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MATHEMATICS FOR ENGINEERS 

PART I 



The Directly-Useful 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 CO-ORDINATES 
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 Directly-Useful 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 subject-matter 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 counter-clockwise direction. Plane 
Trigonometry has its widest application in land surveying, in 
which angles are measured by a right-handed 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 co-ordinates in land surveying is introduced as a good 
instance of the solution of right-angled 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 dividing-head 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 co-ordinates Representation of simple equations 
by straight-line 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 right-angled 
triangles Reading the values of the trigonometric ratios from the 
slide rule Calculation of co-ordinates in land surveying Meaning 
of the terms " reduced bearing " and " whole-circle " 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 mid-ordinate rule. 7. By Simpson's rule. 
8. By graphic integration. 

CHAPTER VIII 
CALCULATION OF EARTHWORK VOLUMES 319 

Volumes of prismoidal solids Volume of a wedge-shaped 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 
log-log 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 constant-volume 
lines. 3. Drawing secondary adiabatics. 4. Plotting the Rankine 
cycle for two drynesses. 5. Plotting the common steam-engine 
diagram for an engine jacketed and non-jacketed. 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# i-25y; 

4 
(40-76) (0 



Again, the use of the lo-inch 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 horse-power." 

B.H.P. ,, ,, " brake horse-power." 

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. ,, ,, " electro-motive 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 = 62-4 Ibs. 
i cu. ft. = 6-24 gallons. 

METRIC TABLE OF LENGTH 

I kilometre (Km.) = 1000 metres . 
I hectometre (Hm.) = 100 
I dekametre (Dm.) = 10 

metre (m.) = 39-37" 

i decimetre (dm.) = -i metre 
i centimetre (cm.) = -01 (2-54 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 = 3-14159 

log<io = 2-3026 Iog 10 e = -4343 

log M N = log.N X -4343 logeN = log 10 N X 2-303 

g = 32-18 ft. per sec. per sec. 

i horse-power = 33000 foot Ibs. per min. = 746 watts. 

Absolute temperature r = /C.+273 or tF.-\-^6i. 

i radian = 57-3 degrees. 

pressure of one atmosphere = 14-7 Ibs. per sq. in. 

i inch == 2-54 centimetres. i sq. in. = 6-45 sq. cms. 

i kilometre = -6213 mile. i kilogram = 2-205 Ibs. 

i Ib. = 453-6 grms. 

The following are the statements of the propositions in Euclid, 
to which reference is made in the text 

Euc. I. 47. In any right-angled 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 

47-83 x 3-142 X 9-41 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 sub-multiples 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 4-81 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 - 

28-4 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 28-4 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 3-142 x 9-41 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 one-third of the strokes in the 
point fraction. 

Example 4. Find an approximation for ^498-4. 

In place of 498-4 write 500, which can be written as 5 x io 2 , 

11 
or as 5 

Then the approximate square root is 2-2 

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 

s-s 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 \_ 

1-5x2 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 8-2 
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. 49-57 x -0243 2. -00517 x -1724 3. ? 

2 3 4 



4. 8-965 x 72-49 x '094 5. -1167 x -0004 x 98-1 x 2710 

4-176 X 25400 _ 

~ 87235 7 - 

11540 x -3276 x 3-142 x -0078 
- 



4-176 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 1614-17 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 -y-b 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 = 1-37. 

4 U 

7. Find the value of vnCv~ n i in terms of p when n =1-41 
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 = 2-718. 

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 2-718. 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 526-3, 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 four-figure tables, and are quite 
full enough for ordinary calculations, but for particularly accurate 
work, as, for example, in Surveying, five- and even seven-figure 
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 526-3 : 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 526-3. 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 526-3 is therefore 2-7212, or, in other words, 526-3 = 10 
raised to the power 2-7212; similarly the log of 52630 must be 
4-7212, because 52630 is the same proportion of a power of 10 
above 10,000 as 526-3 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 whole-number part, settled by the position of 
the decimal point in the number, and called the characteristic or 
distinguisher. The logs of 526-3 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 526-3, 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, 1-7212, 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. 

1-7212 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 1-7212 is adhered to throughout. 

From similar reasoning, log -005263 = 3-7212 (i) 

and log -00005263 = 5-7212 .... (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 3-1553 + a difference of 21 = 3-1574, 
while the log of 1487 is 3-1703 + a difference of 20 = 3-1723. 

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 

70-08 

0009 

500000 



Log. 



4-6102 
1-3728 
1-8456 

4-9542 
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 2-9053." 
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 804-1. 

If we had been asked to find the antilog of 2-0905, the line 
through -09 would havebeen followed and not that through -90, 
and the antilog is found to be 123-1. 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. 


8-1164 


130700000 or i 


307 x io 8 


2549 
i -0062 


1-799 
10-14 




3-8609 


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 right-hand 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 42-76. 



(b) log (-05I7)- 4 -* = 4-2 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 48-21 x 7-429. 

Actual Working Approximation 50x7 = 350. 

Let x = 48-21 X 7-429 

then log x = log 48 -2 1 + log 7-429= 1-6831 + -8709 

= 2-5540 

= log 358-1 from the antilog tables. 
/. x = 358-1. 

V 

Example 10. If C=^, a formula relating to electric currents, 

find the value of C, a current, when the voltage V is 2-41 and the 
resistance R is 28-7. 



Substituting the values of V and R 



.'. log C = log 2-41 log 28-7 
= 3820- 1-4579 



A pproximation. 

2J4 

3 ! 

i. e., -8 -7- 10 or -08 



= 2-9241, 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 = 4-75, 

g=32-2, r= 1-875. 

Substituting the numerical values in place of the letters 
F _ 28 x (4-75) a 



32-2 x 1-875 
Taking logs throughout 
log F = (log 28 + 2 log 4-75) - (log 32-2 

+ log i -8 75) 
= I> 447 2 _ i '579 
1-3534 
2-8006 
= 2-8006 
= 1-0197 
= log 10-47 
' F- 10-47. 



1-7809 
1-7809 



Approximation.. 

3x5x5 \ 
3x2 \ 

i. e., 12-5. 



Explanation. 

log 475= '6767 
.\2xlog 4-75 = 1-3534- 



j? 

Example 12. If/=~r, an equation giving the acceleration pro- 
duced by a force P acting on a weight W, find / when g = 32-2, P = 5-934, 
and m = 487. 



Substituting the numerical values 
, 32-2 x 5-934 

487 

Taking logs 

log/= (log 32 -2 + log 5-934) -log 4' 8 7 
= (i'5079 +7734) -2-6875 
= 2-2813 2-6875 
= T-5938 = log -3924 
/. / = -3924. 



Approximation. 

3x6 \_ 
5 \1 

i. e., 3-6 -^ 10 or -36. 



Example 13. Find the value of -* 

001872 



Let 



_ -05229 
~~ -001872 
then log x = log -05229 log -001872 
= 2-7184 3-2723 
= 1-4461 
= log 27-94 
.'. ^=27-94. 



A pproximation. 

5 Ul 

2 \\ 

i. e., 2-5 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 



47-61 x -0000753 



Taking logs 
log s = log -01154 (log 47-61 + log -0000753) 

= 2 -0622 (l -6777 + 5-8768) 
= 2-0622-3-5545 
= -5077 

= log 3-219 
/. s = 3-219. 



A pproximation. 



4-8 x 7-5 



\\\ 



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 = -X3-i42xr 3 gives the volume of 
a sphere of radius r. Find the volume when the radius r is -56. 



Substituting the numerical values 



= X 3-142 x 



Taking logs 

log V = (log 4 + log 3-142 + 3 log -56) 

-log 3 

= (-6021 + -4972 + 1-2446) - -4771 

= '3439 -'4771 

= 1-8668 

= log -7358 
.'. V = -7358. 



Approximation. 
4x3x6x6x6 

3 Hi 

i. e., 864 -f- 1000 
or -864. 

Explanation. 
log -56 = 1-7482 
3 x log -56 = 1-2446 

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} 
= 1-5926 = log -3913 

.-. *= -3913. 

* We must not divide 5 into 3-9630 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 745-4} (J log -001762 + f log 49-18}. 

Explanation. 
log -2164 = 1-3353 
3 x log -2164 = 2-0059 
lo 745 '4 = 2 -8724 
jxlog 745-4=1-4362 
log -001762 = 3-2460 
\ x log -001762 = 1-5410 
log 49-18 =1-6918 

2 X log 49-18 =3-3836 

.-. ^=-1677. .\fxlog49-i8 = -6767. 



= (2-0059+1-4362) - (1-5410+ -6767) 
= 1-4421 --2177 
= 1-2244 
= log -1677 



Example 18. If x= \/ -- - - 5 find the value of x. 
v 9004 x -0050 

Taking logs 

log x = {(log 29-17 + log -1245) - (log 9004 + log -0856)} 

= & (I-4649+ 1-0951) ~ 03-9544 + 2-9325)} 
= (-5600 - 2-8869) = t(3-673i) = i(8 + 5-673I). 
= 1-7091 = 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 8-8. 






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 = 11-15. 

Substituting values 

F 5 000 x 60 / 60 x 60 5 \ 

8 x -07 1 6 x 11-15 -924! 
= 536000 {53-75 + 5-41} 
= 31-7 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. 85-23x6-917 2. 876-4 x -1194x2-356 

3. 75-42 x-ooo2835 4. _ 4 5. -005376 x -1009 

6 9543 f] 2-896x347-2 g i2-o8x-02H2 

08176 81-48 -01299 

-0005 1Q 4843^29-85 1 154 x -07648 

' -007503 75132 " -009914x36-42 

-9867 x -4693 13 36-87x2-57 

'* -0863 x -1842 ' -085 x 13-77 x -05 

.. 24-23 x -7529 x -00814 ._ -572 x -0086 

3000 x-o 115x45 -2 7" -4539 x -0037 x -059 



16. V94-03 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 ^94-72x853^9" 

('275) s ' (-2347) x 5x10* 



OR (9i'56) 2 x(3-i8 4 )l (4-72) 3 x V26-43 

(2- 3 ) 2 xV8^ 



28. V 7 ' 008 ?^;, 0372 29. 

42-3x1-05 /-O5Oo6\ 3 _. V -6463 x (-086) 



' 



(27-63) 2 x log 10 3-476 
* V('4349) B x 5-007x1 

33. The formula V = nr z l gives the volume of a cylinder. If r = 
= 3-142, / = 12-76 find V, 



AIDS TO CALCULATION 23 

34. Given that L = . Find L when =11-7, ^B = 175-5. 

l z a 

35. If R = 7-+ find R when / = 5-1 and a = 0-87. 

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 ^!= 14-57, ^ 2 = 5'72, ^3 = 4-83. 

37. If v = u+ft and 5 = ^+i// 2 , find values of v and 5 when 
M = 350, / = 27, and t = 4-8. 

38. Find a velocity, v, from 



when g= 32-2, <Z= 0-84, A = 30, /= 5000. 

39. If p= (-7854 x *~) + d, find its value when f s = $, t = o-J5 

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 = alii-i ) 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 2-5 
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 8-72 amps, are supplied at a pressure of 112-5 
volts ? If the efficiency of the motor at this load is 45 %, what is its 
output ? (Output = efficiency x input.) 

44. 2-4 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 motor-cars 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 4-cylinder engine, whose cylinders are 
of 4" diameter, and stroke 8-6" ; 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 
17-83 cu. ins. Find its volume at temperature t C. and pressure H 
where t = 83-7 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 

p-d\. 



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 = 30-7. 

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= 12-73, r=i-5. 

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 4-2; and W, /, a and IT had the values 14-8, 17-56, 
39 and 3-142 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^g-S, m t = !$, t 1 = 2-i, /, = 1*6, =32, d= -126, 
I = 4-83, a = -97 and ?r = 3-142. 

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 = 3-142. 

62. Find the dimensions for the flanged 
cast-iron 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 32-2 ft. per sec. per sec. or 32-2 -. '-^, 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 + 14-7 Ibs. per sq. in., because the gauge 
records the excess over the atmospheric pressure 



.*. total pressure = 79-7 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 139-2 Ibs. 
32-2 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 64-4 



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 87-2 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 87-2 x 30 x io 6 



Calculation 
log d = (log 7+log 2240 + 3 log 288) 

(log 48+ log 87-2+ log 30,000,000) 

= (-845 1 + 3-3502 + 7-3782) 

- (1-6812 + 1-9405 + 7'477 J ) 



= 11.5735- 11-0988 Explanation- 

'4747 = log 2-984 log 288 = 2-4594 



.*. deflection = 2-984 ins. 



A pproximation 

7x2x3x3x3 m 



5x9x3 UUVWU 



or 2-8. 



3 x log 288 = 7-3782. 



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= 1-15 tons, M= 1-2 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*= 2-54 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 1-5". A statement of 
conditions may thus be made, "in the form 

*+ '375 = i-5- 

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 / = 1-5 - -375 = 1-125. 

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 86-2 




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 14-7 = 86-2. 

To each side add 14-7; then 

P = 86-2 -f- I4'7 = iQQ'9 Ibs. per sq. in. 

Thus, 14-7 on the left-hand side becomes + *4'7 when trans- 
ferred to the right-hand 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 4-6 units and the 
relation between/ and s is/ = 255. 

Substituting the numerical value for/ 

4-6 = 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 = 3-ii. 

10 5 

'Example 2. Solve for a, in the equation, 40+17 = 2-509. 

Transposing, to get the unknowns together on one side 

4-2-50 = 9-17. 

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 

F-3, = |C. 

Find the number of degrees C., corresponding to 457 F. 
Substituting for F its numerical value 

457-32 = |C 

' 5= 

Transposing factors 5 and 9, 



236-1 = 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 


86-4 + 32 


118-4 



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 

3W-4-I 

Dissociating knowns and unknowns 
w^yjo = 9'3 
_ i. 3 w = 27-4 



Note that the sign of 1-3 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* 8-1 _ i-gx , 7-21 
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) + (8-1 x 15) = (1-9* x 12) + (7-21 x 20) 
48* 2io#+ 121-5 = 22-8^+144-2 

48* 2IOX 22-8* = 144-2121-5 

or 184-8* = 22-7 



Example 10. The electro-motive 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 = 1-34, 
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 

*(E-V) = R, 

which completes the transposition. 
Substituting the numerical values 

R<= ^siV 1 ' 34 "' 896 ^ = 

= 2-47 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(3y-5)-(7y-l-9)+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 right-hand 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) = (i-j)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) = (i-g)L 

or, since -48 multiplies each bracket, we can take it outside one large 
bracket 

4 8{T A -T B -w-T 8 + T} = (i-ff)L. 

Dividing both sides by L 

(i-q) = ^{T A -TB-W-T 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 2-08 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 2-08 
Total internal resistance = x x -0052 

Total resistance = -0052*+ 1-5 

Current = * M *' ' 
Resistance 

and 100 x -75 = - 



0052* + 1-5* 

Multiplying across, i. e., multiplying throughout by the common 
denominator -0052*+ 1-5. 

75 (-0052*+ 1-5) = z-o8x 
*39# + 112-5 = 2-08* 

112-5 = 2-o8x '39,v = 1-69,1; 
.-. x = 66-6 

Or 67 cells would suffice. 

Exercises 5. On Simple Equations and Transpositions. 

Solve the equations in Exs. i to 6 
1- 5*+7(*~2) = 3-4(*+ 6 ) 

o I , 2 2fl 

2. g+-- 3 = 5-- 

3 4'2ft = 9-58 
' 7-45 4-69 



. . 



52 

-y 3^5 37-5 



= 
1-08 2-95 "*" 9-n 

6. 8-2^ -475(3 -2x) + 2-14(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 = 19-25, 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 = 310-7, 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= 136-4, V = 

97-9, 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 = 6-5, d= 4-7, 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 12-2 ohms? 

SC 

20. If D= +K, find n when D = 500, S = 12, C = 950, and K= 1-5. 

21. The tractive pull P that a two-cylinder 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(161-2) 

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= 1-62. 

27. Find the pitch p of the rivets in a single-riveted 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= 2-53. 

29. The stress / in the material of a cylinder for a steam-engine 

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 = 3-142). 



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, ^=15-3. 



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= 9-5*, 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(r-x) 2W 

ird 3 Trd 3 + ird* 

39. Find the internal pressure p for a thick cylinder from Lamp's 
formula 



where D= 12-74*. 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= 12-4, 
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 /= 0-01(1 + -^,), find v when H = 22-i, 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 = 13-3, TT = 3-142, 
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 = 2-03 sees. 

p = distance of centre of gravity from the centre of suspension = 

3I-43"- 
If h = distance of centre of percussion from centre of suspension 

t=27r*/-; and also k z = ph. 
Find k in inches {n- = 3-142, g= 32-2 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 = 2-5, 
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 5-4 and differ by 2-6? 

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=2-6 (2) 

If equations (i) and (2) are added 

2X = 8-0 

.'. *=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 = 5-4 

and y = 5'4~4 1*4 



x = 4-0 \ 
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 =2-5 
when W= 5; and if =5-5 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 

2-5= 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), 2-5 = i + b 
so that E= -2W+I-5. 

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 = 54-5 when A = 4 ~\ 
V = 48-8 when A = ioj 

Substituting the numerical values, in the general equation 

, n 
54-5 = w+- 

48-8 = m+^ 

Changing the fractions into decimals to simplify the calculation 

54-5 =w+*25n (i) 

48-8 = m+'in (2) 

Subtracting 5-7 = -15^ 

M = f^ = 38 

Substituting this value in equation (2) 
48-8 = w+3-8 
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 1-2)+ (6 xi -44) (i) 

421= (ax -i)+(6x -01) (2) 

To eliminate a multiply equation (2) by 12 and subtract. 
Thus 55212 = i-2a+i'44& 

5052 = i-2a+ -126. 

Subtracting 50160 = 1-326 

6 = 5i6o 
1-32 
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) 

a-b-y = 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 5ia-29& = 371 (4) 



EQUATIONS 



47 



Combining equations (2) and (3), multiplying equation (2) by 5 
and equation (3) by 3 

45a+io& + i5C = 235 

3a- 36-15^ = 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 thermo-electric couple. The corresponding 
values of E and T are 



T (C.) 


IOO 


600 


1000 


E (micro-volts) 


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 = 8-96. 

Substituting for b and c in equation (3) 
450 = a+8g6 29 

a = -417. 
Hence the law of calibration is 

E = 4I7+8-96T--OQ294T*. 

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 = 6-6 4. 48* 27^ = 48 

nn 25 = 2m y$ix = 51 

5. y+i-37=4* 6 - '\x-\V = &*+-&& 

gx-ijy =-49-87 19^+27 = 268 

4X-yy_ , 8. 2a+ 3 6+ 5C=-4- 5 

' ' 3C 7a+ 156 = 62-7 



9. 2p- 55+4^ = -33 10. If E= a+6*+c/ 2 , and also 



P '- for 

3 P-I2* + 2S = 8 9 values 

we 
have 



E 



t 



4-6 



10 



-4-5 



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 = 14-2 when W = 80. 

13. Corresponding values of the volts and amperes (obtained in a 
test on an electric arc) are 

V = 48-75 when A = 4 ; and V = 75-75 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 3-19 when the 
B.H.P. (B) was 2, and 6-05 when the B.H.P. was 5. Find the I.H.P. 
when the B.H.P. is 3-7. {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 2-062", and that on 8" was 
2-444*. 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 5-38 ohms and at 90 is 7-71 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= 30-11 when the length 1=5", and if 
e= 25-6 when /== 8", find the law for this case (Mild steel specimen). 

19. Repeat as for No. 18, when a= 2-12, and /= 3" when e= 59-2 
and 10* when e =24-5 (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 


2O2-2 


570-1 


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 /= 14-8 at 60 F. and /= 13-2 at 400 F. 
E 



50 



MATHEMATICS FOR ENGINEERS 



22. Repeat as for No. 21, the values of /and / (for cast phosphor- 
bronze) being 



/ 


16-06 13-1 

1 


t 


IOO 


400 



23. Given that W= 



Find the law connecting W and p 



if W= 21-11 when p= 80; and also W= 16-56 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 206-8 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 1-311 ft. 
Find the constants, C and K from CS-f-K=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 built-in girder 

=762-5 



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+60-55 = (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= 3-142) 

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 = (A-B)(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 = (A-B)(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., three-term 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) 



(x-5)(x+6) = x*+x-3o ........... (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)2-14(4) -24 = 64+16-56-24 = 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+42-24 = 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*4-3). 



.. 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 

- 



- - 

6xy-i 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 (x-2) (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) 

(4-7) + 3(4-7) ~~ 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 



a-b ~ 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 - (30-46)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. |n-R 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 n-R 2 / irr z l, which gives the volume of a 
hollow cylinder, when IT = 3-142, R= 12-72, ^=9-58, =64-3. 

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*4-355 + 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 left-hand 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 left-hand 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 left-hand 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 right-hand 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 right-hand 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 4-9 = 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-^= 6-88 

.*. x= 7-5 6-88 

= - 7'5 + 6-88 or - 7-5 - 6-83 
= -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 3-64B 4 5I-8B 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 
3-64A 2 5I-8A 900 = o. 



EQUATIONS 63 

Dividing through by 3-64 (the coefficient of A 2 ) and transferring 
the constant term to the right-hand side 
A 2 I4-24A = 247. 

The coefficient of A is 14-24; half of this is 7-12, hence add 7-12*, 
i.e., 50-8 to each side 

i.e., A 2 I4-24A+ (7-I2) 2 = 247+50-8 = 297-8 
or (A 7-i2) 2 = 297-8. 

Extracting the square root throughout 
A 7-12 = db 17-26 
i.e., A = 7-12 17-26 
whence A = 24-38 or 10-14. 

Now A =-B a , so that B 2 = 24-38 or 10-14. Of these values the 
former only is taken, since we cannot extract the square root of a 
negative quantity. 

Thus B 2 = 24-38 or B = 4-94, 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 

right-hand 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 



= 3-616 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 17-4 

10 10 

= 2-54 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 +1-5?+ -32 = o. 

Here A =-4, 6=1-5, C=32. 

-i'5 V2-25- -512 
8 

- 1-5 ^1738 



= -23 or 3-52. 

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 2-22 x io 8 
since ^- = 2-22 and -^4= io 8 - 

Thus -oo 1 5# 2 4- 400^+ (2-22 x io 6 ) = o 

Comparing with the standard form 

A = (1-5 x io- 3 ), B= (4 x io 2 ), C= (2-22 X io) 



66 MATHEMATICS FOR ENGINEERS 



Hence x = ~ (4 x I{)2 ) V(r6 x *o 4 ) (6 x IP' 3 X 2-22 x io 6 ) 

3 x io 3 

- (4 x io 2 ) V(i6 x io 4 ) (1-33 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 2-22 X IO 8 = 13-32 X IO 3 = 1-332 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- 1-33 
3 x io- 3 

_ io 2 (-4 VI4-6;) 



3 x io- 



= (io 6 x 2-61) 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 
= -234 104-3 Vmt. 

Taking the + sign and neglecting the first term, v = 104-3 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^ = 5-47.7. 

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^Uo-f 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. 



= {-32-5 + 31-5}* + {-32-5 -31-5}* 

= (-!) + (-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 = V-8. 

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)* = I2-75+I-26 = 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 ^29-5* 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 #=32-2) 

02 i ^2 

12. If r= -T , find h when f = 15, a = 5-5. 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 = 2-4. 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 = 62-4. 

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 = i-56, 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 1-305 = 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 100-4023 ft. 

Simultaneous Quadratics. Consider the two equations 

= 1-8 (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 

i-8y 

X = - 2 = -ay. 
2 ^ 

Substitute for x, wherever it occurs in equation (2), its value in 
terms of y. Then 

5-6(5-6 -y) = (.gy) = -8iy2 
31-36- 5 -6y = -8iya 
or '8iy 2 +5-6y 31-36 = o. 



= ~ 5 ' 6 V ^'3^ + 4 x -81 x 31-36 
-5-6 i 



Hence, y = 

1-62 



*62 

To find * 

* = -gy and, substituting in turn the two values found for y, 
x= -9 x 10-56 or -9 x 3-67 
= 9-51 or 3-30 

x= -9-51 or 3-30 \ 
and y= 10-56 or 3-67 / 

Example 54. In a workshop calculation for the thickness * of a 
packing strip or distance piece in a lathe the following equations 
occurred 



(9-9)2 = (i-7 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 

98-01 = 2-25 + y 2 +3y+-64 + * 2 +i-6* .... (3) 
98-01 = 3-o625 + y 2 +3-5y + * 2 ........ (4) 



By subtracting equation (4) from equation (3) an equation is 
obtained giving y in terms of x, thus 

o= -8125 -5y+ -64+ i-6x 
or *5y = i'6x '1725 
whence y = 3-2^ -345. 

