radius
/;
COS 
2
6
2
0*
e
(radians)
sin Q
0sin0
Area
I
O
o
o
o
o
2
8
3652'
7344'
1287
960
327
164
4
6
538'
io6i6'
I855
960
895
448
6
4
662 5 '
I 3 2 5 0'
2316
733
1583
792
9
i
84 16'
i68 3 2'
294
199
2741
i37i
12
2
IOI32'
203 4'
3545
'392
3837
1919
**5
~'5
120
240
4186
866
5 52
25.46
17
7
I3426'
2685 2 '
470
'999
5699
285
20
i
1 80
360
6284
o
6284
3142
428
MATHEMATICS FOR ENGINEERS
Plot a curve with h horizontally and areas vertically, as in Fig. 246.
Now volume = area x length
and for a length of 10 ft. and area 3 sq. ft. the volume = 30 cu. ft.
Hence join the origin to the point for which V = 30, A = 3, and mark
this as the line for / = 10. Add other lines for different values of / as
before.
If the chart is to be made perfectly complete, a number of curves
must be drawn in the lefthand portion, one for each separate value of
the diameter. For diam. = 4 ft., ordinates of the curve would be (]
\
\
Valu
46
\
27
Vo/i.
456
2 16 tS 0 .* Q to SO SO 40 SO 60
Fig. 246. Chart giving Volume of Water in Cylindrical Tanks.
i. e., four times those of the curve for d = 2 as already drawn. This
tends to cramp the scale, so that it is preferable to work from the one
curve and to multiply afterwards, remembering that the variation
will be as the squares of the diameters.
E. g., if diam. = 2 ft., h = 46 ft., and / = 5 ft., then vol. = 27 cu. ft.,
the lines for this being shown on the diagram.
But if the diam. = 6 ins., h = 46 x radius, and / = 5 ft., then
vol. = 27 x () 2
= 169 cu. ft.
Again, if h = 172 x radius, diam. = 5 ft., and length = 16 ft., to
find the volume proceed as indicated on the diagram. The volume for
2 ft. diam. is 456, so that the volume for 5 ft. diam.
456 x (1)*= 285 cu. ft.
THE CONSTRUCTION OF PRACTICAL CHARTS 429
The following construction may reasonably be introduced as a
chart :
Example 8. Resistances of 54 and 87 ohms respectively are joined
in parallel ; what is the combined resistance of these ?
This question may be worked graphically in the following manner
Draw OA and OB, Fig. 247, lines making 120 with one another.
Along OA set off a distance to represent 54 ohms, thus obtaining the
point E, and along OB set off OF to represent 87 ohms to the same
scale. Bisect the angle AOB by the line OC.
Join EF to intersect OC at D. Then OD measures, to the same
scale as that used along OA and OB, the combined resistance, and it
is found to be 332 ohms.
Alignment Charts. In these charts two or more variables
are set out along vertical axes, which are so spaced, and for which
the scales are so chosen, that complicated formulae may be evaluated
by the simple expedient of drawing certain crossing lines. Then
for the same connection between the variables, one chart will give
all possible values of all of them within the range for which the
chart is designed. Thus transposition and evaluation of formulae
become unnecessary; and, in fact, the charts can be used in a
perfectly mechanical manner by men whose knowledge of the rules
of transposition is a minimum.
Referring to our work on straight line graphs, we see that the
general equation of a straight line is Y = aX+6. By suitably
choosing the values of a and b we may write this equation in the
form AX+BY = C; and it is with the equation in this form we
wish to deal.
Plotting generally is to most minds connected inseparably with
two axes at right angles : that is certainly the easiest arrangement
of the axes when two variables only are concerned. Suppose,
43 o MATHEMATICS FOR ENGINEERS
now, that three, four, or even eight or nine variables occur; then
our method fails us, and in such a case it is found that vertical
axes only can be used with advantage.
It is not our intention to fill the book with alignment charts,
for examples of these intensely practical aids may be found in the
technical periodicals; what is intended is that the theory of the
Fig. 248. Principle of Alignment Charts.
charts should be grasped, so that any one can construct a chart to
suit his own particular needs and conditions.
Let us consider firstly the simplest case, viz. x{y = c, or,
as we shall write it, +u = c (u and v being adopted for the sake
of clearness, since both the u and the v axes are to be vertical,
whereas axes for x and y are horizontal and vertical respectively).
Draw two verticals AE and BF (Fig. 248) any convenient dis
tance apart, and let AE be the axis of u and BF be the axis of v.
Draw also the horizontal AB, which is to be the line on which the
zeros of the scales along the u and v axes lie.
Assume some value for c and calculate values of and v for
two cases ; set off along AE these values of u to a scale of / t units
per inch, and along BF these values of v to a scale, say, of l z units
per inch. Let AH represent the value of u when v has the value
THE CONSTRUCTION OF PRACTICAL CHARTS 431
represented by BK, and AM the value of corresponding to the
value of v represented by BN. Join HK and MN to intersect at G,
and through G draw a vertical GC, which will be referred to through
out as the midvertical.
Then AH represents the first value of ; call it u 1 ',
and BK represents the first value of v x ; call it v r
Similarly AM and BN represent 2 and v z respectively, and since
u\v = c for all values of u and v, !+*>! = c and 2 +v 2 = c 
AH, AM, BK, and BN are actual distances on the paper, hence
/ x x AH = MJ, /jXAM = u 2 , J 2 xBK = v 1( and / 2 xBN = v 2 .
Substituting in the equations u 1 \v 1 = c and 2 +v 2 == c >
(/ 1 xAH) + (/ 2 xBK)=c ........ (i)
and (/ x xAM) + (/ a xBN) =c ........ (2)
From the figure AH = AM+MH .......... (3)
BK = BNNK .......... (4)
By multiplication of (3) by l t and (4) by l z , we obtain the
equations
AHX/ 1 =(AMX/ 1 ) + (MHX/ 1 ) ....... (5)
BKx* a =(BNx/ 2 )(NKx/ 2 ) ....... (6)
By similar figures
MH AC XTV MHxCB , .
NK = CB' Whence NK= AC~ ..... (7)
Add equations (5) and (6), then
(AHx/,)+(BKx/i) =(AMx/ 1 )+(MHx/ 1 ) + (BNx/ 2 )(NKx/ 2 )
and by substitution for NK its value found in equation (7)
/
~\
MHxCB
AC
*. e., by substitution from (i) and (2)
 x/ t )
Hence Muf^xl) must equal zero, so that either
\ AL/ /
CB ,
MH = o or 'i x/2 = 
432 MATHEMATICS FOR ENGINEERS
Accordingly, since MH is not zero
X'i ........... (8)
Let the lengths AB, AC and CB be represented by m, m 1 and
itoyt 1
w 2 respectively, then equation (8) may be written 1 1 = 
..
m, L j u i W 2 ^i
whence   = , , , and by similar reasoning  = , ,* ,
m /i+/2 m ^1+^2
Any pairs of values of w and v to suit the equation w+w = c
might have been chosen, and the same argument might have been
applied, so that as long as the scales for the u and v axes and the
constant c remain the same, the ratio * will hold, i. e., there can
m z
only be the one midvertical. Also G will be a fixed spot, since it
is vertically over C, and any one crossline satisfying the equation
M+V = c will give the position of G. The length of GC is thus
fixed. Let it represent the constant c to some scale, say the scale
of 1 3 units per inch. A relation between / 3 , /j and / 2 can now be
found.
GC is an actual length, representing c to the scale of 1 3 units
per inch
Substituting in (i) and (2)
(/iXAHJH^xBK) = / 3 xGC
and (/! X AM) + (l t x BN) = / x GC.
Calculate the value of v when u = o, and plot BL to represent
this value ; join AL, then this line passes through G, by the argu
ment already given.
When u = o, v = c, so that BL actually represents c,
or BLx/2 = c
But GCx/s also = c
BLx/ 2 = GCx/ 3 .
AC
By similar triangles = .= x BL x /,
m
' =
THE CONSTRUCTION OF PRACTICAL CHARTS 433
Now  1 = A
m
or /. = /!+/,
. <?., the scale along the mid vertical is the sum of the scales along
the outside axes.
The student of mechanics may be helped by the analogy of
the case of parallel forces. If weights of Wj and W 2 are hung at
the ends of a bar of length /, their resultant W 3 is the sum of the
separate weights, and acts at a point which divides the length into
two parts in the inverse proportions of the weights. Thus, in
Fig. 249, if C is the point of action of the resultant W 3
AC_W a
A _ C _ B A C _ B
L  j I  H Um,4*m.T^ {
*fc ** *' T, *L \
Fig. 249. Fig. 24Qa.
This is exactly the same kind of thing as we have in connection
with the scales along the three axes, for we may replace W^ W 2
and W 8 by l lt 1 2 and l a respectively, and we get the bar loaded as
in Fig. 2490.
We can now proceed to the more general case, viz., that in which
the equation is au\bv = c.
Use may be made of the same diagram (Fig. 248) as that used
for the simpler equation, viz., u+v = c. To do this, however,
the scale of u must be opened out " a " times, and that of v opened
" b " times ; the distance BL, which formerly represented c, now
f*
representing T, since it shows the value of v when u is zero.
Accordingly, if l\ and 1' 2 are the new scales along AE and BF
l\= 1 and ^=4*
a b
Hence,
the scale along GC = Ii+l 2
and * =
*w 2 / x a/!
FF
4 34 MATHEMATICS FOR ENGINEERS
l\ and /' 2 would be the actual scales used. For a general
statement, therefore, we can regard these as /j and l t and the scale
along GC as 1 3 ; so we sum up our results in the forms
m l bl 2
where J lf 1 2 and 1 3 are the actual scales used.
* These results might also be summarised in the following way :
If the general equation is au + bv = c, then the scale of c (along
10'
o
A D B
Fig. 250. Alignment Chart for Equation 41* + 6 = 30.
the midvertical) = " a " times the scale of + " b " times the
scale of v, and the division of AB at C is such that
CB _ a times the u scale
AC ~ b times the v scale
To illustrate by some numerical examples :
Let us first deal with the equation 4M+6v = 30.
To construct a chart for this equation, draw two vertical lines,
as in Fig. 250, fairly well apart, say 6" (this distance being simply
a matter decided by the size of the paper and the degree of accuracy
desired). Number from the same horizontal line scales for and v,
and let the two vertical scales be equal in value, viz.
/j = J a = 2 (units per inch).
THE CONSTRUCTION OF PRACTICAL CHARTS 435
Then the midvertical must be so placed that ^ l = ^
w a/
4X2
or
i.e., MI = xm = x6" = 36*
^ D
or the mid vertical is 36" distant from the axis of .
Also the scale along the midvertical is fixed, since / 3 is given
by /!+&/ 2 , i. e., 1 3 = (4x2) + (6x2) = 20, or i* represents
20 units. If AB is the horizontal on which the zero of the scale
of u and also that of the scale of v lies, number from the point D
the scale along the midvertical, and indicate the marking for the
constant term in the equation, viz., 30.
If u = 75, v = o, and it will be noticed that if a line is drawn
from 75 on the scale of through the point C (30 on the mid
vertical), it intersects the axis of v at the point B, *. e., at the point
for which v = o. Similarly, if v = 5, then u = o, and the line
joining 5 on the axis of v to o on the axis of u passes through the
point C.
Hence if a value of u, say, is given, the value of v to satisfy the
equation 4+6y = 30 can be readily obtained by drawing a
straight line through that given value of u and the point C, and
noting its intersection with the axis of v : e. g., to find the value
of v when = 3 : join 3 on the axis of u to C and produce to cut
the axis of v ; read off this value of v, viz., 3, and this is the solution
required.
As an illustration of the fact that the alteration in the value
of c alone alters the position of the point C on the midvertical
and not the position of the midvertical, let us deal with the equa
tion 4+6v = 18. Working with the same scales, join 45 on the
axis of u to o on the axis of v, since if u = 45, v = o. This line
passes through the point C x numbered 18 on the midvertical.
To find the value of v when u = o, join o on the axis of u to 18
on the midvertical and produce to cut the axis of v in the point 3 ;
then the required value of v is 3.
Example 9. Construct a chart to read values of / in the formula
I "jd+'OO^D, where t = thickness at edge of a pulley rim, d = thick
ness of belt, and D = diameter of pulley, all in inches, d is to range
from i* to 5* and D from 3* to 10*.
Construction of the chart (see Fig. 251). Draw two verticals, say
5* apart (as in the original drawing for Fig. 251). Let values of d be
436
MATHEMATICS FOR ENGINEERS
set out on the lefthand vertical. The range of d being 4*, let 4* repre
sent this value, so that i* = i unit or / x = !. The range of D is 7",
so let 3^* represent this, so that i* = 2 units or /, = 2.
Also a = 7 and b = 005, so that /, = o/x+Wj
= (7Xi) + (005x2)
= 07+ oi = 08
i.e., i" = 08 unit of *, along the mid vertical.
To fix the position of the midvertical
m 2 _ ali _ 7XI _ _7
m^ ~~ bl t ~ 005x2 ~~ i
so that the midvertical is J x 5", i. e., 625 from the axis of d.
Chart Full Size.
Fig. 251. Alignment Chart giving Thickness at Edge of Pulley Rim.
The zero on the t scale will lie on the line joining the zero on the
axis of d to that on the axis of D. We are not, however, bound to
enlarge the diagram to allow this line to be shown ; in fact, in a great
number of cases the line of zeros or virtual zeros is quite outside the
range of the diagram. As a matter of convenience let i on the d scale
and 3 on the D scale be on the same horizontal ; then, since / = 085
when d = i and D = 3, this horizontal will cut the midvertical at
the point to be numbered 085. The scales along the three axes can
now be set out, and the chart is complete.
Use of the chart. To find the value of t when d = 5 and D = 3,
join 5 on the d scale to 3 on the D scale to intersect the midvertical
in the point 365; then the required value of t is 365. Again, if
THE CONSTRUCTION OF PRACTICAL CHARTS 437
E > = 6 and t = 325, the value of d is found by joining 6 on the axis
> to 325 on the axis of t and producing the line to cut the axis of
d in 421 ; the required value of d thus being 421.
To carry this work a step further : Most of the formula encoun
tered in practice contain products, many in addition containing
powers and roots. By taking logs, the multiplications are con
verted to additions, and the methods of chart construction already
detailed can be applied with slight modifications.
To deal with a simple case, by way of introduction :
Chart giving Horsepower supplied to Electric Motor.
Example 10. Construct a chart to give the horsepower supplied
to an electric motor, the amperage ranging from 2 to 12 and the voltage
from no to 240. [Watts = Amps x Volts and H.P. = Watts \
\ 746 ;
Taking initials to represent the quantities
W = AV and H = ^X
746
or 746H = AV.
Taking logs throughout
log 746 + log H = log A + log V.
Let log 746 + log H = C, then if for log A we write K and for
log V we write V, the equation becomes A + V = C, which is of
exactly the same form as an + bv = c, where a = b = i.
Hence / 8 = al t +bl t = li+l t
In order that the scale along the midvertical may be the sum of
the scales along the outside axes, the midvertical must be so placed
that it divides the distance between the outside axes in the inverse
proportion of the scales thereon. By the scales, it must now be clearly
understood that i* represents so many units of logarithms and not
units of the actual quantities.
Slide rule scales will often be found convenient for small diagrams.
If the B scale is used, 986* (the length from index to index) would
represent 2 units (i. e., log 100 log i), whilst if the C scale is used,
986* would represent i unit.
If a log scale is not used, the best plan is to tabulate the numbers,
their logarithms, and corresponding lengths, before indicating the scales
on the diagram. One setting of the slide rule will then serve for the
conversion of the logs to distances, according to the scales chosen.
In this case A varies from 2 to 12, i. e., log A varies from 301 to
10792, a range of about 8 units ; and a fairly open scale will result if
i* = J unit is chosen, t. e., l^ = 2.
MATHEMATICS FOR ENGINEERS
For V the range is no to 240, so that the range in the logs is 20414
to 23802, or about 35 unit ; and accordingly let l t = i.
Then l a = li+l z = 2+I = 3
w a /! 2 2
and  r = =
! j a ! i
In the original drawing (Fig. 252) m, the distance between the
outside axes, was taken as 6"; hence m 1 =
of 6*, i. e., 2", or the
midvertical must be placed 2* from the axis of A.
240
220
200
190
180
170
I6O
150
140
130
110
110
Fig. 252. Chart giving H.P. supplied to Electric Motors.
Preliminary tabulation for the graduation of the outside axes
reads :
For the A axis.
A
2
2 l >
a
V5
4
C
log A
3OI
3Q7Q
4771
5441
6O2I
6OOO
Diff . of logs
o
OQ7
176
243
3OI
^08
Actual distance from\
base line (ins.) . . j"
O
485
88
122
I'S 1
199
6
7
8
9
10
ii
12
7782
8451
9031
9542
10
10414
10792
477
'544
602
653
699
740
778
239
272
301
327
3'5
37
389
THE CONSTRUCTION OF PRACTICAL CHARTS 439
The marking for 2 is first fixed, and then all distances are measured
from that : thus to find the position of the point to be marked 4,
log 4 log 2 = 301, and since i" = 2 units the actual distance from
301
2 to 4 must be ^ , viz., 151*.
For the V axis
V
no
1 20
TJQ
140
T en
160
*5 U
loeV .
2*0414
2*O7Q2
2TT3Q
2*1461
2 1 76 1
Diff . of logs
o
O378
O72S
IO47
T5J.7
1627
Actual distance from\
base line (ins.) . ./
378
725
1*047
1347
1627
170
180
190
200
2IO
22O
230
240
2*2304
22553
2*2788
23010
2*3222
2*3424
2*3617
2*3802
1890
2139
2374
2596
2808
3010
3203
3388
1890
2139
2374
2*596
2*8o8
3oi
3203
3388
The fourth line in the latter table is obtained by division of the
third line by *i, since /, = !.
The scales can now be indicated along their respective axes, and
the midvertical may be drawn. It is not convenient in this par
ticular example to join the zero of each of the outside scales, which
would necessitate the axes being extended to show i on the A scale
and i on the V scale, since log i = o. If such a line were drawn,
however, it would be the line on which the virtual zero of the scale
of H would lie. Then the virtual zero would be ^ since when
A = V = i H = IXI . It is, therefore, the best plan to locate some
746
convenient point on the midvertical to serve as a zero. Thus, join
5 on the A scale to 149*2 on the V scale; and mark the point of
intersection of this line with the midvertical as i, since
H =
440 MATHEMATICS FOR ENGINEERS
For other graduations, tabulate thus :
H
I
15
2
3
8
1 CT H
o
1761
3OI
4771
1^9031
Diff of logs
o
1761
301
4771
0969
Actual distance from i o (ins.)
o
586
I'D
i'59
32
6
5
4
3
2
17782
1699
I602
1477
T30I
2218
301
398
523
699
738
i
132
174
2'33
The fourth line is obtained from the third by division by 3, since
/, = '3. Marking in these numbers along the H axis, the chart is
complete.
Use of the chart. To find the H.P. supplied if the current is 4 amps
and the pressure is 170 volts. Join 4 on the A scale to 170 on the
axis of V : this line passes through 91 on the H axis, and therefore
the required value of H is 91.
Again, if H = 6 and current = 25, what is the voltage? Join
25 on axis of A to 6 on the H axis, and produce the line to cut the
V axis in 179; therefore V = 179.
{It should be noted that the chart is not crowded with figures,
because clearness is desired. Charts to be used frequently, and from
which great accuracy is desired, should be drawn to a much larger scale.}
At a first reading one may be tempted to comment on the
length of calculation necessary to perform what is, after all, a
very simple operation : it must be borne in mind, however, that
(a) a most simple example has been chosen as an illustration, and
(b) a chart once constructed by this method may be used very many
times in a perfectly mechanical way.
So many formulae contain powers, that we must now investigate
the effect of the exponents on the scales, etc., of these charts, and
the modification in the construction due to them.
Flow of Water through Circular Pipes.
Example n. If water is flowing through a pipe of diameter d inches,
at the rate of v ft. per sec., then the quantity Q in Ibs. per sec. is obtained
from
THE CONSTRUCTION OF PRACTICAL CHARTS 441
Transposing
In the log form
. e.,
Q
^34~
log Q  log 34 = log d 2 + log v
log Q  log 34 = 2 log d + log v
Pi g. 253 ._Chart giving the Flow of Water through Circular Pipes
Let
C = log Q  log 34. then log t> + 2 log d = C
or V+aD = C
i. e., in comparison with the standard form, a = i and b = 2.
442
MATHEMATICS FOR ENGINEERS
Assume the range of pipe diameters to be i* to 9*, and the range
of the velocity of flow to be i to 10 ft. per sec. Then the same scale
will be convenient for both axes. Let / x = / a = 7^7, i. e., use the
B scale of the slide rule.
Now
4'93
2X
4'93
IX
4'93
so that if m is taken as 6* (as in the original drawing for Fig. 253),
m = x6*, i. e., 4*, or the midvertical is 2" removed from the axis of v.
3
Also I.  * 1+ W.
+ axjj   607,
or i* = 607 unit along the axis of Q.
Draw the axes of v, Q and d and graduate the outside ones, using
the B scale of the slide rule. In Fig. 253 the I of each scale is on a
horizontal, but it is quite immaterial where the graduations begin.
To select a startingpoint on the midvertical, join 10 on the axis
of v to i on the axis of d, and call the point of intersection with the
midvertical G.
Now, Q = 34^ 2 w, and therefore for the particular values of d and v
chosen, Q = 34Xi 2 x 10 = 34.
G is therefore at the position to represent 34 Ibs. per sec.
The table for the graduation of the midvertical will then be :
Q.
V4
3
2
c
8
IO
loe O .
C32
4.77
3OI
600
on 3
I
Diff . of logs
o
CIS
'231
l67
371
468
Distance above or below G
o
O9
38
27
61
77
15
20
30
40
5
80
IOO
1176
I30I
1477
1602
1699
1903
2
644
769
945
107
1167
i37i
1468
i 06
126
i'55
i'75
192
225
2 4 I
The fourth line is obtained by multiplying the third by 164 or by
dividing it by 607, since l a = 607.
Use of the chart. Find the discharge through a pipe of 9* diam.
when the flow is at the rate of 2 ft. per sec. Join 2 on the axis of v
to 9 on the axis of d, to intersect the axis of Q at 55 ; then the required
quantity is 55 Ibs. per sec.
Again, what diameter of pipe is required if the discharge is 30 Ibs ./sec.
THE CONSTRUCTION OF PRACTICAL CHARTS 443
and the rate of flow is 3 ft./sec. ? Join 3 on the v scale to 30 on the
Q scale and produce the line to cut the axis of d in 54; then the
required diameter is 54*.
To iUustrate the question of scales further, consider the follow
ing cases :
Example 12. Show how to decide upon the scales for the chart
giving the values of T, / and d in the equation T = ^fd*, referring to
the torsion of shafts.
T
The equation may be written   = fd 3 , i. e., 5iT =fd 3 , and by
taking logs throughout
log 51 + log T = log/ + 3 log d.
Write this log/+3logd = C, then F+ 3 D = C (the large letters
being written to represent logs). Thus a = i and 6 = 3.
Hence
if li = 5. sa y. and I* = 2, /, = al^+bl^ = (iX5)+(3X2) = n
w i bl t 3x2 6 6
and ? = * = a = _ or m x = of w.
m t alj. 1x5 5 ii
Similarly for pv n = C, where n may have values such as 9, 137,
141, etc.
Here log p + n log v = log C
t. e., P + wV = C
so that a i, b = n.
Hence l a = (ixlj+(nxl a ) = l t +nl t
m, bl t nl t
and _* = J = _?
W 2 0/ /!
Questions involving more complicated formulae can be dealt
with by a combination of charts. From the above work it will
be seen that when three axes are employed, three variables may be
correlated, or one axis is required for each variable. However
many variables occur, they may be connected together in threes,
so that the graph work is merely an extension of that with the
three axes.
Chart giving Number of Teeth in Castiron Gearing.
Example 13. To construct a chart giving the number of teeth
necessary for strength in ordinary castiron gearing.
Given that T = ~j^r
where T = No. of teeth in wheel, N = revs, per min.
H = H.P. transmitted, p = pitch.
444
MATHEMATICS FOR ENGINEERS
H
so that Tp 3 = C (i)
and also ^p = C ' (2)
i . e., two charts can be constructed, and by suitably choosing the scales
and the positions of the axes the charts may be made interdependent.
For chart (i), let / t = i unit of T and let /, = unit of p,
i. e., use the B scale of the slide rule for both the T and the p axes.
N
5 
Fig. 254. Chart giving Number of Teeth necessary in Castiron Gearing.
Then, since logT+ 3 log/> = log C, /, =
4
4
4'93
Also  =
493^3 BO that m 1 = 3 of m.
_l_ i 4
4'93
Draw two axes for T and p respectively 4* apart, as drawn in the
original. drawing for Fig. 254, and also put the midvertical 3* from
the axis of T. The last is simply a connectinglink between charts
(i) and (2), and therefore no graduations need be shown upon it.
THE CONSTRUCTION OF PRACTICAL CHARTS 445
Along the axis of T mark off readings (using the B scale of the slide
rule) for, say, 6 up to 80 and along the axis of p, readings from i to 8.
Join 2 on the paxis to 30 on the axis of T, and note the point of intersection
with the midvertical ; this must be marked 240, since 30x2' = 240.
For chart (2), we already have the midvertical and its scale. We
must now proceed to find scales for the axes of H and N, t. e., the usual
process is reversed.
Suppose the range of H is 5 to 100 and that of N is 20 to 150, and
we decide to use the same scale for both, say l t .
TJ />
Then ~ N = 79! or log H  log N = log C  log 791
'. e., a = i, whilst 6 = i. If, however, the numbering for N is placed
in the opposite direction to that for H, we may say that b = i.
Hence /, = / 4 +/ 4 = 2/ 4 or / 4 = ? = = 406
also
*
t. e., the midvertical, which has already been drawn, must be midway
between the axes of H and N. For convenience let m l = m t = 4*.
