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Full text of "Arithmetic For Engineers Fourth Edition"

TIGHT BINDING BOOK 



CO CO 

g]<OU 166005 g 



ii4 ARITHMETIC FOR ENGINEERS 



OSMANIA UNIVERSITY LIBRARY 

Call No. T ! 1 Accession No. I ^f O I ^ 

Author 

Title 



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last marked below. 



ARITHMETIC FOR ENGINEERS 



PLEASE SEND FOR DETAILED PROSPECTUSES 
Mathematics for Engineers. By w. N. ROSE, 

B.Sc., Eng. (Lond.), late Lecturer in Engineering Mathe- 
matics at the University of London Goldsmiths' College. 

The two volumes of " Mathematics for Engineers " 
form a most comprehensive and practical treatise on the 
subject, and will prove a valuable reference work, em- 
bracing all the mathematics needed by engineers in their 
practice, and by students in all branches of engineering. 

PART I. 5th Edition. 520 pages. Demy 8vo. Price 
los. 6d. net. 

PART II. 2nd Edition. 432 pages. Price I 3*. 6d. net. 

Metric System for Engineers. By CHARLES 

B. CLAPHAM, Hon. B.Sc. Eng. (Lond.). 

This volume contains a complete discussion of the 
Metric System, and methods of conversion, so that the 
relation between the English and Metric System of 
measurements in industry can readily be understood. 

200 pages, fully illustrated, numerous tables and folding 
charts. Demy 8vo. Price I ^s. 6d. net. 

Line Charts for Engineers. By w. N. ROSE, 
B.Sc. 

Covers the Theory, Construction and use of all forms 
of Line Charts, paying special attention to the widely 
used Nomographs or Alignment Charts, and will prove 
an invaluable aid to all engineers and draughtsmen. 
1 08 pages. Demy 8vo. Price 6s. net. 

CHAPMAN & HALL, LTD. 

II, HENRIETTA ST., COVENT GARDEN, W.C. 2 



The Directly-Useful 



D.U. 



Technical Series 



Founded by the late WILFRID J. LINEHAM, B.Sc., M.Inst.C.E. 
General Editor : JOHN L. BALE. 



Arithmetic for Engineers 

INCLUDING 

SIMPLE ALGEBRA, MENSURATION, LOGARITHMS, 

GRAPHS, TRIGONOMETRY, MfD 

THE SLIDE RULE 



WITH AN APPENDIX ON 

VERNIERS AND MICROMETERS 



BY 

CHARLES B. CLAPHAM 

B.Sc. (IIoNS.) RNG., JT.M.l.hlECH.El 

Lecturer and Demonstrator in the Mechanical Engineering 

Dept., Finsbury Technical College; Author of 

" Metric System for Engineers" 



FOURTH EDITION 



LONDON 
CHAPMAN & HALL, LTD. 

1925 



FRINt'Fn IN GKK\T 
Kll'HAIlP Cl.AV & t-l) 
LiUNt.AV, Sri-J 



BRITAIN I'.Y 
N^, LjMllH), 



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 

THE following work is an endeavour to treat the elementary 
portions of what is usually called " Practical Mathematics " in a 
thorough and practical manner, suitable for elementary students 
of technical schools and for home study. Although a great many 
books on the same subject already exist, the Author has been unable 
to find one dealing with the necessary matter in sufficient detail 
and with a sufficient amount of engineering application to meet 
the needs of his own students. For private study, too, he is of 
the opinion that the existing text-books treat the elementary 
matter in a manner too cursory to really fulfil their object. The 
treatment in the following pages, therefore, has been developed 
from his own lecture notes and class instruction on the subject, 
and mimerous diagrams have been introduced to assist in making 
the work clear. 

Many will think, no doubt, that the title " Arithmetic for 
Engineers " is not sufficiently comprehensive for the matter con- 
tained, but it was found impossible to frame a short original title 
to adequately describe the contents. An examination of the 
examples and exercises should show that the whole of the matter 
is directly useful ; all purely academic work such as Highest 
Common Factor, Recurring Decimals and the like being discarded. 
Wherever possible the examples are truly practical, i. e., are problems 
actually met with in the Drawing Office, Workshop and Laboratory, 
while the data are of correct dimensions. Such examples should 
stimulate the reader's interest in the mathematical work and show 
the applications of the principles to practice ; at the same time 
a little general engineering knowledge will be gained. Even at 
the risk of becoming verbose, all matter is treated at length, every 
principle being followed by worked qxamples. Where slight varia- 
tions of the problem may cause difficulty to the beginner, or where 
special precautions have to be observed on certain points, several 
illustrative examples are given. A case in point is that of the 
extraction of a square root on pp. 102 et seq. For the private student 
this is particularly necessary, as he is often unable to obtain guidance 
in working a new problem which presents features slightly different 
from those in the example worked to illustrate the principle. 

Stress has been laid, in Chap. II, onllimiting the number of 
figures to be given in a result, as class-room experience shows that 
many students will persist in using, and stating in results, far more 
figures than are necessary from the practical point of view, where 



viii AUTHOR'S PREFACE 

an accuracy of I per cent, is often as mfcch as is desired. The method 
of " approximating for a result " shown on pp. 58 et seq. is, it is 
believed, not generally known. In Chap. Ill positive and negative 
quantities, which are of importance in logarithms and higher* work, 
are given special attention, as they often present difficulty, to the 
beginner. The simple equation is treated very metho'dically in 
Chap. IV, with concrete illustrations of the earlier examples, and 
is followed immediately by a similar treatment of the literal simple 
equation, which to a beginner often presents great difficulty, even 
when he is well able to work a similar numerical example. The 
use, only, of logarithms is taken in Chap. VI ; it was hoped to add 
something of the theory, but space would not permit. The Mensura- 
tion in Cnaps. VII and VIII may appear rather extensive, but calls 
for little knowledge beyond the evaluation of a formula or the 
solution of a simple equation. This division of the subject was 
framed to be of use for reference in the Drawing Office, in which 
direction the tables on pp. 312, 313, 345 and 346 should prove 
useful. The " more exact " formula for the circumference of an 
ellipse on p. 260 (due to Boussinesq) is not often seen, but is very 
accurate. It is to be feared that many writers repeat the formula 
n\/2(a 2 -f b z ) as being more exact than fr(a + b), when really it 
is but little better and has only a very limited range of application. 
Chap. IX takes Graphs in an elementary degree, considerable atten- 
tion being paid to the details of setting out and finishing off, in 
which direction many students fail. The last chapter is devoted 
to the Slide Rule, with illustrations of readings and settings, and 
the method of instruction was tested while in manuscript form. 
The chapter was carefully worked through by a novice, who, 
finally, and without other instruction, could use the rule with 
ease and certainty : these pages should be helpful to the private 
student. 

It is hoped that teachers of Practical Mathematics will be saved 
much research work by the numerous classified practical exercises 
throughout the book : answers are given to all these exercises. 

To MR. W. J. LINEHAM, B.Sc., M.I.C.E., the Author tenders his 
sincere thanks for much kindness throughout his career, and also 
for generous assistance and useful criticism in the production of the 
book. 

The Author's gratitude is also tendered to MR. JOHN L. BALE 
for his generous help with the book in all its stages, from the earliest 
research, throughout the manuscript and proof forms to the final 
production. 

The notification of errors, clerical and otherwise, will be gratefully 
appreciated. 

CHARLES B. CLAPHAM 
Goldsmiths' College, 

New Cross, S.E., 

May, iQ/6. 



PREFACE TO THE THIRD EDITION 

IN preparing a third edition of "Arithmetic for Engineers" a 
chapter on Elementary Trigonometry has been added. It is hoped 
that this addition, by filling a gap in the original treatment, will 
render the book of greater value to Technical Schools, in many of 
which a first course of Practical Mathematics includes at least the 
elementary conceptions of this branch of the subject. The new 
chapter covers little more than the meaning of the important 
ratios sine, cosine, and tangent, but is believed to include sufficient 
to show how even this small amount can be of great utility to the 
practical man in drawing office and workshop. The treatment is 
similar to that followed throughout the rest of the book, viz., 
detailed explanations followed by really practical illustrative 
examples and exercises at every step. 

The matter on Verniers and Micrometers, originally added as 
an Appendix to the Second Edition, has been retained as an 
Appendix, as it is rather in the nature of a special application. 

It is a matter of deep regret to the Author that, by the death 
in 1919 of Mr. W. J. Lineham, founder of the D.U. Scries, he has 
been deprived of much valuable assistance and criticism. 

The Author wishes to acknowledge with many thanks the 
assistance of the various readers who have kindly notified him 
as to errors in the first two editions, and he hopes that this useful 
assistance will continue in the future. 

CHARLES B. CLAPHAM. 

Goldsmiths' College, 
New Cross, S.E. 
October, 



PREFACE TO THE FOURTH EDITION 



THE opportunity has been taken to revise one or two sections 
and to add a short account of the finding of logarithms and anti- 
logarithms on the slide rule. About fifty new exercises have 
been added, covering a variety of details including some simple 
applications to wireless and aircraft. 

The notification of errors is gratefully acknowledged; the 
Author will always be pleased to receive further information in 
this direction. 

CHARLES B. CLAPJIAM. 

Finsbury Technical College, 
Leonard St., E.C. 2. 

September, ^925. 



CONTENTS 

PAGE 

INTRODUCTORY i 

Measurement and units Multiples and sub-multiples Abbrevia- 
tions, mathematical signs and terms Tables of measurements. 



CHAPTER I 
VULGAR FRACTIONS 5 

Fractions generally Forms of the vulgar fraction Cancelling 
Addition and subtraction of vulgar fractions Multiplication and 
division of vulgar fractions Compound examples Brackets. 

CHAPTER II 
DECIMAL FRACTIONS 35 

Decimal notation Movement of the decimal point Conversion 
from decimal to vulgar fractions Degree of accuracy and significant 
figures Addition and subtraction of decimals Multiplication and 
division of decimals Conversion from vulgar to decimal fractions 
Compound examples Approximation for result Averages 
Percentages Ratio Proportion . 

CHAPTER III 
SYMBOLS AND THEIR USES . . , 86 

Symbols and formulae Signs of x , -j- , -f- and : Brackets 
Simple evaluation Powers and indices Square root Evaluation 
including square root Other roots Powers and roots of fractional 
expressions Laws of indices : Multiplication, division and powers 
Substitution of symbols Positive and negative quantities 
Addition and subtraction of -|- and quantities Addition and 
subtraction of several terms Multiplication and division of -f- 
and quantities Powers of minus quantities Evaluation in- 
cluding positive and negative quantities Removal and insertion 
of brackets Taking out a common factor Multiplication with 
expressions of two or more terms Evaluating such expressions as 
(a 6) 1 . 

xi 



xii CONTENTS 

CHAPTER IV 

PAGE 

SIMPLE EQUATIONS 157 

Equations generally Equations requiring division, multiphca* 
tion, and combination of the two Equations requiring subtraction 
and addition Equations combining the four rules Equations with 
several terms Equations with brackets Fractional equations 
Equations requiring square root Equations requiring squaring. 

CHAPTER V 
TRANSPOSITION OF FORMULAE 190 

The general solution Equations requiring division, multiplication, 
and combination of the two Equations requiring addition and 
subtraction Combination of all four rules Equations with brackets 
Equations requiring square root Equations requiring squaring. 

CHAPTER VI 
USE OF LOGARITHMS 213 

Introductory Finding logarithms : i. The whole number. 2. The 
decimal part Anti-logarithms Finding anti-logarithms : i. Sig- 
nificant figures. 2. Placing the decimal point Multiplication and 
division by logarithms Compound examples Examples involving 
plus and minus Powers and roots by logarithms Various 
examples. 

CHAPTER VII 
MENSURATION LENGTHS AND AREAS 242 

Measurement of length Conversion and reduction Addition and 
subtraction in length units Simple geometrical terms Simple 
plane figures Perimeter or circumference Circumference of a circle 
ir and its determination by measurement Examples involving 
circumference of circles Circumference of ellipse Measurement 
of angles The protractor Important angles Addition and sub- 
traction of angles Reduction of angles The right-angled triangle 
Proportions of 45 and 6o-3o right-angled triangles Length of 
arc of circle. 

AREA 

Measurement of area Reduction Areas of the simple figures 
Square and rectangle Rhomboid Triangle Trapezium Trape- 
zoid Hexagon Octagon Circle Determination of diameter of 
circle from area Hollow circle or ring Sector of circle Area of 
fillet Segment of circle Ellipse Area of irregular figures Table 
of areas and circumferences of plane figures. 



CONTENTS xiii 

CHAPTER VIII 

PAGE 

MENSURATION (cont.) VOLUMES AND SURFACE AREAS . .314 

Volume Conversion Volume of regular solids Prisms Cylinder 
Hpllow cylinder and tubes Sphere Segment of sphere Pyra- 
mids Frusta of cones and pyramids Calculation of weights. 

SURFACE AREAS 

Cylinder Sphere Cone Square pyramid Frustum of square 
pyramid Frustum of cone Table of volumes and surface areas 
of solids. 

CHAPTER IX 

CURVES OR GRAPHS ........ 347 

Curves and their uses Rectangular co-ordinates Squared paper 
Method of plotting : I. Choice of scales. 2. Setting out scales. 
3. Plotting the points. 4. Drawing in the curve. 5. Further 
information required on the sheet Interpolation Cases where 
origin is not required The straight line Plotting of negative values 
The equation to a straight line Obtaining the equation or law. 

CHAPTER X 
THE SLIDE RULE 385 

Description Division of the scales Method of reading Advice 
as to holding the rule Operations on the slide rule Division 
Multiplication Combined multiplication and division Ratios and 
percentages Square root Squaring Reading logarithms and 
anti-logarithms from the rule. 

CHAPTER XI 
TRIGONOMETRY 410 

The trigonometric ratios Ratios of the angleb 30, 45, and 60 
Ratios of the angles o and 90 Relation between sine and cosine 
of complementary angles The table of trigonometric ratios Use 
of the trigonometric ratios in formulae Problems involving 
simple trigonometry The reciprocal trigonometric ratios Further 
points concerning the table of ratios Important relationships 
between the ratios Graphs of the important ratios. 

ANSWERS 434 

MATHEMATICAL TABLES ....... 454 

APPENDIX 459 

Verniers and Micrometers. 

INDEX 488 



ARITHMETIC FOR ENGINEERS 



INTRODUCTORY 

Measurement and Units. From the practical standpoint our 
calculations are never concerned with numbers alone, but always 
with some properties of materials or machines : simple, such as 
length and weight ; or complicated, such as horse-power and electrical 
resistance. We have always, therefore, to deal with quantities 
which have been measured and which, besides a number, will have 
a name, such as feet, pounds, etc. 

In some of the examples to follow, numbers alone will be used 
when the calculations are only for the explanation of a certain 
mathematical process. A measured quantity is described as being 
so many times larger than some particular " standard " of refer- 
ence. This " standard " is called a unit, a word denoting the 
number I, and which for each particular kind of measurement is 
given a distinctive name. Thus the unit in the table of weights 
is the pound, and in the table of length the yard, and so on. These 
units are fixed by law in the case of the more common measure- 
ments, such as length, weight, etc., and official samples are pre- 
served in Government offices. With more special measurements, 
such as velocity, horse-power, etc., the units are fixed by general 
consent. 

Multiples and Sub-multiples. It is very seldom that we 
find any particular unit suitable for all measurements of a certain 
kind. Thus the yard would be very inconvenient when measuring 
long distances on the earth's surface, for our measurements would 
come to millions of yards. Such numbers, besides being difficult 
to imagine, would render calculation very laborious. Therefore 
to make our measurements of long distances consist of a smaller 
number we employ a larger unit which contains an exact number 
of standard units. Such " large " units are called multiples of 
u 



2 ARITHMETIC FOR ENGINEERS 

the standard. The mile (1760 yards) and the ton (2240^ Ibs.) are 
examples. On the other hand, when measuring very small lengths 
(such as the thickness of sheet metal) the yard would be too large 
a unit, and therefore small units called sub-multiples are adopted. 
Common examples are the inch, of which 36 make I yard, and the 
ounce, of which 16 make i pound. Taking the more special measure- 
ments used by the engineer, the number of multiples and sub- 
multiples is small, while in some cases, as with horse-power, only 
the standard unit is employed. 

Abbreviations, Mathematical Signs and Terms. In 
order to save time and space, and to make our calculations more 
easily read, it is customary to use certain signs for phrases which 
are constantly occurring in mathematical work. These must be 
committed to memory. The more frequent ones are given below; 
others will be introduced as required. 

= stands for " equals " or " is equal to " ; 
+ ,, " plus/' meaning " added to "; 
,, "minus," meaning "subtract or take away"; 
x ,, ,, " multiplied by "; 
-+ " divided by"; 
" therefore." 

Thus the statement 3 2 + 5 = 6 would be read as " three minus 
two plus five equals six," meaning that if 2 is subtracted from 3, 
and 5 added to what is left, the result equals 6. 

Very often a division is not stated with the sign -r, but in the 
form - 2 8 ^, meaning 24 divided by 8. This form is similar to the 
sign ~, the dot representing the numbers, and is then often read 
as " twenty-four over eight." 

The following words have mathematical meanings 

Sum is the result of an addition. 
Difference is the result of a subtraction. 
Product ,, multiplication. 
Quotient ,, ,, ,, ,, division. 
Dividend is a number to be divided. 

An expression is any mathematical statement containing numbers, 
signs, etc. 

If several numbers be multiplied together they are said to be 
factors of the product. 



INTRODUCTORY 3 

Thus since 3 x 4 X 5 = 60, the numbers 3, 4 and 5 are " factors 
of 60." Any number may be split up into factors by knowledge of 
the multiplication tables, but it is seldom necessary to find the factors 
of large numbers. 

Several different factors may be found of the same number. Thus 
besides the above factors of 60, we may also have 20 and 3, 30 and 2, 
2, 10 and 3, etc. This is because some of these factors can themselves 
be split up into factors. Thus 30 x 2 can be written as 6 x 5 X 2, 
and so on. 

The simplest factors of all are those which cannot themselves 
be split up, e. g., 2, 3, 5, 7, n, 13, 17, 19, etc. These are called 
"prime factors." 

If a number contains another number exactly (/. e. t can be 
divided exactly by that number), then it must contain the prime 
factors of that number. Thus 36 contains 12, i. e., 36 can be divided 
exactly by 12; then 36 contains the prime factors of 12, i.e., 2, 
2 and 3, as may be proved by simple division. 

Also the following abbreviations will be met with 

i. e. meaning " that is " ; 

e. g. ,, " for example " ; 

viz. "namely"; 

approx. "approximately" or "very 

nearly " ; 

revs, per min. or r.p.m. ,, " revolutions per minute " ; 
m.p.h. ,, " miles per hour " ; 

h.p. (H.P. also used) ,, " horse-power." 

Previous Knowledge. In the following pages it is assumed 
that the reader is acquainted with the four simple rules of arithmetic 
addition, subtraction, multiplication, and division when applied 
to whole numbers ; and with the simple kinds of measurements, 
such as money, length, weight, time, etc. 

The tables of these measurements are given below for reference 

BRITISH TABLE OF LENGTH 

12 inches (ins.) i foot ; 

3 feet (ft.) =i yard; 

5j yards (yds.) = i rod, pole, or perch; 

40 poles (po.) = i furlong; 

8 furlongs (fur.) = i mile. 



ARITHMETIC FOR ENGINEERS 

WEIGHT (Avoirdupois) 
16 drams = I ounce (oz.) ; 

16 ounces = i pound ; 

28 pounds (Ibs.) = I quarter ; 
4 quarters (qrs.) = I hundredweight (cwt.) 
20 cwt. = i ton. 

also 112 Ibs. = i cwt.; 

and 2240 Ibs. = I ton. 

SQUARE MEASURE 

144 square inches (sq. ins.) = I square foot ; 

9 square feet (sq. ft.) = i square yard ; 

30 J square yards (sq. yds.) = i square pole ; 

40 square poles = i rood ; 

4 roods = i acre ; 

640 acres = i square mile. 

CUBIC MEASURE 

1728 cubic inches (cti. ins.) = i cubic foot ; 
27 cubic feet (cu. ft.) = I cubic yard. 

MEASURE OF CAPACITY 

4 gills = i pint ; 

2 pints = i quart ; 

4 quarts = i gallon ; 

2 gallons = i peck ; 

4 pecks = i bushel ; 

8 bushels = i quarter. 



CHAPTER I 
VULGAR FRACTIONS 

Fractions generally. As the quantities with which we have 
to deal seldom contain an exact number of units, we have to con- 
sider how to deal with parts, or fractions, of a unit. Since I is 
the smallest whole number in our written figures, and we have to 
deal with quantities less than I, it is evident that our ordinary 
system of notation cannot be employed to represent the magnitude 
of such quantities. 

Two systems of representation are in general use, both employ- 
ing the following idea : the unit is divided into a certain number 
of equal parts, and any particular fraction is said to be equal to 
so many of those parts. The number of parts into which the unit 
is divided may be chosen within wide limits, but in practice only 
the most convenient are employed. 

In the Vulgar System this number may be any convenient whole 
number whatever, for example : 8, 13, 100, 346, etc. 

In the Decimal System the number is always i with one or more 
noughts after it, i. e. t 10, or 100, or 1000, etc. 

The decimal system is thus a special case of the vulgar system, 
but appears rather different in practice on account of the different 
method of statement adopted. In Great Britain the vulgar system 
is employed chiefly in connection with money and the everyday 
use of the ordinary weights and measures. The decimal system is 
certainly more valuable for scientific purposes and for the more 
accurate measurements demanded in the workshop, while on 
the Continent it has been adapted, with few exceptions, to all 
measurements. 

Vulgar Fractions. In the vulgar system the following nota- 
tion is adopted : below a short horizontal line is placed the number 
which shows exactly into how many equal parts the unit has been 
divided; above the line is placed the number which indicates to 
how many of these parts our fraction is equivalent. Thus in the 
case of (read as " three-fourths " or " three-quarters ") the unit 

5 



ARITHMETIC FOR ENGINEERS 



has been divided into 4 equal parts, and 3 of them taken. This 
is shown graphically at a, Fig. i, where the length AE represents 
our unit (say i foot, for example) divided into 4 equal parts, AB, 
BC, CD, and DE. The fraction f is then represented by the length 
AD. 



"< 


2_ i ^ 


> 




n JL , 


4- e ' > 






A 


B 


c 


D E 




1 ' 


_ A- 




^ 


1 UNIT ' 







a 




.15 



I 



UNIT 







E rf 


A IB 


c 


D 


, , i | . 


1 1 


I ' 


D $ 1 3 UNITS 



Fig. i. Illustrating Vulgar Fractions. 

The number underneath the line, which indicates the size of 
the equal parts, is called the Denominator ; while that above the 
line, which shows the number of equal parts taken, is called the 
Numerator. Thus in the above example, 3 is the numerator and 4 
the denominator. 



VULGAR FRACTIONS 7 

Similarly b and c, Fig. i, indicate respectively the fractions ^ 
(four fifteenths) and JJ (thirteen twenty -thirds), in which 4 and 13 
are numerators and 15 and 23 are denominators. We shall fre- 
quently refer to the numerator and denominator as the " top " 
and " bottom " of the fraction. 

From the foregoing, the line between the numerator and de- 
nominator is merely a means of distinguishing between them. On 
p. 2 it is stated that this mid-line indicates the division of the 
upper number by the lower one. This is quite in accordance with 
our fractional notation, as may be seen from d, Fig. i. Here the 
distance AE, which represents 3 units, is shown divided into 4 
equal parts on the top side of the heavy line. Beneath the line 
each of the 3 units is shown divided into 4 equal parts, and it is 
easily seen that AB = f of i unit. Therefore the fraction J unit 
is the result of dividing 3 units into 4 equal parts, and hence we 
may consider any vulgar fraction as indicating the result of dividing 
the numerator by the denominator. 

Such of the simpler denominators as quarters (or fourths), 
eighths, and sixteenths, will be familiar to most readers as being the 
usual divisions on the common 12-inch rule, while thirty-secondths 
and sixty-fourths are occasionally marked. In the monetary system 
where the sovereign () is the unit twentieths and two hundred- 
and-fortieths are employed, but are better known by the names of 
shillings (s.) and pence (d.) respectively. 

Other peculiar denominators such as those at b and c, Fig. i, 
are not often met with in practice. 

Forms of the Vulgar Fraction. A true fraction is less than 
a whole unit, and therefore its numerator must always be less than 
its denominator. When this is so the fraction is called a proper 
fraction. Thus f , f , i, T * ff , are proper fractions. 

When a measurement consists of an addition of whole units to 
a fraction of a unit, as, for instance, two inches and three-quarters 
of an inch (commonly " two and three-quarter inches "), it could 
be expressed as 2" + f ". It is customary, however, to omit the 
+ sign and write the whole number immediately in front of the 
fraction, thus, 2". Care should be taken, when writing in this 
form, that the figure representing the whole number is written 
larger than the figures composing the fraction ; otherwise the 2 
may be confused with the 3 or the 4, and the quantity read as *- 
or TJ\. Numbers written in the form 2| are termed mixed numbers. 

In the course of calculation we sometimes arrive at fractions 
in which the numerator is greater than the denominator, a con- 



8 ARITHMETIC FOR ENGINEERS 

dition which cannot denote a true fraction, since we have more 
parts than are contained in our unit. This type of quantity, which 
is known as an improper fraction, is only a statement of a mixed 
number in fractional form, and may be treated in all calculations 
as a proper fraction. Thus - 1 /, |, ^, etc., are improper fractions. 
To convert a mixed number into an improper fraction proceed 
thus : Multiply the whole number by the denominator and add 
the numerator. The result of this is the numerator of the improper 
fraction. The denominator remains the same. The method con- 
sists in finding the total number of equal parts contained in the 
mixed number and stating this in fractional form. Thus in (a), 
Ex. I, below, the denominator is 4, or each equal part is a quarter. 
We therefore convert the 2 into quarters, and as I unit = 4 quarters, 
2 units = 8 quarters. The total number of quarters is then 8 + 3 
~ n, and therefore 2j = n quarters, i. e., - 1 /-. 

Example i. Convert into improper fractions (a) 2j; (b) n^ 

(a) Whole number X Denominator 

2x4=8 
add numerator = 3 

- ,, Sum =11 

2f = V 

This may be stated more mathematically thus, 

2| = * ( 2 _^3) 3 = 8 3 ^ ii 
4 4 4~ 

(b) uA = ( ri x l6 ) + 7 = I767_ - ~ 3 
V ' ie 16 16 ~~ i(7~ 

With a little practice the operation may be performed mentally. 

The reverse operation of converting an improper fraction into a 
mixed number is, naturally, the opposite of the foregoing, thus : _ 
Divide the numerator by the denominator, giving the whole number. The 
remainder of the division is the numerator of the true fractional piece. 

Example 2. Convert into mixed numbers (a) ^ ; (b) ^. 
(a) Numerator ~ denominator = whole number. 

11-7-4 =2 with remainder 3. 

' v = 1 

(6)101-7-12 = 8 with 5 remainder. 



* The use of the bracket signs ( ) is given later. Here it is adopted 
to show definitely that the multiplication of 2 and 4 must be done 
before adding the 3. 



VULGAR FRACTIONS 9 

This also may be done mentally after some practice. A special 
case occurs when the remainder of the division is o : that is, when 
the denominator divides into the numerator an exact number of 
times ; then the improper fraction is equal to an exact whole number, 
e - g-> IF = 2 an d o remaining, /. |- = 2. 

Exercises 1. On Forms of Vulgar Fractions. 

Write out the following numbers, describing each one as a Proper 
Fraction, an Improper Fraction, or a Mixed Number 

1. i W. 8. *. V- 2 - H. H. * *i. ii- 

3. sf. 3i if, tt. iA- * . 6*. i vV- 7&. 

Convert the following mixed numbers into improper fractions 
5. ij, 2J. i A, 3i 2?. 6. 5i, i A. 'ft. 2 A> 3A- 

7. ioj, i5j, nf, i9i 7 3 V 8. loft, 1 1 4. i7f. 2I i H?- 

9. 14?, I5H. 1 7ft, 38J|. 10. I 9 , 4ift. i6ft, 3ift- 

Convert the following improper fractions into mixed n-umbers 

11- f i I- V. V- 12- }i, . S 4 ft , V. V- 

13. Ji, , ?{, :4. V- 14. U. f J. . V. W- 

15. f, Ii, If, W. W- 16. Yi?. W. t. ?B- 

Cancelling : Reduction to Lowest Terms. As a general 
rule it is advisable to keep the magnitude of the figures in any 
calculation as small as possible. By this means we are less liable 
to error, while space, time, and patience are economised. An 
examination of (a), Fig. I, will show that the length AC is dimen- 
mensioned as \ = J, a statement which is easily seen to be true. 
Also from Fig. 2 it can be seen that 

3^6 = 9 ^15 

4 8 12 20 

These fractions may also be written as 

3 = 3__X_2 ==: 3_x_3 ^ 3 X 5 

4 4x2 4x3 4x5 

Thus we may multiply the numerator and denominator of a 
fraction by the same number without altering its value, and similarly 
we may divide both numerator and denominator by the same number 
without altering the value of the fraction. This division is termed 
cancelling, and is of great use in reducing fractions to their lowest 
terms, i. e., making the numerator and denominator as small as 
possible while keeping the value of the fraction the same. 

Cancelling is usually shown by drawing a line through the exist- 
ing numbers, and writing the new numbers immediately above and 
below the old ones. The operation should be performed very 



io ARITHMETIC FOR ENGINEERS 

neatly, so that it is possible to read the original figures afterwards. 
As a simple illustration take the & previously used 



- 

W 4 
4 

The reduction may be performed in a single operation by looking 
for the highest number that will divide into the numerator and 
denominator exactly; but as much time may be spent in such a 
search, it may be advisable to divide in steps, using those numbers 
which can be seen at a glance to be suitable ; c, Example 3, will 



1 ll Z 3| 4-| 


L 3 




* 4- 


1 '1 2| 3| 4 5| 6| 7j 8| 


x e ^ 




^ S 



s <c 7 s 9 o n ig ia >4 ts 



U _ 15 _ >J 

n 20 ~ 

Fig. 2. Illustrating Cancelling. 

illustrate. In connection with this it is useful to remember the 
following points 

A number is divisible by 2 when the last figure is even, i. e. is one 
of the figures o, 2, 4, 6 or 8. Thus 32, 248, 500 are divisible by 2. 

A number is divisible by 5 when the last figure is or 5 ; and is 
divisible by 10 when the last figure is an 0. Thus 25, 40, 2005 are 
divisible by 5. and 40, 2000, 530 are divisible by 10. The division 
by 10 is very easily performed by striking off the last o. Thus 

2000 -r 10 = 200. 

A number is divisible by 3 if the sum of its separate figures (or 
digits) is divisible by 3 ; and is divisible by 9 if the sum of the digits is 
divisible by 9. 

Thus, taking 51, 5 + 1 = 6, which may be divided exactly 
by 3- Therefore 51 is divisible by 3, giving 17 as a result. Taking 
957, we have 9 + 5 + 7 (done mentally) = 21, and this is divisible 



VULGAR FRACTIONS n 

by 3. Therefore 957 is divisible by 3, giving 319 as a result. Taking 
738, the sum of the digits is 18 ; therefore the number is divisible 
either by 3 or 9. 

A number is divisible by 11 when the sum of the odd digits (/' e., the 
first, third, and so on) equals the sum of the even digits ; or, when the 
difference between these sums is exactly divisible by 11. 

Thus in the number 3575 the odd digits (the 3 and the 7) add 
up to 10, while the even digits (the two 5's) also add up to 10. 
The number is then divisible by n, giving 325 as result. The 
totalling of the odd and even digits can be done mentally. 

In the case of the number 9196 the sum of the odd digits is 
9 + 9 = 18, while that of the even digits is 1 + 6 = 7. The 
difference between 18 and 7 is n, which, of course, is divisible 
by ii. Then the number 9196 can be divided by n without 
remainder, giving 836 as result. 

There are tests for divisibility by other numbers, but they are 
somewhat complicated and of doubtful value, the test taking longer 
than cancelling by steps. The rule for 11 is given, as, being a prime 
factor, cancelling by steps is impossible. 

Example 3. Reduce the following fractions to their lowest terms : 

() 15; (*>) Ji; W ; (<*) J&; W H*. 

(a) Cancelling by 3, Jf = j (b) Cancelling by 3, J = *f 

- 7 21 

(c) Cancelling by 6 in one operation, \ = 

i) 
(or by 3 and 2, in two operations). 

(d) Cancelling by 19, the only possible number, j% = J 

(e) This is an improper fraction. It is usually better to reduce to 
a mixed number before cancelling. 

Thus i*J = ij4 = i/ s 

40 

More complicated examples are seldom required. 

Exercises 2. On Cancelling. 

Reduce the following proper fractions to their lowest terms 

1- *. f> A. A. A- 2 - 1*. IS. *. A. M- 

3. M> tt. tt. M, *. jfo, Iff, Hg, fcjfc. 

5. ii, ii J!> , f I- 6. Mi *3J, MS. HJ- 

Simplify the following numbers, reducing them to their lowest 
terms 

7. , V. ft, II- 8- , 3V. ijj, 4- iH- 

9- 2*4. 5f, . s - \V. 181- 10. 7?|. 34, 8. Hi- 
ll. W. fi i8- W, . 12. 33^00, ?f , II?g . 



12 ARITHMETIC FOR ENGINEERS 

Addition of Vulgar Fractions. Addition is only possible 
between quantities of the same kind of measurement. Pounds 
sterling () can only be added to pounds sterling, and feet to feet, 
etc. No one would attempt to add pounds to feet or tons to 
shillings. Quantities of the same kind of measurement, but of 
different units can be added, provided that the sum of the numbers 
is not found. Thus 5 and 3 shillings cannot be added and called 
8 something, but can be stated as 5 35. ; to obtain this result 
there has been no addition of the numbers 5 and 3. 

Similarly when dealing with fractions, the denominators (names] 
may be regarded as units ; and no addition of numerators (numbers) 
can be performed until the denominators or units are the same. 
This is usually stated by saying that the fractions to be added 
must have a common denominator. To take a very simple example, 
let it be required to add J to -j 3 ^. Either the eighths must be changed 
into sixteenths, or the sixteenths into eighths. Now it is seldom, 
if ever, convenient to convert any fraction into one of smaller 
denominator (i. e. larger equal parts). To illustrate this let us 
convert the sixteenths into eighths. 

Then, since ^ = \, Y\ ii eighths, and we may therefore write 

1 j_ _3 ___ i , i* _ 2* 
8 "*" 16 ~~ 8 : "*" 8^ ~ 8 ' 

a form which, with any but the simplest numbers, would become 
hopelessly muddled. 

But let us convert into smaller parts, i. e., convert the eighths 
into sixteenths. 

Then we may write 

JL l '* 213 5 

8i~iO 1C i Tff 169 

a practical proof of this being easily obtained by an inspection of the 
common 12-inch rule. It will be noticed that the numbers which 
compose the fraction -^ (i. e., 5 and 16) are entirely whole numbers. 

Example 4. The tapping size (size cf a hole before being screwed) 
of a certain Whit worth bolt is given as fo" -f- A"- Express this as a 
single fraction. 

A + A 

= i? 4- o U 4 converting the &* into 6 y h ", ^ being & 



Exercises 3. On Addition of Fractions. 

SIMPLE CASES 

1. Drilled bolt holes in pipe flanges are J* larger than the bolts for 
bolts over $" diameter. What sized holes are required for the following 
bolts: (a) J*, W I*, (c) 1 4'? 



VULGAR FRACTIONS 13 

2. For bolts up to j* diameter the drilled holes are ^* larger 
than the bolt. What sized holes are required for the following bolts : 

M &"> (*) r, (o r? 

3. The diameter of the tapping hole for certain Whitworih bolts 
and nuts is given as follows : (a) J" bolt, diameter of tapping hole 
= f H- 3* (b) A" b olt diameter of tapping hole = iV ~f- A- Express 
each of these diameters as a single fraction. 

4. The thickness of the head in certain Whit worth standard bolts 
and nuts is given as follows : (a) J* bolt, thickness of nut = ^ -f- -.> ; 
(b) y bolt, thickness of nut t s ff -f- ^V Express each of these sizes 
as a single fraction. 

Express the following Whitworth bolt-head thicknesses as a single 
fraction in each case : 

5. (a) J" bolt. Head is | -f- > (b} } J" bolt. Head is ft -f ^. 

e. (a) r .. - * + & (*) 13" .. .. .. H + A. 



Express the following Whitworth tapping sizes as a single fraction 
in each case 

8. (a) i" bolt, tapping size is } $ 4- jJ 5 . (6) } J" bolt, tapping size is *$ -f ^ . 

9. (*) r .. i +A- w r ,. ,. M ., J+A. 

10. wr .. M - +& wtr ........ I+A. 

A rather different appearance is presented by the case J + -. 
We cannot convert the quarters into fifths or the fifths into quarters 
without complicating the figures ; but we can obviously give each 
fraction the same denominator by multiplying the 4 by 5, and the 
5 by 4, i. e. multiplying the denominators together, giving 20 in 
each case. Then, remembering that the value of a fraction is 
unaltered if both top and bottom be multiplied by the same number, 
the example may be written thus 

J + i = __ x - 5 + 2 _21 4 
4 5 4~X 5 5 X 4 
= .5 + 8 ^ 13 
20 20 20 

In Fig. 3 this is shown graphically. 

The foregoing method may be applied to any example, but in 
the more difficult cases the labour and figuring may be reduced by 
employing the Least Common Denominator (abbreviated to L.C.D.). 
As an illustration, consider the addition of the fractions ^ and f . 
Working on the previous method, the common denominator would 
be 12 X 8 = 96. Now a little thought wall show that the numbers 
48 and 24 also contain 12 and 8 an exact number of times, so that 
for the purpose of addition the fractions may equally well be con- 
verted into forty-eighths or into twenty-fourths. By using 24 as 



14 ARITHMETIC FOR ENGINEERS 

the common denominator instead of 96, smaller numbers will be 
produced, and the calculation simplified in consequence. This 24 
is then the smallest possible common denominator, and is the Least 
Common Multiple (abbreviated to L.C.M.) of the given denomi- 
nators. Now the L.C.M. of a set of numbers is defined as the least 
number which will contain each of the given ones an exact number of 
times, so, keeping this in view, we may deduce the 24 from the given 
numbers 12 and 8. 







Fig. 3.- Shov/ing that } -f f 



The required L.C.M. is to contain the numbers 12 and 8 exactly, 
and must therefore contain the factors of these numbers (see p. 3). 
Splitting into prime factors, we have 

12 = 2X2X3 8 = 2X2X2 

Now all the factors of 12 will be required, since the 12 itself 
must divide into the L.C.M. But in the case of the 8, two of the 
factors already appear in the 12, viz. 2X2, and need not, therefore, 
be used again. The remaining 2 in the 8 does not appear in the 12, 
and must therefore be counted. Then the L.C.M. of 12 and 8 = 
2x2x3x2 = 24. 

The detailed method of finding an L.C.M. is laid out in the 
following example : 

Example 5. To find the L.C.M. of 8, 15, 4 and 12. 

8 15 \ i* 

8-2x2x2 L.C.M. =2x2x2x3x5 
I 5 = 3X 5 =8x3X5 

12 = ^ x $ X % =120 

Explanation of the Method. Examine the given numbers to 
see if any one will divide exactly into any other one ; if so, cross it 



VULGAR FRACTIONS 15 

out. Thus 4, crossed out, will divide exactly into 8 and 12, so 
that if the L.C.M. contains 8 it must contain 4; we may therefore 
neglect such numbers. Then split each of the remaining numbers 
into its prime factors (see p. 3) as shown. The L.C.M. is to contain 
all the numbers, and so must contain the 8. The factors of the 
first number are then left alone. Now examine the prime factors 
3f the second number (15) and see if any of them already exist in 
the first number. If so, cross these out.* In this case neither 
the 3 nor the 5 appears in the factors of 8, and so are left in. Next 
examine the prime factors in the third number and cross out any 
that still appear in the first and second numbers. Thus the 3 already 
ippears in the factors of 15 and the two 2's already appear in the 
[actors of 8. All these three factors are then crossed out. The 
process is repeated until all the prime factors have been examined, 
>vhen the L.C.M. is obtained by multiplying up all the factors 
>vhich are left (in this case 2x2x2x3x5), giving 120. 

Example 6. Find the L.C.M. of 8, 12, 16, 6 and 18. 
^ 12 16 $ 18 

12 = 2 X 3 X 2 L.C.M. = 2x3X2x2x2x3 

16 = ^X^X2x2 = 144 

18 = \ X ^ X 3. 