Substituting for y in equation (4) 

98-01 = 3-0625 + (3-2* --345) 2 + 3-5(3-2* --345)+** 

= 3-0625+ 10-24**+ -1200-2-208*+ 1 1 -2X I -2075 + x*. 



72 MATHEMATICS FOR ENGINEERS 

Collecting terms, n -24^+ 8-992.* 96-035 = o 



, 8-992 V / (8-992) 2 + (4 X 96-035 x 11-24) 
whence #= ^ .a 

8-992 66-3219 
22-48 

_ 57-3299 nr -75-3I39 
' 22-48 22-48 

= 2-5503 or -3-3503- 

i.e., the thickness of the strip was 2-5503 (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 = 3-2* -345. 

Thus y =-- (3-2 x 2-5503) - -345 or y = (3-2 x - 3'353) - '345 
y = 7-816 or 11-066. 

The positive solutions alone have meaning in this example, so that 
x = 2-5503 and y = 7-816. 

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 1-732 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 2-89. 

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 + 2-236) = 7 x 6-236 

l6- 5 H 

= 3-968. 

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 
4-V2#+3 = 4-V-2 + 3 = 4-1 = 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 - 55-8/5, 

or '535/ 2 +55-8/ - 2704 = o. 

Solving for /, by means of the formula 

f - 55' 8 V3"Q + 579Q 

/a - ^o7~~ 

- 55'8 94'5 = 38-7 Qr _ 50^3 
1-07 1-07 1-07 

= 36-2 or 140-5. 

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, left-hand side = Vi6 7 + 3 V8 +17 

= 3+15 = 1 8, which is the value of the 
right-hand side x = 4 satisfies. 

* Always remember that when squaring a "two-term" 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 * = , left-hand 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*-pq-7i 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 = 22-8, and #=1-5. 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, ^=2-6x60, A=i5-6, a = - ^, =32-2, 

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 one-half 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 one-half 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 one-half 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 one-half 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 one-half the product of the sides including the right angle. 



8o 



MATHEMATICS FOR ENGINEERS 



One further point in connection with the right-angled 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 *= (7-2)(7 + 2) 
or x = 6-7 
so that the width of the stream = 2x = 13-4 ft. Fig. 9. 

Example 2. Find the area of the triangle ABC, Fig. 10. 



Area = base x height 
= $x 14-6 x n -4 
= 83-2 sq. ins. 

C. 



(Note that 11-4 is the perpendicular 
on to AB produced.) 





Fig- 13- 

Example 3. The pressure on a triangular plate immersed in a 
liquid is 4-5 Ibs. per sq. ft. The sides of the plate measure 18-1", 25-3", 
and 17*4" respectively : find the total pressure on the plate. 

Let a = i8'i, b = 25-3, c = 17-4. 

Using these figures, the area will be in sq. ins. 



s = 



18-1 



25-3+17-4 60-8 

" = - - = 30-4 

2 2 

Then A = V3O- 4 (3O- 4 - i8-i)(3O- 4 - 25-3X30-4- 17-4) 

= A/30-4 x 12-3 x 5-1 x 13 
Taking logs throughout 

log A = i{log 30-4 + log 12-3 + log 5-1 + log 13} 
'I-4829 1 
1-0899 
7076 



1-1139 



= 2-1972 



U'3943' 
= log 157-5 
A = 157-5 sq. ins. 



Then total pressure = ^^ X 4-5 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 13-7 knots and then due N. 
for 7 hours at 10-4 knots. How far is it at the end of the n hours from 
its starling-point ? 

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 right-angled 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 = 3-142. 

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 2-014" 
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 17-4", 8-4" and 15-7" 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 0-2*.) 

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 
one-half 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 
1-42 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 = 2-6 x 2-65 = 6-9 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 = 8-74 

This is the area of one side ; but the deposit would be on both sides 
total area = 2 X 8-74 = 17-48 sq. ins. 

and current density = Jj g = '0812 amp, per sq. in. 
or i amp. for 12-3 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 right-angled 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 (many-sided 
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 right-angled triangle, 

(AB) 2 = 26 2 -ig 2 = 7 x 45 = 3'5 

AB = 17-76 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 

= 120-9 sq. ft. 

Area of triangle ABC = x 17-76 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 = 7-92 .) 

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. 

= 8-07 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 five-sided 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 seven-sided 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 semi-circumference 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 1-56"). 

Then area of AOG 
= \ AG X OH = i x 1-5 X 1-56 
= 1*17 sq. ins. 

/. Area of ABCDEFG 
= 7XAAOG = 7x1-17 

= 8-19 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 5-28" 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 24-8", 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 4-28". 

13. The " end fixing moment " for the end A of the built-in 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 1-6 amperes per 100 sq. ins. 

15. An irregular pentagon 
of area 59-08 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 3-1428 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 = -n-r 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 3-5* = 9-62 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 


4-8 


18-1 


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 1-128 and 3-569 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 = 1-286, of which the square root is 1-128. A marking is 

7T 

thus placed at 1-128, 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 

3-569 on the C scale also ; this figure being obtained by extracting 
the square root of 12-86 instead of that of 1-286. 

Some slide rules are supplied with a three-line 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 = ~ = 8-37". 

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(R-rXR+r) or -(D-d)(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 = 4-75", r = 4'75 ~ 3'25 = i'5" 

Hence the area = (R- r)(R + r) = ^(4*75 + i'5)(4'75 ~ 1 '5) 

= x 6-25 x 3-25 
= 63-7 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 = 12-07". 

Products, etc., of TT. Certain relations occurring frequently 
are here stated for reference purposes. 

TT = 3-142 - = -318 = 

ff * = 9-87 4^2 = 39'4 8 sa Y 39'5 4* 1 = 4 -I 9 

o 

TT TT 47T , (often taken 

i = .^236 = -7854 - = 1-256 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 7-13*. 

2. The semi-circumference of a circle is 91-4 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 : 
4-28, 5-19, 6-47, 2-10, -87. 

{Suggestion : area = - ; first find value of constant multiplier 

4T 

(approx. -08). Keep the index of the slide-rule 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 cast-iron 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 = 271-fN, 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 10-51" in diam. Find the VelocRy- 

, . f .... distance moved by effort"! 

Ratio of the machine \ V.R. = ,-r-r- 

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 = 32-2. 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 = 32-2. 

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 18-47". 

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 = 14-7 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 60-5 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 right-angled 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 semi-circumference) 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 one-tenth 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 

2-trr X angle in degrees __ angle in degrees X radius 

or, cure 7= 

360 57-3 

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 = - = 57-3. 

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. 

,, , 2-n-rx 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 
14-8 ft. diam. by an arc of 37-4 ft. ? 

Arc = rd 

. a _ arc _ 37'4 v<2 _ .. . 

6 - -r-^8 x 

Thus the angle required is 5-05 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 57-3, 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 = i-z". 



Then r = 



a z + h z 
2h 



4+ i-44 
2-4 



" 2-4 
= 2-265. 
diam. = 4-53* (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 V225-I 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 = 13-26* 

CO = r-h = 15-6 = 9* 
.*. EF = 13-269 = 4-26* 

or the height of the arc at the 5* mark is 4-26*. 

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 Ti-r 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 
= 10-3 sq. ins. 

Wetted perimeter 
_ arc ACB = ^ x 2 X n- X 6 = ^JT 

= 9-42" 
.*. H.M.D. (usually denoted by tri) 




e 



IQ'3 



= 1-094 



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 = 2-4"; 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 = 1-34*; 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 4-5, find 
the speed of a point on the rim in feet per minute. 

10. Find the area of a sector of a circle of 9-7* diam., the arc of the 
sector being 12-3" 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 2-4" diam. rolls without slipping on the circum- 
ference of another circle of 6-14" 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 = -n-rr.) 
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{l-5(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 semi-circle 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 4-8" and 7-4". Find its 
perimeter and its area. 

According to our notation, viz. as in Fig. 34, ia = 7-4, a = 3-7 

26 = 4-8, 6 = 2-4. 
Then the perimeter = ir(a + b) = 77(6-1) = 19-15" 

77 A/2(a 2 +& 2 ~) = TT V2(i9- 45 ) = 19-58" 
7r{i-5(a + 6)- Vab} = 7r{i-5(6-i)- V^y x 2-4} 



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 semi-circle 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 sub-normal. For the parabola, the length of 
the sub-normal 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 
sub-normal 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 = S-f-s- -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 stress-strain 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- Stress-strain 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 one-half 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 sub-normal 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 bending-moment 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. = 9-94 sq. ft. 
Volume = (30 J) x (10 \) x (15 J) 
= 2 9-5 x 9-5 x 14-75 f . 

1728 
x 9-5 x 14-75 x 6-25 



Capacity = 



1728 

14-9 gallons. 



ions 



or -jd 2 h, where r = radius 



If the weight of water contained is required 
Weight = 14-9 x 10 = 149 Ibs. 

Note. I cu. ft. of fresh water weighs 62-4 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 = 2-rh. 
Total surface = 2-jrrh-\-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. = 4-63" 
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 
2-8 


12-45 
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 2-54 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 

(2-54)3 2 

or weight = -00311 x 453-6 
= 1-409 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 14-2 Ibs. Find the 
density of W.I. (Ibs. per cu. in.) and also its specific gravity if i cu. ft. 
of water weighs 62-4 Ibs. 

3. The weight of a brass plate of uniform thickness, of length 
6'-5" and breadth n" was found to be 79-4 Ibs. If brass weighs -3 Ib. 
per cu. in., find the thickness of this plate. 

4. The sectional area of a ship at its water-line 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", 27-2" and 11-4" respectively. Find the volume of this prism. 

7. Water is flowing along a channel at the rate of 6-5 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 tightly-stretched 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 10-5 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 cross-section 
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 38-7", 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 20-63 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 38-2 Ibs. per sq. in. at a speed of 400 ft. per min. If the horse-power 
developed is 70, find the diam. of the piston. 

r_j p _ Feet per min. x total pressure in lbs.~| 
33000 

19. In a ten-coupled 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 horse-power 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. = 5-5* and stroke = 10*. 

22. The weight of a casting is to be made up from 4-14 Ibs. to 
4-16 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 8-91 grms. 

24. Find the weight, in Ibs. per 100 feet, of electrical conduit tubing 
of external diam. 2" and internal diam. I-872", 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 one-third 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 = 15-6 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 
one-third 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 / "+(Jxi-8") = 4-6" 

.*. total vol. = -x(i-6) 2 X4-6 

4 
= 9-26 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 



.-. / = 3-68*. 

Now R = 3*25, and r 2-13. 
.*. Lateral surface 

= 77 X 3-68(3-25 + 2-13) 

= 7T x 3-68 x 5-38 = 62-2 sq. ins. 
Also 
Volume = {R 2 + y 2 + Rr} 

= 3l 5 {10-54 + 4-53 + 6-92} 



TT X 3-5 X 21-99 o 

-2? 80-5 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 13-7" and 
the height of the pyramid is 9-5*. 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 5-2" side, is 79-6 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 5-12*, the 
base being a regular hexagon of 1-74* 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 62-4 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 11-42" 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 
32-4 ft. and that of the hypotenuse is 55-9 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 19-74 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 one-half 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 1-3750". 

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 lamp-shade 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 semi-circle 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 = ^n-r 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 = ^{Sr-j 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 = - -n-r 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 ^ 

39-27T 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. 7-4*. 

Outside surface = J X 4?r x (3'7) 2 = 85-6 sq. ins. 
Inside surface = \ x 477 x (2'75) 2 = 47-5 ,, 
Area of base = n-(3'7 2 275 2 ) = 19-2 

/. Total surface area = 1523 sq. ins. 

Similar Figures. Similar figures are those^ having the same 
shape : thus a field and its representation on a drawing-board 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 lamp-shade 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 DE-p. 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 
= JxAxAG-JxBxAF 

By substitution from equation (2) 

(AF^ 2 

jfzzJL v A TT 

/ A ,/- \ A /\ Xi.1/ 



_ r(AG)3-(AF)31 

- aii-l / A ^\o 



Factorising the numerator (see p. 53) 

_ JA[(AG- AF)] [(AG) 2 + (AG X AF) + (AF) 2 ] 
(AC) 2 

FAG AF-M i^rAxjAG) 2 AxAGxAF , Ax(AF) 2 ] 
[AG-AF-A] = ^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) 
= 2-8751 + 1-5(2-9294 - 2-9983) 
= 2-8751 1-5 x -0689 
= 2-7717 
.*. 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 9-426" and its 
area is 7-069 sq. ins., then the circumference of a circle of 30" diam. 
will be 9-426 X 10, i. e., 94-26", and its area = 7-069 X I0 2 = 706-9 
sq. ins. 



MENSURATION 



127 



Hence one can form a most useful table, to be used for all sizes 
of circles. 



Diana. 


Circumference. 


Area. 


I 


3-142 


785 


2 


6-283 


3-142 


3 


9-426 


7-069 


4 


I2- 5 66 


12-566 


5 


15-708 


I9-635 


6 


18-850 


28-274 


7 


2I-99I 


38-485 


8 


25-I33 


50-265 


9 


28-274 


63-617 



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.) = 1-1782 
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 7-14* 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 70-35 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 
4-92", 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 84-2 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 gun-metal of external diam. 6* 
was found to be 22-3 Ibs. Find the internal diam., if the gun-metal 
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 3-15 mm. Find the hardness number for the material of the 
plate if the load applied was 5000 kgrms. and hardness number 

= -5 T-J = . (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 2-06" should 
be 2-22". 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 non-parallel 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 52-7 acres. The .chain with which the lines were measured 
was tested immediately after the survey and found to be 100-8 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 10-75 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 right-angled 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 cast-iron 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 
Semi-parabolic 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 anchor-ring 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 


7-21 

7-76 
8-04 


Brass or Gun-metal 
Copper (Cu) ... 
Lead (Pb) ... 
Tin (Sn) 


3 
.32 

HI 

27 


518 

553 
710 
/i6^ 


8-31 
8-87 

n-34 

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. Wrought-iron plate weighs about 10 Ibs., and 
steel 10-4 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. 

Wrought-iron bar or rod weighs about 10 Ibs., and steel 10-4 Ibs. 
per yard for every sq. in. of section. 

Wrought-iron bar or rod, i"diam., weighs 8 Ibs. and steel 
8-2 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 8-2 or 32-8 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 = 15-5 x 8-5 x 3 cu. ins. = 395 cu. ins. 
Vol. below AB = 16-25 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 3-5 = 343 

/. Net vol. = 3i6 

and weight = 316 X -26 = 82-1 Ibs. 

Example 38. Find the weight of a plate for a cast-iron tank. The 
plate (see Fig. 58) is 24* square and f thick; there are 20 ribs, each 
\" x ij" x ij", and 24 bolt-holes in the flanges, each f* square; also 
the flanges are 23!" x J* x ij". 



r -j-j- 




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 = 7-8 

Gross vol. = 281-9 >, 
Subtract for 24 bolt-holes (C) ; 24 x f x f x J = 4-7 

Net vol. = 277-2 
.*. Weight = 277-2 X -26 = 72 Ibs. 



MENSURATION I35 

Example 39 Find the weight of the wrought-iron 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 

= 17-5* 
/. Area = 17-5 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 



= 7-1 



= 44'3 .. 

or the net area of the stamping = 109-7 

Then the volume = 109-7 x 9 cu - i ns - 
and the weight = 109-7 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 8-2 Ibs. per yard (see p. 133) ; therefore 
\" diam. steel rod weighs -^, i. e., 2-05 Ibs. per yard. 

Hence, weight of i link = 5_^> x 2-05 Ibs. 

and weight of 3600 links = 5 '43 * 2-05 x 3600 lbs = III5lbs 

Example 41. Two straight cast-iron 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 Cast-iron 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 2-6 ft. 

.*. Path of centroid = Jx 77X5-2 = 2-04 ft. 
and length of the path of the centroid between the flanges-^ 

% =2-0 4 ft.-(2X") 

5= ;-9oft, - 23-5*, 



MENSURATION 



Area of revolving section = f X4 2 J ( X3 2 ) = 5-5 sq. ins. 

hence the volume of the solid between the flanges = 23-5x5-5 cu. ins. 

= 129 cu. ins. 

Vol. of 2 flanges, each J* thick, 8* external and 3* internal diam. 



= 43-2 cu. ins. 



.*. Gross vol. of bend = 172-2 cu. ins. 

To be subtracted 

Vol. of ten |" diam. holes : 



Diam. 


Length. 


Vol. 


625 


5" 


i'5 



/. Net vol. of bend = 170-7 cu. ins. 

and weight = I7O-7X-26 = 44-4 Ibs. 



Example 42. Find the weight of the wrought-iron 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 


7-25 
7- 2 5 


1282 
462 




- 8 



i 
04 



net volume = 820 cu. ins. 



Fig. 62. Wrought-iron 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- Vrf-aJ = 6- -^36-25 = 6-3-32 = 2-68*. 
Hence add J of 2-68", i, f ,, >f tq the length of the web, 



138 MATHEMATICS FOR ENGINEERS 

For the lower horn B, a 2 = 6", r t = 7-5* 



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. 



= 18-4* 
= n* 

= i8-4X 11X4-5 = 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 one-half the volume of a sphere of diameter i-8d, 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^n-x-g 8 = 1-58 cu. ins. 

vol. of body {Diam. = i", length = 1-125"}= -88 



or 



vol. of i rivet = 2-46 



Number of i* rivets per cwt. = 





2 -46 x -29 



Example 4^. Find the weight of the cast-iron 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 = 6-75", thickness = -75". 

Volume = 12x6-75x75 = 60-75 

B. 4 cylinders, of diam. 1-625* an d total length = 5* 

Volume (obtained from the slide rule) = 10-35 

C. Area of section = semicircle + rectangle 

(5-5x2-5) 



= 10-86 + 1375 = 25-61 
Volume = 25-61x275 
D. Cylinder, diam. = 4*, length = 4* 
Volume 



E. Cylinder, diam. = 4-5*, length = -75 
Volume 



F. 4 cylinders, diam. = 2", total length = i" 
Volume , 



= 70-48 

= 50-30 

= 11-92 

= 3-14 



Gross Volume = 206-94 




Fig. 64. Cast-iron Hanger Bearing. 
To be subtracted 



cub. ins. 
= 28-20 



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 = 157-6 x -26 
= 41 Ibs. 



140 



MATHEMATICS FOR ENGINEERS 

Exercises 19. On Calculation of Weights. 

1. Find the weight of the cast-iron Vee-block shown at (6), Fig. 63. 

2. Find the weight in steel of the crank axle shown in Fig. 65. 



Centre Lir\e 
of Engine. 

' 





P36-9"-- 

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 f-diam. bolt holes. 

4. Determine the weight of a wrought-iron 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 right-angles to the plane of the link. The mean 
lengths of the axes of the link are 4* and 2^". 

6. How many f-diam. snap-headed 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 7-9*. 

8. Required the weight of the cast-iron anchor plate shown in Fig. 66. 

--L, 



ajfef- 
ii_ 



\ 






/ 


/ 


"< 

-8' 


M 




t 

eb 
i 



M 1-6" 



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 cast-iron 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 T-bar. 




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 motor-car rear axle. It 
is made of nickel steel, weighing -291 Ib. per cu. in. Find its weight. 




L-3T- 



Fig. 71. Worm Shaft. 



15. Calculate the weight in cast iron of the half coupling shown in 
Fig. 72. 




Fig. 72. Wrought-iron 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 
mid-ordinates. Then-- 
Area = aver, mid-ordi 
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 / 



2-6S 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 3-32/1 



Lateral = 2irrh 
Two ends = 2irr* 
Whole area = 2jrr(& + r) 



Outer lateral) 
surface / 

Inner lateral) 
surface 

Lateral 



/ 



= 2-irrh 



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. 

RIGHT-ANGLED V-NoxcH 



Head (ft.) H . . 


1888 


2365 


2617 


2878 


3065 


336i 


Quantity flowing 
(Ibs. per min.) Q 


J4I-5 


249-8 


323-5 


411-4 


483-6 


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 = 411-4, 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 V-notch. (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 


3-09 


4-58 


5-67 


6-50 


B.H.P. (Output) 


o 


1-62 


3'33 


4-71 


5-8! 



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 = 2-43 and / = '4 12 - 

Plotting the values of the efficiency so obtained to a base of output, 
a well-defined smooth curve is obtained, as in Fig. 78. 

The efficiencies worked from the experimental figures are 



B.H.P. . . 


O 


1-62 


3'33 


4-71 


5-81 


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 efficiency-output and the input-output 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 Morris-Bastert pulley block gave the 
following results : 



Load lited\ 
(Ibs.) W. j 


27-5 


47'5 


67-5 


87-5 


107-5 


127-5 


147-5 


167-5 


187-5 


Effort re-] 




















quired (Ibs.) 


2-07 


2'5 


3-15 


4-05 


4-52 


5'2 


5-85 


6-4 


7-1 


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 
27-5 on the C scale and -572 is read off on the D scale, so that the value 
of P when W = 27-5, 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, 1-41, 1-82, 2-24, 2-66, 3-07, 3-49, 3-9. 
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 = 1-63, P! = 3-65 

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 453-6 (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 - 



5-77 


r 




















~r 




/ 


























/ 
























J 


7 






4-6 




















/ 


jr 








- co- 
-o 


















/ 


' 










o 
















/ 




























/ 





























/ 


















U?-2-32 
2-10 











































Y\ 






















i-Oi 




> 


/ 


\ 






















/ 






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 


1-05 


2-16 


2-32 


4-6 


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 


406-2 


502-8 



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. Cast-iron 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 


2-367 


5-611 


10-95 


18-94 


30-07 


44'9 


63'9 


87-7 


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 


i-5 


2 


2-5 


3 


Time constant (sees.) . 


2-5 


1-7 


i-4 


i-4 


i-i 


I-I 


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 


20-5 


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. induc-i 
tion density) / 





2400 


4500 


6000 


7100 


7800 


8300 


8500 


8600 


8600 


8700 



(10. Cobalt.) 



H. 


o 


1-55 


3'io 4-65 


6.2O 


775 


12-40 


I5-5 


23-25 


3i 


3875 


46-5 


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 


32-4 


39 


43'4 


47 


50-5 


52-5 


55 


56-8 


58-6 


59-3 


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 


32-5 


4i 


46 


49 


52-5 


54'5 


56-5 


58 


59-5 


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 


Left-hand reaction (Ibs.) 





'4 


8 


1-25 


1-7 


2*1 


2'55 


3 


3'45 


3'9 


5'2 


6-05 


6-5 


7 


R.H. reaction (Ibs.) . . 


7 


6-5 


6*05 


5'6 


5'2 


4-8 


4'3 


4-5 


3H5 


3 


1-8 


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! 


9-75 


1475 


20-25 


2 575 


30-75 


3575 


40-25 


45-25 


49-25 


55-25 



17. Test on a Single Purchase Crab (V.R. = 27). 



W . . 


50-1 


92-1 


137 


1 80 


224 


266 


310 


354 


394 


PI . . 


3-6 


5-35 


7'9 


9-9 


n-8 


13-9 


14-7 


16-9 


19-5 



18. Test on a Screw Jack (V.R. = 60-5). 



W . 


34 


54 


74 


94 


114 


134 


T 54 


174 


194 


214 


234 


PI . 


i-73 


2-85 


3'93 


5-17 


6-19 


7-70 


8-95 


10 


n-3 


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 


6-75 


8-75 


8-75 


10 


13-5 


15 


18-75 


21 


22-5 



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 
18-75 


1-3 

21-5 


1-45 


I'S 


1-65 


i'75 
37'5 


i-77 


1-8 


1-825 


1-85 


Amps. . . . 


4 


13-5 


24'5 


27-5 


32-5 


40-5 


42 


45'5 


47'5 



21. To a base of frequency plot curves giving (a) voltage, (b) current 
taking the following figures : 



Frequency 


40 


43-5 


47 


50 


52 


54 


56 


60 


64 


75 


80 


88 


Current 


5-39 


8-75 


I4'35 


18-67 


1473 


11-66 


9-33 


6-83 


5-19 


3-05 


2-64 


2-14 


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 " stress-strain " 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 4-volt lamp was varied, and the 
candle-power (C.P.) then measured for various values of the voltage, 
the results being as follows : 



C.P. . 


o 


5 


I-O 


!'5 


2 


2'5 


3 


Volts . 





3-03 


3'57 


3-96 


4- 2 5 


4-44 


4'75 


Amps. . 





1-16 


1-29 


1-36 


1-48 


1-63 


1-71 



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 


1-0487 


1-204 


I-37 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 


6-71 


8-i; 


9-Q4. 














I.H.P. . . . 