Then for N the tabulation is as follows :
N
20
on
dO
CQ
60
log N
T 7QI
I '4.77
I '60 2
TfiQQ
I'778
Diff . of logs
o
176
3O I
308
477
Distance from mark for 20
o
432
74
9 8
II7
70
80
90
100
150
2OO
I845
1903
1954
2
2176
230I
544
602
653
699
875
I
i'34
148
161
172
216
246
To obtain the fourth line from the third divide by 406, for / 4 = 406.
Some little trouble may arise in the placing of the marking for 20
conveniently : thus in our case we have marked 20 fairly high up on
the paper.
Join any point on the N axis, say 150, to the 240 on the mid vertical,
and produce this line to cut the axis of H in the point A. We must
now find the reading for A.
C = 240 and C = ^ 9 * , but N = 150
TT 150x240
hence H = * ^* = 4 5'5
Thus we can graduate the axis of H from 455 as zero.
446 MATHEMATICS FOR ENGINEERS
The table for the graduation of the H scale is :
H
4 V5
<;
10
15
20
2<;
loeH
1658
699
i
1176
1301
1398
Diff. of logs. . .
o
'959
658
482
357
260
Distance from A
o
236
162
119
88
64
3
35
40
5
60
80
100
1477
1*544
1602
1699
1778
I903
2
181
114
056
041
120
245
342
445
28
14
i
295
602
84
/ = 406, so that the fourth line is obtained by dividing the figures
in the third line by 406.
Use of the chart. Suppose N = 20, T = 20, p = 5, and the value
of H is to be found. Join 20 on the T axis to 5 on the axis of p ; and
let this line intersect the midvertical at G. Join 20 on the N axis
to G, and produce the line to cut the axis of H in 63. Then the required
value of H is 63.
Again, if H = 15, N = 80, and p = 2, we are to find the value of T.
Join H = 15 to N = 80 to cut the midvertical at D. Join p = 2 to D,
and produce to cut the axis of T in 186 : then the required value of
T is 1 86.
The lines must join values of either T and p or H and N, because
the chart was so constructed.
Exercises 43. On Alignment Charts.
1. Construct a chart giving values of u and v to satisfy the equation
2M+7V = 52, the range of v being 2 to 12.
2. Construct a chart to give values of u and v to satisfy the equation
I2V 6414 = 85, u ranging from 5 to 20.
(To allow for the minus sign, either the mid vertical may be placed
outside the axes of u and v, as for unlike parallel forces, or the number
ing on the u scale may be downward, whilst that on the v scale is
upward.)
3. The thickness of boiler shell necessary if the working pressure
is p Ibs. per sq. in., the diameter of the boiler is d inches, and the allow
able stress is /Ibs. per sq. in., is found from t = &j. Taking the value
of /as 10000, construct a chart to give values of t, the range of diameter
being i'6* to 6 ft., and the pressure varying from 40 to 150 Ibs. per
sq. in. What is the thickness when the diameter is 2 '3* and the
working pressure is 85 Ibs. per sq. in. ? If the thickness is J" and the
diameter is 4'6*. what is the working pressure ?
THE CONSTRUCTION OF PRACTICAL CHARTS 447
4. According to the B.O.T. rule the permissible working pressure
in a boiler having a Fox's corrugated steel furnace is P = ^, where
/ = thickness of plate in sixteenths of an inch and D is the internal
diameter in inches. Construct a chart to give values of P for boilers
of diameters ranging from 2 ft. to 5 ft., the thickness of the shell
varying between fa" and *.
5. The diameter in inches for a round shaft to transmit horsepower
3/rr
H at N revs, per min. (for a steel shaft) is given by d = 2g\/ =^. If
N varies from 15 to 170 and H from J to 10, construct a chart to show
all the diameters necessary within this range.
6. For tinned copper wire the fusing current C is found from
C = 6537<Z 1 ' 403 , where d is the diameter in inches. Construct a chart
to read the diameter of wire necessary if the fusing current is between
22 and 87 amperes.
7. Hodgkinson's rule for the breaking load for struts is
_
1
where d = diameter in inches and L = length in feet, A being a con
stant. Construct a chart to give the breaking load for castiron struts
with rounded ends, the diameters ranging from 2* to 15* and the
lengths from 6 ft. to 20 ft. The value of A for solid castiron pillars
with rounded ends is 149.
8. Construct a chart to give the points on an adiabatic expansion
line of which the equation is pv l ' n = 560, the range of pressure being
147 Ibs. per sq. in. to 160 Ibs. per sq. in.
9. The coefficient of friction between a certain belt and pulley was
32. If the angle of lap varies from 30 to 180, construct a chart to
give the tensions at the ends of the belt, the smaller tension varying
from 50 Ibs. wt. to 100 Ibs. wt. Given that T = te?*, ft being the
coefficient of friction, and 6 being the angle of lap in radians. [Note
that the scales for T and t will be log scales, but that for 6 will be one
of numbers only.]
10. If P = safe load in tons carried by a chain, d = diameter of
stock, and / = safe tensile stress, then for a chain with open links
P =  4 <* 2 /.
If/ varies between 4 and 10 tons per sq. in., and the diameter of the
stock ranges from J* to 2", construct a chart to give the safe load for
any combination of / and d.
CHAPTER XII
VARIOUS ALGEBRAIC PROCESSES, MOSTLY
INTRODUCTORY TO PART II
Continued Fractions. Consider the fraction
i
3+;
or, as it would usually be written as a continued fraction
1 J _g._ 4
2+ 3+ 5+ 7
Its true value would be found by simplification, working from
the bottom upwards; thus, 5+= ^
39
7
39'
131
131
"39"
and
L = *3i
301 301
39 39
131 131
o*
o
o/
02
\N? of Converge nh
X 2 3_L 4
i. e., the true value of the
fraction, known as a continued
fraction, is ^ Fi &' 2 55
301
Conversely, if the resulting fraction ^1 is given, various approxi
mations can be made for it by taking any portion of the continued
fraction, the correct order being maintained.
VARIOUS ALGEBRAIC PROCESSES 449
Thus, for example, a first approximation would be , which is
T Q
too large ; the second approximation is = , which is too
2+ 7
small, but is nearer the correct value.
The approximations, or convergents, are alternately too large
and too small, but the error becomes less as more terms of the
fraction are taken into account. To illustrate this fact a curve is
plotted dealing with the above fraction, in which the ordinates are
errors and the abscissae the numbers of the convergents. (Fig. 255.)
JO J
Occasionally in engineering practice a fraction such as *
occurs which is not convenient to deal with practically, so that
one seeks for some more convenient fraction which is a fair approxi
mation to that given. The following example introduces such a
case :
Example i. A dividing head on a milling machine is required to
be set for the angle 19 25'!* with great accuracy.
For the Brown & Sharpe dividing heads, 40 turns of the crank
make one revolution of the spindle, and there are three index plates
with number of holes as follows
15, 16, 17, 18, 19, 2o^
21, 23, 27, 29, 31, 33V
37. 39, 41. 47, 49
Thus one turn of the index crank would give an angle of ~ , i. e., 9.
4
Evidently two turns will be required for 18, and i25'i* has then to
be dealt with. Expressing this as a fraction of 9, the proportion of
one turn is found.
, _
Now i25'i" =  mms.
. u 5101 5101
hence the fraction of one turn required = ^ x Q x ^ = ^35
We wish to reduce this fraction to its best approximation having
a denominator between 15 and 49, according to the numbers of holes
as above.
Proceed as though finding the G.C.M. of 5101 and 32400.
Thus 5101)32400(6
I794)5 IOI ( 2
281)1513(5
108)281(2
65)108(1
22, etc.
G G
^o MATHEMATICS FOR ENGINEERS
Then the continued fraction is
_i_ j__ JL _i_ JL T 
6+ 2+ 1+ 5+ 2 + l +
ist convergent = g, 2nd = ^, 3 rd = ' 4th = 108' 5th = 235 '
these being found by simplification of the fraction, a method which is
a trifle laborious. The 3rd convergent might have been found from
the 2nd in the following way
Numerator of 3rd convergent = {numerator of 2nd convergent
X denominator of last fraction
added} + {numerator of ist con
vergent x numerator of last
fraction added}.
Denominator of 3rd convergent == {denominator of 2nd convergent
X denominator of last fraction
added} + {denominator of ist
convergent x numerator of last
fraction added}.
In this case
12 I
ist convergent = ^, 2nd = and the next fraction = 
x i) + (i x i) 3
,
3 rd convergent =
x +6 x
Now from the and and 3rd convergents the 4th convergent may be
2 ^
found ; for the 2nd convergent = , the 3rd = , and the next
fraction = .
(3 x 5) + (2 x i) 17
To obtain the 5th convergent : the 3rd convergent = , the
17 I
4th = ^, and the next fraction = 
,, ,, (17 x 2) + (3 x i) 37
.*. the 5th convergent = 7^ : , '. = ^ L
(108 x 2) + (19 x i) 235
For the purpose of the question we require the convergent with
denominator between 15 and 49 : the only one is . Therefore, it
would be best to take two complete turns together with 3 holes on the
i9hole circle. The error in so doing is very small. Thus
5101 , .. , 3
"* = ' I 5744 whilst = 15790
32400 J/ ^ 19
i.e., the error is 46 in 15744 or ^ x 100 %
J 5744
= 292 % too large.
VARIOUS ALGEBRAIC PROCESSES 451
Example 2. Find a suitable setting of the dividing head to give
2i'45*.
No. of turns of index crank = 88 2I/ 45* = 2I2 ? = Q I7<>7
9 2160 ^2160
589
= g2_*
720
Hence 9 complete turns are necessary together with ^? of a turn.
To find a convenient convergent for ^5_? :
720
589)720(1
65)131(2
1)65(65 /. The fraction = L L L
i+ 4+ 2+ 65
The ist convergent = , the 2nd convergent = 4.
so that the 3rd convergent = J4 * 2) + (i x i) = _o_
^ x 2) + (i x i) ii
also the 4 th convergent = / 9 X 6 ? + ( f x ') = 5??
(n x 65) +(5x1) 720
Thus the best convergent for our purpose = , and 27 holes on the
33hole circle would give this ratio.
Therefore, 9 complete turns together with 27 holes on the 33hole
circle are required.
An interesting example concerns the convergents of IT.
Example 3. To 5 places of decimals the value of IT is 314159 :
what fractions may be taken to represent this ?
14159
314159 = 3 ^ ^
J ~ */ ^ */ \
'IOOOOO
14159)100000(7
887)14159(15
5289
854)887(1
33)854(25
194
29)33(1
4
i i
I.e., rr  3 + 
7+ 15+ i+ 25+
452 MATHEMATICS FOR ENGINEERS
22
The ist convergent = 3, the 2nd convergent =
, (22 x 15) +(3x1) 333
and hence the 3 rd convergent = \ 7 x ^ + x J  g
a. X l) + (22 X l) 355
the 4 th convergent = x ^ + \ y x ,/ = ff
the 5 th convergent  J355 x 25) + (333 x i) = 9208
(113 x 25) + (106 x i) 2931
The values of these convergents in decimals are
3, 314286, 314151, 314159 K and 314159 , respectively.
A rule often given for a good setting of the slide rule for multiplica
tion or division by IT is : Set 355 on the one scale level with 113 on the
other, etc. The reason for this is seen from the above investigation;
5^ as a value for TT being far more accurate than, say,
113 7
Partial Fractions. Consider the fractions
 ,  , and their sum.
X 4 2# 7
To find their sum, i. e., to combine them to form one fraction,
the L.C.D. is found, viz., (x 4) (2* 7) or 2x 2 15* + 28; the
numerators are multiplied by the quotients of the respective de
nominators into the L.C.D. , and the results are added to form the
final numerator.
Thus 2 I 4
'
x 4 ' 2x 7 2x 2 15*428
8x 30
The fraction last written may be spoken of as the complete
2 A.
fraction, for which  and  are the partial fractions.
x 4 2* 7
It is often necessary to break up a fraction into its partial
fractions : they are easier to handle, and operations may be per
formed on them that could not be performed on the complete
fraction.
To resolve into partial fractions, proceed in the manner out
lined in the following examples :
Example 4. Resolve , * ~ 3 5 into partial fractions.
2X Z 1$X + 28
8* 30 8* 30 A
(2*7)(* 4 ) (X  4)
where A and B have values to be found.
VARIOUS ALGEBRAIC PROCESSES 453
Reduce to a common denominator, (zx 7) (# 4), and calling
this D
8* 30 _ A( 2 *  7) + B(*  4)
D D
Equating the numerators
8*  30 = A(2*  7) + B(*  4 ).
This relation must be true for all values of x : accordingly let
x = 4, this particular value being chosen so that the term containing
B, vanishes, and one unknown only remains.
Then 3230 = A (8  7) + B( 4  4)
or 2 = A.
Now let the term containing A be made to vanish by writing 3$
in place of x
Then 28  30 = A(7  7) + B( 3 i  4)
 2 =  B
B = 4
/. the fraction =
*4 2 *~7
Example 5. Express ~ as a sum of two or more fractions.
The numerator and denominator are here both of the same degree ;
in such cases divide out until the numerator is of one degree lower
than the denominator.
Now suppose ^ = C with D remainder
then the fraction ~ = C+ ^
Applying to our example, by actual division the quotient =  and
the remainder = ^ : hence the fraction = * f ,/J. m
3 3 6\5 X )
Example 6. (a) Find the sum of
4 7*
(2* + I) 5 (X + I) 8 (X + I)
and (b) resolve ~" 24 * ~T, 12 * ? 2 into P artial fractions.
* f*A* "
(a)
2*+ i 5(^4 i) 2 x+ i
4 X 5^ + I) 2  7Ar(2^r + i)  3 X 5(* + l)(2* + l)
5(2^+ i)(x+ i) 8
2o^ + 40^ + 20  14**  7*  30*'  45* ~ I 5
D
 24**  12* + 5
454 MATHEMATICS FOR ENGINEERS
(6) To resolve ~ 4 ., .% into partial fractions, therefore, it is
necessary to consider the possibility of the existence of (x + i) as a
denominator, in addition to (x + i) 8 , for (x + i) is included in (x + i) 2 .
Let the fraction =
[Bx is written in place of B, so that the numerator shall be of degree
one less than the denominator, i. e., all terms of the numerator, when
over the same denominator, will then be of the same degree.]
Thus the fraction = A(* + ,)* + 5 B*( 2 * + i) + 5 C(a* + x)(* + z)
Equating numerators
 2 4 * a  12* + 5 = A(* + i) 2 + 5B*(2* + i) + 5C(2X + i)(* + i)
Let x = i {i.e., terms containing (#+i) are thus made to vanish}
/. 24+12+5 = + 5 B(i)(i)+o
Let x = {i. e., 2X+i = o}
6+6+5 = A
A = 20.
The numerators must be identically equal, i. e., term for term r
therefore the coefficients of # 2 must be equated.
Thus
24 = A + loB + loC = 20 14 + loC I for A = 20 and B = 2
/. zoC = 30
C = 3
/. the fraction 2
. , .  . ..  _
5(2^+1) 5(*+i) 2 x+i
4 7* 3
Example 7. Resolve . _ ^,~^^ x , x into partial fractions.
Let the fraction = . A . + B * + C
(2X  3) (x* + 5* + 9)
_ A(* 2 + 5 x + 9) + (B* +C)(zx  3)
Equating the numerators
9*i7 = A(# 2 + 5* + 9) + (B* + C)(2x  3)
VARIOUS ALGEBRAIC PROCESSES 455
Let
Then
=  i. e., let 2*  3 = o
A 14
75
Equating the coefficients of x*, and as no terms on the L.H.S.
contain # 2 , its coefficient = o,
o = A + 2B=H + 2B
/. 2B = H
75
B = J7
75
Equating the coefficients of x on the two sides of the equation
_ 7 66
75
~~75~
7* + 383
14
/. the fraction j f  ; r . ?
75 (** + 5* + 9) 75(2*  3)
Limiting Values, or Limits. Let it be required to find the
value of the fraction ^T 1 when x = i.
AX ~^~~ J
^>y 2 O
When x =*= i, ^r '= if x be replaced by i.
4# \x 5 o
We can give no definite value at all to ; it might indicate
anything, and therefore we must find some other method for dealing
with cases such as this.
Let us calculate the value of the fraction F when x is slightly
less than i, say when x has the value 9 :
Then F = i'
When x has a value nearer to i, say 95
456
MATHEMATICS FOR ENGINEERS
Now let us take values of x slightly in excess of i.
~ 212 i
F = . ._ .  = = 2174.
When x = 105,
When x = 11,
F =
441+1055
22 2
46
484+II5
Therefore for values of x in the neighbourhood of i the fraction
has perfectly definite values,
and consequently it is un
reasonable to suppose that
there is no value of F for
x = i. If we plot a curve,
as in Fig. 256, of F against '225
x, we see from it, assuming
that it is continuous (and
there is nothing to negative
this supposition) that the
value of F when x = i is
2222.
We say, then, that the
limiting value of F when x
approaches i is 2222, or
2X2
22 .
L
4* 2 +*5
= 2222.
Fig. 256.
To obtain this value without the aid of a graph we might take
values of x closer and closer to i and see to what figure the value
of F was tending
e. g., when x = 99, F = 2232
when x = 995, F = 2227.
This method, besides being somewhat laborious, is not definite
enough.
As an alternative method : if x does not actually equal i but
differs ever so slightly from it, (x i) does not equal o, and there
fore we may divide numerator and denominator by it.
(x i) 2
Thus
F=
(*i)( 4 *+5) (,
As x approaches more and more nearly to i, this last fraction
becomes more nearly =  and in the limit when x = i, F = .
9 9
Later on we shall see that this method of obtaining a value or
limit by " approaching " it is of great utility and importance,
VARIOUS ALGEBRAIC PROCESSES
457
Example 8. Corresponding values of y and x are given in the
table :
X
3'9
3'94
397
402
405
4'i
y
3042
3104
3I52
3232
3280
3362
Required the probable value of y when x = 4.
When x has values slightly under 4, those of y are increasing
fairly uniformly; thus for an increase of x from 39 to 394 (i. e., 04)
the increase of y is 62, or the rate of increase is , i,e., 155, and
.04
whilst x increases a further 03 unit, y increases 48 unit, or the rate
of change of y compared with x is , i. e., 16. Thtis y is increasing
at a rather greater rate as the value of x increases. This is confirmed
by dealing with values of x greater than 397 : we might tabulate the
differences of x and of y thus :
Change in x.
Change in y.
Rate of change of y.
80
397 to 402, i. e.,
05
80
= 16
'O5
402 to 405, i. e.,
03
48
48 _
2 = 16
03
82
405 to 41, . e.,
05
82
^05"
Therefore, as nearly
as we can estimate, 54
when x has the value 4,
11
*<
y has a value very
slightly over '^ of 80, 53
i. e., slightly more than
48 above its value
\J
JG
o
s
&
f
the value of y when
x = 4 is most probably
3i52+'4 8 t. e, 32. si
This result is further
illustrated by the graph
(see Fig. 257). JQ
:
Example 9. Find
the value of
2# 2 + 1 8*
^
/
+ 28
595
]
w
4 4O5 ^ 
Fig. 257
hen x = 2,
I3* 3 + ?6# a ? 76*  t
458 MATHEMATICS FOR ENGINEERS
TH * R 2(* 2 *2
in F=
+13^38^ 80) ~(x+2)(6x* + x 40)
[(x + 2) is tried as a factor, use being made of the Remainder
Theorem, to which reference is made on p. 55.]
. F _ (*+7) 2+7 . = 5
6# 2 +# 40 24240 1 8
= ^ when x = 2.
Io
(# + a) 4 x*
Example 10. Find the limiting value of     when a = o.
By direct substitution of o for a we again arrive at the indeter
minate form .
o
Proceeding along other lines
F = (x + a)*  x l _ x* + 4* 8 a + 6* 8 a 8 + 4*a 3 + a 4  x*
a a
4# 3 a + 6# 2 a 2 + 4x0? + a*
^ . .
a
If a is to equal o, and the value of F is then required, this value
must differ extremely slightly from the value if calculated on the
assumption that a is infinitely near to o but not exactly so.
If a is not zero, we may divide by it
then F = 4* 3 + 6x*a + 4x0* + a*.
Hence, the limiting value to which F approaches as a is made
nearer and nearer to zero is 4# 3 , for all the terms containing a may
be made as small as we please by sufficiently decreasing a.
L
L(x + a) 4 x* , . ,, , , . , .
5  '  = 4# 3 is the abbreviation recognised for the
statement : " The limiting value to which the fraction ^ a ' ~ x
a
approaches as x approaches more and more nearly to a, is 4**."
Example n. Find the limiting value of ''^j when 6 = o, it being
given that
sin 6 = 6  6  + ^1  . ( 6 te^Z measured:!
6 120 I in radians J
Adopting this expansion
sing _ 6 ^ 120 2
6 ~e~ ~ 6
and L sm ^ _ T as terms containing 2 and higher powers of
0>o 6 6 must be very small compared with i.
VARIOUS ALGEBRAIC PROCESSES 459
This result is of great importance : for small angles we may replace
the sine of an angle by the angle itself (in radians). This rule is made
use of in numerous instances. Thus when determining the period
of the oscillation of a compound pendulum swinging through small
arcs, an equation occurs in which sin is replaced by 6 ; the
change being legitimate since 0, the angular displacement, is small.
Exercises 44. On Continued Fractions, ete.
1. Find the first 4 convergents of 809163. By how much does
the 3rd convergent differ from the true value ?
2. Find the 5 th convergent of  ^ .
3. Convert 4 into a continued fraction. What is the 3rd
convergent ?
4. Express as a continued fraction the decimal fraction 08172.
5. Using the dividing head as in Example i, p. 449, an angle of
59i4 / 5* is required to be marked off accurately. How many turns
and partial turns would be required for this ?
6. Similarly for an angle of 73 2'i9*.
7. Similarly for an angle of 5i9 / 3i'
8. It is desired to cut a metric screw thread on a lathe on which
the pitch of the leading screw is measured in inches. To do this two
change wheels have to be introduced in the train of wheels to give the
correct ratio. If i cm. = 3937*, find the number of teeth in each of
the additional wheels, i. e., find a suitable convergent for the decimal
fraction 3937.
On Partial Fractions.
9. Express 3*+ 8 Q^ a sum or difference of simpler fractions.
x 2 + 7* + 6
1016. Resolve the following into partial fractions
2 H 3* +5 12 *(* + ')
10 ' 6*+i9*+i5 ** ~ 3*  88 " **  3* + 2
6x*  9* + 3Q 14  22**  179* ~ 240
l6 ' (x 5 )(x*+2X 8) ' 6* 3 + is* 2  57*  126
4C 3* + 2 16 2 * ~ 3.
1J>> x 3 + 2x*  x  2 * (x  3)(* 2 + 3* + 3)
On Limiting Values.
17. Find the limiting value of * 2 * + g when x =  I.
L x s + 3*2 ijx + 14
*. ^ ^ ^2 + 2Ar _ 8
19. Show exactly what is meant by the statement
T x*  6ax + 5 g2 _ _ji
^^o x z + gax ioa a n
20. Determine the limiting value of the sum of the series 16, 8,
4, 2, etc.
460 MATHEMATICS FOR ENGINEERS
21. A body is moving according to the law, space = 4 x (time) 3 .
By taking small intervals of time in the neighbourhood of 2 sees., and
thus calculating average velocities, deduce the actual velocity at the
end of 2 sees.
x x 3
22. If e? = i + x H 1 h . . . . ; find the limiting value of
the fraction when x = o.
6 Z 6* 6 3 6 6
23 If cos 6 = i 1 ; and sin 6 = 6 1
1.2 T 1.2.3.4 1.2.3 I2.345
Find i > sin 6, L cos 6, and by combination of these results
0>o 0>
\ ^ . an . Hence show that no serious error is made when calling
9^K> 6
the taper of a cotter the angle of the cotter.
24. Find L **"
Permutations and Combinations. Without going deeply
into this branch of algebra, we can summarise the principal or
most useful rules.
By the permutations of a number of things is understood the different
arrangements of the things taken so many at a time, regard being paid
to the order in these different arrangements.
By the combinations of a number of things is understood the different
selections of them taken so many at a time.
e.g., a firm retains 12 men for their mo tor van service. There
are 6 vans and 2 men are required for each, I to be the driver.
By simply arranging the men in pairs, a number of groups or com
binations is obtained. But if the first pair might be sent to any
one of the 6 vans, i. e., if regard is paid to the arrangement of the
pairs, and if also either of any pair might drive, we get further
arrangements. We are then dealing with permutations.
To make this example a trifle clearer : let the men be repre
sented by A, B, C, D, etc. Then the different selections of the
12, taken 2 at a time, would be A and B, A and C, A and D . . .,
B and C, B and D . . ., C and D . . ., and so on. But A and B
might be in the ist van or in any of the others, so that a number of
different arrangements of pairs amongst the vans would result.
Also A might drive or B might, so that the arrangements in the
vans themselves would be increased. As we might write it for
one van, the different arrangements would be A (driver) and B,
or B (driver) and A.
To find a rule for the number of permutations of n things taken
r at a time.
If one operation can be performed in n ways and (when that
has been performed in any one of these ways), a second operation
VARIOUS ALGEBRAIC PROCESSES 461
can then be performed in p ways, the number of ways of perform
ing the two operations in conjunction will be nxp: e.g., suppose
a cricket team possesses 5 bowlers; then the number of ways in
which a bowler for one end can be chosen is 5. That end being
settled, there are 4 ways of arranging the bowler for the other end.
For each of the 5 arrangements at the one end there can be 4 at
the other end, so that the total number of different arrangements
will be 5x4, i. e., 20.
Suppose a choice of r things is to be made out of a total of n
to fill up r places.
Then the 1st place can be filled in n ways.
For the 2nd place (the ist being already filled) choice can only
be made from (n i) things; hence the number of different ways
in which the ist and 2nd can together be filled is n(n i).