Note. The numbers 8 and 6 are crossed out, being contained in 
[6 and 18 (or 12) respectively. The factors of 12 include two 2's; 
hen two of the 2's forming the factors of 16 are crossed out. The 
>ther pair must be left in, as the L.C.M. must contain four 2's if it is to 
;ontain 16, and in the first line only two 2*s appear. In the third 
ine the 2 and one 3 are cut out. Only one 3 appears in the first two 
ines, and so only one may be cut out. Finally, the L.C.M. is the pro- 
luct of the remaining factors = 144 as shown. 

Exercises 3 (contd.). On L.C.M. 

Find the L.C.M. in each of the following cases 

11. 4, 8, 6. 12. 4, 3, 6, 2. 13. 9, 3, 4, 8. 

14. 16, 9, 8. 15. 12, 10, 5. 16. 14, 21, 7, 2. 

17. 16, 48, 18. 18. 55, ii, 15. 19. loo, 150, 75. 

20. 3. 5 1 * 9> 17- 21. 34, 17, 3. 22. 33, 15, 3. 

23. 125, 25, 150. 24. 280, 14, 150, 50. 

25. 30, i 8o > 45> 9, 5- 26. 3, 17, 51, 34, 9. 

Applying the method to addition problems, the L.C.M. of the 
lenominators is first obtained, and each fraction is converted so 
hat it shall have the L.C.M. for its denominator, i. e. the common 
lenominator is the Least Common Denominator. 

* See Note to Example 6. 



16 ARITHMETIC FOR ENGINEERS 

Example 7. Add together f, ^T, J, and -fa. 

The L.C.D. has been shown in Example 5 to be 120. Dividing 
this L.C.D. 120 by each denominator in turn, gives the number 15, 
which is required to convert the fraction. 

Thus, = 15, and therefore J- = 3 X__y _ . 
and similarly for the others. 

.'. I + A + i + A 

-= 3 _X_L5 4- 2 x ^ 4. I x 3-4- 5 * I0 1 These steps may be 

8 x 15 15 X 8 4 x 30 12 x 10 I omitted or shortened. 

45 , 1 6 , 30 , 50 f See a few lines further 

I2O I2O I2O I2O J On. 

= Hi = I f& or cancelling by 3, = i^ 

In an actual example the working would not be shown as above, 
in which there is a considerable amount of repetition for explana- 
tory purposes. The common denominator need only be written 
once if a long line be placed above it instead of the several short 
lines. Also the actual multiplication of numerators may, in most 
practical cases, be performed mentally. The working would then 
appear in the following form : 

I + A + i + A 

= 45 + 16 -f 30 -f 50 = 141 
i 20 i 20 



Example 8. In connection with an electrical resistance problem 
it was found necessary to evaluate (i. e. t find the value of) the expres- 
sion i 4- J 4~ iV Calculate the required value. 

4, $, 18 

4 = 2X2 

23 18 = ^ X 3 X 3 

.'. L.C.M. = 2x2x3x3 

-36. 

Explanation. -The L.C.M. is quite simple. Taking the first frac- 
tion, 4 into 36 goes 9. Then 9 x 1=9, which is written above the 
long line. Then 3 into 36 goes 12, and 12 x i = 12, which appears 
as shown. Lastly, 18 into 36 goes 2, and 2x1 = 2. Addition of 
the numerators completes the example. 

Where the given quantities include whole numbers, then 
the easier and safer method is to add the whole numbers inde- 
pendently of the fractions. Usually the whole numbers may be 



VULGAR FRACTIONS 17 

added mentally. It is unwise to convert the mixed numbers into 
improper fractions, as unwieldy figures result, and the subsequent 
labour is greatly increased. Where improper fractions are given 
(an unlikely condition), they should be converted into mixed 
numbers. 

Example g. Find the value of3j4-4 + 7i + 6J. 

Add the whole numbers mentally and write the result in front of 
the addition of the fractions, thus 



The expression then = 



24 



L.C.M. =-8x3 

= 2 4 



20 -f ill Ma Y often be done mentally. 



Example 10. The dimensions of the spindle for a sluice valve are 
shown in Fig. 4. Determine the overall (i. e., total) length of the 
spindle. 




Fig. 4. Spindle for a Sluice Valve. 
Writing down the dimensions in order we have 



= 18 4___ 10 t 8 JL_14 
16 



L.C.M. = 16 



Examples like this are of frequent occurrence when dealing 
with drawings, and with practice they can be done mentally. It 
is then necessary to be thoroughly familiar with the ordinary 
measurements of J, T \, etc., and to know at once the simpler con- 
versions such as f = -{ = ! etc - Sucn knowledge is obtained 
by doing a considerable amount of mechanical drawing. Then 
the above example would be done mentally as follows : ioj + f 
is ioj, add 6 gives i6J, add J gives 17$, add 2 gives igf. Leave 
the ^ B till the last as being the most troublesome figure. Then 
19! and } gives 2oJ; and 2oJ and ^ is 20 T V + 

c 



T V = 



i8 



ARITHMETIC FOR ENGINEERS 



Exercises 3 (contd.). Addition of Fractions with L.C.M. 

Find the value of the following : 

27. i + f + ft- 23. + A + A. 29. A + & + J- 

30. J + ! + &. 31. A + I? + A- 32. 9i + } + 3i- 

33. U + t + A- 34. ij + 5 J 6 + jpa + t- 35. ft + ij + A + 5l- 

36. In finding the current required for some electric glow lamps 
fed in parallel, it was necessary to obtain the value of the expression 

ifo H- lib + fs> Find this value - 

37. Find the overall length in inches of the lathe mandrel shown 
at a, Fig. 5. Convert the result into feet and inches. 

38. b, Fig. 5, shows an eccentric strap bolt. Find its overall length 
in inches. 




Fig. 5. Exercises on Addition of Vulgar Fractions. 

39. Find the overall length in feet and inches of the piston rod 
shown at c, Fig. 5. 

40. Find the length of the cross-head bolt, shown at d, Fig. 5, from 
the underside of the head to the point. 

41. The vice screw shown at e, Fig. 5, is to be cut from the solid 
bar. Allowing an extra inch of length for holding, etc., what length 
of bar must be cut off ? 

42. Find the overall length of the armature shaft for an electric 
motor, shown at/, F*ig. 5. 

Subtraction of Vulgar Fractions. This involves no new 
principle, the method of finding the L.C.D., and the separate treat- 
ment of whole numbers, being adopted as in addition. An example 
is the best explanation. 



VULGAR FRACTIONS 



Example n. Find the value of 5^ 2. 

12 = 2x2x3 



_ , 
3 



- 2 I 
- 9 - 3A 



24 



.'. L.C.M. = 2x2x3X2 

= 2 4 

Explanation. Taking the whole numbers first, 5 2 = 3. L.C.M. 
= 24. Then 12 into 24 goes 2, and 7 X 2 = 14. Similarly 8 into 24 
goes 3 and 3X3 = 9. Then 9 from 14 = 5. 

Example 12. From 20 JJ take 13 fa. 

30 = 2 X 3 X 5 
, 22 ~ 35 



= 7 60 

= 6 82 - 35 = 
60 



12=^X^X2 

L.C.M. = 30 x 2 
= 60 



Explanation. As before, the whole numbers are dealt with first, 
followed by the finding of the L.C.M., and the conversions as shown. 
A special point arises in the second line. We cannot take 35 from 22, 
but it must be remembered that the statement means 7$$ fg, which 
of course is possible. The subtraction may be performed by con- 
verting a unit into sixtieths, and adding this unit to the fraction $J, 
as shown in the last line, giving gg, from which g can be subtracted. 

A short combination of addition and subtraction presents but 
little difficulty, and is illustrated by the following example 

Example 13, Evaluate 15! 3^ + I J- 
Whole numbers 15 
Expression = 13 



36 



4 9 $ 

L.C.M. = 4X9 
= 36 




Fig. 6. Rock Drill Piston. 



Example 14. Part of a rock drill piston is shown in Fig. 6, and by 
a draughtsman's oversight a dimension has been omitted. Assuming 
that all the other figures are correct find the value of the missing 
dimension. 

It is evident from the picture that if the sum of the small lengths 



20 ARITHMETIC FOR ENGINEERS 

be found, and subtracted from the " overall " distance, this will give 
the value of the missing quantity. 

Then ij + 3i + 5i + 4 = i" L.1+_LJ = 10 J. 
Then missing dimension 13! loj 



.*. Missing dimension = 2f" 
Note. With practice this type of example can also be done mentally. 

With a more lengthy combination it may not be easy for a 
beginner to deal with the whole numbers mentally. Then it is 
best to find the sum of the whole numbers to be added, and the 
sum of the whole numbers to be subtracted ; the final subtraction 
being simple. The numerators of the converted fractions may be 
dealt with in the same way. 

An example will illustrate 

Example 15. Find the value of jf + 2j 5/2 3^ -f g\\. 
Whole numbers : 2 5 3 + 9=11 8 = 3. 



r 


3 


5 


+ 


36 


35 9 4- 44 


12 


^ 12 30 
== 2 X 2 X 


\ 
3 


^ 










60 


= 


3 


130 


60 


_i_4 


= 3*8 


30 

L.C.M. 


= % X J 
= 12 X 


* X 5 

5 = 


60 


= 4fr r 4^5 















Note. Taking the whole numbers first, 2 and 9 are to be added, 
and 5 and 3 to be subtracted. Combining these pairs gives n 8 3, 
as shown. After converting to the L.C.M. , all the plus numerators 
are added, giving 130, and all the minus numerators, giving 44. Then 
the final subtraction gives 86. 

Exercises 4. On Subtraction of Vulgar Fractions. 

1. The correct size of tapping hole for a certain screw is ij". If 
a fa" drill is the nearest size available, by how much will the hole be 
too large ? 

2. By how much will a hole be too large if a \%" hole be drilled 
when the correct size is f \" ? 

3. A f hole is drilled for tapping when the size should be gj". 
What is the difference in size, and is the hole larger or smallei than it 
should be ? 

Find the value of the following : 

4. * - i. 5. | - J. 6. fa - i 7. ft - ft. 

8. i* - |. 9. 2j - ft. 10. M - A- H. 4 ~ 28- 

12. 2-| - i|. 13. i ft - f 14. 9 ft - 6tf . 

15. From 3^ take 2j. 16. From i{ j take ^. 

17. From ij take ^ 7 . 18. Subtract 23 J from 3oJ. 

19. Subtract 4}J from 5?J. 20. Subtract JJ from i?. 



VULGAR FRACTIONS 



21 



Find the value of the following : 

21. 3i - * + U- 22. 3 - | + 3aV 23. tf +J - M- 

24. ft ~ f + i?,. 25. 2$ - ii - i ft. 26. 4 ft -f 5j - 7ft. 

27. 12 - 3 + Ji - I- 28. M - 2 i 3 o - i + 5i- 

29. 3ft - ft + 18 -f 2} - J. 30. i-/ a + | 2J5 -f 3^ 

31. Find the value of 3ft ij and take the result from i 

32. .. ,. * I - I 

33. Take the result of ift -f- from 3-j%- 

34. 9l + ft ., I3S- 



ft- 




I. A may be irscreased to suih; 
or- 2. Rackirxg raay be puhal~p. 



o 



V) 



Fig. 7. Exercises on Subtraction of Vulgar Fractions. 



35. From 2ft take the result of f -f f . 

36. 2i| 3 + 4i 4- 2f. 

37. 37l .. .. 3* + 2i| -f- ft. 

38. From the result of 2 ft .p 2 - take i}J. 

39. ft + }" &. 

40. * - A T^J. 

41. a, Fig. 7, shows a screw from a lathe head-stock. Find the 
length of the part C. 

42. The valve spindle for a stop valve is shown at b, Fig. 7. Find 
the length of the plain portion marked A. 

43. A milling machine spindle is shown at c, Fig. 7, where by a 



22 



ARITHMETIC FOR ENGINEERS 



draughtsman's oversight the length A of the front bearing has been 
omitted. Find this value from the other dimensions. 

44. Find the value of the missing dimension B from the drawing 
of the electric motor shaft shown at d in Fig. 7. 

45. An electric motor and a centrifugal pump, built by different 
makers, are to be assembled on a single bedplate as shown at e, Fig. 7, 
which also gives dimensions from centre line to base. 

(a) Supposing a special bedplate is to be made to suit, find what 

the dimension A must be. 

(b) Supposing that a standard bedplate, in which the dimension 

A is ij",be used, find what packing thickness is required under 
the motor feet. 

Multiplication of Vulgar Fractions. Multiplication is a 
convenient method of performing a long series of additions of 
equal quantities. Thus, 5 x 4 is a shorter way of stating both 
5+5 + 5 + 5 (*'*, 5 added 4 times) and 4 + 44-4 + 4 + 4 



B 



P 










"1 



~r 



N 



Large Area A F G M 
represents i UNIT. 
Then : 

Area A D J M = f , 
and shaded area 

= -2 of A D J M 

.'5 n f 3 

~ 4 OI 5" 



M L K J H . G 

Fig. 8. Showing that f x | -** T V 

(i. e. t 4 added 5 times). The multiplication of a fraction by a whole 
number may be shown the same way. 

Thus, |X5 = | + f + j + f + | 

= 3 + 3 + 3 + 3 + 3 = 3 _XJ 

4 4 

= = 3l 

So that the result is obtained by multiplying the numerator 
by the whole number. 

The multiplication of two fiactions, e. g. t f X f , is very con- 
veniently shown graphically, as in Fig. 8. The multiplication 
sign (x) may be replaced by the word " of," for f x f is f taken f 
of a " time " ; the result will naturally be less than |. In Fig. 8, 



VULGAR FRACTIONS 23 

the square A F G M represents i unit, which is divided by vertical 
lines 13 L, C K, D J, and E H, into fifths. Then the area A D J M 
represents the fraction J. Similarly the horizontal lines R U, 
Q T, and P S, divide both the unit and the area A D J M into 
4 equal parts. Therefore area A D N P is of area A D J M, 
i. e., f of f of a unit. The two sets of lines cut the unit square 
into 20 small figures such as H G S W, and in the area A D N P 
there are 9 such figures 

.'.? of 3 =9- 

4 5 20 

But the 9 is the product of the numerators 3 and 3, and the 20 is 

the product of the denominators 4 and 5. 

Hence the rule for the multiplication of vulgar fractions 
Multiply the numerators together to give the new numerator, and the 

denominators together to give the new denominator. 

Reduction to lowest terms should follow, but in many cases 

the cancelling may be more conveniently performed before multi- 

plying up. Thus in the following we first cancel by 3: 

3 

$ x 7 = 3 X 7 = 21 

16 \\ 16 X 4 64 

4 

The method is applicable to any number of fractions. When 
mixed numbers occur, it is usually advisable to convert them into 
improper fractions. 

Example 16. Find the product of , if, 2^ and ^. 
f X If X 2j X A 

= I x y x 



Cancelling being performed with n, 4 and 3. 

Example 17. If i kilogramme is 2j Ibs., find the number of Ibs. in 
235 kilogrammes. 

Evidently 235 kilogrammes = 235 times as much as i kilogramme. 
Then as i kilogramme = 2j Ibs. 

235 kilogrammes = 2j x 235 Ibs. 

47 

= V X W 
V X 47 517 Ibs. 

The cancelling should always be shown neatly, so that the 
original figures may be easily read. Care in this direction enables 
the working to be checked through, for possible errors, by any one. 
Examples involving heavy cancelling are met with when calculating 



24 ARITHMETIC FOR ENGINEERS 

the speeds of various combinations of gear wheels, such as occur 
in the feed motions of machine tools, variable speed gears, etc. 
In examples of this kind, special care must be taken and plenty of 
space allowed above and below the expression so that the cancelling 
may not be too confined. 

Example 18. The following figures were obtained when calculating 
the feed in inches of a drilling machine, for one revolution of the drill. 
Complete the calculation. 

3 X JO X 7^ X -fo X 16 x J 

i i 

* i x i 

ys/f v ix \/ i \/ i \/ l# \/ 

Ho x ^ x 7* x XV x i x 2 

i i 

== .} Q x ft = Jo inch per revolution. 

It must be noted that " multiplication " does not necessarily 
imply an " increase." If the multiplier is itself a fraction, then 
the resulting product is only a piece of the quantity multiplied, in 
which case a decrease results. 

Exercises 5. On Multiplication of Vulgar Fractions. 

Find the value of 

1.1X7- 2. ^ x 5. 3. .& X 3- 4. 15 x 4- 

5. f x j. 6. j x ,V 7. 3? x jj. 8. {? x ,. 

9. i? X }. 10. ft X 2ft. 11. 3-2 X if,. 12. 5* x i ^. 

13. y x il xi 14. A x i? x A. 

15. 2j X I A X I X J. 16. J X ft X 5t X Ij. 

17. Find the value of ij X ij X ij- X ij X il. 

The following multiplications refer to certain spur wheel trains. 
Complete the calculations. 

18. Yi a X \^ X V- 19. 500 x 

20. U x 88 x Sf. 21. V&- x g 

The following multiplications give the feed in inches per revolution 
of the mandrel in an 8" self-acting lathe. Complete the calculation. 

22. Greatest surfacing feed = S| X fe X f X J. 

23. Lowest surfacing feed = t x 'dV x i x J- 

24. Greatest traversing feed = ^| X ^ X ??, X 14 X f . 

25. Lowest traversing feed = gj X ^ X ifg X 14 X |. 

26. The speed of a certain lathe spindle, when driving through the 
back gear, is given by the value of the expression 

140 X ft X jf X Jf r.p.m. 
Find the value required. 

27. The feed of a drilling machine, in inches, for one revolution of 
the drill spindle, is given by the figures : ? J X JJ X ^ X^jX i6x f. 
Complete the calculation. 



VULGAR FRACTIONS 25 

28. (a) The finest measurement that can be made with a certain 
micrometer is J s of the " smallest scale division " ; if the latter is ^ of 
an inch what is the finest measurement ? 

(b) When the instrument is fitted with a " vernier " we can measure 
^Q of the result of (a). What is then the smallest measurement ? 

29. The ordinary mile contains 5280 ft., and the nautical mile is 
equal to I 3 5 o ordinary miles. How many feet are there in a nautical 
mile ? 

30. If i nautical mile be considered equal to 1} ordinary miles, 
how many ordinary miles are there in a distance of loj nautical miles ? 

31. One mile contains 1760 yds. If a " kilometre " is of a mile 
how many yards are there in i kilometre ? 

32. If i cubic metre = i^ cubic yds., find how many cubic yards 
there are in 260 cubic metres. 

Division of Vulgar Fractions. The division of one 
quantity by another is the determination of the number of times 
that the second quantity is contained in the first. The result 



| urn i ' 


1 


a 


3 


+ 


5 


G 


7| 8| 


J *--! * 




J z 


si * 


5 I 6 


?l e 


a| ,o 


ul 11 


131 H| 151 16 



i ur> 


JIT 




1 


,,,. .., U 


! | j 1 1 I I 
i I ! i j j j 






3 'S 


J 


8 * x 8 


^i 


\-k V"\m&<z> . ._ ^ 





Fig. 9. Illustrating Division of Vulgar Fractions. 

could be obtained by continued subtraction, but such a process 
would be extremely laborious. 

First take the case of dividing a fraction by a whole number, 
e - !*:" 3- T ne number of parts in the fraction is 15 and these 
are to be divided into 3 equal groups ; then obviously each group 
must contain 15 3 = 5 parts, and each part being y^-, the result 
is then -/ ff , so that the numerator is divided by the whole number. 
But this method becomes rather inconvenient when the numerator 
is not exactly divisible by the whole number, e. g. t f -r 2. Accord- 

2J. 

ing to the foregoing the result would be -, obviously an incon- 
venient form. A better statement of the result is obtained if the 



26 ARITHMETIC FOR ENGINEERS 

denominator be multiplied by the whole number, giving -f s . The 
truth of the method is evident from the upper diagram in Fig. 9. 

The second, and more general, case is the division of a fraction 
by a fraction. As an example take -j\~ f , i. e. t " how many times 
is f contained in f 1 " First divide T V by 3, i. e., the numerator 

only, giving -^ . But the actual divisor is 8 times smaller 
10 x 3 

than 3, and therefore the result must be 8 times larger than the 
previous one of / ff , that is 

9 . 3_ 9 v Q 

" /- ~T "n ~~ ~r A O 

16 8 16 x 3 

3 i 

fy x 3 3 , 
= * = ^ il 

^ X ^ 2 * 

2 I 

The proof is seen in the lower diagram in Fig. 9. 

It can be seen that the f has been turned upside down and 
now multiplies the 1 ^ r . Hence the rule for dividing one fraction 
by another : Invert the divisor, and proceed as in multiplication. 
Where mixed numbers occur, it is advisable, as when multiplying, 
to convert these into improper fractions. 

Cancelling should be used where possible. 

Example 19. Divide i^ by if. 

T 5 fS 

"T ~ I ff 

3 

== 17^ I 7 == \\ v $ ^3 
12-9 \\ 

This step usually omitted. 
Example 20. Find the value ofijxj-f-2g. 

I*Xi-r2f 

_ * x 7 x _5 _ 35 
- 2 X 8 X i - 6^ 

4 

Example 21. A large pipe-joint is to have a number of bolts evenly 
spaced on a circle whose length right round is 754"'. The bolts are to 
be spaced 3$ * from centre to centre. How many bolts must be used ? 

A glance at Fig. 10 will show that there are as many bolts as spaces. 



VULGAR FRACTIONS 27 

Then if we find how many times the 3| is contained in the 75 J we 
shall have the number of spaces and therefore the number of bolts. 
Then, 75* ~ 3* 

2 

*5 _i_I5 _i5i v i_ 
~ ' ~ 



or, as we cannot use a fraction of a bolt, 
say 20 bolts. 

Example 22. An engineering drawing 
is made to a scale of " J* to i ft." (i. ., 

every i ft. on the job is drawn as " on the Distance round This 
drawing). Find what fraction of full size "Pitch Circle- 11 is 75 2" 
this represents (i. e., find what fraction of an 
inch on the drawing represents \" on the job). lg ' I0 * 

The quantities must be changed into the same units before dividing. 
Now " J" to i ft." means that i ft. is shown as \" , also, changing to 
inches, i2 /r are shown as |*. 

/. i* will be J* -f- 12 

= f x A = iV'- 

4 

Thus i" on the job appears as fa" on the drawing, or a scale of |* 
to i ft. = ^ full size. 

Example 23. It is desired to make measurements from a certain 
drawing of which the " scale " is not known. A dimension figured 
as I2J" measures 2J". What is the scale ? 

I2J" is shown as 2 1". 
Then i" will be shown by 2j* -r- 124-" 



2 3 
Then the drawing is J full size or 2" to i ft. 

It should be noted that division does not necessarily produce 
a decrease, as might at first be inferred from the word " divide/' 
When we divide by a fraction, i. e., by something less than i, then 
the result of the division may be considerably greater than the 
number divided. Thus, if 3 be divided by J the result is 12, the 
reason being that J is contained 4 times in i, and therefore 12 times 
in 3 ; and as a proof 12 X J = 3. 



28 ARITHMETIC FOR ENGINEERS 

Exercises 6. On Division of Vulgar Fractions. 

Find the value of the following 

1. I -r ,!-,. 2. fr - r ,V 3. .?., -r- JJ. 4. if -> yV 

5. 3 A - y - 6. 1& -r- 3?,- 7. 4 J -r i?. 8. i j s -- JJ. 

9. ^ -r 14- 10. 5t -r 9- 

11. The top joint of an oil-separator shell is to have 40 bolts evenly 
spaced round the pitch circle, and the length of this circle is 123 J*. 
Find the " pitch " of the bolts (i. e., the distance from, the centre of 
any bolt to the centre of the next. Fig. 10 will make this clear). 

12. The length round a boiler ring on the pitch line of the rivets 
is 1 8 '-9", and the rivets are to be 2 J" pitch. (Fig. 10 will show what is 
meant.) How many rivets will be required? (Note. Work in inches.) 

13. The size of the teeth in a spur wheel are often given by the 
" diametral pitch," i. e., the number of teeth that the wheel has for 
every inch of its diameter (e. g., a wheel of 50 teeth and 5" diameter 
has 10 teeth per inch of diameter). Find the diameter of a wheel 
having 35 teeth and 2j teeth per inch of diameter. 

Find the diameters of the following spur wheels 

14. To have 162 teeth, and 12 teeth per inch of diameter. 

15. 55 2j 

16. 22 2| 

The following figures are taken from tests on toothed wheels. Find 
in each case the " pressure per inch of width/' i. e., pressure ~ width. 

17. Pressure on teeth 270 Ib. ; width i \" . 

18. 434 1 2"- 

19. ,, 416 ,, \\". 

20. 1638 -2\". 

What fractions of full size do the following scales represent : 

21. " to i ft. 22. iV' to i ft. 23. \" to i ft. 
24. I" to i ft. 25. ii" to i ft. 26. $\" to i ft. 
27. 4" to i ft. 28. T V to i ft. 29. J" to i ft. 

On referring to certain drawings it is loiind that the " scales " to 
which they are made are not stated. Find what fraction of full size 
the drawings are made in each of the following cases : 

30. A dimension figured as 8" measures 2". 

31. ioj" I". 

32. 6" 2jr /r . 

33. or ,. ir- 

34. 7 r .. 2ir- 

35. 45T 3H?"- 

36. ,, ior iji". 

Various Examples. Brackets. The study of the four 
simple rules applied to Vulgar Fractions is usually followed by 
the working of compound examples, i. e., those in which addition, 
multiplication, etc., are mixed up together in various ways. Such 
examples, however, seldom occur in practice, as decimal work is 
much more convenient, but one or two cases of a simple character 
will be taken as they serve to revise the foregoing principles of 



VULGAR FRACTIONS 29 

vulgar fractions, and these principles are of some importance in 
certain parts of algebraic work. 

They also provide a convenient opportunity for introducing 
another sign, the bracket, of which there are three common forms, 
thus : (the round), the {curly}, and [the square]. These are used 
to denote that the numbers and quantities contained within the 
two pieces of the bracket are to be treated as one quantity only. 
They may not be removed or inserted anywhere without obeying 
certain laws, which are given in the algebraic section, where brackets 
are of considerable importance. 

For the present it may be taken that the operations within 
any bracket must be carried out and a result obtained, before the 
remainder of the problem can be proceeded with. 

As a simple illustration of their use, let it be required to take 
the difference between J and J, from . Using brackets, this may 
be concisely stated thus : f (\ J). 

The statement is by no means the same as -J- i J. 

In the first case In the second case 

i (z 4) 4 2 8 

oo* 041. 

==: A - jl __ 3 - _____ 3_ . __ - JL 

488 o 8> 

which shows a considerable difference. 

It should be noted that the bar of division also acts as a bracket. 

c \ Q 
Thus, in the case of - (-> the 6 and the 8 must be considered as 

one quantity. Cancelling is not permissible where the terms of the 
division contain + and signs, unless all the numbers on either side 
of the line be divided at the same time. For instance, if it be desired 
to cancel the above expression by 2, both the 6 and 8 must be 

divided by 2, thus 

3 4 



8 
and not either of the following 

)$ + ! or 

10 x4 

8 
Thus(i) =" = i& and 

Now 5~ evidently = J-J-, which can be cancelled to |, thus proving 
the first system of cancelling to be correct. 



30 ARITHMETIC FOR ENGINEERS 

With compound examples, like the following, it will be found 
that the working presents a much better appearance, and can easily 
be re-read at any time, if the various operations are carried out 
in their proper places in the given expression. The steps in the 
next examples should be easily followed with but little explanation. 

7 

Example 24. Find the value of . 1 |T 

1 3 +io 



= ? 49 = -1 v 3 = * 
15 ' 30 ^ X ^ 7 

1 7 "" 

Explanation. The first operation must be to evaluate the bottom 
line, then the ^ can be divided by the result. The addition may be 
done in a separate place and the result collected. 

a 

Example 25. Simplify the expression -.-* . 

rs + tr 

Expression = ~r^7 inverting the 3 and 

- y -^ X 4 

^ A 3 cancelling by 2 

6 

9 

= t = 3 v ill = 'i 7 
A" ^7 *8 
4 

Where + and signs are found mixed up with X and 4- signs, 
the operations of multiplication and division must be performed before 
those of addition and subtraction, unless the existence of brackets 
indicates some other order. The order of the signs is quite immaterial. 
The reason for the rule is as follows : A sign of multiplication (and 
division is merely multiplication inverted) only stands between the 
components or factors of a number, and does not denote the exist- 
ence of two separate numbers. As an illustration consider the 
expression 5+3X4- Now it must be remembered that the 
sign X is only an abbreviation for a long addition, and 3 X 4 in 
the above means 4 + 4 + 4 (or 3 + 3 + 3 + 3). We may not 
then follow any method which neglects this meaning. 

Then 5 + 3 x 4 

= 5 + 4 + 4 + 4 = 17 



VULGAR FRACTIONS 31 

To obtain this result without replacing the multiplication by its 
lengthy meaning, the multiplication must be done first, as follows 

5 + 3X4 
= 5 + 12 = 17 

Following the wrong scheme of taking the signs in order (which 
may appear the natural way at first sight) we should have 

5 + 3X4 

= 8 X 4 = 32 

But in so doing we have neglected the meaning of 3 X 4 as given above. 
The signs of X and -~ may be looked upon as strong signs, and those 
of + and as weak signs, so that the former have to be obeyed first. 
Should it be desired to give the expression 5 + 3x4 the mean- 
ing of 8 X 4, then brackets must be used thus (5 + 3) x 4, thus 
indicating that both the 5 and the 3 are to be multiplied by 4. 
Of course, the existence of a fraction, in any form, makes no differ- 
ence to the above principles. 

Example 26. Find the value of i + f X & + ft. 

i i 

1 ^ ^ q 
The expression s= j+C x f + I 5 

i 7 

I+I+- 9 - 

2 ^ 7 ^ 16 

= 5 6 + l6 + 6 3 135 _ 23 

112 112 ~~ *!* 

Note. The multiplication must be done first. 
Example 27. Simplify the expression (J + ) X & -f ft. 
The expression = *_+_4 X + *"* 

9 Thcn 



. """ 4 16""" 16 

3 
Example 28. Find the v^lue of i T V~V 

The expression = i -- 1 <* ^ 

"To"" 

i 

==I ~S ==I ~^ x! 7 

3 __ 4 



ARITHMETIC FOR ENGINEERS 



of cuitmq-off tool 
' 1 



Example 29. Cast-iron piston rings, fa" wide, are to be turned from 
a rough casting. The width of the cutting-off tool is fa*, and 3* is re- 
quired at one end for holding in the chuck. Allowing y for facing up 
the end of the pipe, find the least possible length of pipe from which 

to cut i dozen rings (see Fig. n). 

After facing, which will finish one 
side of the first ring cut off, each part- 
ing will produce one ring. Therefore 
there will be 12 partings. 

Total length required == Chuck grip 
+ total width of partings -f- total width 
of rings -f- allowance for finishing 

Clld '_ 3 , 1 




Fig. II. 



= 3 + VM-Y- + i 
=-. 13!, say 14 in. 



Example 30. A mortar is to be composed of i part of cement to 
3 parts of sand, by weight. Find how much sand and cement arc 
needed to make 3 cwt. of this mortar. 

Using i Ib. of cement, say, we must use 3 Ibs. of sand, as there are 
to be 3 parts of sand to i of cement ; the total weight of the mortar 
is then 4 Ibs. Thus the weight of cement is of the total weight (i.e., 
i Ib. cement in 4 Ibs. of mortar), and the weight of the sand is | of the 
total weight (i. e., 3 Ibs. sand in 4 Ibs. of mortar). 

The actual working may be as follows 
Cement = i part 
vSand = 3 parts 



/. Mortar --= 4 parts 

Then cement J- of mortar J- 
and sand = i ^= J 



3 cwt. = 
3 ,. = 



cwt. 



Total (as proof) = 3 cwt. 

Nute. The method of working is always the same, no matter how 
many constituents there may be in the mixture. 

Exercises 7. Compound Examples. 

Find the value of the following expressions 



1 1H 

* 8| 
5 ' p-^ 

.!+-: 



2. 

6. ^ 

10. ? 



?* 



3. A- 

s T i 



-i 9 * 

2-*V 



7. 



i V i' 



11. 



A + iV 

2\ I* 



8. 3 
s 



V 

X 



VULGAR FRACTIONS 



33 



Simplify the following expressions : 
12. 2 - ij x i^. 13. (2j - ij) X i,V 

14- iV X | + f X f . 15. A X (I + ft X . 

16. f of | -r * + f . 17. t of f - (* + |). 

18. 3i + 2* -^ i - A- 19- (3i + 2i) -r- (J - i 



20. 



- i*. 



21. 



22. - x 



23. Special set screws are to be cut from bar stock in an automatic 
machine (see a, Fig. 12). The overall length of the screw is iA*. 
The width of the cutting-oif tool is J". and the bar stock is in 5 -o* 



li^^^.l.-*. __O I Ug_ . ti J If ,i - -, j.,.. tT -jr ___ 

16^ 1 16 HBr^ ne ^ | ^*" 




Lor\c 



Fig. 12. Illustrating Exercises 7. 



lengths. How many screws are obtainable from each bar, and how 
many bars will be required for an order of 12 gross of screws ? (Note. 
I gross 12 dozen = 144.) (Hint. Remember that if, say, 20 J 
screws can be cut from i bar, only 20 screws are really obtainable ; 
but if, say, 55J bars are required, 56 must be bought.) 

24. It is required to turn 9 handles each 5^' long, as shown at 6, 
Fig. 12, from a mild steel bar. Allowing -" extra on each handle for 
cutting-oif, etc., find what length of bar must be cut off for the 9 handles. 

25. Some sleeves, as at c, Fig. 12, are to be cut from a piece of 
wrought-iron pipe, in the lathe. If we allow -^" extra on each sleeve 
for parting off and i" extra at the end for holding when cutting off 
the last one, what length of pipe is required for 16 such sleeves ? 

26. Pillars for stop valves, as shown at d, Fig. 12, are to be turned 
from bars. Allowing an extra \? on each one for cutting off, how many 
pillars can be cut from a bar 5 / -6" long ? 

27. Electric contact fingers, J" wide, are to be sawn in a power 
hack-saw, from extruded bars each 6'-6* long. (See e, Fig. 12.) If 
the saw-cut is j^" wide, how many fingers can be cut from each bar ? 



34 ARITHMETIC FOR ENGINEERS 

28. Special nuts arc to be turned from the bar as follows : Seven f * 
thick, and fifteen " thick. Allowing an extra f s " on each nut for 
parting off, find the total length of bar required. 

29. Forty-eight angle stiffeners for a plate girder are required 40 J" 
long, and are sheared from the long angle bar without waste. What 
length of angle will be required ? If the stock angle bar is in 45 ft. 
lengths, how many lengths will be needed ? 

30. A concrete is to be made of cement i part and coke breeze 
4 parts by weight. How much of each constituent is required for 
15 cwt. of concrete ? 

31. What weight of each metal must be used to make 150 Ibs. of 
each of the following solders, assuming nothing lost in the melting ? 

(a) Lead 3 parts; tin 2 parts. 

(b) Tin 4 parts ; lead i part ; bismuth i part. 

32. Black gunpowder is composed of nitre 15 parts, charcoal 3 parts, 
and sulphur 2 parts, by weight. Find the quantities of each material 
required to make 60 Ibs. of gunpowder. 

33. The white metal for fusible plugs is made of varying propor- 
tions, according to the temperature at which it is required to melt. 
Find what weight of each metal is required to make i cwt. (112 Ibs.) of 
each of the following alloys : (a) tin 2 parts, bismuth i part ; (b) lead 
i part, tin 4 parts, bismuth 5 parts ; (c) lead 5 parts, tin 3 parts, 
bismuth 8 parts. 

34. Find what weight of each material is required to make the 
following quantities of foundry sand : (a) 2 cwt. of facing sand con- 
sisting of 6 parts old sand, 4 parts new sand, and i part coal dust ; 
(b) 10 cwt. of floor sand consisting of 6 parts old sand, 2 parts new 
sand, and part coal dust. (Give results in Ibs.) 

35. A filling material, for cast iron, which will expand on solidifying, 
is composed of lead 9 parts, antimony 2 parts, and bismuth i part. 
How much of each metal must be used to make 30 Ibs. of filling ? 

36. Find what weight of each material is necessary to make 75 tons 
of concrete, to consist of cement i part, sand 3 parts, and aggregate 
(broken brick, etc.) 6 parts. 

37. When helical springs are used in compression the last three- 
quarters of the coil at each end is flattened to provide a flat seating; 
these flattened portions do not form part of the effective coils. Eighteen 
such springs are required, each having 7 effective coils, the average 
length of a coil being 3f". Calculate the total length of wire in a 
complete spring and also the total length of wire required for the 
1 8 springs, allowing 2" per spring as wastage in cutting off. 

Find the values of the following expressions. 

38. I 39. * 40. JJL(LiL 



m _ 140 
41. 31 + J> 20 ' 42. (J x &) -i- (f+ 58)- 48. 



CHAPTER II 
DECIMAL FRACTIONS 

Decimal Notation. The ordinary method of numbering is 
built upon a scale which uses only ten signs, and is therefore called 
decimal notation. The ten different signs employed, which are 
called digits or figures, are o, i, 2, 3, 4, 5, 6, 7, 8 and 9; and all 
numbers are expressed with these only, by employing multiples 
of the unit in which 10 of each kind make I of the next larger 
variety. As examples of these multiples we have the ten, which 
is a set of 10 units ; the hundred, a group of 10 tens, *. e. 9 
10 X 10 100 ; the thousand, a group of 10 hundreds, i. e., 
100 X 10 = 1000 ; and so on. 

The value of each digit in a number (i. e., the multiple to which 
that figure refers) is indicated by its position in the number. Thus 
the extreme right-hand digit is the " units " figure, the next to 
the left the " tens " figure, the next the " hundreds " figure, and 
so on. 

For example, the number 15 means i set of 10 units + 5 single 

units (i.e., i X 10 + 5) 
the number 746 means (7 X 100) + (4 x 10) + 6, and 

5280 (5 x 1000) + (2 x 100) + (8 x 10) + o 

From the above examples it can be seen that, starting from the 
extreme left with the most important figure (i. e., the one of the 
largest multiple) and proceeding to the right, every multiple is j l 
of its left-hand neighbour, until finally we arrive at the simplest 
unit. Now proceeding to divide the unit, in order to deal with 
fractional quantities, it is only natural that we should make the 
first sub-multiple ^ of the unit, and the next -^ of that (i. e., 
^ of jV j J<j), and so on. Then, if the vulgar system of writing 
down such fractions were employed, we should have for denomi- 
nator a number such as 10, 100, 1000, etc., i. e., a number composed 

35 



ARITHMETIC FOR ENGINEERS 



of i with one or more noughts following it.* Fractions having 
such denominators are called " decimal fractions/' but the denomi- 
nator is omitted in writing down the fraction. The numerator 
only is written, the denominator being indicated by the position 
of the figures in the numerator. A dot (), called a decimal point, 
is used to distinguish between whole numbers and the numerators 
of decimal fractions; figures to the left of the point are whole 
numbers, and those to the right are fractions. The first place of 
figures to the right of the point is the T \j- place, the second the y^ 
place, the third the -nn^ place, and so on. 

All the foregoing points are illustrated in the diagram below, 
which shows the composition of a mixed number when expressed 
decimally. 



Multiples 



Sub-multiples 





o 
o 


o 
o 


8 



H 


M 




-13 


-IS 


E 


g 









o 


H 










H 


""r 


IHJO 






o 

M 


w 










. 





. 


. 





















+. 











JS 






-a 










o 
Pk 


. 


. 


(/) 








OJ 


TJ 


t/5 


, 


. 


i 





1 




crt 






O 


a 

03 


' 


t 


. 


'3 


c/> 


ft 


a 


O 






H 

a 


3 

s 


a 


s 


.-S 





'S 
"S 


a 


s 


H 
a 






H 


H 


ffi 


H 





H 


H 


ffi 


H 


H 










8 


1 


5 





3 


2 

- -~ 


7 







Whole Number 



Fraction 



300 



The number 815-327 means 815 + T 3 ^ + T rr + 

But, by the rules of Chap. I, T ^ - T |^, and - 
so that the number 815-327 = SiS^V 

Besides omitting the denominator in writing, it is also omitted 
in reading a decimal fraction : the figures following the point are 
simply read in order. Thus the above quantity 815*327 is read as 
eight hundred and fifteen point three two seven. As further examples 

39*37 reads thirty-nine point three seven and 39 + ^ + ioV 



* Numbers such as these (i.e., those composed of i followed by one 
or more noughts) are called " powers " of 10 : see Chap. III. 