4'5 


7^7 


10-66 


11-69 


12-95 


Oil per hour (Ibs.) 


i'5 


2-37 


3-63 


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 


13-12 


iQ'54 


9-83 


8-85 


8-15 


6-57 


1-84 



27. Results of an efficiency test on a small motor gave the follow- 
ing :-r- 



Output (watts) . . . 


6-46 


24-2 


33-8 


37'5 


40 


55-3 
138 


61-5 | 64-9 


77'i 


9* 


"7 


Input (watts) . . . 


57-6 


82-4 


IO2 


104-2 


107-2 


142-2 


141-1 


162-4 


187-5 


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 = 2-15, at 2.30 o'clock and also 
at 3.30 voltage = 2-06, at 6.30 voltage = 1-87 and at 9 o'clock voltage 
= 1-72. Another cell was charged at a uniform rate from 2 o'clock to 
7 o'clock, the voltage rising from 1-75 to 2-38. Assuming that the 
discharge was uniform, find the time at which the cells had the same 
voltage. 



Co-ordinates. 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 co-ordinates, 
that along the horizontal being usually called the abscissa, while 
vertical distances are called ordinates. In representing a point by 
its co-ordinates 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 (1-5, i) 

Point C is ( 3-2, o) 
Point Dis ( 1-4, 2-3). 



D.|... 

{-1-4, 2-3) 



I 



-^ 



-4 -3 



-2 



-1 



(I-5.-I) 



-2 



oAa 
(-4,-: 



-3 



(4,-3: 



Fig. 81. Co-ordinates of Points. 



To fix the position of a point in space it would be necessary 
to state the three co-ordinates, 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* 

= -12-4*, 



From (2) 
From (3) 



y 5 -- 1-6 -8* 



y = 2-4 -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 = 1-68*. (2) y = 8*. (3) y = -2-48*. 

and tabulate the numerical values : 



(i) 



* 


i-6 -8* 


y 


-4 


1-6+3-2 


4-8 


2 


1-6+ 1-6 


3-2 


O 


i-6 o 


1-6 


2 


i-6 1-6 i o 


4 


1-6-3-2 


-1-6 




* 


2-4 -8x 


y 


-4 


-2-4+3-2 


8 


2 


-2-4 + 1-6 


- -8 


o 


2-4 


-2-4 


2 


2-4 i-6 


-4-0 


4 


-2-4-3-2 


-5-6 



These lines are parallel (see Fig. 83) and cross the y axis, (i) at 1-6, 
(2) at o, and (3) at 2-4, or the values of b in the three cases are 1-6, 
o and 2 -4 respectively. 



164 



MATHEMATICS FOR ENGINEERS 



Solution of Simultaneous Equations by a Graphic 
Method. Knowing that a first-degree 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 co-ordinates 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 = 6-33 1-67*. 
Table of values reads : 



165 



X 


6-33- 


i-6jx 


y 


- 3 


6-33 + 


5 


n-33 


o 


6-33- 




6-33 


4 


6-33- 


6-68 


- '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 Co-ordinates, 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 (1-2, o). Indicate each point clearly. 

2. Join up the four points (-10, 10) ; (5, 10) ; (15, -2-5) ; and 
(10, 2-5) in the order given, and find the area in sq. units of the 
figure so formed. 

3. On the same diagram plot the points (1-4, 2500); (-75, 374) ' 
(-1-82, 1140); (-32, 4816). Indicate clearly the scales chosen. 

4. Plot the straight line $x-8y .= 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 = 4-28 4-06*. 



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 = 1-45^+ 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 4-5* 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 = 57-3.) 

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 = 2-5 V + 75. 

What is the slope of the resulting graph ? 
Solve graphically the equations in Exs. 12 to 16 : 

12. yn 6n = 6-6 13. 48* 27^ = 48 
nn 25 = am. y$ix = 51. 

14. y+i-37=4* 15. 7#+3y=io 

gx-ijy =-49-87. 35*-6y = i. 

16. y = 1-4* 3 
2-6# y = 13. 

17. The co-ordinates 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 starting-point, 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 = - ; ^- = = 15-8, .'. a = 15-8. 

increase in I 5 

Intercept on axis of W through I = o is 40 units, .'. b = 40. 
Thus the equation is W = 15-81+40- 

Second Method, or Simultaneous Equation Method 
Select two convenient points on the line, not too close together 
e. g., W = 167-5 \ 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 167-5 = 8a + b (i) 

87-5 = 3 + b (2) 

Subtracting 80 = $a 

whence a = 16. 

Substituting in equation (2) b = 87-5 48 = 39-5 

.*. 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 horse-power is known as a Willans' line (named after 
Mr. Willans, who first put the results of steam-engine 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 


i-6 3 


2-13 


2-63 


3-25 


3-75 


4-25 


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 
3-8 = 6oa + b 
7= 5+& 



(i) 



a = -0564 

6 = -7 - -282 = -418 
! = -0564 W + -418. 



or 



2-82 



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 



2-2 



16-35 

W 



e. g., if W = 50, efficiency = rj 



2-2 + 
396. 



16-35 

5 



Example 10. The following are the results of a test on a 6-ton 
Hydraulic Jack (V.R. = 106). 



Load (Ibs.) 1 600 


IO2O 


1445 


I88 5 


2320 


2740 


3210 


3625 


4010 


Effort (Ibs.) ii 


17 


22 


27-9 


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 = 18-9, Pj = 28, , = ^ = -675} 
To find the maximum efficiency, i. e., the efficiency at 6 tons load. 



Also 



P = V 

PI = 5'6+,^4-W - 

P 

P, 



2oo 



5-6+-OH2W . . (See Fig. 87) 
i 

5'6 -OII2W 

" - 7 + -oo9 44 W 



or efficiency at any load = 



593 
W 



+ 1-19 



Then for the efficiency at 6 tons load we must write 6 x 2240 for 
W, hence 

maximum efficiency = * 



-^3_ + i-i 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 


9-5 


"5 


13-5 


!5'5 


17-5 


P (Ibs.) . 


1-25 


2-25 


2'75 


375 


4'25 


5-25 


6-25 


7-25 



This was for the case of wood on wood. 

2. Recalculate but for cast iron on cast iron (dry). 



W (Ibs.) . 


33 


53-3 


63-2 


72-9 


93'2 


"3 


P (Ibs.) . 


n-3 


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, hard-rolled phosphor-bronze. 



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 


3-33 


6-71 


8-35 


9'94 


I 


4'5 


7-27 


10-66 


11-69 


12-95 



{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 


1-108 



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 


4-9 


1-56 


2-1 


2-7 


3'4 


4 


4'5 


5'i 


5-3 


6-25 


6-82 



9. 



T 


o 


1 200 


2400 


3600 


4800 


6OOO 


7200 


e 


o 


34 


67 


I'O2 


1-36 


I-7I 


2-O6 



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 


162-5 


164-5 



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 ) 


1-039 


i-i 


1-141 


1-198 


1-32 


I- 357 


1-402 



14. 



t 


17-1 


25'4 


30-3 


36-2 


41 


49'4 


61-3 


67 


74-3 


8o'i 
i'532 


93-8 


R, 


1-214 


1-259 


1-285 


1-317 


I-34I 


1-369 


144 


1-473 


1-505 


1-622 



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 


587-6 


650 


779-3 


803-3 


95l'4 


975-1 


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 = 3-29 x slope 
of line : obtained by plotting t 2 horizontally and / vertically.) 



/ 


34'5 


30 


28 


25 


21 


16 


12 


t 


1-87 


1-76 


1-67 


1-6 


1-49 


1-26 


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.) . 





36-5 


56-5 


96-5 


136-5 


176-5 


216-5 


256-5 


296-5 


3I6-5 


Deflection (d in.) 





12 


198 


34 


5i 


63 


'79 


925 


1-07 


1-17 



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 straight-line 
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 


45-21-9 


15 


2 


4 


20 14 9 


- 3 


I 


i 


5-7-9 


ii 


O 


o 


o +0-9 


- 9 


I 


i 


5+7-9 


3 


2 


4 


20+14 9 


25 


3 


9 


45 + 21-9 


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 
- 18-75 + 10 + -61 
-12 + 8 + -61 


- 14-39 
- 8-14 

- 3-39 


4 
- 3 

2 

I 


9 
4 

I 
o 


- 6-75 + 6 + -61 
_ 3 + 4 + -61 

- 75 + 2 + ' 6l 
o + o + -61 


- -14 

+ 1-61 
+ 1-86 
+ 61 


I 
2 

3 


i 

4 
9 


- '75 - 2 + ' 6l 
-3 -4+ -6i 

6-75 6 + -61 




-2-14 
- 0-39 
12-14 





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 = 2-22 : 




Fig. 89. Curve of 
4? = - 3* 2 - 8 * + 2 ' 



therefore x = -82 or 2-22 gives the two solutions of 5* 2 +7# 9 = o. 
In like manner the roots of 75** 2#+-6i are 2-95 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 
right-hand 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 right-hand 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+DA-f-DA = 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 = + 4-04 units, EA = n-i units and OA = 5 units 
DA 



or the roots are 



and 



EA 
OA' 



t. e., 



i. e. 



5 
ii-i 



or -81 



or 2-22. 



Example 14. Solve by the 
same means the equation 

1'5X* + 4* -f 1-22 = O. 

First, change the signs 
throughout to make the co- 
efficient of x z positive, i. e., the 
equation becomes 

I'5# 2 4* 1-22 = O. 

Set out, in Fig. 91, OA = 1-5 
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 left-hand rotation about O) ; 

EA = + 4- 45 , OA =1-5. 




Fig. 91. Solution ot Quadratic 
Equation. 



INTRODUCTION TO GRAPHS 
= -28 



179 



Then the roots are 7=r-r- = 
OA 



and 



EA 



1-5 



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 4-8 = in 
while for the determination of cube roots the process is reversed ; 
thus v'so = 3-1 and -^200 = 5-85. 

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 -8-3 + 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 


54-81 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 second-degree equation, 
two bends for a third-degree 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 = 1-5, 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 uniformly-dis- 
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 
1-35 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* 


Y-fio 5 


y 


o 


O 





o 


+ 


o 


o 


5 


25 


125 


625 


i-35Xio 5 - -i5xio s + -o63Xio 5 


1-26 


0105 


10 


IOO 


1000 


I0 


5-4 X io 5 1-2 Xio 5 -f -i xio 5 


4'3 


0358 


15 


225 


3375 


50600 


I2-I4XI0 5 4'05Xio 5 + -5o6xio 5 


8-6 


0716 


20 


400 


8000 


160000 


21-6 xio 5 9-6 xio 5 +i-6 xio 5 


13-6 


'"33 


25 


625 


15630 


390000 


33-7 Xio 5 -i8'75Xio 5 +3-9 xio 5 


18-85 


-I 57 


30 


900 


27000 


810000 


48-6 xio 5 32-4 xio s +8-i 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. 

Turning-points 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 turning-points 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 turning-point, 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 = 1-33 (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 turning-points. 



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+32-16+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 turning-point is in the neighbourhood of 2 and 
another in the neighbourhood of i, therefore take as additional 



INTRODUCTION TO GRAPHS 



185 



values for x, 2-5, 1-5, -5 and 1-5. Thus the subsidiary table 
reads : 



X 


x z 


x 3 + 2X 2 4* + 7 


y 


-2-5 


6-25 


- 15-63 + 12-5 + 10 + 7 


13-87 


- i'5 


2-25 


- 3-38 +4'5 +6+7 


14-12 


'5 


25 


13 + '5 -2 + 7 


5-63 


i'5 


2-25 


3-38 +4-5 -6+7 


8-88 



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 = 8-4 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 


o-o 


57 


325 


o-ooo 


5 


1-07 


1-14 


438 


I'O 


i'57 


2-46 


406 


i'5 


2-07 


4-27 


352 


2'O 


2'57 


6-6 


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 


1-07 


I-I45 


4367 


6 


1-17 


1-37 


4383 


8 


i'37 


1-88 


4263 


i-o 


i'57 


2-46 


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 = 32-2. {The belt is 4* wide and i* thick.} 

Find the speed at which the greatest horse-power is transmitted 
under these conditions : find also the maximum horse-power 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 


1-728 x io 6 


42OOO 2 1 IOO 


20900 


140 


2-352 x io 6 


49OOO 28700 


20300 


1 60 


4-096 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 


1-16 x io 6 


36800 14400 


22400 


97 


913000 


33930 - 11310 


22620 








224-OO 



9Q 



197-6 



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 97-6. Maximum value of 

H! is 22615, i.e., the maximum H.P. = y $ = 20-6. Hence 



I IOO 



we con- 



clude that the greatest H.P. is transmitted at a speed of 97-6 ft. per 
sec. and that the greatest H.P. transmitted is 20-6. 



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 = 4-15* -23*2 + 1.94. 

3. The centrifugal force on a pulley rim running at v ft. per sec. is 
found from T - . If w = 3-36 and g = 32-2, 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 = ^ + 8-5. 

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 6-095*. 

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* - 1-54 = o. 

17. (3 x io 6 * 2 ) + (2-8 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 = 3-42* -ix 2 . 

19. The following values were given for the B.H.P. and I.H.P. for 
different values of the valve cut-off. Find the cut-off when the engine 
uses least steam, (a) per I.H.P. hour; (b) per B.H.P. hour. 



Cut-off .... 


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 gas-engine tests. 



is IQX cu. ft. 

4 

which gives the greatest volume of water. 



P-vHn nf ait > 


11-7 


10-43 


9-13 


774 


5-38 


4-40 


3-60 


3-14 


gas J 


Gas per I.H.P.) 
hour J 


3i'9 


22 


20-8 


19 


21-6 


24-8 


29-8 


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 non-condensing engine running at 400 revs, per min. the 
following results were obtained ; 



Ratio of expansion r 


4 


4'4 


4-8 


5'2 


5-6 


6-0 


8 


Ibs. of steam per 1 
I.H.P. hour / 


2075 


20-48 


20-35 


20-16 


20 


20-32 


23-14 



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 horse-power 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 (V-i2) 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 1-07* 1-48*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 8-82 = o. 37. $op 3 + 4 = 2$p $p*. 

38. Find the turning-points 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 built-in 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- 1-305 = 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. 


16-8 


3i'i 


SI'S 


9<>'5 


157 


203 


230 


256-5 


259-2 


233 


196 


128-5 



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 self-evident. 

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 4-6 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 
4-6 - 



Substituting this value in the second case 

A='23X^ = '^f =3-68 ft. 

Example 2. The weight of shafting varies directly as its length 
and also as its cross section. If i yard of wrought-iron 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 
horse-power H is proportional to the cube root of the horse-power. 
If a shaft of 1-5* 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, 2-82, 5-2, 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 



= 5-2 

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 direct-acting 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 (i-33) 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~ = 1-526 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 3-4" weighs 47-8 Ibs. ; what will be the weight of a 
sphere of the same material, of which the radius is 4-17* ? 

2. The candle-power (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 2-235 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 8-7 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 2-14 ohms. Find the 
diam. of the wire of which 128 cms. had a resistance of 8-33 ohms. 

8. The power in an electric circuit depends on the square of the 
current and also on the resistance. The power is 15-34 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 kilowatt-hour.) 

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 1-845 ohms. 
Find the resistance of 1-94 metres of platinoid wire of diam. -028 cm. 

10. The number of teeth T necessary for strength in a cast-iron 
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 6-90 and the absolute tem- 
perature is 468; find the absolute temperature when the pressure is 
8-92 kgrms. per sq. in. and the volume is 1-39. 

14. In some experiments on anti-rolling tank models, the number 
of oscillations per min. of a model of length 10-75 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 130-4 cu. ins. and the absolute temperature 
is 641, find the absolute temperature when the pressure is 41-3 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 one-half 

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 + (n-i}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 one-half 
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 = 



(n-i) 



^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 z-6 and 0-4 
i.e., insert 4 terms between 1-6 and 9-4 equally spaced so that together 
with the terms given they form an A.P. 




9-4 



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 = 1-6 

and 6th term = 9-4 

but the 6th term = a + 5^ and a = 1-6 

1-6 + sd = 9-4 

and $d = 7-8 or d = i -56. 

Hence the means are 3-16, 4-72, 6-28, and 7-84. 

The graphical construction would be quicker in this instance. 
Referring to Fig. 103, draw a vertical through i on the horizontal 
scale to represent 1-6, and a vertical through 6 to represent 9-4; 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 12-5 



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^a-i 
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) 

-(l-r-J or (r--l) 
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 



1-r 

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 28-4, 284, 2840 

(i. e., in a G.P. having the common ratio = io), 

then their logs are i'4533, 2-4533, 3-4533 

(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. 7-1, 13-3, 25-1, and 47-3. 

To check this by calculation 

a = 2, and ar* = 89. 

Hence, by division r 6 = = 44-5. 

Taking logs 6 log r = i -6484 

.*. log r = -2747- 
Now log av = log 2 + log r = -3010 + -2747 = -5757 

.'. ar = 3-763. 

Also log ar z = log ar + log r = -5757 + -2747 = -8504 
/. ar 2 = 7-084. 

Similarly, the other means are found to be I3'34, 2 5' Il > an d 47-26. 

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 -^ = 1-4 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. 

= 1-964 ohms. 



starting current 1-4 x 80 

The problem now is to insert 7 geometric means between 1-964 
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 1-964 (at the right-hand end of the rule) : 
similarly the point Q should be indicated, the distance gQ representing 
133 (at the left-hand end of the rule). Join PQ. 




I 234-56789 
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 1-403, 1-002, -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 = P-f 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(i-fr), 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 pulley-block 




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 velocity-ratio 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) = 2-613. 

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 +(n-i)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 " straight-line 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 = 500-80-20 
= 400, . e., its value as a working machine, and the depreciation 
fund would then stand at 20. 

At end of 2nd year, value = 500-80-40 = 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 , = D-f-fD+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} = P-R 

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(Soo-8o) 



(i-03) 21 i 
x 420 



Explanation. 
Let x = (i-o 3 ) 2 i 

log X =21 log 1-03 

73^ = 21 x -0128 

= -2688 

x = 1-857 

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 = 14-7 

2nd = (2 X 147) + (-03 X 147) 

= 29-84 
3rd = (3 x 147) -f- (-09 X 14-7) 

+ (-0009 x 14-7) 
= 45-44. 

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 
three-quarters of the depreciation has to be allowed for yearly, 
i. e., 14-7 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+(P-KP) X K 

= P(2K-K 2 ) 

= P{i-(i-2K+K 2 )} 



At the end of the 3rd year 

depreciation fund = KP+K(P-KP)+K(P-KP)(i-K) 
= P{K+K-K 2 -f-K-2K 2 +K 8 } 
= P{ 3 K- 3 K 2 +K 8 } 
= P{i-(i-3K+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 P-R = P-P(i-K) 

or P(i-K)* = 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 



=-{9031-1-699} 
= 7959 

21 

= " '0379 
= 1-9621 

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 2-8 and 10-9. 

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 1-8, 1-4, i . . . must be taken 
so that the sum of them is 67-2 ? 

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 1-30 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, 1-2, 1-44 ... 

13. Find the sum to infinity of the series, 40, 10, 2-5 ... 

14. Insert 3 geometric means between ij and 6|. 

15. Calculate the sum of 15 terms of the series 5, 6-5, 8-45 . . . 

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, 9-6 ... 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 21-08 Ibs. and that for 90 is 27-38 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 2-4; 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 back-gear. 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 2-718 (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. 2-3026 or 2-303 approximately. 

Hence, from the rules given above 

log e N = 2-303 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* 48-72, log,. and log e -00234. 



log, 48-72 = log, (4-872 x 10) = log e 4-872 + log* 10 

= 1 '5 8 35 + 2-3026 
= 3-8861 

log ' 4Ti = log ' Je? = loge 5 ' 7 ~ loge 4 ' 61 
= 1-6233 ~~ 1-5282 

= -0951 



FURTHER ALGEBRA 2I9 

log, -00234 = log, ^24 = log, 2-34 - log, 1000 

= log, 2-34 - 3 log, 10 
= -8502 3 x 2-3026 
-= ^8502 6-9078 
= 9424 

It will be observed that for each power of ten in the number 2-3026 
has to be added or subtracted as the case may be. 

Example 15. Without tables of natural logs, find values of 



log, 9-63, log, -?, and log, -2357. 

log, 9-63 = Iogj 9-63 x 2-303 
= -9836 x 2-303 
= 2-266 

x 2-303 . : ' 

= 2-303 {Iog 10 i7i 7 - Iog lo453 } 
= 2-303{3-2347 - 2-6561} 
= 2-303 x -5786 

log,-23 5 7 = logi -2357 x 2-303 
= 1-3724 x 2-303 

Separating the two distinct parts 

= (f x 2-303) + (-3724 x 2-303) 
= - 2-303 + -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 (2I-25) 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 = 1-6932 = 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) ~ l-1 



Let- *= 

Then log x = 2*1 x log -1473 = 2-1 x 1-1682 
= ( 2-1 x I) + ( 2-1 x -1682) 

= + 2-1 - -3532 = 1-7468 

= log 55-82 
x = 55-82 

Example 18. Evaluate {^3-187} -*-* 

Let x = {log* 3 -i87}- 2 

_ y-'ost where y = log 3-187. 

The value of y must first be found. 
From the tables log* 3-187 = 1-1591 
Hence y = 1-1591 and x = (1-1591) OM 

Now log x = -024 x log 1-159 

= -024 x -0641 

= -001538 = 1-998462 or 1-9985 
= log -9965 
x = -9965 



Example 19. Evaluate -||- 

Let x = this fraction. 
Then- 
log x ={J log 42-17 - 2 log -0145} {2 log 8-91 - -116 log 58-27} 
=ii log 42-17 + -116 log 58-27} {2 log 8-91 + 2 log -0145} 



FURTHER ALGEBRA 22 j 



= (-5417 + -2046) - (1-8998 + 4-3228) 

B -7463 2'2226 

= 2-5237 = log 333-9 



= 333-9 



Explanation. 
log 42-17 = 1-6250 
X log 42-17= -5417 



log 58-27 =1-7654 
116 x log 58-27= -2046 



log 8-91= -9499 
2 x log 8-91 = 1-8998 



log -0145 =2 -1614 
2 X log -0145 =4-3228 



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 + 1-2 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 = 2-93 and n = 1-3. 

Substituting values C = 85 x (2-93) 1 -* 
In the log form log C = log 85+1-3 log 2-93 

= 1-9294+ (1-3 x -4669) 
= 1-9294 + -607 = 2-5364 
C = 343-9 

Example 21. The insulation resistance of a length / inches of 
fibre-covered 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 



= 1-343 megohms. 



is a ratio, and .*. r t and 



?! may be in any units so 
long as both are in the 



same 



log* 



06 

1-25 



= logi-25 -log, -6 
= -2231 1-4892 
- 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 = 11-5, 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 11-5 X (-5I7) 1 ' 5 
Then log x = log -27 + log 11-5 + 1-5 log -517 

= 1-4314 + 1-0607 + (1-5 X I7I35) 

= 1-4314 + 1-0607 1-5 + 1-0703 

= -0624 
x = 1-154 

Also y = bW'* = 28 x (-5I7) 2 ' 5 

Then log y = log 28 + 2-5 log '5 1 ? 

= 1-4472 + (2-5 x 17135) 

= 1-4472 - 2-5 + 17838 



Alternative Method 
of Setting Out. 



V = 5-383 
Q = x + y 



+ 5'383 = 6-537 



No 


Log 


517 

ii'5 

27 


i'7 I 35 
i'5 


35675 
7135 


1-0703 + 
i'5 - 


i -5703 
1-0607 

I-43I4 


I-I54 


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 115-1 Ibs. per sq. in. absolute pressure, 
expands adiabatically to a pressure of 20-8 Ibs. per sq. in. absolute : 
find its final dryness. 

From the steam tables, r t (corresponding to pressure 115-1 Ibs. per 
sq. in.) = 799 F. absolute temperature, and r (for pressure = 20-8 
Ibs. per sq. in.) = 691 F. absolute temperature. 

Also the respective latent heats are L t = 879 and L = 954. 

Then if q-i = i, since the steam is originally dry 



-725{i-i + -145} 
-725 x 1-245 
-903- 



Explanation. 



2-303(2-9025 - 2-8395) 

2-303 x -0630 

145 



Example 24. For an air-lift pump for slimes (a mixture of water 
and very fine portions of crushed ore, of specific gravity = 1-1013) 
the formula for the horse-power per cu. ft. of free air can be reduced to 

H.P. = -015042 {P*(p-|) "- p i} 
Find H.P. when P! = 12-5, P, = 15- 



Substituting values 
H.P. = .oi50 4 2{i 5 (^ 5 ) ?1 - 12-5} 

= -015042 {15 x -8786 - 12-5} 
= -015042 (13-18 12-5) 
= -01023 



Explanation. 