The ist, 2nd and 3rd together can be filled in n(n 1)( 2)
ways, and so on, so that all the r places can be filled in n(n i)
(n 2) ... to r factors.
When there are 3 factors, the last = (n 2) = (n 3+1)
When there are 4 factors, the last = (n 3) = (n 4+1)
.*. When there are r factors, the last = (n r+i)
.*. The number of permutations of n things taken r at a time
For shortness this product is often written n r .
If n places are to be filled from the n things the number of
possible ways
= n P n = n n = (i) (n 2) .... ( +2)(n n+i)
= n(n I)(M 2) .... 2.1
i.e., is the product of all the integers to n : this is spoken of as
factorial n and is written \n or n \
Thus " factorial 4 " = [4 = 1.2.3.4 = 24.
To find the number of combinations of n things taken r at a
time, written B C f : Obviously n C r must be less than n P r , because
groups of things may be altered amongst themselves to give different
permutations. For groups of r things, the number of different
arrangements in each group must be \r_ (r things taken r at a
time) ; hence the number of permutations must = [r X the number
of combinations
or P r =
= n(ni)(n2) .... (n r+i)
I 2.3 .... r
462 MATHEMATICS FOR ENGINEERS
If both numerator and denominator are multiplied by \n r
i. e., by 1.2.3 ( ')
then
nr _n(n i) .... (n r+i)x(n r) .... 2.1
from which we conclude that
C r = n Cn r
a result often useful.
The number of permutations of n things taken n at a time when
\n
b of them are alike and all the rest are different = Hr
\L
The number of permutations of n things taken r at a time
when each thing may be repeated once, twice, . . . . r times in
any arrangement = n r .
The total number of ways in which it is possible to make a
selection by taking some or all of n things = 2 rt i.
Example 12. Find the values of 8 P 2 , 9 C 3 and 15 C U .
P, = 6(6  i) = 30
9, =
[3 123
c u = I5u or III 154
III HI 14 If
15.14.13.12.
= J ^ = 1365.
1.2.3.4 ^
When w and r are nearly alike (as in this last case) and n Cr is
required, we use the form n C r = n Cn_r ; and the arithmetical work
is thus reduced.
Example 13. There are six electric lamps on a tramcar direction
board ; find the number of different signs that may be shown by these.
If the lamps all show the same coloured light, the question resolves
itself into finding the total possible arrangements of 6 lamps when any
number of them are lighted.
Thus if 6 lamps are on, there is only one arrangement possible.
If 5 lamps are on, these can evidently be placed amongst the six places
in six different ways ; or, in other words, the number of arrangements
in this case is 6 C B or 'C^ ["Q. = n Cn_ f ]. If 4 lamps only are to be
switched on, the possible arrangements will be *C t , i. e., 6 C, . e., 15.
Similarly the numbers of arrangements for the cases of 3, 2 and i
VARIOUS ALGEBRAIC PROCESSES 463
lamp on will be fl C,, 6 C, and 6 C X respectively : hence the total number
of different arrangements giving the different signs will be
I + 6 C S + 8 C 4 + C S + 6 C, + "Ci = 1 + 6+15 + 20+15 + 6 = (.
This result could also have been obtained by making use of the
rule given on p. 462 for the total number of ways in which it is possible
to make a selection by taking some or all of n things.
Thus total = 2 n i = 2' i = 641 = 63.
If the lamps had been of different colours the number of different
signs would be greatly increased, since the different sets of the above
could be changed amongst themselves.
Example 14. Twelve change wheels are supplied with a certain
screwcutting lathe ; find the number of different arrangements of
these, 4 being taken at a time, viz. for the stud, pinion, lathe spindle,
and spindle of leading screw.
In this case the order .in which the wheels are placed is of conse
quence ; hence we are dealing with Permutations.
As there are 12 to be taken, 4 at a time, the total number of arrange
ments = 12 P 4 = 12.11.10.9 = 11880.
The Binomial Theorem. By simple multiplication it can be
verified that
(x+a)* = x z +2ax+ a*
(x+a)* =
(x+a)' =
It is necessary to find a general formula for such expansions;
(x+a) is a twoterm or binomial expression, and the expansion
of (x+a}* is performed by means of what is known as the Binomial
Theorem. For simple cases, such as the above, there is no need
for the theorem, but for generality it is desirable that some rule
should be found. The expansion of (x+ a)' 3 could certainly be
found by writing it as , * , and then performing the division,
an endless series resulting, but it would be a painfully laborious
process.
Suppose the continued product of (x+a)(x+b)(x+c) .... to
n factors is required, n being a positive integer.
The ist term is obtained by taking x out of each factor, giving x.
The 2nd term is obtained by taking x out of all brackets but
one, and then taking one of the letters a, b, c .... out of the
remaining bracket. The 2nd term thus = x n ~ l (a+b+c+d ....
to n terms).
The 3rd term is obtained by taking x out of all brackets but
two, and combining with the products of the letters a, b, c . . . .
taken two at a time.
464 MATHEMATICS FOR ENGINEERS
The 3rd term thus = x*~ z (ab + ac + ad + . . . .+ be + . . . .
to, say, P terms).
p is then the number of combinations of n letters taken two at
a time
.,
[2 1.2
so that the 3rd term is found.
In the same way any particular term may be found.
Example 15. Write down the value of the product
ist term = x* (i. e., x is taken out of each bracket).
2nd term = x 3 { 2+4+6 7} = x 3 (x being taken out of all brackets
but one).
3 rd term = ^(2)x(+ 4 )+(2)x(+6)+(2)x(7)
= # 2 {8 12+14+24 2842} = 52**.
4 th term = *{( 2 )(+ 4 )(+6)+(2)(+6)(7)+(+ 4 )(+6)(7)
+ (2) (+4) (7)}
= #{48+84168+56} = 76*.
5 th term = (2)(+ 4 )(+6)( 7 ) = 336.
7) = **+* 3 5 2 * 2 7
Now let
b = c = d= . . . . = a, then (x+a)(x\b)(x{c) .... to
n factors, becomes (#+)".
Then ist term of the expansion
= x n
the 2nd term of the expansion = x*~*(a + a f a . . . . to n terms)
= nx n ~ 1 a
the 3rd term of the expansion = x n ~ z (a 2 }a 2 \a 2 ... to n C a terms)
n(n i) _
= J  'x n *a?
1.2
Similarly, the 4th term of the expansion
n(ni)(n2)
. ^\ _ /_\ _ / v W ~ >/7 **
1.2.3
= x
. . .
l.J
Thus the indices of x and a together always add up to n, that
of x decreasing by one each term. The numerical coefficients can
be remembered in a somewhat similar fashion; the numerator
VARIOUS ALGEBRAIC PROCESSES 465
having a factor introduced which is one less than the last factor
in the preceding numerator, whilst the denominator has an addi
tional factor one more than the last factor in the preceding de
nominator, i. e., a kind of equality is preserved.
The proof here given is of an elementary character, and only
applies when n is a positive integer, but it can be proved that the
theorem is true for all values of n, integral or fractional, positive
or negative.
To find an expression for any particular term in the expansion :
The 3rd term = 2 x n ~ 2 a 2 , i.e., is distinguished by the 2's
throughout, and is on that account called term (2+1) or T ( J + D
The I4th term is thus written T (1J + 1)
Putting the terms in this form we are enabled to write down at
a glance, i. e., without full expansion, any particular term desired.
e. g., the 6th term = T (6+1) = ^ x 5 a 5 .
is
The (r + i) th term is usually taken as the general term, and
it is given by
or
Example 16. Find the 8th term of the expansion of (x 2y) 10 .
Here n = 10 ^
x = x I in comparison with the standard form.
and a = 2y )
Hence T 8 = T (7+1) = ^V '(*?)'
V(_ 2 j,)i [for "C 7 = "C lt _ 7 = "CJ
Li
= i^!*s
Example 17. Expand (a 3&) 1 to 4 terms.
[Whenever n is fractional or negative the expansion gives an
infinite series, and therefore it is necessary to state how many terms
are required.]
Comparing with the standard form
x = a
a = (36)
*
H II
466 MATHEMATICS FOR ENGINEERS
Hence the expansion
1.2.3
*  a
fx Jx
Example 18. Expand f 3m  j to 3 terms.
Here x = 3m, a = , n = 4
Hence the expression
1.2
4.3~ 8 m 5 , 4 X 5 X
 *  
5 2 x 25
, ,
Sim* I2I5W 6 3645m 6
The method of setting out the work in these examples (Nos. 17
and 18) should be carefully noted ; the brackets inserted helping to
avoid mistakes with signs, etc. Thus in the evaluation of n(n i)
when n = 4 one is very apt to write down the result straight
away as 4X 3, whereas its true value is ( 4)( 4 i),
i. e., f 20.
Example 19. In the Anzani aero engine the cylinder is " offset,"
t. e., the cylinder axis does not pass through the axis of the crank shaft,
but is " offset " by a small amount c. The length of the stroke is given
by the expression V(l + r)* c* V(l r) 2 c 2 , where / = length of
connectingrod and v = length of crank. Show that
stroke = zr\ i + J , 2 _ 2 [
Dealing with the expression V(l + r) z c 2 , we may rewrite it as
{(/ + y) 2 c 2 }* and then expand by the binomial theorem.
Thus {(I + r)*  c 2 }* = {(/ + r) 2 }* + {(l + r) z }l~ l x (c 2 )
(/ + r) . f terms containing as factors
the fourth and higher powers of c; these terms being negligible, since
c\ c 6 , etc., are very small.
VARIOUS ALGEBRAIC PROCESSES 467
In like manner it can be shown that
Hence
stroke =(< + r) 

Comparing this result with the length of the stroke of the engine if
not offset, we see that there is small gain in the length of the stroke ;
the increase being the value of rc a i / 2 r 1 .
Use of the Binomial Theorem for Approximations.
Let us apply the Binomial Theorem to obtain the expansion for
(!+*).
Writing i in place of x, and % in place of a, in the standard
form
/ \n i n(ni) , n(n 1)( 2) , .
(i+#) n = i+nx\ s  'x 2 \ *  '  'x 3 \ . . . ,
1.2 1.2.3
If x is very small compared with i, then x z , x 3 , and higher
powers of x will be negligible in comparison. Hence
(l+jr) B = l+flx when x is very small.
Example 20. In an experiment on the flow of water through a
pipe the head lost due to pipe friction was required. The true velocity
was 10 f.p.s., but there was an error of 2 f.p.s. in its measurement.
What was the consequent error in the calculated value of the head
lost, given that loss of head oc (velocity) 2 ?
Let He = calculated loss of head.
H = Kv 2 = K(io + 2)* {v being the measured velocity}
= K x io a (i + 02) 2
Making use of the above approximation
H = iooK(i + 02 x 2)
= iooK(i + 04)
But true head lost = K x io 8 = looK
error = iooK x 04 or 4 %.
Example 21. Find the cube root of 998.
998 = 1000 2 = 1000(1 002)
/. cube root of 998 = 998* = iooo*(i 002)*
= io (i J x 002)
= io (i 0007) = 9993
468 MATHEMATICS FOR ENGINEERS
Example 22. Find the value of 1005*.
1005 = 1000 (i+ 005)
1005* = iooo 4 (i+oo5)* = iooo[i+(4Xoo5)]
= IO la XI'O2.
With a little practice one can mentally extract roots or find
powers for cases for which these approximations apply
e. g., Vcfi = 99
For 98 differs from 100 by 2, hence its square root differs by
\ of 2, *. e,, i from 10.
Similarly, (iO3) 3 = 109 very nearly.
Further instances of approximation are seen in the following :
(i+*)(i+y) = i+x+y+xy = i+x+y
when x and y are small
(i\x)(i\y)(i\z) = i+x+y+z when x, y and z are small
Example 23. Find the value of g85 X 5 ' 8
1004
F _ 1000(1 '015) x 5(1 + 'Qi6)
1000(1 + 004)
= 5(1 015 + 016 004) = 4985.
Example 24. If / = measured length of a base line in a survey
L = correct or geodetic length, i. e., length at mean
sealevel
h = height above mean sealevel at which the base
line is measured
and r = mean radius of the earth
Then ^ = JL,
/ r + h
und it is required to find a more convenient expression for L.
T lr I
L = ' Whence L 
since h is very small compared with r.
VARIOUS ALGEBRAIC PROCESSES 469
Exponential and Logarithmic Series.
(i\"*
i+ )
w/
i) JL + m(mi)(m2) x 1
1.2 W a . 1.23
f ^ f V ~\
m\ w/ w\ wA m)
~~
1.2.3
= I + I+ ^^ + 1.2.3 +
Suppose now that m is increased indefinitely, then  etc.,
nt tn
become exceedingly small, and may be neglected.
Hence when m is infinitely large
/ . i\ m . i , i
(i+is) ..i + i + j5 + 5 +....
This is the case of compound interest with the interest very
small but added to the principal at extremely short intervals of
time. The letter e is written for this series
[If it is any aid to the memory this statement may be written
1,1,1,1, i
= \o + i + J2 + J3 +' ']
(I\ mx
i\ ) would be e* if m were infinitely large.
.
But
i\ m ' i . mx(mxi) i . mx(mxi)(mx2)
 L 
X X
= T.+X++.+ .... (when m is very large)
2 3
47 o MATHEMATICS FOR ENGINEERS
To obtain a more general series, i.e., one for a*, where a. has
any value whatever, let a = e k , so that log* a = k.
Then a 1 = e**
The series for e kx can be obtained from that for e* by writing
kx in place of x.
llv\* lbv\Z
Then a* =
and substituting for k its value we arrive at the important result
This is known as the Exponential Series.
A further series may be deduced from this, by the use of which
natural logarithms can be calculated directly ; common logarithms
being in turn obtained from the natural logs by multiplying by the
constant 4343.
For let a = I + y
Then by employing the exponential series
It is now required to obtain a series for log e ( i f y), which can
be done by equating coefficients on the two sides.
The lefthand side may be expanded by the Binomial Theorem,
giving
x(x i)(x 2)y 3 .
   '
Now x occurs in every term except the first, and the coefficient
of x in the second term = y.
v 2
The third term is %(x 2 y z xy z ) ; and the coefficient of x is
2
The fourth term is ^(^y 3 3# 2 y 3 f 2#y 3 ) ; and the coefficient
, . y 3
of x is
3
yZ y3 ,y4
Hence the coefficients oi x = y f  f
and, equating the coefficients of x on the two sides
10ge(l+^)=^^+^^+ ...... (I)
which is known as the logarithmic series.
In the form shown, however, it is not convenient for purposes
of calculation, because the righthand side does not converge
VARIOUS ALGEBRAIC PROCESSES 471
rapidly enough; and a huge number of terms would need to be
taken to ensure accurate results.
In the expansion for log,(i+y) let y be replaced by y;
then
log, (iy) =  y  y ~ __ ....... ( 2 )
Subtracting the two series, . e., taking (2) from (i)
log* (i+y)  log, (iy) =
but log, (i+y)  log, (iy) = log, 
hence log, 2j = 2 (y + + ^ + ....)
Now let L_ ! be denoted by , i. e., m my= n+ny
nin
or y = 
w+n
+K i w m of"" i 1/ni n\ 3 I/in /i\ 5
then log,, = 2 { s ^ s ( ; + (   + . . . ,
n Im+n 3\fn+/i/ 5V/n+n/
which is a series well adapted for the calculation of logs.
Example 25. To calculate log, 2.
Let m = 2, n = i, and thus y = J
then log, 2 = 2J + (x a) 4 (~ x ~5j + [
= 6930
(which is one wrong in the 4th decimal place; and this error would
have been remedied by taking one more term of the series) .
An equally convenient series would be obtained by writing
, i+y . i
for T i. ' e  y =
, . ., 
i y' 2w+i
Then
Thus if
= 6930 as before.
and this latter form is slightly easier to remember.
472 MATHEMATICS FOR ENGINEERS
To obtain log, 3 let n = 2.
Then
= 40546
but log*  = log, 3 log* 2
2
40546 = log, 3 6931
log, 3 = 10986.
Again, log 4 = 2 X log, 2 and log, 5 can be obtained by using
the series for log,  when n = 4, and the value of log, 4, so
that a table of natural logs could be compiled : in fact, this is the
way log tables are made.
The corresponding common logs are found by multiplying the
natural logs by 4343.
Example 26. The " modified area " A, a term occurring in con
nection with the bending of curved beams, is given by
for a rectangular section of breadth 6 and depth d.
Show that this can be written as
R _ . might be written as  j and is therefore of the form,
Hence
R in this formula is the radius of curvature of the beam, and hence
if the beam is originally straight R = oo , so that ^ = o and the expres
sion for A reduces to bd, i. e., the area of the section.
VARIOUS ALGEBRAIC PROCESSES 473
Exercises 45. On the Binomial Theorem, etc.
1. Write down the 5th term in the expansion of (a 6) 7 .
2. Expand (20 + $c) 11 to 4 terms.
3. Find the aoth term of the expansion of (3* y) 23 .
* T? j ( m 2\ 8 ,
4. Expand (  I to 4 terms.
5. Write down the first 5 terms of the expansion of (a 2)"*.
6. Find the 7th term of (i  j^"
7. Expand to 3 terms (2 # 2 )*.
8. Expand to 4 terms ($a + 4c)~*.
9. Write down the 3rd term of (a 26)"*
10. Expand ^/ i ^ sin 2 # to 4 terms, and hence state its
/ /length of connectingrod \
approximate value when I ,, \ is large.
a \ length of crank )
On Permutations and Combinations.
11. In the Morse alphabet each of our ordinary letters is repre
sented by a character composed of dots and dashes.
Show that 30 distinct characters are possible if the characters are
to contain not more than 4 dots and dashes, a single dot or dash being
an admissible character.
12. Find the number of ways in which a squad of 12 can be chosen
from 20 men.
(a) When the squad is numbered off (i. e. t each man is distinguished
by his number).
(6) When no regard is paid to position in the line.
13. Find the values of 15 C 1S , 12 P 4 , 5 P .
On Approximations.
14. Use the method of p. 467 to obtain the value of (996)*.
15. Evaluate ^^ * 2 'f 3 3 X '" 8 by the same method.
\'997l
16. State the approximate values of
(a) (ioo2) 8 ; (6) (9935) 7 : ( c ) (i  oo6) 5 ;
(d) (10 + 17) X 995 X 4044
/I tan 4*\ a ,
17. The maximum efficiency of a screw = ( ) , where $
\l ~}~ i tin j <p/
is the angle of friction, t. e., tan = p. Show that this may be written
in the form T ~ ** if u is small.
i +/*
On Series.
18. Find series for the expression cosh x, i.e., (^ ^ J
and for sinh x, i.e., (   j
474 MATHEMATICS FOR ENGINEERS
19. Find, by means of a series, the value of log e 4 correct to 3 places
of decimals.
T>
20. Express ^  as a series. What is the approximate value of
K. + y
this fraction when y is small compared with R ?
21. A cable hanging freely under its own weight takes the form
%
of a catenary, the equation of which curve is y = c cosh , c being the
. horizontal tension
value of the ratio weight per foot run
Express y as a series, and thence show that if the curve is flat it
TT iv2
may be considered as a parabola, having the equation y = \ jj
22. By substituting 5 for x in Newton's series
calculate the value of re correct to 3 places of decimals.
Determinants. When a long mathematical argument is being
developed, as occurs for example when certain aspects of the
stability of an aeroplane are being considered, it frequently happens
that the coefficients of the variable quantities become very involved ;
and in such cases it is often convenient to express the coefficients
in " determinant " form. This mode of expression is also utilised
for the statement of some types of equations, for by its use the
form of equation and its solution are suggested concisely and the
attention is not distracted from the main theme of the working.
Thus when dealing with the lateral stability of an aeroplane in
horizontal flight the equation occurred
AX 4 f BX 3 + CX 2 f DA + E = o
where A, B, C, etc., were all solutions of other equations and hi
some cases rather long expressions. For example, A had the form
a z b z c 4 and E was equal tog' sin Q(l i n 1 / 2 w 2 ) S cos ^i n 3~ n z^)
To avoid writing these expressions in their expanded form, they
were expressed thus
A =
c
c 2 b*
and E = g sin 6
g cos 6
and it will be shown that from these " determinant " forms the
expansions may easily be obtained.
Before proceeding to illustrate the use of determinants it is
necessary to define them and to show how they may be evaluated.
VARIOUS ALGEBRAIC PROCESSES
475
Let
D =
a b c
d f g
h k I
then D is called the determinant of the quantities a b c . . . I,
and a determinant of the third order since there are three columns
and three rows; its value being found according to the following
plan :
The letter a occurs both in the first row and in the first column :
take this letter and associate it with the remaining columns and
rows, thus
[It will be observed that
/ g
k I
is a determinant of the
second order and it is termed the minor determinant of a.]
Then the value of the minor of a is found by multiplying f by I
and subtracting from it the product k by g.
a
Thus
= l k and
= a(flsk} =A.
In like manner the minor containing the products of b is
b
and for c
g
h I
d f
h k
= b(dl  gh) = B
= c(dk fh) = C.
Then the value of the full determinant
= D = A  B + C = a(fl  gk)  b(dl  gh) + c(dk fh).
To avoid the minus sign before the second term the letters
might be written out as follows
a b c
f
h k I h k
and the one sequence could be maintained, thus
D = a(fl  gk) + b(gh  dl) + c(dk  hf)
Similarly, for a determinant of the fourth order
D =
m
r
476
MATHEMATICS FOR ENGINEERS
g h k
m n p
r s t
 b
f h k
I n p
q s t
f g k
I m p
q r t
d
f g h
I m n
q r s
each of these determinants of the third order being evaluated in the
manner previously explained.
Example 27. Evaluate the determinant
235
6 4 2
319
D = 2[ 3 6 _ ( _ 2)] _ 3[ _ 54 _ ( _ 6)] + 5 [_6l2]
= 76 4 144 90=130.
Example 28. Evaluate the determinant
D =
2 41 2
3653
122 3
4 82 4
653
4
353
+ I
3 63
42
3 65
223
123
I 2 3
I 2 2
824
424
4 84
4 82
6)5(82 4 ) + 3(4i6)} 4 {3(86)5( 4 i2)
+ 3(~28)}
+ i{ 3 (82 4 )6( 4 i2) + 3 (8 4 8)}
4 2{ 3 ( 4 i6)6(28) + 5 (8 + 8)}
= 224224 4 4 = 
It will be observed that all the numbers in the second column are
the same multiple of the corresponding numbers in the first column;
and it can be proved that when this is the case the determinant is
equal to zero.
Example 29. A number of equations in a long investigation reduced
to the determinant form
* + 15  '3 30
6 #45 100*
o i x* 4
= o
Express this in the form necessary for the solution of the equation.
The determinant = (* + I5){(# + 5)(* a + 9*) + iox}
+ 3{6* 2  54*} 3O{ 06}
= ** 4 I4'i5* 3 + 5728*2 4. 9.87* + 18
and thus the equation is
** 4 I4I5* 3 4 5728*2 4 987* 4 18 = o.
VARIOUS ALGEBRAIC PROCESSES
477
Solution of Simultaneous Equations of the first degree
by the determinant method. Equations containing two or more
unknowns may be readily solved by setting them in a determinant
form and proceeding according to the following scheme :
To solve the equations 5* 4y = 23
Write the equations as $x 4y 23 = o
and set out in the determinant form
x y i
5 4 23
375
the last column containing the constants.
x v i
Then
= y =
x minor " y minor i minor
'' 20 + 161
whence
and
35 + 12 * "25 + 69 ' 35 + 12
x = 3 and y = 2.
Example 30. Solve the equations
^ax cy = 6 a
$bx + 2ay = a*
x and y being the unknown quantities.
Set out thus
y I
c b 2
20, a 2
x
4 a
Then
ca
whence
and
8a 2
_ a z c + 2ab 2
* ~ 8a 2 + tfc
y= 8a 2 + 3&c.
Example 31. Solve the equations
2a 56 + 4^ = 28
o + lib $c = 41
3a 2& c = 3
473
MATHEMATICS FOR ENGINEERS
Set out thus
Then
Thus
a b c i
25 428
i ii 5 41
3  2  i  3
b c
I
a minor b minor c minor
i mnor
 5(15 + 41)  4(  33 + 82)  28(  ii 10)
i
15)
233)
6
and
2(15 + 41)  4 (  3  123)  28(  i + 15)
c
2(  33 + 82) + 5(  3  123)  2 8(  2  33)
112
I
112
whence
Exercises 46 On Determinants.
Evaluate the determinants in Nos. 14.
1.
3.
5'4  6
3  5
354
15 25 2
3 75
2.
R 2
when R = 36
R! = 72
R 2 = 710
R 3 = 220
4.
2351
3246
84 35
2 I 6 2
5. A coefficient C in an equation was expressed as
C =
U0 } Zn
M u
Evaluate this when Z w = 3, M w = 25, Mg = 200, Z 3 = 9,
U = I, X u = !, Xg = 5, M M = O, KJB = 20, XK, = 2, Z tt = I.
6. Solve the equation a 53=0
2 a + 2 i
3 16 4
Using the determinant method solve the equations in Nos. 710.
7. n# 4y = 31 8. 8a b = 20
2X + 37 =28 ioa + 76 = 71
9. 4* 5y + 7* = 14 10. za + 36 + 5c = 45
# 7 5.2=11 96 ioa = 393
ANSWERS TO EXERCISES
Exercises 1
1. 12
2. 0009
3. 150
4. 60
5. 108
6. 12
7. 000225
8. 285
9. 009
10. 052
11. 93
12. 161
13. 27
14. 9
15. 22
16. 56
17. 031
18. 74
19. 25
20. 63
Exercises
2
*&' v> ^
11 I 2 *7
', 9V; ~t 2. 2; i
y 32' 8 '
384; 958*
3 "8 7x /^
' 4 7
5ii
f%^T
8ix*y 2 *
.
i9& !