DECIMAL FRACTIONS 37 

But as ^ = jv, a shorter statement in the vulgar form is 39 $$ . 
Similarly, 62-425 reads sixty-two point four two five, and = 62 

+ 4 I 2 I 6 A/> 425 

TTF -T TTTTF ~T TfftfTT ~ 2 TTJtf tf- 

When a whole number alone is written down, the decimal point 
is omitted, as there is no fraction to consider. If a number such 
as 15-00 is obtained as the result of any calculation, then the dot 
and the noughts may be omitted, and the figure written simply as 
15, since -oo is, of course, o. But when a proper fraction has to 
be written down, then the dot is absolutely necessary, as its omission 
will convert the fraction into a whole number. Thus -305 is the 
fraction -^nnr* an( ^ ^ the dot were omitted the figures would read 
305 ! For this reason a nought is sometimes written in front of 
the point thus, 0-305, so that, should the dot be lost, the appear- 
ance of the figures o 305 would indicate its intended existence. 
The number -305 may be read as either " nought point three nought 
five " or simply " point three nought five. 1 ' 

It should also be noted that noughts may be added to the right 
of the figures of the fraction, and to the left of the figures of the whole 
number, without affecting the value of the number. Thus 3-1400 
is only 3-14, the o's merely stating that no y^ths or ^i^ths 
exist, which is equally well shown by writing nothing whatever 
beyond the 4. For similar reasons 0078-54 = 78-54. 

Movement of the Decimal Point. It is a feature of the 
decimal system that any number may be multiplied or divided by 
10, or 100, or 1000, etc., without any actual operation on the figures 
themselves beyond moving the decimal point. As an example 
take the number 39-37. This = 39^, which may be expressed 
in the improper form -^nnr- H the decimal point be moved one 
place to the right the figures will appear as 393-7, which = 393^ 
= m*-. Converting to ^ths we have - 3 -|^, which = Stfp- X 10, 
so that 393-7 is 10 times 39*37. The converse is, of course, equally 
true. Thus 39-37 is ^ of 393-7. Hence the rule : to multiply or 
divide a number by 10 move the decimal point 1 place to the right or 
left respectively. 

Since 100 = 10 x 10, then to multiply or divide by 100 the 
above rule should be obeyed twice. Similarly, to operate by 1000, 
which = 10 X 10 x 10, the decimal point must be moved three 
times. Hence we have a more general statement of the above rule : 
To multiply or divide a number by any power of 10,* move the decimal 
point to right or left respectively, 1 place for each in the multiplier. 

* See foot-note on p. 36. 



38 ARITHMETIC FOR ENGINEERS 

As examples : 2-5 X 10 =25 (the dot is not needed final]) 7 ). 

2-5 ~io = -25 

7854 x 100 = 78-54 

3-1416 x 1000 = 3141-6. 

When a whole number is to be operated upon in this manner, 
and apparently has no decimal point, e. g., 1760, then it must be 
remembered that the dot really exists immediately to the right of 
the units figure, i. e., after the o. Then places may easily be counted 
off when moving the dot to the left. Thus 1760 -f- 100 = 17-60. 
When moving the dot to the right the operation is that of adding 
a o for every time that the number is multiplied by 10. Thus 
1760 x 100 = 176,000. 

Similarly, when moving the decimal point to the left and no 
more existing figures are available, noughts may be introduced as 
necessary. Thus 3-14 -r 100 "0314, for we may consider 3-14 
as 03-14. 

Also -0625 -f- 10 = -00625. 

Exercises 8. On Movement of Decimal Point. 

What is the result of 

1. () 5'5 X 10. (b) 5-5 10. (c) 1-34 X ioo. 
(d) 1-34 4- ioo. (e) -153-6 x 1000. (/) 453-6 H- 1000. 

2. (a) -ooi 1 18 x 1000. (b) -0000087 x 10,000. (c) n>ooWo- 
(d) 35 x 1,000,000. 

3. If i kilowatt = 1000 watts, convert the following into watts 
(a) 35 kw. (b) 7-3 kw. (c) -08 kw. 

(Hint. Multiply by 1000.) 

4. Convert the following into kilowatts 

(a) 29,500 watts. (b) 231 watts. (c) 305,000 watts. 

(Hint. Divide by 1000.) 

5. If i volt 100,000,000 absolute units, convert the following 
into absolute units 

(a) 2 -7 volts; (b) -005 volts ; (c) -000003 volts. 

(Flint. Multiply by 100,000,000.) 

6. If i Joule = 10,000,000 ergs, convert the following into Joules 
(a) 42,600,000 ergs, (b) 550,000 ergs. 

Conversion from Decimals to Vulgar Fractions, This 
has already appeared in the section on decimal notation, and only 
a few points remain to be noted. It will be seen, from the examples 
given in the section referred to, that the denominator consists of i 
followed by as many o's as there are figures in the numerator ; for 
example, in the case of 39-37 the denominator (ioo) contains two 
noughts, one for each of the figures behind the decimal point. 
Similarly with ^425 there are three noughts, i. e., the denominator 
is 1000. Hence we have the following rule : For the numerator 



DECIMAL FRACTIONS 



39 



write the figure or figures behind the decimal point. For the denomin- 
ator write 1 for the decimal point and a for every figure following it. 
Finally reduce the fraction to its lowest terms. 

Example 31. Convert the following decimals into vulgar fractions 
(a) -2; (b) -5; (c) 0-3; (d) -45; W -125- 

(a) Numerator = 2. Denominator i followed by one nought, 

i. e., 10, as there is one figure behind the decimal point. 
/. -2 = j 2 ; cancelling by 2 gives . 

(6) -5 =! 5 o: .. 5 

(c) 0-3 = 1%- ; no reduction possible. 

9 

(d) -45 = I'\M>; cancelling ^ == J Q . 

JO 

(e) -125 = fiffa ; cancelling by 125, or 3 times by 5, gives ^. 

In cases where the decimal contains a o, this must be counted 
as a " figure behind the decimal point/' 

Thus -305 = -roVci = -V<r cancelling by 5. 

This rule must be noticed particularly when a o or o's follow 
immediately after the decimal point. 

Thus 025 = 1 Q i oV= i wir iV cancelling by 25 or by 5 and 5 again. 

Similarly -0036 = ^VV = roVW = WW cancelling by 4. 

(The steps ^Vo anc ^ voVoV aDOve arc only introduced for ex- 
planation and need not be put in the working.) 

Since many of the simpler fractions are frequently used in the 
vulgar form, it is desirable to remember the decimal equivalents of 
the more common ones. The following table gives the decimal 
equivalents for every ^ up to i. Certainly the equivalent for 
every should be memorised, and it is advisable to learn also the 
equivalents for the intermediate sixteenths. 

DECIMAL EQUIVALENTS. 



Fraction. 


Decimal 
Equivalent. 


Fraction. 


Decimal 
Equivalent. 


Fraction. 


Dec-mal 
Equivalent. 


T? S ' ' 


03125 


3 
B 


375 


V?. 


71875 


,'* - 


0()25 


Ji - 


40625 


I 


75 


3, " ' ' 


09375 


./,> 


4375 


l - 


78125 


1 

8 


125 


M 


46875 


U - 


8125 


& 


15625 


1 
li 


5 


P - 


84375 


ft 


1875 


..- 


53125 


7 
3 


875 


Tts 


21875 


10 '"* ' ' 


5625 


3S 


90625 


4 


25 


18 - - 


'59375 


8 


'9375 


A ... 


28125 


5 
y 


625 


31 ... 


96875 


A 


3125 


;5J 


65625 


i 


I '00 


M 


'34375 


| U - 


6875 







40 ARITHMETIC FOR ENGINEERS 



Exercises 9. On Conversion of Decimals to Vulgar 

Fractions. 

Convert the following decimal fractions into the vulgar form 

1. (a) -65; (b) -45; (c) -425. 

2. (a) -028; (b) -875; (c) -3264. 

3. (a) 2-205; (b) 1-34; (c) -305. 

4. (a) -964; (b) 3-142; (c) -00256. 

5. Convert into vulgar fractions : (a) -515625; (b) '390625. 

Degree of Accuracy. Significant Figures. Mathematic- 
ally there is no limit to the number of figures which may be written 
on either side of the decimal point, but a few only are required for 
practical problems. As some operations with decimals produce 
large numbers of digits in the results, it is desirable to consider 
how many figures are really necessary in any mixed number. All 
the quantities with which we have to deal are connected with 
measurement ; the accuracy with which a quantity can be stated 
depends upon the nature of the measurements relating to that 
quantity. Thus we might say with truth that the size of a steel 
ball, taken from a ball bearing, was 1-2498" ; machines and instru- 
ments which can measure i^J^n/' (-0001") in a distance of ij* are 
fairly common, and it is also very desirable to know the size of the 
ball accurately in order to secure smooth and safe running of the 
bearing. But to say that the length of a liner's voyage is 2536-4978 
miles would be utterly absurd, for the actual distance travelled 
would depend so much on the state of the weather, ocean currents, 
etc., etc., that we could not be sure that even the last figure in the 
whole number (the 6) was correct, much less any of the succeeding 
figures. We could not give, with confidence, the distance any 
more accurately than 2540 miles. Evidently, then, there is no use 
in stating the figures 6-4978, and still less in using them in any 
calculation in which the above distance is required. A quantity 
such as this 2540 miles is, then, only an approximate, or nearly 
correct, measurement. When describing the accuracy to which 
this distance has been measured we say that it is " correct to 3 
significant figures" the digits 2, 5 and 4 being the " significant 
figures " ; that is to say, it is as near as possible to the truth while 
only containing 3 significant figures. The o only helps to denote 
the various multiples of the figures 2, 5 and 4, i. e., it fixes the 
position of the decimal point and does not itself signify any 
number. 



DECIMAL FRACTIONS 41 

All the digits i, 2, 3, 4, 5, 6, 7, 8, and 9 are significant figures 
in all cases. The digit o is only a significant figure when it has a 
higher digit on both sides of it, as in the case of the number '305. 
Then the o is significant, distinguishing the number -305 from '035 
or '350. Thus, taking the following numbers : 

In -305 the sig. figs, are 305. In 6o65 the sig. figs, are 6065. 

-035 35. 6080 608. 

-035 ., 35- -000341 341. 

n 2938 2938. 30,000,000 fig. IS 3. 

Most of our measurements fall under this " approximate " 
heading, and consequently are only stated " correct to so many 
significant figures/' The actual number of significant figures 
depends upon the nature of the measurement. Thus in giving 
the horse-power of a ship's engines 2 significant figures are usually 
all that can be relied upon, e. g., 12,000 h.-p. ; but when stating the 
length of a base-line used for the large surveys of Great Britain, 
7 or more significant figures are given, e. g.> 26405*78 ft. ; and in 
accurate chemical work it is quite possible to work to 7 significant 
figures. No definite rules can be given for the number of significant 
figures desirable in any measurement or result. It is a matter of 
experience, and the reader must be guided by the style of answer 
given to the various examples, both worked and set, throughout 
this book. It may be taken, as a general rule, that 3 or 4 significant 
figures are quite sufficient for most engineering calculations. 

As a matter of interest it may be noted that calculations on the 
ordinary slide rule (Chap. X) are performed with the " significant 
figures " only, the position of the decimal point being fixed later. 

The adjustment of a quantity to any required number of sig- 
nificant figures is governed by the idea that any error introduced 
thereby shall be as small as possible. Thus if the 2536 be required 
correct to 3 significant figures it should be written 2540, and not 
2530. The difference between 2536 and 2540 is 4, whereas between 
2536 and 2530 it is 6. Hence less error is introduced by using 2540 
rather than 2530. But the number 2533 correct to 3 significant 
figures is 2530 and not 2540, the difference between 2533 and 2530 
being 3, whereas between 2533 and 2540 it is 7. 

Hence we have the rule : increase the last figure retained by 1 

when the rejected figures are over 5000 but leave it unaltered 

when the rejected figures are less than 000 When one figure 

only is to be rejected and is 5 exactly, either of the above may be 
applied. 



42 ARITHMETIC FOR ENGINEERS 

Example 32. Express the following numbers correct to 2 significant 
figures: (a) -333; (b) 1-666; (c) -01734; (d) -7071; (e) 10-01 , 

(/) 2 35-5; fe) i'57 8 ; W '995 r - 

( a ) *33 as 3 is less than 5; (6) 1-7 as 66 is greater than 50 ; 
(c) -oi7as3 4 ,, 50; (d) -71 71,, 50; 
(e) 10 as 01 05; (/) 240,, 55,, ,, 50; 
(g) 1-6 as 708 is greater than 500; (h) i-o as 51 is greater than 50. 
Note. In case (h) only one sig. figure can be given. 

Example 33. Express the following numbers correct to 3 significant 
figures: (a) 3744; (6) 39'37i W 2-7183; (d) 2-3026; (e) 1-897; 



(a) 3740; (b) 39-4.' ( c ) 2-72; (d) 2-30 (i.e.. 2-3); 
(*) 1-90 (i.e.. 1-9); (/) 4880. 

. In (d), since only 3 figures are required, all figures to the 
right of the o (the third significant figure) must be eliminated, leaving 
2-30. The last o has no use, so that our result is 2-3. This contains 
only 2 significant figures but cannot be avoided since the third figure 
is a o. 

In (e), since 3 figures are required, the digit to the right of the 9 
must go out, but being a 7, i. e., over 5, must be allowed for by adding 
I to the 9. This gives 1-90, i. e., 1-9, or only 2 figures, which, however, 
cannot be avoided. 

Example 34. Express the number 3-1415928* correct to 
(a) 5 significant figures ; (b) 4 significant figures ; 
(c) 3 significant figures. 

(a) 3-1416; (b) 3-142; (c) 3-14- 

It should be noted that, after adjusting, the last significant 
figure is not reliable. Where several numbers which have been 
thus adjusted are used in a calculation there is a possibility of 
introducing errors beyond those of measurement. To avoid this 
it is usually advisable to carry our calculations to i figure more 
than is required, and to adjust to the required number finally. 
Thus a problem requiring an answer to 2 significant figures should 
be worked to 3, and the answer reduced to 2 finally. 

Exercises 10. On Significant Figures. 

What are the significant figures in the following numbers : 
1- 39-37: 5280; 2-205; 30-45; -0365. 

2. 29,850,000; 10,500,000; -0807; 1083; 454. 

3. -00000066; 606-5; 'OoniS; 32-09; 1-5708. 

4. 10-9; '0000087; -0000105; 600-9; 2000. 



* This particular number, which is very important in connection 
with circles, has been calculated to 700 decimal places ! 



DECIMAL FRACTIONS 43 

Express the following numbers correct to 3 significant figures 

5. 453*6; 30-48; -4343; 62-425; -001118. 

6. 14380; 1393; 32-182; 600-9; 3749-9. 

Express the following numbers correct to 2 significant figures : - 

7. -0807; ^282; 14-7; -0646; 5-37. 

8. 2-205; 606-5; *i6o4 ; 29,850,000; '1145. 

9. The Imperial Gallon contains 277-274 cubic inches. Express this 

(a) correct to 4 significant figures ; (b) correct to 3 significant figures. 

10. The British pound weight contains 453-5924 grammes. Express 
this (a) correct to 5 significant figures ; (6) correct to 3 significant 
figures. 

Express the following numbers correct to (a) 5 significant figures ; 

(b) 4 significant figures; (c) 2 significant figures 

11. -142857. 12. 2-71828. 

Addition and Subtraction of Decimals. Since ten is the 
(< rate of exchange " on each side of the decimal point, addition 
and subtraction with decimal fractions are carried out in exactly 
the same way as with whole numbers, provided that all the decimal 
points are kept in a vertical line, and the various decimal figures 
representing the same decimal place are directly beneath each 
other. That is to say, tens may only be added to tens and units 
to units, and similarly tenths may only be added to tenths and 
hundredths to hundredths, etc. 

Thus, finding the total of 25-3, 12-57 an< ^ '^ 2 

25-3 
12-57 

O-02 

38-49 

d c b a 

Explanation 
Column a, 2nd place of decimals, 2 -{- 7 =9, write 9. Nothing to carry. 

b, ISt 6 -{- 5 -[-3 1 4, ,, 4. Carry i. 

,, C, units I -f- 2 -[- 5 ~ 8, ,, 8- Nothing to carry. 

d,tens i-\-2 = 3, 3. 

Result = 38-49. 

Apart from mistakes in common addition, the only blunders 
likely to arise are from not writing the figures and decimal points 
in distinct separate columns, when figures of the wrong columns 
may get added together. Therefore great care should be given to 
the writing down of the figures. 



44 ARITHMETIC FOR ENGINEERS 

Example 35. When considering the strength of a certain steel 
column it was necessary to add together the quantities 6-48, 0-312, and 
2 '3 7. Find the value required. 

6-48 

0-312 

2*37 

! ! The 2 may be neglected. 

9-162 Then result = 9*16. 

In subtraction, the number to be taken away is written below 
the number from which it is to be taken, the subtraction being 
performed in the ordinary way. The previous remarks on the 
decimal point and the neat spacing of the figures should be noted. 

Thus in taking 1*875 from 2-327 

2-327 
1-875 



452 = Result. 

Example 36. When measuring the difference of pressure between 

two points of a pipe in which water is flowing, a mercury U gauge 

(see Fig. 13) is sometimes employed. The top surface of the mercury 

stands opposite -817 ft. on the scale, and the bottom surface opposite 

199 ft. What is the " difference of level/' i. e. t the distance marked " ? " 



Top reading = -817 

Bottom ,, = -199 




Difference of level = -618 ft. 



Fig. 13. 



Example 37. In connection with the strength of a cast-iron rail 
section certain areas had to be measured with a planimeter.* The 
following three pairs of readings were taken, the area in each case being 
the difference of the pair of figures. Find these areas (sq. ins.). 

(a) 19-45, 12-25; (&) 57' 6 3 5 6 '5 2 ; ( c ) 7'7 6 > 6 9'09- 

(a) 19-45 (&) 57* 6 3 (c) 77 6 

12-25 56-52 69-09 

7-20 sq. ins. i-n sq. in. 1-67 sq. in. 

* An instrument for measuring accurately, and with very little 
calculation, the area of an irregular figure. 



DECIMAL FRACTIONS" 



45 



Example 38. In finding the bending moment at a certain section 
of a girder it was necessary to take 62-2 ft. tons from the sum of 3*4, 
45-6 and 37'6 ft. tons. Find this value. 

3 4 ft. tons 
45* 6 



Sum 



86-6 
62-2 



Difference = 24-4 ft. tons 



Result = 24*4 ft. tons. 



Example 39. A portion of a surveyor's levelling book is shown. 
The height of the place called G above the place called A is found by 
subtracting the sum of column 3 from the sum of column 2. Fine? 
this height, all the figures being in feet. 



Point. 


Back Sight. 


Fore Sight. 


A 


7-52 




B 


4-09 


4-41 


C 


9'75 




D 


4-14 


I I2 


E 


2-01 


2 '73 


F 


1-99 


4-69 


G 




8-78 


Totals . 


29-50 


26-89 



Deduct Total Fore Sights 26-89 

(*. e. t Col. 3) 

2*61 



Result : Place G is 2-61 ft. 
above place A. 



The engineer has seldom to add up long columns of figures. 
A good deal of the decimal addition will be only with two or three 
numbers at a time, and hence the method of writing down in vertical 
columns, shown in the foregoing examples, is often omitted as 
being laborious. With a little practice one is soon able to sort out 
the last decimal place (say the fourth) in each number to be added, 
and to perform the addition of those particular figures mentally; 
then pass on to the next larger decimal place (say the third) 
and repeat the process ; and so on until the example is complete ; 
similarly with subtraction. 



4 6 



ARITHMETIC FOR ENGINEERS 



The following examples will serve as illustrations 

Example 40 In a problem on a retaining wall it was necessary to 
find the sum of the distances 1*19 ft. and 2-33 ft., and then to subtract 
2-83 ft. from this sum. Find the resulting measurement. 

We have 1-19 + 2-33 2-83 
= 3-52 - 2-83 
= -69 ft. 

Example 41. The expression 92*1 -f '56 ( 2 5 -f- i?'9) relates to 
the flow of water through a pipe. Find the value required. 

92-1 -f- -56 (25 -f 17-9) 
= 92-66 42-9 
= 49-76, say 49-8. 

Exercises 11. On Addition and Subtraction of Decimals. 

1. Find the sum of 10-5, 63-58 and 25-2. 

2. Add together 67-35, 85-62, 105-14 and 5-7. 

3. Add together 125-5, 67*825, 1-5625, 13-125 and 12-03125. 

4. The total head against which a centrifugal pump has to work 
is made up of three heads : loss of head in pump 5-5 ft., loss by friction 
in the pipe 93-6 ft., and height through which water has to be lifted 
250-5 ft. Find the total head. 

5. In a laboratory experiment the following weights (Ibs.) were 
placed on a certain hook : -7, 2-9, 3, 3, 2-9, 2-9, 3. 3, 2, 2, 1-5, 1-5, i-i, 
2, 3, 1-5. Find the total weight on the hook. 

6. The lengths of the lines in a traverse survey were as follows : 
13-63, 3-8, 6-97, 9-41, 3*48, 6-1 1, 9-66, 5-08, 6-84, 2-07 chains. Find 
the total length of line surveyed. 

7. In calculating the bending moment on a continuous girder the 
expression 5 -j- 3-75 -f- 1-06 -f- 1-42 was obtained. Finish the calculation. 

8. The following table gives the chemical analysis of some gaseous 
fuels. The combustible constituents are hydrogen, methane, and 
carbon monoxide. Determine the total number of parts of combustible 
matter in each gas. 



Gas. 


Carbon 
Monoxide. 


Hydrogen. 


Carbon 
Dioxide. 


Methane. 


Nitrogen. 




parts 


parts 


parts 


par ts 


parts 


Dowsoii Gas 


25-07 


i8'73 


6-57 


62 


49-1 


Mond Gas . . 


10-00 


23-00 


I5-00 


3-00 


49-00 


Natural Gas 


41 


1-64 


2 5 


93-7 


3MI 



9. Find the difference between 68-75 an( i 59'&5' 

10. Find the result of 125-3 68-45 26-9. 

11. In an experiment on linear expansion, the following were some 
figures (millimetres) taken for brass and steel 

(a) Brass. Initial reading -5 Final reading 2-3 

(b) Steel. ,, ,, 3*215 ,, ,, 4*225 

Find the actual expansion in each case (i. e., the difference between each 
pair of figures) . 



DECIMAL FRACTIONS 



47 



12. The following figures were taken in a delicate experiment on 
the electrolysis of copper sulphate 

Weight of plate before depositing . . . 63-8203 grammes 

,, ,, after ,, ... 64-5091 ,, 

Find the weight of copper deposited on the plate. 

13. In an experiment on the latent heat of steam the following 
weights were measured- 

Weight of beaker (empty) 264 -9 grammes 

,, ,, -V cold water .... 1428-6 ,, 
,, ,, ,, -f- cold water -f- steam condensed 1480-5 ,, 

Find (a) weight of cold water used; (b) weight of steam condensed. 

14. Find the difference (a) between '3575 and -5463, and (b) between 
"3575 an d '6453, figures which occurred in measuring the quantity of 
water flowing over a V -notch. 

15. Kxtreme sizes of a shaft and hole, nominally 3" diameter, when 
made to limit gauges are given below 

Maximum diameter of hole = 3-0035 in. 
Minimum ,, ,, ,, 3-0000 ,, 

Maximum ,, ,, shaft = 2-9965 ,, 

Minimum ,, ,, ,, = 2-9930 ,, 

Find the extreme clearances which can occur, i. e , find the difference 
in diameter between the largest shaft and the smallest hole; and 
between the smallest shaft and the largest hole. 

16. and 17. The following tables arc from a surveyor's level book. 
Find the difference between the sums of columns FS and BS in each 
case (which gives the difference of level between the first and last 
points). 

16. ~~ 17. 



Pt. 


B.S. 


F.S. 


A 


23 




B 


3'4 T 


13-96 


C 


4-18 


9\57 


D 




4'95 



Pt. 


B.S. 


F.S. 


A 


2-475 




13 


3-37 


3*9 I 


C 


2-15 


4") 9 


D 




6-21 


E 


2-13 


5-33 


F 




2-15 



18. Figures from a surveyor's level book are given. If the levelling 
is correct the total of column 11 will equal the total of column F. Any 
difference is an error. Find the totals of columns R and F and find 
what error exists, if any. (All measurements in feet.) 



Pt. 


B.S. 


F.S. 


R. 


F. 


A 


7*33^ 









B 


10-265 


9'455 





2-119 


C 


3'44i 


5'55 


5-210 





D 


4-60 


3-3^5 


076 





E 


6-61 


9-885 





5-285 


A 





4-49 


2'12 





48 ARITHMETIC FOR ENGINEERS 

19. Referring to the last exercise, the sum of column B.S should 
equal the sum of column F.S. Any difference is an error. Find what 
error, if any, exists. 

20. Find the value of 3*75 1-42 + I '5> figures obtained in calcu- 
lating the bending moment on a beam. 

21. One of the reactions of a continuous girder was calculated to be 
2 -5 -f- I% 5 "h 2*5 -27, all in tons. Complete the calculation. 

22. Find the value of the expression (connected with a bending 
moment) 12-7 47 15-55 + 12-3 1-75. 

Multiplication of Decimals. Let us consider the multiplica- 
tion of two decimals, e. g., & X '2. Convert each decimal into a 
vulgar fraction, but omit the cancelling. Then '8 becomes ^ and 2 
becomes T 2 ^. The expression -8 x *2 may now be written as ^ x ^. 

Multiplying (without cancelling) we have 

TV x TV ^ iVV *' e -> * tenth + 6 hundredths, 

or *i6 in the decimal form. It should be noticed that the figures 
in -16 are simply the product of the figures in -8 and -2 ; also that 
the total number of decimal places in the quantities multiplied is 2 
(i in '8 and I in 2) and there are 2 decimal places in their result -16. 
Again, let us consider 2-45 X 1-3: 

V.AZ _ <7 45 _ 2 iff 

* 45 * T<rff i<7i7 
also 1-3 = if\ = {. 
Hence we have ^g X | J = \\% = 3-185. 



Again notice that the digits in 3-185 are the product of the digits 
in 2 '45 and 1-3 ; also the total number of decimal places in the numbers 
multiplied is 3 (2-45 has 2 decimal places and 1-3 has i), and there 
are 3 decimal places in the result 3*185. 

Any example may be similarly treated and the same points will 
be seen. Hence we have the rule for the multiplication of decimals 

Multiply the figures together without considering the decimal point. 
Count up the total number of decimal places in the quantities multiplied, 
and starting from the right, point off this number of figures in the result 
obtained. 

To illustrate this rule, and also the method of writing down 
the work, consider the second example just taken, i. e., 2-45 X 1-3. 
Either number may be taken as the multiplier ; but usually it will 
be found easier to take the one with the fewer digits by which to 
multiply, in this case 1-3. Set down the quantities as in ordinary 
multiplication ; the decimal point need not be placed in any particular 
position. Multiply, disregarding the decimal point (i. e. t as in ordinary 
multiplication). 



Thus 



DECIMAL FRACTIONS 49 

Mental work to fix the decimal point 



2*45 . , . , .2 decimal places 
1-3 . . . . .1 decimal place 

735 

245 _ 

3-185 . .... 3 decimal places. 

This gives the figures 3185. 

Now fix the decimal point thus : Count the number of decimal 
places in 2*45 (two) and add this to the number of decimal places 
in 1*3 (one), making 3 decimal places in all. Starting from the 
right in the product 3185 count off 3 digits and place the decimal 
point : thus point off the 5, the 8 and the i, and then write the 
dot. Thus the product of 2-45 and 1*3 is 3-185. 

Example 42. If i cubic foot of water weighs 62-5 Ibs. and cast 
iron is 7-2 times heavier than water, what is the weight of i cubic foot 
of cast iron ? 

i cu. ft. of cast iron will, of course, weigh 62-5 X 7*2 Ibs. 
62-5 i decimal place 



1250 
4375 
450'QO 2 decimal places. 

The total number of decimal places in 62-5 and 7-2 being two, two digits 
are pointed off from the right in the product 450-00. 
.'. i cu. ft. of cast iron weighs 450 Ibs. 

Occasionally, when a large number of decimal places are required 
to be pointed off in the result, it will be found that there are not 
sufficient figures available. Then a nought must be supplied for 
every required figure which does not exist in the product obtained. 
The following example will illustrate. 

Example 43. In finding the area of a small circle it was necessary 
to multiply together the numbers -785 and -0039. Find this value. 

Again the number with the fewer multiplying figures is taken as 
the multiplier, i. e. t -0039. 

785 . . 3 decimal places 

0039 4 

7065 
2 355 



0030615 . . 7 decimal places. .*. Result = -00306 



50 ARITHMETIC FOR ENGINEERS 

The multiplication of 785 by 39 gives the figures 30615 as result. Seven 
decimal places being required and five figures only existing, two noughts 
must be inserted to the left of the 3, in order to complete the seven 
decimal places. The answer then becomes '0030615. The last two 
figures may be neglected and the result given as '00306. 

The foregoing method can be used for the multiplication of 
more than two numbers. In these cases a large number of figures 
will be obtained in the result, and many will have to be discarded 
as useless. Much labour can be saved, and many useless figures 
avoided, by reducing the result each time to 3 significant figures before 
using the next multiplier. In the following example the multiplica- 
tion is worked in ordinary " long hand " manner, and also by the 
method just suggested. It will be seen that there is a big saving 
in labour, and the difference in the results is not worth considering. 
To 3 significant figures the results are the same. 

Example 44. Evaluate 1670 x 2-75 X 3-73 X 6-78. 

Long hand working. Working when each result is 

reduced to 3 sig. figs. 

1670 1670 

275 2-75 



8 35 835 

11690 11690 

334 3340 



373 
say 4590 to 3 sig. figs. 



137775 373 

321475 



17130-025 1377 

6-78 



17120-70 



137040200 

say 

119910175 y 

102780150 



116141-56950 I368 

^ 119700 

102600 



115938-00 
to 3 sig. figs. = 116000 to 3 sig. figs. = 116000 



DECIMAL FRACTIONS 51 

It can be seen from the above that the multiplication of decimals 
becomes very long and tedious when more than two numbers are 
multiplied together. There are " contracted " methods of multi- 
plying which are sometimes advocated, but their value is very 
doubtful, and they are probably never used in practice. The 
majority of engineering multiplication (and division) can be carried 
out with sufficient accuracy by logarithms (see Chap. VI), which 
are very simply and quickly used and certainly should be known 
by any one attempting calculation seriously. 



Exercises 12. On Multiplication of Decimals. 

(Note. Nos. i to 10 are set to test the student's accuracy of 
multiplying; the answers should be given both fully and to 3 sig. 
figs.) 

I. Multiply 29-5 by 1-38. 2. Multiply 235 by 1-72. 
3. -057 13-9- 4. 15-95 .. 7 8 5- 
5. 3-14 ,, 19-6. 6. ,, -0087 ,, -054. 
7. ,, 13400 ,, -0036. 8. ,, 191*5 ,> '0986. 
9. -0046 ,, 21-2. 10. ,, 1114 19*25. 

II. If a cruiser's speed is 27 knots, what is this in miles per hour? 
(i knot = I -15 miles per hour.) 

12. The speed of a cargo boat is loj knots. The speed of a tramcar 
is 12 miles per hour. Which vehicle is the faster ? 

13. The calculation of the *' moment of inertia" of a cast-iron rail 
section (a figure used in connection with the strength of the 
section) gave the figures 2 X 1*78 X 2-7 X 2-7. Calculate the required 
value. 

14. In calculating the weight per foot of a certain angle bar 
the following figures appeared : 7*11 X 12 X '28 Ib. Finish the 
calculation. 

15. The following calculation was necessary to convert the readings 
of a water meter into Ibs. per min. : 60 x 62-5 X -026. Find the 
required value. 

16. The calculation for the weight per foot of a large wrought-iron 
pipe produced 10-25 X '7^5 X 3-4 Ibs. Find the required value. 

17. The working volume of an engine cylinder was given by the 
expression 36 X 3*14 X 24 cu. ins. Find the required value. 

18. Find the value of the following figures, which give the amount a 
certain rail will expand under certain conditions : 360 X 25 X -0000087 
inches. 

19. Pressures in connection with compressed air are frequently 
spoken of as " atmospheres," i atmosphere being 14*7 Ibs, per sq. in. 
Find the number of Ibs. per sq. in. in 35 atmospheres. 

20. Find the number of Ibs. per sq. in. in 32 J atmospheres. 

21. The area of the steam cylinder in a steam-driven air compressor 



52 ARITHMETIC FOR ENGINEERS 

is to be 1*38 times the area of the air cylinder. Find the steam-cyhnder 
area when the air-cylinder area is 113 sq. ins. 

22. Find the total resistance of 2 miles of electric cable whose 
resistance per yard is -008 ohms (i mile = 1760 yards). 

Division of Decimals. Let us first consider the case where 
the divisor is a whole number, e. g., 1175 -r 5- Express 1175 as 
an improper vulgar fraction without cancelling, and it becomes 
-VoV"' Divide by 5 according to the method stated in Chap. I, 
for the division of vulgar fractions, i. e. t divide the numerator. In 
this case divide 1175 by 5 = 235. 

Thnn TT-TS ~ < _ IITB ^ __ 235 

men n 75-7-5 flro - -r 5 100* 

Expressed decimally the result is 2*35. Now it will be seen that* 
the digits in 2-35 have been obtained by dividing the digits in 1175 
by 5 in the ordinary manner ; also that the whole number of the 
quotient is complete when the whole number in the 1175 is 
finished with, i. e., 2 is obtained when u has been divided. Or, 
when the decimal point, is reached in 1175 it is also reached in 
the 2-35. 

Hence to divide a decimal by a whole number 
Divide the digits in the ordinary manner, and place the decimal point 
In the result when the decimal point is reached in the number being 
divided. 

Then, the working of the above appears thus 



The decimal point is reached when n has been divided by 5, and 
therefore comes after the 2 in the result. 



Now take the general case of a mixed number divided by a 
mixed number, e. g. f 8-672 -7- 271. By a simple adjustment the 
example can be converted into one of the foregoing nature and 
worked in a similar manner. 

8*672 
Let us write the example in the form of a fraction, thus ' . 

Now on p. 9, when considering cancelling, it was shown that 
a fraction may be multiplied top and bottom by the same 



DECIMAL FRACTIONS 53 

number without altering the value of the fraction. Applying 
this here let us multiply our fraction, both top and bottom, 
by loo. 

8-672 X 100 867-2 (See p. 37, on movement 

Then we have ^ x IOQ = ~^~ of dec imal point.) 

The example is now similar to the first one taken in this section, 
with the exception that " long " division is more convenient. 

(32 Explanation. Divide in ordi- 

813 nary way. Place decimal point in 

- quotient when it is reached in num- 

54 2 her divided, i.e., 271 into 867 goes 

54 2 3, and since decimal point is then 

reached in 867-2, it is placed after 
the 3 in quotient. 



A rough test is available: 8-672 is nearly 9, and 2-71 is nearly 3 ; dividing 
9 by 3 gives 3, which is in agreement with our result 3-2. 

If the divisor contains three decimal places then multiplication 
by 1000 would be necessary, instead of 100, to convert to a whole 
number. That is, the decimal point would have to be shifted three 
places to the right in both numbers, and so on. 

From this we have the following rule for the division of 
decimals : Convert the divisor into a whole number by shifting th* 
decimal point to the right. Move the decimal point in the dividend th* 
same number of places, also to the right, and then perform the division 
in the ordinary way. The decimal point must be placed in the quotient 
when it is reached in the number divided. All cases may be dealt 
with by this rule. 

The following examples and remarks serve to illustrate certain 
special points that arise. When the divisor is already a whole 
number (a case of frequent occurrence) then no shifting of the 
decimal point is necessary. 

It will frequently be found, when doing the actual dividing, 
that there are not sufficient figures in the dividend to " bring down." 
In this case a nought may be brought down each time, since a 
number like 34-5 may, of course, be written as 34-500. The following 
example illustrates. 



54 ARITHMETIC FOR ENGINEERS 

Example 45. Divide 3-45 by 1*2. 

Move the decimal point in 1*2 one place, making 12 

3'45 34'5 

Then divide 

I2)34'5(2*875 Note. 12 into 34 goes 2. Deci- 

24 mal point now being reached in 

34*5 place it after the 2 in quo- 

IO 5 tient. Continue as in ordinary 

96 division. 

90 
84 

60 
Go 

Hence 3-45 -r 1-2 = 2-875. 

It will soon be found when dividing, that very few examples 
will work out " exactly, 1 ' as have those taken up to the present. 
This is especially so with examples of a practical nature, and in 
these cases the division should not be carried too far. Thus if 2375 
be divided by 3*14, the answer is found to be 7*5636942, while yet 
leaving a remainder ! It would be ridiculous to give a result of 
anything like this accuracy. In the first number (2375) there 
are only 4 significant figures, while the second (3*14) only 
contains 3. 

We may therefore reckon with safety that 3 significant figures 
would be sufficient in the answer ; but the division should be carried 
to 4 significant figures in order that the third may be as correct as 
possible. Thus the division can be stopped when the quotient 
reads 7-563. The answer is then stated as 7-56. 

The general rule of 3 or 4 significant figures in the final 
answer may be followed in all cases, unless other instructions are 
given. 

When shifting the decimal point in the dividend it will some- 
times be found that sufficient figures do not exist, in which case 
noughts may be added as desired. This is shown in the following 
case 

Example 46. Divide 231*4 by i'938. 

The dot has to be moved three decimal places in 1*938, but only 
one place is available in 231-4. Hence two noughts must be added to 



DECIMAL FRACTIONS 55 



make up the three places, and the example appears as ^J-j-ijp, the 
decimal point in the numerator standing after the second o. 

1 938)23 1400 (119-40 
1938 
3760 
1938 



18220 
17442 



< When this point is reached the decimal 



77 8 point is placed in quotient. 



280 
Result = ii9*4' 

Note. Four sig. figs, are given in result as there are four in each 
of the given numbers. 

When the number divided is a small fraction it is possible that 
even after shifting the decimal point there will be no whole number. 
The procedure in this case is shown by the following example 

Example 47. Divide -01193 by 2*3. 
Making the divisor 23 the dividend becomes -1193. 



"5 



43 

23 

- Result = -00519. 

200 _ 

184 

- (This can be checked roughly by 
i Co multiplying -005 by 2-3.) 

138 



Explanation. The first thing met with in the dividend is the decimal 
point. Therefore write the dot in the quotient, for if a fraction be 
divided by a whole number the result must be a fraction. Try if the 
first figure in the dividend can be divided by the divisor. If so divide, 
if not put a nought in the quotient and then try the first two figures, 
and so on. Thus 23 into i won't go ; put a nought after the decimal 
point. 23 into n won't go; put a second nought. 23 into 119 goes 
5 times. Put 5 in the quotient and continue as in ordinary division. 



ARITHMETIC FOR ENGINEERS 



Exercises 13. On Division of Decimals. 



1. Divide 219-5 DV 5- 
3. 413-8 22-75. 
5. 1*897 .. 21-35. 
7. ,, -0076 ,, -016. 
9. 621 ,, -31. 



2. Divide 32-73 by 7. 
4. 135 i. 4'36. 
6. ,, -158 ,, -0065. 
8. -098 11-57. 
10. ,, 33000 117. 



11. If 14-7 Ibs. per sq. in. = i atmosphere, find how many atmospheres 
in the following : (a) 165 Ibs. per sq. in. ; (6) 90 Ibs. per sq. in. 

12. The following figures are extracts from the results of an experi- 
ment to determine the coefficient of discharge of a rectangular notch. 

Actual discharge 



The coefficient is the value 
coefficient in each case. 



Theoretical discharge" 



Calculate the 



Actual discharge. Cu. ft. per sec. . 


0372 


0531 


1085 


Theoretical discharge. Cu. ft. per sec. 


0571 


0829 


1750 


Coefficient ... .... 

















13. If i gallon of water == 10 Ibs., and 2-205 Ibs. = i kilogramme, 
find how many kilogrammes there are in i gallon of water. 

14. If i kilogramme 1000 grammes, and 453*6 grammes = i lb., 
find how many Ibs. there are in i kilogramme. 

15. If i kilogramme = 2-205 Ibs., and i cwt. = 112 Ibs., find how 
many kilogrammes go to i cwt. 

16. If 746 watts i horse-power, find how many horse-power = 
i kilowatt (i. e., 1000 watts). 

17. The reaction of a plate girder (i. e,, the upward force at the 
support) is 97 tons, and i square foot of surface is required for every 
20 tons of load. Find how many square feet are required, 

18 to 20. In certain experiments the quantity of water delivered 
by a pipe in a certain time was measured with the results below. Find 
in each case the quantity ot water passing through in i minute. 

18. 100 Ibs. ot water passed in 4-55 minutes. 

19. ioo ,, ,, 2-37 

20. 94 >. 3 '02 

21. Calculate the resistance per foot (i. e., ohms for i foot) of an 
electric cable which is 520 yards long, and has a resistance of 20*7 
ohms. 

22. If 62-5 Ibs. of water occupy a volume of i cubic foot, find the 
volume occupied by i Ib. of water. 

Conversion of Vulgar Fractions to Decimals. In Chap. I, 
p. 7, it is shown that a vulgar fraction is the result of dividing 
one number by another. If now this division be performed as 
just described for decimals, a decimal answer will be obtained, 
and hence our vulgar fraction will have been converted into a 



DECIMAL FRACTIONS 57 

decimal fraction. Therefore to convert a vulgar into a decimal 
fraction : Divide the numerator by the denominator. 