/I2-5Y 71 

Let AT = (^P) 

log x = -71 (log 12-5 - log 15) 
= -71(1-0969 - 1-1761) 
= -71 x -0792 
= - -0562 
= I-9438 

... 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 = 1-37, and p = 22. 

Substituting values 22 x v 1 - 37 =146 
Then log 22 + 1-37 log v = log 146 

x '37 log v log J 46 log 22 

log 146 log 22 
log v = - * * 

_ 2-1644 1-3424 _ -822 _ 
1-37 ~ i'37~~ 

v = 3-981 

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 = 4-7 and I = 12. 

Transposing for d 

OOO4U 1 ' 87 / 

* = -j- 

n 

Taking logs of both sides 

1-4 \ogd = log -0004 + 1-87 log v + log / log h 

= log -0004 + 1-87 log 4-7 + log 12 log -87 
f4-6o2i 

J 1-2568 1-9395 Explanation. 

1-0792 

log 4-7 = -6721 



12-9381 
2-9986 



1-87 x log 4-7 = 1-2568 



Then- logrf = ^6 _ - 2 + -9986 = - 1-0014 
1-4 1-4 1-4 

= - 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 3-0257 2-8202 
or log x = - 

-^=685 

.. x = 4-842 which is thus the value of ^ 

Hence v e = 4-8421;,. 

and v e v e = 4-842^ 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 21-42 ; 3-18; -164. 

i 1871 

2. Find the values of log, -007254; log* 72-54; 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 = 4-71 and r = 5-13. 

Evaluate Exs. 6 to 14. 

6. (24-91)- ' 7l 7. (-1183)" 8. 



(-0054)' 1 ' 
9. (3-418)" x (-4006)- 3-4 10 . S*2Lj3*5J>:* 

11. (-04I05)- 2 ' 3 12. (-3724)- 2 ' 43 

(lo ge i-62)3 X (log 10 325-6)- 247 



14. 1-163 x (-0005) 7 ' 76 -i- V(log 10 2i-67)- 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 = 32-2 and h = 49-5. 

17. The collapsing pressure P Ibs. per sq. in. for furnace tubes with 
longitudinal lap-joints may be calculated from Fairbairn's formula 

p= 7-363 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 = 14-7 and R = 4-6. 

21. Find H, a hardness number, from 

i6PD-* 
' n(2d) 
Given that D = 24, d = 5, P = 58, n = 2-35. 



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,, = 4-33 ^ 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 = 12-5, 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= 1-41 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. = 1-5, D = diam. of 
cross section in ins. = 2-1, 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 perfectly-jacketed 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 = 32-2, 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.) = 3-33 (L 
If the breadth L = 5-4, the head H = -4, and n = 2, find Q. 

36. If * = ^| , find * when m = 2-16, v = 1-65. 

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 = 89-6 C. 

38. Recalculate, when P = 7200 Ibs. per sq. foot and t = 138-2 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 => 1-84. 



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 = 1-56, r = 14-3 and n = 1-4. 

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(T-fl 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 
2-257 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 = 58-7. 

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 /, = 19-5 : find p a . 

49. In the law connecting pressures and temperatures of a perfect 
gas, find p, from the equation 



having given that = 1-37, 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 = 8-91 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 self-inductance of the circuit respec- 
tively, and V is the voltage. If R = 350, L = 5-5 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 26-43, find the value of R. 

56. Calculate the efficiency of a Diesel Engine from the formula 



^CH 



where n = 1-41, r c = compression ratio = 13-8 and r e = expansion 
ratio = 7-4. 

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 = 1-13. 

59. If T = te* 9 (the letters having the same meanings as in Example 
40) : 6 = 2-88 radians, p. = -15 and t = 40, find the value of T. 

T 

60. Similarly if 6 = 165, and j = 1-78, find /i. 

61. In the expansion of a gas it is given that pv n = c, and that 
p = 107-3 when v 3 : and p = 40-5 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 = 5-9. 

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 = 40-15 bh? (Ibs. per min.). 



ft 


k 


Actual Discharge 
(Ibs. per minute). 


Theoretical 
Discharge. 


c. 


i-75 


829 


35 






i'75 


I- 4 I 


79 






i'75 


1-81 


112-6 







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. 


2-325 


1-075 


I-2 5 


88-8 






3'34 


2-09 


1-25 


109-6 






4'4I5 


3-165 


I-2 5 


133 






6-1 1 


4-86 


I-2 5 


156-6 







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 14-13 when velocity was 10-23, 
and loss was 6-31 when velocity was 6-76, 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 -*v-X 



If * (hydraulic gradient) = ^ , find this value. 

2040 

70. When a disc revolves in air the H.P. lost in air friction varies 
as the 5-5 power of the diameter of the disc and the 3-5 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 


1-26 


V ft./sec. . . 


10 


25 



Find a formula connecting F and V. 



CHAPTER VI 
PLANE TRIGONOMETRY 

Trigonometric Ratios. If the ordinary 30 : 60 set-square 
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 set-square 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 3-46" 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 
right-angled 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 co-ratio 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 = co-sine of its complement B. 
tan B = co-tan 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 



*** * J-J.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 co-ratio 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 = 1-2741 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 = 28-6363 

cot 5 = 11-4301, 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' = -68584-0008 = -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 = 1-9626, whilst 
tan 63 30' is written 0057, and this means 2-0057, 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 3-6*. 



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 = 4-2, 6 = 7-8 and c = 6 ; find A. 

Substituting the numerical values 

_ 7 -8'+6 2 -4-2 2 _ 60-84+36-17-64 
'2x7-8x6' 93-6 

= .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 1-283 
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 1-7, 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 = 2-43, 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 = 5-108, 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 = 47-7 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! = 4-5, R z = 3-75, 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 = 32-2. 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 power-factor of a motor = cos <f> = ' ' ~] 
If H.P. is 4-78 find <f>, the angle of lag of the current. 

21. If P = effort on crosshead of a steam engine, T = crank-pin 

connecting rod 
effort, 6 crank angle, n = -- - ; and if P = 450 Ibs., = u 






Sm 2 * 



and 6 = 1-5 radians ; find T, from T = P/sin 6 + . 

I Vw 2 sin 2 6} 

{Hint. Sin 171-9 = sin 8-1} 

22. Calculate the value of y = Re~ K ' sin (wt + d) when R = 3-5, 
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 = 3-515, n = 20, X = -0867, C = 42-05, 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 = 4-5. 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 
alternating-current 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 52-96" 
from the tank bottom at the start and 14-86* 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 = pr-f- = -Q 
OA 28-91 

a = 60 56' or 



H 



28-91 14-86 = 14-05 
= -4859 = cos 60 56' 

60-93 

57'3 



1-063 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 = 28-91, 6 = u AOB = 2a = 2-126, 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 = EG-OE = 52-96 - 28-91 = 24-05 






and 



ft = 33 43' or -588 radian 
sin 2,3 = sin 67 26' = -9234. 



-'83.8 -cos 33' 43" 



Hence area of DHC = 



- -923) 



= 105-9 sq. ins. 



Area of the whole circle = - x 57-8i 2 = 2625 sq. ins. 

area of ABCD = 2625533-8105-9 = 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 right-angled 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 = 422-6' 

= sin 62 = -8829 
900 

BC = 900 x -8829 = 79^4' 

CE = BD = I200-AB = 777-4' 



^g = tan 40 = -8391 
EF = 777-4 x -8391 = 652-4 
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 left-hand 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 left-hand 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 right-hand 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 18-5 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 17-6 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 1-6. 

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 42-4 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 = 62-8 (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 = 93-8 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 Co-ordinates 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 co-ordinates 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 co-ordinates, with the object 
of introducing an accuracy of drawing which is impossible if the 
field-book 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., co-ordinates). 

Quadrant bearings. The co-ordinate axes being chosen as just 
stated, viz. North-South and East-West, 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. Whole-circle Bearings. 



For example, in Fig. 119, the whole-circle 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 
co-ordinates. Find the reduced bearings of BC and CA, the co-ordinates 
of the points B and C, and also the area of ABC. 

Right-hand 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 = i8o-62 -3 4 = 84 S.W. 

To calculate the co-ordinates 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 

= 3-8496 + I-I943 = 3-0439 
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 

= 3-8496 + T-994 6 = 3-844 2 
.". AD = 6985 links, which is the latitude 
B north of A 

Hence the co-ordinates 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 

= 3-888 5 + 1-7476 = 3-6361 
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 = 3-8885 + 1-9186 

= 3-8071 
= 6413 links, which is the difference of 

latitude between B and C. 

Thus the co-ordinates 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 = ADEF-ABD-BEC-ACF 

= (5432 x 6985)- (x 6985x1106) 



= (37-95 x io)-(3-87X io)-(i3-88x io 6 )-(i-5 53 x io 

= 18647000 sq. links 
Dividing by ioo 2 , = 1864-7 S( l- chns. 
Dividing by 10, = 186-47 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 = 3-9823 
cos 87 = -0523 and log cos 87 = log -0523 = 2-7185 

Example g. From the following co-ordinates 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 4501-2 
Southing 20-1 


Westing 56-1 
Easting 4788-1 


Reduced level 249-2 
Reduced level 329-2 



248 



MATHEMATICS FOR ENGINEERS 



By plotting the points A and B from the co-ordinates given, their 
actual positions are represented. Complete the triangle ABC by 
drawing AC vertically and BC horizontally (see Fig. 121). 

Then AC = 4501-2 + 20-1 = 4521-3 

and BC = 4788-1 + 56-1 = 4844-2 



56-1 



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 
4521-3 
i.e., log tan CAB = log 4844-2 

- log 452 1 -3 
= 3-68523 - 3-655 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' 
= 4844-2 

sin 46 19' 
log AB = log 4844-2 log sin 46 19' 

= 3-68523 - 1-85924 = 3-82599 
= log 6698-7 
AB= 6698-7 



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 6698-7; and the height is 80. 



In the triangle ABB 1 : 



tan B'AB = - 
6698-7 



and log tan B 1 AB = log 80 log 6698-7 = 1-90309 3-82599 

= 2-07710 
= 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* 

= 6699-3- 



PLANE TRIGONOMETRY 



249 



Exercises 28. On the Solution of Right-Angled Triangbs, and the 
Calculation of Co-ordinates. 

In the following Examples i to 7, ABC is a triangle right-angled 
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 = 27-37", A = 54, find a and 6. 

6. 6 = 72 -5", A - 3 8i, find a and c. 

7. c = 23-4", 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 sea-level) 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 4-7, 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. " Set-over " of Lathe 
Tailstock. 



Fig. 123. Brown and Sharpe 
Worm- thread. 



15. Calculate the " set-over " of the tailstock of a lathe for turning 
a taper (the angle being 9 and the length of job 15-2). 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 co-ordinates 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 50-5 N.E. 

19. In Fig. 126 calculate the co-ordinates 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 1-14 chains. The angle CAB was then ob- 
served by a box sextant and found to be 71 54'. Calculate the length 
ofCB. 

21. The co-ordinates 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 circle-bearing 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 cup-shaped 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 right-hand 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 left-hand 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 

= 258-i8o = 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 right-hand 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 right-hand 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 (i8o-A) = 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 (i8o-A) = + 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 \ Q-h 

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 = 2-1445 

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. 

3.199-2; 34i5'; 9842 3 '. 4. 4 ; u-62; -85; 1-16 radians. 

5. 1194; 2-45 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 half-circumference 
and is thus 180.] 

7. Find a value of A between o and 180 if 

cos A - b * + cZ ~ a * and b = 9-8' 

\*\Jj X*. _ _ _ 

zbc c = 6-4* 

a = 14-45* 

2 V2 _|_ 7/2 -zoT-T 

8. The equation cot 6 = - ^y - relates to the design of 

water turbines. If V = 53-4, u = 10, H = 100, e = 32-2, find 6 (between 
o and i So ). 

9. As for the preceding question, but taking V = , n = V8o, 
g = 32-2 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 (A-f-B), since A+B+C = 180; 

or alternatively 

B-C 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 (i8o-C) = 
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 
= 3-2425 + 1-9331 - 3-2823 
= ^'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' 
= 3-2425 + 1-9717 ~ 1-8933 
= 3'3 2 9 
a = 2093* 



Example 15. Solve the A ABC completely when a = 12-6*. 
b = 17-8", 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 17-8*; 
this is the side b. With centre C and radius = 12-6* (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 = 17-8, CB = CB' = a = 12-6 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'=i8o-B 



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 12-6 

In the log form 

log sin B = log 17-8 + log sin 40 log sin 12-6 
= 1-2504 + i -8081 1-1004 
= 1-9581 = 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 12-6 +_log sin 74 47' - log sin 40 
= 1-1004 + 1-9845 - i -8081 
= 1-2768 
c = 18-91* 

In the second case 

log c - log 12-6 + log sin 25 13' - log sin 40 
= 1-1004 + 1-6295 i -8081 
= -9218 
c = 8-352 

Grouping the results 

B = 6.si3' or 114 47'. C=7447' or 2SiV. c=i8-gi" or 8-352" 
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. 

_ 12-6 x sin 74 4 7* 

sin 40 

Set the cursor over 12-6 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 = 18-9. 

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 right-angled 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 8-4 = 44 

and L ACD -55 

Thus /.ACB=i8o-55 = 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 right-angled 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 



= go-a 



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 starting-points of the curved road. 

Examples on the use of Cosine Rule. 

Example 19. In the triangle ABC (Fig. 144) find L C when a = 4-45', 
b = 7-85', and c = 11-94'. 




C= 11-94- 
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 = (4-45) 2 +(7-85) 2 -(n-94) 2 
2ab 2 X 4-45 X 7-85 

19-8 + 61-5 142-5 
69-8 

61-2 

^ = "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 = 4-45, 6 = 7-85, c = 1 1 -94 

S = 12-12 

C _ /(s-b)(s~a) _ / 4 -2 7 x 7-67 

^ 2 ~ V S (S - C) ~ V 12-12 X -18 

.-. log tan - = J 



^{(-6304 +-8848) 



4-27 + log 7-67) - (log 12-12 + log -18)} 

(1-0835 + 1-2553)} 

= -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 half-minute 
since the rule gives as the direct result the value of a half-angle.] 



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 +(2-94) 2 

(2 x 5 -93 x 2 -94 x cos65) 
= 35-1+8-62 

-(2X5-93X2-94X-4226) 



= 5-39' 



= 5-93". c = 2-94", B = 65, 




a. = 5- 93" 
Fig. MS- 



PLANE TRIGONOMETRY 



267 



Example 21. Two forces, of 47-2 Ibs. and 98-4 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 = 47-2, AD = 98-4, t-ADC = i8o-6 3 < 
(AC) 2 = (AD) 2 +(DC) 2 -(2xADxDCxcos 117) 

= (98-4)*+ (47'2) a - (2 x 98-4 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 = 5-93*, c = 2-94" 
and B = 65. 

Area = ac sin B = \ x 5-93 x 2-94 x sin 65 
= i x 5-93 x 2-94 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 5-93, 5-39 and 2-94 respectively (compare 
Example 20). 

Thus- 5= 5-93 + 5-39 + 2-94 = ^ 

and area = ^7-13 x 1-20 x 1-74 x 4-19 = 7-91 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+a-b)(a+b-c)(a+b+c) 

= 2(s a) x 2(s b) x 2(s-c) x 2s 



i. e., sin C = ~y' S (s a)(s-b)(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 four-figure log tables or the slide 
rule allow. 

When extremely careful observations have been made it is 
advisable to employ five- or even seven-figure 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 = 5-93, c 2-94, B = 65. 
To find b 

From the above 

b = (c+a) cos 6 

zVca B 
and sin 6 = - cos - 

j. e., sin 



2^2-94 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 2-94 + log 5-93} + log cos 32 \- log 8-87 

= 1-8998 = log sin 52 33'' 
/. = 52 33' 
Then b = 8-87 x cos 52 33' 

In the log form 

log b = log 8-87 + 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 = 5-2", B= 7 8i. 2. a=79-5",C = 5 i 32',B = 4 7 3 6'. 
3. C = 26 50', 6 = 8-86", c = 5-68". 4. 6 = 5-97", C = 6 4 18', A = 75. 
5. c = 9-2, a = 10-31, = 46. 6. 6 = 6-ift,,c = 9-3ft.,A = 73i6'. 
7. a =124-4, 6 = 93-7, c = 99-3. 8. a = 13-7*. 6= 10-5*, 0=130. 
9. a = 4-27", A = 29, b = 5-86*. 10. c = 688o, B = 30, b = 5141. 
\D 

cross-section o/'u/'V'e 




Fig. 149. Solution of Triangles. 

11. A = 50 50', 6 = 922-4, c = 1003-8. 

12, B = 353o', 6 = 38-6, c= 4 3-57. 13. = 21-8,6= 15-7,0 = 47 32'. 

14. c = 32-7, 6 = 39-4, 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 3-7 tons, 
2-275 tons an d 3' 02 5 tons respectively. Find the angles of this triangle. 

18. Forces of 21-6 and 19-7 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 3-voltmeter 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 sea-level. 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 sea-level. 

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 = 517-75 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 V-threaded 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 water-carrying 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 6-5* 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 .~ uTT-n 

= 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. 
(left-hand side) ; the other by R.H.S. (right-hand 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 (A-B) 

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 right-angled 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 two-term 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 right-hand 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 -f-6 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* = 360-6 

and tan c = |^ = 1-5 = tan 56-3 

c 56-3 = ^ radians = -983 radian 

/ J 

V = 360-6 sin (6ool + '983) 

or, as it might be written, V = 360-6 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 

Sm-S 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 == 1-62, 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^ + 2-7 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 = 4-6, / = 60, L = -02 and I = 13-8. 

Formula for the Ratios of the Multiple and Sub-multiple 
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 
= l-2sin 2 A J 
2 tan A 



tan 2A = 



1 - tan 2 A 



If the ratios of the half-angles 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 = 



i-tan 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 = 3tanA-tanA 
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 = 1-1106 

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 slide-rule 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 half-angles 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 = 1-526 

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 3-4* radius at 
the bottom ; and other dimensions are as shown. 




B 
329-KB 



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 = 1-704 + x - 1-375 = '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 = (6-8 + *)* . . . . (4) ' (*- EDC * a ri g ht an 6 le ) 



also 
and 
Connecting (i) with (3) 



329 + x 

y 

i. e., y* 



1-36 (-329 + *) 
1-8496 (-329 + 



Hence from (4) 
whence 



(6-8 + x]x = 1-8496 (-329 + *) 

8496*' - 5-583* + '20021 - O 



284 MATHEMATICS FOR ENGINEERS 



+ 5*583 V3I-I699 -6804 
so that x = J J - T? 

1-6992 

or the required value of x = - = -0361*. 

1-6992 

Now- 



y -329 + x 
so that y = -4965 

also tan a = ^>- = 6-8482. 

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 = - 



i-tan z - 

2 

i. e., 6-8482 = 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 = 2-75 2 X '4323 = 1-8854*. 

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 left-hand side in this last 
formula.] 

Now 

A = ' ' '*\ L i.e.,= % sum of the two angles, 

and B = ( A +B)-(A-B) { 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 (A-B) = 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 C-sin 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(A-B)} ..... (5) 

eosA sinB = J{sin(A+B)-sin(A-B)j ..... (6) 

cosA cosB = {cos(A+B)+eos(A-B)} ..... (7) 

sinA sinB = {cos (A-B) -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 = 8-j, {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 
= 7-47- 

R.H.S. = 8-5 {cos 92 cos 204} 
= 8-5 { - cos 88 + cos 24} 
= 8-5 {- -0349 + -9135} = 8-5 x -8786 = 7-47. 

/. 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 = 3-5 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 15-7 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 
5-72 and 8-41, 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 = 98-87, sin 6 = -2753, K = -75, S = -69, E = 4-88 and Q = 9-55. 

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 = 3-5 

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 

4-606 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 i8o-64i5', 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 connecting-rod. 

Y 

The velocity is a maximum when cos0+j-cos20 = 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 8-485 

4 
= ~ 16-485 or -485 

4 4 

= 4-1212 or -12 12, which are the values of cos 6. 

But cos 6 cannot = - 4-1212, 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 connecting-rod 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 2-5 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 = 12-6. 

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. 2-35 sin # 1-72 cos x = -64. 

19. The velocity of a valve actuated by a particular Joy valve 
gear is maximum when 

i'2p* cos pt + i-8p* 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" + *) = 1-195 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 half-axes 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 semi-axes 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; 1.2.3-1.2.3.4.5 

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 = J-T-T- 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 4-5 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 8-3 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 = - = i-iSix 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 =1-012. 

Thus *- 1176 = 1-012 and e-' 01 "' which is the reciprocal of ' 01 
is -9883. 



294 



Then 



and 



MATHEMATICS FOR ENGINEERS 

2 + -988 



cosh -01176 = 
sinh -01176 = 



2 
1-012 



= 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 one-half 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 2-f &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 = 5-7 

r* = 3 2 + (~5'7) 2 = 9 + 32-4 = 4iH 
r = 6-44 

and tan = ^2 =* i-g 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+1-5; were added to the vector 4 + -6;', the resultant vector 
would be 2 4+i"5/+-6;, i.e., 2+2-1;. The addition is really 
simpler to perform by the graphic method, thus : OB at (c) 
Fig. 157 represents the vector quantity 2+1-5; 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+2-1; since OF = 2 units measured 
in a negative direction and FE = 2-1 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 
= rr-L (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 star-connected 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;) = 1-949 + 1-998; (by actual multi- 
plication). 

= 2-791 (cos 45 43' + ; sin 45 43') 
y = -9 -f- -i8j = -9179 (cos 11 19' + j sin 11 19') 
.'. C = Py = 2-563 (cos 57 2'4-y sin 57 2') or, alternatively, 
t-395 + 2-1497'. 

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 + ! sinh-oi5. 

3. Find the true length of a cable weighing 1-8 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 , 2n-L , 

alternating current circuit is given by * ^ x tan -1 p where 

n = number of cycles per second, L = self-induction 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 = 1-2, find the. 
lag in seconds. 



PLANE TRIGONOMETRY 299 

8. If cosh y = 1-4645, 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 = i-8 

.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 = 4-5 and b = 2. 

15 If C = 5-4 (cos 62 + j sin 62) and y = x-8 (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 mid-ordinate 
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 tracing-point 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 starting-point 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 8-44 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 tracing-point P, and draw a circle with this radius 
on a sheet of tracing-paper. 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 right-handed 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 
left-handed 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 right-handed 
direction, and this becomes -f- x if the tracing is in a left-handed 
direction. The true area = x + zero circle area, and the tracing 
is performed in a left-handed 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 tracing-point 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 
one-half 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 tracing-paper with 
a number of parallel lines exactly J" apart, i. e., i chain apart 
according to the particular scale chosen. Place the tracing-paper 



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\ = 2-5 

y n = 2-0 
sum of first and last = 2-25 



y 2 = 4-40 

ys = 5 >I0 

y\ = 5-34 

y& = 4-13 

y^ = 3-83 

y s = 3-80 

Vg = 3'63 



i(yi + yu) = 2-25 

Sum = 39-88 

Width of one division of the base = h = i" 
Area = i x 39-88 



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 
mid-ordinates as distinct from the extreme ordinates through the 
actual points of section. The average of the mid-ordinates 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 mid-ordinates are 3-66, 4-90, 5-24, 5-24, 4*4, 3-92, 
3'8, 3'73, 3'3 and 2-13 ins. respectively, and the average = - = 4-032. 



AREAS OF IRREGULAR CURVED FIGURES 309 

Hence the area = 40-32 sq. ins., as against the previous result of 
39-88 sq. ins., showing a difference of i%. 

The mid-ordinate 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 Mid-ordinate 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 
mid-ordinates 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 
Mid-ordinate (ins.). 


Area of Strip 
(sq. ins.) 


Sum of Areas of Strip 
(sq. ins.) 


AB 


1-8 


3'3 


5'94 


5'94 


BC 


4 


5-02 


2-OI 


7'95 


CD 


'3 


4'35 


I-3I 


9-26 


DE 


5 




1-89 


11-15 


EF 


4 


3-85 


i'54 


12-69 


FG 
GH 


i'3 
6 


3'54 
2'53 


4-60 
1-52 


17-29 
18-81 


HJ 

J K 
KL 


i'3 
6 

8 


3-6 
4'54 
3'4 


4-68 
2-72 
2-72 


23H9 
26-21 

28-93 


LM 


I-O 


2-31 


2-31 


31-24 



Thus the total area = 31-24 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 Mid-ordinate 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= 38-5 

Vs = 34'5 y= 30-5 
Vw= 24 



ist = y l = o 
last = y u = o 

Sum =o 



Sum = 237-5 
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 set-square, 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 mid-ordinates (shown dotted) are erected for every division. 
Next the tops of the mid-ordinates 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 " velocity-time " 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" = 2-5 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+oc-H 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 cross-section at various 
distances through an irregular solid be noted or estimated, and 
ordinates be erected to represent these cross-sections 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 cross-section, 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 mid-section are 24^" and 17^" and the areas of the three 
sections are 

A = TT X 14 X 10 = 14077 sq. ins. 