6*343
9
7. 141^ 8.
a a~a~
9. 16
10
jj Pt V 2 Pl v l
/Cf, 1 " might be written \
1 #if i n x fj 1 ", etc. J
~ in
Exercises
3
1. 5895
2. 2465
3. 02138
4. 5703
5. 0005423
6. 116700
7. 1234
8. 1963
9. 06664
10. 1924
11. 2444
12. 2914
13. 1618
14. 00009506
15. 4964
16. 3114
17. 0001382
18. 6874
19. 02231
20. 2777
21. 3642
22. 1669
23. 00001509
24. 3352
25. 001155
26. 3841 x io 6
27. 2017
28. 2421
29. 4814
30. 721 ix io 14
31. 07041
32. 9718
33. io
34. 858
35. 5418
36. 3275
37. 2204; 1369
38. 329
39. 198
40. 5400
41. 46
42. 50077
43. 1315 ; 591
44. 356
45. 36; 392
46. 0935
47. 2221
48. 400
49. 80072
50. 1434
51. 506
52. 479
53. 5130
54. 1572
55. 46,500
56. 2 8 1
57. 35 x io 5
58. 1392
59. 1016 x io 6
60. 284
61. 128
62. t = &, *  ft, B  x&, T  ft,  7
Exercises
4
1. Ibs. per sq. in. 2. 168 3.
279
4. 544
f 1
I2ttt^
5. 2785
6. Ibs. 7.
Stress = 
75
g
10. Incorrect as
wriHon TT P = '
Sir 1
cu. ft. per sec.
' 396000
12. 746
48o MATHEMATICS FOR ENGINEERS
Exercises
5
1.
i 7 a
2.
 2285
3.
362
4 7'54
5.
6
93
6.
75
7.
7 d
ioa 2
b
8 +
t
9.
301 x
io 10.
()
L =
(T
9+T
*li
(b) 933
11.
1205
12.
i
A Ti
1
13.
501
14. E = 2
5C
15.
I
97
16.
R
AJU
1265
17. 167
18.
q
= l{
^(A,^)
+ A
.A}
19.
n cells.
20.
2285
21.
i5
43
22.
d=
3/I274P/
V
f
23.
6
'9
24.
foj
4915
\
25.
()
2Wh \
Z\ TT
5j o XV t
467
J
(&)
Xl/C
294
I ' : VV 1
26.
3
27.
24
28.
80
29. 1493
30.
3
3
31. Ibs.
ft.
must be first brought to Ibs. ins
. ; 646 ins
32. 200,000 33. 257 34. 273 ft.
J ~\ k 2 1 , \ ^
(*)= 26.6 7 1 W'^j^l 3? ^ C =
85> ,M ^_ 36 ' W 00675 f 37 ' (W E _ 9CK
(6) ^  24 J  3K +
38. r + ^ 39. 7oolbs./ a * 40. 137
2 It>
41. A's speed =10 m.p.h. ; B's speed = 15 m.p.h. 42. 631
43. M = 2214; H = 17 M
{Multiply equations together ; thus, = x MH = M 2 }
5  57 ' 7 " D =
lx) c(2.h
47. y ^
Exercises 6
2. a=7 3. w=i8 4. * =
&= 2 n = 26 y =
5.
9.
* = 124 6. # = 12
y = 359 y = 20
p = 9 10. E = 4
5 = 7
7. * = 5
y = 3
8. a = 12
6 = 5.7
c= 48
11. a = 1160,
b = 69000
~ 4 12. E = I2W+46
13. V = 42+^
14.
1 = 953B+ 1284: 481
15. a = 916
16. R = 4714
6 = 191
a = 00707
17
T. = TTTC '?/ ' nfi'7
1 R * 7 '
\/area
ANSWERS TO EXERCISES 481
._ ioi6\/area .
length + 97 20. E = i6 4 i + 7 3 o 9 T+ooo326T
21. /= i48ooooi38(*  6o) a 22. /= i6ioooo26(/  6o) a
23. W = 828 +
n* . , i7oo/ total steam per hour \
24. w = 103 + y ( w =   ^  and must be first \
1 V ' calculated /
25. 8610 of iron; 7000 of copper 26. = 998; K=i
27. A = 120, B = 140, C = 160 28. m = 4471, m t = 3154
29. 3111 tons of saltpetre ; 3889 tons of ginger.
Exercises 7
1. (X+22)(X 4) 2. (*!!)(* 8)
3. (#5)(* 21) 4. (2a5& 2 )( 4 a a + 25&* + ioa& 2 )
5. (8*  1 1) (3* + 4) 6. ( 5 a  36) (56  a)
7. (a + 96) (a  56) 8. (3*  7 y)(^x  i 5 y)
9. (8 + *)(n 3*) 10. 2(5W an) (2m  5**)
11.  g  (9&r/ i6/ 3 + 5/Ar a 24^r 3 ) 12. ^
13. (94* + 32 1) (x  3) 14.
15. 6a z b(a + 2c)(ga  25c) 16. (20  36 + 4c)(2a 36
17. 8(2C 2 + Ja 3 6) ( 4 c + ia 6 6 2  a 3 6c 2 ) 18. {R 2 + Rr + r 2 }
19. (* + 7 )(*  i) (2*  5) 20. (^  i)( 3 / + 7 )(2p + 5)
21. 199 x 23(2 + 6 4) = 18308 22. 14,130
23. 12 (*  3) (*  4) (* + 2) 24.  a ."^
66 J cTT
9 r 4(^+ 2) o R 3(3* a ~ 4* 6)
^ 5  5 1^^) ^' 2(^2^8)
21 (a 6) 2g 560  327* + 99* 2  I20* 3
' 4(9  I4&) ' 20(3*  5) (3*+ io)(2*  7)
ort ii __
29. or 407 30. 37'8 31.
32. 3*(* + 9)(*  7) ; (8  9*) (3 + 8x) ; (5*
33. (x + 8)(x  i}(x 2 + 7X + 26). {Hint : Let X = x* + jx + 6.}
Pas(i8s+2 5 ) n _ _
34 ' 35 * U t;
3(6 + )
Exercises 8
1. 4 or i 2. 25 or 3 3. 413 or 113
4. 4 or i 5. 283 or 83 6. 278 381;
7. 3 8. 423 or 243 9. 125 1219;
10. 2421 x io 5 or 2379 x io 5 11. 275 or 457
12. 2898 or 103 13. 389 x io 4 or 297 x io*
14. 575 or 565 15. 23 (18 has no meaning here)
II
482 MATHEMATICS FOR ENGINEERS
nr
16. /, = 
_ ab VaW  24t 2 v 2 + Satgv
** u ~
6t
19. 120 or 133. (Divide all through by 75 x io 6 first)
20. 155 or 32 21. 845 (2845 h as no meaning here)
22. v g$Vmi 23. 5 or 75 24. 80 or 90
25. 2ii and 789 of span from one end 26. 13 or 1042
27. 1475 28. 655 or 305 29. 100 ft.
Exercises 9
1. *= 7 or3 1 2. o = ^or 3 I 3. p = T 3\
y =  or 2 6 9 J y = * or V q = 2)
4. x = 2\ 5. a = 6 or i \ 6. 363 or 23
y = 5/ b = 28 or 2) 7. 68* 8. 5
9. 765 (the work is shortened by dividing by 256 straight away)
10. 197 11. 274 12. I 13. ^ 14. 922 15 737
16. m = 01277; n = 0026 17. 39 or 158 18. 6763
Exercises 10
1. 911 nautical miles 2. 2263* 3. 707
5. 3 1 '9* 6. (a) i in 125; (b) i in 1246 7. 15 ft.
8. AB= 1237'; BC= 1215"; AC = 2163" 9. 724
10. 26 11. 405 grey; 340 red 12. 480 13. 512
14. 132 15. 236" 16. 6110 Ibs. 17. 866"; 307
18. 511 tons 19. 6580* 20. ^3 175. od.
21. 724 sq. ins. 22. (b) 1225; ( c ) 4'> (<0 5"86
Exercises 11
1. 344 sq. ins. 2. 102*; 521 sq. ins. 3. 6 sq. ins.; 24 ins.
4. 2374 5. 1282 sq. ins. 6. 536
7. 585 sq.ft.; 1807 ft. 8. 1375 sq. ft. 9  3'6i ft.
10. 3015 sq. ft. 11. 1325 sq. ins. 12. 477 sq. ins.
13. 695 14. 52 amp. 15. 642 ins.
16. 913 sq. ins.
Exercises 12
1. 224" 2. 291 ft. 3. 161 sq. ft
4. 146; 214; 333; 35; 06 5. 251* 6. 1571"
7. 29 ft. 8. 18" 9. 1380 ft. per sec.
10. 3123* 11. 892* 12. 66 13. 970 14. 5935
15. 2713 sq. ins. 16. 3'2 J* 17. 196 sq. in.
18. 12080 Ibs. 19. 4816 Ibs. 20. 376 miles 21. 128*
22. 1665 sq. ins. 23. 1010 sq. ft. 24. 0294 ohm 25. 268
26. 24378*; 66 and 36 27. 214 sq. ins. 28. 60 ohms
ANSWERS TO EXERCISES 483
Exercises 13
1 c i = 475*; h = 36* 2. c, = 4472*, h = 20*
3. 7 = 628"; A = 4'97* 4. 612*; 207 sq. ins.
5. 906 sq. ins.; 60 6. iii'; 367'
7. 2n*; 333*; 129 sq. ins.; 56* 8. 131
9. 1686 10. 298 sq. ins. 11. 6076 12. 58*
13. B = A/fRT 14. 141 15. i '26' 16. 375*
Exercises 14
1. 45"; 36^"; 848" 2. 18"; i"; 72:3"; 55" from base
3. 248"; 12" 4. 473"; 8"; 18"; 2514"; 2591*; 2557'
5. 6630*; 3'88*; 296"; 2975*; 293* 6. 3340*
7. 304 ft. tons; 338 ft. tons 8. 623 yds.
9. 3*; ii* 10. 7000
Exercises 15
1. 689 cu. ft. 2. 28; 775 3. 3125* 4. 144
5. 833 6. 513 cu. ft. 7. 691 ; 25900
8. 411 9. 52600 10. 4794 cu. ft.; 6715153.
11. 2385 sq.ft.; 4017 sq.ft.; 193 cu. ft. 12. 70
13. 53 (watt = volts x amps) 14. 3185 Ibs. 15. 245 Ibs.
16. 9 17. 852 sq. ft. 18. 14* 19. i 5 '6"
20. 508000 21. 0006 22. 174*
23. 183 x io 10 ohms 24. 137 25. 1255*
Exercises 16
1. 593 cu. ins.; 321 sq. ins.; 1355 ins. 2. 204*
3. 200 sq. ft. 4. 134 cu. ins. 5. 1592 Ibs.
6. 261 ft.; 581 sq. ft. 7. 4'O3"; 1069" 8. 367 cu. ins.
9. 14520 sq. ft. ; 70420 cu. ft. 10. 773'8; 9674 Ibs.
11. 1738 sq. ins. ; 2343 12. 2135 cu. ft. 29,890 Ibs.
13. 241 tons 14. 389"
15. 155 cu. ins. ; 402 Ibs. 16. 559 cu. ins. ; 243 sq. ins.
17. 463* 18. 2*; 5*; 612*
19. 105 sq. ins.; 138 sq. ins. 20. U59sq. ins.; 2530 cu. ins.
Exercises 17
1. 160 sq. ins.; 191 cu. ins. 2. 83; 518 Ibs. 3. 88 cu. ins.
4. 8610 Ibs. 5. 1033 6. 544 cms.
7. 1004 sq. ins. ; 151 cu. ins. 8. 42* 9. 636 10. 559 sq. ft.
11. 0941* 12. 1440 sq. yds. 13. 1082* 14. 15,520 sq. ft.
15. 759* from vertex 16. 1046 sq. ins. 17. 147; 245
18. 5351 acres 19. 77* 20. 161, 473, 276, 276 sq. ins.
21. 406 Ibs. 22. 72 ft. 23. 13*
Exercises 18
1. 175 sq.ft.; 100 cu. ft. 2. 326 sq. ins.; 244 cu. ins.
3. 847 sq. ins.; 1057 cu. ins. 4. 136 sq. ins.; 982 cu. ins.
5. Paraboloid = J cylinder 6. 902 cu. ins. 7. 202 Ibs.
484 MATHEMATICS FOR ENGINEERS
Exercises 19
1. 644 Ibs.
2. 2630 Ibs.
3. 1278 Ibs.
4. 960 Ibs.
5. 272 Ibs.
6. 372
7. 116 tons
8. 171 Ibs.
9. i 84 Ibs.
10. 761 Ibs.
11. 455 Ibs.
12. 1025 tons
13. 508 Ibs.
14. 1955 lbs 
15. 282 Ibs.
16. 93*5 Ibs.
17. 359 Ibs.
18. 258 Ibs.
19. 647 Ibs.
Exercises 20
1. 7'; 745 tons 2. 30800; 650; 25000 3. 420 Ibs. per min.
4. 3400; if 5. 1440 8. 339; 55 !! 11850 tons
12. 317 13. ssmins.; 45 14. 63mins.; 42
22. 54000 lbs./D* 24. 8% low 28. 410 o'c.
Exercises 21
2. 250 4. Slope = 375; intercept = 2375 8. 578"
11. Slope = 25 if V is plotted along horizontal
12. m= 18, n= 26 13. x = i, y o 14. x = 124, y 359
15. x = 43, y = 233 16. x = 318, y =  475 1755
Exercises 22
1. 392 2. 31 3. 302 x io 6 4. 179 x io 6
5. I = 8576 + 471 6. d l = 84^ 03 7. di = 95^ 07
8. T== 5170 +7 9. T = 35300 10. R = 78V + 86
11. R = 772V+645 12. R = 25V + 75 13. R = i, a = 004
14. R = 1125, a = 00452 15. I = 00232* 96
16. 32 17. 2925 x io 6
Exercises 23
1. Vertex downward 2. Vertex upward
7. Total weight = 50^ + 5/ 2 14. 6 or i 15. 267 or 35
16. 144 or 765 17. Divide throughout by io 4 : 922 or 12
18. 171 19. (a) 49"; (6) 55" 20. # = 364'; h =183'
21. 77 air to i of gas 22. 55 23. e 55, effy. = 5
24. 1523 knots; 948 25. 40; 69 26. 21
27. Assume some value for v : u =  ;i 28. 833; 
29. 2" 30. 2 rows of 8 31. 6 34. i, 2 or 15
35. 2, 225 or 3 36. 12, 46 or 16
37. 2, 5 or 8 38. max at x = 3, min" at x = + 2
39. x=ziil 40. 1475 41 5'6
Exercises 24
1. 881 Ibs.
4. 1089 tons
8. 23 pence
12. 22 knots
16. 01088'
2. 237
3. v = 663 Vr; 1195
5. 10970 Ibs.
6. 5" 7. 028 cm
9. 1375 ohms
10. 246 11. 80
13. 841
14. 427 15. 533
M
17. Cost oc ~
ANSWERS TO EXERCISES
485
Exercises 25
1.
28; 72
2. 200>
3.
105 4.
3'7. 4'6 ... i
5.
12, 15, 18 or 9,
165, 24 6. 24
7. &
[ 155. 6d. ;
377 los. od.
8.
10; 8160
9. 155 ^
10. 592 ft. ;
1 6 sees.
11.
315 p.m.
12. 2074
13.
5333
14.
2, 3. 44
15.
8352 .
16. 55
17.
10081 ; 821
18.
20 Ibs.
19.
752, 1 8, etc.
20.
a = 2
b = o, c =
= i;
8040
21.
25 days
22.
983 in.
23.
4284; 6116; 8
734:1248
178;
25H3
3631 ', 5*'
84; 7403; 1055
Exercises 26
1.
306; 11569; 2192
2.
5071; 428; 14
3. o; 081 ; 194;
285; 55
775
4.
1301
5. 923
6.
09877
7. 00005445
8.
4612
9. 264
10.
i 086
11. 1546
12.
1103
13. 07784
14.
33 x io 26
15. 26, 560
16.
472
17. 754
18.
370
19. 382
20.
123; 109 21. 401
22.
1518
23. 0336
24.
475
25. 0391
26.
528 27. <t> w =
= '39* $ =
= I796
28.
4000
29. 638
30.
243 %
31. 296
32.
325
33. 103
34.
357 35. 448
36.
000334
37. 384
38.
851 39. 444
40.
715 41. 65
42. j= 1875; T =
= 445, * = 237
43.
Pv 1 ' 06 = 392 44. 147
45. o or 728 46. 216
47.
o or 1368
48. 0955
49.
4815 50. 033
51.
V = 222VH
52. 01895
53. 46 54. 5380
55.
i48xio 8
56. 605
57.
144 58. 7965
59.
616
60. 2
61. n 1405; C =
502 62. 81300 63. 47610
64. 1115
65.
66.
Thl. Disch.
Cd
Thl. Disch.
Cd
53
66
1205
.731
118
672
1547
708
183
727
171
658
22O4
711
67. r = 00356^ 5 68. h = I538*; 1  8
f Take logs of both equations and solve as a pair of simul^
by. 3 07 1 taneous equations.
70. 841 71. F = 00277V 1 ' 9
1.
2.
5.
8746;
9756;
17*13'
7. 3342'
12. 528
00305
455
17.
21.
Exercises 27
"3443 > '5 2 3 I '2309
9641; 16139; 2826 3. 382 4. 397
6. 252 ; o (the case of wattless current)
8. 1433 9. 20 10. 28 6' 11.
13. 1340 14. 6750 15. 3150 16.
18. 61,200 ft. 19. 234 20. 268'
22. 8923 23. 16850 24. 25!'
181
825
486 MATHEMATICS FOR ENGINEERS
Exercises 28
1. a = 1165"; 6 = 4347" 2.6 = 872"; c = 1483*
3. a = 483"; 6 = 435* 4. a = 6673"; c = 748;
5. a = 2214"; b = 1608* 6. a = 5766"; c = 9263"
7. a = 208"; 6=1072" 8. 6037 ft. ; 2927 ft.
9. 8 8' 10. 306 ft. 11. 78 12. 14 4i'
13. 2" 14. (a) 14 16'; (6) 20 53' 15. 238"
16. 2856' 17. AB = 5 oxAD
18. A = o, 50 R.B. of BC = 355 S.E.
B = 339, 778 R.B. of CD = 825 S.W.
C = 746, 205 R.B. of DA = 23 N.W.
D = 152, 13 Area = 2700 n'
19. A = 10, 20 R.B. of BC = 67 S.W.
B = 1905, 148 R.B. of CA = 18 N.W.
C = 1258, 1206 Area = 29
20. 3 chns. 49 links 21. 235 i' 22. 736 ft.
23. 2901" 24. 121*
Exercises 29
1. 8988
.6157
6157
4384 ;
+.7880 ;
7880
20503
.7813
+ 7813
2. 9903
6157
+ 8480
+1392 ;
+ 7880 ;
5299
7813
i 6003
3. 3289
3242
9953
9444 ;
+ 9460 ;
0979
+ 3482
3427
+ 1017
4. 7570
8m
7513
6534 ;
+5850 ;
6600
+ 11585
13865
11383
5. 9I35
6374
9218
4067 ;
+ 7705 ;
3877
+22460
+ 8273
23772
7. 124 36'
8. 120
55'
10. i4946' or 2
>ioi4'
9171
3987
22998
6. 7265; oo ; 1625
9. 1 1 9 30'
Exercises 30
1. c = 489", A = 34 25 / , C = 67 5 '
2. A = 8o52', 6 = 5946", c = 6304"
3. B = 4446' or I35i4', A = io824' or i756', a = 1193* or
387'
4. B = 4042', a = 884", c = 825"
5. A = 5343' or I26i7', B = 8oi7' or 743', 6 = 1261 or 172
6. a = 954 ft., B = 37 4 7', C = 685 7 '
7. A = 8o6', B = 48i8', C = 5i36'
8. c = 2197", B = 2i29', A = 283i'
9. B = 4 i 4 2'or I38i8', C=io9i8'or I2 4 2 / , c = 831" or 194*
10. C = 42 or 138, A = 108 or 12, a = 9779 or 2138
11. C = 6940', B = 59 3o', a = 830
12. A = I0333' or 527', C = 4 o57' or i393', a = 6462 or 6311
ANSWERS TO EXERCISES 487
13. A = 863i', B = 4557', c = 1611
14. C = 43V, A = 8i2i', a = 4728; area = 637
15. 1946 ft. 16. 8 30'
17. 55. 87, 38 18. 1875 Ibs.; 58
19. OB = 1224 chns., OC = 3236 chn., BE = 7673 chn.,
CF = i 667 chns.
20. 5636' 21. Jib = 262 ft.; tie = 174 ft. 22. 1191 ft.
23. 647^.; 374ft. 24. 532 ft. 25. 2083 ft.
26. AB = 2983 links ; 767 links 27. BG = 7496 chns.; CH = 7414
28. AB = 5274"} 29. AB (horiz.) = 6075 yds. \
DC = 4753 I Hnks Diff. of level = 1293 yds./
BC = 7746 J
30. r = 4733' 1 31. AP = 983' (AC = nSo'U
BE = BF = 1267' h BP = 9 6 7'
CF = CG = 4733' J CP = 919' J
32. I233f.p.s.; 29 36' 33. diam. = 506 34. 1064; 93'
35. 906 to i 36. 603 sq.ft.; 422 Ibs. 37. 746"; 1065"
38. 103"; 145" 39. 244 sq.ft.
40. 2286; 2912 41. 1725 sq. ins.
Exercises 31
1. cos A = 893, tan A = 504
3. cos (A+B) = 442, sin (AB) = 298
4. tan (A+B) = 367, tan (AB) = 536
^  tan a . 6 p =
' i + n tan a ' 3 2nrpfji
7. P = W (sin a + /i cos a) 8. 1162
9. 499 sin (5* + i) 10. 2385 sin (50* 576)
" R = os'' 12 ' ' I89: I0 42 '
13. E = I2i6sin (i2onf + 1022)
Exercises 32
1. cos 2 A = 566; tan 2 A = 1455
2. sin 2A = 7962; cos 2A = 605 3. (i + cos 28)
4. sin 2B = 2 cos B\/icos a B; 731
A A
5. sin  = 161 ; cos = 987; sin 3A = 8236
2 2
A
6. cos 4A = 616; tan = 114 7. 25(1  cos 4*)
8. 785 (sin 189  sin 131) = 785 ( sin 9  sin 49)
9. 2 sin gt {cos 6t + sin it} 10. sin A = 261 ; tan A = 270
11. 505 x sin 2a; 53
12. (a) 2cos323o' sin I53o'; (6)  2 cos 423o' cos 383o';
(c) 24 sin 50 sin 45
13. 2591; 6373 14. 333 or 125
15. 9'4 {'995  cos U"/'
A
cos  = 991
2701
J
MATHEMATICS FOR ENGINEERS
Exercises 33
1. 30 or 150 2. 45. ^S ' 225 or 315
3. 120 or 240 4. 88' or 188 8', I5326' or 33326'
5. 120 or 240 6. 538', 191 32', I26 5 2' or 34828'
7. 30 or 150, 210 or 330 8. 45, 7i34'. 225 or 2 5 i 3 4 '
9. 30 or 150 10. 3545' or I44i5'
11. 45, 225, i6i34' or 34i34' * 2  45 or 225
13. 45, 225, i826' or I9826' 14. 30, 150, 210 or 330
15. o or 45 16. o or 120 17. 27^' or 2433o'
18. 4856' or I5639' 19. I46 ig or 326I9'
20. o; 180; 2o56' or 339 V 21. 57'
Exercises 34
2. 50018 3. 151 ft.
5. E cosh x *J  + ^rV sinh x lj ^
8. 93 9. i852' 11. 86 12. 194
15. 1928 + 2298;
Exercises 35
1. 318 2. 517 Ibs. per sq. in. 3. 3835 sq.ft.; 575 cu. ft.
4. 765 sq. ft. 5. 7231 sq. ft. 6. 269 ft.
7. 430 sq. ft.; 2190 cu. ft. per sec. 8. 6850 sq. ft.
9. 730 10. 872 sq. ins. 11. 605 Ibs. per sq. in.
Exercises 36
1. 168750 cu. yds. 2. 8350 tons 3. 244000 tons
4. 44920 sq. yds. 5. 521 x io 6 galls. 6. 40 ft.; 5143 ft.
7. 12020 cu. yds. 8. 966 ft.; 619 ft.; 11600 cu. yds.
9. 265 ft.; 1766 ft.; 335 ft.; 2235 ft .' 275 ft.; 1833 ft.; 184,
325, 202 sq. ft.; 1375 cu. yds.
10. 2825 anc * 43'3 ft from the centre line
Exercises 37
14. The table of values would be arranged thus :
1. 12552; 21293
4. 4054 ft. ; 1566 ft.
6. io45' 7. 00383
14. 4409
6
log sin 6
cos 6
i 84 cos 6 I = A
A log sin 6 + log P
log/>
P
15. Treat ( i z ) as a constant multiplier
iioo\ &*)
16. Values of R and V are as follows :
V
10
20
30
40
55
R
25
321
474
69
961
146
17. 29
18. Values of v and 77 are as follows :
Y
2
3
5
7
10
12
1
962
968
961
947
934
932
ANSWERS TO EXERCISES 489
19. latus rectum = 25; vertex is at (275, 842)
20. 427 tons per sq. in. ; 23
Exercises 38
1. 404 6. 622 Ibs.
8. Plot E = cosh#(# ranging from o to 500 Vgr) and then alter
both scales
Exercises 39
1. Amplitudes: 8; 2; 518; 116; 91
T> J If 27r /
Periods: ; ; 02; 0102; 369
23
2. Amplitude = 4 .ic r. j .r 2jr o
Period = if Period as for cosine curve :  or 120
Exercises 40
1. {Assume some convenient value for /}. x = 403^
2. * = 5*3 3. i 221 4. 458 5. 36 or 217
6. 19 or 245 7. 279 8. 143 or 333
9. 266 10. 537 (308) 11. / = 3 5 L 12. 79
13. 449 rad. (257) 14. 1042, 13 15. 5523'
16. 1484' 17. 29 18. 6005* 19. 796'34*
20. 187. (Plot the curves y l = cosh x and y a = = sec x
and note the point of intersection.) 21. 634 ft.