Example 48. Convert the following vulgar into decimal fractions : 

() i; (^ i; to A 

(c) 16)7-0(0-4375 
(a) 4)1-00 64^ 

* 25 " " 4 6 8 

' A ^ '4375: 

(fc) 8)1-00 r*o ---"* 

~~. "'* ~ I2 ~* 

or 8 J o J 8o 

= * of -25 = '125 

With the more uncommon vulgar fractions that appear, the 
division may not work right out without remainder. Then the 
rule of 3 or 4 significant figures may be followed. 

Example 49. The value of tr,* a constant of great use in calculations 
with circles, may be given by the fraction 3i\V Express this decimally. 

113)16-0(0-1415 



47 
45* 

* 3\ 1 A = 3-14 to 3 sig. figs, 
1 80 *" - 



Example 50, In a certain experiment the following intervals of 
time were read with a stop-watch 

(a) i min. 42 sec. ; (b) 2 min. 35 sec. ; (c) i min. 13 sec. 
For the purposes of calculation it is desired to state these times as 
minutes only. Find the necessary values (i. e. t convert the seconds to 
decimals of a minute). 

i min. = 60 sec. 

(a) i min. 42 sec. = ijft min. = i/^ cancelling by 6 = 1-7 mins. 

(b) 2 min. 35 sec. = 2gJ min. = 2^ cancelling by 5 

2^ = 2-583 min. or 2*58 to 3 sig. figs. 

(c) i min. 13 sec. = ijg min. = 1-217 mm - or I<22 * 3 s ig- fig 3 - 



* Greek letter "pi." 



58 ARITHMETIC FOR ENGINEERS 

Exercises 14. On Conversion of Vulgar Fractions into 

Decimals. 

Convert the following into decimal fractions 

1. f. 2. &. 3. V. 4- i!- 5. J|. 

6. &. 7. A. 8. fl. 9. ft. 10. 3l \V 

11 to 15. In certain experiments measurement of time had to be 
taken in minutes and seconds. Convert the seconds into decimals of 

a minute in the following cases (i.e., convert the fraction into 

a decimal, since i min. = 60 sec.) 

11. 18 s.ec. 12. 34 sec. 13. n sec. 14. 54 sec. 15. 13 sec. 

16 to 20. The " pitch " of a screw thread is > ~, j- per inch. 

Find the pitch as a decimal for the following cases : 

16. 24 threads per inch. 17. 18 threads per inch. 

18. 9 threads per inch. 19. 4-5 threads per inch. 

20. 3 '5 threads per inch. 

21 to 24. Find the pitch, in decimals, of the following Whitworth 
Gas Threads : 

21. ii threads per inch. 22. 14 threads per inch. 
23. 19 threads per inch. 24. 28 threads per inch. 

25. The " go in " end of an internal limit gauge is -0004" smaller 
than the nominal size, and the " not go in " end is ooi2" larger than 
the nominal size, the latter being if". What are the actual sizes of 
the ends ? 

Compound Examples. Approximation for Result. 

Some of the calculations that occur most frequently in engineering 
problems are the evaluation of expressions containing multiplica- 

j i- t- 3*28 x 58*7 x 58-7 . , 

tion and division, such as ~~r L * Ihese may be 

32 x 600 J 

worked out on the lines of the previous paragraphs. Cancel where 
possible. Then the product of the numbers above the line may 
be obtained; also the product of the numbers below the line. 
Finally the division is performed. The number of figures in the 
calculations is often very large, and, in consequence, there are 
plenty of opportunities for making blunders and slips in the working. 
The most common mistake is that of wrongly placing the decimal 
point, and in many cases the blunder may pass unnoticed. Some- 
times the nature of the problem gives a check on the accuracy 
of the result, e. g., if on calculating the thickness of a boiler plate 
a result of 7-5* was obtained, then there is evidently a mistake in 
the working; ^\" boiler plate is obviously absurd, and an examina- 
tion of the calculations would no doubt show that the result should 
be *75*> which is reasonable. But in cases where there is no 
such guide, and, in fact, in all problems, it is advisable to get a 



DECIMAL FRACTIONS 59 

rough idea of the result before doing the actual detailed working. 
To those who use a slide rule, or intend to do so, this " approxi- 
mating for a result " is of great use, for the slide rule really works 
with the significant figures only, independently of the position of 
the decimal point. 

The general method of approximating is to replace the actual 
figures by some near convenient round numbers, so that, using 
these, the approximate calculation becomes a very simple one, and 
may often be done mentally. The following example will illustrate : 

Example 51. Find the approximate value of ?53_2?_1!?. 

6 3 

The 293 may be called 300^ This is mental work and is 
4*9 5 r only written here as an 

,, 63 ,, ,, ,, 60 j explanation. 

Then our approximate expression would be ^~. ^ and cancel- 



ling by 60. 



5 



I 

Hence our result must be about 25 ; it may be anything from, say, 
20 to 30, but must have two figures in the whole number. 

The actual result is 22-8 to 3 significant figures. Now supposing 
our detailed working had been bungled, and a result of, say, 2-28 or 
228 had been obtained : the mistake would have been shown by our 
approximation of 25. 

While the method just shown is suitable for simple examples, 
it is not accurate enough for more complicated ones. Also, for 
those readers who are not accustomed to much mental calculation, 
a more systematic method is desirable. 

Considering the expression -9-- ---- 5 2 9 --- 7 

b F 31 X 1-8 

let us shift the decimal point in each of the compoi_ ..... ________ 

until a units figure only stands in front of the point (as in the r8) ; 
at the same time multiplying or dividing by the necessary number of 
tens to preserve the value of each number. Thus in the top line 

19*5 becomes 1-95 x 10 
529 5-29 x 10 x 10 

and -67 ~ 7 - 
' " 10 

while in the bottom line 

31 becomes 3*1 X 10 
and the r8 remains unchanged, being already in the form required. 



60 ARITHMETIC FOR ENGINEERS 

Using these " converted numbers/ 1 we may rewrite the given 
expression as 

(1-95 x 10) X (5-29 x io X 10) 
(3-1 x io) x r8 

For convenience, let us separate the tens into one expres- 
sion (b), and the remaining numbers into another (a), thus 

^95_x_5!19 * J?7 x IQ x I0 x IQ 

~ 3-1 x 1-8 ~ io x io 



(a) (b) 

Expression (b), by cancelling, evidently reduces to io. 
Now replace each number in (a) by its nearest whole number, 
i. e. t write 2 in place of 1*95 and 3 in place of 3*1, and so on. 
Then we have 

-^ X5 * 7 XIO 
3 X X 



(a) (b) 

- f x io 

= 12 (very nearly) x it) 

= 120 

which is then an approximate answer to the given expression. The 
actual answer, it may be noted, is 123-9. 

In practice this method may be considerably abridged. Thus 
in expression (a) the " nearest whole numbers " may be obtained 
at one operation from the values in the given expression, the inter- 
mediate step being a mental one. For instance, taking the 19*5, the 
btep 1*95 is performed mentally, the 2 only being written down ; again, 
taking the "67. the step 6'7 is mental, 7 only being written. Also, 
the result of expression (a) is not required with great accuracy ; 
thus in the foregoing -./- is called "12 very nearly/' since the ex- 
pression - 3 / would give 12. A certain amount of caution in this 
direction is necessary, however, otherwise some of the " approxi- 
mate " results may be too far from the truth. In expression (b) 
a stroke (i) may very conveniently be used to represent each io, 
while the X signs may be omitted, since the only operations in- 
volved are those of multiplication and division.* These strokes 
may be cancelled in the usual way, and every stroke remaining 

* It must be clearly understood that this method of approximating 
cannot be applied to expressions involving -f and signs. 



DECIMAL FRACTIONS 61 

after the cancelling will, of course, signify a 10. Instead of expres- 
sion (b) in the foregoing we should then have-^-, the remaining 

stroke on the top line indicating that the result of expression (a) is 
to be multiplied by 10; similarly a stroke beneath the line would 
indicate division by 10. These " strokes " are perhaps best placed 
at the right-hand side of the sheet, and will be shown in this position 
in the succeeding examples, under the heading of " Tens/' 

In connection with these " strokes " (i. e., tens), it is seen from 
the detailed working already shown, that when a " converted num- 
ber " contains a multiplying 10 the stroke must appear on the 
same side of the division bar as the number itself; but when con- 
taining a dividing 10, the stroke must appear on the other side of the 

bar. Thus, in the case of the -67 ( which = ~ \ the stroke must 

be under the bar, while the number itself is on top, i. e.> the stroke 
is on the other side. Note also that in a number containing a mul- 
tiplying 10 the decimal point is shifted to the left when obtaining 
the " nearest whole number/' and in one containing a dividing 10, 
to the right. Considering all these points, we obtain the following 
rules for approximation 

Taking each number in turn, shift the decimal point 
mentally until there is a units figure only in front of the 
point. Find the nearest whole number to the quantity 
so obtained, and write it in a new expression (called (a) in 
the foregoing working) ; at the same time place a stroke 
in an auxiliary expression (called (b) in the foregoing work- 
ing) for every place that the decimal point has been shifted, 
according to the following rules 

When the point is shifted to the left place the stroke (or 
strokes) on the same side of the division bar as the number 
occupies in the original expression; when shifted to the 
right place the stroke (or strokes) on the opposite side of 
the line. 

Work out the simple expression (a) and cancel the 
" strokes " so far as possible. For every remaining 
stroke, multiply or divide the result of (a) by 10, according 
to whether the strokes are above or below the line. This 
gives an approximate result. 

// is important to notice that this method, although lengthy in 
description, is very quick and easy in actual use. 



62 ARITHMETIC FOR ENGINEERS 

Example 52. The expression ^ amperes had to be 

' J * -000328 X 1200 * 

evaluated in connection with an experiment on electrolysis. Find the 
required value. 



Approximation 



____4 __ 
3 X 1-2 

4 



= I very nearly. 



Tens 



no strokes remaining, 

/. no multiplication or 



divisions by 10. 
Therefore the result is about i ampere. 

Explanation of Approximation. 

Top line. Convert -39 to 3-9 and write 4. Decimal point having 
been moved one place to right, place a stoke below the line. 

Bottom line. Convert -000328 to 3-28 and write 3. Decimal point 
moved 4 places to right, place 4 strokes above the line. 

Convert 1200 to 1-2 and write i. 

The strokes all cancel out, so that the approximate answer is the 
result of the first expression, i. e. t about i. 

Working in full 

000328 -39 _ 390 

1200 -394 ~ 394 

394)3900(0-9898 
354 6 



393600 3540 

3152 

3880 
.'. Result is '99 amperes, which 6 

agrees with the approximate answer. 

3340 

Example 53. Find an approximate answer to each of the following 
calculations. Given also the significant figures in the result (obtained 
by a slide rule, for example), state, from an inspection of the approxi- 
mate answer, what the actual results must be. 

W ~T-^~ Si & fi S s - are *3- 



Approximation Tens. 

* XJL - i _ . 2 < 

'A v fc * -5 
4 A ^ 

25 = approximate result 
.'. Actual result must be -23, this being the nearest number to 
25 that can be made from the sig. figs. 23. 



DECIMAL FRACTIONS 



Approximation 



5 X 33 X 12 c . -. 

^ ^ ~ Sig. figs, are 209. 

35 X 144 X 3-14 X 6 b b J 

Tens 



_ 5 
"" 



= about J = -2 
.". an approx. result = -2 X 100 = 20 

.'. Actual result must be 20-9. 



\\\ 

Two strokes above. 

/ Multiply by 10 X 10 

= IOO. 






Approximation 

'- = about 5 

i X 1-5 D 

/. an approx. result = - = 
11 jooo 

.'. Actual result = -00434. 



arc 434. 

Tens 

Tlf 
/. divide by 10 X 10 X 10 

= 1000. 



Note. When a number, after being "converted," is nearly 1-5, as 
with the 144 at (b) above, it is best to call it 1-5 rather than i or 2, to 
avoid an answer which is too approximate (see also c above) . 

Example 54. Determine, without actually working out the expres- 
sions, whether the answers given below are reasonably correct. 
, N -000267 
- 



-^0852 



= ' 314 



Tens 

J1L. 

= J = -3 UU 

.*. divide by 10. 

/. Result is about ~ = -03, and answer given is reasonably correct. 



Approximating 



4 X 7-14 X 60 
" 4 rx~6"28 X 2-1 



Approximating 



Tens 
1 



.'. multiply by 10. 



/. Result is approximately 2 X 10 = 20, and answer given is 
reasonably correct. 

(c) 1-41 x 182 x -00034 x i -08 x 187 x 187 = 33200. 



Approximating 

1X2X3x1x2x2= about 24. 

.*. Result is about 24 x 100 = 2400. 



Tens 
11UU 



UU 
/. multiply by 100. 



64 ARITHMETIC FOR ENGINEERS 

.*. Given result is incorrect, being 10 times too large, and should 
be 3320. 

(j\ '00307 X '0123 x 8 02 x ro2 

' 



Approximating 

3X1 X 8xi 



I 



24 about 



= -26 

Tens 
\\ 

\\rn~ 

.". divide by 1000. 
/. Given result is incorrect, and should be '026. 

Example 55. A calculation for the indicated horse-power (I.H.P.) 
of a petrol engine reduced to the figures 



.'. Result is about - -024. 

Trn 



Tens 



+ ^ ,03 ^ _sj>* -iA_^_^. F i nd the I.H.P. 
33000 X 3 

Cancel two noughts in 800 and 33000, making 8 and 330. 
Approximation 

Z_ x _?_ x ^ X ^ X 8 _ 
~$ X ^ 

.*. divide by 100. 

* Approximate result is - - ~ 4'48, t. e., about 4j H.P. 
Working in full, expression *^ ^ 5'7^ I.H.P. 



Example 56. On testing a certain coal gas for its calorific value 

(i.e., heating value) the figures ^ ' ~- were obtained. Approxi- 

454 x ' I 4 

mate for the answer, and, given that the significant figures are 6009, 
state the calorific value (in B.T.U. per cu. ft.) correct to the units figure. 



Approximation 



Tens 



U 

.*. multiply by 1000. 
.*. Approximate result = -6 X 1000 = 600 and actual result 600-9. 
.'. Calorific value to units fig. = 601 B.T.U. per cu. ft. 

Exercises 15. On Approximating. 

1 to 8. Determine, without actually working out the expressions, 
whether the answers given in the following are reasonably correct. 

2. ^6_ = 6o6oo . 



. . . _ 

00136 -00341 x -7 

X 9 X 

32 X 

9*87 



3 I '3 I x 33QQQ X 150 = . 4 X 9 X 38 x 3-14 = 

6400 X 32-2 J 5< " 32 X 144 X 6 55 ' 



K 

5 - - = ' 494 ' 



< 3-aa X 900 

7. 2 x- 7 8 5 x 6-25x140=1375. s. *' 75 x x 8o ' 4 



DECIMAL FRACTIONS 65 

9 to 18. Find an approximate answer to each of the following 
expressions. Given also the significant figures in the results (obtained, 
say, from a slide rule), state, from an inspection of your approximate 
answers, what the actual results are (i. e. t put the decimal point in its 
correct place among the given figures). 

9 v.jT-^TTg Si S- fi 6 s - in answer are 478. 

10. I000 x "5 X -5 

3-25 X 12 " " " ^' 

11 10 X IIP X_2640 

33000 " " " 



_ __ 

30 x 1-06 X 10-56 " " " 7I4 * 

?i2<_3-i4_ x 1-5 X 24 

- - 4 x - 3l ., 214. 