B = 77 x 10-5 x 7-5 = 7 8 '75 7r .. 
M = 77 x 12-25 x 8*75 = 107-277 ,, 



Volume = -{A + B + 4 M} = ^{140 + 78-75 + 428-8} 

= 277 x 647-6 = 4070 cu. ins. 
.*. Capacity = 4 ' gallons = 14-7 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 62-8 and 20-5 sq. ft., the section 
mid-way between is 36-7 sq. ft. and the length of the prismoid is 15 ft. 
Find the average cross-section and the volume. 



AREAS OF IRREGULAR CURVED FIGURES 317 

4. The mid-ship section of a vessel is given, the height from keel 
to deck being 19$ ft. ; and the horizontal widths, at intervals of 3-21 ft 
are respectively 46-8, 46-2, 45-4, 43, 36-2, 26-2 and 14-4 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 half-ordinates from bow to stern were -6, 2-85, 9-1, 15-54, 18, 
18-7, 18-45, I 7' 6 I 5'i3 an d 6-7 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 


2-8 


3'6 


5 


6-2 


77 


8-9 


10-3 


12 


Velocity (ft."l 
per sec.) . J 


37'3 


31-5 


27-5 


25H 


22 -4 


20'3 


18-2 


16-9 


15-8 


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 cross-section 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 


5-6 


6 


5'7 


4-8 


4'7 


4'5 


4 


3 



Find the area of the cross-section. If the average velocity of the 
water normal to the cross-section is 5-1 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 candle-power (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 


1-8 


2'5 


3'5 


4-2 


4'9 


5'4 


5-8 


6 


Candle power . 


o|n5 

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 


I-IQ2 


1-28 


1-88 


2-475 


3-487 



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 

cross-section 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. = 314-8 cu. yds. 





Fig. 171. Fig. 172. 

Example 2. Calculate the weight of clay removed in making the 
simple wedge-shaped 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. 

= 2032-6 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 wedge-shaped 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 = 62-65 

sin L BAD = . 
62-65 

but sin L BAC = 

sin ( 1 80 -BAD) 
= sin L BAD 
and hence 

sin ,1 BAC =7^ 
62-65 

Also 

AC = 250 62-65 
= I8 7'35- 




Fig. 173. Wedge-shaped Excavation. 



Then area of triangle ABC = BA . AC sin L BAC 



= \ x 78 x 187-4 x 



60 



= 7000 sq. ft. 



62-65 

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 = 862-5 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 " half-widths " 
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 = MK-MG = - - - 



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 = HE-HB = m-a 

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 DFA-CBE 

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 half-widths, 
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 I-25 

n-a _ 36- 30 = g f 

5 1-25 S 

mna 2 (60 x 36) 



1-25 



= 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 = 97-5- 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< 

* i-5 3 
BS = BQ - SQ = 51-66 - 17-22 = 34-44 ft. 

v di 

n = ~-(a+hs) = ^(20+45) = 48-75 ft. 

f~T~S D 

PD = 48-75 - 20 = 28-75 ft - 

^ = , whence PR = = 6-39 
PD 4-5 4*5 

AP = PD tan 6 = *^p- 19-17 ft - 

AR = AP + PR = 19-17 + 6 '39 = 25-56 ft- 



326 MATHEMATICS FOR ENGINEERS 

We can now proceed to find the volume of the solid SRFE. 

A = xBSxBE = 1x34-44x150 = 2583 sq. ft. 
B = xARxAF = -1x25-56x100 = 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 34-44x77-5 
= 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 

= 2X25-56x28-75 

= 367-5 s q- ft - 

The height = 100 ft., and hence 

Volume = -x X 367-5 cu. yds. = 453-6 cu. yds. 

3 / 

.*. Total volume = 2805 + 2471 + 453-6 = 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 road-making 
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^6QL-94O--H i050-^-772-V584'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)] = 501-75, since h = 9 
L = 560 

/. Volume = =^{o + 1287 + (4 x 501-75)} = 307400 cu. ft. 

For the portion abfe 
A = 1287 

B = 15^0 + (15 X if)] = 994. since h = X 5 
M= i6-5[40+ (16-5 X if)] = 1136, since h = 16-5 
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+ (7-5 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 + (8-5 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=8-5[40 + (8-5 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 = (1-829 1-209) 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 602-4 sq. yds. Then the net 
volume of earth removed = 602-4 x 1000 = 602400 cu. yds. 

Cutting with Unequal Sides, in Varying Ground. 
First Method. 

To find the average cross-section 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 half-widths, etc., for A; and 
w 2 , 2 , h. z , and r 2 the corresponding values for B. 



Then- Area of A = 
For the mid-section M 



m = 



, 
and 3 = 



33 - 
and the area of M = - "-| 

/. Average cross-section 
A+B+4M 

6 
_ m 1 n l 






6s 



6a 2 



6s 



6s 



330 



MATHEMATICS FOR ENGINEERS 



Example 9. Find the average cross-section of ground with twisted 
surface, when the formation width is 20 ft. and the side-slopes 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 1-5)} = 30-3 ft. 
{10 + 18} = 26-05 ft. 



20 1-5 
2O 



For the section B 



10 



+ (6 x 1-5)} = 16-5 ft. 



- 10 - i 

* = , I o+ I . 5 ( I + 9) = 22-4 ft. 
Hence the average cross-section 

_ (30-3 x 26) + (16-5 x 22-4) + (46-8 x 48-4) 600 

9 
= 313-6 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 


2-6 


4-6 


6-6 


C 


4'i 


6-4 


8-7 


D 


4-0 


6-2 


8-4 


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 


13-7 


169-5 


C 


42-2 


15-7 


280 


D 


39-9 


15-55 


260 


E 


10 


10 


O 



Then the volume = {o + o + 4(169-5 + 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 half-widths 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 


67-8 


65H5 


30 



332 



MATHEMATICS FOR ENGINEERS 



Applying Simpson's rule 

Area of land required = {30 + 3 + 4 (55 '7 + 6 5'45) + (2 x 67-8)} 
= 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 cross-sections at intervals of, say, 40 ft. The area of 
each cross-section 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 5-083 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 1-633, 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 1-633 = 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 wedge-shaped 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 



- ZOOO-yds. 

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 side-slopes 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 side-slopes 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 half-way 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 side-slopes 
are ij to i, and the surface of the ground falls i in 8. Calculate the 
half-breadths 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 5-6 ft., 8-4 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 cross-section ; 
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 = 2-2* 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 = 2-2A 1 ' 75 , values of x ranging 
from o to 4. 

y 2*2#^* 7 ^ 
/. log y = log 2-2 + 1-75 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 


i-75 log * + log 2-2 


logy 


y 





00 


00 + -3424 


00 





5 


1-699 = '3 01 


- -527 + -3424 


1-8154 


654 


i-o 


o 


o + -3424 


3424 


2-2 


1 '5 


1761 


308 + -3424 


6504 


4-47 


2-O 


3010 


527 + -3424 


8694 


7-40 


2'5 


3979 


696 + -3424 


1-0384 


10-92 


3-0 


4771 


835 + -3424 


1-1774 


15-04 


3'5 


'544 1 


952 + -3424 


1-2944 


19-7 


4-0 


0021 


1-054 + "3424 


1-3964 


24-91 



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 = 2-2* 1 ' 75 and disregarding the factor 2-2 until the 
end 

logy = i -75 log x 
log (log y) = log 1-75 + 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 = ^1-75 ; an d supposing the value of y is required when x = 2-5. 

Set the index of the C scale level with 2-5 on the log log scale ; 
move the cursor until over the power, 1-75- on the C scale : then the 
reading on the log log scale (4-96) is the value of 2-5"". Multip icatwn 
by 2-2 for y = 2-2*, can be done with one setting of the rule afte 



338 



MATHEMATICS FOR ENGINEERS 



X 


% vn 


y = 2-2* 1 - 75 


2'5 


4-96 


10-92 



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 3-37, 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 = 1-41 : 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 = 1-41 

log p = log 2500 1-41 log v. 
The arrangement of the table is as follows : 



9 


logy 


log 2500 1-41 log v 


log/> 


P 


4 


602 


3.398- -847 


2-55 1 


356 


7 


845 


1-191 


2-207 


161 


10 


o 


1-41 


1-988 


97-3 


14 


146 


1-615 


1-783 


60-7 


18 


255 


1-770 


1-628 


42-5 


20 


301 


1-834 


1-564 


36-6 


24 


380 


1-945 


1-453 


28-4 


27 


43i 


2 -O2 


1-378 


23-9 


30 


477 


2-O82 


1-316 


20-7 



(b) Expansion of superheated steam; n = 1-3. 
Values of v and log v are as above ; and the table is completed as 
shown : 



log 2500 


1-3 logu 


log/) 


P 


3-398 - 


783 


2-615 


412 




1-097 


2-301 


200 




3 


2-098 


125 




49 


1-908 


80-9 




632 


1-766 


58-3 




691 


1-707 


5 '9 




794 


1-604 


40-2 




860 


1-538 


34'5 




920 


1-478 


30-1 



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 


92-6 


83-3 


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)1-111 

= I-(-4226) l ' ul 

Let x = (-4226) 1 ' 111 

then log* = i -i 1 1 x log -4226 = i-in x 1-6259 

= i -i 1 1 + -696 

= 1-585 

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 = 1-37. 

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 



2-25 
Subtracting 5 = B (^-5 

= B (-0625 --4444) 
5 = 382B 

or B -=-'3-x. 



342 



MATHEMATICS FOR ENGINEERS 



Also 
Hence 



5 = A + (-0625 x - 13-1) = A --818 

A = 5-818. 

p = ">-8i8 5. a = 5-818 -f- i 

f */ y ' 3 J yZ 



[Note that (p + q) = 11-636 = 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: 



1-5 



3-5 



Fig- 193- Curves of Radial and Hoop Stresses. 

right angles to the paper must elongate by an amount proportional to 
(P + q)- If the cross-section 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 


" 


13-1 


I3 .! 


-Sifi I3>1 * 


r* 


J r* ~ g 


5 OI r i P 


1-5 


2-25 


5-818 


11-636 





2-O 


4 


3-275 


9-093 


2'543 


2-5 


6-25 


2-095 


7-9I3 




3'O 


9 


1-455 


7-273 


4-363 


3'5 
4-0 


12-25 
16 


1-068 
818 


6-886 
6-636 


4-750 
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 {i-o3 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 1-03" = x; then log* = wlog 1-03 = -oi28 



also 



14-7 

^ = 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 


1-126 


126 


61-7 


8 


1024 


1-266 


266 


130 


JO 


128 


1-343 


343 


1 68 


12 


1536 


1-424 


424 


207-5 


16 


2048 


1-603 


603 


296 


21 


2688 


1-857 


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 half-major and half-minor 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 right-hand 
side becomes unity. 

Thus 



2O 



so that 



12 

and 

and 



a z = 20, and a = 4-472 
fc 2 = 12, and b = 3-464. 
Hence the range of x is from 4-472 to +4-472, 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 one-half of the ellipse, the table of values reads 



X 


X 2 


12 -6# 2 


y2 


y 


o 





12 O 


12 


3-464 


I 


I 


12 '6 


n-4 


3-38 


2 


4 


12 2-4 


9-6 


3-10 


3 


9 


12 - 5'4 


6-6 


2-55 


4 


16 


12 9-6 


2-4 


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 co-ordinates of P 
then x = MP =/ 2 sin 0, and y = OM = / x cos e 



X V 

y- = sin 6, f-ss 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+n-45; 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, -11-45- 

The equation is 

(*+7) 2 = 



5* 2 +7*+2-45-H'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(*+i-4*-i-8) 



= 5(*+7) 2 -'45 
= (*+7) 2 
which equation is of the form 4#Y = X a 

where 4* = T> Y = y-HMS. 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 = 4-9 

and & = Vg-6 = 3-1 

If a rectangle be constructed by verticals through x = 4-9 and 
+ 4-9, and horizontals through y = 3-1 and + 3-1, 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 = 48-2*2 
5 y z = 2*2-48 
y 2 = -4# 2 -9-6 



y = V'4# 2 9-6 

i. e., an expression is found for y in terms of x, 
The table of values reads : 



X 


x z 


4*2 9-6 


y z 


y 


4'9 


24 


9-6 9-6 


o 


o 


5'0 


25 


10 9-6 


4 


-632 


5'5 


30-3 


I2'I 9-6 


2-5 


1-58 


6-0 


36 


14-4 - 9-6 


4-8 


2-19 



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 co-ordinates, 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 = PM-ML = p-q 
PQ = PM+MQ = p+q 
PR h . V & 

PL 



= = 



also 



P-9 = \ 

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 = 5-j6x 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 = 4-2* 1 ' 83 + '3i.* 3 ' 47 . {x to range from -2 to 3-5.} 

4. Plot, from y = -5 to y = +'5, a curve to give values of C, 

when C = 1-69 (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 = 4-5. 

6. For a steam engine, if x = mean pressure (absolute) expressed 
as a percentage of the initial pressure (absolute), and y = cut-off 
expressed as a percentage of the stroke, then 

* = y(5-6o5-iogy)- 

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^T-rA |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 f-J 

If n = 1-41, 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 

2log---6i8 

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 = 32-2, 6 = angle of lap in radians. 

16. Aspinall gives as a rule for determining the resistance to motion 
of trains 



T 65-82 
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 three-stage air compressor with spray 
injection is given by 



where n = 1-2 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 


2-7183 


7-3891 


20-08 


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 


7-389 




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 = 3-5 : its slope is measured and found to 
be 33, and this is seen to be the value of y when x = 3-5. 

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 self-induction (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\ 








* \ 
^ \ 


C-SOe" 60 * 

? 






V 

\ 








> 

3 


V 






& 


X 






\ 


X 


^_ 






/OI25 


^ 


[ l^ 



Consl-r-? Seal* * 

O 77-uo Sco/e 025 -O375 "OS 

Fig. 202. " Dying-away " 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 
= 1-699 80 y. -4343' 
= 1-699- 34 -74/ 



t 


1-699- 3474' 


logC 


C 




0025 
005 
etc. 
05 


i -699 o 
1-699- 1-737 


1-699 
1-962 


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 3-14. (n radians = 180.) 



100 




4o 

5 1 i-S ~ -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 

= 1-6021 + -3 x -43430 
= 1-6021 + -13030 



6 


1-6021 + -1303$ 


logT 


T 


o 


1-6021 + o 


1-6021 


40 


5 


+ -0652 


I -6673 


46-5 


i-o 


+ -1303 


1-7324 


54 


i'5 


+ '*955 


1-7976 


62-8 


2-O 


+ -2606 


1-8627 


72-9 


2'5 


+ -3258 


1-9279 


84-7 


3-0 


+ -3909 


1-9930 


98-4 


3-14 


+ -4180 


2-O2OI 


104-7 



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 2-6 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 = = H-42. 

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 11-42 x X on construction scale 

or X = - = 2-19 
11-42 

so that no values of X need be taken beyond, say, 2-2. 



THE PLOTTING OF DIFFICULT CURVE EQUATIONS 359 

Taking values of X from o to 2-2, the values of cosh X are found 
from Table XI, thus : 



X 





'5 


i-o 


i'5 


2-O 


2-2 


cosh X = Y 


i 


1-128 


1-543 


2'352 


3-762 


4-568 



The curve is now plotted, as in Fig. 204, and then for unity on the 
construction scales 11-42 must be written, and the 25 ft. is marked off 
on either side of the centre line. The sag is read off as 39-3 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 = 3-5 x 15-9 = 55-6 Ibs. 

Exercises 38. On the plotting of Curves representing Exponential 

Functions. 

1. Plot, for values of x from -8 to 2-9, the curve y = 2e~*. Find 
its slope when x = 1-6. 

2. Plot the curve y = -250-$* from x = o to x = 15. 

3. Plot, from # = 5 to #=+3, the curve y = -021 x 1-62*. 

4. If C = C e~ at , C = 14-6, a = 410, and t ranges from o to -023, 
represent by a graph the change in C (the dying-away 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 2-18 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 = 48-7(1 e~i*). If R = -56, L = -008, plot a current-time 
(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 



-1-0 




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 = 4-5 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 1-571), 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 


i-i 


i'4 


I-57I 


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 4-5, i. e., i on the original scale now 

reads 4-5 ; 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 = 4-5 sin looirt 


o 





O 


o 


o 


OOI 


3H 


18 


309 


1-39 


OO2 


628 


36 


588 


2-645 


003 


9 4 2 


5< 


809 


3-64 


004 


1-256 


**l 


951 


4-275 


005 


I-57 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 right-hand 
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 = 4-5 sin iooirt by this means. 

Radius of circle = 4-5, 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 connecting-rod 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 = 4-5 sin (loont i-i) for a complete period. 



366 



MATHEMATICS FOR ENGINEERS 



Let us reduce the equation to a form with which we have already 

dealt; thus 

C = 4-5 sin (looirt i -i) 

C I, i-i \ 
= sm loon-! t 

4-5 \ I007T/ 

= sin ioo?r(/ -0035) 



j. e., C = sin loon-Ti = sin 1 

C 

where T l = t -0035, T = loou-T, and C = - 
4 5 

*$ r/vWw? 


3-6 


, I 


^ AX ' 5 




^ 


\, 















^0 


Mi 


s > 


/ 




\ 














1-8 & 


5.1 


L / 






\ 


Vu C=4-5 sin dooTfl- 1-1} 







O -o 


| 


i/ 








\ 












?2 '/ 


00* 


36 


O8 .-Ol < 


12 \< 


14- O 


16 


10 < 


2 Scale for I 


-OO35 -V 
9 /, 


2 ^ 


/O 


1-57 ' 

1 


>co/e jQ 


T' \ 
\ 


. 








/ 




/ 






OO5 


' 


\ 








/ 


27 / 


/ 

1 


g 










\ 






/ 




?'/ 


1 

I 














\ 




/ 




4-5 


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 


i-o 


60 


1-732 


80 


5-671 


85 


n-43 


90- 


+ 00 



X 


y = tan x 


90 + 


00 


95 


- n-43 


100 


- 5-67I 


120 


- 1-732 


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 89-99. Similarly 90+ would 
indicate 90-01, 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 screw-jack 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 Screw-jack. 

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 sun-dial, 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 one-half 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 


i-54 


7854 


707 


2-828 


960 


819 


3-276 


i-i 


891 


3-564 


i'4 


985 


3*94 


I-57I 


I'D 


4-0 



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 


i-i 


891 


446 


7 


i'4 


985 


493 


7854 


I-57I 


I-O 


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 2-96. Thus 
AB = 2-96 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(&#-f-c), 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*+ 2-4), showing two 
complete waves. 

Let y = yixy, where y t = e~* x and y, = sin (5*4-2-4). 

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* + 2-4) = 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 starting-line. 

y, = sin 5X, the period being -^ = 1-256 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 


I-57I 


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 2-4). 

Curve (i) yi = "". For one complete wave of curve (2) 
v = 1-256, 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 


1-2 


1-6 


2-O 


2-4 


2-8 


3-0 


\x or X 


o 


2 


4 


6 


8 


1-0 


1-2 


1-4 


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 = 1-2 ; 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 = 1-06, 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 

-,-}*+l-256 f -\x v fl'256 

'-~ = - ~ - = * 1 ' 256 = 3-5 (approx.) 



.'. logarithmic decrement = log 3-5 = 1-253. 

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 right-hand 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 
connecting-rod (or valve-rod) being many times as long as the crank. 
The travel of the valve is to be i|*. 

6. Plot the curve s = 2-83 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^ = 4-72 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- 

, ,. ,, connecting-rod 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 = 1-2 cos pt 1-8 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^ = 2-8 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 


4-48 


i-o 


3 


20-09 


1-5 


4'5 


90 


2 


6 


404 



X 


5* 2 + I 7 


y 


o 


0+17 


17 


5 


1-25+ 17 


18-25 


i 


5 +17 


22 


i'5 


11-25 + J 7 


28-25 


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 i-o and 1-5. 

Values of x are next taken between i-o and 1-5 ; thus : 



X 


3* = X 


Vi 


5 *2+I7 


y\ 


i-o 
I-I 


3-0 
3'3 


2O-I 
27-1 


5 + J 7 
6-05 + 17 


22 
23-05 



Therefore, the solution is evi- 
dently between i-o and i-i; 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 = 1-032, and therefore 
x = 1-032 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 




1-02 104. 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 7-45 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 64-4 x d* 
_ 40 x 5280 x 9-48 x 9-48 , 

350 x 64-4 
= 8 3 8/. 



o. 



from which d 6 4-19^ -35 

To solve this equation, we know that no negative values need be 
taken ; hence as a first approximation 

Let y = d 6 4-19^ -35 



Then 



d d- 


- 4-igd- -35 


y 


o o 

I I 

2 64 


- o - -35 
- 4-19 - '35 
- 8-38 - -35 


- '35 
-4-54 
55-27 



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 = 1-5, y = (i'5) 6 -(4'i9Xi-5)--35 = 4'76- 
Thus the required value of d is between i and 1-5. 
If d = 1-3, y = -98, and we see that the required value is between 
i -3 and 1-5. Plot the values of y for the values of d 1-3 and 1-5, 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 2-67*, 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 = 2-67 and r = ^ L 



. 6 1-25 . i'25 

sin = *. t. e., r = ., 

2 r ' sini 

2 
2-67 1-25 



1-5 

Fig. 218. Solution of Equation 
giving Diameter of Pipe. 

2-67 



or 



sin - = ^d = -480. 

2 2-67 



Making our first approxi- 
mation, taking 6 from o to 



6 


d_ 


6 

sin - 


4680 




2 


2 







O 





o 


5 


25 


247 


234 


i-o 


5 


479 


468 


1-5 


75 


682 


702 


2-O 


i-o 


842 


936 


2-5 


1-25 


945 


1-17 


3-0 


I- 5 


-998 


1-403 


3-14 


i-57 


i-o 


1-47 




0T* 

Fig. 219. Length of Metal in 
Corrugated Sheet. 



we see that the solution must lie between 6 = i-o and d = 1-5. 



When 6 = 1-2, 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 = i-o to 1-5; we note the point of intersection to 
be at 6 = 1-19. .*. 6 = 1-19 radians or 68-2. 



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 = 3-2, /= 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*= 1-8. 
(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 + 42-5/ + 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 + 4-47^ 

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 two-thirds full. 

Construction of PV (pressure-volume) and r<f> (temper- 
ature-entropy) 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 14-7 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 


7-16 


10-50 


20 


26-8 



THE PLOTTING OF DIFFICULT CURVE EQUATIONS 383 

Horizontals through 100 and 14-7 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 - 2-303(log 10 r-log lo4 93) 



T 


Iog 10 r-log, 493 


2-303 x column (2) = log, - 
493 


62O 
660 
7OO 

750 

780 


2-7924 2-6928 
2-8195 2-6928 
2-8451 2-6928 
2-8751 2-6928 
2-8921 2-6928 


0996 x 2-303 = -229 
1267 x 2-303 = -292 
1523 x 2-303 = -351 
1823 x 2-303 = -42 
1993 x 2-303 = -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 


I-6I5 


200 


660 


974 


1-475 


240 


700 


947 


1-353 


290 


750 


912 


1-215 


320 


780 


891 


1-142 



Hence the scale for entropy must be chosen so that the largest 
value may be shown, viz. 1-844, which is obtained by adding 1-615 

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 


14-7 


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 14-7 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 14-7 Ibs. 
pressure the volume of i Ib. weight of steam = 26-8 cu. ft., while at 
10 Ibs. pressure the volume of i Ib. weight is 38-4 cu. ft. Consequently 
if only 26-8 cu. ft. of steam are present at the lower pressure instead 

of the 38-4 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. 26-8 cu. ft. 
At 5 Ibs. pressure, volume of i Ib. weight = 72-4 cu. ft.; hence 

, 26-8 

wf = - ws = 'ZJiws 
72-4 

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. Temperature-entropy 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 = 4-44 
p = 13 when v = 26-8 
also log p + n log v = log C 

Thus log 100 + n log 4-44 = log C 

log 13 + n log 26-8 = log C 
or 2 + -6474% = log C 

and 1-1139 + 1-4281^ = 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. 

= 1-035 + 'I? where q is the initial dryness. 
For BC q = i and therefore n = 1-035 + ** = i'i35 
For EF q = -332 and therefore n = 1-035 + -0332 = 1-068. 
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 -fr-4 

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 1-0 1-2 I 4 1-6 1-8 

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 


fj-f.+Lj-Jt, 


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 6-9 % 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., = 93-1% = -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 = 14-7, r = 461 -f 62 = 523, 
and the volume (read off from the steam tables) = 13-14 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 13-14, hence 



*. e., 
so that the point B is fixed. 