Exercises 41
2. 454 3. 3341 560 7. 1043; io??
9. 22. [Hint. Let <*> = a + br ; also <t> = log J + g ^ 437 ~ T ^ ; and
4OX T
solve for q.]
Exercises 42
1. W = 47 + 6o5A 2. /i= 2 + oo4\/t/" 3. W = 328^
4. m =  4 o 5 +^H 5. /=oi 4 82 6. W=ri<P + i8
7. S = logi/ 1 ' 51 8. H = 0955^ u 9. T = 435 ' 238
10. <Z=i"2v7 11. T = 5 4 i'079 12. h = 0724W 1 ' 8
13. T= 129 x io 7 . 2 ' 46 14. Q = 6nH 48 15. y = 224\/H
16. / = io/ a 17. a = 1205, b = 535 18. a = 1320, 6 = 544
19 ' h = g.gx'io' 20< W = ' 87 ' C = 2 5
21. a = 149, b = 58, c = 02
22. E = I + OI32T 00000583T 2
23. E = 15 + OO795T 'Ooooo2iT a
24. A = 1446 27631; + OI384U 2
25. R = 160 I64V+4V 2 26. v = 3195 +'452D 
27. a = 10, b = 277 28. 2 29. 3 30. 4
31. y = i85 26 * 82. Q = isH 2 ' 5
490 MATHEMATICS FOR ENGINEERS
Exercises 43
3. 115*; 926 Ibs./n*
5. Write equation
log d log 29= JlogH JlogN or D = JH  N
Exercises 44
1. 8, 8^, 8f, 8^ ; 009% too low 2.
i i i i. 31
10+ 2+ 15+ i+ i+ 7' 325
4 x _I_ JL JL_
12+ 4+ 4+ 1 +
5. 6gf, say 6 complete turns with 19 holes on 33 hole circle
6. 8 complete and 2 holes on 18 hole circle (approx.)
7. 10 holes on 17 hole circle 8. 50 to 127
91.2 ^ A 4 j j
* *
x+i ' x+6
2* +3 3* +5 x ii
13 3 , 5 2
4
X I X 2
4 5
'*+4 #5 #2
2 15 5 . i
2X+ 7 x 3 3(.
X + 2) * 6(AT i) 2(# + i)
F 17 6 18 7
3(* + 2)
io 6 a 8 c 3
32 21. 48
22. i 23. I 24. b*
5
Exercises 45
A m 8 vn'n t jm*n z i^m 5 n 3
7 I 7 2 55*> /
256 40 100 125
8 I 8c 8oc 2
'
3 .
a 2 sin 2
12. (a) 20.19.18 . . . 10.9; (6) 125970
13. 105; 11880; 120 14. 984 15. 2074
16. (a) i oi x io 15 ; (b) 9545; (c) 73; (d) 4092
18. cosh x = i + r^ + ~ + ... ; sinh x = x + p + + . . f
~~ y y* y 3 jT ~~
Exercises 46
1. 45 2. 5904 3. o 4. 1728 5. 795
6. 4372 or 2449 T. x = 5, y 6 8. a = 15, & = 8
9 * = 5, y = 4, z = 2 10. a = 12, & 57, c = 48
MATHEMATICAL TABLES
TABLE I. TRIGONOMETRICAL RATIOS
Angle.
Chord.
Sine.
Tangent.
Cotangent.
Cosine.
De
grees.
Radians.
o
o
o
i
1414
15708
90
i
2
3
4
0175
0349
0524
0698
017
035
052
070
0175
0349
0523
0698
0175
0349
0524
0699
572900
286363
190811
143007
9998
9994
9986
9976
1402
1389
1377
1364
15533
15359
15184
15010
89
88
87
86
5
0873
087
0872
0875
114301
9962
I35I
I4835
85
6
8
9
1047
1222
1396
1571
105
122
140
157
1045
1219
1392
1564
1051
1228
1405
1584
95I44
81443
7U54
63138
9945
9925
9903
9877
1338
1325
1312
1299
14661
14486
14312
I4I37
84
83
82
81
10
1745
174
1736
1763
567I3
9848
1286
13963
80
ii
12
13
H
1920
2094
2269
2443
192
2O9
226
244
1908
2079
2250
2419
1944
2126
2309
2493
5I446
47046
433I5
40108
9816
9781
9744
9703
1272
1259
J245
1231
13788
13614
13439
13265
79
78
77
76
IS
2618
261
2588
2679
37321
9659
1218
13090
75
16
17
18
19
2793
2967
3142
3316
278
296
313
33
2756
2924
3090
3256
2867
3057
3249
3443
34874
32709
30777
29042
9613
9563
95"
9455
1204
1190
1176
1161
12915
12741
12566
12392
74
73
72
7i
20
3491
347
3420
3640
27475
9397
1147
12217
70
21
22
23
24
3665
3840
4014
4189
364
382
399
416
3584
3746
3907
4067
3839
4040
4245
4452
26051
24751
23559
22460
9336
9272
9205
9135
II33
1118
1104
1089
12043
11868
11694
11519
69
68
67
66
25
4363
433
4226
4663
21445
9063
1075
II345
65
26
27
28
29
4538
4712
4887
5061
450
467
484
501
4384
454
4695
4848
4877
5095
5317
5543
20503
19626
18807
18040
8988
8910
8829
8746
i 060
1045
1030
1015
11170
10821
10647
64
3
62
61
30
5236
518
5000
5774
17321
8660
IOOO
10472
60
31
32
33
34
54"
5585
5760
5934
534
551
568
585
5150
5299
5446
5592
6009
6249
6494
6745
16643
16003
15399
14826
8572
8480
8387
8290
985
970
954
939
10297
10123
9948
9774
59
58
57
56
35
6109
601
5736
7002
14281
8192
.923
9599
55
36
37
38
39
6283
6458
6632
6807
618
635
651
668
5878
6018
6157
6293
7265
7536
7813
8098
I3764
13270
12799
12349
8090
7986
7880
7771
908
892
877
861
9425
9250
9076
8901
54
53
52
51
40
6981
684
6428
8391
11918
7660
845
8727
5
41
42
43
44
7156
733
7505
7679
700
717
733
749
6561
6691
6820
6947
8693
9004
9325
9657
11504
11106
10724
10355
7547
7431
73M
7193
829
813
797
781
8552
8378
8203
8029
49
48
47
46
45
7854
765
7071
I'OOOO
IOOOO
7071
765
7854
45
Cosine
Cotangent
Tangent
Sine
Chord
Radians
Degrees
Angle
491
492
MATHEMATICAL TABLES
TABLE II. LOGARITHMS
1
2
3
4
5
6
7
8
9
123
456
789
10
0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
4 9 13
4 8 12
17 21 26
16 20 24
30 34 38
28 32 37
11
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
4 8 12
4 7 11
15 19 23
15 19 22
27 31 35
26 30 33
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1106
3 7 11
3 7 10
14 18 21
14 17 20
25 28 32
24 27 31
13
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
3 7 10
3 7 10
13 16 20
12 16 19
23 26 30
22 25 29
14
1461
1492
1523
1653
1584
1614
1644
1673
1703
1732
369
369
12 15 18
12 15 17
21 24 28
20 23 26
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
369
368
11 14 17
11 14 16
20 23 26
19 22 25
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
358
358
11 14 16
10 13 15
19 22 24
18 21 23
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
368
267
10 13 15
10 12 15
18 20 23
17 19 22
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
267
257
9 12 14
9 11 14
16 19 21
16 18 21
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
247
246
9 11 13
8 11 13
1C 18 20
15 17 19
20
3010
3032
3054
3076
3096
3118
3139
3160
3181
3201
246
8 11 13
16 17 19
21
22
23
84
3222
3424
3617
3802
3243
3444
3636
3820
3263
3464
3655
3838
3284
3483
3674
3856
3304
3502
3692
3874
3324
3522
3711
3892
3345
3541
3729
3909
3365
3560
3747
S927
3385
3579
3766
3945
3404
3598
3784
3962
246
246
246
345
8 10 12
8 10 12
7 9 11
7 9 U
14 16 18
14 15 17
13 15 17
12 14 16
25
8979
3997
4014
4031
4048
4065
4082
4099
4116
4133
235
7 9 10
12 14 15
26
27
28
29
4150
4314
4472
4624
4166
4330
4487
4639
4183
4346
4502
4654
4200
4362
4518
4669
4216
4378
4533
4683
4232
4393
4548
4698
4249
4409
4564
4713
4265
4425
4579
4728
4281
4440
4594
4742
4293
4456
4009
4757
236
236
235
134
7 8 10
689
689
679
11 13 15
11 13 14
U 12 14
10 12 13
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
134
679
10 11 13
31
82
83
34
4914
5051
5185
5315
4928
5065
5198
5328
4942
5079
5211
6340
4955
5092
5224
5353
4969
5105
5237
5366
4983
5119
5250
5378
4997
5132
5263
5391
5011
6146
6276
5403
5024
5159
5289
5416
6038
5172
5302
5428
134
134
134
134
678
678
668
568
10 11 12
9 11 12
9 10 12
9 10 11
S5
5441
5453
6465
6478
6490
5502
5514
6527
6539
5551
124
567
9 10 11
36
37
38
39
5563
5682
6798
6911
5575
5694
5809
5922
5587
6705
5821
5933
5599
5717
5832
5944
5611
5729
5843
5955
5623
5740
5855
5966
6635
5752
5866
6977
5647
6763
6877
5988
5658
5775
6888
5999
5670
5786
5899
6010
124
123
123
123
567
567
567
457
8 10 11
8 9 10
8 9 10
8 9 10
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
123
456
8 9 10
41
42
43
44
6128
6232
6335
6435
6138
6243
6345
6444
6149
6253
6355
6454
6160
6263
6365
6464
6170
6274
6375
6474
6180
6284
6385
6484
6191
6294
6395
6493
6201
6304
6405
6503
6212
6314
6415
6513
6222
6325
6425
6522
123
123
123
123
456
456
456
456
789
789
789
789
45
6532
6542
6551
6561
6571
6580
6590
6599
6609
6618
123
456
789
46
47
48
49
6628
6721
6812
6902
6637
6730
6821
6911
6646
6739
6830
6920
C656
6719
6839
6928
6665
6758
6848
6937
6675
6767
6857
6946
6684
6776
6866
6955
6693
6785
6875
6964
6702
6794
6884
6972
6712
6803
6893
6981
123
123
123
123
456
455
445
445
778
678
678
678
50
6990
G998
7007
7016
7024
7033
7042
7060
7059
7067
123
345
678
MATHEMATICAL TABLES
TABLE II. (contd.)
493
1
2
3
4
5
6
7
8
9
1 2 3
4 5 6
789
51
7076
7084
7093
7101
7110
7118
7126
7156
7143
7162
52
53
54
7160
7243
7324
7168
7261
7332
7177
7259
7340
7185
7267
7348
7193
7275
7356
7202
7284
7364
7210
7292
7373
7218
7300
7380
7226
7308
7388
7235
7316
7396
l 2 a
123
123
345
346
346
677
667
667
55
7404
7412
7419
7427
7435
7443
7461
7459
7466
7474
132
346
667
56
7482
7490
7497
7505
7513
7530
7528
7636
7643
7551
67
58
59
7559
7634
7709
7566
7642
7716
7574
7649
7723
7582
7657
7731
7589
7664
7738
7597
7672
7745
7604
7679
7752
7612
7686
7760
7619
7694
7767
7627
7701
7774
122
112
113
346
344
344
667
667
567
60
7782
7789
7796
7803
7810
7818
7825
7833
7839
7846
61
62
63
64
7853
7924
7993
8062
7860
7931
8000
8069
7868
7938
8007
8076
7875
7945
8014
80S2
7882
7952
8021
8089
7889
7959
8028
8096
7896
7966
8035
8102
7903
7973
8041
8109
7910
7980
8048
8116
7917
7987
8055
8122
112
112
112
113
344
334
334
334
566
566
666
656
65
8129
8136
8143
8149
8156
8162
8169
8176
8182
8189
112
334
666
63
67
68
69
8195
8261
8325
8388
8202
8267
8331
8395
8209
8274
8338
8401
8215
8280
8344
8407
8222
8287
8351
8414
8228
8293
8357
8420
8235
8299
8363
8426
8241
8306
8370
8432
8248
8312
8376
8439
8254
8319
8332
8445
112
112
113
112
334
334
334
234
556
666
456
466
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
112
234
466
71
72
73
74
8513
8573
8633
8692
8519
8579
8639
8698
8525
8585
8645
8704
8531
8591
8651
8710
8537
8597
8657
8716
8543
8603
8663
8722
8549
8609
8669
8727
8555
8615
8675
8733
8561
8621
8681
8739
8567
8627
868(5
8745
113
113
113
112
234
234
234
234
455
466
456
456
75
8751
8766
8762
8768
8774
8779
8785
8791
8797
8802
112
233
466
76
77
78
73
8808
8865
8921
8976
8814
8871
8927
8982
8820
8876
8932
8987
8825
8882
8933
8993
8831
8887
8943
8998
8837
8893
8949
9004
8342
8899
8954
9009
8848
8904
8900
9015
8854
8910
8965
9020
8859
8915
8971
9025
112
112
112
113
233
233
233
233
455
445
446
446
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
112
233
446
81
82
83
84
9085
9138
9191
9243
9090
9143
9196
9248
9096
9149
9201
9253
9101
9154
9206
9258
9106
9159
9212
9263
9112
9165
9217
9269
9117
9170
9222
9274
9122
9175
9227
9279
9128
9180
9232
9284
9133
9186
923S
9289
112
112
112
112
233
233
233
2 33
445
446
445
445
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
112
233
446
86
87
88
89
9345
9395
9445
9494
9350
9400
9450
9499
9355
9405
9455
9504
9360
9410
94CO
9509
9365
9415
9465
9513
9370
9420
9469
9518
9376
9425
9474
9523
9380
9430
9479
9528
9385
9435
9484
9533
9390
9440
9489
9538
112
Oil
Oil
Oil
233
223
223
223
445
344
344
344
90
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
Oil
223
344
91
92
93
91
9590
9638
9635
9731
9595
9043
9689
9736
9600
9647
9094
9741
9605
9652
9699
9745
9609
9657
9703
9750
9614
9661
9708
9754
9619
9666
9713
9759
9624
9671
9717
9763
9628
9675
9722
9768
9633
9680
9727
9773
1 1
Oil
Oil
Oil
223
223
223
223
344
344
344
344
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9813
Oil
223
344
96
97
98
99
9823
9868
9912
9956
9827
9872
9917
9961
9832
9877
9921
9965
9836
9881
9926
9969
9841
9886
9930
9974
9845
9890
9934
9978
9850
9894
9939
9983
9854
9899
9943
9987
9859
9903
9948
9991
9863
9908
9952
9996
1 1
Oil
Oil
Oil
223
223
223
223
344
344
344
334
494
MATHEMATICAL TABLES
TABLE III. ANTILOGARITHMS
1
2
3
4
5
6
7
8
9
123
456
789
00
1000
1002
1005
1007
1009
1012
1014
1016
1019
1021
001
111
222
01
02
03
04
1023
1047
1072
1096
1026
1060
1074
1099
1028
1052
1076
1102
1030
1054
1079
1104
1033
1057
1081
1107
1035
1059
1084
1109
1038
1062
1086
1112
1040
1064
1089
1114
1042
1067
1091
1117
1045
1069
1094
1119
001
001
001
Oil
111
111
111
112
222
232
222
222
05
1122
1125
1127
1130
1132
1135
1138
1140
1143
1146
Oil
112
222
06
07
08
09
1148
1176
1202
1230
1161
1178
1205
1233
1163
1180
1208
1236
1156
1183
1211
1239
1159
1186
1213
1242
1161
1189
1216
1246
1164
1191
1219
1247
1167
1194
1222
1260
1169
1197
1225
1253
1172
1199
1227
1256
Oil
Oil
Oil
Oil
112
112
112
112
222
222
223
223
10
1259
1262
1265
1268
1271
1274
1276
1279
1282
1285
Oil
112
223
11
12
13
14
1288
1818
1349
1380
1291
1321
1352
1384
1294
1324
1365
1387
1297
1327
1358
1390
1300
1330
1361
1393
1303
1334
1365
1396
1306
1337
1368
1400
1309
1340
1371
1403
1312
1343
1374
1406
1315
1346
1377
1409
Oil
Oil
Oil
Oil
122
122
122
122
223
223
233
233
15
1413
1416
1419
1422
1426
1429
1432
1435
1439
1442
Oil
122
233
16
17
18
19
1445
1479
1514
1549
1449
1483
1517
1562
1452
1486
1521
1556
1455
1489
1524
1560
1459
1493
1528
1663
1462
1496
1531
1567
1466
1500
1635
1570
1469
1603
1538
1674
1472
1507
1542
1578
1476
1510
1546
1581
Oil
Oil
Oil
Oil
122
122
122
192
233
2 3 S
233
333
20
1585
1689
1592
1596
1600
1603
1607
1611
1614
1618
Oil
122
333
21
22
23
24
1623
1660
1698
1738
1626
1663
1702
1742
1629
1667
1706
1746
1633
1671
ino
1750
1637
1675
1714
1754
1641
1679
1718
1758
1644
1683
1722
1762
1648
1687
1726
1766
1652
1690
1730
1770
1656
1034
1734
1774
Oil
1 1
Oil
1 1
222
222
222
222
333
333
334
334
25
1778
1782
1786
1791
1795
1799
1803
1807
1811
1816
Oil
222
334
26
27
28
29
1820
1862
1905
1950
1824
1866
1910
1954
1828
1871
1914
1959
1832
1875
1919
1963
1837
1879
1923
1968
1841
1884
1928
1972
1845
1888
1932
1977
1849
1892
1936
1982
1854
1897
1941
1986
185S
1901
1945
1991
Oil
Oil
Oil
Oil
223
223
223
223
334
334
344
3 4 i
30
1995
2000
2004
2009
2014
2018
2023
2028
2032
2037
Oil
223
344
31
32
33
34
2042
2089
2138
2188
2046
2094
2143
2193
2051
2099
2148
2198
2056
2104
2153
2203
2061
2109
2158
2208
2065
2113
2163
2213
2070
2118
2168
2218
2075
2123
2173
2223
2080
2128
2178
2228
2034
2133
2183
2234
Oil
Oil
Oil
119
223
233
223
233
344
344
344
445
35
2239
2244
2249
2254
2259
2265
2270
2276
2280
2286
112
233
445
35
37
38
89
2291
2344
2399
2455
2296
2350
2404
2460
2301
2355
2410
2466
2307
2360
2415
2472
2312
2366
2421
2477
2317
2371
2427
2483
2323
2377
2432
2489
2328
2382
2438
2495
2333
2388
2443
2500
2339
2393
2449
2506
112
112
112
112
233
233
233
233
445
445
445
455
40
2512
2518
2523
2529
2535
2541
2547
2563
2559
2564
112
234
465
41
42
43
44
2570
2630
2692
2754
2576
2636
2698
2761
2582
2642
2704
2767
2688
2649
2710
2773
2594
2655
2716
2780
2600
2661
2723
2786
2606
2667
2729
2793
2612
2673
2735
2799
2618
2679
2742
2805
2624
2685
2748
2812
112
112
112
112
234
234
334
334
455
456
456
456
45
2818
2825
2831
2838
2844
2851
2858
2864
2871
2877
112
334
556
46
47
48
49
2884
2951
SO20
3090
2891
2958
3027
3097
2897
2965
3034
3105
2904
2972
3041
3112
2911
2979
3048
3119
2917
2985
3055
3126
2924
2992
3062
3133
2931
2999
3069
3141
2938
3006
3076
3148
2944
3013
3083
3155
112
112
112
112
3 S 4
334
344
344
566
556
566
666
MATHEMATICAL TABLES
TABLE III. (contd).
495
1
2
3
4
5
6
7
8
9
123
456
789
50
3162
3170
3177
3181
3192
3199
3206
3214
3321
3223
113
344
567
51
52
53
54
3236
8311
3388
3467
3243
3319
3396
3475
3251
3327
3404
3483
3258
3334
3412
3491
3266
3342
3420
3499
3273
3350
3428
3508
3281
3357
3436
3516
3289
3365
3443
3524
3296
3373
3451
3532
3304
3381
3459
3540
1 3 2
133
133
132
345
345
345
346
567
667
667
667
55
3548
3656
3565
3573
3581
3589
3597
3606
3614
3623
133
346
677
56
57
58
59
3631
3715
3802
3890
3639
3724
3811
3899
3648
3733
3819
3903
3656
3741
3828
3917
3664
3750
3S37
3926
3673
3758
3846
3936
3681
3767
3855
3945
3690
3776
3864
3954
3698
3784
3873
3963
3707
3793
3883
3972
133
133
123
123
345
346
446
466
678
678
678
678
60
3981
3990
3999
4009
4018
4027
4036
4046
4055
4064
123
466
678
61
62
63
64
4074
4169
4266
4365
4083
4178
4276
4375
4093
4188
4285
4385
4102
4198
4295
4395
4111
4207
4305
4406
4131
4217
4315
4416
4130
4227
4325
4426
4140
4236
4335
4436
4150
4246
4345
4446
4159
4256
4355
4457
133
123
133
123
466
466
466
456
789
789
789
789
65
4467
4477
4487
4498
4608
4619
4529
4639
4550
4660
123
456
789
66
67
68
69
4571
4677
4786
4898
4581
4688
4797
4909
4592
4699
4808
4920
4603
4710
4819
4932
4613
4721
4831
4943
4624
4732
4842
4965
4634
4742
4853
4966
4645
4753
4864
4977
4656
4764
4875
4989
4667
4775
4887
5000
123
123
123
133
456
457
467
567
7 9 10
8 9 10
8 9 10
8 9 10
70
6012
6023
6035
6047
5058
C070
6082
6093
5105
6117
134
667
8 9 11
71
72
73
74
5129
5248
6370
6495
5140
6260
5383
5508
5152
5272
5395
6621
6164
5284
6408
5534
6176
5297
6420
5546
5188
5309
6433
5559
6200
6321
6445
6572
5212
6333
5458
5585
5224
6316
5470
6598
5236
6358
5483
6610
134
124
134
134
567
567
568
568
8 10 11
9 10 11
9 10 11
9 10 12
75
5623
6036
6649
5662
6676
6689
5702
5715
6728
6741
134
678
9 10 12
76
77
78
79
6754
6888
6026
6166
6768
5902
6039
6180
6781
6916
6053
6194
5794
5929
6067
6209
6808
5943
6081
6223
6821
5957
6095
6237
6834
6970
6109
6252
6848
6984
6124
6266
5861
5993
6138
6281
6875
6012
6152
6295
134
134
134
134
578
678
678
679
9 11 12
10 11 13
10 11 13
10 11 13
80
6310
6324
6339
6353
6368
6383
6397
6413
6427
6442
134
679
10 12 13
81
82
83
14
6457
6607
6761
6918
6471
6622
6776
6934
6486
6637
6792
6950
6501
6653
6808
6966
6516
6668
6823
6982
6531
6683
6839
6998
6546
6699
6855
7015
6561
6714
6871
7031
6577
6730
6887
7047
6592
6745
6902
7063
335
235
235
236
689
689
689
6 8 10
11 12 14
11 12 14
11 13 14
11 13 15
85
7079
7096
7112
7129
7146
7161
7178
7194
7211
7228
236
7 8 10
12 13 15
86
87
83
89
7244
7413
7586
7762
7261
7430
7603
7780
7278
7447
7621
7798
7295
7464
7638
7816
7311
7482
7656
7834
7328
7499
7674
7852
7345
7616
7691
7870
7362
7534
7709
7889
7379
7551
7727
7907
7396
7568
7745
7925
336
236
345
245
7 8 10
7 9 10
7 9 11
7 9 11
12 13 15
12 14 16
12 14 16
13 14 16
CO
7943
7962
7980
7998
8017
8036
8054
8072
8091
8110
346
7 9 11
13 15 17
91
92
93
4
8128
8318
8511
8710
8147
8337
8531
8730
8166
8356
8551
8750
8185
8375
8670
8770
8204
8395
8590
8790
8222
8414
8610
8810
8241
8433
8630
8831
8260
8453
8650
8851
8279
8472
8670
8872
8299
8492
8690
8892
246
246
246
246
8 9 11
8 10 13
8 10 12
8 10 12
13 15 17
14 15 17
14 16 18
14 16 18
95
8913
8933
8954
8974
8995
9016
9036
9057
9078
9099
246
8 10 12
15 17 19
f6
97
98
99
9120
9333
9550
9772
9141
9354
9672
9795
9162
9376
9594
9817
9183
9397
9616
9840
9204
9419
9638
9863
9226
9441
9661
9886
9247
9462
9683
9908
9268
9484
9705
9931
9290
9506
9727
9954
9311
9528
9750
9977
246
247
247
267
8 11 13
9 11 13
9 11 13
9 11 14
16 17 19
15 17 20
16 18 20
16 18 20
496 MATHEMATICAL TABLES
TABLE IV. NAPIERIAN, NATURAL, OR HYPERBOLIC LOGARITHMS
Number. I
1
2
3
4
5
6
7
8
9
01
1*6974
7927
8797
9598
0339
1029
1674
2280
2852
3393
02
23906
4393
4859
5303
5729
6i37
6529
6907
7270
7621
03
04
7960
10837
8288
1084
8606
1325
8913
1560
9212
1790
9502
2015
9783
2235
0057
245
0324
2660
0584
2866
05
3068
3267
346i
3651
3838
4022
4202
4379
4553
4724
06
07
4892
6433
5057
6575
5220
6715
538o
6853
5537
6989
5692
7"3
5845
7256
5995
7386
6i43
7515
6289
S 43
Mean Differences.