3-14 X 24 X 1000 

- ~~~~ 



15 ? x goo x 32-2 

* 9 : 67rx^7~x~ : 6 " " " 44 ' 

16 4 X 3-14 X -18 X_7\5 X_72_'5 

342 

17. - 5 35 -_. 5 ~~ ., 245. 



12 

7 x -0625 x 2240 
6 X 232 



18.- z-TT^-n 705. 



In some examples expressions appear with the four simple 
arithmetical rules in various combinations. Then the best method 
of working is to preserve the general form of the expression 
throughout the various steps, as was adopted in Chap. I, p. 30. 
The following example illustrates : 

Example 57. In calculating the safe eccentric load on a stanchion, 
the following expression was arrived at : 

-- 6- x~6- tons - Complete the calculation. 
1 + ^6~x~ir 3 6 

The first step necessary is to evaluate the right-hand term under the 
line. 

tons () 6 '5 X 6-5 = 42-3 



- - 
6*5 X 6-5 

5-36 ~ 



. 
+ ~6 <*> 5-36 X 5-36 = 28-7 



tons = 96*4 tons 



66 ARITHMETIC FOR ENGINEERS 



Exercises 16. Compound Examples. 

1. The height of a mercury barometer is 30". The height of the 
water barometer in feet is found by multiplying this figure by -yj 

Find the required height. 

2 and 3. In finding the centre of gravity of two rolled sections the 
following expressions had to be evaluated. Find the values required. 

2 . I^LX 57 inches< 3. 1*7*. 5L5_ inches . 

7-19 3*6 

4. The current to pass through an electro-magnet was calculated 

to be ^ - %~ amperes. Complete the calculation. 

5. The resistances in an electric circuit, which is to carry a current 
of 4-5 amperes, are 3-5 ohms, '75 ohms, and 2-38 ohms. The voltage 
(volts) required is equal to " current (amperes) X total resistance 
(ohms)." Find (a) total resistances; (b) voltage required. 

6. The bending moment at a point on a continuous girder was 

given by the expression -- - . Find the required value. 

7. A stream is found to deliver 48 cu. ft. of water per sec. with a 
total fall of 1 20 ft. Find the horse-power of the stream, which is 

it. 48 X 62-5 X 120 
given thus : - --- J -- . 
b 550 

8. Find the thickness of plate for a steam boiler 6'-6" diameter, to 
work at 100 Ibs. per sq. in. pressure, from the following expression : 

100 X 78 . , 
----- / - inches. 
2 x 9000 x -7 

9. A dynamo is to supply 600 .lamps, each of 16 candle-power. 
Each lamp takes 1*2 watts for every candle-power, (a) How many 
watts must the dynamo supply? If the voltage is no, the current 

watt s 
(amperes) supplied is --- -- - (b), what current is supplied? 

10. The tensile test of an iron wire showed that the " Modulus of 

Elasticity "= 3 4 - Ibs. per sq. in. Complete the calculation 

J '105 X -00104 r *i r 

(3 sig. figs, will do). 

11. The following figures were obtained in a torsional test on 

wrought iron : IO ' 2 X ^ 8 X 5 Jl 3 X 5I ' 8 Ibs. per sq. in. Find the 
300 X *3oo 

required value. 

12. A coal gas was tested for its " calorific " or " heating " value. 
When -146 of a cubic foot of gas was burnt, the temperature of 4-9 Ibs. 
of water rose from 51 F. to 69 F. Calculate the calorific value (in 
B.T.U.) which equals Lbg^of.wate L X rigejntemperature 

' ^ cu. ft. of gas burnt 

13. Find the value of the expression 6-48 -j- '3125 -f- 4*68 x '506, 
figures obtained when calculating the strength of a stanchion. 

14. Calculation in connection with a column gave the expression 






- ------ tons. Complete the calculation. 



DECIMAL FRACTIONS 67 

15. Evaluate the expression 2 x (21*6 + 246*5) + 427, which 
refers to the strength of a built-up girder. 

16. The bending moment on a girder, at a certain point, was found 
to be given by the expression 2S^J^ + 26 + 32) tons ft . 

4O 

Complete the working. 

17. Find the value of the expression 5 x (9*5 + 8-8 + 5*6) + 15 X 
(7-8 -f 6-7) tons ft., which is the bending moment on a bridge truss. 

18. One of the reactions of a continuous girder was found to be 

5 X 20 , '5 X IO . 16-45 , 16-45 !V75 t^- t_ AI_ i 

_J_^ -- ^ J> ----- -- 2 ^ -- 5^ --- 2-L2 tons. Finish the working. 

2 2 20 10 b 

19. The weight of a proposed plate girder was estimated to be 

_85L2LZ2_x_i4_ tons. Find its amount. 
1400 x 7 70 x 14 

20. In calculating the safe eccentric load on a column the following 
expression was arrived at: 

___4 ---- tons. Complete the calculation. 

T + 3'15_X_5_ P 
"*" 2-82 x 2-87 

21. Find the value of the expression 130 -f- '85 (606-5 -7 x 170). 



22. Evaluate _ 4 lbs per sq in ^ which occurred 

in a steam engine calculation. 

23. The following figures were obtained when calculating the 
theoretical amount of air required to burn i Ib. of oil- 

4*35(2-26 -f 1*2) Ibs. Find the required value. 

24. Complete the following calculation, which refers to the mean 
effective pressure in a steam engine 

75 X -67 X i'45 17 Ibs. per sq. in. 

25. Find the value of the expression 37'5 xjfo _ 37'5 X 18 f 

44 
figures relating to a bending moment. 

26. Find the weight of a roof truss, given by the following figures : 



Averages. When a quantity has various values at different 
times it is often necessary to state a value that shall give a general 
idea of the size of the quantity. Let us suppose that a boiler in 
the heating apparatus of a building had consumed the following 
quantities of coal during a certain month : ist week n tons, 
2nd week 6J tons, 3rd week 12^ tons, 4th week 8 tons. If it is 
desired to state how much was generally burnt per week at that 
season of the year, it is hardly fair to quote any one of the actual 
numbers, since each number only applied to one particular week. 
Thus, the 12^ tons was burnt in a cold week, while the 6J was 
used when the weather was mild ; generally, the amount burnt 
was more than the 6J and less than the I2j. Now let us imagine 



68 ARITHMETIC FOR ENGINEERS 

that all the four weeks had been equally cold, so that an equal amount 
of coal was burnt each week; also that the total amount burnt 
under the supposed conditions is equal to the total amount actually 
burnt. Then we consider the total amount to have been spread 
equally over four weeks. 

The total amount burnt = n + 6J + *2j -f 8 = 38 tons, and, 
dividing this equally among four weeks, the consumption for one week 
would be - S 4 8 ~ = 9^ tons. This supposed constant consumption is 
called the average value or mean value. A common way of making 
the statement would be, " On the average the consumption is 9^ 
tons per week/' meaning that the consumption per week is in the 
neighbourhood of g\ tons. From the above we obtain the rule : 

To find the average of a series of quantities, find the sum of all the 
quantities and divide by the number of quantities. 

There are many applications of the use of the average, 
as will be seen from the examples and exercises in this section. 
Particularly, when testing or experimenting to find the value of 
a certain quantity, the experiment is repeated three or four or more 
times and slightly different results will be obtained owing to errors 
of instruments, etc. Then the average of all the measured results 
is found, and we may safely assume that this average is nearer the 
actual truth than any of the measured values, as the errors have 
been spread over the whole set of numbers. 

Example 58. A wrought-iron shaft under a torsion (twisting) 
test had its diameter measured in several places, the values being 
625", '62I", -626", -623", -624", 622"', '623', -62I*. Find the average 
diameter of the shaft. 

625- 
621 
626 
623 

624 

622 Average diameter = 623"' 

623 """"" 

621 



8)4-985 
6231 

It should be noticed that the accuracy of the average should not 
exceed the accuracy of the given figures. 



DECIMAL FRACTIONS 



69 



A case of common occurrence to the engineer is the " averaging " 
of an indicator diagram from a steam or gas engine for the purpose 
of finding the indicated horse-power. Roughly an indicator dia- 
gram is of the form shown on the left of Fig. 14, and as will be seen 
in Chap. VII, the idea of " averaging," i. e. t finding the average 
height, is to get a figure equal in area to the diagram, but with a 
constant or unvarying height as on the right of Fig, 14. Certain 




This hei 



lines n 



HV is t"He 
of Irfce dotted 
Irhe. diareuvt. 



Pig. of same avca 
ID of consVanV' 
and avea 



diagram 

The figures ave e^ual m 
Fig. 14. "Averaging " an Indicator Diagram. 

lines (called ordinates) are drawn, as indicated by the dotted lines, 
and measured, and the average of these heights is found. 

Example 59. The ordinates, in inches, of an indicator diagram 
from a steam engine, were as follows : -47, '72, '82, '83, '76, '60, ^44, 
28, '2O, and -08. Find (a) the average ordinate. One inch of ordinate 
represents 70 Ibs. per sq. in. Find (6) the average pressure in Ibs. per 
sq. in. (called the mean effective pressure, or M.E.P.). 



Oj 

.S 

*s ( 



'47 
72 
82 

83 
76 
60 

'44 
28 

20 



(a) Average ordinate = '52* 



i* of ordinate = 70 Ibs. per sq. in. 

/. 52 // of ordinate = 70 x '52 = 36-4 Ibs. per sq, in. 



(b) M.E.P. = 36*4 Ibs. per sq. in. 



10)5*20 
52 

Example 60. The readings of temperature in the table on p. 70 were 
made in an experiment on the calorific value of coal gas. Find (a) the 
average inlet temperature, (b) the average outlet temperature, and. 



7 o 



ARITHMETIC FOR ENGINEERS 



(c) the rise in temperature (i. e. t average outlet temperature, average 
inlet temperature). 



Inlet temp. "C. 


Outlet temp. C C. 


9H7 


30-62 


9-46 


31-^3 
3I-80 
3I-90 




32-6I 
32-82 


9-42 


33-13 

33-22 


3)28-35 12)385-88 


9-45 32T57 




Exercises 17. On Averages. 

1. Three separate experiments were made to determine the " re- 
fractive index " of a certain kind of glass, the results being 1-52, i'5Q5 
and 1-49. Find the average value. 

2. Experiments to determine the "coefficient of discharge" of a 
V -notch gave the following results ; -575, -574, '580, -581, -581, '581. 
Find the average value. 

3. In measuring the internal resistance of an accumulator the 
experiment was performed five times with the following results : -025, 
028, -03, -0286, -0275 ohms. Find the average resistance. 

4. The diameter of an iron wire measured in several places for a 
tensile test was as follows : -0365, '0346, -0365, '0364, -0358, -0368, 
0365, '0348 in. Find the average diameter. 

5. The following values of the coefficient of friction between two 
wood surfaces were obtained in an experiment : '417, '49 '367, 395, 
37, -389, -403 and -415. Find the average value. 

6. During a boiler test the following readings were taken on the 
water gauge measuring the supply of feed water to the boiler : 31*6, 
3i-7. 3i-7. 3i-6, 33, 34'5. 32'4> 26-5, 20-0, 23-8, 24-0, 24-0, 24-5, 247, 
24-5 Ibs. per minute. Find the average value of the supply. 

7 to 9. The following figures were taken during the test of a steam 
electric plant ; the steam pressure, voltage, and current being kept as 
steady as possible. 

Steam pressure (Ibs. per sq. in.) : 96, 100, 99, 100, 100, 98, 102, 
102, 102, 103. Voltage : no, in, 111-5, IO 9 I0 9'5' I0 9'5 IIJ II2 , 
in, no. Current (amperes) : 220, 223, 223, 230, 229, 224, 225, 217, 
220, 200. 

7. Find the average steam pressure. 

8. Find the average voltage. 

9. Find the average current. 



DECIMAL FRACTIONS 71 

10. The following readings were taken on an anemometer at differ- 
ent parts of the ashpit, in measuring the air supplied to a furnace : 
57 580, 620, 650, 660, 600, 600, 460 cu. ft. per min. Find the average 
value. 

11. In an experiment to find the specific resistance of a piece of 
wire the following readings of diameter were taken : '465, '463, -460, 
460, '461, '460, -462, '464 millimetres. Find the average diameter. 

12. Tests on a number of materials, to determine the ratio of shear 
to tensile strength gave the following results : -67, '64, -66, -72, -81, 
82, -88, -85, -85, -87, -83, -73, -91, -59, -90, -98. Find the average 
value. 

Percentage. The words " Percentage " and " so much per 
cent." are frequently met with, a percentage being another way 
of expressing a fraction (or division) whose denominator is 100. 
Thus the statement " 5 per cent/' (written as 5%) means 5 parts 
out of 100 parts, *'. e. t T ^ in the vulgar form. Cancelling reduces 
this to -$ ; or it may be expressed in the decimal form -05, shifting 
the point two places to the left to divide by 100. Both forms are 
useful. From the above we obtain the following rules : 

1. To convert a percentage into a decimal fraction, shift the decimal 
point two places to the left. 

2. To convert a percentage into a vulgar fraction, write, the percentage 
as numerator, and write 100 as denominator ; then reduce to the lowest 
terms. 

Example 61. Convert the following percentages into decimal 
fractions: (a) 30%, (b) 37'5% W 1-13% (<*) -028%. 

(a) Shifting decimal point two places to left (to divide by 100), 
30% = -3 

(b) Similarly 37-5% = -375 

(c) 1-13% = ' OI I3 

(d) -028% = -00028 

Example 62. Convert the following percentages into vulgar frac- 
tions : (a) 45%, (6) 12*%, (c) 33*% (d) 26-25%, (e) -oz 5 %. 

() 45% = J 4 5 ' Cance in g by 5 = A 

i 

(6) 4% = j - | -=- 100 = X JL ^ 

4 

(c) 331% =fg = 33} ,. I00 = \Si x _1_ =i 

26*2^ 262^ 

(d) 26-25%= -j^ = ^ Q - Cancelling by 25 and 5 = fj 

(e) -025% = 5. == 25 - Cancelling by 25 = ,A 6 

v ' J /0 IOO IOOOOO o / O ^gOQ 



72 ARITHMETIC FOR ENGINEERS 

It is often required to find a certain percentage of a quantity, 
meaning that we are to find a certain fraction of the quantity. 
To do this, express the percentage as a fraction and multiply by the 
quantity. 

Example 63. An electric motor is to work at 30 h.-p. under ordinary 
full load conditions. It is required, however, to deliver 25% overload 
(i. e. t 25% extra) for a certain period. Find the greatest h.-p. the motor 
must be capable of giving. 

Full load = 30 h.-p. 

Overload = 25% = -^ x 3Q = 7-5 

2 

.'. Greatest h.-p. = 30 4- 7*5 = 37'5 h.-p. 

The vulgar form is the more useful when the percentage is equiva- 
lent to a simple fraction with I as numerator; e. g., 20% = , 
I2i% = J, etc. In these cases, to multiply as per rule we merely 
have to divide by the denominator of the fraction ; thus the above 
can be done mentally : 25% = J, therefore divide the 30 h.-p. by 
denominator 4. For other cases the decimal form is better, as 
multiplying by a decimal is easier than by a complicated vulgar 
fraction. It is useful to memorise the equivalents of the simpler 
percentages, which are given below. 

i% = r = -oi 25% -J--25 

5% = A = '05 33i% = i = '3333 

10% = T V =-i 50% =i = -5 

12*% = i = -125 66|% - - -6667 

20% = * - -2 75% = i - 75 

100% of course = j = i 



Example 64. The catalogue price of a certain machine is y los. 
The makers announce that " owing to the increased cost of raw material, 
all prices are advanced io%." An agent who buys the machine under 
these new conditions is allowed 20% discount. Find the actual price 
he pays. 

Note. The figures in brackets thus [ ] are only to assist the 
beginner and do not appear on the actual bill. 

s - *. 

Catalogue price = 7 10 o 

10% advance [= ^ of 7 105.] = 15 o 

New price = 850 

20% discount [= J of 8 55.] = i 13 o 

Price paid [= difference] = 6 12 o 



DECIMAL FRACTIONS 73 

Explanation. 10% of 7 105. = ^ of 1505. = 155. 20% of 
{ 5 s - = & f 8 5 5 - Dividing 8 by 5 gives i and 3 over. 3 ~ 
605. and adding in the 5 gives 655. This divided by 5 = 135. Thus 
20% of 8 55. = i 135. 

A common error in such examples as the last is to say that 
" 10% advance with 20% discount is equal to 10% discount." 
This is not so, since the 10% and the 20% refer to different quantities : 
the 10% to the old price, and the 20% to the new price : the 20% 
cannot be used until the old price has been found. 

Example 65. In steel-makers' catalogues it is stated that the 
actual weights of rolled steel section bars may vary 2j% on either 
side of the listed weights. Calculate the greatest and least weights of 
a channel bar listed at 19-3 Ibs. per foot. 

Listed weight = 19-3 Ibs. per ft. 

2. *> 

2}% allowance = x 19-3 

= -482 Ibs. per ft. 
/. Greatest weight = 19*3 + '48 = 19*78 Ibs. per ft. 

and least = 19-3 '48 = 18-82 Ibs. per ft. 

Conversion to a Percentage. It is frequently necessary 
to express some quantity as a fraction of another, and the most 
usual way is to express it as a percentage. Suppose that the 
nominal full load of an alternator is 60 kilowatts, but at some par- 
ticular time it is called upon to deliver 75 kw., an increase of 15. 
Now this in itself is not much guide as to how much the machine 
has been overloaded, as the overload allowable depends on the 
size of the machine. Thus an alternator designed for 6000 kw. 
could easily deliver another 15 kw. ; but it would be impossible to 
get another 15 kw. out of a machine designed for, say, 20 kw. In 
each case the actual increase is the same, but as a fraction of the full 
load is very different ; this fraction is the important figure. Taking 
the big machine 15 kw. expressed as a fraction of 6000 kw. is yJo^ ; 
with the small machine 15 as a fraction of 20 is i. Now to avoid 
the confusion of having vulgar fractions with different denominators, 
the fractions are expressed as a percentage, i. e. t they are converted 
so that the denominator is always 100. 

From Rule 2 on p. 71, it follows that to convert a vulgar fraction 
into a percentage we must multiply by 100. Converting the above 
fractions to percentages we have 

i 5 

_il x iqq = i% or -25% overload, and ^ x ^ = 75% overload. 

4 * 



74 ARITHMETIC FOR ENGINEERS 

Then 15 kw. increase on 20 kw. is an overload of 75% but an in- 
crease of 15 on 6000 kw. is only *25% overload. Thus we have the 
rule : To express one quantity (A) as a percentage ol another quantity 

(B) write (A) over (B) as a vulgar fraction, multiply by 100, and evaluate. 

^ 
Thus A expressed as a percentage of B is ^ X 100. 

Examples where this occurs are many and include : efficiencies 
of all kinds, errors of instruments and experiments, changes of 
speed and load, extensions of test bars, the composition of alloys, 
etc., etc. 

Example 66. An ammeter, when tested, was found to read 4-63 
amperes, the true value being 4-5 amperes. Find the percentage 
error (i. e. t express the error as a percentage of the true value). 

Apparent value == 4-63 
True value = 4*50 

/. Error = -13 too high 

.\ % error = ~ 3 X I0 = 2-89% too high, 
4-5 _____ 

i. e. t for every 100 amperes indicated the error is 2-9 amperes. 

Example 67. The following figures were obtained in a tensile test 
of a piece of steel 

Original length 5*. Original area of cross section -994 sq. in. 

Final length 6'O$". Final area of cross section -407 sq. in. 

(a) Express the total extension as a percentage of the original 
length, (b) Express the contraction in area as a percentage of the 
original area. 

(a) Total extension = Final length original length. 

= 6-05 5 = i -05" 

/. Extension as % of original length = -- 5 * *?_ = 21% 

(b) Contraction in area = Original Final area = -994 '407 = 
587 sq. in. 

.'. Contraction as % of original area = --5..Z. x IQO = 59-05%, 

994 
say, 59% 

Actually, the statements made are not as complete as in the 
above two examples. Thus in Example 66, the usual statement 
is " find the percentage error/ 1 and the figure of which the error 
is to be a percentage would not be stated. Then the student may 
be in doubt as to whether he ought to express the -13 amperes as a 



DECIMAL FRACTIONS 75 

percentage of the true value 4-5, or the apparent value 4*63. A 
slight knowledge of the particular work with which the example 
is connected is useful in this respect, but the following will help 

Instrument and experimental errors are stated as a percentage 
of the true value. 

With changes of load and speed, the error is stated as a percentage 
of the normal or full load, or speed. 

Extensions arid^contractions of test bars are stated as percentages 
of the original length or area. 

With composition of alloys, etc., the quantity of each material 
is given as a percentage of the total quantity. 

Example 68. A sample of coal weighing 3-35 Ibs. was found on 
analysis to contain : Carbon 2-52 Ibs., hydrogen -234 lb., oxygen -197 lb., 
nitrogen -133 lb., and the remainder ash. Find the percentage com- 
position of the coal (i. e. t express the weight of each constituent as a 
percentage of the total weight). 

Carbon . . . . 2-52 Ibs. Total coal ... 3-35 Ibs. 

Hydrogen .... -234 lb. Total (less ash) . 3*084 ,, 

Oxygen . . . . -197 ,, /. Ash = difference = *266 lb. 

Nitrogen . . . . -133 ,, 

Total (without ash) . 3*084 Ibs. 



Carbon 
Hydrogen 
Oxygen 
Nitrogen 
Ash 


2-52 X ioo 

~ TaJ 

234 x ioo 


3'35 ' 
_ 2 3*4 


75-2% 

5-88% 
3*97% 
7'95% 


3'35 
197 X ioo 


3'35 
19-7 


3'35 
133 X ioo 


3'35 
13*3 


3'35 
266 x ioo 


3'35 
26-6 


3*35 


3'35 



Total 99-99% 

The sum of the percentages of all the constituents should be 
100%. Any serious difference would indicate a mistake some- 
where. Due to slight approximations, etc., small differences may 
arise, but if, as in the above, the percentages total 99-99%, the 
figures can quite well be accepted. The exact difference allowable 
depends on the delicacy of the experiments, etc. 

Example 69. The chemical analysis of a " Self-hardening " steel 
gave the following results : Carbon -63%, chromium 1-04%, manganese 
05%, silicon -15%. Find the percentage of iron " by difference " 
(t. e., subtract the total % of matter other than iron, from 100%). 



76 ARITHMETIC FOR ENGINEERS 

This example does not involve any finding of percentages, but is 
merely an addition dealing with percentages. 

Carbon . . . '63% 100-00 

Chromium . . 1*04,, 1-87 

Manganese . . -05 ,, - - 

Silicon . . . -15 ,, 98-13% Iron 

'. Iron " by difference " = 98-13 



Exercises 18. On Percentages. 

1. A firm engaged in the manufacture of valves, makes the following 
extra charges for drilling holes in the flanges of cast-iron bodies : 

i y holes 45. per dozen holes ; i" holes 25. 6d. per dozen. 
For cast-steel bodies they charge 50% extra. Find the rates for the 
above holes in cast steel. 

2. If 5% extra is charged for packing a job for export, find the 
charge for packing a job costing 50. 

3. If the estimated cost of a certain job is 26s. and a profit of 12^% 
is to be added, find the selling price of the job. 

4. The ordinary price of a certain clutch is 4. Due to an 
increase in cost of material the price is increased 5%. Find (a) the 
new charge. A customer is now allowed a discount of 25%. Find 
(b) the actual amount he is to pay. 

5. A wrought-iron shaft will only transmit 70% of the h.-p. that the 
same-sized steel shaft will transmit at the same speed. If the h.-p. 
of a certain steel shaft is 256, what is that of the same-sized wrought- 
iron shaft ? 

6. A square foot of wrought-iron plate i" thick weighs 40 Ibs. If 
steel is 2% heavier than wrought iron, find the weight of the same- 
sized sheet of steel. 

7. The " working volume " of a steam engine cylinder is 5*96 cu. ft. 
The " clearance " volume is 8% of the working volume. Find (a) this 
clearance volume in cubic feet; (b) the total volume in cubic feet 
(i. e., clearance -f working). 

8. A variation in weight of 2% either side of the standard weight 
is allowed in electric conductors. The standard weight of 1000 yds. 
of 37/13 conductor is 2900 Ibs. Find the maximum and minimum weights 
allowable. 

9. The Admiralty test load on a \" crane chain is 3 tons. At the 
Elswick Works the test load is 10% higher. Find the Elswick test load. 

10. It is found that an engine governor keeps the speed between 
486 and 474 revolutions per minute. Find the change of speed and 
express it as a percentage of the average speed. 

11. At 40 F., i cu. ft. of water weighs 62*43 Ibs., and at 400 F. 
weighs 53*63 Ibs. Find the decrease in weight, and express it as a 
percentage of the greatest weight. 

12. The correct value of " g " (the acceleration due to gravity) in 
the metric system at London is 981 units. An experiment gave the 
value 992. Calculate the percentage error (i. e,, express difference as a 
percentage of the true value). 

13. A pump was rated by its makers at 300 gallons per minute. 
On test it filled a tank, holding 750 gallons, 4 times in 9 min. 27 sec. 



DECIMAL FRACTIONS 



77 



Find (a) the test rating of the pump (i. e., the gallons delivered in i min. 
on test), and (b) the difference between the makers' and the test ratings. 
Express this as a percentage of the makers' rating, saying whether it is 
above or below. 

14 to 16. The composition of some alloys is given below. Find in 
each case the percentage composition. 

14. Bell metal : 16 parts copper, 5 parts tin. 

15. Soft gun metal : 16 parts copper, i part tin. 

16. Ajax plastic bronze : 13 parts copper, i part tin, 6 parts lead. 

17. A solution of electrolyte for copper plating is composed of I Ib. 
copper sulphate, i Ib. sulphuric acid, and 10 Ibs. water. Calculate the 
percentage composition of the solution. 

18. A quantity of exhaust gas is known to consist of the following : 
Water -18 cu. ft., carbon dioxide -24 cu. ft., nitrogen '852 cu ft., oxygen 
08 cu. ft. Express the quantity of each constituent as a percentage of 
the total quantity. 

19. The following figures were calculated from an analysis of a 
sample of flue gas : Carbon dioxide 1-54, oxygen 1-81, nitrogen 11-43. 
Express these figures as percentages of the total. 

20 to 23. The chemical analyses of various irons and steels are given 
in the accompanying table. Find the percentage of iron in each case, 
by difference. 



No. 


MATERIAL 


Carbon 
% 


Silicon 
% 


Sulphur 
% 


Phos- 
phorus % 


Manga- 
nese % 


Iron 

% 


20 


Locomotive Boiler 
















Plate (Mild Steel) 


I 4 8 


024 


039 


022 


562 




21 


Tram Rail (Mild 
















Steel) .... 


739 


347 


O^O 


028 


720 




22 


Cast Iron 


3-23 


1-46 


I 4 


638 


5<> 




23 


Wrought Iron . 


04 


17 


013 


31 


OQ 





24 to 26. Engine governors are designed to prevent the engine speed 
from changing more than a certain amount when the load is removed 
or applied suddenly. The variation allowed depends upon the kind 
of work 

For cotton spinning the change is not to be more than 2% above 
and below normal speed. 

For electric traction the change is not to be more than 1*5% above 
and below normal speed. 

For machine shop work the change is not to be more than 3% above 
and below normal speed. 

Find the greatest and least speeds in the following cases 

24. Normal speed 200 revs, per min., for cotton spinning. 

25. Normal speed 180 revs, per min., for electric traction. 

26. Normal speed 85 revs, per min., for machine shop drive. 

27. A salt solution is made by adding 65 grammes of salt to 250 
grammes of water. Find the percentage of salt in the solution (i. e. f 
salt as a % of the total). 

28. A mixture for " blueing " steel is composed of water 10 Ibs., 
hypo 2 oz., sugar of lead 2 oz. Calculate the percentage composition 
by weight of the mixture. (Note. Work in ounces.) 



78 ARITHMETIC FOR ENGINEERS 

Ratio. It is often necessary to state the number of times 
that one quantity contains another, and this is usually given in 
the form of a ratio. Thus, if the diameter of a steam engine cylinder 
is 12" and the stroke 18", the stroke is evidently ij times the dia- 
meter, since 18" = 1-5 X 12". Now this fact of the stroke being 
equal to i J diameters is expressed by the words " the Ratio of stroke 

to diameter is 1*5 to 1." The 1*5 is evidently the value of = S -j- 

i. e., j-S - -v-5-. Thus, the ratio of one quantity (A) to another 
2nd quantity ^ j \ i 

(B) is the number of times that (A) contains (B). 

It should be clear that we cannot give a ratio between two 
quantities of different measures. Thus, we cannot say that the 
ratio of 27 tons to 9 miles is 3 to i ; certainly 27 tons is not 3 times 
9 miles. Hence the measures must be alike. Further, the value of 
the ratio cannot be worked out without bringing both quantities 
to the same unit. Thus, consider the ratio of 2 ft. to 6 in. Obviously 
this is not or J or i to 3, for 2 ft. is not one-third of 6 in. But, 

converting the 2 ft. into inches (24*) the ratio is ^ or- or 4 to i, 

meaning that 2 ft. is 4 times 6 in., which is certainly true. 

Thus we obtain the rule : To find the ratio of one quantity (A) to 
another quantity (B), reduce both to the same units and divide the first 

A 
quantity by the second, i. e. ^. 

Example 70. The " grate area " of a boiler is 22 sq. ft., and the 
heating surface is 1365 sq. ft. Find the ratio of heating surface to grate 
area. 

The first-named quantity in the ratio required is heating surface, 
so that heating surface is to be divided by grate area. 



Then, ratio of heating surface to grate area = ^--^ = 62*1 

or, say, 62 to i. 

Example 71. A locomotive crank is 13" long and the connecting 
rod is 6 ft. 2 in. long. Find the ratio of the length of connecting rod 
to length of crank. 

Convert to the same units, preferably inches. Then the rod is 74* 
long. The required ratio is " connecting rod to crank," so that con- 
necting rod is to be divided by crank. 

Then ratio of connecting rod to crank = J J = 5-69, or, say, 5-7 to i. 

Slight variations in the manner of stating a ratio are met with. 
The word " to " is frequently replaced by two dots, thus (:), so that 



DECIMAL FRACTIONS 79 

the statement " 4 : I " is read as " 4 to i." Sometimes only the 
result of the division is stated, the words " to I " being under- 
stood. Thus a ratio 57 means a ratio of 57 to i. 

It should be noticed that two ratios always exist between two 
quantities. Thus, considering our original case of stroke and 
diameter we can have 

(a) The ratio of stroke to diameter, which = ^ = - 
x ' dia. 12 

or i -5 to i 
i. e., stroke = i*5 times diameter. 

(b) The ratio of diameter to stroke, which = r- = 
v ' stroke 18 

or -667 to i 
i. e. t diameter is "667 times stroke. 

In order to avoid any doubt as to which is required the form 
of statement with the words "of " and "to" should be carefully 
kept ; a statement such as " the ratio between stroke and diameter " 
is not sufficiently clear and should be avoided. Then the quantity 
first mentioned will be the numerator of the ratio fraction, e. g., 
in (a) the stroke is first mentioned and therefore stroke value is 
above diameter value in the fraction. 

Also the tw r o ratios are "reciprocal," i. e. t either one is the other 
inverted. 

Thus stroke to diameter = = -5 and diameter to stroke = 

12 I 

$ = = '667 ; one can always be obtained from the other. 
18 1*5 

Example 72. The following figures refer to a steam-driven air com- 
pressor : Area of steam cylinder =113 sq. ins.; area of air cylinder = 
93*5 S( l' ms> Calculate the ratio of air cylinder area to steam cylinder 
area and give also the reverse ratio of steam cylinder to air cylinder. 

Ratio air cylinder area to steam cylinder area = 93_5 

= '827, i. e. t -827 to i 
Then the ratio steam cylinder to air cylinder = i to '827 



Example 73. The efficiency of any machine is the ratio of the out- 
put to the input. A generator gives an output of 27*6 kilowatts for an 
input of 51 h.-p. Find the efficiency. 

(Note i kilowatt = 1*34 h.-p.) 



8o ARITHMETIC FOR ENGINEERS 

Reduce to same units, preferably horse-power. 

Then, since i kw. = 1-34 h.-p. 

27-6 kw. =s 27-6 x 1-34 = 37 h.-p. 

Efficiency = Ratio of output to input = f = '725 

Note. In the case of efficiency it is usual to state only the decimal 
result as shown, omitting the words " to i." In many cases also this 
result is converted to a percentage, which for the above would be 
72-5%. 

Ratios are used when speaking of the relative sizes of the cylinders 
in a compound or multi-stage expansion steam engine. The case 
of the compound engine, with only 2 cylinders, may be similarly 
treated, but with triple and quadruple expansion 3 and 4 cylinders 
have to be dealt with, and 2 or 3 ratios are combined into one short 
statement. 

Example 74. The triple-expansion engines of the White Star liner 
Olympic have the following cylinder areas : High Pressure (H.P.) 
2290 sq. ins.; Intermediate Pressure (I. P.) 5542 sq. ins., and Low Pres- 
sure (L.P.) 7390 sq. ins. Find the "cylinder ratios " (viz. ratios of the 
cylinder areas), taking H.P. as i. 

There will be two ratios ; and H.P. area will be the 2nd quantity in 
each case. 

The area ratios are : 1. I. P. to H.P. 2. L.P. to H.P. 

1. Ratio of I.P. to H.P. = |^ - 2-42 : i 

2. Ratio of L.P. to H.P. - ~ = 3^3 i 



Instead of writing these results as two, they are combined into 
one by giving " the ratio of L.P. to I.P. to H.P/' ; thus 3-23 : 2-42 : i, 
meaning that if the H.P. area be i, then the I.P. area is 2-42 and the 
L.P. area 3*23. Then the ratio of trie L.P. to the I.P. should be 

according to this, ^^ = 1*335. This is proved by taking the actual 



areas given. Then ratio of L.P. to I.P. = ^ 1-335 as before. 

The dots (:) are invariably used when giving cylinder ratios. 

Ratios connected with certain triangles are very important, and 
are the foundation of a special branch of mathematics known as 
" Trigonometry/ 1 (See Chap. XI.) 



DECIMAL FRACTIONS Si 



Exercises 19. On Ratios. 

Find the ratio of connecting rod length to crank length in the follow- 
ing engines : 

1. (a) A petrol motor ; crank 2 J", connecting rod n*. 

(b) A marine engine ; crank 30", connecting rod 5'-!*. 

2. (a) A high-speed electric-lighting engine; crank 4", connecting 
rod i'-S\ 

(b) A power-driven air-compressor; crank 2\", connecting rod 14". 
The " diagram factor " of a steam engine is the ratio of the actual 
mean pressure to the theoretical mean pressure. Find the diagram 
factor for the following cases : 

3. A marine engine; theor. mean press. 36 Ibs. per sq. in.; actual 
mean press. 23 Ibs. per sq. in. 

4. Corliss engine ; theor. mean press. 29 Ibs. per sq. in. ; actual mean 
press. 25 Ibs. per sq. in. 

5. The " buckling factor " of a column is the ratio of its length 
(inches) to a measurement known as its "least radius of gyration" 
(inches). Find the buckling factor : (a) when length 14 ft., and 
least radius of gyration = 4*13"; (b) when length = 38 / -6 /y , and least 
radius of gyration 4-1.* 

6. Find the ratio of length to diameter of a Lancashire boiler : (a) 
when length = 24 ft. and diameter 7 / -6 y/ ; (b) when length = 21 ft. 
and diameter = 6 '-6*. Note work in feet. 

7. Find in each of the following boilers the ratio of heating surface 
to grate area : 

(a) Heating surface 720 sq. ft. ; grate area 24-75 sq. ft. 

(b) Heating surface 1616 sq. ft. ; grate area 53-6 sq. ft. 

8. Values for the tensile and shearing strength of two metals are given 
below. Find in each case the ratio of shear strength to tensile stress : 

(a) Copper: Tensile 14 tons per sq. in.; shear n-6 tons per sq. in. 

(b) Hard rolled bronze: Tensile 26-9 tons per sq. in.; shear 16-06 
tons per sq. in. 

9. The results of tests giving the elastic limit stress and breaking 
stress (in tons per sq. in.) of various metals are given below. Find for 
the two materials ratio of the elastic stress to breaking stress in each case. 

(a) Gun steel. Elastic stress 24-8. Breaking stress 46-9. 

(b) Lowmcor iron. Elastic stress 12-5. Breaking stress 22*1. 

10. At atmospheric pressure *oi6 cu. ft. of water, when evaporated, 
becomes 26-37 cu. ft. of steam. Find the " relative volume " (i. e. t the 
ratio of steam volume to water volume). 

11. A two-stage air-compressor has the following dimensions : 
Area of H.P. steam cylinder, 314 sq. ins. ; Area of H.P. air cylinder, 

380 sq. ins. 

Area of L.P. steam cylinder, 1385 sq. ins. Area of L.P. air cylinder, 
1018 sq. ins. 

Find the ratio of L.P. to H.P. (a) for steam end ; (b) for air end. 

12. The areas of the cylinders in a triple-expansion marine engine 
are as follows : H.P. 962 sq. ins ; I. P. 2290 sq. ins. ; L.P. 6082 sq. ins. 
Find the ratio of L.P. : I.P : H.P. 

13. The quadruple expansion engines of an Atlantic liner have the 
G 



82 ARITHMETIC FOR ENGINEERS 

following areas : H.P. 1097; I.P. No. I, 1905; I. P. No. 2, 4390; and 
L.P. 9852 sq. ins. Find the cylinder ratios (i.e., L,P. : 2nd I.P. : ist I.P. : 
H.P.). 

14. The following figures refer to two American locomotives : 

(a) Firebox heating surface 231 sq. ft.; tubes heating surface 
3193 sq. ft. 

(b) Firebox heating surface 294-5 S( l- **; tubes heating surface 
3625 sq. ft. 

Find for each case the ratio of total heating surface to the fire- 
box heating surface. (Note. Total heating surface covers tubes and 
firebox.) 

15. The efficiency of any simple lifting machine is the ratio of 
the theoretical effort to the actual effort. Find the efficiency if the 
theoretical effort is -59 Ib. and the actual 1-43 Ibs. 

16. and 17. The " velocity ratio " of any machine is the ratio of the 
movement of the effort to the movement of the load. Find this ratio 
in each of the following cases : 

16. Weston pulley block. Effort moves 30"; load moves 2-5". 

17. Geared capstan. Effort moves 31 '-5"; load moves 18". 

18. The "expansion ratio" in a simple steam engine is approxim- 
ately the ratio of the initial pressure to the final pressure. Find the 
expansion ratio in the following cases : 

(a) Initial pressure 72 Ibs. per sq. in. ; final pressure 18 Ibs. per sq. in. 

(b) Initial pressure 43 Ibs. per sq. in ; final pressure 17-5 Ibs. per sq. in. 

19. If a cast-iron propeller costs ^24, then the same sized propeller 
would cost : (a) in steel, ^38, (b) in Delta metal 115, (c) in gun metal 
130, (d) in manganese bronze 135, (e) in aluminium bronze 145, 
(/) in phosphor bronze ^170. Find the ratio of the cost of each of these 
materials to the cost of the cast-iron propeller. 

Proportion. When two ratios have the same value, the four 
quantities composing the ratios are said to be " proportionals/' 
or " in proportion." Thus the ratio of 3 to 2 is f or 1-5. Similarly 
the ratio of 12 to 8 is - 1 / which also equals 1-5. Then evidently the 
ratio of 3 : 2 = the ratio of 12 : 8 and the numbers 3, 2, 12 and 8 are 
said to be in proportion. Thus 4 quantities are in proportion when 
the ratio of the ist to the 2nd is the same as the ratio of the 3rd 
to the 4th. The equals sign is commonly replaced by four dots, 
thus (: :), and the above result is stated as 3 : 2 : :i2 : 8, and is read 
shortly as "3 is to 2 as 12 is to 8." A more useful form is 
obtained by stating the ratios as vulgar fractions, thus f = -\ 2 . 
Although each ratio must have the same units and measures, they 
need not be the same in the two ratios, since each ratio is only a 
number. 

Proportion has its use when, knowing the ratio between two 
quantities, and the value of one of them, we wish to find the value 
of the other quantity ; that is, knowing three of the quantities in 
the proportion we wish to find the fourth. 



DECIMAL FRACTIONS 83 

Thus, supposing the ratio of the H.P. diameter to the L.P. 
diameter in a compound steam engine is to be 3 : 5, and the L.P. 
diameter is 32" ; and it is desired to find the H.P. diameter. 

Then the statement H.P. : L.P. : : 3 : 5 is a proportion, meaning 
that if the H.P. is, say, 3" dia. the L.P. is 5" dia., and if the H.P. is, 
say, 6" dia. the L.P. is 10" dia., or, whatever the actual size the 
H.P. is f of the L.P. 

Then since in this case L.P. = 32" 

H.P. = of 32 = 19*2" or, say, 19^" dia. 

Now if our answer is correct the ratio of the calculated H.P. to 
the L.P. should be 3 : 5 = or -6 to i. 

Ratio of 19^ (H.P.) to 32 (L.P.) = -^~ = -6 to i, which proves 

the work. 

Proportion is best treated by the simple equation (see p. 164), 
when any one of the quantities can be easily obtained. But from 
the foregoing calculation we can deduce a rule which will suit all 
ordinary cases. It is seen in the above that H.P. = L.P. x . 
Now numbering the terms in the proportion H.P. : L.P. 1:3:5, 

from left to right as No. i, 2, 3, and 4, then No. i == ^^ No> 3 . 

Should the given ratio be written so that the required quantity is 
No. 2, then the ratio should be reversed as in Ex. 77. 

Example 75. Brass is an alloy of copper and zinc; how much 
copper must be mixed with 80 Ibs. of zinc when the ratio of copper to 

zinc is 7:3? 

Copper : Zinc : : 7 : 3 

-, Zinc x 7 
Then copper = - 

gg- x -Z = ,867 Ibs. 

When the given ratio is stated as " so much to i," then the i will 
have no effect on the result, so that only one operation, instead of 
two as above, is required. 

Example. 76. The ratio of the lengths of connecting rod to crank 
in a steam engine is to be 4-5 (i. e., 4-5 : i). Find the length of the 
connecting rod if the crank is 8". 

Connecting rod : Crank : : 4*5 : i 

Then connecting rod = 



84 ARITHMETIC FOR ENGINEERS 

Example 77. The efficiency (i. e., the ratio of the output to the 
input) of a transmission gear is '87. Find the input if the output is 
required to be 35 horse-power. 

output : input : : -87 : i 

Input (which is here the 2nd quantity) is required ; then the state- 
ment must be changed to 

input : output : : i : '87. 
Then input = <"*P^ X i __3S = 40-3 h .-p. 

Example 78. In a triple expansion engine the ratios of the cylinder 
areas (H.P. : I.P. : L.P.) are to be as 2 : 5 : 13. The L.P. area is 2463 
sq. ins. Find the areas of the other two cylinders. 

Considering first the I. P., we have the proportion 

I.P. : L.P. : : 5 : 13 
Then I.R- k*2L3 = ^XJ, = ^ sq> ^ 

Now take the H.P. ; we have another proportion 
H.P. : L.P. : : 2 : 13 

TT _, L.P. X 2 2463 X 2 
/. H.P.= = = 379 sq. ins. 



Exercises 20. Proportion. 

1. Pewter consists of tin and lead, the ratio of tin to lead by weight 
being 4:1. If a founder only has in stock ij cwt. (168 Ibs.) of tin 
and desires to use the whole of it in making pewter, how much lead 
will be required ? 

2. A quantity of sulphuric acid is to be diluted so that the final 
solution contains water and acid in the ratio of 9 : i. How much water 
must be mixed with gallon of acid ? 

3. The ratio of the L.P. cylinder area to the H.P. cylinder area in a 
compound air compressor is to be 2*25 : i. If the L.P. area is 113 sq. ins. 
find the area of the H.P. cylinder. 

4. If gun metal weighs heavier than cast iron in the ratio of 505 to 
450, find the weight of a casting in gun metal to replace one in cast 
iron of the same size weighing 14-5 Ibs. What is the increase in weight ? 

5. The cylinder diameters of a triple expansion steam engine are to 
be in the ratio 3 : 5 : 8 (i. e., H.P. : I.P. : L.P.) and the L.P. is to be 42^ 
diameter. Find the diameter of the H.P. and I.P. cylinders. 

6. The areas of the cylinders of a quadruple expansion marine engine 
are to be in the ratio 1:2-2: 4-6 : 10 (i.e., H.P. : ist I.P. : 2nd I.P. : 
L.P.). If the area required in the L.P. is 7543 sq. ins. find the areas of 
the other three cylinders. 

7. The ratio of the H.P. area to the L.P. area in a hydraulic intensifier 



DECIMAL FRACTIONS 85 

is to be 35 : 800. If the H.P. cylinder is required to be 154 sq. ins. area, 
find the area of the L.P. cylinder. 

8. A solution for pickling castings is to be made of sulphuric acid 
and water in the ratio of 4-5 parts water to i part acid. How much 
water is required for 30 gallons of acid ? 

9. Another pickling solution is composed of hydrofluoric acid and 
water in the ratio of i : 35. What quantity of water must be mixed 
with 5 gallons of acid ? 

10. If the ratio of copper to tin in bell metal is 16 : 5, how much tin 
must be alloyed with 12 cwt. of copper, assuming no loss in the melting ? 

Exercises 20a. Miscellaneous Examples involving 
Decimals. 

1. Find the value of the expression 4 " ' -j- ' 

2. Find the value of 27(17-4 -f- 2-5) -f- 30 x 6-5 x 11*75. 

3. An Avro aeroplane has a total wing area of 330 square feet, 
an engine of no horsepower, and when fully loaded weighs 1820 pounds. 
Calculate (a) the loading per square foot of wing area, (b) the loading 
per H.P. 

4. A variable electrical condenser has 22 fixed plates, each 0-032" 
thick, spaced uniformly o-i" apart. The two supporting end plates 
are 0-104" thick and are each |" beyond the outside fixed plates. Cal- 
culate the overall height of the condenser. 

5. The loss of liquid by evaporation in a liquid air container is 
required to be not greater than 10% of the full charge in a period of 
24 hours. An actual container on test lost i\ Ib. in 6J hours. If the 
full charge is 50 Ib. of liquid, calculate the percentage loss in this case. 

6. The wing span of a twin-engined aeroplane is 67 ft. 2 ins., and 
the chord of the wings is 10 ft. 6 ins. Calculate the " aspect ratio," 
i.e., the ratio of span to chord. 

7. A batch of compressed gas cylinders, all of the same size, was 