14-7x13-14x1461., 
3-14 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 = 41-1 and 
Va _ 13-14, the value of the constant is 41-1 x 13-14 = 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 13-14 = 39'42- 

CD is vertical ; and also 

T D 



540 x 523 _ 
"f 1461 x 39-42 



so that the position of D is fixed. 



39 2 



MATHEMATICS FOR ENGINEERS 



The constant for the isothermal DA is 4-91 X 39-42 = 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 2-303 (Iog 10 r -Iog 10 523) 
= -39(log 10 r-log 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 r-10g lo5 2 3 


39 x column 


(2)=* 


7OO 


2-8451 - 


2-7185 


049 




IOOO 


3*0 - 


2-7185 


1096 




I2OO 


3-0792 - 


2-7185 


1405 




1461 


3-1647 - 


2-7185 


174 





The position of c is fixed, since the work done = r ^ 



and thus the distance be = 



log,, r 



774 
53-2 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 = 14-7, 

the volume = 13-14 cu. ft., and r = 523 : hence p v v = 14-7 x 13-14 = 193. 

The point A, at pressure 60, is on the isothermal through C; then 

/>Af A = P<,v = 193 

whence V A = -^ = 3-22 



40-- 




20 - 



10 



Fig. 226. PV Diagram for Joule Engine. 

For the adiabatic AD pv 1 '* 1 = K (say) 

so that K = 60 x 3-22 1 ' 41 

log K = log 60 + 1-41 log 3-22 = 1-7782 + (1-41 x -5079 

= 2-4943 
K = 312-1. 

Hence points on the line AD may be found from pv l ' tl = 312-1. 
The pressure at D = 14-7, and the volume = V 14^7" 8 '73 2 - 
Also- t^ = P^ 



3-22 x 60 



= 347 . 7 F. absolute. 



394 



MATHEMATICS FOR ENGINEERS 



For the adiabatic CB, the constant=/> u 1 " 41 =i4-7 x i3-i4 l ' 41 = 555-1 ; 
and thus this line may be drawn. 

Substituting the 
values of p, U A , T A , p B , 
1^(4-845) in the equa- 
tion 

P*v p t v A 7OO 

t- 5 - 2 = -, T B is found 

TB TA . 

to be 787-7 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. 



347-7 
787-7. 



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 14-7 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= 1-035 + ' J <7> <7 being the initial 
dryness. 

8. Draw constant-dry ness lines for dryness fractions of -2 and -3 
respectively. 

9. Calculate the dryness fraction for which the constant-dryness 



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 
horse-power, 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 
" straight-line " 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 straight-line 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 straight-line 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 straight-line 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 
straight-line 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 = 24-2 + b 
b = 9-2 

Y=. 3 o2X+(- 9 . 2 ) 
i. e., y = '$O2X 3 9-2. 

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 = - 9-2. .*. Y = -302X4- ( 9-2) 
and y -yzx* 9-2. 



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 


59-7 


58 


56 


53-8 


52-2 


Si'4 


A 


i-95 


2-46 


3 


3'44 


3-96 


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 


53-8 


52-2 


5i-4 


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 = 64-5 when 

Inserting values 64-5 = b + -456 . 



V = 52 when X = -15"! 
5 = -45/ 



(i) 



Subtracting 



12-5 = -y 

c = 41-7 
and by substitution in (i) 

64-5 = 6+1875 
whence b = 45-75 

= 



V = 45-75 + 



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 


2-46 


3 


3-44 


3-96 


4*99 


5-95 


7 


W = AV 


130-5 


155 


179-1 


199-4 


221-5 


269 


3ii 


359-8 



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 co-ordinates 
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 
4-78 
4-76 

4-74 

E 

4-72 
4-70 
4-ee 
4-66 




08 



62- 




4-4 4-6 4-8 5-O 5-2 5-4 

Fig. 233. Endurance Tests on Mild Steel Rods. 



5-6 



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 . . 


4-655I 


4-6767 


4-6875 


4-6902 


4-7168 


47324 


4 - 7540 


4-7686 


4-7793 


4-8116 


R = log R . . 


5-6232 


53489 


5-3166 


5-2700 


5-1093 


4-93I5 


4-8388 


4'6532 


4-6021 


4-3655 



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 = 4-68 when K = 5-36 

F = 4-76 when R = 4-74 

4'76 = A+4-7 4 n ' . . . . (i) 

4-68 = A+536 (2) 

08 = -62 



and 
Inserting values 



Subtracting 

whence 



n = 



08 
62 



= 129 



Substituting 129 in place of n in equation (i) 
4-76 = A + (- -129 x 4-74) 

A = 5-37 

but A = log a and therefore a = antilog of 5-37 = 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. 




4-OG5 4-77 



1-2 




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) . . / 


1-2 


2-09 


2-50 


2-925 


3-i8 


4-065 


4-46 


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 = 2-718, 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. 


3-7076 


3-7709 


3-8513 


3.9294 


2-0086 


2-0934 


2-1703 



In plotting the values of M it should be remembered that 3-7076 
is 3 + '7076, and that therefore the marking for 3-7076 on the 
vertical scale is above that for 3, to the extent of -7076 unit. 



2-2 _ 




5-8 . 



3-7 

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 = 2-13 when t = 65 \ 

and M = 3-73 when t= 115) 

we substitute these values in the equation M = A + Ct. 

Thus 2-13 = A+ 656 (i) 

3-73 = 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. 



2-13 = A+(- 1-195) 
A = 1-325 

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 well-defined 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 
thermo-electric 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 7-49- 

E = 150 + 7'49T OOI54T 2 . 

6ooo_ 




000 



/ooo _ 



Zoo 4oo 600 800 /ooo 
Fig. 236. Calibration of a Thermo-Electric Couple. 

To find an expression for T, solve the quadratic 

ooi 54 T 2 - 7'49T+ (150- E) = o. 
_ _ 7-49 V$6 '00616(150 E) 



Thus 



00308 



= 2430 325^/55-08 + -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 


1-064 


1-635 


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 62-4 

= -415 cu. ft./sec. 
Area of 4* diam. pipe = -0873 sq. ft. 

Velocity = _ = 4 . 75 ft./sec. 



Similarly, when Q = 2138, v = 6-54, and when Q = 2690, v 8-22. 
Now h = cv 3 ~ n d- n 

log A = logc+ (3 -n) \ogv-nlog_d f d = _'3333\ 

= log c + (3 ri) log v n x i -5228 \log d Y-5228J 
= 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 = 5-8 
h = -047 when v = 7-3 

and substituting in (i) we have the equations 

2-6721 = log c+ (3- ) x -8633 + - 4 7 7 n ...... ( 2 ) 

2-477i=logc+(3-) X 7634+-477W < , , , , , (3) 



THE DETERMINATION OF LAWS 

Subtracting -195 = (3 w) -0999 

= -3 -in 

In = -3 - -195 = -105 
n = 1-05. 

Substituting in (2) 

2-6721 = log c + (1-95 x -8633) + (-477 x 1-05) 

= log c + 1-682 + -501 
log c = 4-489 
.'. c = -0003083 

Hence h = -0003083 



411 



rfl-05 




OE 

4-5678 

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 


3-66 


3 


588 


6-14 


4 


735 


9-18 



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. 

6-24 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 = 8-5, h = -057/ 

XT v n f d = '5 \ 

Now- A = te^r I log d = 1-699 [ 

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/* + 1-672 + -361 
log p, = 4723 
p. = -0005284 

j,l-8 

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 


68-4 


A (amps) 


8-5 


5 


9-2 


8 


9-0 


10-5 


6-5 



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 


1-27 


1-83 


2-49 


3-25 



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 


1-312 


1-64 


1-968 


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 


1-2 


i'5 


3'4 


6-0 


9'4 


13-4 


18-1 


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 cast-iron pedestals 
for various diameter of shaft d : 



d(it.) . 


i 


i 


\ 


1 


i 


2 


W (Ibs.) 


18-005 


18-017 


18-138 


18-464 


19-1 


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 


2-4 


4-4 


6 


7-6 


1 1 -2 


I 5 -6 


20-4 


Time (sees.) 


2 


4 


6 


7 


8 


i-o 


1-2 


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 


24-9 


30-2 


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) 


10-4 


12-53 


15-41 


19-2 


23-67 


29-28 


35-58 


42-49 


5 I-2 



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 


307-5 


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 


12-5 


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 


8-04 


11-67 


14-43 


17-41 


I9-90 


h 


3-03 


. 6-n 


9-07 


12-21 


15-62 



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 


1-29 


1-76 


2-4 


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 . 


2-2 


1-8 


i'4 


i-i 


8 


6 


Quantity Q 


8-9 


8-03 


7-23 


6-4 


5'5 


4-85 



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 


1-87 


1-76 


1-67 


1-61 


1-49 


1-27 


i-ii 


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 


72-1 


71-7 


70-7 


70-4 


70-5 


69-7 


69-2 



80 


82 


84 


86 


88 


90 


92 


94 


67-8 


68-4 


67-7 


67-2 


67-2 


66-6 


66-3 


66-2 



Find the values of a and b in the equation R = ^ + 6. 

18. The following are results of a test on a loo-volt 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 


1-04 


i'ii 


1-2 


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 


1-645 


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 


16-8 


18-75 



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 


3-84 


4'5 


6-32 



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 


3-152 


5-036 


5*773 


6-382 



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. 


47-9 


52-9 


54 


55'5 


57-6 


62-5 


64-2 


70-3 


75 


79 



The law fitting these results has the form A = a+bV+cV 2 ; find 
this law. 

25. Given the equation R = a+&V-fcV 2 , and a table of the 
corresponding values of R and V, find the values of a, b and c. 



R 


o 


9-3 


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.) 


3-195 


3-23 


3-253 


3-261 


3-252 


3-228 


3-181 




7 


8 


9 


3-127 


3-059 


2-976 



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 


2-5 


3 


3-5 


4 


4'5 


y 


13-28 


15-04 


^SS 


I9-80 


23-11 


26 


30-5 


34'4 



In Exercises 28 to 30 the law is T = 20^. 
28. 



29. 



SO. 



T 


22-2 


24-66 


28-86 35-56 





524 


1-047 


1-833 2-880 






T 


23-4 


27-38 


34-66 


47H4 





524 


1-047 


1-833 


2-880 






T 


24-66 


30-42 


41-64 


63-26 


6 


524 


I-0 4 7 


1-833 


2-880 



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 


2-30 


3-10 


4 


4-92 


5-91 


7-2 


y 


33 


39-i 


50-3 


67-2 


85-6 


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.) 


I-I2 


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- 
2-7 --,L 




1-52 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 2-07 : then '^38 = 2-07. 




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 = 1-41 units of length 
(i. e., actual distances) : join ON and produce to cover the given 
range. Then for v = 5, w 1 ' 41 = 9-7, 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 horse-power transmitted 
by cast-iron 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 Cast-iron Wheels. 



200 



Again, if the pitch is ij*, what speed is necessary if 3^ H.P. is to 
be transmitted ? Draw a vertical through 3-5 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 Crank-shaft 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 6-12 ins. Again, if M = 20000, and T = 30000, then D = 3-57, 
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 62-4 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 62-4 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 
20-4^2, the maximum value being 20-4x36= 735; and draw the 
curve having the equation y = 20 -4^, a table for which is : 



60 x 62-4 



d 


o 


i 


2 


3 


4 


5 


6 


20-4^2 


o 


20 -4 


81-6 


183-6 


326 


5io 


735 



thus obtaining the curve OA. 



THE CONSTRUCTION OF PRACTICAL CHARTS 425 

In the right-hand 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 rivet-heads) 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 left-hand 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 right-hand 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 = 4-5. A perpendicular through this point cuts OX t in 
the point for which L+D = 38-5, but as D = 4-5, 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 cross-section 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 cross-section 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 


0-sin0 


Area 





I 


O 


o 





o 


o 


o 


2 


8 


3652' 


7344' 


1-287 


960 


327 


164 


4 


6 


538' 


io6i6' 


I-855 


960 


895 


448 


6 


4 


662 5 ' 


I 3 2 5 0' 


2-316 


733 


1-583 


792 


9 


i 


84 16' 


i68 3 2' 


2-94 


199 


2-741 


i-37i 


1-2 


2 


IOI32' 


203 4' 


3-545 


-'392 


3-837 


1-919 


**5 


~'5 


120 


240 


4-186 


-866 


5- 52 


2-5.46 


1-7 


7 


I3426' 


2685 2 ' 


4-70 


-'999 


5-699 


2-85 


2-0 


i 


1 80 


360 


6-284 


o 


6-284 


3-142 



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 left-hand portion, one for each separate value of 

the diameter. For diam. = 4 ft., ordinates of the curve would be (-] 





\ 







\ 



Valu 



46 



\ 



2-7 



Vo/i. 



456 



2 1-6 t-S -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. = 2-7 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. = 2-7 x (|) 2 
= -169 cu. ft. 

Again, if h = 1-72 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 45-6, so that the volume for 5 ft. diam. 



45-6 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 33-2 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 mid-vertical. 

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 = BN-NK .......... (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 mid-vertical. 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) 

(/iXAHJ-H^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 U-m,-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 mid-vertical) = " 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 mid-vertical must be so placed that ^ l = ^ 

w a/ 



4X2 



or 



i.e., MI = xm = x6" = 3-6* 

^ D 

or the mid- vertical is 3-6" distant from the axis of . 

Also the scale along the mid-vertical 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 mid-vertical, and indicate the marking for the 
constant term in the equation, viz., 30. 

If u = 7-5, v = o, and it will be noticed that if a line is drawn 
from 7-5 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 mid-vertical 
and not the position of the mid-vertical, let us deal with the equa- 
tion 4+6v = 18. Working with the same scales, join 4-5 on the 
axis of u to o on the axis of v, since if u = 4-5, v = o. This line 
passes through the point C x numbered 18 on the mid-vertical. 
To find the value of v when u = o, join o on the axis of u to 18 
on the mid-vertical 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 left-hand 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 

= (-7X-i) + (-005x2) 
= -07+ -oi = -08 

i.e., i" = -08 unit of *, along the mid -vertical. 
To fix the position of the mid-vertical 
m 2 _ al-i _ -7X-I _ _7 
m^ ~~ bl t ~ -005x2 ~~ i 
so that the mid-vertical 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 mid-vertical 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 mid-vertical 
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 Horse-power supplied to Electric Motor. 

Example 10. Construct a chart to give the horse-power 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 mid-vertical may be the sum of 
the scales along the outside axes, the mid-vertical 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, 9-86* (the length from index to index) would 
represent 2 units (i. e., log 100 log i), whilst if the C scale is used, 
9-86* 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 
1-0792, 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 2-0414 
to 2-3802, 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 



mid-vertical 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 


1-22 


I'S 1 


1-99 
















6 


7 


8 


9 


10 


ii 


12 


7782 


8451 


9031 


9542 


1-0 


1-0414 


1-0792 


477 


'544 


602 


653 


699 


740 


-778 


2-39 


2-72 


3-01 


3-27 


3'5 


37 


3-89 



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., 1-51*. 



For the V axis 



V 


no 


1 20 


T-JQ 


140 


T en 


160 












*5 U 




loeV . 


2*0414 


2*O7Q2 


2-TT3Q 


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 


1-347 


1-627 



170 


180 


190 


200 


2IO 


22O 


230 


240 


2*2304 


2-2553 


2*2788 


2-3010 


2*3222 


2*3424 


2*3617 


2*3802 


1890 


2139 


2374 


2596 


2808 


3010 


3203 


3388 


1-890 


2-139 


2-374 


2*596 


2*8o8 


3-oi 


3-203 


3-388 



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 mid-vertical 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 mid-vertical 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 mid-vertical as i, since 

H = 



440 MATHEMATICS FOR ENGINEERS 

For other graduations, tabulate thus : 



H 


I 


1-5 


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 


1-7782 


1-699 


I-602 


1-477 


T-30I 


2218 


-301 


398 


-523 


-699 


738 


i 


1-32 


-1-74 


-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 = 2-5, what is the voltage? Join 
2-5 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 mid-vertical 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 starting-point on the mid-vertical, join 10 on the axis 
of v to i on the axis of d, and call the point of intersection with the 
mid-vertical G. 

Now, Q = 34^ 2 w, and therefore for the particular values of d and v 
chosen, Q = -34Xi 2 x 10 = 3-4. 

G is therefore at the position to represent 3-4 Ibs. per sec. 

The table for the graduation of the mid-vertical 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 


1-176 


I-30I 


1-477 


1-602 


1-699 


1-903 


2 


644 


-769 


945 


1-07 


1-167 


i-37i 


1-468 


i -06 


1-26 


i'55 


i'75 


1-92 


2-25 


2- 4 I 



The fourth line is obtained by multiplying the third by 1-64 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 5-4; then the 
required diameter is 5-4*. 

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., 5-iT =fd 3 , and by 
taking logs throughout 

log 5-1 + 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, 1-37, 
1-41, 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 Cast-iron Gearing. 

Example 13. To construct a chart giving the number of teeth 
necessary for strength in ordinary cast-iron 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 Cast-iron Gearing. 



Then, since logT+ 3 log/> = log C, /, = 



4 



4 



4'93 



Also - = 



4-93^3 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 mid-vertical 3* from 
the axis of T. The last is simply a connecting-link 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 mid-vertical ; this must be marked 240, since 30x2' = 240. 

For chart (2), we already have the mid-vertical 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 mid-vertical, 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 


T-fiQQ 


I'778 














Diff . of logs 


o 


176 


3O I 


308 


477 














Distance from mark for 20 


o 


432 


74 


9 8 


I-I7 



70 


80 


90 


100 


150 


2OO 


I-845 


1-903 


1-954 


2 


2-176 


2-30I 


544 


602 


653 


699 


875 


I 


i'34 


1-48 


1-61 


1-72 


2-16 


2-46 



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 45-5 as zero. 



446 MATHEMATICS FOR ENGINEERS 

The table for the graduation of the H scale is : 



H 


4 V5 


<; 


10 


15 


20 


2<; 
















loeH 


1-658 


699 


i 


1-176 


1-301 


1-398 
















Diff. of logs. . . 


o 


-'959 


-658 


482 


-357 


260 


Distance from A 


o 


2-36 


1-62 


-1-19 


-88 


64 



3 


35 


40 


5 


60 


80 


100 


1-477 


1*544 


1-602 


1-699 


1-778 


I-903 


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 mid-vertical 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 mid-vertical at D. Join p = 2 to D, 
and produce to cut the axis of T in 18-6 : then the required value of 
T is 1 8-6. 

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 
I-2V -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 horse-power 

3/rr 
H at N revs, per min. (for a steel shaft) is given by d = 2-g\/ =^. 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 cast-iron 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 cast-iron pillars 
with rounded ends is 14-9. 

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 
14-7 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 
i9-hole 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 
33-hole circle would give this ratio. 

Therefore, 9 complete turns together with 27 holes on the 33-hole 
circle are required. 

An interesting example concerns the convergents of IT. 

Example 3. To 5 places of decimals the value of IT is 3-14159 : 
what fractions may be taken to represent this ? 

14159 
3-14159 = 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, 3-14286, 3-14151, 3-14159 -K and 3-14159 , 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*4-28 
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 32-30 = 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. 

~ 2-12 -i 

F = . ._ . - = = -2174. 



When x = 1-05, 
When x = 1-1, 



F = 



4-41+1-05-5 

2-2 2 



46 



4-84+I-I-5 

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 


3-97 


4-02 


4-05 


4'i 


y 


30-42 


31-04 


3I-52 


32-32 


32-80 


33-62 



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 3-9 to 3-94 (i. e., -04) 

the increase of y is -62, or the rate of increase is , i,e., 15-5, 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 3-97 : we might tabulate the 
differences of x and of y thus : 



Change in x. 


Change in y. 


Rate of change of y. 








80 


3-97 to 4-02, i. e., 


05 


80 


= 16 

'O5 


4-02 to 4-05, i. e., 


03 


48 


48 _ 

-2- = 16 
03 








82 


4-05 to 4-1, . 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 
3i-52+'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 


5-95 

] 

w 


4 -4-O5 ^ - 

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 8-09163. 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 
10-16. 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 I-2.3-4-5 

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 1-2-3 
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 
screw-cutting 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 two-term 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 28-42} = 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 
connecting-rod 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) = 9-993- 



468 MATHEMATICS FOR ENGINEERS 

Example 22. Find the value of 1005*. 

1005 = 1000 (i+ -005) 
1005* = iooo 4 (i+-oo5)* = iooo[i+(4X-oo5)] 

= 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 = 9-9 

For 98 differs from 100 by 2, hence its square root differs by 
\ of -2, *. e,, -i from 10. 

Similarly, (i-O3) 3 = 1-09 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) = 4-985. 

Example 24. If / = measured length of a base line in a survey 

L = correct or geodetic length, i. e., length at mean 

sea-level 
h = height above mean sea-level 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(m-i)(m-2) x 1 



1.2 W a . 1.2-3 



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 left-hand 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 right-hand 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, (i-y) = - y - y ~- __ ....... ( 2 ) 



Subtracting the two series, . e., taking (2) from (i) 
log* (i+y) - log, (i-y) = 



but log, (i+y) - log, (i-y) = log, - 



hence log, 2-j = 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 = 1-0986. 

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 connecting-rod \ 

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 4-044- 

/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(fl-sk} =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 [_6-l2] 
= 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(-8-2 4 ) + 3(-4-i6)}- 4 {3(8-6)-5(- 4 -i2) 

+ 3(~2-8)} 
+ i{ 3 (-8-2 4 )-6(- 4 -i2) + 3 (-8 4 8)} 

4 2{ 3 (- 4 -i6)-6(-2-8) + 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 -|- 5-4*} 3O{ -06} 
= ** 4- I4'i5* 3 + 57-28*2 4. 9.87* + 1-8 
and thus the equation is 

** 4 I4-I5* 3 4- 57-28*2 4 9-87* 4 1-8 = 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 

2-5 428 
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) 



2-33) 



-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 
1-5 2-5 2 

-3 75 



2. 



R 2 



when R = 3-6 
R! = 7-2 
R 2 = 710 
R 3 = 220 



4. 



2351 
3246 

8-4 3-5 

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 = 2-5, 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 1-6 -4 

Using the determinant method solve the equations in Nos. 7-10. 

7. n# 4y = 31 8. 8a b = 20 

2X + 37 =28 ioa + 76 = 71 

9. 4* 5y + 7* = 14 10. za + 36 + 5c = 4-5 

# 7 5.2=11 96 ioa = 39-3 



ANSWERS TO EXERCISES 



Exercises 1 


1. 1-2 


2. -0009 


3. 150 


4. 60 


5. 10-8 


6. 1-2 


7. -000225 


8. 28-5 


9. -009 


10. -052 


11. 93 


12. -161 


13. 2-7 


14. -9 


15. 22 


16. -56 


17. -031 


18. 7-4 


19. 2-5 


20. 6-3 




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. 1-41^ 8. 


a a~a~ 


9. 1-6 


10 


jj Pt V 2 Pl v l 


/Cf, 1 -" might be written \ 
1 #if i n x fj 1 -", etc. J 




~ i-n 




Exercises 


3 




1. 589-5 


2. 246-5 


3. -02138 


4. 57-03 


5. -0005423 


6. 116700 


7. 12-34 


8. 19-63 


9. -06664 


10. 1-924 


11. 244-4 


12. 29-14 


13. 1618 


14. -00009506 


15. 49-64 


16. 3-114 


17. -0001382 


18. -6874 


19. -02231 


20. -2777 


21. 3642 


22. 1-669 


23. -00001509 


24. -3352 


25. -001155 


26. 3-841 x io 6 


27. 20-17 


28. -2421 


29. 4-814 


30. 7-21 ix io- 14 


31. -07041 


32. 971-8 


33. io 


34. 85-8 


35. 5-418 


36. 32-75 


37. 220-4; 1369 


38. 3-29 


39. 1-98 


40. 5400 


41. 4-6 


42. 500-77 


43. 1-315 ; -591 


44. 356 


45. 36; 39-2 


46. -0935 


47. 22-21 


48. 400 


49. 80-072 


50. 1-434 


51. -506 


52. -479 


53. 5130 


54. 15-72 


55. 46,500 


56. 2 -8 1 


57. 3-5 x io 5 


58. 1-392 


59. 1-016 x io 6 


60. -284 


61. -128 


62. t = &, * - ft, B - x&, T - ft, - 7 




Exercises 


4 




1. Ibs. per sq. in. 2. 1-68 3. 


2-79 


4. 5-44 






f 1 


I2ttt^ 


5. 2-785 


6. Ibs. 7. 


Stress = - 
75 




g 


10. Incorrect as 


wri-H-on T-T P = ' 


Sir 1 


cu. ft. per sec. 