08
7769
7893
8015
8i37
8256
8375
3492 8007
8722
8835
09
8946
9057
9166
9274
938i
9487
9592
9695
9798 9899
123
456
789
10
ooooo
0100
0198
0296
0392
0488
0583
0677
0770
0862
11
0953
1044
"33
1222
1310
1398
1484
157
1655
1740
9 17 26
35 44 52
61 70 78
12
1823
1906
1989
2070
2151
2231
2311
2390 2469
2546
8 1624
32 40 48
56 64 72
13
2624
2700
2776
2852
2927
3001
3075
31483221
3293
7 1522
30 37 45
52 59 67
14
3365
3436
3507
3577
3646
371
3784
38533920
3988
71421
283541
48 55 62
15
455
4121
4187
4253
43i8
4383
4447
45"
4574
4637
6 13 19
26 32 39
45 52 S 8
16
4700
4762
4824
4886
4947
5008
5068
5128
5188
5247
6 12 18
24 30 36
42 48 55
17
53o6
5365
5423
548i
5539
5596
5653
571
5766
5822
6 ii 17
24 29 34
4 46 52
18
5878
5933
5988 6043
6098
6152
620662596313
6366
5 " 16
22 27 32
38 43 49
19
6419
6471
6523
6575
6627
6678
672967806831
6881
5 II5
2O 26 31
36 41 46
20
6931
6981
73i
7080
7129
7178
7227
7275 7324
7372
5 IOI 5
20 24 29
34 39 44
21
7419
7467
75H
756i
7608
7655
7701
7747 7793
7839
5 914
1923 28
33 37 42
22
7885
793
7975
8020
8065
8109
8154
81988242
8286
4 913
18 22 27
31 36 40
23
8329
8372
8416
8459
8502
8544
8587
8629 8671
8713
4913
17 21 26
30 34 38
24
8755
8796
8838
8879
8920
8961
9002
9042 9083
9123
4 8 12
16 20 24
29 33 37
25
9163
9203
9243
9282
9322
936i
9400
9439
9478
9517
4 8 12
16 20 24
2731 35
26
9555
9594
9632
9670
9708
9746
9783
9821
9858
9895
4811
15 19 23
26 30 34
27
9933
9969
0006
0043
0080
0116
0152
0188
0225
0260
47"
15 l8 22
26 29 33
28
10296
0332
0367
0403
0438
0473
0508
0543
0578
0613
4 7"
14 18 21
25 28 32
29
0647
0682
0716
0750
0784
0818
0852
0886
0919
0953
371
14 17 20
242731
30
0986
1019
1053
1086
1119
"5i
1184
1217
1249
1282
3 7 10
13 16 20
23 26 30
31
I3H
1346
1378
1410
1442
*474
1506
1537
1569
1600
3 6 10
13 16 19
22 25 29
32
1632
1663
1694
1725
1756
1787
1817
1848
1878
1909
369
12 15 18
21 25 28
33
1939
ig6c
2OOO
2030
2060
2090
2119
2149
2179
2208
369
12 15 18
21 24 27
34
2238
2267
2296
2326
2355
2384
2413
2442
2470
2499
369
12 1417
2O 23 26
35
2528
2556
2585
2613
2641
2669
2698
2726
2754
2782
368
II 1417
20 22 25
36
2809
2837
2865
2892
2920
2947
2975
3002
3029
3056
3 5 8
ii 14 16
19 22 25
37
3083
3110
3137
3164
3i9i
3218
3244
3271
3297
3324
3 5 8
ii 13 16
19 21 24
38
3350
3376
343
3429
3455
348i
3507
3533
3558
3584
3 5 8
10 13 16
18 21 23
39
3610
3635
3661
3686
37"
3737
3762
3788
3813
3838
3 5 8
1013 15
18 20 23
40
3863
3888
3913
3938
3962
3987
4012
4036
4061
4085
257
10 12 15
17 2O 22
41
4110
4134
4 J 59
4183
4207
4231
4255
4279
433
4327
257
IO 12 14
17 19 22
42
4351
4375
4398
4422
4446
4469
4493
45i6
454
4563
257
9 12 14
16 19 21
43
4586
4609
4633
4656
4679
4702
4725
4748
477
4793
257
9 " 14
16 18 21
44
4816
4839
4861
4884
4907
4929
495i
4974
4996
5i9
247
9 " 13
16 18 20
45
504 1
5063
5085
5107
5129
5i5i
5173
5195
5217
5239
247
9" 13
15 18 20
46
5261
5282
5304
5326
5347
5369
5390
5412
5433
5454
246
9 " 13
15 17 !9
47
5476
5497
55i8
5539
556o
558i
5602
5623
5644
5665
246
8 ii 13
15 1719
48
5686
577
5728
5748
57 6 9
5790
5810
5831
5851
5872
246
8 10 12
14 16 19
49
5892
5913
5933
5953
5974
5994
6014
6034
6054
6074
246
8 IO 12
14 16 18
50
6094
6114
6134
6i54
6174
6194
6214
6233
6253
627^
246
8 IO 12
14 16 18
MATHEMATICAL TABLES
TABLE IV (contd.)
497
51
52
53
54
55
56
57
53
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
10
7048 70
7228 7246 7263 7281
745 7422 7440,7457 7475 7492
16292 6312 6332 6351 6371 6390 6409 6429 6448 6467
648/65o6'6525!6544 6563 6582
6677:6696,6715673467526771
f  '
648/65o6'6525!6544 6563 6582 6601 662OJ6639 6658
69196938:69566975699370117029
66 7084 7102 7120 7138 7156 7174:7192,7210
7750 7767 7783 7800 7817 7834
7918 7934 7951 7968 7984 8001
62 8278 8294 8310 8326 8342 8358 8374^390
8437 8453 8469 8485 8500 8516 8532 8547
8083 8099 8116 8132 8148 8165 8181 819718213 8229
82 4 682 '
84068421
8 503'8579 8594J86ioJ8625 8641
871887338749)876418779 8795 8810 8825884o8856
887188868901
902190369051
6
8
9
Mean Differences.
123456 789
246
246
679o68o86827J68462 4 6
246
245
750975277544;75i
769977167733'
7851 7868 7884 7901
8017 8034 8050 8067
7299731773347352,7370,738712 4 5
245
235
235
235
235
235
235,
235
235:
235
234
234
1 3 4
i 3 4
i 3 4
i 3 4
i 3 4
i 3 4
134
8976 8991 9006
9066*9081 9095 9110 9125 9140^155
9169 9184 9199 92139228 9243 9257 9272 9286,9301
9315 9330,9344,935919373 9387 942 94i,943i 9445
9459 9473 9488,9502 9516 953 9545 9559 9573 9587
9601 9615^629 9643 9657 9671 9685 9699^7139727
9741 9755!97 6 9'9782!9796]98io 9824 9838,985119865
loooi
0412 0425^438 0451,0464
oi49'oi62 0176 0189 0202 0216 0229 0242 0255 0268
0281 029510308 0321(0334 0347 0360 0373 03860399
0477 0490 0503 0516 0528
U^X^ \J^4.^) \J^^V,\JI^ J.
0541 0554 l 0567Jo58o'o592 0605 061810631 0643^0656
066906811069410707,071907320744
0794 0807:0819 0832 0844 o  ~ Q<; ~'
1033
1748
1861
1972
2721
28242834
2
23026
KK
0919 0931 0943 0956 0968 0980 0992
1041 1054
1163 1175
1282
1401 1412 1424
151815291541
1066 IO78IO9O IIO2
1187 1199 I2II
I759I77
18721883
1983 1994
1552
1645 1656 1668 1679 1691
1782
1894
. 075707690782
857 0869 O882 0894 O9O6
III4
1223 1235
1342 1354
1436144814591471
1564
1679
I793I804
1905 1917 1928 1939 I950I96I
2083 2094 2105 2116 2127 2138 2149 2159 2170 2181
2192 22O3 221
4 222512235 2246 2257 2268 2279 2289
2300 23IIJ23222332 2343 2354 2365 2375 2386 2397
2481 2492 2502
2513 2523 2534 2544 2555 2565 2576 2586 2597 2607
261826282638
2844 2854
2732 2742 2752 2762 2773 2783 2793 2803 2814
925 2935 2946 2956 2966 2976 2986 2996 3006 3016
15761587
I7O2
18161827
2OO6 2OI7 2O28 2O39 2050 2O6l 2072
2649 2659 2670 268O 2690 2701
1005
1126 113
1305
1483!
1599 1610 1622
171317251736
2865 2875 2885 2895 2905 2915
1029
2711
134
i 3 4
i 3 4
i 3 4
i 3 4
i 3 4
i 2 4
i 2 4
i 2 4
i 2 4
2 4
2 3
2 3
2 3
2 3
2 3
2 3
2 3
2 3
2 3
2 3
2 3
2 3
2 3
8 IO 12
8 IO 12
8 9 n
7 9 ii
7 9 ii
7 9 ii
7 9 10
7 9 10
7 8 10
7 8 10
7 8 10
6 8 10
6 8 10
689
689
689
679
679
679
679
678
678
5 7 8
5 7 8
5 7 8
5 7 J
578
568
568
567
567
5 6 7
567
567
567
567
567
56?
467
467
4 5 I
456
456
456
456
456
456
456
498
MATHEMATICAL TABLES
TABLE V. NATURAL SINES.
i
I
Q
0'
00
6'
01
12'
02
18'
03
24'
04
30'
05
38'
06
42'
07
48'
08
54'
09
Mean Differences.
1' 2' 3' 4' 5'
oooo
0017
0035
0052
0070
0087
0105
OI22
0140
0157
3 6 9 12 15
1
0175
0192
0209
0227
0244
0262
0279
0297
0314
332
3 6 9 12 15
2
349
0366
0384
0401
0419
0436
454
0471
0488
0506
3 6 9 12 15
3
0523
054 1
0558
0576
0593
0610
0628
0645
0663
0680
3 6 9 12 15
4
0698
0715
0732
075
0767
0785
0802
0819
0837
0854
3 6 9 12 14
5
0872
0889
0906
0924
0941
0958
0976
0993
IOII
1028
3 6 9 12 14
6
1045
1063
1080
1097
"15
1132
1149
1167
1184
1201
3 6 9 12 14
7
1219
1236
1253
1271
1288
1305
1323
1340
1357
1374
3 6 9 12 14
8
1392
1409
1426
1444
1461
1478
1495
1513
1530
J 547
3 6 9 12 14
9
1564
1582
1599
1616
i633
1650
1668
1685
1702
1719
3 6 9 12 14
10
1736
1754
1771
1788
1805
1822
1840
1857
1874
1891
3 6 9 ii 14
11
1908
1925
1942
1959
1977
1994
2OII
2O28
2045
2062
3 6 9 ii 14
12
2079
2096
2113
2130
2147
2164
2181
2198
2215
2233
3 6 9 ii 14
13
2250
2267
2284
2300
2317
2334
2351
2368
2385
2402
3 6 8 ii 14
14
2419
2436
2453
2470
2487
2504
2521
2538
2 554
2571
3 6 8 ii 14
15
2588
2605
2622
2639
2656
2672
2689
2706
2723
2740
3 6 8 ii 14
16
2756
2773
2790
2807
2823
2840
2857
2874
2890
2907
3 6 8 u 14
17
2924
2940
2957
2974
2990
3007
3024
3040
3057
3074
3 6 8 ii 14
18
3090
3107
3123
3MO
3156
3i73
3190
32O6
3223
3239
3 6 8 ii 14
19
3256
3272
3289
3305
3322
3338
3355
3371
3387
3404
3 5 8 ii 14
20
3420
3437
3453
34 6 9
3486
352
35i8
3535
3551
3567
3 5 8 ii 14
21
3584
3600
3616
3633
3649
3665
3681
3697
37M
373
3 5 8 ii 14
22
3746
37 62
3778
3795
3811
3827
3843
3859
3875
3891
35 ? ii 14
23
3907
3923
3939
3955
3971
3987
4003
4019
435
45i
3 5 8 ii 14
24
4067
4083
4099
4H5
4131
4M7
4163
4179
4195
4210
3 5 8 ii 13
25
4226
4242
4258
4274
4289
4305
432i
4337
4352
4368
3 5 8 ii 13
28
4384
4399
4415
4431
4446
4462
4478
4493
4509
4524
3 5 8 10 13
27
454
4555
4571
4586
4602
4617
4633
4648
4664
4679
3 5 8 10 13
28
4695
4710
4726
4741
4756
4772
4787
4802
4818
4833
3 5 8 10 13
29
4848
4863
4879
4894
4909
4924
4939
4955
4970
4985
3 5 8 10 13
30
5000
5015
5030
5045
5060
5075
5090
5105
5120
5135
3 5 8 10 13
31
5150
5165
5180
5195
5210
5225
524
5255
5270
5284
2 5 7 10 12
32
5299
5314
5329
5344
5358
5373
5388
54 2
5417
5432
2 5 7 10 12
33
5446
54 6 i
5476
5490
5505
5519
5534
5548
5563
5577
2 5 7 10 12
34
5592
5606
5621
5635
5650
5664
5678
5693
577
5721
2 5 7 10 12
35
5736
5750
5764
5779
5793
5807
5821
5835
5850
5864
2 5 7 9 12
36
5878
5892
5906
5920
5934
5948
5962
5976
5990
6004
2 5 7 9 12
37
6018
6032
6046
6060
6074
6088
6101
6115
6129
6i43
257 9 12
38
6157
6170
6184
6198
6211
6225
6239
6252
6266
6280
257 9 ii
39
6293
6307
6320
6334
6347
6361
6 374
6388
6401
6414
247 9 ii
40
6428
6441
6455
6468
6481
6494
6508
6521
6534
6547
247 9 ii
41
6561
6574
6587
6600
6613
6626
6639
6652
6665
6678
247 9 ii
42
6691
6704
6717
6730
6743
6756
6769
6782
6794
6807
2 4 6 911
43
6820
6833
6845
6858
6871
6884
6896
6909
6921
6934
246 8 ii
44
6947
6959
6972
6984
6997
7009
7022
7034
7046
759
246 8 10
45
7071
7083
7096
7108
7120
7133
7145
7157
7169
7181
246 8 10
MATHEMATICAL TABLES
499
TABLE V. (contd.)
8
0'
6'
12'
18'
24'
30'
36'
42'
48'
54'
Mean Differences.
1
00
01
02
03
04
05
06
07
08
09
1' 2' 3' 4' 5'
45
7071
7083
7096
7108
7120
7133
7145
7i57
7169
7181
246 8 10
46
7i93
7206
7218
7230
7242
7254
7266
7278
7290
7302
246 8 10
47
7314
7325
7337
7349
736i
7373
7385
7396
7408
7420
246 8 10
48
7431
7443
7455
7466
7478
7490
75oi
75i3
7524
7536
2 4 6 8 10
49
7547
7559
757
758i
7593
7604
7615
7627
7638
7649
24689
50
7660
7672
7683
7694
7705
7716
7727
7738
7749
7760
24679
51
7771
7782
7793
7804
78i5
7826
7837
7848
7859
7869
2 4579
52
7880
7891
7902
7912
7923
7934
7944
7955
7965
7976
24579
53
7986
7997
8007
8018
8028
8039
8049
8059
8070
8080
23579
54
8090
8100
8m
8121
8131
8141
8151
8161
8171
8181
2357 8
55
8192
8202
8211
8221
8231
8241
8251
8261
8271
8281
23578
56
8290
8300
8310
8320
8329
8339
8348
8358
8368
8377
23568
57
8387
8396
8406
8415
8425
8434
8443
8453
8462
8471
23568
58
8480
8490
8499
8508
8517
8526
8536
8545
8554
8563
23568
59
8572
8581
8590
8599
8607
8616
8625
8634
8643
8652
3467
60
8660
8669
8678
8686
8695
8704
8712
8721
8729
8738
3467
61
8746
8755
8763
8771
8780
8788
8796
8805
8813
8821
3467
62
8829
8838
8846
8854
8862
8870
8878
8886
8894
8902
3457
63
8910
8918
8926
8934
8942
8949
8957
8965
8973
8980
3456
64
8988
8996
9003
9011
9018
9026
9033
9041
9048
9056
3456
65
9063
9070
9078
9085
9092
9100
9107
9114
9121
9128
2456
66
9135
9143
9150
9157
9164
9171
9178
9184
9191
9198
2356
67
9205
9212
9219
9225
9232
9239
9245
9252
9259
9265
2346
68
9272
9278
9285
9291
9298
9304
93"
9317
9323
9330
2345
69
9336
9342
934 8
9354
936i
9367
9373
9379
9385
9391
2345
70
9397
943
9409
9415
9421
9426
9432
9438
9444
9449
2345
71
9455
9461
9466
9472
9478
9483
9489
9494
9500
9505
2345
72
95"
95i6
9521
9527
9532
9537
9542
9548
9553
9558
2334
73
9563
9568
9573
9578
9583
9588
9593
9598
9603
9608
2234
74
9613
9617
9622
9627
9632
9636
9641
9646
9650
9655
2234
75
9659
9664
9668
9673
9677
9681
9686
9690
9694
9699
1234
76
973
9707
9711
97*5
9720
9724
9728
9732
9736
9740
1233
77
9744
9748
975i
9755
9759
9763
9767
977
9774
9778
233
78
9781
9785
9789
9792
9796
9799
9803
9806
9810
9813
223
79
9816
9820
9823
9826
9829
9833
9836
9839
9842
9845
i 223
80
9848
9851
9854
9857
9860
9863
9866
9869
9871
9874
122
81
9877
9880
9882
9885
9888
9890
9893
9895
9898
9900
122
82
9903
9905
9907
9910
9912
9914
9917
9919
9921
9923
O 122
83
9925
9928
993
9932
9934
9936
9938
9940
9942
9943
O 112
84
9945
9947
9949
9951
9952
9954
9956
9957
9959
9960
112
85
9962
9963
9965
9966
9968
9969
997 1
9972
9973
9974
O O I I I
86
9976
9977
9978
9979
9980
9981
9982
9983
9984
9985
I I I
87
9986
9987
9988
9989
9990
9990
9991
9992
9993
9993
O O I I
88
9994
9995
9995
9996
9996
9997
9997
9997
9998
9998
O O O O O
89
9998
9999
9999
9999
9999
IOOO
IOOO
IOOO
IOOO
IOOO
O O O O O
90
IOOO
KK2
5oo
MATHEMATICAL TABLES
TABLE VI. NATURAL COSINES
0'
6'
12'
18'
24'
30'
36'
42'
48'
54'
Mean Differences.
9
q
00
01
02
03
04
05
06
07
08
09
1' 2' 3' 4' 5'
IOOO
ooo
I'OOO
IOOO
ooo
IOOO
9999
9999
9999
9999
o o o o o
1
9998
9998
9998
9997
9997
9997
9996
9996
9995
9995
o o o o o
2
9994
9993
9993
9992
9991
9990
9990
9989
9988
9987
O O I I
3
9986
9985
9984
9983
9982
9981
9980
9979
997 s
9977
I I I
4
9976
9974
9973
9972
9971
9969
9968
9966
9965
9963
O I I I
5
9962
9960
9959
9957
9956
9954
9952
9951
9949
9947
I I I 2
6
9945
9943
9942
994
9938
9936
9934
9932
9930
9928
O I I I 2
7
9925
9923
9921
9919
9917
9914
9912
9910
9907
9905
O I I 2 2
8
993
9900
9898
9895
9893
9890
9888
9885
9882
9880
O I I 2 2
9
9877
9874
9871
9869
9866
9863
9860
9857
9854
9851
O I I 2 2
10
9848
9845
9842
9839
9836
9833
9829
9826
9823
9820
II223
11
9816
9813
9810
9806
9803
9799
9796
9792
9789
9785
II223
12
9781
9778
9774
977
9767
97 6 3
9759
9755
9751
9748
II233
13
9744
974
9736
9732
9728
9724
9720
9715
9711
9707
II233
14
973
9699
9694
9690
9686
9681
9677
9673
9668
9664
II234
15
9659
9655
9650
9646
9641
9636
9632
9627
9622
9617
12234
16
9613
9608
9603
9598
9593
9588
9583
9578
9573
9568
12234
17
9563
9558
9553
954 s
9542
9537
9532
9527
9521
95i6
12334
18
95"
9505
9500
9494
9489
9483
9478
9472
9466
9461
12345
19
9455
9449
9444
9438
9432
9426
9421
94*5
9409
943
12345
20
9397
9391
9385
9379
9373
9367
936i
9354
934 s
9342
12345
21
9336
9330
9323
9317
93"
934
9298
9291
9285
9278
12345
22
9272
9265
9259
9252
9245
9239
9232
9225
9219
9212
12346
23
9205
9198
9191
9184
9178
9171
9164
9157
915
9143
12356
24
9135
9128
9121
9114
9107
9100
9092
9085
9078
9070
2456
25
9063
9056
9048
9041
9033
9026
9018
9011
9003
8996
3456
26
8988
8980
8973
8965
8957
8949
8942
8934
8926
8918
3456
27
8910
8902
8894
8886
8878
8870
8862
8854
9846
8838
3457
28
8829
8821
8813
8805
8796
8788
8780
8771
8763
8755
34 6 7
29
8746
8738
8729
8721
8712
8704
8695
8686
8678
8669
13467
30
8660
8652
8643
8634
8625
8616
8607
8599
8590
8581
13467
31
8572
8563
8554
8545
8536
8526
8517
8508
8499
8490
23568
32
8480
8471
8462
8453
8443
8434
8425
8415
8406
8396
23568
33
8387
8377
8368
8358
8348
8339
8329
8320
8310
8300
23568
34
8290
8281
8271
8261
8251
8241
8231
8221
8211
8202
2357
35
8192
8181
8171
8161
8151
8141
8131
8121
8111
8100
23578
36
8090
8080
8070
8059
8049
8039
8028
8018
8007
7997
23579
37
7986
7976
7965
7955
7944
7934
7923
7912
7902
7891
24579
38
7880
7869
7859
7848
7837
7826
7815
7804
7793
7782
24579
39
7771
7760
7749
7738
7727
7716
775
7694
7683
7672
24679
40
7660
7649
7638
7627
7 6l 5
7604
7593
758i
757
7559
24689
41
7547
7536
7524
7513
75oi
7490
7478
7466
7455
7443
2 4 6 8 10
42
7431
7420
7408
7396
7385
7373
736i
7349
7337
7325
246 8 10
43
7314
7302
7290
7278
7266
7254
7242
7230
7218
7206
246 8 10
44
7193
7181
7169
7157
7H5
7133
7120
7108
7096
7083
246 8 10
45
7071
759
7046
734
7022
7009
6997
6984
6972
6959
246 8 10
MATHEMATICAL TABLES
TABLE VI (contd.)
501
j
w
p
Q
0'
00
6'
01
12'
02
18'
03
24'
04
30'
05
36'
06
42'
07
48'
08
54'
09
Mean Differences.
1' 2' 3' 4' 5'
45
7071
7059
7046
7034
7022
7009
6997
6984
6972
6959
246 8 10
46
6947
6934
6921
6909
6896
6884
6871
6858
6845
6833
246 8 ii
47
6820
6807
6794
6782
6769
6756
6743
6730
6717
6704
246 9 ii
48
6691
6678
6665
6652
6639
6626
6613
6600
6587
6 574
2 47 9 ii
49
6561
6547
6534
6521
6508
6494
6481
6468
6455
6441
2 4 7 9 ii
50
6428
6414
6401
6388
6374
6361
6347
6334
6320
6307
247 9 ii
51
6293
6280
6266
6252
6239
6225
6211
6198
6184
6170
257 9 ii
52
6157
6i43
6129
6115
6101
6088
6074
6060
6046
6032
2 57 9 12
53
6018
6004
5990
5976
5962
5948
5934
5920
5906
5892
2 5 7 9 12
54
53/8
5864
5850
5835
5821
5807
5793
5779
5764
5750
2 5 7 9 12
55
5736
572i
5707
5693
5678
5664
5650
5635
5621
5606
2 5 7 10 12
56
5592
5577
5563
5548
5534
5519
555
5490
5476
546i
2 5 7 10 12
57
5446
5432
5417
5402
5388
5373
5358
5344
5329
53M
2 5 7 10 12
58
5 2 99
5284
5270
5255
5240
5225
5210
5195
5180
5165
2 5 7 10 12
59
5150
5135
5120
5105
5090
5075
5060
5045
5030
5015
3 5 8 10 13
60
5000
4985
4970
4955
4939
4924
4909
4894
4879
4863
3 5 8 10 13
61
4848
4833
4818
4802
4787
4772
4756
4741
4726
4710
3 5 8 10 13
62
4695
4679
4664
4648
4633
4617
4602
4586
457i
4555
3 5 8 10 13
63
454
4524
4509
4493
4478
4462
4446
4431
44i5
4399
3 5 8 10 13
64
4384
4368
4352
4337
4321
4305
4289
4274
4258
4242
3 5 8 ii 13
65
4226
4210
4i95
4179
4163
4147
4i3i
4"5
4099
4083
3 5 8 ii 13
66
4067
4051
435
4019
4003
3987
3971
3955
3939
3923
3 5 8 ii 14
67
3907
3891
3875
3859
3843
3827
3811
3795
3778
3762
3 5 8 ii 14
68
3746
3730
3714
3697
3681
3665
3649
3633
3616
3600
3 5 3 ii 14
69
3584
3567
3551
3535
35i8
3502
3486
3469
3453
3437
3 5 8 ii 14
70
3420
3404
3387
3371
3355
3338
3322
3305
3289
3272
3 5 8 ii 14
71
3256
3239
3223
3206
3190
3173
3156
3MO
3123
3107
3 6 8 ii 14
72
3090
374
3057
3040
3024
3007
2990
2974
2957
294
3 6 8 ii 14
73
2924
2907
2890
2874
2857
2840
2823
2807
2790
2773
3 6 8 ii 14
74
2756
2740
2723
2706
2689
2672
2656
2639
2622
2605
3 6 8 ii 14
75
2588
257 1
2554
2538
2521
2504
2487
2470
2453
2436
3 6 8 ii 14
76
2419
2402
2385
2368
2351
2334
2317
2300
2284
2267
3 6 8 ii 14
77
2250
2233
2215
2198
2181
2164
2147
2130
2113
2096
3 6 9 ii 14
78
2079
2062
2045
2028
2OII
1994
1977
1959
1942
1925
3 6 9 ii 14
79
1908
1891
1874
1857
1840
1822
1805
1788
1771
1754
3 6 9 ii 14
80
1736
1719
1702
1685
1668
1650
1633
1616
1599
1582
3 6 9 12 14
81
1564
1547
1530
1513
1495
1478
1461
1444
1426
1409
3 6 9 12 14
82
1392
1374
1357
134
1323
I35
1288
1271
1253
1236
3 6 9 12 14
83
1219
I2OI
1184
1167
1149
1132
i"5
1097
1080
1063
3 6 9 12 14
84
1045
1028
IOII
0993
0976
0958
0941
0924
0906
0889
3 6 9 12 14
85
0872
0854
0837
0819
O8O2
0785
0767
0750
0732
0715
3 6 9 12 14
86
0698
0680
0663
0645
0628
0610
0593
0576
0558
054 i
3 6 9 12 15
87
0523
0506
0488
0471
0454
0436
0419
0401
0384
0366
3 6 9 12 15
88
0349
0332
0314
0297
0279
0262
0244
0227
0209
0192
3 6 9 12 15
89
0175
0157
0140
OI22
0105
0087
0070
0052
0035
0017
3 6 9 12 15
90
oooo
502
MATHEMATICAL TABLES
TABLE VII. NATURAL TANGENTS.