weighed after manufacture, the weights being as follows : 10 Ib. 6 oz., 
10 Ib. 2 oz., 10 Ib. 7 oz., 10 Ib. 8 oz., 10 Ib. 4 oz., 10 Ib. 7 oz., 10 Ib. 5 oz., 
10 Ib. 7 oz. Calculate (a) the average weight of a cylinder, (b) the 
variation in weight (i.e., the difference between the heaviest and lightest 
cylinders) as a percentage of the average weight. 

8. Calculate the total weight of the instrument equipment of 
an aeroplane, which requires the following : 2 Air Speed Indicators ; 
i Pressure Head ; 56 feet of Air Speed Tubing ; 2 Altimeters ; 2 Cross 
Levels; i Compass Type A; i Compass Type B; i Revolution 
Indicator; i Flexible Drive for latter, 10 feet long; i Radiator Thermo- 
meter ; 2 Pressure Gauges ; 2 Watches. The weights of the various 
components are as follows : 

Air Speed Indicator . 1-45 Ib. Compass Type A . 4-5 Ib. 

Pressure Head . . . -25 Ib. Compass Type B . . 6-7 Ib. 

Air Speed Tubing . . -035 Ib. Revolution Indicator . 2*7 Ib. 

per foot. Flexible Drive ... -26 Ib. 

Altimeter . . . . -94 Ib. per foot. 

Cross Level .... -5 Ib. Radiator Thermometer . 1-25 Ib. 

Watch -375 Ib. Pressure Gauge . . . '625 Ib. 



CHAPTER III 
SYMBOLS AND THEIR USES 

Symbols and Formulae. To avoid repetition in connection 
with the weights and measures in every day use, it is customary to 
use some letter or sign instead of the full name of a unit. Thus 
" pounds sterling " is replaced by the sign , " dollars " by $, 
" pounds " (in weight) by Ibs., etc., while every one understands 
that 5 means 5 pounds sterling, or that 15 Ibs. means 15 pounds 
in weight. This system is extended to other uses, and signs are 
employed to denote particular measured quantities. The branch 
of calculation dealing with these signs, or symbols, as they are called, 
is usually known as Algebra. 

The engineer frequently has to indicate that certain quantities 
or measurements must be added or multiplied, or generally operated 
upon in various definite ways ; and often this has to be done without 
giving any actual figures. If words be employed, then cumbersome 
sentences are obtained, and calculation would be very slow and 
laborious. But if some sign, say a letter, be used to stand for each 
quantity dealt with, then any relationship, however complicated, 
may be easily expressed. For example, taking the statement 

Voltage = Current x Total resistance 

Instead of writing this out every time it is required, a shorter 
statement may be obtained by using letters to stand for the three 
quantities involved. Let us represent " Voltage " by the letter V, 
"Current" by C, and "Total resistance" by R, using the initial 
letters in each case, a method usually adopted where possible. We 
may then write 

V = C X R 

Knowing what V, C and R indicate, then this statement conveys 
as much information as the former written sentence. The new 
statement is known as a Formula (plural : Formulae) and the 
various letters in it are called Symbols. If the value of current 
and resistance be given for some particular case, we can calculate 

86 



SYMBOLS AND THEIR USES 87 

the corresponding value of V by " substituting " the given values 
for C and R, and working out the expression so obtained. 

Thus, if C = 4'5 amperes, and R = 6-63 ohms 
then V = C X R 

= 4'5 x 6-63 = 29-8 volts 

Letters Used. Many kinds of signs may be used, but the 
simplest are the letters of the alphabet, both small and capital. 
The latter form should be printed and not written, as they are 
then clearer, and the formulae are more easily read. Letters of the 
Greek alphabet, a (alpha), ft (beta), TT (pi), etc., are also used. 
Further symbols may be obtained by attaching a " suffix " to a 
letter. Thus a^ (read as " a one ") and T, may be used to denote 
different quantities from that represented by a and T respectively ; 
but suffixes are best avoided, as being small they are liable to be 
forgotten, when confusion would result. Of course, the same symbol 
should never be used to denote two different things in the same example. 

It is unfortunate that there is no universal system of symbols 
for use in engineering work. Thus one man will use / for stress, 
another q, and so on, while in the same book t will be adopted to 
denote temperature, as well as thickness. This is confusing, as 
the same formula has to be considered in two or three different 
forms. Usually each person keeps some particular letter for each 
quantity in all his own calculations, and it is necessary to say in 
each example what the various letters stand for. If this precaution 
is not adopted considerable confusion may arise, wrong values being 
substituted and the like. It is noteworthy that an attempt has 
been made to standardise the symbols for use in Reinforced Concrete 
calculations, and in Electrical work; and in most engineering 
calculations there are certain letters employed by every one for 
certain quantities, e.g., IT (pi) for a constant used in circle calcula- 
tions, " g " for the " acceleration due to gravity/' etc. 

Signs of x and -f-. When dealing with letters, the multiplica- 
tion sign x is almost invariably omitted, the letters being written 
close to one another, so that C X R would appear as CR. This is 
perfectly understood for letters, but it must be remembered that 
the same scheme cannot be employed for numbers. Thus 46 cannot 
mean 4x6. The original intention was that a dot (.) should be 
placed between the symbols multiplied ; thus C X R would appear 
as C.R, but the dot being very small, it is not a difficult matter to 
forget to write it, especially as its omission does not alter the value 
of the expression. The dot is, however, sometimes used with 



88 ARITHMETIC FOR ENGINEERS 

numbers ; thus 4.6 means 4x6, but its use in such a manner is not 
advisable. 

The algebraic method just mentioned may be adopted whenever 
a letter and a figure are to be multiplied together. 

Thus 30 (read as " three a ") means 3 x a, i. e. t 3 times a t 
= a + a -\- a 

or 3 + 3 + 3 + 3+ until there are " a " threes. 

Of course in this last case we cannot say how many threes there 
are unless we know the value of a. 
Similarly wl = w x / 

i. e., I + / + / + I . . . w times 
or w + te>+ w + w . . . I 

A common example of this nature occurs in connection with a beam 
supporting a " uniformly distributed load/' i. e., a load spread 
evenly over some distance. At a, Fig. 15, the load is spread evenly 






Load = ur lb pe-v 





\ \ 


- ur \bs ptr foot 






i i 


I 1 


\ \ 


i i 


] I 


1 I 


r 






1 1" 

fl. 


"\ 


l FT 




L_ 


b F 



Fig. 15. Uniform loading of Beams. 

over a portion of the beam. If the intensity of the loading, i. e. t 
the load upon i ft. length, be w Ibs per ft.., then 

on i ft. the load will be w Ibs. 

2 2.W 

j* >* ii i> O"' i 

4 . 4^ 

/ ft. the total load will be Iw Ibs. or wl Ibs. 

Then if w were ij tons per ft., and / were 18 ft., the total load 
would be ij X 18 = 27 tons. 

At b t Fig. 15, the total load = wL, where L = span. This 
statement is true for any kind of beam, with any span and any 
intensity of loading. 

The system may be extended to three or more quantities. Thus 
2irrn means 2 x n x r x n. 

The symbols need not be written in any particular order, since 
the order in which a multiplication is carried out does not affect 
the result. Thus 2irrn is the same thing as irn2r or rznw. It is 



SYMBOLS AND THEIR USES 89 

usual, however, to retain a particular order in any formula, as it 
is then more easily remembered. It is also customary to write 
the figures before the symbols. In expressions of this type the 
number in front of the symbols is called a coefficient, and is, of 
course, a constant quantity. When an expression contains no co- 
efficient, then a " I " must be understood and not nought. Thus 

ab means lab 

wly ,, iwly and so on. 

The division sign ~ is seldom used, the fractional method of 
writing being preferred. This gives a more compact form of state- 
ment than the sign 4-, as can be seen from a comparison of the 
following expressions : 

4 3_ w hi cn equals 66 x 4 X 36 ~ (100 x 11-3) 

IOO X 11*3 

and - which equals TWN ~ 33000. 
33000 H 

WL 

Similarly # denotes that the result of W x L is to be divided by 

the result of a x A. 

Signs of + and : Brackets. These signs are used in 
algebra in exactly the same manner as in arithmetic. Thus s + L 
means the addition of the quantities represented by s and L. Simi- 
larly W w + b means that w is to be subtracted from W and 
then b added to the result. In any "expression" (i.e., mathe- 
matical statement) the signs of + and separate the various 
quantities into groups of symbols called " terms/' Thus in the 

Pi; 2 P t> 2 

expression H + ^ -f- , the terms are H, ^ and . 
F G ^ 2g' G 2g 

The remarks made in Chap. I on brackets (p. 29) and on the 
order in which multiplication and addition operations are to be 
performed (p. 30) must be observed when dealing with symbols. 
When brackets appear thus (d i)(r + 2), then the results of the 
brackets are to be multiplied together. 

Simple Evaluation. The evaluation of formulae by sub- 
stitution, i. e., finding the value of the expression when certain 
values are given to the symbols, is an important piece of work, 
and some simple examples will be given here. In all cases the 
reader is advised, when working through the exercises, to copy 
closely the method of stating the work shown in these examples. 

* A, Greek letter "delta." 



go ARITHMETIC FOR ENGINEERS 

The first step should merely be the substitution of the given values 
in place of their symbols. 

Example 79. If E is the voltage supplied to an electric circuit, 

EC 

and C is the current in amperes, then the power supplied is kilo- 
watts. Find the power supplied when E no volts and C = 25 
amperes. 

EC no x 25 

_ ^_ . - __3 with values substituted 

IOOO IOOO 

= 2 '75 kilowatts 



Example 80. The M Modulus of Elasticity " in tons per sq. in. 

WL 
of a bar of metal is given by the formula , where W = load on 

J #A 

bar in tons : L = length of bar in inches : a = area of cross section 
in sq. ins. : and A = extension in inches. Find the Modulus when 
W = 5 tons, L = 8", a = -44 sq. in. and A -0067 in. 



WL ^ 5 



X -0067 



00295 



Approximation. 



2 



5 



j m 

-. if = i - 4 



about 14000, 
= 13600 tons per sq. in. 

Example 81. The voltage required to maintain an electric arc is 
a + fcL -| -- -t where a, b, d and e are constants found by experi- 

v^ 

ment : L == length of arc in millimetres : and C = current in amperes. 
If a = 39, b = 2'i, d = 11-7 and e = 10-5, find the volts necessary 
for an arc 5 millimetres long with a current of 10 amperes. 

Voltage required = a + bL + 




39 + I0 . 5 

49-5 + 6-42 
= 55'92, say 55-9 volts. 

Note. The X operations must be done before the + operations. 
Also the division line is a bracket; hence the addition above it must 
be carried out before dividing by 10. 



SYMBOLS AND THEIR USES 91 

Example 82. The horse-power (h.-p.) per cylinder of a petrol engine 
can be calculated from the formula : h.-p. = '45 (d -f s)(d 1*18) 
where d = dia. of cylinder in inches, and s = stroke. Find the h.-p. 
for an engine of 3 -23" dia. and 5" stroke. 

h.-p. = -45 ( d + s )( d " i'i8) = -45 (3*23 -f 5)<3'23 1-18) 

= -45 X 8-23 X 2-05 
= 7-6 h.-p. 



Exercises 21 . On Simple Evaluation : x and -r . 

1. If C is the current in amperes in an electric circuit, and R the 
resistance in ohms, then the voltage necessary = CR. Find the voltage 
if C = 35 amps, and R = 2*48 ohms. 

2. The weight (grammes) of metal deposited in electro-plating = eCt 
where C = current in amperes, t time in seconds that current flows, 
and e is a number depending on the metal deposited. Find the weight 
of copper deposited when C = 22*5 amps., t = 1800, and e '000328 
for copper. 

3. The formula Dwr refers to the " power " of a lathe. Find the 
" power " in each of the following cases : 

(a) Small lathe : D == 8, w = i J and r = 7; 

(b) Large lathe : D = 38, w = 3 and r 14. 

4. The speed of a belt in feet per minute is given by the formula 
TrDN, where ir = 3-14, D = dia. of pulley in ft., and N = revs, per min. 
of the pulley. Find the speed when D = 3-5 ft. and N = 140 revs. 

5. The formula ~Eaat refers to the force exerted by a metal bar 
when heated. Find the force exerted (Ibs.) by a wrought-iron bar if 
E = 25,000,000, a = 3*1, a = -0000066 and t = 300. 

6. Find the value of the expression 2*wLI (a formula used in con- 
nection with alternating electric currents) when IT = 3-14, n = 87, 
L = -02, and I 65. 

7. If a load of W Ib. stretches a bar a distance of A inches, then 

FA 
the work done in inch-pounds is equal to . Find the work done 

if F == 24000 and A = '0075". 

8. The area of a certain geometrical figure is \lh sq. ins. when / = 
length in inches and h = height in inches. Find the area when / = 9 
and h = 4*8. 

9. Find the value of the formula Iw^w^ which relates to a gyro- 
scope, when I = i '5, w l = '0785, and w 2 == 210. 

10. If W tons is the total distributed load on a girder and / is the 
length of the girder in feet, then the " Bending Moment " in tons 

f ee t JLL:. Find the bending moment for a girder where W = 75 tons 
and I = 22'5 ft. 

2 f 

11. The number of poles required in an alternator is - J - where 

/ = frequency of current and n = revs, per sec. Find the number 
when / = 60 and n = 2-5. 



92 ARITHMETIC FOR ENGINEERS 

A h 

12. The formula ~- occurs in connection with the strength of 

A 

beams. Find its value when A = 3-6, A r = 1-67, h = 5-5. 

13. The resistance of any electrical conductor is ohms, where 

/ = length, a = its area and 5 a figure depending on the material. 
Find the resistance where / = 36000, a = -00181, and 5 = -00000066. 

14. The expression =- gives the minimum thickness of a hydraulic 

pipe, where p = pressure in Ibs. per sq. in., r radius in ins., and/ = safe 
stress in material (Ibs. per sq. in.). Find thickness / when p 750, 
r = 2'5", / = 2500. 

prJ-J 

15. Find the value of the expression ~- t which relates to a centrifugal 

pump, when g = 32, H = 12-5, V = 27-5, and v = 23. 

A 

16. The formula K = . refers to electrical condensers. Find K 

4?r( 

when k == 2'2, A = 9-75, TT = 3-14 and t = -15. 

17. When measuring resistances by a Wheatstone Bridge we have 

the formula x = - ^~. Find the value of x when R = 287, Rj 1000, 

K 2 
and R 2 = 10. 

18. The H.P. of a steam engine is given by the formula when 

p = mean effective pressure, a = cylinder area, L = length of stroke 
and N number of working strokes per min. Find the H.P. when 
p = 22-5, L = 1-75, a = 113 and N = 220. 

19. When measuring the B.H.P. of an engine we have the formula 

2?rRNW 

B.H.P. = - - where TT = 3-14, R = radius of brake arm (ft.), 

N revs, per min., and W = load supported (Ibs). Find the B.H.P. 
for a petrol engine when R = 2-5, N 1250, and W = 5-3. 

20. The B.H.P. of a three-phase motor is given by the formula 
1-732^7? ^ w here p = power factor, 77 = efficiency, E == voltage, 

C = current. Find the B.H.P. of a motor when p '8, 17 = -93, 
E = 107, C = 35. 

Exercises 22. On Simple Evaluation, x and -f- with 

4- and . 

1. If the thickness of plate in which a hole is to be punched is /in. 
and d is the diameter of the punch, then the diameter of the hole re- 
quired in the bolster is d -j- ^ . Find what size of hole is required if a 

J" dia. hole is to be punched in a J" plate. (N.B. Work in decimals.) 

2. The diameter of tapping hole for a Whitworth bolt is d 1-28 p, 
where d = bolt dia., and p pitch of screw. Find this value when 
d == -75" and p = !". 

3. The total heat of steam (H) at *F. is very nearly 1082 -f- ^l. 
Find H when / = 345. 

4. The latent heat of steam at *F. is very nearly 1114 -7*. 
Find the latent heat when t = 330. 



SYMBOLS AND THEIR USES 93 

5. The total heat in i Ib. of wet steam is S -f #L, where S = sensible 
heat, x = dryness, and L = latent heat. Find the total heat when 
S = 310, x '93 and L == 896. 

6. The maximum speed allowable (ft. per min.) when cutting cast iron 
with high-speed tools is given approximately by the following formulae : 
Soft cast iron 115 1300 a; hard cast iron 40 4000, where a area of cut. 
Calculate the speed in the following cases : (a) cutting soft cast iron, area 
of cut = -0124 sq. in.; (b) hard cast iron, area of cut = -046 sq. in. 

7. The circumference of an ellipse may be obtained approximately 
from the formula i-8aD -f 1-315^, where D = long axis, and d = short 
axis. Find circumference when D = 2-5 and d = 1*5. 

8. The dimensions of American countersunk- 
head screws are calculated from the following 
formulae (sec Fig. 16) : D = zd -ooS ; 

d -008 , . . H 

-015 ; A = 3 




where d = diameter of screw. Calculate the 
values of D, H, w and h for a screw " dia. 

9. The piston speed of a petrol motor is 
given as 600 (r -f- i) ft. per min., where r ratio 
of stroke to diameter. Find the speed if 
v = 1-25. 

10. If E is the E.M.F. of a battery, r its own resistance, and R the 
resistance through which it is supplying current, then the current C 

is given by the formula C = T^T- Find C when E = 2-5, R = 5-7, 

and r = O2. 

11. When measuring the flow of water in a stream by a notch 

the following formula is met with : C = -57 H -- r^. Find the value 
of C if B = 3 and W = 15. 5/ IoW zR 

12. The " velocity ratio " in a Wcston pulley block is ^ _ where 

R is the large radius, and r the small radius, of the compound pulley. 
Find the velocity ratio if R = 3-5 and r = 3*25. 

13. If C is the temperature on the Centigrade scale, the corresponding 
Fahrenheit temperature is % C -J- 3 2 - Find this value if C = 85. 

14. The current delivered by a number of cells connected in series 

is -Ty^V amperes, where n = number of cells, E = E.M.F. of i cell, 

L\. -j- tTY 

Y = internal resistance of i cell, and R = outside resistance. Find the 
current delivered by 57 cells if E = 2-7, R = 2*3 and r = '03. 



15. The coefficient of friction in a pipe carrying water is '0075 i + ^ 
where d is diameter in inches. Find this value when d = 9". 

16. The expression h{ ^ l -j gives the cross-section of a chimney 

under certain conditions. Find the value of the expression when 
h = 100, d = -0761 and d l = -0414. 

17. A formula for the allowable working stress under live loads is 

- . Calculate the value of the expression when / = 6, W = 7, 



and w = 2. 



94 ARITHMETIC FOR ENGINEERS 

18. The expression . T_ , occurs in certain investigations on 
the strength of materials.* Find its value when h = 15, a = 2-5, and 

19. The weight of a girder before designing may be obtained from 
the expression J^ i tons. For a particular plate-web girder w = 50, 

r = 12, c = 1300, / = 45, / = 6-5. Find its weight in tons. 

20. If a wire rope contains n wires, each d ins. dia., then the diameter 

of the rope is approximately -f- yd. Calculate this diameter if 

n = 72 and d -064. 

21. The expression (p + '38) (v -41) relates to the expansion of 
steam. Find its value when p = 165 and v = 2-7. 

22. The rating formula used by the Society of Motor Manufacturers 
for a petrol engine is 'if)^d(d i)(r -f 2) H.P. per cylinder, where 
d = cylinder dia. in ins., and r ratio of stroke to diameter. Calcu- 
late this H.P. when d 3-23 and Y = 1-55. 

\yrjr f s i c \i 

23. The expression - 1 ~, /J is used in connection with a 

L,nt) 

drop test for materials. Find its value when W = 103, F = 5, s = 2-8, 
c = -57, L = 36, / = 2 and 6 == i. 

24. The expression / (i ec) tons is a formula relating to long 
columns. Find its value when/ = 6, e == -0053 and c = 95. 

25. Cochran's evaporation formula is E = 26(G L^ , 

where E Ibs. of water evaporated per hour, G = grate area in sq. ft., 
and H = heating surface in sq. ft. Find E for a boiler where G = 14 
and H = 250. 

Powers and Indices. When we wish to indicate that a quan- 
tity is to be multiplied by itself, as for example in certain calculations 
with circles where the diameter has to be multiplied by the diameter, 
we could write " d d " if d stood for the diameter, meaning of course 
d X d. But in such cases, when the quantities multiplied together 
(i. e., the factors) are the same, a shorter method of statement is used. 
The symbol for the quantity is only written down once, and a small 
figure is placed at its right-hand top corner to indicate the number 
of symbols which are to be multiplied together. Thus the fore- 
going statement " d d " would be written as d 2 , to indicate that two 
d's are multiplied together. This small figure is called an index, 
or "exponent/' while the quantity represented by a symbol to- 
gether with an index, is said to be a " power " of the symbol ; thus 
d 2 is a power of d. The index ( 2 ) is used when two of the same 
quantities are to be multiplied together, 3 when three of the 
same quantities are to be multiplied together, and so on. 

* See p. 87 for use of dot as a multiplication sign. 



SYMBOLS AND THEIR USES 95 

Thus we have 

d 2 = d x d, i. e. t two d's multiplied together, 

(Read as " d squared " or " d to the second power.") 

If d = 3, then d 2 = 3 X 3 = 9. 

Similarly x 2 = % X x, and so on. 

d 3 = d x d x d, i. e., three ^'s multiplied together, 
(read as " d cubed " or " d to the third power/') 

If d = 3, then d* = 3 x 3 X 3 = 27. 

R 4 =RXRxRxR (read as " R to the fourth," meaning " R 
raised to the fourth power"). 

The index must not be confounded with a suffix, mentioned on 
p. 87. The index is placed at the top, but the suffix at the bottom 
of a symbol ; also the actual figure of an index has a definite mathe- 
matical meaning, while that of a suffix has none. 

It must be clearly understood that an index is not the same as a 
coefficient. Thus % 3 is not the same as 3*. 

For # 3 = % x x x x , i.e., three x's multiplied together, but 
3* = % -f- % -j- x t i. e., three x's added together. 

Let x = 4, 

then x 3 = 4 3 = 4 X 4 X 4 = 64 
but 3x = 3x4 = 12 

showing a considerable difference. 
Similarly d 5 is not 5^. 

For let d = 3, then d* = 3* = 3 X 3 X 3 X 3 X 3 (i. e., the 
product of five 3's) 

= 27 x 9 = 243 
whereas $d = 5 X 3 = I 5- 

The indices met with in practical problems are neither numerous 
nor large. The index 2, representing the " square," is the most 
common. It occurs in all calculations dealing with areas of circles, 
and in many others besides. It is useful to remember the " squares " 
of the simpler numbers as given in the table on p. 101. 

The indices 3 and 4 are less common, but still find a prominent 
place in connection with the strength of shafts and beams, among 
other things. 

The 5th power is found in hydraulic work. Indices beyond 5 
are not common, and are usually only introduced for convenience 
in calculation. Mathematically an index may be of any form and 



96 ARITHMETIC FOR ENGINEERS 

size whatever, but only the whole number form will be considered 
in this book. 

Certain peculiarities met with are worth noting. Thus i raised 
to any power whatever = i. 

For example i 2 = i x i = i 

i 8 =1x1x1x1x1 = i, etc. 
Similarly o raised to any power = o. 

Also a number raised to the first power is not altered, since the 
first power of any quantity is just the single quantity. Thus 
x l = x and 3 1 3. For this reason the index i is not written. 

If 10 be raised to any power such as we have been dealing with, 
the result is always a number composed of a i and some noughts, 
the number of noughts being the same as the index. 

Thus lo 1 = = io, i. e. y i nought 

10 2 10 x 10 = 100 2, noughts 

10 3 = io x io x io = looo ,, 3 

10 4 = io x io x io x io = 10000 ,, 4 

Where large numbers such as 10,000,000 are required in formulae, 
it is customary to replace them by their corresponding power of io, 
to simplify the appearance of the number. Thus the formula for 
the Electromotive Force (E.M.F.) of a dynamo is stated as 
PFNZ, u . PFNZ u A ... . 

Yo ~~ V ltS ' meanm 100,000,0^ volts ' A11 that ls nece ssary 
is to count the number of o's after the i, which gives the index 
to be written above the io. 

Expressions of the forms 3r 2 , a*b, etc., are of frequent occurrence, 
and it must be noted that the index applies only to that symbol against 
which it is placed, unless brackets be used. 

Thus 3r 2 (read as " 3 r squared ") means 3 times the square of r = 
3 X r X r. 

It does not equal 3? x 37, which would be the square of y t and 
would have to be written as ($r) 2 and read as " 3? in brackets, 
squared/ 1 or " 37, all squared." The bracket indicates that every- 
thing inside it must be considered as one quantity. Similarly a 2 6 
(read as " a squared b ") means the " square of a " multiplied by 
b, and therefore = a x a x b. 

It does not mean ab x ab, which would be the " square of ab" 
t.e. t (ab)*. 

Similarly 51* = ^xrxrxr; 
But (5r) 3 = 5r x 5' X 5r 



SYMBOLS AND THEIR USES 97 

The difference in meaning may be further shown if we use some 
figures. Thus, let r = 2. 

Then $r* * 5 X 2 3 

= 5x8= 40. 
But (5?) 3 = (5 x 2) 3 

= IO 3 = IOOO. 

Taking the statement (5?) 8 = $r X $r X 5r, this, of course, = 
5XfX5XfX5Xf. Since the multiplication may be carried 
out in any order, we may write 5X5X5XfXrxr, 

'. e., 5 3 x f 3 = I25r 3 . 

Hence to simplify an expression such as (5r) 3 raise each quantity 
inside the bracket to the power outside. Then evaluate the powers of 
the numbers. 

.'. (5') 8 = 5 3 ^ = 125;*. 

When all the quantities inside the brackets are symbols, then only 
the first sentence of the above rule can be obeyed. Thus (ab) 2 = 
a 2 b 2 and nothing further can be done. The expression a 2 b 2 is no 
simpler than the one (ab) 2 when it stands alone ; but when it forms 
part of a larger expression then the form a 2 b 2 is to be preferred, as 
other simplifications may then be carried out. 

Collecting, for comparison, some of the foregoing statements 
showing the factors in full 

a 2 b = a X a X b 
ab 2 = a X b X b 
(ab) 2 = abxab = axaxbxb~ a 2 b*. 

Then if a = 3 and & = 4 

a 2 b = 3 2 x 4 = 9 X 4 = 36 
ab 2 =3 X 4 2 = 3 X 16 == 48 
(ab) 2 = a 2 b 2 = 3 2 x 4 2 = 9 X 16 = 144. 

Example 83. Find the value of the following expressions : (a) f 4 , 
(b) v, (c) 4 2 , (d) ( 4 a) 2 , (e) z<5p*, (f) (2- 5 p) 2 t when the symbols have the 
following values : a = 3, r = 4, p 2-5. 

(a) r 4 = 4* = 4 X 4 X 4 X 4 = 256. 

(6) 4' = 4 X 4 - 16- 

(c) 4 a = 4 X 3 s == 4 X 9 = 3^. 

(d) ( 4 a) 2 = (4 X 3) 2 = 12* = 144. 

or = 4 2 a 2 = i6a 2 == 16 X 9 = 144. 

(e) 2-5/> 2 = 2-5 X 2-5 2 = 2-5 X 6-25 = 15-625. 
(/) (2'5^) a = (2'5 X 2- 5 ) 8 = 6-25* = 39-0625. 

U 



9 8 ARITHMETIC FOR ENGINEERS 

bd* 
Example 84. Find the value of the expression - . a formula 

relating to the strength of beams, when 6 = 3, d = 5. 
bd 3 3 X 5 3 $ X 125 

. __-_^ ~L i ^L ai2 | N. 

4 
Example 85. The formula *tt gives the safe stress in tons per 

sq. in. in a column, under certain conditions. Find the value of the 
expression when ir 3-14, E = 13400 and C = 95. 

Approximation . 



3x3X1 UU 
5x1x1 UU 

= = 2 approx. 



3 -4^X^13^00 
~5~xT5 S 

s^L^JLM ? 1321^ 

-y^ 9^5 " ~ iJiw Top line = I32I24 

. - say 132100 

2-93 tons per sq. in. _ ,, , 

-23 K * Bottom line = 45125 

say 45100. 

P v 2 
Example 86. The expression H-j-i-H is very important in 

hydraulics. Find its value when H = 15, P 2200, G = 62-5, v = 4, 
and g = 32. 

H -h 7* H = 15 4- ^ 4- - 

Cr 2g ^ 62-5 2 X 32 

= J 5 + 35^ + ^ 

= J 5 + 35'2 + -25 = 50-45- 

Example 87. The strength of copper alloys at high temperatures 
is given by Unwin's formula : / = a b(t 60) 2 . Find the strength 
(tons per sq. in.) of rolled brass, when i = 500, a = 24*1, and b = -000028 

/ = a b(t 6o) 2 = 24-1 -000028(500 6o) a 
= 24-1 -000028 X 440 * 
= 24-1 -000028 X 193600 

= 24-1 5-42 

= 18-68, say, 18-7 tons per sq. in. 

Exercises 23. On Evaluation with Powers. 

Write out the factors of the following : 
1. 4 3 , io\ 3*, -2 3 , - 5 a , io 5 . 2. r 2 , Z>*, a 3 , h*. 

Find the values of the following expressions when a = 3, 6 = 2, 
R = 4, and p = 1-5. 

H. (i) a'; (ii) ^a. 4. (i) R'; (ii) 3 R. 6. (i) 2p; (ii) p*. 
6. (i) a 4 ; (ii) 40. 7. (i) 56; (ii) 6 s . 



SYMBOLS AND THEIR USES 99 

Write out the factors of the following expressions : 

8. (i) c 2 r; (ii) cr 2 ; (Hi) c 2 r*; (iv) (cr) 2 : (v) What do you notice 
about (ih) and (iv) ? 

9. (i) 2M 2 ; (ii) 3M 2 ; (iii) (aM) a ; (iv) (3M) a . 

10. (i) a*r; (ii) a? 3 ; (iii) aV 3 ; (iv) a 3 * 2 . 

11. (0 3 2 ; (ii) 3np 2 ; (Hi) 3" 2 /> 2 . 

12. (i)3(/>) 1 ; (ii)(3) 2 />; (i") (3") 2 . 

Find the values of the following expressions when T = 3, e = 2, 
# = 3, V 10, C = *5, w = 12, / = -i. 

13. (i) T 2 C; (ii) TC 2 ; (iii) T 2 C 2 ; (iv) (TC) 2 . 

14. (i) 4 C 2 ; (ii) 2C 2 ; (iii) ( 4 C) 2 ; (iv) ( 2 C) 2 . 

15. (i) e*x\ (ii) ex*\ (iii) * 2 * 3 ; (iv) e*x*. 

1G. (i) ?>wH\ (ii) iwl 2 ; (iii) 3 w 2 / 2 ; (iv) $(wl}\ 

17. (i) (2V) 3 C; (ii) (2VC) 3 ; (iii) 3W; (iv) 3 PV. 

18. The area of a circle is given by the formula -785 D 2 if D is its 
diameter. Find the area (sq. ins.) if D = 2 5". 

19. If Y is the wave resistance of a ship's model when towed in an 
experimental tank and k is the number of times the ship is bigger than 
its model, then the wave resistance of the ship itself is k 3 r. Find this 
value if k 40 and r = 1-76 Ibs. 

20. Find the value of the expression A 2 & 2 , which relates to the 
strength of a beam, when A 2 = 2-22 and h = 5-7. 

21. Find the value of the expression J I o> 2 , which refers to flywheels, 
when I = 47-2 and = 35-7. 

C 2 

22. The safe load in tons on a white manilla rope is where C = 

F 30 
girth in inches. Find the safe load for a rope of i J" girth. 

23. The H.P. of a petrol motor (i cylinder) may be roughly esti- 

d 2 s 
mated from the formula - > where d = diameter in ins., and s = stroke 

in ins. What is the H.P. when d 3-3 and 5 = 3*74? 

bh* 

24. Find the value of the expression , which refers to the strength 

of beams, when b = 7-5 and h = i-i. 

25. The " kinetic energy " (i. e. t energy of motion) of a body weigh- 

vuv 2 
ing w Ibs. and moving at v ft. per sec. is where g = 32. Modern 

rifle ammunition carries a bullet weighing -0249 lb., which leaves the 
muzzle with a velocity of 2440 ft. per sec. Find the kinetic energy of 
the bullet (ft. Ibs.). 

26. The H.P. of a low-pressure fan for ventilation is given by the 

Q 3 
expression '0000115 g- 4 where Q = cu. ft. of air per sec., and D = dia. 

of fan in ft. Find the H.P. required for a fan 4-5 ft. dia., delivering 
250 cu. ft. per sec. 

27. Henderson's formula for the rating of a petrol engine is 
2d 2 (f H- i) where d = dia. in ins., and r = stroke ratio. Find the 
rating when d = 4*5, and r 1*15. 

23. The copper loss in a transformer is C^ R t + C a 2 R 2 where 



ioo ARITHMETIC FOR ENGINEERS 

G! and R, are current and resistance of primary winding, and C 2 and 
R, current and resistance of the secondary. Find the loss when Cj = 
20, C 2 =i, R! = -08, and R 2 = 1-8. 

29. White and Poppe's formula for the weight of a piston and con- 
necting rod in a petrol engine is -037 (d + 1*9)* Ibs. Calculate the weight 
for a case where d = 4-02*. 

30. Burls' formula for a purpose similar to that in Ex. 29 is 
o8d s (i -f *i5f) + 1-5 Ibs. Calculate the weight for a car engine when 
d = 3 > i5 / " and r = 1-5. 

31. The greatest stress in a thick cylinder under internal pressure 

R2 _L. Y 2 

is p R2 _ a . A cylinder has R = 9 and r = 6 while p = 750. Cal- 

culate the greatest stress. 

32. When measuring the power in an alternating current circuit 
by the 3-voltmeter method we have 

Power = ~V 2 - Vj 2 - V a a watts. 



- Vj 2 - V a a ) 



Calculate the value of the power when C = 6-5 amperes, V = 218 volts, 
V, == 68 volts, V 2 = 156 volts. 

33* The formula J la -- j refers to radial-arm bogies on locomo- 
tives. Find the value (ft.) when a = 12 ft. and b = 6-75 ft. 

Square Root. Consider the statement d x d = d 2 . We have 
already seen that d 2 is called the " square " of d, and that if the 
value of d be known then the value of d 2 can be obtained. Thus if 
d = 4, then d 2 = 4 X 4 = 16. 

Now in many cases our calculations will give us the value of 
an expression like d 2 , from which we have to obtain the value of d. 
Thus we may require the value of d when we know that d 2 = 16, 
that is to say we require the value of some number which if multi- 
plied by itself will give 16 as the product. Now we know that 
4 X 4 = 16, hence 4 is the number required in this case. Then 
4 is said to be the " square root " of 16. The square root of any 
quantity A is that quantity which, if multiplied by itself, gives 
A as product. 

Thus i is the square root of I because I X I = I 
2 4 2x2=4 

i* 3 * i >* * 9 3 X 3 == 9 

4 .. l6 4X4 = 16 

Similarly, from our previous work on powers, we know that 

d X d = d 2 , therefore d is the square root of d 2 
ab X ab = (ab} 2 ab (ab} 2 

or a 2 b 2 . 



SYMBOLS AND THEIR USES 



101 



In formulae the words " square root of " are replaced by the sign <\/ . 
The horizontal line extends over all the quantities of which the 
square root has to be taken. 

Thus we write ^/z$ = 5 because 5 X 5 =25 

V/ioooo = 100 because 100 x 100 = 10000 
= ab ab X ab = 



The square roots of certain numbers, such as those of 4, 9, 16, etc., 
are found in the common multiplication tables, and should therefore 
be known after a moment's thought. Some others, not usually 
remembered, such as \/ 22 5 === I 5* V3 2 4 = *&> etc., are useful, and 
are included in the following table : 

TABLE OF CERTAIN SQUARES AND SQUARE ROOTS. 



Number. 


Square Root. 


Number. 


Square Root. 


Number. 


Square Root. 


I 


I 


25 


5 


169 


13 


2-25 


i'5 


36 


6 


196 


H 


4 


2 


49 


7 


225 


15 


6*25 


2-5 


64 


8 


256 


16 


9 


3 


81 


9 


289 


17 


12-25 


3'5 


IOO 


10 


324 


18 


16 


4 


121 


ii 


361 


19 


20-25 


4-5 


144 


12 


400 


20 


Square. 


Number. 


Square. 


Number. 


Square. 


Number. 



The squares are in the left-hand columns reading from the head- 
ings at the bottom; square roots are in the right-hand columns, 
reading from the headings at the top. 

The square root of a number such as 56 is not given above, as 
it is not a whole number. 

Now 7 X 7 = 49, which is less than 56 
8 x 8 = 64 greater 56 

Hence V$6 is between 7 and 8, i. e., it is 7 and a decimal. Such 
square roots as these can be calculated or " extracted " by a kind 
of division as given below. The method can only be applied to 
decimal numbers. If it be required to find the square root of a 
number containing a vulgar fraction, then usually this must be 
first converted into a decimal. To illustrate the method consider 
the following example. (The mathematical explanation of the 
method is omitted as being somewhat lengthy.) 



102 ARITHMETIC FOR ENGINEERS 

Example 88. Extract the square root of 22*09. 

Write the number down with a bracket to its right-hand side ; 
the answer is to be placed to the right of this bracket. Mark 
off the figures (or digits) in pairs (called periods) by dashes, 
starting from the decimal point and working to the right and 
left. 

Thus I 22*09 1 ( 

Consider the extreme left-hand period, the 22. Find mentally or 
from the table the highest whole number whose square is not greater 
than 22 ; in this case it is 4. Place this number (4), in the answer 
and its square (16) under the period 22. Subtract the square 16 
from the period 22, giving 6 as a remainder. The next thing in 
the number is the decimal point. Therefore place the decimal 
point in the answer, which, as in ordinary division of decimals, 
is done when all the figures in the whole number have been dealt 
with. Bring down the next period or pair of figures, i. e., 09, to the 
side of the remainder 6, giving 609 as a new number to be 
divided. 

Thus '22 -09' (4 

16 

609 

Place a vertical line to the left of the working. Double the 
partial answer 4, making it 8, and write this 8 opposite the re- 
mainder 609, on the left of the vertical line. 



Thus 



8 



1 22 -09' (4- 
16 

609 



Divide the 8 into the first two figures of the remainder, 60. 
It will go 7 times. Write this 7 in the answer, and also at the 
side of the 8 to the left of the vertical line, giving 87. 



Thus 

16 



87 



609 



Now multiply this new number 87 by the last figure put in the 
answer, 7, and write the product under the remainder 609. Now 
87 X 7 = 609. No subtraction can be made, and the example has 



SYMBOLS AND THEIR USES 103 

worked right out. The complete working should now appear 

thus: 

'22-09' (4-7 
16 

609 
609 



Then, V22-09 = 4*7. 

This and the following examples should be proved by actual 
multiplication, i. e., squaring. 

The above method, if followed strictly, will give the square root 
of any number whatever. Certain points arise which require a 
little explanation. 

When the example does not work out without a remainder, as 
does the above, then it is usually sufficient to work to 4 or 5 
significant figures and to give the result correct to 3 or 4. 

The extreme left-hand or starting period will often contain 
only one figure, as in the case of 583*75, which, when marked off 
in pairs, will appear as 5 I 8375 I . Here proceed exactly as before, 
treating the figure 5 as a period; i. e., finding the highest number 
whose square is not greater than 5, in this case 2. 

When it is found that there are not sufficient figures in the 
number to " bring down," then noughts may be supplied to the 
right of the decimal point. Thus, 597 may be marked off as 
5 l 97*oo'oo l , the number not being affected by such adding of 
noughts. It is not absolutely necessary to write the noughts, but 
care must be taken that they are " brought down " in pairs. To 
avoid the very probable error of adding only a single o to a re- 
mainder, it is safer to write the additional noughts, and point them 
off as shown. 

Should the last or right-hand period only contain one figure, 
then a single nought should be added to complete the period. 
Thus 692*5 should be marked off as 6^2 -so 1 . It is advisable to 
write a few additional noughts as required. 

When dividing the first figure of a divisor into the first two 
figures of a dividend, the former may go exactly into the latter, 
or very nearly so. Thus, in Example 89, which follows, taking the 
first division, 4 into 29 goes 7 times almost exactly. In such cases 
it will usually be found that this result is not suitable on account 
of the resulting product being too large, and a smaller number will 
have to be taken. This is shown in the explanations to Examples 



104 ARITHMETIC FOR ENGINEERS 

89 and 92. To avoid writing any figures which may have to be 
discarded, it is advisable to jot down each product in a place re- 
served for scribbling and to test its correctness before writing it in 
the complete working of the example. 

Example 89. Find the square root of 692 '5. 



46 



523 



5261 



292 

276 .... After obtaining this product 



1650 



place the decimal point. 



8100 



Proof by Squaring. 
26-3 
26*3 

"789 

1578 
526 



5261 

691-69 

.'. \/692 -5 = 26-3. 

Explanation and Notes. The highest number whose square is not 
greater than 6 is 2. Hence 2 is placed in answer and 2 2 = 4, under the 6. 
Subtraction gives 2. The next period to bring down is the 92, which 
is placed at the side of the 2 ; the new dividend is thus 292. Double 
the partial answer 2 making 4, and write the 4 on the left opposite the 
292. Dividing 4 into 29 gives 7, but this is found unsuitable, thus : 
placing the 7 in the answer and in the new divisor to the right of the 4, 
we should have 47 X 7 = 329, which is bigger than 292. Hence the 
next smaller number must be taken, which is 6. Then 6 is written to 
the right of the 2 in the answer, and also to the right of the 4 on the 
left, so that the new divisor is 46; then 46 X 6 = 276, which appears 
beneath the 292. Subtraction gives 16. The next thing in our given 
number is the decimal point, which is now written in the answer. The 
third period is the 50, which is brought down to the right of the 16, 
making 1650. Double the partial answer 26, making it 52, which ap- 
pears on the left of the 1650. Dividing 5 into the 16 gives 3. The 
3 is written after the 52 (making the new divisor 523), and also in the 
answer after the decimal point. Multiplying 523 X 3 gives 1569. 
Subtracting gives 81. The next period is oo. Doubling the 263 gives 
526, which is placed on the left. Dividing 5 into 8 gives i. Then i is 
put after the 526 and after the 263 ; and then 5261 X I = 5261. The 
answer to three significant figures is 26-3. As a proof, 26-3 is squared, 
giving 691*69 (shown at the side). The error due to using only three 
significant figures is -81, or about -i% of the original quantity, which 
error is so small as to be completely ignored. 

Great care must be taken when a divisor will not go at all into 



SYMBOLS AND THEIR USES 



105 



one of the dividends. Then a nought must be placed in the answer, 
as in the ordinary division of decimals. 

Example 90. To find ^940. 

9 1 40-00' (30-659 



606 
6125 
61309 


4000 

3636 


36400 
30625 


577500 
551781 



== 3O-66 



Notes. The first period gives 3 in the answer, and leaves no re- 
mainder. Bringing down the second period, the new dividend is 40. 
Doubling the 3, the new divisor is 6. Now 6 into 40 will go 6. But 
placing this with the first 6, giving a divisor of 66, the product would 
be 66 X 6 which is evidently much larger than 40. Even i is too 
large, 61 X i giving 61. Therefore the result of the division in this 
case is o, and this o must be placed in the answer. Bringing down the 
next period oo, proceed as before. 

When the number of which the square root is to be extracted 
is a decimal fraction, then the decimal point is the first thing met 
with, and is then the first thing placed in the answer. After this 
the decimal point may be disregarded, the example being worked 
out as for a whole number, remembering that the " pointing off/' 
must commence from the decimal point. 

When dealing with a decimal fraction which commences with 
one or more noughts, care must be given to the number of noughts 
which are to appear in the answer, as in the next example. 

Example 91. To find <\/ -000972. 

oo'o9' 72*00' (-03117 
9 



'000972 = '0312 



61 


72 




61 


621 


1 100 




621 


6227 


47900 




435^9 



io6 ARITHMETIC FOR ENGINEERS 

Note. Here the decimal point is first met with; hence it is the 
first thing put in the answer. The first period is oo ; hence a o is put 
in the answer. Next period 09, and from there the example is similar 
to previous ones. 

A common mistake, in cases such as Example 91, is to obtain 
the wrong number of noughts following the decimal point in the 
result. The number of noughts may be checked by the following 
rule : 

For every complete period of noughts following the decimal 
point, there must be a nought after the decimal point in the result. 

Thus, in the above cases, there is i complete period of noughts in 
oo'o9'72, and therefore there must be one nought in the result, as 
shown. 

The square root of a whole or mixed number is always smaller 
than the number itself (except with i, for Vi = i). But with 
fractions the square root is always larger than the fraction itself. 
Thus, in Example 91, ^'000972 = -0312, the root being much 
larger than the number. Squaring the result will prove the truth 
of the statement. 

It may be found that after subtraction, a remainder is obtained 
which is larger than the divisor. This would indicate a blunder in 
an ordinary division, yet it is quite possible in square root. The 
following example illustrates : 

Example 92. Extract the square root of 284*5. 
2'84'50 I (16-867 



26 



328 



3366 



33727 



184 

156 Decimal point is placed when this 

product is obtained. 

2850 ^ 

_ ^284-5 = 16-87 
22600 
20196 

240400 
236089 



Notes. On bringing down the second period the new dividend 
is 184. Dividing 2 into the 18 the result is 9 exactly, which is dis- 
carded as being too large. Also both 8 and 7 give products too high. 
Then 6 must be taken, and we have 26 X 6 = 156; 156 from 184 leaves 
28, which is larger than the divisor 26. Now, in ordinary division this 



SYMBOLS AND THEIR USES 107 

would indicate that the figure 6 in the answer was too small, but here 
it must be accepted, as 7 is too large. This peculiarity is due to the 
last figure in the answer at each step being placed in the corresponding 
divisor; thus we have 26 and 6, using the 6, but would have 27 and 7 
if the 7 were used. The twofold increase is sufficient to give a product 
larger than our dividend. 

When dealing with large numbers which end in several noughts, 