' 396000 


12. 746 









48o MATHEMATICS FOR ENGINEERS 



Exercises 


5 










1. 




i 7 a 


2. 


- 2-285 


3. 


3-62 




4- 7'54 




5. 


6 


93 


6. 


75 


7. 


7 d 
ioa 2 


b 


8 -+ 


t 


9. 


30-1 x 


io 10. 


() 


L = 


(T- 


9+T- 


-*li 


(b) 933 






11. 


1205 


12. 


i 


A Ti 


1 


13. 


5-01 




14. E = 2- 


5C 


15. 


I 


97 


16. 


R 


AJU 




126-5 






17. 1-67 




18. 


q 


= l{ 


^(A,-^) 


+ A 


.-A} 


19. 


n cells. 








20. 


22-85 


21. 


i5 


43 




22. 


d= 


3/I-274P/ 


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. 


2-4 


28. 


80 




29. 1493 




30. 


3 


3 


31. Ibs. 


ft. 


must be first brought to Ibs. ins 


. ; 6-46 ins 





32. 200,000 33. 2-57 34. 27-3 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) ^- - 2-4 J - 3K + 

38. r + ^ 39. 7oolbs./ a * 40. 1-37 

2 It> 

41. A's speed =10 m.p.h. ; B's speed = 15 m.p.h. 42. 6-31 

43. M = 221-4; 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=i-8 4. * = 

&= 2 n = 2-6 y = 



5. 
9. 


* = 1-24 6. # = 12 

y = 3-59 y = 20 

p = 9 10. E = 4- 

5 = 7 


7. * = 5 

y = 3 


8. a = 1-2 
6 = 5.7 
c= 4-8 
11. a = 1160, 
b = 69000 


~ 4 12. E = -I2W+4-6 


13. V = 42+^ 


14. 


1 = -953B+ 1-284: 4-81 


15. a = -916 


16. R = 4-714 






6 = -191 


a = -00707 


17 


T. = TTTC '?/ ' nfi'7 


1 R * 7 ' 


\/area 



ANSWERS TO EXERCISES 481 

._ ioi-6\/area . 

length + 97 20. E = -i6 4 -i + 7- 3 o 9 T+-ooo326T 

21. /= i4-8--ooooi38(* - 6o) a 22. /= i6-i--oooo26(/ - 6o) a 

23. W = 8-28 + 



n* . , i7oo/ total steam per hour \ 

24. w = 10-3 + -y- ( w = - - ^ - and must be first \ 

1 V ' calculated / 

25. 8610 of iron; 7000 of copper 26. = 99-8; K=i 

27. A = 120, B = 140, C = 160 28. m = 44-71, m t = 31-54 

29. 311-1 tons of saltpetre ; 388-9 tons of ginger. 

Exercises 7 

1. (X+22)(X- 4) 2. (*-!!)(* -8) 

3. (#-5)(* 21) 4. (2a-5& 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. 2-5 or 3 3. 4-13 or 1-13 

4. 4 or -i 5. 2-83 or --83 6. --278 -381; 

7. 3 8. 4-23 or -2-43 9. --125 1-219; 

10. 2-421 x io 5 or 2-379 x io 5 11. 2-75 or 457 
12. 28-98 or 1-03 13. 3-89 x io 4 or 2-97 x io* 

14. 57-5 or 56-5 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 13-3. (Divide all through by 75 x io 6 first) 

20. 155 or 32 21. -845 (2-845 h as no meaning here) 
22. v g$Vmi 23. 5 or 7-5 24. 80 or 90 

25. -2ii and -789 of span from one end 26. 13 or 10-42 
27. 1-475 28. 6-55 or 3-05 29. 100 ft. 

Exercises 9 

1. *= 7 or-3 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. 3-63 or 2-3 

y = 5/ b = 2-8 or 2) 7. -68* 8. 5 

9. 765 (the work is shortened by dividing by 25-6 straight away) 
10. 197 11. 27-4 12. I 13. ^ 14. 9-22 15 7-37 
16. m = -01277; n = -0026 17. 3-9 or 15-8 18. 6-763 

Exercises 10 

1. 91-1 nautical miles 2. 22-63* 3. 70-7 

5. 3 1 '-9* 6. (a) i in 12-5; (b) i in 12-46 7. 15 ft. 
8. AB= 12-37'; BC= 12-15"; AC = 21-63" 9. 72-4 

10. 26 11. 405 grey; 340 red 12. 480 13. 51-2 

14. 132 15. 2-36" 16. 6110 Ibs. 17. 8-66"; -307 
18. 5-11 tons 19. 65-80* 20. ^3 175. od. 

21. 7-24 sq. ins. 22. (b) 12-25; ( c ) 4'> (<0 5"86 

Exercises 11 

1. 3-44 sq. ins. 2. 10-2*; 52-1 sq. ins. 3. 6 sq. ins.; 2-4 ins. 

4. -2374 5. 12-82 sq. ins. 6. -536 

7. 58-5 sq.ft.; 18-07 ft. 8. 137-5 sq. ft. 9 - 3'6i ft. 

10. 301-5 sq. ft. 11. 13-25 sq. ins. 12. 47-7 sq. ins. 

13. 69-5 14. -52 amp. 15. 6-42 ins. 

16. 9-13 sq. ins. 

Exercises 12 

1. 22-4" 2. 29-1 ft. 3. 161 sq. ft 

4. 1-46; 2-14; 3-33; -35; -06 5. 2-51* 6. 1-571" 

7. 2-9 ft. 8. 18" 9. 1380 ft. per sec. 

10. 31-23* 11. 8-92* 12. 66 13. 970 14. 5935 

15. 27-13 sq. ins. 16. 3'-2 J* 17. -196 sq. in. 

18. 12080 Ibs. 19. 4816 Ibs. 20. 3-76 miles 21. 12-8* 

22. 16-65 sq. ins. 23. 1010 sq. ft. 24. -0294 ohm 25. 2-68 

26. 24-378*; 66 and 36 27. 214 sq. ins. 28. 60 ohms 



ANSWERS TO EXERCISES 483 

Exercises 13 

1- c i = 475*; h = -36* 2. c, = 44-72*, h = 20* 

3. 7 = 62-8"; A = 4'97* 4. 6-12*; 2-07 sq. ins. 

5. 9-06 sq. ins.; 60 6. i-ii'; 3-67' 

7. 2-n*; 3-33*; 1-29 sq. ins.; -56* 8. 13-1 

9. 1686 10. 29-8 sq. ins. 11. 6076 12. 5-8* 

13. B = A/fRT 14. 141 15. i '-26' 16. -375* 

Exercises 14 

1. 4-5"; 36^"; 8-48" 2. 1-8"; i"; 7-2:3"; -55" from base 
3. 2-48"; 12" 4. 473"; 8"; 1-8"; 25-14"; 25-91*; 25-57' 

5. 66-30*; 3'88*; 29-6"; 29-75*; 29-3* 6. 3340* 

7. 30-4 ft. tons; 3-38 ft. tons 8. 623 yds. 
9. 3*; ii* 10. 7000 

Exercises 15 
1. 689 cu. ft. 2. -28; 7-75 3. -3125* 4. 144 

5. -833 6. 5-13 cu. ft. 7. 6-91 ; 25900 

8. 41-1 9. 52600 10. 47-94 cu. ft.; 6715153. 
11. 23-85 sq.ft.; 40-17 sq.ft.; 19-3 cu. ft. 12. 70 

13. -53 (watt = volts x amps) 14. 31-85 Ibs. 15. 245 Ibs. 

16. 9 17. 852 sq. ft. 18. 14* 19. i 5 '-6" 

20. 508000 21. -0006 22. 1-74* 
23. 1-83 x io- 10 ohms 24. 137 25. 12-55* 

Exercises 16 
1. 593 cu. ins.; 321 sq. ins.; 13-55 ins. 2. 20-4* 

3. 200 sq. ft. 4. 13-4 cu. ins. 5. 1592 Ibs. 

6. 26-1 ft.; 581 sq. ft. 7. 4'O3"; 10-69" 8. 367 cu. ins. 

9. 14520 sq. ft. ; 70420 cu. ft. 10. 773'8; 967-4 Ibs. 
11. 173-8 sq. ins. ; 234-3 12. 213-5 cu. ft. 29,890 Ibs. 

13. 241 tons 14. -389" 

15. 155 cu. ins. ; 40-2 Ibs. 16. 559 cu. ins. ; 243 sq. ins. 

17. 4-63* 18. 2*; 5*; 6-12* 

19. 105 sq. ins.; 138 sq. ins. 20. U59sq. ins.; 2530 cu. ins. 

Exercises 17 

1. 160 sq. ins.; 191 cu. ins. 2. 8-3; 518 Ibs. 3. 8-8 cu. ins. 

4. 8610 Ibs. 5. -1033 6. 5-44 cms. 

7. 100-4 sq. ins. ; 151 cu. ins. 8. 4-2* 9. 636 10. 559 sq. ft. 
11. -0941* 12. 1440 sq. yds. 13. 1-082* 14. 15,520 sq. ft. 
15. 7-59* from vertex 16. 104-6 sq. ins. 17. 14-7; 2-45 

18. 53-51 acres 19. 77* 20. 16-1, 47-3, 27-6, 27-6 sq. ins. 

21. 406 Ibs. 22. 72 ft. 23. 1-3* 

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.; 98-2 cu. ins. 

5. Paraboloid = J cylinder 6. 90-2 cu. ins. 7. 2-02 Ibs. 



484 MATHEMATICS FOR ENGINEERS 

Exercises 19 



1. 6-44 Ibs. 


2. 2630 Ibs. 


3. 1278 Ibs. 


4. 960 Ibs. 


5. 272 Ibs. 


6. 372 


7. 1-16 tons 


8. 171 Ibs. 


9. i -84 Ibs. 


10. 761 Ibs. 


11. 45-5 Ibs. 


12. 10-25 tons 


13. 5-08 Ibs. 


14. 19-55 lbs - 


15. 28-2 Ibs. 


16. 93*5 Ibs. 


17. 3-59 Ibs. 


18. 258 Ibs. 


19. 6-47 Ibs. 





Exercises 20 

1. -7'; 74-5 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. 4-10 o'c. 

Exercises 21 

2. 250 4. Slope = -375; intercept = 2-375 8. 5-78" 

11. Slope = 2-5 if V is plotted along horizontal 

12. m= 1-8, n= 2-6 13. x = i, y o 14. x = 1-24, y 3-59 

15. x = -43, y = 2-33 16. x = 3-18, y = - 4-75 17-55 

Exercises 22 

1. -392 2. -31 3. 30-2 x io 6 4. 17-9 x io 6 

5. I = -8576 + 4-71 6. d l = -84^ -03 7. di = -95^ -07 
8. T== 51-70 +7 9. T = 35300 10. R = 78V + 86 

11. R = -772V+64-5 12. R = 2-5V + 75 13. R = i, a = -004 

14. R = 1-125, a = -00452 15. I = -00232* 96 

16. 32 17. 29-25 x io 6 

Exercises 23 

1. Vertex downward 2. Vertex upward 

7. Total weight = 50^ + 5/ 2 14. 6 or i 15. 2-67 or 3-5 
16. 1-44 or 7-65 17. Divide throughout by io 4 : 9-22 or -12 
18. 17-1 19. (a) 4-9"; (6) 5-5" 20. # = 3-64'; h =1-83' 
21. 7-7 air to i of gas 22. 5-5 23. e 55, effy. = -5 
24. 15-23 knots; 948 25. 40; -69 26. 2-1 

27. Assume some value for v : u = - ;i 28. 8-33; - 

29. 2" 30. 2 rows of 8 31. 6 34. i, 2 or 1-5 

35. -2, 2-25 or 3 36. 1-2, 4-6 or 1-6 

37. -2, -5 or -8 38. max- at x = 3, min"- at x = + 2 

39. x=-ziil 40. 1-475 41- 5'6 

Exercises 24 

1. 88-1 Ibs. 
4. 10-89 tons 

8. 2-3 pence 

12. 22 knots 

16. -01088' 



2. 2-37 


3. v = 66-3 Vr; 1195 


5. 10970 Ibs. 


6. 5" 7. -028 cm 


9. 13-75 ohms 


10. 246 11. 80 


13. 841 


14. 4-27 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, 


16-5, 24 6. 24 


7. & 


[ 155. 6d. ; 


377 los. od. 


8. 


10; 8160 


9. 15-5 ^ 


10. 592 ft. ; 


1 6 sees. 


11. 


3-15 p.m. 


12. 2-074 


13. 


53-33 




14. 


2, 3. 44 


15. 


835-2 . 


16. 5-5 


17. 


10-081 ; -821 


18. 


20 Ibs. 


19. 


7-52, 1 8, etc. 




20. 


a = 2 


b = o, c = 


= i; 


8040 


21. 


25 days 




22. 


983 in. 




23. 


4-284; 6-116; 8- 


734:12-48 


17-8; 


25H3 


36-31 ', 5*' 


84; 74-03; 105-5 


Exercises 26 


1. 


3-06; 1-1569; 2-192 


2. 


5-071; 4-28; 1-4 


3. o; -081 ; -194; 


285; -55 


775 


4. 


1-301 




5. 923 


6. 


09877 


7. -00005445 


8. 


4-612 




9. 264 


10. 


i -086 


11. 1546 




12. 


11-03 


13. -07784 


14. 


3-3 x io- 26 


15. 26, 560 


16. 


47-2 


17. 75-4 


18. 


370 


19. 38-2 




20. 


123; -109 21. -401 


22. 


1518 


23. -0336 




24. 


475 


25. -0391 


26. 


528 27. <t> w = 


= '39* $ = 


= I-796 


28. 


4000 


29. -638 


30. 


24-3 % 


31. -296 




32. 


325 


33. -103 


34. 


357 35. 4-48 


36. 


000334 


37. 38-4 


38. 


8-51 39. 4-44 


40. 


71-5 41. -65 


42. j= 1-875; T = 


= 445, * = 237 


43. 


Pv 1 ' 06 = 392 44. 1-47 


45. o or 7-28 46. 2-16 


47. 


o or 1-368 


48. -0955 


49. 


481-5 50. -033 


51. 


V = 222VH 


52. -01895 


53. 4-6 54. 5380 


55. 


i-48xio 8 


56. -605 


57. 


1-44 58. 7965 


59. 


61-6 


60. -2 


61. n 1-405; C = 


502 62. 81300 63. 47610 


64. 1-115 


65. 








66. 










Thl. Disch. 


Cd 






Thl. Disch. 


Cd 




53 


66 






120-5 




.731 




118 


672 






154-7 




708 












183 




727 




171 


658 






22O-4 




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. 8-41 71. F = -00277V 1 ' 9 



1. 

2. 
5. 



8746; 
9756; 
17*13' 
7. 3342' 
12. 52-8 
00305 
455 



17. 
21. 



Exercises 27 

"3443 > '5 2 3 I '2309 

9641; 1-6139; -2826 3. 3-82 4. -397 

6. 252 ; o (the case of wattless current) 

8. 143-3 9. 20 10. 28 6' 11. 

13. 1340 14. 6750 15. 3150 16. 

18. 61,200 ft. 19. 23-4 20. 268' 

22. -8923 23. 16850 24. 25!' 



181 

825 



486 MATHEMATICS FOR ENGINEERS 

Exercises 28 

1. a = 11-65"; 6 = 43-47" 2.6 = 8-72"; c = 14-83* 

3. a = 48-3"; 6 = 43-5* 4. a = 66-73"; c = 74-8; 

5. a = 22-14"; b = 16-08* 6. a = 57-66"; c = 92-63" 

7. a = 20-8"; 6=10-72" 8. 6037 ft. ; 2927 ft. 

9. 8 8' 10. 30-6 ft. 11. 78 12. 14 4i' 

13. 2" 14. (a) 14 16'; (6) 20 53' 15. 2-38" 

16. 2856' 17. AB = 5 oxAD 

18. A = o, 50 R.B. of BC = 35-5 S.E. 
B = 33-9, 77-8 R.B. of CD = 82-5 S.W. 
C = 74-6, 20-5 R.B. of DA = 23 N.W. 
D = 15-2, 13 Area = 2700 n' 

19. A = 10, 20 R.B. of BC = 67 S.W. 
B = 19-05, 14-8 R.B. of CA = 18 N.W. 
C = 12-58, 12-06 Area = 29 

20. 3 chns. 49 links 21. 235 i' 22. 73-6 ft. 
23. -2901" 24. -121* 

Exercises 29 



1. -8988 


-.6157 


-6157 


-4384 ; 


+.7880 ; 


7880 


-2-0503 


-.7813 


+ 7813 


2. --9903 


-6157 


+ 8480 


+1392 ; 


+ 7880 ; 


-5299 




7813 


i -6003 


3. --3289 


-3242 


-9953 


-9444 ; 


+ 9460 ; 


-0979 


+ 3482 


-3427 


+ 10-17 


4. --7570 


-8m 


7513 


6534 ; 


+5850 ; 


6600 


+ 1-1585 


-1-3865 


1-1383 


5. --9I35 


-6374 


9218 


4067 ; 


+ 7705 ; 


3877 


+2-2460 


+ 8273 


2-3772 


7. 124 36' 


8. 120 


55' 


10. i4946' or 2 


>ioi4' 





9171 

3987 

2-2998 

6. 7265; oo ; -1625 



9. 1 1 9 30' 

Exercises 30 

1. c = 4-89", A = 34 25 / , C = 67 5 ' 

2. A = 8o52', 6 = 59-46", c = 63-04" 

3. B = 4446' or I35i4', A = io824' or i756', a = 11-93* or 
3-87' 

4. B = 4042', a = 8-84", c = 8-25" 

5. A = 5343' or I26i7', B = 8oi7' or 743', 6 = 12-61 or 1-72 

6. a = 9-54 ft., B = 37 4 7', C = 685 7 ' 

7. A = 8o6', B = 48i8', C = 5i36' 

8. c = 21-97", B = 2i29', A = 283i' 

9. B = 4 i 4 2'or I38i8', C=io9i8'or I2 4 2 / , c = 8-31" or 1-94* 

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 = 64-62 or 6-311 



ANSWERS TO EXERCISES 487 

13. A = 863i', B = 4557', c = 16-11 

14. C = 43V, A = 8i2i', a = 47-28; area = 637 

15. 19-46 ft. 16. 8 30' 

17. 55. 87, 38 18. 18-75 Ibs.; 58 

19. OB = 1-224 chns., OC = -3236 chn., BE = -7673 chn., 
CF = i 667 chns. 

20. 5636' 21. Jib = 26-2 ft.; tie = 17-4 ft. 22. 1191 ft. 
23. 647^.; 374ft. 24. 53-2 ft. 25. 2083 ft. 

26. AB = 2983 links ; 767 links 27. BG = 74-96 chns.; CH = 74-14 
28. AB = 527-4"} 29. AB (horiz.) = 607-5 yds. \ 

DC = 475-3 I Hnks Diff. of level = 129-3 yds./ 

BC = 774-6 J 

30. r = 473-3' 1 31. AP = 983' (AC = nSo'U 

BE = BF = 126-7' h BP = 9 6 7' 

CF = CG = 473-3' J CP = 919' J 

32. I233f.p.s.; 29 36' 33. diam. = -506 34. 106-4; 93' 

35. 9-06 to i 36. 60-3 sq.ft.; 422 Ibs. 37. 7-46"; 10-65" 

38. 10-3"; 14-5" 39. 244 sq.ft. 

40. 2286; 2912 41. 17-25 sq. ins. 

Exercises 31 

1. cos A = -893, tan A = -504 

3. cos (A+B) = -442, sin (A-B) = -298 

4. tan (A+B) = 36-7, tan (A-B) = -536 

^ - tan a . 6 p = 

' i + n tan a ' 3 2nrpfji 

7. P = W (sin a + /i cos a) 8. 1-162 

9. 4-99 sin (5* + i) 10. 238-5 sin (50* -576) 



" R = -os'' 12 ' ' I89: I0 42 ' 

13. E = I2i-6sin (i2on-f + 1-022) 

Exercises 32 

1. cos 2 A = -566; tan 2 A = 1-455 

2. sin 2A = -7962; cos 2A = -605 3. |(i + cos 28) 

4. sin 2B = 2 cos B\/i-cos a B; -731 

A A 

5. sin - = -161 ; cos = -987; sin 3A = -8236 

2 2 

A 

6. cos 4A = -616; tan = -114 7. 2-5(1 - cos 4*) 

8. 7-85 (sin 189 - sin 131) = 7-85 (- 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. 25-91; 63-73 14. -333 or- 1-25 
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. 5-0018 3. 151 ft. 

5. E cosh x *J - + ^rV sinh x lj ^ 

8. -93 9. i852' 11. -86 12. 19-4 
15. 1-928 + 2-298; 

Exercises 35 

1. 318 2. 51-7 Ibs. per sq. in. 3. 38-35 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. 8-72 sq. ins. 11. 60-5 Ibs. per sq. in. 

Exercises 36 

1. 168750 cu. yds. 2. 8350 tons 3. 244000 tons 

4. 44920 sq. yds. 5. 5-21 x io 6 galls. 6. 40 ft.; 51-43 ft. 

7. 12020 cu. yds. 8. 96-6 ft.; 61-9 ft.; 11600 cu. yds. 

9. 26-5 ft.; 17-66 ft.; 33-5 ft.; 22-35 ft -.' 27-5 ft.; 18-33 ft.; 184, 
325, 202 sq. ft.; 1375 cu. yds. 

10. 28-25 anc * 43'3 ft- from the centre line 

Exercises 37 

14. The table of values would be arranged thus : 



1. 1-2552; 2-1293 
4. 40-54 ft. ; 156-6 ft. 

6. io45' 7. -00383 
14. 44-09 



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 


2-5 


3-21 


4-74 


6-9 


9-61 


14-6 



17. 2-9 



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 = 2-5; vertex is at (2-75, 8-42) 

20. 4-27 tons per sq. in. ; 23 

Exercises 38 
1. -404 6. 62-2 Ibs. 

8. Plot E = cosh#(# ranging from o to 500 Vgr) and then alter 
both scales 

Exercises 39 

1. Amplitudes: 8; -2; 51-8; -116; -91 

T> J If 27r /- 

Periods: ; ; -02; -0102; 36-9 
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. 4-58 5. -36 or 2-17 
6. 1-9 or -2-45 7. 2-79 8. -143 or -333 

9. 2-66 10. 5-37 (308) 11. / = -3 5 L 12. 7-9 
13. 4-49 rad. (257) 14. 10-42, 13 15. 5-523' 

16. 1-484' 17. 2-9 18. 6-005* 19. 796'34* 

20. 1-87. (Plot the curves y l = cosh x and y a = = sec x 

and note the point of intersection.) 21. 6-34 ft. 

Exercises 41 

2. -454 3. -3341 560 7. 1-043; i-o?? 

9. -22. [Hint. Let <*> = a + br ; also <t> = log -J- + g ^ 437 ~ T ^ ; and 

4OX T 

solve for q.] 

Exercises 42 
1. W = 47 + 6o-5A 2. /i= -2 + -oo4\/t/" 3. W = 3-28^ 

4. m = - 4 o 5 +^H 5. /=-oi 4 82 6. W=ri<P + i8 

7. S = lo-gi/ 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= 1-29 x io- 7 . 2 ' 46 14. Q = 6-nH- 48 15. y = 224\/H 

16. / = io/ a 17. a = 1205, b = 53-5 18. a = 1320, 6 = 54-4 

19 ' h = g.gx'io' 20< W = ' 87 ' C = 2 5 

21. a = 14-9, b = -58, c = -02 

22. E = -I + -OI32T -00000583T 2 

23. E = -15 + -OO795T 'Ooooo2iT a 

24. A = 144-6 2-7631; + -OI384U 2 

25. R = 160 I6-4V+-4V 2 26. v = 3-195 +'452D - 
27. a = 10, b = -277 28. -2 29. -3 30. -4 
31. y = i85- 26 * 82. Q = i-sH 2 ' 5 



490 MATHEMATICS FOR ENGINEERS 

Exercises 43 
3. -115*; 92-6 Ibs./n* 
5. Write equation 

log d- log 2-9= 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. 2-074 
16. (a) i -oi x io 15 ; (b) -9545; (c) -73; (d) 40-92 

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 2-449 T. x = 5, y 6 8. a = 1-5, & = 8 

9- * = 5, y = 4, z = 2 10. a = 1-2, & 5-7, c = 4-8 



MATHEMATICAL TABLES 



TABLE I. TRIGONOMETRICAL RATIOS 



Angle. 


Chord. 