Q
0'
00
6'
01
12'
02
18'
03
24'
04
30'
05
36'
06
42'
07
48'
08
54'
09
Mean Differences.
1' 2' 3' 4' 5'
oooo
0017
0035
0052
0070
0087
0105
OI22
0140
oi57
3 6 9 12 15
1
0175
0192
0209
0227
0244
0262
0279
0297
0314
0332
3 6 9 12 15
2
0349
0367
0384
0402
0419
437
454
0472
0489
0507
3 6 9 12 15
3
0524
54 2
559
577
0594
0612
0629
0647
0664
0682
3 6 9 12 15
4
0699
0717
734
0752
0769
0787
0805
0822
0840
0857
3 6 9 12 15
5
0875
0892
0910
0928
0945
0963
0981
0998
1016
1033
3 6 9 12 15
6
1051
1069
1086
1104
1122
"39
"57
"75
1192
1210
3 6 9 12 15
7
1228
1246
1263
1281
1299
1317
1334
1352
137
1388
3 6 9 12 15
8
1405
1423
1441
1459
T 477
1495
1512
1530
1548
1566
3 6 9 12 15
9
1584
1602
1620
1638
1655
1673
1691
1709
1727
1745
3 6 9 12 15
10
1763
1781
1799
18*17
1835
1853
1871
1890
1908
1926
3 6 9 12 15
11
1944
1962
1980
1998
2016
2035
2053
2071
2089
2IO7
3 6 9 12 15
12
2126
2144
2162
2180
2199
2217
2235
2254
2272
2290
3 6 9 12 15
13
2309
2327
2345
2364
2382
2401
2419
2438
2456
2475
3 6 9 12 15
14
2493
2512
2530
2549
2568
2586
2605
2623
2642
266l
3 6 9 12 16
15
2679
2698
2717
2736
2754
2 773
2792
2811
2830
2849
3 6 9 13 16
16
2867
2886
2905
2924
2943
2962
2981
3000
3019
3038
3 6 9 13 16
17
3057
3076
3096
3"5
3134
3153
3172
3I9T
3211
3230
3 6 10 13 16
18
3249
3269
3288
3307
3327
334 6
3365
3385
3404
3424
3 6 10 13 16
19
3443
3463
3482
3502
3522
3541
356i
358i
3600
3620
3 7 io 13 16
20
3640
3659
3679
3699
37*9
3739
3759
3779
3799
3819
3 7 io 13 17
21
3839
3859
3879
3899
3919
3939
3959
3979
4000
4O2O
3 7 io 13 17
22
4040
4061
4081
4101
4122
4142
4163
4183
4204
4224
3 7 io 14 17
23
4245
4265
4286
4307
4327
4348
4369
4390
44"
4431
3 7 i 14 17
24
4452
4473
4494
4515
4536
4557
4578
4599
4621
4642
4 7" 14 18
25
4663
4684
4706
4727
4748
4770
4791
4813
4834
4856
4 7 ii 14 18
26
H877
4899
4921
4942
4964
4986
5008
5029
5051
5073
4 7 " !5 18
27
5095
5"7
5139
5161
5184
5206
5228
5250
5272
5295
4 7 n, 15 18
28
5317
534
5362
5384
5407
5430
5452
5475
5498
5520
4 8 ii 15 19
29
5543
5566
5589
5612
5635
5658
5681
574
5727
5750
4 8 12 15 19
30
5774
5797
5820
5844
586 7
5890
59M
5938
596i
5985
4 8 12 16 20
31
6009
6032
6056
6080
6104
6128
6152
6176
6200
6224
4 8 12 16 20
32
6249
6273
6297
6322
6346
6371
6395
6420
6445
6469
4 8 12 16 20
33
6494
6519
6 544
6569
6594
6619
6644
6669
6694
6720
4 8 13 17 21
34
6 745
6771
6796
6822
6847
6873
6899
6924
6950
6976
4 9 13 17 21
35
7002
7028
7054
7080
7107
7133
7 r 59
7186
7212
7239
4 9 13 18 22
36
7265
7292
7319
7346
7373
7400
7427
7454
7481
75 08
5 9 14 18 23
37
7536
7563
7590
7618
7646
7673
7701
7729
7757
7785
5 9 14 18 23
38
7813
7841
7869
7898
7926
7954
7983
8012
8040
8069
5 9 14 19 24
39
8098
8127
8156
8185
8214
8243
8273
8302
8332
8361
5 io 15 20 24
40
8391
8421
8451
8481
8511
8541
8571
8601
8632
8662
5 io 15 20 25
41
8693
8724
8754
8785
8816
8847
8878
8910
8941
8972
5 io 16 21 26
42
9004
9036
9067
9099
9131
9163
9195
9228
9260
9293
5 ii 16 21 27
43
9325
9358
939i
9424
9457
9490
9523
9556
9590
9623
6 II 17 22 28
44
9657
9691
9725
9759
9793
9827
9861
9896
993
9965
6 ii 17 23 29
45
I OOOO
0035
0070
0105
0141
0176
O2I2
0247
0283
0319
6 12 18 24 30
MATHEMATICAL TABLES
503
TABLE VII. (contd.)
g
9
Q
0'
00
6'
01
12'
02
18'
03
24'
04
30'
05
36'
06
42'
07
48'
08
54'
09
Mean Differences.
1' 2' 3' 4' 5'
45
oooo
0035
0070
0105
0141
0176
0212
0247
0283
0319
6 12 18 24 30
46
0355
0392
0428
0464
0501
0538
0575
0612
0649
0686
6 12 18 25 31
47
0724
0761
0799
0837
0875
0913
0951
0990
1028
1067
6 13 19 25 32
48
1106
"45
1184
1224
1263
1303
1343
1383
1423
1463
7 13 20 27 33
49
1504
1544
1585
1626
1667
1708
1750
1792
1833
1875
7 14 21 28 34
50
1918
1960
2OO2
2045
2088
2131
2174
2218
2261
2305
7 14 22 29 36
51
2349
2393
2437
2482
2527
2572
2617
2662
2708
2753
8 15 23 30 38
52
2799
2846
2892
2938
2985
3032
3079
3127
3i75
3222
8 16 24 31 39
53
3270
33i9
3367
34i6
3465
35M
3564
3613
3663
3713
8 16 25 33 41
54
3764
3814
3865
39i6
3968
4019
4071
4124
4176
4229
9 17 26 34 43
55
4281
4335
4388
4442
4496
4550
4605
4659
47i5
4770
9 18 27 36 45
56
4826
4882
4938
4994
5051
5108
5166
5224
5282
5340
10 19 29 38 48
57
5399
5458
5517
5577
5637
5697
5757
5818
5880
594i
10 20 30 40 50
58
6003
6066
6128
6191
6255
6319
6383
6447
6512
6577
ii 21 32 43 53
59
16643
6709
6775
6842
6909
6977
745
7"3
7182
7251
ii 23 34 45 56
60
17321
7391
7461
7532
7603
7675
7747
7820
7893
7966
12 24 36 48 60
61
18040
8115
8190
8265
8341
8418
8495
8572
8650
8728
13 26 38 51 64
62
18807
8887
8967
9047
9128
9210
9292
9375
9458
9542
14 2 7 41 55 68
63
19626
9711
9797
9883
997
0057
oi45
0233
0323
0413
15 29 44 58 73
64
20503
0594
0686
0778
0872
0965
1060
"55
1251
1348
16 31 47 63 78
65
21445
1543
1642
1742
1842
1943
2045
2148
2251
2355
17 34 51 68 85
66
22460
2566
2673
2781
2889
2998
3109
3220
3332
3445
18 37 55 73 92
67
23559
3673
3789
3906
4023
4142
4262
4383
4504
4627
20 40 60 79 99
68
2475I
4876
5002
5129
5257
5386
5517
5649
5782
59i6
22 43 65 87 108
69
26051
6187
6325
6464
6605
6746
6889
7034
7179
2326
24 47 7 1 95"9
70
27475
7625
7776
7929
8083
8239
8397
8556
8716
878
26 52 78 104 131
71
29042
9208
9375
9544
97H
9887
0061
0237
0415
0595
29 58 87116145
72
30777
0961
1146
1334
1524
1716
1910
2106
2305
2506
32 64 96 129 161
73
32709
2914
3122
3332
3544
3759
3977
4197
4420
4646
36 72 108 144 180
74
75
34 8 74
3732I
5io5
7583
5339
7848
5576
8118
5816
8391
6059
8667
6305
8947
6554
9232
6806
9520
7062
9812
41 81 122 163 204
46 93 139 186 232
76
40108
0408
0713
1022
1335
1653
1976
2303
2635
2972
77
4'33i5
3662
4 OI 5
4374
4737
5107
5483
5864
6252
6646
78
47046
7453
7867
8288
8716
9152
9594
0045
6504
0970
79
5 I 44 6
1929
2422
2924
3435
3955
4486
5026
5578
6140
80
567I3
7297
7894
8502
9124
9758
0405
To66
1742
2432
81
63138
3859
4596
535
6122
6912
7720
8548
9395
0264
Mean differences are
82
7II54
2066
3002
3962
4947
5958
6996
8062
9158
6285
no longer suffici
ently accurate,
83
81443
2636
3863
5126
6427
7769
9152
0579
2052
3572
since the ilitler
84
85
95M
n43
9677
n66
9845
1191
1002
1216
IO2O
1243
1039
1271
1058
1300
1078
133
1099
1362
II2O
I395
ences vary cou
siderably along
each line.
86
1430
1467
1506
I546
1589
1635
1683
I734
1789
1846
87
1908
1974
2045
2120
2202
2290
2386
2490
2603
2727
88
2864
3014
3182
3369
3580
3819
4092
447
4774
5208
89
57 2 9
6366
7162
8185
9549
1146
1432
1910
2865
5730
90
oo
504
MATHEMATICAL TABLES
TABLE VIII. LOGARITHMIC SINES.
V
0'
6'
12'
18'
24'
30'
36'
42'
48'
54'
Mean Differences.
8>
6
00
01
02
03
04
05
06
07
08
09
1' 2' 3' 4' 5'
oo
32419
5429
7190
8439
9408
O2OO
0870
1450
1961
1
22419
2832
3210
3558
3880
4i79
4459
4723
497i
5206
2
5428
5640
5842
6035
6220
6397
6567
6731
6889
7041
3
7188
7330
7468
7602
773i
7857
7979
8098
8213
8326
4
8436
8543
8647
8749
8849
8946
9042
9i35
9226
93i5
16 32 48 64 So
5
9403
9489
9573
9655
9736
9816
9894
9970
0046
OI2O
13 26 39 52 65
6
10192
0264
0334
0403
0472
0539
0605
0670
0734
0797
ii 22 33 44 55
7
0859
0920
0981
1040
1099
"57
1214
1271
1326
I 3 8l
10 19 29 38 48
8
1436
1489
1542
1594
1646
1697
1747
1797
1847
1895
8 17 25 34 42
9
1943
1991
2038
2085
2131
2176
2221
2266
2310
2353
8 15 23 30 38
10
2397
2439
2482
2524
2565
2606
26 47
2687
2727
2767
7 14 20 27 34
11
2806
2845
2883
2921
2959
2997
3034
3070
3107
3M3
6 12 19 25 31
12
3179
3214
3250
3284
33i9
3353
3387
3421
3455
3488
6 ii 17 23 28
13
3521
3554
3586
3618
3650
3682
3713
3745
3775
3806
5 ii 16 21 26
14
3837
3867
3897
3927
3957
3986
4 OI 5
4044
4073
4IO2
5 10 15 20 24
15
4130
'4158
4186
4214
4242
4269
4296
4323
4350
4377
5 9 14 18 23
16
4403
4430
4456
4482
4508
4533
4559
4584
4609
4 6 34
4 9 13 17 21
17
4659
4684
4709
4733
4757
4781
4805
4829
4853
4876
4 8 12 16 20
18
4900
4923
4946
4969
4992
5015
5<>37
5060
5082
5104
4 8 ii 15 19
19
5126
5M8
5170
5192
5213
5235
5256
5278
5299
5320
4 7 ii 14 18
20
5341
536i
5382
542
5423
5443
5463
5484
5504
5523
3 7 10 14 i?
21
5543
5563
5583
5602
5621
5641
5660
5679
5698
5717
3 6 10 13 16
22
5736
5754
5773
5792
5810
5828
5847
5865
5883
59oi
3 6 9 12 15
23
5919
5937
5954
5972
5990
6007
6024
6042
6059
6076
3 6 9 12 15
24
6093
6110
6127
6144
6161
6177
6194
6210
6227
6243
3 6 8 ii 14
25
6259
6276
6292
6308
6324
6340
6356
6371
6387
6403
3 5 8 ii 13
26
6418
6 434
6449
6465
6480
6495
6510
6526
6541
6556
3 5 8 10 13
27
6570
6585
6600
6615
6629
6644
6659
6673
6687
6702
2 5 7 10 12
28
6716
6730
6744
6759
6 773
6787
6801
6814
6828
6842
2 5 7 9 12
29
6856
6869
6883
6896
6910
6923
6937
6950
6963
6977
2 4 7 9 ii
30
6990
7003
7016
7029
7042
755
7068
7080
793
7106
2 4 6 9 ii
31
7118
7131
7M4
7156
7168
7181
7 J 93
7205
7218
7230
2 4 6 8 10
32
7242
7254
7266
7278
7290
7302
73H
7326
7338
7349
2 4 6 8 10
33
7361
7373
7384
7396
7407
74i9
7430
7442
7453
7464
2 4 6 8 10
34
7476
7487
7498
7509
7520
7531
7542
7553
7564
7575
24679
35
7586
.7597
7607
7618
7629
7640
7650
7661
7671
7682
24579
36
7692
7703
7713
7723
7734
7744
7754
7764
7774
7785
23579
37
7795
7805
78i5
7825
7835
7844
7854
7864
7874
7884
23578
38
7893
7903
7913
7922
7932
794i
7951
7960
7970
7979
23568
29
798^
7998
8007
8017
8026
8035
8044
8053
8063
8072
23568
40
8081
8090
8099
8108
8117
8125
8i34
8i43
8152
8161
13467
41
8169
8178
8187
8195
8204
8213
8221
8230
8238
8247
3467
42
8255
8264
8272
8280
8289
8297
8305
8313
8322
8330
34 6 7
43
8338
8346
8354
8362
8370
8378
8386
8394
8402
8410
3457
44
45
8418
8495
8426
8502
8433
8510
8441
8517
8449
8525
8457
8532
8464
8540
8472
8547
8480
8555
8487
8562
3456
2456
MATHEMATICAL TABLES
505
TABLE VIII. (contd.)
j
ff
0'
6'
12'
18'
24'
30'
36'
42'
48'
54'
Mean Differences.
Q
00
01
02
03
04
05
06
07
08
09
1' 2' 3' 4' 5'
45
46
47
48
49
18495
8569
8641
8711
8778
' OO 00 00 00 OO
IVJ V) ONOl Ol
OO HI 4> VI O
4>. 00 OOVJ W
8510
8584
8655
8724
8791
8517
8591
8662
8731
8797
OO 00 OO CO 00
OOv} O>Ol Ol
o oo o\\o to
4>. (X>\O OOOi
8532
8606
8676
8745
8810
OO 00 OOO 00
OOVI ON O>Ol
M Ol OO HI 4*
VJ HI OO OO O
8547
8620
8690
8758
8823
8555
8627
8697
8765
8830
8562
8634
8704
8771
8836
12456
12456
12356
12346
1 2 3 4 5
50
8843
8849
8855
8862
8868
8874
8880
8887
8893
8899
1 2 3 4 5
51
52
8905
8965
8971
8917
8977
8923
8983
8929
8989
8935
8995
8941
9000
8947
9006
8953
9012
8959
9018
1 2 3 4 5
1 2 3 4 5
53
9023
9O29
9035
9041
9046
9052
9057
9063
9069
9074
12345
54
9080
9085
9091
9096
9101
9107
9112
9118
9123
9128
1 2 3 4 5
55
9134
9139
9144
9149
9155
9160
9165
9170
9175
9181
1 2 3 3 4
56
9186
9191
9196
9201
9206
9211
9216
9221
9226
9231
1 2 3 3 4
57
9236
9241
9246
9251
9255
9260
9265
9270
9275
9279
12234
58
9284
9289
9294
9298
9303
9308
9312
9317
9322
9326
12234
59
"9331
9335
9340
9344
9349
9353
9358
9362
9367
937 1
11234
60
'9375
9380
9384
9388
9393
9397
9401
9406
9410
9414
11234
61
9418
9422
9427
9431
9435
9439
9443
9447
945i
9455
11233
62
'9459
94 6 3
9467
947 1
9475
9479
9483
9487
9491
9495
1 233
63
'9499
9503
9507
95X0
9514
9522
9525
9529
9533
1 233
64
'9537
9540
9544
9548
9551
9555
9558
9562
9566
9569
i 223
65
'9573
957 6
9580
9583
9587
9590
9594
9597
9601
9604
i 223
66
9607
9611
9614
9617
9621
9624
9627
9631
9634
9637
i 223
67
9640
9643
9647
9650
9653
9656
9659
9662
9666
9669
i 223
68
9672
9675
9678
9681
9684
9687
9690
9693
9696
9699
122
69
9702
9704
9707
9710
9713
9716
9719
9722
9724
9727
O 122
70
9730
9733
9735
9738
9741
9743
9746
9749
975i
9754
O 122
71
'9757
9759
9762
9764
9767
9770
9772
9775
9777
9780
O I I 2 2
72
9782
9785
9787
9789
9792
9794
9797
9799
9801
9804
I I 2 2
73
9806
9808
9811
9813
9815
9817
9820
9822
9824
9826
I I 2 2
74
9828
9831
9833
9835
9837
9839
9841
9843
9845
9847
I I I 2
75
9849
9851
9853
9855
9857
9859
9861
9863
9865
9867
I I I 2
76
9869
9871
9873
9875
9876
9878
9880
9882
9884
9885
I I I 2
77
9887
9889
9891
9892
9894
9896
9897
9899
9901
9902
O I I I I
78
9904
9906
9907
9909
9910
9912
9913
9915
9916
9918
01 II
79
9919
9921
9922
9924
9925
9927
9928
9929
9931
9932
O I
80
'9934
9935
9936
9937
9939
994
9941
9943
9944
9945
O O I
81
9946
9947
9949
995
9951
9952
9953
9954
9955
9956
O O I
82
9958
9959
9960
9961
9962
9963
9964
9965
9966
9967
O I
83
9968
9968
9969
997
9971
9972
9973
9974
9975
9975
OOO I
84
9976
9977
9978
9978
9979
9980
9981
9981
9982
9983
O I
85
9983
9984
9985
9985
9986
9987
9987
9988
9988
9989
O O
86
9989
9990
9990
9991
9991
9992
9992
9993
9993
9994
O O
87
'9994
9994
9995
9995
9996
9996
9996
9996
9997
9997
O O
88
'9997
9998
9998
9998
9998
9999
9999
9999
9999
9999
o o o o o
89
9999
9999
oooo
oooo
oooo
oooo
oooo
oooo
oooo
OOOO
o o o o o
90
O'OOOO
506
MATHEMATICAL TABLES
TABLE IX. LOGARITHMIC COSINES
V
0'
6'
12'
18'
24'
30'
36'
42'
48'
54'
Mean Differences.
&
Q
00
01
02
03
04
05
06
07
08
09
1' 2' 3' 4' 5'
5*OOOO
oooo
oooo
oooo
oooo
oooo
oooo
oooo
oooo
9999
O O
1
19999
9999
9999
9999
9999
9999
9998
9998
9998
9998
O O O
2
9997
9997
9997
9996
9996
9996
9996
9995
9995
9994
O O O O O
3
4
9994
9989
9994
9989
9993
9988
9993
9988
9992
9987
9992
9987
9991
9986
9991
9985
9990
9985
9990
9984
O O O O O
00000
5
9983
9983
9982
9981
9981
9980
9979
997 s
9978
9977
O O O I
6
7
9976
9968
9975
9967
9975
9966
9974
9965
9973
9964
9972
9963
9971
9962
997
996i
9969
9960
9968
9959
O O I I
001 I
8
9953
9956
9955
9954
9953
9952
9951
9950
9949
9947
001 I
9
9946
9945
9944
9943
9941
994
9939
9937
9936
9935
001 I
10
9934
9932
993 1
9929
9928
9927
9925
9924
9922
9921
001 I
11
9919
9918
9916
9915
9913
9912
9910
9909
9907
9906
Oil I
12
9904
9902
9901
9899
9897
9896
9894
9892
9891
9889
Oil I
13
9887
9885
9884
9882
9880
9878
9876
9875
9873
9871
Oil 2
14
9869
9867
9865
9863
9861
9859
9857
9855
9853
9851
Oil 2
15
9849
9847
9845
9843
9841
9839
9837
9835
9833
9831
Oil 2
16
9828
9826
9824
9822
9820
9817
9815
9813
9811
9808
O I I 2 2
17
9806
9804
9801
9799
9797
9794
9792
9789
9787
9785
O I I 2 2
18
9782
9780
9777
9775
9772
9770
9767
9764
9762
9759
O I I 2 2
19
9757
9754
975i
9749
9746
9743
9741
9738
9735
9733
O I I 2 2
20
9730
9727
9724
9722
9719
9716
97 J 3
9710
9707
9704
O I I 2 2
21
9702
9699
9696
9693
9690
9687
9684
9681
9678
9675
O I I 2 2
22
9672
9669
9666
9662
9659
9656
9653
9650
9647
9643
II223
23
9640
9637
9634
9631
9627
9624
9621
9617
9614
9611
II223
24
9607
9604
9601
9597
9594
9590
9587
9583
958o
957 6
II223
25
9573
9569
9566
9562
9558
9555
9551
9548
9544
9540
II223
26
9537
9533
9529
9525
9522
95i8
9514
95io
9507
9503
II233
27
9499
9495
9491
9487
9483
9479
9475
9471
9467
9463
II233
28
9459
9455
945i
9447
9443
9439
9435
9431
9427
9422
II233
29
9418
9414
9410
9406
9401
9397
9393
9388
9384
9380
II234
30
9375
9371
9367
9362
9358
9353
9349
9344
934
9335
II234
31
9331
9326
9322
9317
9312
9308
9303
9298
9294
9289
12234
32
9284
9279
9275
9270
9265
9260
9255
9251
9246
9241
12234
33
9236
9231
9226
9221
9216
9211
9206
9201
9196
9191
12334
34
9186
9181
9175
9170
9165
9160
9155
9149
9144
9i39
12334
35
9134
9128
9123
9118
9112
9107
9101
9096
9091
9085
12345
36
9080
9074
9069
9063
9057
9052
9046
9041
9035
9029
12345
37
9023
9018
9012
9006
9000
8995
8989
8983
8977
8971
12345
38
8965
8959
8953
8947
8941
8935
8929
8923
8917
8911
12345
39
890"
8899
8893
8887
8880
8874
8868
8862
8855
8849
12345
40
8843
8836
8830
8823
8817
8810
8804
8797
8791
8784
12345
41
8778
8771
8765
8758
8751
8745
8738
8731
8724
8718
12356
42
8711
8704
8697
8690
8683
8676
8669
8662
8655
8648
12356
43
8641
8634
8627
8620
8613
8606
8598
8591
8584
8577
12456
44
8569
8562
8555
8547
8540
8532
8525
8517
8510
8502
12456
45
8495
8487
8480
8472
8464
8457
8449
8441
8433
8426
I345 6
MATHEMATICAL TABLES
507
TABLE IX (contd.)
1
!
0'
00
6'
01
12'
02
18'
03
24'
04
30'
05
36'
06
42'
07
48'
08
54'
09
Mean Differences.