whose roots are whole numbers ending in noughts, care must be 
taken towards the end of the working that the decimal point is 
placed correctly. 

Example 93. Extract the square root of 7,290,000. 
7' 29' oo 1 oo (2 700 
4 



47 



329 /. \/ 7290000 = 2700 



329 



In this case the decimal point, although not seen, is on the extreme 
right of the number, hence mark off in pairs from the right. The first 
and second periods present no difficulty. The subtraction leaves no 
remainder and the figures in the answer up to this stage are 27. The 
third period is oo, hence a o is placed in the answer. The fourth period 
is also oo, hence another o is placed in the answer. This brings us to 
the end of the number, i. e. t to the decimal point (not placed) and the 
point comes after the 2700. The result, 2700, can be proved by actual 
squaring. 

The three following square roots should be memorised for future 
use ; and, as an exercise, the reader should verify their accuracy 

^2 = 1-414 >/3~= 1-732 J^ = 707. 

Exercises 24. On Square Root. 

Extract the square roots of the following numbers. (As these 
exercises from i to 17 are set as a test of the student's grasp of the 
method, they should be worked right out.) 

1. 11-56. 2. 240-25. 3. 4-41. 4. 841. 

5. 84-2724. 6. 110-25. 7. 1624-09. 8. 11881. 

9. -6084. 10. -0729. 11. -009025. 12. i -0201. 

13. -00000961. 14. 252-81. 15. 2490-01. 16. 102,400. 

17. 6,760,000. 

Work out the following square roots to 4 decimal places and state 
the answer to 3 : 

18. 22-7. 19. 2-61. 20. 2558. 21. 3682. 
22. 763-1. 23. 583-7. 24. -09075. 25. -1267. 



ro8 ARITHMETIC FOR ENGINEERS 

Evaluation, including Square Root. When square roots 
appear in a formula the algebraic method of expressing multiplication 
and division is adopted. 

Thus -i8\/D means -18 X Square Root of D, 
-~ means 850 -f- Square Root of P. 

When the square root of a single quantity only is required in 
an expression (as in the two cases shown), the root is first found 
and can then be combined with the other quantities. But when 

the root sign covers a complicated expression as in A/- L we 

have the rule : 

The value of any expression under a root sign must first be obtained 
before attempting to extract the root. 

When writing out an expression containing a square root, care 
should be taken that only the proper quantities are covered, as 
the inclusion or omission of any quantities will alter the value of 



the expression considerably. Thus taking ~ and - t the first 

means that a must be divided by b and the square root of the 
quotient then extracted ; but the second indicates that the square 
root of a must first be found, and then divided by b. 

Similarly, Va + b is by no means the same as Va + b, but 
carelessness in writing the sign might turn the first expression into 
the second. If possible, it would be better to write b + Va instead 
of Va + b. Also Va b (meaning *fa x b) is best written as 'b^fa, 
as the latter form, unlike the former, cannot be mistaken for Vab. 

Example 94. The permissible current in an electric cable may be 
calculated from Kennelly's rule, C = 560 *Jd 3 where C = current in 
amperes and d = diameter of conductor core in inches. Calculate the 
current which may be carried by a conductor having a core '16* dia. 



C = 560 V^ 3 



= 560 \/ -004096 

tas 560 x -064 =* 35-84 amps. 



i6 3 = -004096 
Extraction of * 



oo 1 40' 96(^064 



124 



36 

496 
496 



The value of -i6 8 is first found. Then the square root of 
the result is extracted as shown. Finally, the root is multiplied by 560. 



SYMBOLS AND THEIR USES 



109 



Example 95. The velocity of steam in the nozzle of a steam turbine 
is given by the formula V = \/2gJ(H 1 H 2 ). Find the velocity 
when HJ = 1194, H 2 = 1130, and J and g are constants, being 778 
and 32*2 respectively. 



V = V2 X 32*2 X 778(1194 1130) 

= ^64 -4 x 778 x 64 

= N/3, 207,000 

= 1791 ft. per sec. 



' 



349 



358 



3' 26' 70' 00(1790* 
I 

220 
189 

3I7 



2900 



Next figure would be 7 or 8. 
No need to work out. 

Example 96. The outside diameter of a hydraulic pipe is given by 
the formula 



when D = outside diameter in inches. 
d inside ,, ,, 

/ = safe stress in the metal, in Ibs. per sq. in. 
p = pressure of the water, 

Find the value of D when d = 4", / 2800 Ibs. per sq. in., and p 
750 Ibs. per sq. in. 



/ p 


i-73'oo' (1-315 
i 


/28oo + 75 


V 2800 750 


4 ^2<r 


23 


73 


4NAT 7 73 

4 X I-3I5 = 5-26* 


261 


400 
261 


? $ to the next largest J* 


2625 


13900. 



Exercises 24 (contd.). On Evaluation with Square Root. 

26. If an observer is situated with his eye h ft. above sea-level 
the distance of the apparent horizon in nautical miles is 
Calculate this value when h = 38 ft. 



no ARITHMETIC FOR ENGINEERS 

27. When calculating the size of steam pipe required for a steam 
engine the velocity of the steam (ft. per sec.) is given by the expression 

~^~ where P = pressure in Ibs. per sq. in. Calculate the velocity if 
p = 130 Ibs. per sq. in. 

28. The expression g- VF gives the diameter (ins.) of a steam engine 

piston rod where D = cylinder diameter and P = boiler pressure. 
Calculate the rod diameter when D = 33" and P = 150 Ibs. per sq. in. 

29. The Lanchester formula for petrol engine rating is f ^d 2 \/r where 
d = cylinder diameter in inches and v = stroke ratio. Find the rated 
H.P. of an engine when d = 3-5* and r = 1*285. 

30. Find the value of the expression A 'o6\/A, which relates 
to chimney stacks, when A = 12-6. 

31. In locomotive boilers the thickness of the firebox plates is 

/P 

calculated from the formula T = \/ i, where P = working pressure 

in Ibs. per sq. in. and T = thickness in sixteenths of an inch (i. e. t if T 
is 9, plate is &" thick). Calculate the thickness required for a boiler 
to work at 175 Ibs. per sq. in. 

32. The thickness_pf a steam engine cylinder may be calculated 

from the formula - + "015!^ when D cylinder diameter in ins. 

Calculate the thickness of a cylinder 33" dia. Give your actual result 
and also the result to the next ^ 6 ". 

33. Find the value of the expression Jd^, which relates to the 
inertia of links in machines, when d = 12 and d 2 = 21. 

34. If / is the length of a pendulum in ft. then the time of a double 
swing in seconds is 2ir\/ - where * = 3-14. Calculate the time 
when / = 3-26 ft. 

35. The permissible current in an electric conductor may be obtained 
from the formula 138^^ amps, where d = dia. of core in centimetres. 
Calculate the current allowable when d = -64 cms. 

36. The expression \// 2 r 2 refers to an engine governor. Find 
its value when / = 1-25 and r = -85. 

37. The expression B -f \/B 2 -+- T 2 occurs in shaft calculations. 
Calculate its value when B = 4-55 and T = 5*73. 

38. If a trolley wire / ft. long be hung between two poles L ft. apart, 
then the dip at the centre of the span in feet is -6i2\/L/ L 2 . Calcu- 
late tjie dip for a wire 120-1 ft. long hung between two standards 120 ft. 
apart. 

p /L S / 

39. The expression \J ^ relates to the movement of a 

2 1^ 2 

slide valve. Find its value when T = 5-5, L = i, S = -75 and / = '25. 

Other Roots. Besides square root we have to deal with 
Cube Root, Fourth Root, Fifth Root, etc., all of which involve an 
idea similar to that of square root. 



SYMBOLS AND THEIR USES m 

Thus, 2 is the cube root of 8 because 2x2x2 = 8, 

i. e. t because 8 is the cube of 2 ; 
3 is the cube root of 27 because 3 X 3 X 3 27. 
Similarly 3 is the fourth root of 81 because 3x3x3x3 = 81, 
i. e. t because 81 is the fourth power of 3. 

Thus, the cube root (or fourth or fifth root, etc.), of any number 
A, is such that when 3 (or 4 or 5) roots are multiplied together the 
result is A. 

In formulae the root sign V is used in conjunction with a 
small figure placed immediately outside, near the top, to denote 
whether the root is cube, fourth, etc. 

Thus 'v 7 signifies cube or third root 

tf fourth root 

t/~~ fifth root 

When writing these signs the little figure must be small and 
close to the root sign. If this is not done there is a possibility of it 
being taken for a coefficient. 

Then, ^125 = 5 because 5 x 5 X 5 = 125 

= ab because ab X ab x db = (ab}* = a 3 6 3 
= 10 because 10 x 10 x 10 x 10 10000 
= D because DxDxDxDxD = D 6 

Therefore %% = a , if a 3 = x 

i. e., a X a x a x 
and tfy n t if n 4 y 

i.e.,nxnxnxn y 

Roots beyond the fifth are not common in practical problems. 
It is possible to extract these " higher " roots by arithmetical 
methods somewhat similar to that followed for a square root, but 
the methods are so much more complicated and laborious that they 
are never used. 

Most engineering pocket-books give tables of square and cube 
roots for numbers i to 1000, in sufficient detail for practical work. 
Any root can be quickly and easily extracted by Logarithms. The 
actual method is dealt with in Chap. VI. Examples involving cube 
or other roots cannot be evaluated until the reader can use loga- 
rithms, and will be left until then. 



H2 ARITHMETIC FOR ENGINEERS 

Powers of Fractional Expressions. Consider the fraction 

-. If this is to be squared then it may be indicated in the form 

//\a /a 

( J . It is not correct to write , as the index only refers to the 

symbol against which it is placed. If the value of / be known 
the easier method of working is to find the value of - and then 

square it. But often the value of / is not known, and a different 

/A 2 
form is required from ( ) , to enable further working to be carried 

out - //v / / 

Now(-) - JXJ 

Carrying out the multiplication of these two vulgar fractions 
in the ordinary way, we should have for the numerator I x I = / 2 , 
and for the denominator 3x3 = 9. 



Therefore to square a fractional expression, the numerator is squared 
to give a new numerator and the denominator is squared to give a 
new denominator. 

A similar treatment can be applied to other powers. 

~ 
Thus- 

All this is contained in the following: 

To raise any fractional expression to any power, raise both numerator 
and denominator to that power, to give a new numerator and a new 
denominator. Then simplify where possible. 

Example 97. Remove the brackets from the following expressions 
and simplify where possible. 



(E \ a E* 
= J = pa (No simplification possible) 



SYMBOLS AND THEIR USES 113 

o~3 

Note here that in certain cases the form would be retained as 

27 

being useful for cancelling purposes. 



60 / 60* 3600 

Exercises 25. On Powers of Fractional Expressions. 

Remove the brackets from the following expressions, and simplify 
where possible : 

*.)' 3. (I)' 4. (5)' , 



7. (-) 8. IT) 9.11) 10- 

('? " 

/^ \a / v \ 3 

16. (|) 17. () 18. 

Roots of Fractional Expressions. With regard to roots, 
exactly similar laws apply. When it is required to denote the 

D 2 . /D 2 

square root of, say, this must be written as A/--, care being 

taken that the whole of the fraction comes under the root sign 

/D* D t D D D 2 

Now A/ = because X -- = 

\ 4 2 224 

Note that 2 = V^ and D = 



We might write /--, although this form has no value here, yet 
V 4 

in many cases it would be employed. 

Hence we have the law: 

To extract a root of a fractional expression find the root of both 
numerator and denominator to give a new numerator and a new 
denominator. Then simplify where possible. 



9 Vg 3 
Similar remarks to the foregoing apply to other roots. 



Tb^^.W-i 



Where a root sign covers a product, the expression may be 
written as the product of the roots of the factors. 

Thus V^bc = vVx Vb X Vc 

I 



ii4 ARITHMETIC FOR ENGINEERS 

This point is of use when some of the factors have an index 
the same as the root number. 



Thus ^fcfb = V^ X 

= a\/6 since yV = a 

The form a\/b is much simpler than \/a 2 b 

Similarly ^"278^ = -^27 X j/S~X % 
= 3 X ^S X r = 3/ 

This may also occur in fractional forms. 

Thus- ii^-VpS-P- 
lhus V 16 ~ ^/i6 ~ 2 



/D*S~"~ 
Example gS. The expression \J --- is given for the diameter of 

4.OOO 

the exhaust pipe of a steam engine, where D = dia. of cylinder and 
S =s piston speed. Reduce to a simpler form. 



x /4 6 3' 2 

The result can either be left in this form or, as multiplication is 
usually more direct than division, it can be slightly modified thus 



Example 99. The expression \ relates to the flow of water in 

pipes. In all cases g is a constant quantity and is 32. Reduce to 
a simpler form. 



x~~3? _ A /6? 

-V- 



Exercises 26. On Roots of Fractional Expressions. 

Jn each of the examples Nos. i' to 5 the expressions are to be 
simplified so far as possible. 

1. \/~|r~ which gives the speed in revs, per min. of an engine 
governor of height H ft. 

2. VfRT relating to the wheel base on a tramway track. 



SYMBOLS AND THEIR USES 115 

i d^p 

3. \/ "~r giving the steam cylinder diameter in a direct-acting 

v '75P 
stearn pump. 

/ D 2 P 

4. /Y/ which gives the diameter of a piston rod. 

5. *j'i2r relating to masonry arches. 

6. The velocity of a body falling h ft. is */2gh where g = a 
constant due to gravity = 32*2. Substituting this value, simplify 
the expression. 

7. In metric units the value of g in the above formula is 981. 
Reduce the formula by using this value. 

8. In connection with water flowing from orifices we have the 

expression \/~ Simplify this if g = 32. 

9. The diameter of a gas engine valve is ^J -y- where S piston 

speed, V = gas velocity through valve, and D = cylinder diameter. 
Taking S as 300 and V as 4000 reduce to a simpler form. 

10. The velocity of steam issuing from a turbine nozzle is 

Simplify this if J = 778 and g 32*2. 



11. If g = 32, simplify the expression *j2g x 2*3^ which refers 
to the velocity of water under pressure p. 

12. The formula \J --^~ refers to the measurement of voltage 

by an electrometer. TT and g are constants, TT being 3*14, while g is 
981. Substituting these values, reduce to a simpler form. 

13. The diameter of a firebox stay bolt is given by the expression 

v r 8? * n ^ ocomo ^ ve bil ers with copper stays / = 3500 and 
5 = 1 6. Substituting values, simplify the expression. 

14. The formula A/ - i refers to the thickness of 

firebox plates. For locomotive boilers S = 16. Substitute this value 
and simplify. 

Laws of Indices : Multiplication. When multiplying ex- 
pressions containing symbols, it will often be found that the same 
kind of quantities have to be multiplied together, as for example 
a 2 X a 3 . Now by the ordinary system they would appear as a 2 *? 3 ; 
but a shorter expression may be employed as shown below. 

a 2 = a X a and a 3 * = a X a X a 
Hence a 2 X a 3 means axaxaxaxa 
i. e. t 5 a's multiplied together. 

Now this may be written as a 5 

Hence a 2 X a 3 a* 



n6 ARITHMETIC FOR ENGINEERS 

Note that the 5 = 2 + 3, i. e. t the sum of the indices. Therefore the 
indices are not multiplied as might be expected for a multiplication, 
but are added. 

Thus a 2 X a 3 = a 2 * 3 = a 5 

A further proof may be obtained if figures be used. Thus, let 
a = 3, then a 2 = 9 and # 3 = 27. 

Then a 2 X a 3 = 9 x 27 = 243 
But if a = 3, a 5 = 3 5 = 243 

Similarly r* X ?* r z+ * = r 7 

Hence the rule : To multiply together two or more powers of the 
same quantity, write down the quantity and add the indices. 

This does not, of course, apply to cases of different quantities. 
Thus in a 2 X & 3 the indices cannot be added or interfered with in 
any way. The product a 2 X 6 3 can only be written as"fl 2 & 3 . 

When one of the quantities is of the first power, then it must 
be remembered that, although not written, its index is I. 

Thus p X p* = p*+* = p* 

Similarly x 2 x x = x 2+l = # 3 

and n x n = n 1+1 = n 2 

The general rule for multiplication of indices applies to any 
number of powers of the same quantity. 

Thus r 2 X r X r 3 = r* since the sum of the indices 2, 1, and 3 = 6. 

(Remember the index i for r). 
The addition of the indices may usually be done mentally. 

Example 100. Simplify the following expressions: 

(a) h x A 3 ; (b) x 2 x x* X x; (c) c X c*\ (d) io 3 x io 5 

(a) h X A = A 1 * 8 = / 

(6) X 2 X AT* X X = tf 2 *** 1 = x^ 

(c) c x c 4 = c x + = c^ 

(5) io 3 x io 5 = io 3 + 5 = u 

Example 101. Simplify the following: 

(a) r 8 x / X / 2 ; (6) 2 H X H/>; (c) a&* x a*c. 

(a) r 2 X / X / 2 = y 2 / 1 * 2 = r 2 /! 

(6) /> 2 H X H/> = 2 X /> X H x H = p z <-* X H 1 + 1 = /> 3 H 2 

(c) afc 2 X a 3 c -= a X a 3 x 6 2 X c = a*6 2 c 



SYMBOLS AND THEIR USES 117 

Exercises 27. On Laws of Indices : Multiplication. 

Multiply: 

1. a 3 by a 2 . 2. / by / 2 , 3. N 2 by N. 

4. d by d*. 5. r* by r. 6. a 4 by a 3 . 

7. H 5 by H. 8. 10 by 10*. 9. io 2 by io 5 . 
Simplify the following expressions: 

10. wr X w. 11. px x x. 12. b 2 X 6 X ft 2 . 

13. D x D 4 x D. 14. R 2 x R X R. 15. io x io 2 x io. 

16. io 3 x io X io 2 . 17. / 2 T X /. 18. RE 3 x R 2 E. 

19. irn X irtt 2 . 20. pd 2 x p X d 3 . 21. #* x #y x yv. 

Laws of Indices : Division. In a similar manner the division 
rf powers of the same quantity may be treated. 

6 5 
Thus, take-p. Now > 5 means 5 b's multiplied together, and & a 

neans 2 b's multiplied together. 

Hence g = A*.* X & X * X b 

b 2 b X b 

6 5 
Cancelling leaves only 3 b's on top, i. e., v 2 - = ^ 3 

But the index 3 is the result of taking the index 2 from the 
ndex 5, 

7,5 

- 



Hence we have the rule : To divide a power by another power of 
the same quantity, write the quantity and subtract the index of the 
livisor from the other index. 

Remember that the indices are not divided but are subtracted. 



Similarly = _ 

J p* p X p X r 

^> 4 
or ~ = /> 4 ' 3 = p l t written as p. 

Example 102. Simplify the following: 

H 3 io' , a 5 , A a 2 R 

(a) TP (6) 55" (C 



(a) H 3 - 2 = H^ = H (6) = I 08 ' 2 = 21 

-> X R- 

aR' 



nS ARITHMETIC FOR ENGINEERS 

Exercises 28. On Laws of Indices : Division. 

Simplify the following : 

*w4 S5 /2 T?3 

1 2 - a ~- 4 

*' w 3 S *' t R 2 

O. 5" 6. 7 i. ~ o O ~ ~~ 

io J ioc^ a 2 ?? y 



a Z2L- 10 >: 2 ' 2 n io'R I !*. 4 . 

y ' />d 1U ' rz0 11- io*K io 3 x io a 

Laws of Indices : Powers. Occasionally we require to find 
the value of an existing power raised to a power of itself, e. g. the 
square of a 3 or (a 3 ) 2 . Now just as x 2 = x X x so (a 3 ) 2 = a 3 X a 3 , 
the a 3 as a whole being like the x. 

Now a 3 X a 3 = a 3 * 3 = a 6 . Thus (a 3 ) 2 = 6 where, of course, 
6 = 3x2., Similarly (a 2 ) 4 = a 2 X a 2 X a 2 X 2 = a 8 ; and 8 = 2x4. 

The actual operation on the indices of the quantity in the brackets 
is addition, but at the same time it is essentially a multiplication. 
Hence the rule : To raise a power of a quantity to some power of itself 
write the quantity and multiply the indices. 

This must not be confounded with the rule for the adding of 
indices. 

Example 103. Simplify the following expressions: 

(a) (w 2 ) 3 ; (b) (io 4 ) 2 ; (c) (a 2 r 2 ) 3 ; (d) (np 2 ) 2 ; (e) (a 2 b*)* 

(a) (w 2 ) 3 i.e., n 2 X n 2 X n 2 = w 2x3 = * 

(b) (io*) 2 = io lx2 = icy; 

(c) (aV 2 ) 3 = a 2x3 r 2x3 = a e r e 

(d) (np 2 ) 2 = w 2xl /> 2X2 = n 2 /)* 



Exercises 29. On Laws of Indices : Powers. 

I. Find the 4th power of a 3 . 2. Find the 3rd power of a 3 . 
3. 3rd io 2 . 4. ,, ,, square of N*. 

5. ,, 5th ,, D 2 . 6. cube of R 4 . 
Simplify the following: 
7. (io 3 ) 2 8. (io 2 )' 9. (R 3 ) 3 10. (w 5 ) 2 

II. (r 2 ) 2 12. M 2 ) 3 13. (c 2 d)* 14. (c 2 d 2 )* 
15. (a 2 /? 3 ) 2 16. (ap 3 ) 3 17. (io 2 x io 3 ) 2 18. (m 3 />) 4 

Substitution of Symbols. It is often necessary to evaluate 
a formula when the given values are themselves symbols ; the result 
obtained is another formula satisfying certain conditions. Much of 



SYMBOLS AND THEIR USES 119 

this work requires a knowledge of the rules in the previous sections. 
Many of the expressions obtained will be of the vulgar fraction 
form, and cancelling is of great assistance. The method is best 
seen from the following examples. The student should, first of all, 
merely substitute. It is only with considerable practice that the 
processes of substituting and evaluating can be combined. 

Example 104. In connection with the bending of beams the 
expression fL/t occurs. Under certain conditions L has the value 

and h = --. Find by substituting and simplifying, the value of 
the given expression under these conditions. 



Given expression = L7t 



wl * substituting ~ for L 

and - for A 



Note. After substituting the new values and cancelling, the frac- 
tions can be multiplied up in the ordinary way. 

Example 105. The Marine Motor Association formula for the H.P. 
rating of a petrol engine is 3-25 A where A = area of the exhaust valve. 
Now A = -78552* where 5 = dia. of exhaust valve; and, on the 
average, 5 = -5^, where d = dia. of the motor cylinder. By substi- 
tuting first for A and then for 5 find the rated H.P. in terms of d, 
the dia. of the cylinder. 

H.P. = 

substituting -78552 for A 
-d for 5 




v 



Example 106. The expression - - relates to the inertia of moving 
engine parts. Find its value when v o>r.f 



Given expression = 



substituting &>r for D 



5 is a Greek small letter " delta/' 
w is a Greek small letter " omega." 



120 ARITHMETIC FOR ENGINEERS 

Example 107. Find the value of the expression - z when r = 

The given expression = ^ 7^x2 (writing ^ in place of r j 

\' 



^ ? _ 9 (inverting the divisor and 
$* ~ P cancelling by p) 

p 



Exercises 30. On Substitution of Symbols. 

Find the values of the following expressions, Nos. i to 8, when 

n* 
a = 2tt and x = . 

1. ax* 2. a*x 3. a z x* 4. -- 



9. If E is the voltage a ad C the current in an electric circuit, the 
power supplied is EC. Find another expression for the power : (a) when 

E = CR; (b) when C = ~ 

10. In connection with beams the expression ax 2 appears. In a 
particular case a = bh and x =- . Find the value of the given expres- 
sion under these conditions. 

11. Evaluate ax 2 for the case where a = |BH and x = f B. 

12. If v = 3*14, find the value of Try 2 when r ==--. 

vP 

13. Find the value of the expression PV when V = - -. 

14. The formula -^- relates to beams, b being the width and h the 

depth. Find its value for a beam whose depth is twice the width 
('.*., h = 2b). 

15. The expression^,- gives the thickness of a pipe under internal 
pressure. When used for water supply it is more convenient to replace 

>, by - and r by ; if / = 2800, convert the formula to give the 
r J< 2-3 2 

thickness in terms of H and d. 



16. Simplify the expression ? x when x | - . 



SYMBOLS AND THEIR USES 121 

17. The expression ~ is of importance in connection with con- 
tinuous girders. Find its value for the special case where S = - - 
and y = . 

' 2 

18. If Q is the charge of electricity supplied to an electric condenser 
and V = the potential, the energy = JQV. Find the value of this 

when V = jj, K being the capacity. 

7T 2 EI 

19. The formula -j^/r-' relates to long columns. A more useful form 
is obtained by substituting A 2 for I. Perform this substitution. 

20. The fraction was met with in a problem on beams, where 

A = and B = -y- . Substitute these values and simplify. 

21. The expression JIw 2 relates to fly-wheels. Find its value if 
W 2 __ 

22. The volume of a ball is $irR* where R = radius. Using 
diameter, R = . Substitute this value and simplify the formula if 
TT = 3-142. 

23. The expression is connected with beams. For a particular 

case I = ~ and y = . Substituting these values reduce the given 

expression. 

24. The volume of a ring is *Da where D = mean diameter and 
a area of cross-section. If the ring is of circular section wire, 

a = * d 2 if d is diameter of wire. If if = 3-14 find a formula for the 

4 
volume in terms of D and d. 

25. In connection with masonry structures we have the expression 

K 2 R 

-,-. For a certain case K = and d == R. Find the value of the 
d 2 

given expression under these conditions. 

26. Find the value of the expression in the last example for the 

, D , . D 
case where K = and d = . 
4 2 

27. Connected with the bending of a beam we have the expression 

~L For a certain case M = gB 2 H and A = JBH. Substitute these 

values and simplify. 

Pv 

28. A formula giving the deflection of a beam is ^;. Find the 

W/ 2 / 

value of this expression when P = g-, y = -, and H = El. 

Positive and Negative Quantities. The signs + and 
are also used in Algebra to denote "positive" and " negative" 



122 



ARITHMETIC FOR ENGINEERS 



Fveezmg 



quantities. Up to the present we have only considered quantities 
greater than o. While in the ordinary weights and measures of 
everyday life, quantities " less than nothing " do not occur, yet 
in engineering science they are often to be found. As an example 
let us consider the Centigrade or Continental Thermometer, which 
is used in all scientific measurements. The Zero or o on this is 
the temperature of freezing of water, and temperatures above this 
are numbered I, 2, 3, 4, ... and so on, as far as may be desired. 
But we often have to consider temperatures colder than that of 
freezing ; for example, the atmospheric temperature in winter, when 
the thermometer mercury will stand at some point below the zero, 
say 8 (read as " 8 degrees ") as indicated at A in Fig. 17. Now the 

temperature cannot be said to be 
" 8 degrees " since this is used to denote 
8 divisions above zero. We can say 
11 8 degrees of frost " or "8 degrees 
below zero/' but neither of these 
methods is sufficiently compact for use 
in calculation. But if we call the 
temperatures above zero, positive or plus 
quantities, and denote them by the 
addition sign +, then we may call the 
temperatures below zero, negative or 
minus quantities, and denote them by 
the subtraction sign . Thus a tem- 
perature " above zero/ 1 as at B in 
Fig. 17, is o + 8 degrees. The o itself 
not having any value, is omitted, the 
reading appearing as + 8 degrees (read 
as "plus 8 degrees 1 '). Similarly a temperature " below zero/' as 
at A, is o 8 degrees. As before, the o is omitted and the reading 
is called 8 degrees (read as " minus 8 degrees "). These forms are 
now quite suitable for calculation. Although not apparent when 
we write + 8 or 8, the meaning of add and subtract still holds 
good, the numbers really meaning + 8 and 8. Similarly if 
we consider changes of temperature, a rise may be called + and 
therefore a fall will be . 

Again, suppose we take the case of direct stress in a bar, which 
may be tension or the effect of pulling apart ; or compression which 
is the effect of pushing together. The two forms are quite opposite 
in character, as a bent bar subjected to a tension will straighten, 
but if subjected to a compression will become more bent. Now if, 








\^^\ 




1 1 







1 1 









5 




. 


G. 








FlG 



SYMBOLS AND THEIR USES 123 

when measuring stresses, we call the tension +, the compression 
must be called . Supposing now that we have to find the stress 
in some part of a machine or structure, and we cannot say from 
an inspection what kind of stress exists there. If the answer to 
our calculation comes out +, we know it is a tension; if it is 
a compression. 

Further, positive or + quantities may be regarded as " income/ 1 
and negative or quantities as " expenditure." Many other 
examples may be found : thus with alternating electric currents 
the voltage is + when in one direction, and when in the other : 
if the extension of a helical spring is +, the compression is : if 
the height of a mountain is +, taking the sea-level as the zero of 
heights, then the depth of the sea is . 

It is customary when a positive number or letter stands alone, 
or is the first thing met with in an expression, to omit the + sign. 
Thus 5 means -f- 5 and x means + x. Similarly with a compound 

wl* , wl* r ,. ., ,, , 

expression, -~- means ~| ~. In the middle of any expression 

the + sign must appear. The minus sign, however, must always 
be included, no matter what the position of the quantity. 

Addition of + and Quantities. This may be illustrated 
by reference to changes of temperature on a thermometer, with 
the aid of Fig. 18. Actual readings of temperature will be 

+ when above the zero 
and .*. below 

Considering changes of temperature (i. e., distances through 
which the mercury moves) 

a Rise of temperature will be + 
and .'. a Fall ,, ,, 

Now supposing the temperature is 8 and it rises 5 (i. e., changes 
+ 5), then obviously the final temperature is 13. Thus we start 
at 8, change 5, and finish at 13. Then, 

Start + Change = Final, 
a statement true for all temperatures and for rises as well as falls. 

I. Addition of Quantities with the Same Sign. 
(i) Both quantities +. This has already been taken as 8 + 5 

= 13. 

(ii) Both quantities . Let a temperature be 2 (i.e., 2 
below zero) and let it fall 6 (i. e. t the change is 6). From a, 



I2 4 



ARITHMETIC FOR ENGINEERS 



Fig. 18, it can be seen that the final temperature is 8 below zero, 
i. e. t 8. Then, since 

Start + Change = Final 
- 2 + (- 6) = - 8 

Note that in both (i) and (ii) the result is the sum of the actual 
figures, and the sign of the result is the same as the sign of the 
numbers. 




3 


L 


<5tarKrtj 
Tmf>.-Ha 


- 


1 A 


i 


10 
5 


Chang e 
-"7 


: 


iu 


- 




Then Final- 


^_ 




- 




Tcwp. is 5* 


- 




tmmm 







- 


o 





5 







5 


--10 

V A 




-Ito 




ie +(-?) s 

FIG. 18. Illustrating the Addition of + and Quantities. 

2. Addition of Quantities with Unlike Signs. 

(i) Let the temperature be 12, and let it fall 7 (i. e., the change 
is 7) ; then it can be seen from b, Fig. 18, that the final tempera- 
ture is 5. Then, since 

Start + Change = Final 
12 + (- 7) = 5 

(ii) Let the temperature be 10 and let it fall 16 (i. e., the change 



SYMBOLS AND THEIR USES 



125 



is 16). Then from % Fig. 18, it can be seen that the final 
temperature is 6 below zero, i. e. t 6. Then, since 

Start + Change = Final 
10 + ( 16) = -6 

Note that in both (i) and (ii) the result 
is the difference of the actual figures, e. g. t 
in (i), 5 = 12 7; and that the sign of 
the result is the sign of the larger number, 
e. g., in (i) sign of result 5 is the sign of 
the 12. Hence we have the rules : 

To add two quantities 

1. When the signs are the same, add the 
numbers, and write the sign in front of the 
sum. 

2. When the signs are different, take 
the difference of the numbers, and write the 
sign of the larger number in front of the 
difference. 



in 

. ^ 


Staffing . 


"""^r 


Y^ 

t f\ 




IV 


1 


r 


- 


\\) 





5 




Change 





5 


- 






-16 


. 




MB 


O 









O 





5 


1 




JB- 


5 


_ 


ir* 


Then Finals 


i 

m\/\ 



Example 108. Find the value of the 
expression p -j- q for the following cases : 
(a) p = 10, q = 5 ; (b) p = 4, q = 2 ; 
(c) p = - 7, q = - 3 ; (d) p = 3, 9 = 5 ; 
(e) p = 6, q = - 15. 








-6 



FIG. iSc. 



P + 
p + q 



5 = 



4 + (- 



15 



(c) 



= -7+ (-3) = 



+ 2 . 



10 . 



+2 



(e) p + q = 6 + (- 15) = -9- 



Ordinary case. 

Signs are different, and thus the 
difference of 4 and 2 is found. The 
sign is that of the larger number (4) . 

Signs are the same, so that the 
numbers are added, and the same 
sign ( ) given to the sum. 

Signs are different, therefore the 
difference of the numbers is taken, 
and the sign of the larger number 
(7) is prefixed. 

Signs are different, therefore the 
difference is found, and the sign is 
that of the larger number. 



The whole idea throughout this section has been addition, al- 
though in some cases subtraction has been performed between the 
numbers. This is due to the signs of the quantities, and although 



126 ARITHMETIC FOR ENGINEERS 

not used in ordinary arithmetic, is very important in scientific work. 
The result of an " addition " carried out by this consideration of the 
signs or nature of the quantities is called an Algebraic Sum, and 
does not necessarily produce an increase in the size of the quan- 
tities. An arithmetical sum is simply an addition of the figures, 
disregarding the signs. 

Example 109. (a) Add together $ab, 6ab, i^ab, and ab. (b) Prove 
the result to be correct when a = 2 and 6 = 3. 

(a) (b) Proof 

6a6 = 6x2x3 = 36 
= 13 X 2 X 3 = 78 
a& = 1X2x3= 6 



Sum = 25^6 150 

25 ab = 25 X 2 X 3 = 150. 

Notes. The beginner had better write the terms in a vertical column, 
as indicated; but with a little practice this may be dispensed with. 
Th? letters ab may be regarded as a unit like feet, etc., and thus the unit 
of the answer will be the same as that of each term, i. e. t ab. (Note 
that ab is really i ab.) Proof. Substitute the value of a X 6 in each 
term and evaluate, giving 150 as the sum of the quantities. Substitut- 
ing the values of a and b in 25 ab we obtain 150, thus proving the 
algebraic addition to be correct. 

With positive and negative terms it is usually advisable to sort 
out all the positives and all the negatives, add separately, and 
finally make an addition of the positive and negative results. 

Example no. (a) Add together $r z f, r 2 f, 9^ 2 /, zr 2 f, zr 2 /. 
and r 2 /. 

(b) Prove the result correct when r = 2 and / = i. 

Positive Quantities. Negative Quantities. 



8r 2 f = Sum of + Quantities I2r 2 f 

I2r 2 f= 

Result of Addition. 



Note. Adding 8f 2 /and I2r 2 f gives 4? 2 /by Rule 2, p. 125. 



SYMBOLS AND THEIR USES 127 

Proof. The sum of the 4- and quantities need not be ques- 
tioned, the method having been proved in the previous example. 

8r*f = 8 x 2 2 X i = 32 

I2r 2 f = (12 x 2 2 X i) = 48 

Sum = 16 

Now 4? 2 / = (4 X 2 2 X i) = 16, thus proving the foregoing result. 
This addition and subtraction of positive and negative numbers 

is very important in connection with calculation by Logarithms 

(Chap. VI). 

Exercises 31. On Addition of + and Quantities. 

Find the value of the expression T -f- ^ m the following cases : 

1. T = 5, * = 3- 2. T = 7, t = - 4. 3. T = - 9. < = - 6. 

4. T = 14, t = 4. 5. T = 2, t = 19. 
Find the value of the expression / -\- m in Exs. Nos. 6 to 10. 

6. / = -7, m = 1-2. 7.1= -4, m = -6. 8. I = 1-9, m = -3. 

9. / = -3, m = -8. 10. / -15, m = 1-4. 
11. Add 4- -3570 to 4- -5160. 12. Add 4- -2180 to -3560. 
13. ,, "2250 
15. ,, -0050 
17. 4- i 
19. - 1-505 
Add together : 

21. 3, - 5, - 7, 4- i, - 2, 4- 12. 22. - 5, 4- 6, 4- 2. 
23. 2, 7, i. 24. 4, 4- 6, - 5. 

25. - 3, - 2, 4- 5- 26. - i, 4- i, + 7- 

27. 7, - 2, ~ 5, 4- i. 28. - 3, - i, + 2, - 6. 

29. 14, 7, 6, 4- 3 7- 30. i, 4- 2, 4- 6, 5, 9. 

Complete the following additions : - 
31. dab 32. x 2 y 33. gr z 34. lApm 



-7910. 


14. 


- -7750 


4- -1050. 


4- 1-05. 


16. 


,, 2 


.. + -305- 


-046. 


18. 


,,4-3 


,, 2-05. 


-61. 


20. 


,,4-2 


,. - 1-6 5 . 



Add together the following: 
35. 4a*c, 3V, 2 c, -f 2a 2 c. 
35, _ ^t;/, 3^/, + wl, -j- 14^, 
37,, - 7/> 2 > - 3^ 2 , P 2 , 7P 2 , p 2 - 

38. + 5<2, 3^, d, 6d, + 2^. 

39. 2-5M, 3w, i5W, 7'5n. 

40. 2-8u>, $w t 5-75^, 3*22ie; 4- 



Subtraction of -f and Quantities. This we will also 
examine by considering changes of temperature, with the aid of 
Fig. 19. In this case, knowing the starting and final temperatures, 



10 



Starttng 
Temp. 1 



Final Temp,, 



Then 
change is-9 



ic (a 



3-(j-ie)--9 

rar 



IO 




Final "Tcrnf 





Temp 5 

Then Hie 
10 change is 17 





10 





Fig. 19. Illustrating the Subtraction of -f and Quantities. 

128 



SYMBOLS AND THEIR USES 129 

we shall find the size and nature of the change. The ordinary 
arithmetical subtraction is a case of both quantities being +, and 
in all cases the quantity to be subtracted is less than the other. 
Algebraically these conditions need not exist. 

Let the temperature on our thermometer be 7, and suppose 
it rises to 18, what is the change in temperature ? Evidently the 
temperature has risen 11 ; then the change algebraically is + n. 

Thus, 18 7 = n, or in words, 
Final Start = Change, 

a statement true for all temperatures, no matter whether they rise 
or fall. 

I. Subtraction of a + Quantity. 

Consider the case where the first number is smaller than the 
second. 

(i) Let the temperature be 12, and let it change to 3 (final 
temperature). From a, Fig. 19, it is seen that the change is a fall 
of p, i. e. t 9. Then since 

Final Start = Change 
3 - (+ 12) = - 9 

The + sign may be omitted before the 12 ; hence the above state- 
ment may be written as 3 12 = 9, and when the idea of oppo- 
site nature or opposite direction is considered the statement is 
evidently true, since to pass from 12 to 3 the mercury must fall 
p. To perform the operation algebraically we notice that the 
result 9 is also the result of 3 + ( 12), reversing the two signs 
before the 12. Evidently 3 (+ 12) == 3 + ( 12), from which 
we conclude that " the subtraction of + 12 is equivalent to the addi- 
tion of 12." We will examine this for another case. 

(ii) Let the temperature be 8 above zero, and let it change to 
6 below zero. Then from b t Fig. 19, the change is a fall of 14 
(i. e. t 14). Then as- 
Final Start = Change 
- 6 - (+ 8) = - 14 

But 14 is also the result of 6 + ( 8), or tfie subtraction of 
+ 8 is equivalent to the addition of 8. 

Both cases show us that the subtraction of a + quantity is equivalent 
to the addition of a quantity. 

Denoting the first number by a and the second by b we can write : 
a - (+ b) = a + (- b) = a - b 

K 



130 



ARITHMETIC FOR ENGINEERS 



2. Subtraction of a Quantity. 

(i) Let our temperature be 5 below zero and let it change to 
12 above zero. Then from c, Fig. 19, the change is evidently a 
rise 0/77 (or change = + 17). Then as 

Final Start = Change 
12 - (- 5) = + 17 

Notice that + 17 is also the result of 12 + 5, which we may write 
as I2 -f (-f- 5). Then 12 ( 5) = 12 + (-f 5) whence we conclude 

that the subtraction of 5 is equivalent to 
adding -f 5. The truth of this is evident 
from c, Fig. 19 for certainly the mercury 
has risen 77. 

As a further illustration : If we call a 
" possible event " a + event, then an 
" impossible event " must be a event. 
Now suppose we say that an event is 
(< not impossible/ 1 which corresponds to 
( ), then we mean that it is possible, 
i. e. t it is +. 

We will examine the above for two 
other cases. 

(ii) Let the temperature be 8 below 
zero, and let it become 5 below zero. 
Then from d, Fig. 19, the change is 
evidently 3 rise. Then as 

Final Start = Change 
- 5 - (- 8) = +3 




-3 



Fig. ige. 



But + 3 is also the result of 5 + 8, 
i.e., =5 + (+ 8); hence, as before, the 
subtraction of 8 is equivalent to 
adding + 8. The " addition " is, of course, algebraic. 

(iii) Let the temperature be 5, and let it change to 8- 
Then the change should be 8 ( 5) = 8 -f- (+ 5) = 3, 
or a fall of 3. From e, Fig. 19, this is evidently true. Hence we 
conclude that the subtraction of a quantity is equivalent to the 
addition of a + quantity, or, if a is the first number and b the one 
to be subtracted 



(- b) = a + ( + b) - a + b 



SYMBOLS AND THEIR USES 



In both our conclusions we notice that a change of operation (i. e. t 
addition replacing subtraction) accompanies a change of sign. 
Hence we have the rule : To take one quantity from another, change 
the sign of the quantity to be subtracted, and proceed to add algebraically. 



Example in. (a) Take $wl 2 from 8ze// 2 ; (b) ?wl* from 
(c) 3*>/ 2 from ^wl z ; (d) 6wl 2 from gwl 2 . 



(a) 



8wl* 



Difference = 



(b) 



jwl* 



Difference = 



Difference 





Difference = 



Change sign of $wl* to and add. 
Sum of 8 and 5 is 3. 

The " unit " wl* remains. 

Change sign of ywl 2 to and add. 
Sum of 2 and 7 is 5. 



Change sign of 
Sum of 4 and 3 is 7. 



+ and add. 



Change sign of 6wl* to + an d add. 
Sum of 9 and + 6 = 3. 



Exercises 32. On Subtraction of + and Quantities. 

Find the value of the expression W w in the following cases : 



1. W = n, 0; = 5. 
4. W= i, w = 10. 

7. w = 2,w = -f 12. 



2. W = 4, w == 7. 

5. W = 3, w 2. 
8. W = 4, w = *- 3. 



3. W = 7, w = 3. 
6. W = 15, w = -}- 3. 
9.W = 3,; = 3. 



10. 


Subtract 1-5 


from 3-7. 


11. 


Subtract 1*2 from 


- 2-4. 


12. 


. -'7 


1-6. 


13. 


i'9 


4-6. 


14. 


2-6 


"3- 


15. 


- 2-18 


- 1-28. 


16. 


tt -3-13 


.. 3-59- 


17. 


2-29 


1-16. 


18. 


5'35 


-1-56. 


19. 


358 


~ *759. 


20. 


7'2 


5*5- 


21. 


- - 1*355 >, 


2-746. 


22. 


2-38 


tt 2-05. 


23. 


x *357 


3-359- 


24. 


3 


-2-155. 


25. 


2-35 


5- 


26. 


ywl 


i^wL 


27. 


I9^/" 


7 rf. 


28. 


,, i2ad 


2 i$ad*. 


29. 


_ c 2 p 


3C 2 ^>. 


30. 


**y 


tt ~ xy. 


31. 


5'5*y tt 


4*25#y 



132 ARITHMETIC FOR ENGINEERS 

Addition and Subtraction of Several Terms. 

The addition and subtraction of expressions containing several 
terms is subject to the same general rule that applies to vulgar and 
decimal fractions, and in fact to everything: i. e. t that only things 
of the same name or kind may be directly added together. Thus, 
in adding the expressions a 3 2a 2 b + ab 2 and a 2 b + zab 2 fc 3 
care must be exercised in distinguishing the terms containing a 2 b 
from those containing ab 2 . The distinction between these has 
already been given on p. 96. 

In the earlier examples, as with decimals, the expressions may 
be written in separate lines, separate columns being arranged for 
terms of the same nature ; but, unlike decimals, each column is 
complete in itself, there being no carrying to think of. Thus we 
shall have a column for a 3 , another for a 2 b, another for & 3 , and so on. 
It should be remembered that the order of the terms in any expres- 
sion may be altered at will, provided that the correct signs be retained : 
thus a 3 2a 2 b + ab 2 is equivalent to 2a 2 b + ab 2 + a 3 , etc. 

Example 112. Add together the expressions a 3 2a 2 b + ab 2 , and 
- a*b + tab* - b 3 . 

a 8 2a 2 b -f ab 2 

- a*b + tab* b* 

Sum = a 3 sa 2 6 + $ab 2 b 3 
W W (3) U) 

Explanation. Write the first expression down (the order given will 
usually be suitable), keeping the terms well spaced. Write the terms of 
the second expression beneath this, placing them beneath those of the 
same kind in the first line. Thus a 2 b in the second expression is written 
beneath the second term in the top line 2a 2 6; also + 2a& 2 is placed 
beneath -f- ab 2 . The term b 3 must make a new column, there being 
no similar term in the top line. When all are arranged the addition 
may be proceeded with. It must be " algebraic," i. e., regarding sign. 

Column (i) a* is the only term : write a 3 in lower line. 

(2) 2a 2 b and ia 2 b, added give $a 2 b. 

(3) + iab* and + zab 2 , added give + $a b 2 . 

(4) 6 s is the only term : write b 3 . 
/. Final result is a a 3a 2 6 + 3a6 2 6 3 . 

When several terms, including + and signs, have to be added 
in a column, the addition may be performed continuously after a 
little practice, mentally retaining the sum from step to step with- 
out separating the + from the quantities. 



SYMBOLS AND THEIR USES 133 

Example 113. Add together the expressions 3/> 2 <? + 5<7 2 P> 
3P 2q*, 2pq + 2p*q* p, 6p*q -f 2^ 7? 2 4- 3/>. 

+ 5? a - P 
2? a -f 3/> 

p 2pq + 2/>V 
- 73* + 3P + 2 P<1 



Sum 



Explanation. The first expression is written down as given. With 
the other expressions, like terms are placed beneath each other as shown ; 
and if there is no like term, a new column is made. 

Column of p 2 q : 6 and -f 3 = 3 /. write $p*q in result. 

,. q* : - 7> - 2 and + 5 = - 4 /. - ^ 

P : + 3 + 3 i i = 4- 4 ' + 4 > . 

,, pq :-f 2 and 2 = .'. ,, nothing ,, 

,> PV - only -}- 2^> 2 g 2 exists, .*. write -j- 2p 2 q z in result. 

.'. Result of addition = 3p 2 q 4<7 2 + 4^ + 2/> 2 ^ 2 or any other 
arrangement of these terms with their proper signs. 

Example 114. From 45 pr take 35 + 2pr 37. 
45- pr 

35 -f 2/>X 3? 



Difference => s 3^7 + 



Explanation. The expressions are arranged in columns of s, pr, 
and r. 

Column of s : 45 3$ = is written as s. 
,, ,, pr : change sign of 2pr and add, 
i.e., pr 2pr = $pr. 
tt ,, r : change sign of 37 and add, 

i. e. t o -}- 37 = -\- 3 r - 
/. Result of subtraction s $pr + 3?. 

At a later stage, when dealing with the removal of brackets, 
it will be necessary to perform addition without arranging the terms 
in columns. Then the various like terms are picked out by looking 
along the expression. The following example will illustrate : 

Example 115. Simplify the expression 

3a 2 10 -f 4a 2 -f 10 2a 2 




134 ARITHMETIC FOR ENGINEERS 

Start from one end, say the left, and take all the terms with a* and 
add mentally. Thus, 30* and 40* = jo 2 , ja* and 20? = $a 2 , which 
completes the terms in a 2 . Then take all the terms of another kind, 
here numbers only, and repeat : 10 and -f 10 = o. Then-f- 3 alone 