Sine. 


Tangent. 


Co-tangent. 


Cosine. 








De- 
grees. 


Radians. 





o 


o 


o 








i 


1-414 


1-5708 


90 


i 

2 

3 

4 


0175 
0349 
0524 
0698 


017 
035 
052 
070 


0175 
0349 
0523 
0698 


0175 
0349 
0524 
0699 


57-2900 
28-6363 
19-0811 
14-3007 


9998 
9994 
9986 
9976 


1-402 
1-389 
1-377 
1-364 


1-5533 
1-5359 
1-5184 
1-5010 


89 
88 
87 
86 


5 


0873 


087 


0872 


0875 


11-4301 


9962 


I-35I 


I-4835 


85 


6 

8 
9 


1047 

1222 
1396 
1571 


105 

122 
140 
157 


1045 
1219 
1392 
1564 


1051 
1228 
1405 
1584 


9-5I44 
8-1443 
7-U54 
6-3138 


9945 
9925 
9903 
9877 


1-338 
1-325 
1-312 
1-299 


1-4661 
1-4486 
1-4312 
I-4I37 


84 
83 
82 
81 


10 


1745 


174 


1736 


1763 


5-67I3 


9848 


1-286 


1-3963 


80 


ii 

12 
13 
H 


1920 

2094 

2269 

2443 


192 
2O9 
226 
244 


1908 
2079 
2250 
2419 


1944 
2126 
2309 
2493 


5-I446 
4-7046 
4-33I5 
4-0108 


9816 
9781 
9744 
9703 


1-272 
1-259 
J-245 
1-231 


1-3788 
1-3614 

1-3439 
1-3265 


79 
78 
77 
76 


IS 


2618 


261 


2588 


2679 


3-7321 


9659 


1-218 


1-3090 


75 


16 
17 
18 
19 


2793 
2967 
3142 
3316 


278 
296 
313 

33 


2756 
2924 
3090 
3256 


2867 
3057 
3249 
3443 


3-4874 
3-2709 
3-0777 
2-9042 


9613 
9563 
95" 
9455 


1-204 
1-190 
1-176 
1-161 


1-2915 
1-2741 
1-2566 
1-2392 


74 
73 

72 

7i 


20 


3491 


347 


3420 


3640 


2-7475 


9397 


1-147 


1-2217 


70 


21 
22 
23 

24 


3665 
3840 
4014 
4189 


364 
382 
399 
416 


3584 
3746 
3907 
4067 


3839 
4040 
4245 
4452 


2-6051 
2-4751 
2-3559 
2-2460 


9336 
9272 
9205 
9135 


I-I33 
1-118 
1-104 
1-089 


1-2043 
1-1868 
1-1694 
1-1519 


69 
68 
67 
66 


25 


4363 


433 


4226 


4663 


2-1445 


9063 


1-075 


I-I345 


65 


26 
27 
28 
29 


4538 
4712 
4887 
5061 


450 
467 
484 
501 


4384 
454 
4695 
4848 


4877 
5095 
5317 
5543 


2-0503 
1-9626 
1-8807 
1-8040 


8988 
8910 
8829 
8746 


i -060 
1-045 
1-030 
1-015 


1-1170 

1-0821 
1-0647 


64 

3 
62 

61 


30 


5236 


518 


5000 


5774 


1-7321 


8660 


I-OOO 


1-0472 


60 


31 

32 

33 
34 


54" 
5585 
5760 
5934 


534 
551 
568 
585 


5150 
5299 
5446 
5592 


6009 
6249 
6494 
6745 


1-6643 
1-6003 
1-5399 
1-4826 


8572 
8480 
8387 
8290 


985 
970 
954 
939 


1-0297 
1-0123 
9948 
9774 


59 
58 
57 
56 


35 


6109 


601 


5736 


7002 


1-4281 


8192 


.923 


9599 


55 


36 
37 
38 
39 


6283 
6458 
6632 
6807 


618 
635 
651 
668 


5878 
6018 
6157 
6293 


7265 
7536 
7813 
8098 


I-3764 
1-3270 
1-2799 
1-2349 


8090 
7986 
7880 
7771 


908 
892 
877 
861 


9425 
9250 
9076 
8901 


54 
53 

52 

51 


40 


6981 


684 


6428 


8391 


1-1918 


7660 


845 


8727 


5 


41 
42 
43 
44 


7156 
733 
7505 
7679 


700 
717 
733 
749 


6561 
6691 
6820 
6947 


8693 
9004 
9325 
9657 


1-1504 
1-1106 
1-0724 
1-0355 


7547 
7431 
73M 
7193 


829 
813 
797 
781 


8552 
8378 
8203 
8029 


49 
48 
47 
46 


45 


7854 


765 


7071 


I'OOOO 


I-OOOO 


7071 


765 


7854 


45 








Cosine 


Co-tangent 


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 
SO-20 
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 




0-1 


1*6974 


7927 


8797 


9598 


0339 


1029 


1674 


2280 


2852 


3393 




02 


2-3906 


4393 


4859 


5303 


5729 


6i37 


6529 


6907 


7270 


7621 




0-3 
0-4 


7960 
1-0837 


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 
0-7 


4892 
6433 


5057 
6575 


5220 
6715 


538o 
6853 


5537 
6989 


5692 
7"3 


5845 
7256 


5995 
7386 


6i43 
7515 


6289 
S 43 


Mean Differences. 


0-8 


7769 


7893 


8015 


8i37 


8256 


8375 


3492 8007 


8722 


8835 








0-9 


8946 


9057 


9166 


9274 


938i 


9487 


9592 


9695 


9798 9899 


123 


456 


789 


1-0 


o-oooo 


0100 


0198 


0296 


0392 


0488 


0583 


0677 


0770 


0862 








1-1 


0953 


1044 


"33 


1222 


1310 


1398 


1484 


157 


1655 


1740 


9 17 26 


35 44 52 


61 70 78 


1-2 


1823 


1906 


1989 


2070 


2151 


2231 


2311 


2390 2469 


2546 


8 1624 


32 40 48 


56 64 72 


1-3 


2624 


2700 


2776 


2852 


2927 


3001 


3075 


31483221 


3293 


7 1522 


30 37 45 


52 59 67 


1-4 


3365 


3436 


3507 


3577 


3646 


371 


3784 


38533920 


3988 


71421 


283541 


48 55 62 


1-5 


455 


4121 


4187 


4253 


43i8 


4383 


4447 


45" 


4574 


4637 


6 13 19 


26 32 39 


45 52 S 8 


1-6 


4700 


4762 


4824 


4886 


4947 


5008 


5068 


5128 


5188 


5247 


6 12 18 


24 30 36 


42 48 55 


1-7 


53o6 


5365 


5423 


548i 


5539 


5596 


5653 


571 


5766 


5822 


6 ii 17 


24 29 34 


4 46 52 


1-8 


5878 


5933 


5988 6043 


6098 


6152 


620662596313 


6366 


5 " 16 


22 27 32 


38 43 49 


1-9 


6419 


6471 


6523 


6575 


6627 


6678 


672967806831 


6881 


5 II5 


2O 26 31 


36 41 46 


2-0 


6931 


6981 


73i 


7080 


7129 


7178 


7227 


7275 7324 


7372 


5 IOI 5 


20 24 29 


34 39 44 


2-1 


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 


2-3 


8329 


8372 


8416 


8459 


8502 


8544 


8587 


8629 8671 


8713 


4913 


17 21 26 


30 34 38 


2-4 


8755 


8796 


8838 


8879 


8920 


8961 


9002 


9042 9083 


9123 


4 8 12 


16 20 24 


29 33 37 


2-5 


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 


2-7 


9933 


9969 


0006 


0043 


0080 


0116 


0152 


0188 


0225 


0260 


47" 


15 l8 22 


26 29 33 


2-8 


1-0296 


0332 


0367 


0403 


0438 


0473 


0508 


0543 


0578 


0613 


4 7" 


14 18 21 


25 28 32 


2-9 


0647 


0682 


0716 


0750 


0784 


0818 


0852 


0886 


0919 


0953 


371 


14 17 20 


242731 


3-0 


0986 


1019 


1053 


1086 


1119 


"5i 


1184 


1217 


1249 


1282 


3 7 10 


13 16 20 


23 26 30 


3-1 


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 


3-3 


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 


3-5 


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 


3-7 


3083 


3110 


3137 


3164 


3i9i 


3218 


3244 


3271 


3297 


3324 


3 5 8 


ii 13 16 


19 21 24 


3-8 


3350 


3376 


343 


3429 


3455 


348i 


3507 


3533 


3558 


3584 


3 5 8 


10 13 16 


18 21 23 


3-9 


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 


4-1 


4110 


4134 


4 J 59 


4183 


4207 


4231 


4255 


4279 


433 


4327 


257 


IO 12 14 


17 19 22 


4-2 


4351 


4375 


4398 


4422 


4446 


4469 


4493 


45i6 


454 


4563 


257 


9 12 14 


16 19 21 


4-3 


4586 


4609 


4633 


4656 


4679 


4702 


4725 


4748 


477 


4793 


257 


9 " 14 


16 18 21 


4-4 


4816 


4839 


4861 


4884 


4907 


4929 


495i 


4974 


4996 


5i9 


247 


9 " 13 


16 18 20 


4-5 


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 


4-7 


5476 


5497 


55i8 


5539 


556o 


558i 


5602 


5623 


5644 


5665 


246 


8 ii 13 


15 1719 


4-8 


5686 


577 


5728 


5748 


57 6 9 


5790 


5810 


5831 


5851 


5872 


246 


8 10 12 


14 16 19 


4-9 


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 
7-2 
7-3 
7-4 
75 

76 

7-7 
7-8 
79 
80 

8-1 
8-2 
83 
84 
8-5 

8-6 
8-7 
8-8 
8-9 
90 

91 
92 
93 
94 
95 

96 
97 
98 
99 
10 







7048 70 

7228 7246 7263 7281 

745 7422 7440,7457 7475 7492 



1-6292 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 6|82 ' 

84068421 

8 503'8579 8594J86ioJ8625 8641 

871887338749)876418779 8795 8810 8825|884o|8856 

887188868901 
902190369051 



6 



8 



9 



Mean Differences. 



123456 789 



246 
246 

679o68o8|6827J6846|2 4 6 
246 
245 



75097527|7544;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 

2-3026 

KK 



0919 0931 0943 0956 0968 0980 0992 



1041 1054 

1163 1175 

1282 

1401 1412 1424 

151815291541 



1066 IO78|IO9O 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 23IIJ2322|2332 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' 
0-1 


12' 
02 


18' 
0-3 


24' 
0-4 


30' 
05 


38' 
06 


42' 
0-7 


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 


0-0 


0-1 


02 


0-3 


0-4 


0-5 


0-6 


0-7 


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 


I-OOO 


I-OOO 


I-OOO 


I-OOO 


I-OOO 


O O O O O 


90 


I-OOO 























KK2 



5oo 



MATHEMATICAL TABLES 
TABLE VI. NATURAL COSINES 






0' 


6' 


12' 


18' 


24' 


30' 


36' 


42' 


48' 


54' 


Mean Differences. 


9 

q 


0-0 


0-1 


02 


0-3 


0-4 


0-5 


06 


0-7 


08 


09 


1' 2' 3' 4' 5' 





I-OOO 


ooo 


I'OOO 


I-OOO 


ooo 


I-OOO 


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 


9-593 


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' 
0-1 


12' 
0-2 


18' 
0-3 


24' 
0-4 


30' 
0-5 


36' 
0-6 


42' 
0-7 


48' 
0-8 


54' 
0-9 


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' 
0-1 


12' 
02 


18' 
0-3 


24' 
0-4 


30' 
0-5 


36' 
0-6 


42' 
0-7 


48' 
0-8 


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' 
0-1 


12' 
02 


18' 
03 


24' 
0-4 


30' 
05 


36' 
0-6 


42' 
0-7 


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 


1-6643 


6709 


6775 


6842 


6909 


6977 


745 


7"3 


7182 


7251 


ii 23 34 45 56 


60 


1-7321 


7391 


7461 


7532 


7603 


7675 


7747 


7820 


7893 


7966 


12 24 36 48 60 


61 


1-8040 


8115 


8190 


8265 


8341 


8418 


8495 


8572 


8650 


8728 


13 26 38 51 64 


62 


1-8807 


8887 


8967 


9047 


9128 


9210 


9292 


9375 


9458 


9542 


14 2 7 41 55 68 


63 


1-9626 


9711 


9797 


9883 


997 


0057 


oi45 


0233 


0323 


0413 


15 29 44 58 73 


64 


2-0503 


0594 


0686 


0778 


0872 


0965 


1060 


"55 


1251 


1348 


16 31 47 63 78 


65 


2-1445 


1543 


1642 


1742 


1842 


1943 


2045 


2148 


2251 


2355 


17 34 51 68 85 


66 


2-2460 


2566 


2673 


2781 


2889 


2998 


3109 


3220 


3332 


3445 


18 37 55 73 92 


67 


2-3559 


3673 


3789 


3906 


4023 


4142 


4262 


4383 


4504 


4627 


20 40 60 79 99 


68 


2-475I 


4876 


5002 


5129 


5257 


5386 


5517 


5649 


5782 


59i6 


22 43 65 87 108 


69 


2-6051 


6187 


6325 


6464 


6605 


6746 


6889 


7034 


7179 


2326 


24 47 7 1 95"9 


70 


2-7475 


7625 


7776 


7929 


8083 


8239 


8397 


8556 


8716 


878 


26 52 78 104 131 


71 


2-9042 


9208 


9375 


9544 


97H 


9887 


0061 


0237 


0415 


0595 


29 58 87116145 


72 


3-0777 


0961 


1146 


1334 


1524 


1716 


1910 


2106 


2305 


2506 


32 64 96 129 161 


73 


3-2709 


2914 


3122 


3332 


3544 


3759 


3977 


4197 


4420 


4646 


36 72 108 144 180 


74 
75 


3-4 8 74 
3-732I 


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 


4-0108 


0408 


0713 


1022 


1335 


1653 


1976 


2303 


2635 


2972 




77 


4'33i5 


3662 


4 OI 5 


4374 


4737 


5107 


5483 


5864 


6252 


6646 




78 


4-7046 


7453 


7867 


8288 


8716 


9152 


9594 


0045 


6504 


0970 




79 


5 -I 44 6 


1929 


2422 


2924 


3435 


3955 


4486 


5026 


5578 


6140 




80 


5-67I3 


7297 


7894 


8502 


9124 


9758 


0405 


To66 


1742 


2432 




81 


6-3138 


3859 


4596 


535 


6122 


6912 


7720 


8548 


9395 


0264 


Mean differences are 


82 


7-II54 


2066 


3002 


3962 


4947 


5958 


6996 


8062 


9158 


6285 


no longer suffici- 
ently accurate, 


83 


8-1443 


2636 


3863 


5126 


6427 


7769 


9152 


0579 


2052 


3572 


since the ilitler- 


84 
85 


9-5M 
n-43 


9-677 
n-66 


9-845 
11-91 


10-02 

12-16 


IO-2O 

12-43 


10-39 
12-71 


10-58 
13-00 


10-78 
13-3 


10-99 
13-62 


II-2O 
I3-95 


ences vary cou- 
siderably along 
each line. 


86 


14-30 


14-67 


15-06 


I5-46 


15-89 


16-35 


16-83 


I7-34 


17-89 


18-46 




87 


19-08 


19-74 


20-45 


21-20 


22-02 


22-90 


23-86 


24-90 


26-03 


27-27 




88 


28-64 


30-14 


31-82 


33-69 


35-80 


38-19 


40-92 


44-7 


47-74 


52-08 




89 


57- 2 9 


63-66 


71-62 


81-85 


95-49 


114-6 


143-2 


191-0 


286-5 


573-0 




90 


oo 























504 



MATHEMATICAL TABLES 
TABLE VIII. LOGARITHMIC SINES. 



V 


0' 


6' 


12' 


18' 


24' 


30' 


36' 


42' 


48' 


54' 


Mean Differences. 


8> 

6 


0-0 


0-1 


02 


0-3 


0-4 


0-5 


06 


0-7 


0-8 


0-9 


1' 2' 3' 4' 5' 





oo 


3-2419 


5429 


7190 


8439 


9408 


O2OO 


0870 


1450 


1961 




1 


2-2419 


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 


1-0192 


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 


0-3 


0-4 


0-5 


0-6 


0-7 


08 


0-9 


1' 2' 3' 4' 5' 


45 

46 
47 
48 
49 


1-8495 
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 


0-0 


0-1 


0-2 


0-3 


0-4 


0-5 


0-6 


0-7 


0-8 


0-9 


1' 2' 3' 4' 5' 




5*OOOO 


oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


9999 


O O 


1 


1-9999 


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' 
0-1 


12' 
0-2 


18' 
0-3 


24' 
0-4 


30' 
05 


36' 
06 


42' 
0-7 


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 


2-9403 


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' 
0-1 


12' 
0-2 


18' 
0-3 


24' 
0-4 


30' 
0-5 


36' 
0-6 


42' 
0-7 


48' 
0-8 


54' 
09 


Mean Differences. 


1' 2' 3' 4' 5' 





-00 


3-2419 


5429 


7190 


8439 


9409 


02 oo 


0870 


M50 


1962 




1 


2-2419 


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 


I-02I6 


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' 
0-1 


12' 
0-2 


18' 
0-3 


24' 
04 


30' 
05 


36' 
0-6 


42' 
0-7 


48' 
0-8 


54' 

0-9 


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 


1-0580 


0669 


0759 


0850 


0944 


1040 


1138 


1238 


I34i 


1446 


16 32 48 64 81 


86 


I-I554 


1664 


1777 


1893 


2012 


2135 


2261 


2391 


2525 


2663 




87 1-2806 


2954 


3106 


3264 


3429 


3599 


3777 


3962 


4155 


4357 




88 1-4569 


4792 


5027 


5275 


5539 


5819 


6119 


6441 


6789 


7167 




89 1-7581 


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 


1-1052 


9048 


1-0050 


1002 


0997 


2 


1-2214 


8187 


I-020I 


2013 


1974 


3 


1-3499 


7408 


1-0-153 


3045 


2913 


4 


1-4918 


6703 


1-0811 


4108 


3799 


5 


1-6487 


6065 


1-1276 


5211 


4621 


6 


1-8221 


-5488 


1-1855 


6367 


537 


7 


2-0138 


4966 


1-2552 


7586 


6044 


8 


2-2255 


4493 


1-3374 


8881 


6640 


9 


2-4596 


4066 


I-433I 


I-0265 


7163 


10 


2-7183 


3679 


I-543I 


I-I752 


7616 


11 


3-0042 


3329 


1-6685 


1-3357 


8005 


1-2 


3-3201 


3012 


1-8107 


1-5095 


8337 


13 


3-6693 


2725 


1-9709 


1-6984 


8617 


1-4 


4-0552 


2466 


2-1509 


1-9043 


8854 


15 


4-4817 


2231 


2-3524 


2-1293 


9051 


16 


4-9530 


2019 


2-5775 


2-3756 


9217 


1-7 


5-4739 


1827 


2-8283 


2-6456 


9354 


1-8 


6-0496 


1653 


3-1075 


2-9422 


9468 


19 


6-6859 


1496 


3-4 J 77 


3-2682 


9563 


20 


7-3891 


1353 


3-7622 


3-6269 


9640 


2-1 


8-1662 


1225 


4-M43 


4-0219 


9704 


22 


9-0251 


1108 


4-5679 


4-457I 


9758 


23 


9-9742 


1003 


5-0372 


4-937 


9801 


24 


11-0232 


0907 


5-557 


5-4662 


9837 


25 


12-1825 


0821 


6-1323 


6-0502 


9866 


26 


13-4638 


0743 


6-7690 


6-6947 


9890 


2-7 


14-8797 


0672 


7-4735 


7-4063 


9910 


2-8 


16-4446 


0608 


8-2527 


8-1919 


9926 


29 


18-1741 


0550 


9-1146 


9-0596 


9940 


30 


20-0855 


0498 


10-068 


10-018 




31 


22-1980 


0450 


11*122 


11-076 


9959 


32 


24-5325 


0408 


12-287 


12-246 


9967 


33 


27-1126 


0369 


13-575 


I3-538 


9973 


3-4 


29-9641 


334 


14-999 


14-965 


9978 


35 


33-II55 


0302 


16-573 


16-543 


9982 


36 


36-5982 


0273 


I8-3I3 


18-285 


9985 


3-7 


40-4473 


0247 


20-236 


20-211 


9988 


3-8 


44-7012 


0224 


22-362 


22-339 


9990 


39 


.49-4024 


O2O2 


24-711 


24-691 


9992 


4-0 


54-5982 


0183 


27-308 


27-290 


9993 


4-1 


60-3403 


Ol66 


30-178 


30-I62 


9995 


4-2 


66-6863 


0150 


33-351 


33-336 


9996 


43 


73-6998 


0136 




36-843 


9996 


4-4 


81-4509 


0123 


40-732 


40-719 


9997 


4-5 


90-0171 


OIII 


45-OI4 


45-003 


9997 


4-6 


99-4843 


OIOI 


49-747 


49-737 


9998 


4-7 


109-9472 


0091 






9998 


4-8 


121-5104 


0082 


60-759 


60-751 


9999 


4-9 
50 


134-2898 
148-4132 


0074 

0067 


67-149 

74-210 


67-141 
74-203 


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 

mid-ordinate rule, 308 

planimeter, 300 

Simpson's rule, 310 

trapezoidal rule, 307 
Asymptotes of hyperbola, 108. 349 



Bearing, Reduced, 244 
, Whole-circle, 245 
Binomial theorem, 463 
Boussinesq's rule for the perimeter of 
an ellipse, 105 



Calculation of co-ordinates 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 
Co-ordinates, 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, 3-76 

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 right-angled 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 
Log-log scale on the slide rule, 337 

If 

Mantissa of logarithm, 14 
Maximum and minimum values, 183 
Mensuration, 79 et seq. 
Mid-ordinate 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 drawing-board, 176 

Quadratic expressions, Plotting of, 1 74 
Quadrilateral, Area of irregular, 87 
, Centroid of, 130 

R 

Radian, 99 

Ratios of multiple and sub-multiple 
angles, 279 

, Trigonometric, 232 

Rectangle, Area of, 79 
Reduced bearing, 244 
Remainder theorem, 55 
Reservoir, Volume of, 332 
Rhombus, area of, 85 
Right-angled 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 
, Log-log 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 
Sub-normal 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 

, Right-angled, relation between 

sides of, 80 



Triangles, Solution of, 255 et seq. 
Trigonometric equations, 287 

ratios, 232 

from slide rule, 242 

from tables, 234 

Turning-points 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 wedge-shaped excavation, 321 

of zone of sphere, 120 

W 

Wedge-shaped 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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A Treatise on Hand -Lettering for Engineers, 
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Mechanical Drawing 

By the late WILFRID J. LINEHAM, Hons. B Sc., M Inst.C.E., 
M.I.Mech.E., M.I.E.E., Author of "A Text-book of Mechanical 
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Arithmetic for Engineers 

By CHARLES B. CLAPHAM, lions. B Sc. (Eng.), Lecturer in 
Engineering and Elementary Mathematics at the University 
of London, Goldsmiths' College. Second Edition. Demy 8vo. 
7/6 net. 

A Treatise on Mechanical Testing 

By R. G. BATSON, M.Inst.C.E., M.I.Mech.E., and J. H. HYDE, 
A.M. Inst.C.E., M.I.A.E. 

Volume I. Testing of Materials. Demy 8vo. i8/- net. 
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Demy 8vo. i8/- net. 

Theory and Practice of Aeroplane Design 

By S. T. G. ANDREWS, B.Sc. (Eng.), and S. F. BENSON, B.Sc. 
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Geometry for Builders and Architects 

By J. E. PAYNTER. Demy 8vo. 421 pages, 376 figures, and 
numerous examples and exercises with answers. i5/- net. 

Metric System for Engineers 

By CHARLES B. CLAPHAM, B.Sc. (Eng.), Author of " Arithmetic 
for Engineers." Demy 8vo. 200 pages fully illustrated ; 
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Electrical Engineering Testing 

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Text -Book of Mechanical Engineering 

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The eleventh edition of this standard treatise contains 1244 
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The Strength of Materials 

By EWART S. ANDREWS, B.Sc. Eng. (Lond.). A Text-Book for 
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Elementary Strength of Materials 

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The Theory and Design of Structures 

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Slide Rules : and How to Use Them 

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Concrete Work 

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Deals extensively with the following subjects : Reading blue prints, 
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materials, selection and disposition of equipment, organization of gang, 
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concrete and protection of work, etc. 451 pages, 224 figures, 20 full- 
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Graphical and Mechanical Computation 

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Alternating -Current Electricity 

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