1' 2' 3' 4' 5'
45
^8495
8487
8480 8472 8464
8457
8449 ' 8441
8433
8426
1 3 4 5 6
46
8418
8410
8402
8394 9386
8378
8370 8362
8354
8346
13457
47
8338
8330
8322
8313 8305
8297
8289 8280
8272
8264
1 3 4 6 7
48
8255
8247
8238
8230 8221
8213
8204 8195
8187
8178
** i /
13467
49
8169
8161
8152
8M3
8134
8125
8117
8108
8099
8090
13467
50
8081
8072
8063
8053
8044
8035
8026
8017
8007
7998
23568
51
7989
7979
7970
7960
795i
794i
7932
7922
79i3
7903
23568
52
7893
7884
7874
7864
7854
7844
7835
7825
7815
7805
23578
53
7795
7785
7774
7764
7754
7744
7734
7723
77 J 3
773
23579
54
7692
7682
7671
7661
7650
7640
7629
7618
7607
7597
24579
55
7586
7575
7564
7553
7542
7531
7520
7509
7498
7487
24679
56
7476
7464
7453
7442
7430
7419
7407
7396
7384
7373
246 8 10
57
736i
7349
7338
7326
7314
7302
7290
7278
7266
7254
246 8 10
58
7242
7230
7218
7205
7193
7181
7168
7*56
7M4
7*31
246 8 10
59
7118
7106
7093
7080
7068
7055
7042
7029
7016
7003
246 9 II
60
6990
6977
6963
6950
6937
6923
6910
6896
6883
6869
2 4 7 9 ii
61
6856
6842
6828
6814
6801
6787
6773
6759
6744
6730
2 5 7 9 12
62
6716
6702
6687
6673
6659
6644
6629
6615
6600
6585
2 5 7 10 21
63
6570
6556
6541
6526
6510
6495
6480
6465
6449
6434
3 5 8 10 13
64
6418
6403
6387
6371
6356
6340
6324
6308
6292
6276
3 5 8 ii 13
65
6259
6243
6227
6210
6194
6177
6161
6144
6127
6110
3 6 8 ii 14
66
6093
6076
6059
6042
6024
6007
5990
5972
5954
5937
3 6 9 12 15
67
5919
5901
5883
5865
5847
5828
5810
5792
5773
5754
3 6 9 12 15
68
5736
57 J 7
5698
5679
5660 5641
5621
5602
5583
5563
3 6 10 13 16
69
5543
5523
5504
5484
5463 5443
5423
5402
5382
53^1
3 7 i H 17
70
534i
5320
5299
5278
5256
5235
5213
5192
5170
5M8
4 7 ii 14 18
71
5162
5104
5082
5060
5037
5 OI 5
4992
4969
4946
4923
4 8 ii 15 19
72
4900
4876
4853
4829
4805 4781
4757
4733
4709
4684
4 8 12 16 20
73
4 6 59
4 6 34
4609
4584
4559 4533
4508
4482
4456
443
4 9 13 17 21
74
'443
4377
4350
4323
4296 4269
4242
4214
4186
4158
5 9 14 18 23
75
4130
4102
4073
4044
4015
3986
3957
3927
3897
3867
5 10 15 20 24
76
3837
3806
3775
3745
37*3
3682
3650
3618
3586
3554
5 ii 16 21 26
77
3521
3488
3455
3421
3387 3353
3319
3284
3250
3214
6 ii 17 23 28
78
3179
3143
3107
3070
3034 2997
2959
2921
2883
2845
6 12 19 25 31
79
2806
2767
2727
2687
2647 2606
2565
2524
2482
2439
7 14 20 27 34
80
2397
2353
2310
2266
2221
2176
2131
2085
2038
1991
8 15 23 30 38
81
1943
1895
1847
1797
J 747
1697
1646
1594
1542
1489
8 17 25 34 42
82
1436
1381
1326
1271
1214
"57
1099
1040
0981
0920
10 19 29 38 48
83
0859
0797
0734
0670
0605
0539
0472
0403
0334
0264
ii 22 33 44 55
84
0192
OI2O
0046
9970
9894
9816
9736
9655
9573
^489
13 26 39 52 65
85
29403
9315
9226
9135
9042
8946
8849
8749
8647
8543
16 32 48 64 80
86
8436
8326
8213
8098
7979
7857
773i
7602
7468
7330
87
7188
7041
6889
6731
6567
6397
6220
6035
5842
5640
88
5428
52O6
497 1
4723
4459
4179
3880
3558
3210
2832
89
2419
I96l
M50
0870
O2OO
9408
8439
7190
5429
2419
90
00
5 o8 MATHEMATICAL TABLES
TABLE X. LOGARITHMIC TANGENTS.

E
be
Q
0'
00
6'
01
12'
02
18'
03
24'
04
30'
05
36'
06
42'
07
48'
08
54'
09
Mean Differences.
1' 2' 3' 4' 5'
00
32419
5429
7190
8439
9409
02 oo
0870
M50
1962
1
22419
2833
3211
3559
3881
4181
4461
4725
4973
5208
2
543 1
5643
5845
6038
6223
6401
6571
6736
6894
7046
3
7194
7337
7475
7609
7739
7865
7988
8107
8223
8336
4
8446
8554
8659
8762
8862
8960
9056
9150
9241
933i
16 32 48 64 81
5
9420
9506
9591
9674
9756
9836
9915
9992
0068
oi43
13 26 40 53 66
6
I02I6
0289
0360
0430
0499
0567
0633
0699
0764
0828
II 22 34 45 56
7
0891
0954
1015
1076
"35
1194
1252
1310
1367
1423
10 20 29 39 49
8
1478
1533
1587
1640
1693
1745
1797
1848
1898
1948
9 17 26 35 43
9
1997
2046
2094
2142
2189
2236
2282
2328
2374
2419
8 16 23 31 39
10
2463
2507
2551
2594
2637
2680
2722
2764
2805
2846
7 14 21 28 35
11
2887
2927
2967
3006
3046
3085
3123
3162
3200
3237
6 13 19 26 32
12
3275
3312
3349
3385
3422
3458
3493
3529
3564
3599
6 12 18 24 30
13
3634
3668
3702
3736
3770
3804
3837
3870
3903
3935
6 II 17 22 28
14
3968
4000
4032
4064
4095
4127
4158
4189
4220
425
5 10 16 21 26
15
4281
43ii
4341
43?i
4400
443
4459
4488
4517
4546
5 10 15 20 25
16
4575
4603
4632
4660
4688
4716
4744
477i
4799
4826
5 9 14 19 23
17
4853
4880
4907
4934
4961
4987
5014
5040
5066
5092
4 9 13 18 22
18
5118
5143
5169
5195
5220
5245
5270
5295
5320
5345
4 8 13 17 21
19
537
5394
5419
5443
5467
549i
55i6
5539
5563
5587
4 8 12 16 20
20
5611
5634
5658
5681
574
5727
5750
5773
5796
5819
4 8 12 15 19
21
5842
5864
5887
5909
5932
5954
5976
5998
6020
6042
4 7 " 15 19
22
6064
6086
6108
6129
6151
6172
6194
6215
6236
6257
4 7 ii 14 18
23
6279
6300
6321
6341
6362
6383
6404
6424
6445
6465
3 7 10 14 17
24
6486
6506
6527
6547
6567
6587
6607
6627
6647
6667
3 7 10 13 17
25
6687
6706
6726
6746
6765
6785
6804
6824
6843
6863
3 7 10 13 16
26
6882
6901
6920
6939
6958
6977
6996
7015
7034
753
3 6 9 13 16
27
7072
7090
7109
7128
7146
7165
7183
7202
7220
7238
3 6 9 12 15
28
7257
7275
7293
73ii
7330
7348
7366
7384
7402
7420
3 6 9 12 15
29
7438
7455
7473
7491
7509
7526
7544
7562
7579
7597
3 6 9 12 15
30
7614
7632
7649
7667
7684
7701
7719
7736
7753
7771
3 6 9 12 14
31
7788
7805
7822
7839
7856
7873
7890
7907
7924
7941
3 6 9 ii 14
32
7958
7975
7992
8008
8025
8042
8059
8075
8092
8109
3 6 8 ii 14
33
8125
8142
8158
8i75
8191
8208
8224
8241
8257
8274
3 5 8 ii 14
34
8290
8306
8323
8339
8355
8371
8388
8404
8420
8436
3 5 8 ii 14
35
8452
8468
8484
8501
8517
8533
8549
8565
8581
8597
3 5 8 ii 13
36
8613
8629
8644
8660
8676
8692
8708
8724
8740
8755
3 5 8 ii 13
37
8771
8787
8803
8818
8834
8850
8865
8881
8897
8912
3 5 8 10 13
38
8928
8944
8959
8975
8990
9006
9022
9037
9053
9068
3 5 8 10 13
39
9084
9099
9H5
9130
9146
9161
9176
9192
9207
9223
3 5 8 10 13
40
9238
9254
9269
9284
9300
9315
9330
9346
936i
9376
3 5 8 10 13
41
9392
9407
9422
9438
9453
9468
9483
9499
95H
9529
3 5 8 10 13
42
9544
9560
9575
9590
9605
9621
9636
9651
9666
9681
3 5 8 10 13
43
9697
9712
9727
9742
9757
9773
9788
9803
9818
9833
3 5 8 10 13
44
9848
9864
9879
9894
9909
9924
9939
9955
9970
9985
3 5 8 10 13
45
O'OOOO
0015
0030
0045
0061
0076
0091
0106
0121
0136
3 5 8 10 13
MATHEMATICAL TABLES
509
TABLE X. (contd.)
V
b
of
Q
0'
00
6'
01
12'
02
18'
03
24'
04
30'
05
36'
06
42'
07
48'
08
54'
09
Mean Differences.
1' 2' 3' 4' 5'
45
oooo
0015
0030 0045
0061
0076
0091
0106
OI2I
0136
35 8 10 13
46 0152
0167
0182
0197
O2I2
0228
0243
0258
0273
0288
35 8 10 13
47 0303
0319
0334
349
0364
0379
0395
0410
0425
0440
35 8 10 13
48 0456
0471
0486
0501
0517
0532
547
0562
0578
0593
35 8 10 13
49 0608
0624
0639
0654
0670
0685
0700
0716
0731
0746
35 8 10 13
50
0762
0777
793
0808
0824
0839
0854
0870
0885
0901
35 8 10 13
51
0916
0932
0947
0963
0978
0994
IOIO
1025
1041
1056
35 8 10 13
52 1072
1088
1103
1119
"35
1150
1166
1182
1197
1213
35 8 10 13
53 1229
1245
1260
1276
1292
1308
1324
1340
1356
1371
35 8 ii 13
54 1387
1403
1419
1435
i45i
1467
1483
1490
1516
1532
35 8 ii 13
55 1548
1564
1580
1596
1612
1629
1645
1661
1677
1694
35 8 ii 14
56 1710
1726
1743
1759
1776
1792
1809
1825
1842
1858
35 8 ii 14
57 1875
1891
1908
1925
1941
1958
1975
1992
2OO8
2025
36 8 ii 14
58 2042
2059
2076
2093
2IIO
2127
2144
2161
2178
2195
36 9 ii 14
59 2212
2229
2247
2264
228l
2299
2316
2333
2351
2368
3 6 9 12 14
60 2386
2403
2421
2438
2456
2474
2491
2509
2527
2545
36 9 12 15
61 2562
2580
2598
2616
2634
2652
2670
2689
2707
2725
3 6 9 12 15
62 2743
2762
2780
2798
2817
2835
2854
2872
2891
2910
36 9 12 15
63 2928
2947
2966
2985
3004
3023
3042
3061
3080
3099
36 9 13 16
64 3"8
3i37
3i57
3176
3196
3215
3235
3254
3274
3294
3 6 10 13 16
65 33I3
3333
3353
3373
3393
3413
3433
3453
3473
3494
3 7 10 13 17
66 3514
3535
3555
357 6
3596
3617
3638
3659
3679
3700
3 7 10 14 17
67 3721
3743
3764
3785
3806
3828
3849
3871
3892
39M
4 7 ii 14 18
68 3936
3958
3980
4002
4024
4046
4068 4091
4ii3
4136
4 7 ii J 5 19
69 4158
4181
4204
4227
4250
4273
4296 4319
4342
4366
4 8 12 15 19
70 4389
4413
4437
4461
4484
4509
4533
4557
458i
4606
4 8 12 16 20
71
4630
4 6 55
4680
4705
4730
4755
4780
4805
4831
4857
4 8 13 17 21
72
4882
4908
4934
4960
4986
5013
5039
5066
5093
5120
4 9 13 18 22
73
5147
5174
5201
5229
5256
5284
5312
5340
5368
5397
5 9 14 19 23
74
5425
5454
5483
5512
5541
5570
5600
5629
5659
5689
5 10 15 20 25
75
5719
5750
5780
5811
5842
5873
5905
5936
5968
6000
5 10 16 21 26
76
6032
6065
6097
6130
6163
6196
6230
6264
6298
6332
6 ii 17 22 28
77
6366
6401
6436
6471
6507
6542
6578
6615
6651
6688
6 12 18 24 30
78
6725
6763
6800
6838
6877
6915
6954
6994
733
7073
6 13 19 26 32
79
7113
7 X 54
7*95
7236
7278
7320
7363
7406
7449
7493
7 14 21 28 35
80
7537
758i
7626
7672
7718
7764
7811
7858
7906
7954
8 16 23 31 39
81
8003
8052
8102
8152
8203
8255
8307
8360
8413
8467
9 17 26 35 43
82
8522
8577
8633
8690
8748
8806
8865
8924
8985
9046
10 20 29 39 49
83
9109
9172
9236
9301
9367
9433
95i
9570
9640
9711
ii 22 34 45 56
84
9784
9857
9932
0008
0085
0164
0244
0326
6409
0494
13 26 40 53 66
85
10580
0669
0759
0850
0944
1040
1138
1238
I34i
1446
16 32 48 64 81
86
II554
1664
1777
1893
2012
2135
2261
2391
2525
2663
87 12806
2954
3106
3264
3429
3599
3777
3962
4155
4357
88 14569
4792
5027
5275
5539
5819
6119
6441
6789
7167
89 17581
8038
8550
9130
9800
0591
1561
2810
4571
758i
90
00
5 io MATHEMATICAL TABLES
TABLE XI. EXPONENTIAL AND HYPERBOLIC FUNCTIONS
X
e*
.*
cosh x '
sinhx
tanhx
2
2
e*+e *
1
11052
9048
10050
1002
0997
2
12214
8187
I020I
2013
1974
3
13499
7408
10153
3045
2913
4
14918
6703
10811
4108
3799
5
16487
6065
11276
5211
4621
6
18221
5488
11855
6367
537
7
20138
4966
12552
7586
6044
8
22255
4493
13374
8881
6640
9
24596
4066
I433I
I0265
7163
10
27183
3679
I543I
II752
7616
11
30042
3329
16685
13357
8005
12
33201
3012
18107
15095
8337
13
36693
2725
19709
16984
8617
14
40552
2466
21509
19043
8854
15
44817
2231
23524
21293
9051
16
49530
2019
25775
23756
9217
17
54739
1827
28283
26456
9354
18
60496
1653
31075
29422
9468
19
66859
1496
34 J 77
32682
9563
20
73891
1353
37622
36269
9640
21
81662
1225
4M43
40219
9704
22
90251
1108
45679
4457I
9758
23
99742
1003
50372
4937
9801
24
110232
0907
5557
54662
9837
25
121825
0821
61323
60502
9866
26
134638
0743
67690
66947
9890
27
148797
0672
74735
74063
9910
28
164446
0608
82527
81919
9926
29
181741
0550
91146
90596
9940
30
200855
0498
10068
10018
31
221980
0450
11*122
11076
9959
32
245325
0408
12287
12246
9967
33
271126
0369
13575
I3538
9973
34
299641
334
14999
14965
9978
35
33II55
0302
16573
16543
9982
36
365982
0273
I83I3
18285
9985
37
404473
0247
20236
20211
9988
38
447012
0224
22362
22339
9990
39
.494024
O2O2
24711
24691
9992
40
545982
0183
27308
27290
9993
41
603403
Ol66
30178
30I62
9995
42
666863
0150
33351
33336
9996
43
736998
0136
36843
9996
44
814509
0123
40732
40719
9997
45
900171
OIII
45OI4
45003
9997
46
994843
OIOI
49747
49737
9998
47
1099472
0091
9998
48
1215104
0082
60759
60751
9999
49
50
1342898
1484132
0074
0067
67149
74210
67141
74203
9999
9999
INDEX
A 2 B 2 , factors of, 52
A 3  B 3 , , 53
A 3 + B 3 , ,; , 53
Abbreviations, i
Abscissa, 159
Addition formulae in Trigonometry
273
Adfected quadratic, 60
Algebraic fractions, Addition of, 57
, Multiplication of, 56
Alignment chart with four variables,
443
charts, Choice of scales
for, 434
involving powers of the
variables, 440
, Principle of, 429
Allowance for depreciation, 211
" Ambiguous " case in the solution of
triangles, 260
Amplitude of sine functions, 361
Amsler planimeter, 300
Angle of elevation, 239
of regular polygon, 88
Angles of any magnitude, Ratios of,
251
Angular velocity, 363
Annulus, Area of, 93
Antilogarithms, 16
Approximation, by use of the
binomial theorem, 467
for the area of a circle, 92
for products and quotients, 6
for square roots, 8
for the volume of a cylinder, 1 1 1
Arc, Height of elliptic, 105
, Height of circular, 97
, Length of circular, 98
Area of annulus, 93
of circle, 90
of ellipse, 104
of fillet, 132
of indicator diagram, 87
of irregular polygon, 87
of irregular quadrilateral, 87
of parabolic segment. 106
of parallelogram, 84
of rectangle, 79
Area of regular polygon, 88
of rhombus, 84
of sector of circle, 101
of segment of circle, 101
of trapezoid, 85
of triangle, 79, 80, 267
Areas of irregular curved figures, by
averaging boundaries, 305
computation scale, 306
counting squares, 305
graphic integration, 312
midordinate rule, 308
planimeter, 300
Simpson's rule, 310
trapezoidal rule, 307
Asymptotes of hyperbola, 108. 349
Bearing, Reduced, 244
, Wholecircle, 245
Binomial theorem, 463
Boussinesq's rule for the perimeter of
an ellipse, 105
Calculation of coordinates in land
surveying, 244
Cardan's solution for cubic equation?,
67
Catenary, 217, 292, 357
Ccntroid, Definition of, 129
Centroids, Positions of, 130
Characteristic of a logarithm, 14
Charts, Alignment, 429
Correlation, 419
Intercept, 421
Circle, Arc of, 98
Area of, 90
Chord of, 97
Circumference of, 90
Sector of, 101
Segment of, 101
Coffin averager and planimeter, 303
Combinations, 460
Complement of an angle, 233
Complex quantities, 294
Compound interest, 208
periodic oscillations, 369
5*2
INDEX
Computing scale, 306
Cone, Frustum of, 117
, Surface area of, 116
, Volume of, 116
Constant heat lines, 387
volume lines, 384
Constants, Useful, 4
Construction of regular polygons, 88
Continued fractions, 448
Convergents of ir, 451
Coordinates, Calculation of, 244
, Plotting of, 159
Correlation charts, 419
Cosine rule for the solution of tri
angles, 256
Cubic equations, Solution of, 67
Curves of type y = ax n , 336
y = ae bt , 352
y = e~ ax sin (bx + c), 373
Cutting, Section of, 321
, Volume of, 324
Cylinder, Surface area of, in
, Volume of, in
D
Definitions, i
Depreciation allowance, 211, 343
Determinants, 474
Determination of laws, 396
Difference of two squares, Factorisa
tion of, 52
Dividing head problem, 449
E
Efficiency curves, Plotting of, 151
Ellipse, Area of, 104
, Equation of, 344
, Height of arc of, 105
of stress, 345
, Perimeter of, 105
Embankment, Section of, 321
, Volume of, 326
Equation of time, 370
of a straight line, 162
Equations, Cubic, 67, 181
, Graphic solution of, 376
, Quadratic, 61, 176
, Quadratic, with imaginary
roots, 67
, Simple, 31
, Simultaneous, 43, 46, 164
, Simultaneous quadratic, 70
, Surd, 74
, Trigonometric, 287
 to conic sections, 344
equilateral triangle, Area of, 82
Equivalent acute angle, 252
Ericsson engine diagrams, 394
Euclid, Propositions of, 4
Expansion curves for gases, 338
Exponential series, 470
Factor theorem, 55
Factorisation. Method of, 51
Factors, i
Fathom, 3
Fillet, Area of, 132
, Centroid of, 130
Formula for solution of cubic equa
tions, 67
of quadratic equations, 64
Fractions, Addition of algebraic, 57
, Continued, 448
, Multiplication of algebraic, 56
, Partial, 452
Frustum of cone, 117
Function, 2, 161
Graph of a sine function, 359
tangent function, 366
Graphic integration, 312
solution of equations, 376
of quadratic equations, 1 76
of simultaneous equations,
164
Graphs, Introduction to, 148
of quadratic expressions, 174
Guldinus, Rules of, 129
H
Homogeneous equations, 73
Hyperbola, Definition of, 108
, Equation of, 348
Hyperbolic functions, 290
Hypotenuse of rightangled triangle,
So
Imaginary quantities, 67
Independent variable, 161
Indices, 10
Intercept charts, 421
Inverse trigonometric functions, 297
j, Meaning of, 67
Joule engine diagrams, 393
Knot, 3
K
Latus rectum of parabola, 106
Laws of machines, 166
of type y = a + , 398
y = ax n . 401
y = ae*, 405
INDEX
513
Laws of type y = a + bx + ex*, 407
y = a + bx", 408
y = b(x + a)", 409
y = a f fee"*, 409
y = ax"z m , 410
L.C.M., Finding the, 51
Length of chord of a circle, 97
Limiting values, 455
Logarithm, Definition of, 12
Logarithmic decrement, 375
 equations, 224
series, 470
Logarithms, Napierian, 216
of trigonometric ratios, 247
, reading from tables, 13
Loglog scale on the slide rule, 337
If
Mantissa of logarithm, 14
Maximum and minimum values, 183
Mensuration, 79 et seq.
Midordinate rule, 308
N
Napierian logs. 13
, Calculation of, 216, 471
, reading from tables, 216
Parabola, Area of segment of, 106
, Definition of, 106
, Equation of, 347
: . Length of arc of, 106
Parabolic segment, Centroid of, 130
Parallelogram, Area of, 84, 268
Partial fractions, 452
Period of sine functions, 361
Permutations, 460
Planimeter, Use of the Amsler, 300
, Use of the Coffin, 303
Polygon, Area of irregular, 87
, Area of regular, 88
. Construction of regular, 88
Prism, Surface area of, no
, Volume of, no
Prismoidal solid, Volume of, 319
Products of IT, 94
Progression, Arithmetic, 201
, Geometric, 205
PV diagrams, 381 et seq.
Pyramid, Frustum of, 117
, Surface area of, 115
, Volume of, 115
Q
Quadrant of circle, Centroid of, 130
Quadratic equations, Solution of, by
completion of square, 61
Quadratic equations. Solution of.
by factorisation, 61
, , by graphs, 176
. , by use of formula, 63
, , on the drawingboard, 176
Quadratic expressions, Plotting of, 1 74
Quadrilateral, Area of irregular, 87
, Centroid of, 130
R
Radian, 99
Ratios of multiple and submultiple
angles, 279
, Trigonometric, 232
Rectangle, Area of, 79
Reduced bearing, 244
Remainder theorem, 55
Reservoir, Volume of, 332
Rhombus, area of, 85
Rightangled triangle, Relation be
tween sides of, 80
, Solution of, 239
" Roots " of a quadratic equation, 61
" s " rule for area of triangle, 80
Sector of circle. Area of, 101
Segment of circle, Area of, 101
Semicircular arc, Centroid of, 130
area, Centroid of, 130
perimeter, Centroid of, 130
Series, 200
, Exponential, 470
for calculation of logs, 471
, Logarithmic, 470
Similar figures, 122
Simple harmonic motion, 365
Simpson's rule, 310
Sine curves, Plotting of, 359 et seq.
rule for the solution of triangles,
256
Slide rule, Area of circle by, 92
, Loglog scale on, 337
. Reading of logs from, 17
, Reading of trigonometric
ratios from, 242
. Special markings on, 17
, Uses of, for plotting log
quantities, 403, 419
, , in solution of tri
angles, 261
, , Volume of cylinder
by, in
Solution of triangles, 255 et seq.
Sphere, Surface area of, 120
, Surface area of zone of, 120
, Volume of, 120
, Volume of segment of, 121
, Volume of zone of, 120
Square measure, 3
514
INDEX
Sterling engine, Diagrams for, 390
Subnormal of parabola, 106
Sum curve, 312
Surd equations, 75
Surds, Rationalisation of denomin
ators of, 74
Surface area, for cuttings and em
bankments, 331
of cone, 116
of cylinder, 1 1 1
of frustum, 117
of prism, no
of pyramid, 115
of sphere, 1 20
Surveyor's measure, 87
Table of areas and circumferences of
circles, 127
of areas and circumferences of
plane figures, 144, 145
of earthwork slopes, 319
of signs of trigonometric ratios,
253
of volumes and surface areas of
solids, 146, 147
of weights of earths, 319
of weights of metals, 132
Tables of weights and measures, 3
Terms, i
Transposition of a factor in an equa
tion, 33
of term in an equation, 32
T< diagrams, 381 et seq.
Trapezoid, Area of, 85
, Centroid of, 130
Trapezoidal rule for area of irregular
curved figure, 307
Triangle, Area of, 79, 267
, Lettering of, 80
, Rightangled, relation between
sides of, 80
Triangles, Solution of, 255 et seq.
Trigonometric equations, 287
ratios, 232
from slide rule, 242
from tables, 234
Turningpoints of curves, 183
U
Units, Investigation for, 26
Variation, 193
Vectors, 295
Velocity ratio of machine, 169
Volume of cone, 116
of cylinder, in
of frustum of cone or pyramid,
"7
of prism, no
of prismoidal solid, 319
of pyramid, 115
of reservoir, 332
of segment of sphere, 121
of sphere, 120
of wedgeshaped excavation, 321
of zone of sphere, 120
W
Wedgeshaped excavation, Volume
of, 321
Weights and measures, Table of. 3
, Calculation of, 132 et seq.
of earths, Table of. 319
of metals. Table of, 132
Whole circle bearing, 245
Zero circle of planimeter, 96, 302
Zone of sphere, Surface area of, 120
Volume of, 120
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