remains. 

When a term has been added in, it is advisable to tick it off 
before proceeding to another. This serves to show what terms 
have been dealt with, and indicates when all terms are included. 

This method is sometimes known as " collecting-up like terms/' 

Example 116. Collect up like terms in the expression 750 3/ + 
5 /a _ t + 24 _ 3.5/2 + 4 . 7/< 

V V v/ V V V V 

75 - 3* + 5* - < + 24 ~ 3-5' 1 + 4-7* = 774 + '7< + 1-5**- 

Taking the numbers first, 750 + 24 = 774. Next taking terms in 
t : 3* and t = 4*, then 4* and -f 4-7* = -f- "jt. Now taking 
terms in t 2 (quite distinct from those in t) : -f- 5^ and 3-5^ = -f i-$t 2 . 

Exercises 33. On Addition and Subtraction of 
Several Terms. 

(The first few examples are arranged in columns to accustom the 
student to the kind of example before setting out his own work.) 

Find the sums of the following : 

1. 2 a 3pq + q* 2. ar 2 ar r* 

+ pq 2q 2 zar 2 -f ^ar 



3. / -f w 2 l* 4. 3m 3 2m 2 -f m 

yv -f 3/ 2W 2 1 2 m* ^m 

410 2m 3 -f m 2 -f- 2m 



5. Add a 8 -f a 2 fc + ab* to afe - ab* - Z? 3 . 

6. Add p 3 - p 2 q + #>g 2 to 2 ? - ^ 2 + q 3 . 

7. Add together the following expressions : 

#* -f 3^ 5 y 4- y 2 J ~ 2A; 3 y 3 + y 2 ; ^ 4 $x*y 3^ ; 3^* x 

8. Add together : 

4W/ 17 + /*; 3/ 8 + 4 ; w 2 -f- I 2 w/; 3w 8 5. 
Collect up like terms in the following 

9. 3? 2 10 -f 12 + 2f 2 r 2 . 

10. ii J 1 + 2< 15 ii* 1 I. 

11. -75/ + 3^ i '3^ ' y - 

12. 2-75y 2 + 1-457 37* 27 + 7. 

13. w 1 n 5 -f 17 3n -f 4*1 -f- 6n f . 



SYMBOLS AND THEIR USES 135 

14. \a + a - Ja + d. 

15. c 2 - \c - ic 2 + it*. 

16. g* 2 - 10 + J* 2 + AT. 

17. t* - x + 3 * 2 - <zx - Id. 

18. in + - + \n + * 

Add together the following expressions : 

19. bd 2 2b 2 d 6 -f 3<2; $b 2 d -f 36 2<Z; ibd* 76 d\ 
5 b 2 d + 2b. 

20. 7* - 3 

-f 25 + 6/t. 

Multiplication and Division of + and Quantities. This 
may be investigated by numbers and then applied to symbols. 

i. Multiplication of two Plus Quantities, e. g., (-f- 3) X (+ 7). 

This is the common case already known and used, and written 
as 3 x 7- 

Now 3X7 means 3 sevens added together. 

For use in other cases we will consider that these 3 sevens are 
added to o, thus o +7 + 7 + 7 = + 21 

Hence (+ 3) X (+ 7) = + 21 

Similarly with letters (+ a) X (+6) means a times + b, or + b 
added a times to o, 

i.*., 0+& + & + 6 + 6+ . . . until the letter b is written a times 
= db. 

2. Multiplication of a Positive and a Negative Quantity. 
*. g-, (+ 3) X (- 7) 

Now in a similar manner to i, (+ 3) x ( 7) means 3 times 
7, or 7 added 3 times to o, i. e., o + ( 7) + ( 7) + ( 7) 

= 7 7 7 (See case (ii), p. 123.) 

= 21 

Hence (+ 3) X (- 7) = - 21. 

Similarly with letters, (+ a) x ( b) means b added a times to o, 
i. e. t o + ( 6) 4- ( b) + ( b) -f .... until the quantity b is 
added a times to 0, 

= ft b b . , . until the number of " b's " is a 
= - ab. 



136 ARITHMETIC FOR ENGINEERS 

3. Multiplication of Two Negative Quantities. 

* g; (~ 3) X (- 7). 

Now if (+ 3) X ( 7) means 7 added 3 times to o, then ( 3) 
X ( 7) must mean 7 subtracted 3 times from o, since the sign 
of + and have directly opposite meanings. 

Thus (- 3 ) x (- 7) = o - (- 7) - (- 7) - ( - 7) 

= + 7 + 7 + 7 ( b Y ca se (in), p. 130) 

= +21 

Hence (- 3) x (- 7) = + 21. 

Similarly with symbols : ( a) x ( 6) means b subtracted 
a times from o. 

That is, o ( 6) (- 6) ( b) . . . . until the symbol ( b) is 
written " a " times. 

== _!-&-t-&-|-&_{-&-f- ... until the number of " b's " is a 

= ab. 

Collecting our results we have the following : 

From i : (+ 3) X (+ 7) = + 21 or (+ a) X (+ b) = + ab 
(- 3) X ( 7) = + 21 or (- a) X ( b) = + ab 
(+ 3) X (- 7) = - 21 or (+ a) x (- b) = - ab 

whence it is seen that the same signs (i. e., two positives or two 
negatives) give +, and that two different signs (i. e. t a + an d a ) 
give ; or, as usually given by the " Rule of Signs " : 

Like signs give plus. 
Unlike signs give minus. 

This rule is equally true for division, as may be seen from the 
following : 

(i) The division of a positive quantity by a positive quantity 
needs no explanation, being the ordinary arithmetical case. This 
is a question of " like " signs. 

(ii) Division with unlike signs. 

From case 2, p. 135, we have the statement (-f- 3) X ( 7) = 
21. Now if we divide 21 by + 3, the result must be 7 to 
agree with our multiplication. 

21 Minus ,,. 

... __ aa _ 7f0r _ H _ =Mino8 . 



SYMBOLS AND THEIR USES 137 

From case 3, p. 136, we have the statement ( 3) X (7) = + 21. 
Now dividing +21 by 3, the result must be 7 to agree with 
the multiplication. 

+ 21 Plus ,,. 
.'. = 7, or .r-p = Minus. 

3 ' Minus 

Evidently then unlike signs, no matter in what order, give minus. 

(iii) Division with two minus signs, i. e. t a case of " like " signs. 

From case 2, p. 135, we have as before (+3) X ( 7) = 21. 
Now if we divide the 21 by the 7, then our result must be 
+ 3 to ensure agreement with the multiplication. 

21 , Minus 

.". = + 3, or ^ = Plus. 

7 Minus 

Now combining (i) and (iii), since the result is of the same sign, we 
have as for multiplication 

Like signs give plus. 
Unlike signs give minus. 

When more than two quantities, with varying signs, have to 
be multiplied together the product of them all is found in the usual 
way, after which a consideration of the signs will readily show the 
sign of the final product. Thus, in the case of 6X5X 3x4, 
the first sign is and the next -(-, giving ; then this mul- 
tiplied by the belonging to the 3 gives a + ; and finally this + 
multiplied by the + of the 4 gives + as result; thus the final 
product is + 360. 

A similar effect to the Rule of Signs is obtained from the action 
of two suspended bar magnets. Thus, two unlike poles, i. e., a 
North and a South, will always attract each other; but two North 
poles brought together will have the same effect as two South poles 
brought together, for they will repel each other. This action is 
usually summed up in the phrase : Unlike poles attract and like 
poles repel. 

Exercises 34. On Multiplication and Division of + 
and Quantities. 

Find the value of the expression aR for the following cases : 
1. a - 4, R - 5. 2. a = - 4 , R = - 5. 

3. a = 4, R = 5. 4. a == 4, R = 5. 

5. a = 3-5, R = '6. 6. a = 4-9, R = i-i. 

7. a = - 2, R = 1-563. 8. a = - 1-5, R = - -65 



138 ARITHMETIC FOR ENGINEERS 

Find the value of the expression alx for the following cases : 
9. a =s 2, / = 3, x = 4. 10. a = - 2, / = 3, x 4. 

11. a = 2, / =3 3, 4? = 4. 12. a = 2, / = 2, # = 4. 

13. a = i, / = 2, # = 5. 14. a = 1-5, / == 3, x = i'5- 

15. a = 1-5, /= 6, # = i. 

Find the value of the expression - for the following conditions : 

16. p = 10, q = 2-5. 17. p = 10, = 2-5. 
18. /> = 10, q = 2*5. 19. = 10, q ~ 2-5. 
20. p = - 7'5. ? = 3- 21. p = - 7-5, ? - - 4 . 

Powers of Minus Quantities. We may now consider the 
value of expressions such as ( 3) 2 , ( r) 3 , i. e. t the powers of 
negative quantities. A power applied to a minus quantity has the 
same meaning as when applied to a plus quantity, thus ( 3) 2 
means ( 3) X ( 3). Its value is, then, by the preceding 
paragraph, + 9. 

... (- 3) 2 = + 9. 

Also, (- 3 ) 3 = (- 3) X (- 3) X (- 3) 

(+ 9) X (- 3) . 
= 27 by preceding paragraph 

.-. ( 3) 3 = - 27. 

Again, (- 3 ) 4 - (- 3) 3 + l = (- 3) 3 X (- 3) 
= (_ 27 ) x (- 3) 
= +8i 
/. (- 3) 4 = + 81. 

Continuing in this way we should find that all the even powers 
(i. e. " squares," 4th powers, 6th powers, etc.) are positive, while 
all the odd powers (i. e. t " cubes," 5th powers, etc.) are negative. 

This is equally true with symbols. 

Thus (- r}* = (- r) X (- r) = + r 2 . 

It must be remembered that ( r) 2 is not the same thing as 
- r 2 . The first means the " square of r" and the second, 
" the square of r." 

Thus ( r) 2 means ( r) X ( r) = + r 2 , whilst r 2 means 
(r X r), or just the subtraction of r 2 . 

Example 117. Find the value of x* for the following cases: 
()*=, 6; (&)#= 3 r; (c) x = - 2r*. 

(a) x* = (- 6) 2 = (-6) X (- 6) = + 36 

(b) x* = (- sr) 2 = (- y) X (- 3f) = + gr* 

(c) x* = (- 2f 2 ) 2 = (~ 2r 2 ) X ( 2f 2 ) = + 4 r 4 



SYMBOLS AND THEIR USES 139 

Exercises 35. Powers of Quantities. 

If n = 2 find the values of the following expressions : 

1. n a . 2. n s . 3. n 4 . 4. n a . 5. n 8 . 

6. _ n i 7. n s. g. w 4 . 9. w 5 . 10. 2w 8 . 

Find the value of x* in the following cases : 
11. x = 4. 12. # = 4. 13. # == 47. 14. # = 4f. 

Evaluation, including Positive and Negative Quantities. 

We will here take a few actual examples where positive and 
negative quantities are used. 

Example 118. The width of the expansion valve in a Meyer valve 
gear is given by the expression a -f S R where a, S and R represent 
certain sizes of the gear. Find the width (inches) for the following 
cases : 

(!) a = 1-7, S = 3-27, R = -5; (2) a = 1-5, S =* 3, R = 75. 

(1) a + S - R = 1-7 + 3-27 - -5 = 4-47* 

(2) a + S-R = i- 5 + 3- (- 75) 

= i'5 + 3 + '75 = 5*25* 

Example 119. The latent heat of Ammonia as used in refrigeration 
is 566 *8/, where t = temperature F. Calculate the latent heat when 
the temperature is 10 F. 

566 -St = 566 -8 X 10 

= 566 - (- 8) = 566 + 8 - 574 

Note. -82 is to be subtracted, and since t is 10, then -8 of t is 
8 of 10 or 8. Now subtracting 8 is adding + 8, hence the 
result. 

Example 120. In connection with an apparatus for drawing large 
circular arcs we have the expression r -^ft--^- - Find the value of 
this expression when r = 5, b = 4*5, and R = 25. 

) = 5(5 + 4-5 + (~ 



_ = 
R - r - 25 - 5 

= 5 X{- 1515}^ -J77-5 _ 8 
- 30 - 30 i. 

Note. After substituting, the expression 5 + 4-5 + ( 25) = 9-5 -f 
( 25), or, 9*5 25 = I 5*5> Finally the division with like signs 
gives a + result. 



140 ARITHMETIC FOR ENGINEERS 

Exercises 36. On Evaluation with -f and Quantities. 

I to 4. In heat engine work it is frequently necessary to know the 
" absolute temperature," which = / + 461, where / = Fahrenheit tem- 
perature. Find the absolute temperature when t has the following 
values : 

I. 250. 2. 10. 3. 25. 4. o. 

5 to 8. The absolute temperature, if / is Centigrade temperature, 
is / -|- 273. Find the absolute temperature when t has the following 
values : 

5. 10. 6. 5. 7. o. 8. 241. 

9 and 10. Find the value of the expression a -f- S R (which 
refers to a certain steam engine expansion valve) for the following cases : 

9. a = 1-25, S = 2-75, R = o. 10. a = 1-06, S = 3-5, R = r8. 

II to 14. If F. is the Fahrenheit temperature, the corresponding 
Centigrade temperature is (F 32). Calculate the temperature C 
when F has the following values ; 

II. 15. 12. o. 13. - 2. 14. 32. 

15 to 17. If C is the Centigrade temperature, the corresponding 

Fahrenheit temperature is gC -f- 32. Calculate the temperature F 
when C has the following values : 

15. 15. 16. o. 17. 37-5. 

18 to 22. The expression is important in connection with 

epicyclic gearing. Calculate its value for the following cases : 
18. n = 5, a = 3, m = i. 19. = 4, a = i, m o. 

20. n 4, a 2, m = o. 21. n = 2*5, a = o, m = 1-5. 

22. n = 5-6, a = 2-8, m = i. 

23 and 24. The latent heat of sulphur dioxide, a refrigerating agent, 
is 175 -27* when t = Fahrenheit temperature. Calculate the latent 
heat when / has the following values : 

23. o. 24. 20. 

-r> 

25 to 28. The expression A ^ relates to thick cylinders under 

pressure. Find its value for the following cases : 

25. A = 8-48, B = 212, r =4-5. 26. A = 8-48, B = 212, r = 5. 
27. A = 3-6, B = - 57-5, r = 4-5. 28. A = 3-6, B = - 57-5, r = 5 -5. 

29 and 30. Find the value of the expression -JT~_ ^~ for the 

following cases : 

29. r = 6, b = 6, R = 55. 30. r = 7-5, b = 7, R = 20. 

31. The formula v = u -f at is important in questions on velocity. 
Find v when u = 5, t = 3-5, and a = 2. 

32 to 35. The expression - -refers to the acceleration of a moving 

body. Find its value for the following cases : 

32. v = 25, u = 12, t = 2-5. 33. v = 6, w = 22*5, * = 3-5. 
34. v = 10, u = 5, * = 1*5. 35. v = 12, w = 15, t = 1-5. 
36 to 39. In solving " quadratic " equations an expression of the 

form b 2 4ac has to be evaluated. Find the value of this expression 
in the following cases : 



SYMBOLS AND THEIR USES 141 

36. a 3, b = y5, c = 2'5- 37. a = 2, b i, c i6. 
38. a = i, 6 = 4*5, c = 2. 39. a i, 6 = 35, c = i8. 

40 to 44. An expression of the form i a i so occurs when 

solving quadratics. Find its values for the following cases : 

40. a = i, b = 5, R = i. 41. a = 2, b = i, R = 37. 

42. a = 2, 6 = i, R == 3-7. 43. a = 15, b = 6, R = 13-5. 
44. a = 15, b 6, R = 13-5. 

The Removal of Brackets. The meaning and use of a 
bracket have been introduced in Chap. I, p. 29. They are em- 
ployed in a similar manner when dealing with symbols, and then 
require some further consideration. In arithmetic a bracket can 
always be removed, and an expression simplified, by completing 
the operations within the bracket, as in Example No. 27, etc. 
But with an algebraic expression very frequently the quantities 
within the bracket cannot be worked out, and yet it is required 
to simplify the expression by removing the brackets. The removal 
may easily be carried out if certain adjustments be made, depend- 
ing on two things : (i) the sign immediately in front of the bracket, 
and (2) any quantity by which the bracketed expression is to be 
multiplied. To illustrate and prove the " adjustments " we will 
take expressions with numbers only, and work them out in two 
ways : (i) as in ordinary arithmetic, keeping strictly to the mean- 
ing of the bracket, and (2) by omitting the brackets and making 
" adjustments." 

First, consider the sign in front of the bracket : 

I. Plus sign in front, e.g., (i) 9 + (7 + 4). 
Evaluating the expression within the bracket first, we have 

9 + (7 + 4) 
= 9 + ii =20 

Now let us write the given expression and simply omit the 
brackets. Then we have 

9 + (7 + 4) 

= 9+ 7 + 4 

= 16 + 4 = 20 

which agrees with the former. 

Now take the expression (ii) 9 + (7 4). 

Evaluating the expression within the bracket first, we have 

9 + (7 - 4) 
= 9 + 3 =12. 



142 ARITHMETIC FOR ENGINEERS 

Now do as before, i. e., write the given expression without the 
brackets. Then we have 

9 + 7-4 
= 16 4 = 12 

which also is in agreement with the first value. 

Hence we see that when the sign immediately in front of the 
bracket is +, we may " remove the brackets " by simply omitting 
them. 

2. Minus sign in front, e. g., (i) 9 (4 + 3). 
Evaluating the expression within the bracket first, we have 

9 - (4 + 3) 
=9- 7 =2 

Now evidently the meaning of the given expression is that both 
the 4 and the 3 are to be taken away from the 9, so that if we wish 
to omit the brackets the expression should be written as 

9~4-3 
= 5 -3=2 
which agrees with the former result. 

/. 9 - (4 + 3) = 9 - 4 - 3 

Note that in the given expression the sign of the 4 is + (not 
written, but understood), but in the altered expression is . Also 
the sign of the 3 is + originally, and becomes when the brackets 
are removed. Thus the sign of the quantities inside the brackets 
have been changed. 

Now take the expression (ii) 9 (4 3) 

Evaluating the portion within the brackets first, we have 

9 - (4 - 3) 
= 9 i =8 

Now, following the method of the preceding paragraph, let us 
omit the brackets and change the signs of the quantities inside, 
i. e. t change the 4 into 4 and the 3 into -f 3. Then we have 

9 - (4 ~ 3) 
= 9- 4 + 3 
=5+3=8 

which again agrees with the former answer. 

.'. 9 - (4 - 3) = 9 ~ 4 + 3 



SYMBOLS AND THEIR USES 143 

Hence the rule : To remove brackets from an expression whilst re- 
taining its original value ; if the sign immediately in front of the bracket 
is + , simply remove the brackets : if it is , change the signs of all 
quantities within the brackets when removing them. 

Representing these results with symbols 

(a + (b + c) = a + b + c\ + sign in front 

\a + (b c) = a + b c) no change. 

(a (b -{- c) =' a b c\ sign in front 

\a (b c) = a b -\- C) signs changed. 

Secondly, consider the quantity multiplying the bracketed 
expression, and let us take the case of 8 + 5(6 + 3). It should 
be noted that the algebraic method of indicating multiplication is 
adopted here. This is usually followed with brackets, the given 
expression of course meaning 8 + 5 X (6 + 3). 

Also it should be remembered that the multiplication of the 
quantities in the brackets by 5 must be done before the 8 is added 
(see p. 30). Working on the ordinary method of evaluating the 
quantities in brackets first, we have 

8+ 5(6 + 3) 
= 8+ 5x9 
= 8 + 45 =53 

Now the sum of the 6 and 3, i. e., both the 6 and the 3, have to 
be multiplied by 5, so that multiplying by 5 before evaluating the 
bracket we have 

8 + 5(6 + 3) 

= 8 + 5x6 + 5x3 (this step may be done mentally) 
= 8 + 30 + 15 =53 which agrees with the former result. 

Similarly with the expression 8 + 5(6 3) 



Ordinary Method 



Removing Brackets 



8 + s (6 - <0 8 + 5(6-3) 

=8+5x6-5x3 



Hence we see that when there is a multiplier with a plus sign 
immediately outside a bracketed expression the brackets may be 
removed if all the quantities inside the bracket be multiplied. 

When a minus sign stands in front of the bracket and its mul- 
tiplier, then both the multiplication and the change of sign must 
be performed when removing the brackets. 



144 ARITHMETIC FOR ENGINEERS 



Thus 



15 - 4(3 - 2) 



= 15 4x3 + 4x2 

= 15 12 +8 

= 3 +8 = 



Proof, working on ordinary 

method 
15 - 4(3 - 2) 
= 15 - 4 X i 
= 15-4 =11 



Combining all the foregoing rules we may represent them with 
symbols, thus 

a + b(c + d) = a + bc + bd 

a + b(c d) = a + be bd 

a b(c + d) = a be bd 

a b(c d) = a be + bd 

Example 121. The " latent heat of steam " is found by experiment 
to be 966 -7 (t 212). Reduce the expression to a simpler form. 

966 -7 (/ 212) Note. Remove brackets, chang- 

966 7/ -f 7 X 212 ing sign inside, and multiply all 

= 966 '7* 4- 148-4 terms inside by -7. " Collect up " 

= 1114-4 *7* the two numbers. 

Example 122. Simplify the expression r + f (i r), which occurs in 
connection with the strength of materials. 

r + f (i - r) 



Example 123. Simplify the expression 2(^r + 3) (x 5) + 
3(1 -f 2X) - 7. 

2(* + 3) - (^ - 5) + 3(i + 2*) - 7 
= 2# + 6 #- r -5-f 3 -f - 6* 7 = 7^ 4- 7- 

When an expression contains brackets which enclose other 
brackets, start by removing the inside brackets, then remove the 
next pair, and so on until all have been removed. At each stage 
the quantities within the innermost brackets should be reduced as 
much as possible by collecting. 

Example 124. Simplify the expression 3{2 + ?<(r i) 3(r 6)}. 

3(2 + 2(r - i) - 3(r - 6)} 

V \f V V v/ 
= 3{2 -f 2r 2 3^ 4- 18} 

r} = 54 - 3 r 



SYMBOLS AND THEIR USES 145 

Example 125. The E.M.F. of the Clark JStandard Cell is given as 
!*434 t 1 ~ *ooo8(/ 15)] where / = temperature C. Simplify the 
expression. 

1-434 [i -ooo8(/ - 15)] 
i '434 [i -ooo8/ 4- *oi2] 
= i'434 [1*012 -ooo8/] 
= 1*451 001147* 

Exercises 37. On Removal of Brackets. 

Remove the brackets from the following expressions and simplify 
as far as possible : 

1. H - (H - M). 2. (H + C) (H - M). 

3. 2* + (3 - x). 4. 4 - ( 2 - p). 

5. C + 4 (C - i). 6. 25 - 3(d + i). 

7. 2w z w(i w). 8. 2 (a i) -f 3(1 a). 

9. r(2 -r) - 2(r* - r). 10. 4 (H + 3) - 2(2 -H) + H. 

11. A -f (w - i) A. 12. - / - (> 3 -f /) -f 3/> 3 - 

13. v 2 (v t f 2 ), which relates to the efficiency of a Pelton wheel. 

14. 3(w i) 4, referring to balancing of reciprocating engines. 

15. '133(1 *oo2 it), in connection with the effect of temperature 
upon fluid friction. 

16. 1 4500 (C -}- 4'28H), giving the " heating value " of oil. 

17. 99 ( i -T ), which refers to the mean pressure in a petrol 

engine. 

/ W \ 

18. ( P p )R, an expression in connection with a worm-geared 

pulley block. 

19. a -f- i (b a), relating to the strength of a dam. 

D 

20. r (/! / 2 ), which appeared in a problem on measuring the 
^2 

internal resistance of a cell. 

21. W W f i -jj, a formula appearing in a problem on continuous 
girders. 

22. x + J ( x j, which refers to the shear stress in beams. 

23. c 1 -\~ (i c z ), an expression occurring in a problem on the 
strength of materials. 

24. EC C (E e), in connection with an electric motor. 

25. i -f- *ooo2 (t 15), giving the E.M.F. of the Hibbert i volt cell 
at temperature t Centigrade. 

26. 100 f i H J -f 461, which gives the absolute initial tempera- 
ture of the charge in a scavenging gas engine. 

27. 167 "2j(t 32), relating to the latent heat of sulphur 
dioxide, a refrigerating agent. 

28. s -f 966 -f '48(/> 32), an expression referring to the heat 
lost in the exhaust gases of an oil engine. 

L 



146 ARITHMETIC FOR ENGINEERS 

29. 1114 *7/ -f- (/ -^32), which gives the total heat of i Ib. ot 
steam at temperature t Fahrenheit. 

30. '44* -f- (i*2X *5# 2 ), an expression which occurred in a problem 
on the strength of materials. 

31. 1-019 {i -f -00003(2 20)} volts, which is the E.M.F. of the 
Weston cadmium cell at temperature t Centigrade. 

32. "J2 { i -0007^ 60)}, giving the specific gravity of petrol at 
temperature / Fahrenheit. 

33. 14500 {C { 4'28(H + JO)}, a formula giving the heating value 
of coal when the chemical composition is known. 

34. i^fl [(212 t) -f- 966 -{ -48 (T 212)], which refers to the loss 
of heat due to moisture in coal. 

Insertion of Brackets. It is sometimes necessary to insert 
a bracket in a given expression for the purpose of further calcula- 
tion. Thus, suppose we have the expression CR + Cr and it is 
desired to use the symbol C in some particular way. For ease of 
later calculation the C, now appearing in two terms, should only 
appear once. The C is a factor of each term in the expression ; it 
is therefore said to be " common to the two terms " or it is a " com- 
mon factor/* When we adjust the expression (without, of course, 
altering its value) so that this symbol C occurs only in one place, 
we are said to " take out " the common factor C from the two 
terms. When doing this, brackets must be inserted, and exactly 
similar rules must be obeyed as when removing brackets; the 
operation is, in fact, exactly opposite to that of removal. 

The expression CR -f- Cr would appear as C(R -f- r) if we take 
out the common factor C. Removing the brackets from this 
expression according to the rules given on p. 143, we obtain the 
original expression. 

. Taking the formulae given on p. 144 and reversing the order of 
statement, we have 

2. a + be bd = a + b(c d) 

3. a be bd = a b(c + d) 

4. a be + bd = a b(c d) 

Thus, when inserting brackets, we must conform with the methods 
given on p. 143, for their removal. Hence we have the rule : When 
brackets are put in after a + sign, the signs then coming within the 
bracket remain unchanged ; when brackets are put in after a sign, 
the signs then coming within the bracket are changed. 

An inspection of the four cases shown will make this clear. Thus, 
in i and 2, where the bracketed quantity follows the + sign in 



SYMBOLS AND THEIR USES 147 

front of be, the sign between c and d is unchanged ; but in 3 and 4 
the bracket is inserted after the sign in front of bc t and the sign 
between c and d is changed. 

Taking out a Common Factor. The terms to be put inside 
the bracket are obtained by dividing the terms of the given ex- 
pression by the common factor. Where a term in the given expres- 
sion does not contain the common factor under consideration, then, 
of course, this term is not included in the operation of inserting 
brackets, but is merely repeated. 

Take case 2 in the previous section 

a + bc bd = a + b(c d) 

The only x ">r which is common to two terms (or more) is b\ 
hence b is t^ Tactor to be taken out. The first term a does not 
contain b, and therefore can only be repeated. Then the second 
term be will supply the first term inside the bracket. Now be 

divided by b, i. -,-> equals c, which is then put inside. Then the sign 
is written (not changed as the sign in front is +) and the third 
term bd is considered. Similarly, -j- = d, which is then placed 

after the sign ; the bracket is then closed. The actual division 
can usually be done mentally. 

Example 126.- Take out the common factor in the expression 
PV 4- PV. 

PV + PVA = PV(i + k) 

Note. The common factor in this case consists of a product of two 
symbols P and V ; it is however treated in exactly the same manner 
as a single symbol. 

PV 
Remember that = i 

Unless a certain common factor is specified to be taken out, 
then the factor taken should be the largest possible. Thus, in 
Example 127, which follows, R is a common factor, and might 
perhaps be needed. But R 2 is also common, and being larger than 
R, is therefore taken. 

Example 127. Take out the common factor from the expression 
^RZ __ R 2 . (This expression occurs in connection with thick cylinders, 
such as guns and hydraulic pipes.) 

/R 2 - pR* 
= R(/ - p) 



148 



ARITHMETIC FOR ENGINEERS 



Example 128. Take out the common factor from the expression 
2vV 2V a , relating to water turbines. 

Here 2 appears in each term and also V. Hence aV is the common 
factor. 



= 2 V(v - V) 



2V 



Example 129. Take out the common factoi from the expression 

t/*A* t/ 2 

(This occurs in connection with the Venturi water meter.) 

a 2 2g 2g 

Note here that v 2 is common on the top line, and 2g on the bottom. 

t 2 
Hence - - is the common factor. 



Note that 



Example 130. Take out the common factor # in the expression 
4/8 _ 5^/2 ^_ ^2^ which refers to moving loads on continuous girders. 



4 /3 _ 



_ yl) 



In this case the signs are changed, as the brackets have been 
put in after a sign. The term 4/ 3 does not contain the common 
factor, and is therefore merely repeated. When there are two 
common factors in the expression (e. g., % and I 2 in the last example), 
the factor to be taken out will be mentioned in the exercise work 
here. In ordinary calculation where the taking out of factors is 
merely a step in the work, the nature of the calculation must decide 
the actual factor to be taken out. 

Exercises 38. On Taking out a Common Factor. 

Rewrite the expressions given in Nos. i-n, in each case taking out 
the common factor. 

1. a ea> which occurs in a problem on epicyclic gearing. 

2. aR ar, referring to a machine for drawing large arcs. 

3. cA + mca, a formula relating to reinforced concrete. 

4. M -f M-, in connection with the bending of beams. 



SYMBOLS AND THEIR USES 149 

Kl 
5. K Y' relating to a coil spring. 

JL> 
6. h 7&, giving the outside diameter of a wire rop. 

7. 2M/ t -f- 2M/ 2 , a formula referring to continuous girders. 

8. a 2 -}- zab, which relates to a machine for drawing large arcs. 

9. Shr 4/* a , referring to the area of a geometrical figure. 

w z r* 

10. w*r y , a formula in connection with the inertia of engine 

parts. 

x* 

11. x y, which occurred in a problem on beams. 

In Nos. 12-21 the given expressions are to be rewritten, taking out 
the common factor mentioned in each case. 

12. h - relating to continuous girders. (Common factor ?. ) 

44 \ 4 ' 

W3 

13. 5/ 2 , another continuous girder expression. (Common 
factor / 2 .) 

bd* 

14. -f- bdx z , a formula in connection with beams. (Common 

factor bd.) 

. _ nPV npv , . ,. ii. /r- 

15. ^_ ~, giving the work done in air compression. (Common 

factor . ) 

n i) 

i^V/w Wxfv' 

16. Wy - -~-\ 73 - , an expression occurring in connection 

4* 4^ 

with rolling loads on continuous girders. (Common factor Wy.) 

17. a z b -f ab 2 . (Common factor ab.) 

18. 3p*r 2p-r. (Common factor p*r.) 

1^' A~~ ^"AT ' an ex P ress i n relating to the volume of a 



certain solid. (Common factor -r .\ 

\ A a ) 



x x x 

20. h -7 , referring to the bending of a cantilever. 

(Common factor lx z .) 

21. (x + a)l +(x + a) 1 -^. (Common factor (x -f a).) 

x 3 

22. Take out the common factor in the expression given in 

No. 20. 

23. In the expression x* -f- ax + ex + ac, take out the common 
factor in the first two terms, and also that in the last two terms. Then 
take out the common factor in the resulting expression. 



150 ARITHMETIC FOR ENGINEERS 

24. Simplify the expression , ~ which refers to bridge girders. 
(Hint. Take out common factor in bottom line. Then cancel.) 

25. The expression -HB 3 - HB 3 + 8 HB 3 relates to the bending 

7 25 75 

of beams. Take out the common factor HB 3 and simplify. 

26. Simplify the expression b h H (a 6), which refers to the area 

of a certain geometrical figure. (Note. First remove the brackets ; 
then take out the common factor h and reduce.) 

Multiplication with Expressions of Two Terms. It is 

sometimes necessary to multiply together two quantities, each of 
which is represented by an expression containing 2 terms, e. g., 
a b multiplied by a -f c. The operation would be indicated 
thus : (a - b)(a + c). It must be noted that it is incorrect to 
write the statement as a 6 X a + c ; in this case only the a and b 
in the middle of the expression are to be multiplied. But when 
we say " multiply the expression a b by the expression a + c," 
it is intended that the whole of the expression a b is to be multi- 
plied by the whole of the expression a -f- c. Hence brackets must 
be used to denote that each expression is taken as a whole. These 
symbolic expressions with several terms and brackets are exactly 
similar to the compound expressions containing vulgar fractions, 
as given on p. 33, etc. 

The method is to take one expression, say the a b, and multiply 
it, first by one term, say a of the other expression ; then by the 
other term c and add the results together; just as in multiplying 
33 x 24 we multiply the 33 first by 4 giving 132, and then by 20 
giving 660, and add the two results together, giving 792. 

The usual method of laying out the work is as follows : 

a -b 

a + c 

a 2 ab 

+ ac be 



a 2 ab-\- ac be = Product. 

Explanation. Multiply the whole of the top line (i. e., the ex- 
pression a b), by the symbol a in the expression a + c : a times 
a = a a , which is written beneath the line under a + c. Now + a 
multiplied by b = ab which is written after the a 2 . This 
completes the multiplication by the a. Now multiply the whole 



SYMBOLS AND THEIR USES 151 

of the top line by the c : (+ c) x ( + a) = ac. This must be 
written in another column, as there is no existing column for ac. 
Then (+ c) X ( b) = be. Another column has to be made 
for this, as shown. This completes the multiplication, and now 
the products are added up. In this particular case there is no 
actual addition to do, as each column contains only one term. Then 
the result is a 2 ab + ac be', the order of the terms does not 
matter so long as the signs are correct. 

Example 131. Simplify the expression (b xR)(a 2x) + axR, 
which occurred in an investigation on retaining walls. 

(b - xR)(a - 2x) + axR b - xR 



= ab- axR - *xb -f 2^R + a*R ab ~ a * R 

2xb 

= ab 2xb -f- 2# 2 R 

ab axR 2xb 



a 



The multiplication is exactly similar to the previous one. When 

" collecting up the like terms " in the second line, note that axR 

and -f a#R = ; tick off the terms when collected so that all are 
included. 

Complicated examples in this kind of work are not very com- 
mon. The most important application is given in the following 
section : 

Evaluating such Expressions as (a b) 2 .* 

1. (a + 6) 2 , i. e. t " the sum of two quantities, squared," or, 
" the square of the sum of two quantities." 

2. (a 6) 2 , i. e. t " the difference of two quantities, squared," 
or, " the square of the difference of two quantities." 

3. (a + b}(a b), i. e.> " the sum of two quantities, multiplied 
by the difference of the same two quantities," or, " the product 
of the sum and difference of two quantities." 

Such expressions often occur, and the reader should be able to 
write them down at once when required. This operation is fre- 
quently called " expanding the expression." 

It must be noticed that (a + b) 2 is not equal to a 2 + t 2 , and that 
( a _ 5)2 j s no t equal to a 2 6 2 , as might be imagined at first 
sight. 

* The statement (a 6) a is a short way of saying " (a -f- b) 2 and 
(a - &)." 



152 ARITHMETIC FOR ENGINEERS 

i. (a + 6) 2 = (a + b)(a + b). 



a + b 
a + b 



+ ab + b* 



Multiply by + a 

a x a = a 2 ; b X a ab 
Multiply by -f- b 

a X b = + a& ; written beneath 
the previous one 

b X b = fc a 



Let us call " a " the first term and " b " the second term. Then 
it is seen that our result contains 

+ a 2 which is the square of the first term. 

+ b 2 second term. 

-f 2ab twice the product of the two terms. 

Then the result is best summed up in the following sentence : 

The square of the sum of two quantities = the square of the first 
term -f the square of the second term + twice the product of the 
two terms. 

Thus, in writing out the value of (p -f- x}*, 

-|- Square of first term = Square of p = -f- 3 
-f second term = x = -f x* 
+ Twice product of the two terms == -\-2px 

/. (p + x y = p* + *px + x* 

2. Difference of two quantities, squared. 
(a - b) z = (a- b)(a - b) 



a-b 
a -b 

2 ab 
- ab + b* 

a* 2ab + b 2 



Multiply by -f a 

axa==a 2 ; --&x-f-a= ab 

Multiply by b 

a X b = ab] written beneath the 
previous one 



Note that in this case we have 

I- a 2 which is the square of the first term. 

+ 6 a ,, second term. 

zab which is minus twice the product of the two terms. 



SYMBOLS AND THEIR USES 153 

Therefore 

The square of the difference of two quantities = the square of 
the first term + the square of the second term twice the product 
of the two terms. 

Thus, in writing out the value of (n y) 2 

-f- Square of first term = Square of n = w* 
+ ,, second term = ,, y = y 2 
Twice product of the two terms = - zny 
.*. (n y)* = n 2 zny -f y 2 

3. Sum of two quantities multiplied by the difference of the 
same two quantities. 

(a + b ) (a - b) 

a + b \ 

a b 



a 2 + ab 



ab-b* 



b' 1 



Multiply as before 

It will be seen that the sum of 
+ ab and ab = o, since -f- i 
neutralizes i 



Note that here there are only two terms, thus 

-f- a 2 which is the square of the first quantity 

b 2 which is minus the square of the second quantity 

Therefore 

The sum of two quantities multiplied by the difference of the same 
two quantities = the squart of the first term the square of the second 
term. 

Hence in writing out the value of (N -f n)(N n) 

+ Square of first term = Square of N = N a 

,, ,, second term = Square of n = n 1 
/. (N + w)(N - n) = N 2 M a 

It should be noted that the order of the terms in the two brackets 
should preferably be the same, e. g., (N + n)(N n) and not 
(n + N)(N n). The latter is equally correct, but might lead to 
confusion as to which is the " first term." Actually the first term 
is the one which comes first in the subtraction. 

The advantage of remembering the rules in the above manner 
is that any two-term expression may be dealt with. Should one of 
the terms be a number, then some reduction may be made after 
having written out the result according to the above rules. 



154 ARITHMETIC FOR ENGINEERS 

Thus in evaluating the expression (x 3) 2 , an example of 
case 2, 

-f Square of first term = Square of x = x 2 
,, ,. second term = ,, ,,3 = 9 

Twice product of the two terms = 2 X 3 X # = 6* 
.*. (* - 3) a = ** - 6* + 9 

With a little practice the result may be written down almost directly, 
the squaring being done mentally. 

When the terms contain more than a single symbol it is ad- 
visable to work in two steps. First write out the " squares of 
terms/ 1 etc., without any reduction. Then take each term of the 
result and simplify where possible. 

Example 132. Expand ($m -}- ny)*. 

Result = Square of first term + Twice product + Square of second term 
.". (3^ + ny) 2 = (3m) 2 + 2 x 3 X ny + (ny) 2 

Now reduce terms where possible. 
Thus, (3m) a = 3 2 m a = gw 1 

2 X 3m X ny ~ 6mny 
(ny) 2 = n 2 y 2 

Hence result = gm 2 -f 6mny -f- n 2 y* 

Working directly without the intermediate step, there is a greater 
possibility of error. With practice, of course, it is quite possible to 
obtain the answer directly. 

/ P\* 
Example 133. Find the value of If -- ) 

P /P\ a 



Example 134. Simplify the expressions 
(a) (-i) 2 ; (b) 



(a) (n i) 8 = n 2 2 x i X n 

= n 2 in -\- i 



SYMBOLS AND THEIR USES 155 

Example 135. Simplify the expression 

V 2 _ y2 _ ^ v __ v) 2 , which relates to water turbines. 

After expanding (v V) 2 , the result must be enclosed in a bracket 
to indicate that it is all to be subtracted. 

v z _ v 2 (v - V) 2 

1,2 _ v 2 (v 2 2vV + V 2 ), expanding the square 
= v 2 V 2 v 2 4- 2vV V 2 , removing the bracket 
2V 2 , collecting like terms. 



The expression 2vV 2V 2 may also be stated in the form 2V (v V). 

The case of (a + b) (a b) = a 2 6 2 is often of use when 
reversed, or stated in words 

The difference of the squares of two quantities may be expressed 
as the product of the sum and difference of the two quantities. 

Or, M 2 - N 2 = sum x difference = (M + N)(M - N) 
This form is useful, as it may enable further calculation to be 
done in a simpler manner, or it may make cancelling possible. For 
example, in certain calculations connected with fly-wheels, expressions 
like N 2 w 2 occur when N and n have values which are nearly 
equal. Then the form (N + n)(N n) makes the evaluation of the 
expression an easy matter. The following example will illustrate. 

Example 136. Find the value of N 2 n 2 , which relates to the 
energy in a fly-wheel, when N = 250 and n = 247. 



w 2 = (N + w)(N - n) 
= (250 -{- 247) (250 - 247) 
= 497 X 3 



Proof. 
2 5 2 = 62500 



247 2 = 61009 
1491 

Note. If we had taken 25o 2 247** the working would be less 
easy, as squaring these numbers is comparatively a long process. 

As an example of its use to effect cancelling and further simplifi- 
cation of a problem, the following example may be considered: 



2 V(y V] 

Example 137. The expression a^tl'v 2 occurs in connection 

with the efficiency of hydraulic turbines. Reduce to a simpler form. 

2V(wjT_Y) _ __ ?V(v ~ V) 
t/a _ V a ~ '(v + V)(fT^V) 

_ 2V cancelling (v V) in top and 
v + V bottom 

Note. By converting the f 2 V 2 into the form (v -f V)(t/ V), 
the term (v V) may be cancelled. 



156 ARITHMETIC FOR ENGINEERS 

Exercises 39. On Expansion of Squares of Two 
Terms, etc. 

Multiply : 

1. x 3 by 2x -f i. 2. x 2 4 by # 2 -f 7. 

3. / -f x by 2/ x. 4. 20 -f y by a ir. 

5. 2s + 3/> by s . 6. fc 3 3& 2 by 6 -f 2. 

7. a 2 fl& -}-- b 2 by a -f 6. 8. a 2 -f- a& -f- b 2 by a 6. 

Expand the following expressions : 
9. (2* -}- y) 2 . 10. (AT - 2 y) 2 . 11. ( 3 - 2/>) 2 . 

12. (/> - 4 ) 2 . 13. (/ + Av) 2 . 14. -f 2 



15. *-. "-(-- 17. (iv -2)-. 

18. ( 3 m -H i-2/O 1 . 19. (| 4- /)*. 20. (in - - 5 ) 2 . 

Simplify the following expressions : 

21. (2 - l)(z + 1). 22. (M + 3 )(n - 3). 23. ( j - *)^ + *) 

+ , - *. 25. ^ - *y + -. 26. 1 - + 



27. Simplify the expression (/ -f z) (I 2 $z 2 ) (I 2 z~}z, which 
occurs in a problem on moving loads over suspension bridges. 

By using factors find the value of N 2 n 2 in the following cases: 

28. N == 100, n = 98. 29. N == 450, n = 445. 
30. N = 130, n = 127. 31. N 144, n 141. 

32. In connection with a differential hydraulic accumulator it was 
necessary to evaluate D 2 d 2 when D = 14 and d = 12. Find 
the required value by using factors. 

33. With reference to the lengths of the sides of right-angled tri- 
angles we have the statement b 2 = a 2 c 2 . Find by the " expansion 
method " the value of b 2 when a = 9-625 and c = 3-125. (Note the 
shortness of this method compared with squaring.) 

PV -bv 

34. Simplify the expression PV -f - PV which relates to 

the power required for air compression. (Hint. Reduce all terms to 
the common denominator n i, remove brackets from the resulting 
numerator and simplify, then take out common factor.) 

35. For the area of a certain geometrical figure which has certain 
dimensions, a t c, A and h, there is an approximate formula \ch\ there 
is also another approximate formula |A/?. By adding 3 7 ^ of the first 
expression to ^ of the second expression, a third formula is obtained 
which is found to be very accurate. Obtain this formula and in it 

substitute for A the expression - . Simplify to give a formula 

3 
for the area in terms of the dimensions a, c and /*. 



CHAPTER IV 
SIMPLE EQUATIONS 

Equations Generally. Speaking generally an equation is a 
short statement that two things are equal, the sign = being used, 
e. g., 2 f -6" = 30". But the word equation is usually applied to a 
statement of equality containing various symbols, from which 
something has to be calculated. 

The statement on p. 86, V = CR is, of course, an equation, 
but it is also called u a formula for V in terms of C and R," because 
the value of V can be obtained directly by simple arithmetic, after 
substituting values for C and R. Now supposing that we know 
the values of C and V, and desire to find the corresponding value of 
R. Let V = no volts and C = 250 amperes; then substituting 
these values in the above formula we can write 

no = 25oR 

Now this statement is called an equation in one unknown quantity, 
since it contains only one symbol whose value is not yet known. 
From it the value of R can be calculated, but other operations 
than those of simple arithmetic will have to be performed. The 
object of this chapter is to show how to proceed in various cases. 
The quantities contained in any equation are divided into two 
classes : known quantities, i. e. t those whose numerical values are 
known, as the V and C above; and unknown quantities, i. e., those 
whose values have to be calculated, as R above. The process of 
finding the value of the unknown quantity is known as " solving 
the equation," and the final result is called the solution. When 
describing the operations to be performed we shall speak of the 
" left-hand side " of the equation, meaning all written to the left 
of the equals sign, and the " right-hand side/' meaning all on the 
right of the sign, 

The all-important thing that must be kept in view when dealing 
with equations of any type is that the two sides must always be equal. 
To suit our own convenience the sides may be considerably changed 

157 



158 ARITHMETIC FOR ENGINEERS 

in appearance, but the values must remain the same. As soon as 
the equality is disturbed the statement is no longer true. 

An equation can be looked upon as a balance or pair of scales, 
when just in equilibrium (see Fig. 20). The actual things in the 
scale pan may be very different, but the weight of each side must 
be the same. The weight in the pair of scales corresponds to the 
value in our equation, and the two scale pans correspond to the 
two " sides " of the equation. We can operate upon the weights 
in the scale pans in various ways without destroying the balance, 
but whatever is done to one side must be done to the other. Thus 
if we add 2 Ibs. to one side we must add 2 Ibs. to the other ; if 3 Ibs. 
be taken from one side, 3 Ibs. must also be taken from the other 
side. Also if the weight on one side is multiplied or divided 
by 4 exactly, the same must be done to the other side. Exactly 
the same idea applies to equations. 

Simple Equations. Equations are divided according to the 
highest power of the unknown quantity which they contain. When 
only the first power is present then the equation is a Simple Equation, 
e. g., 966 = 1114 7/, where / is the unknown quantity, and is only 
of the first power. When the second power, or square, is the highest 
power of the unknown present then the equation is said to be a 
Quadratic Equation, e. g., x 2 + 5* + 18 = 12 where x is the unknown 
and x 2 , i. e. the second power, is the highest power present. Only 
simple equations will be taken in this book, but in all equations 
the principles of the simple equation apply. 

The solving of an equation requires manipulation of the quantities 
involved, and therefore a knowledge of the preceding chapters. 
The most common operations which have to be performed are 
those of multiplication, division, addition, and subtraction. These 
will be taken singly and in combination. 

Equations requiring Division. As a simple example con- 
sider the equation 

$w = 168 

To give this a real meaning let w represent the weight in pounds 
of a small bag of cement. Then the equation can be represented 
as in Fig. 20, where on one scale pan we have 3 bags of cement 
(all the same size), and 168 Ibs. on the other pan just balancing. 
The 168 Ibs. has been made up of three 56-lb. weights. 

Then evidently if three sacks weigh three 56-lb. weights, one 
sack will weigh one 56-lb. weight. 

That is, w = 56 = J of 168 



SIMPLE EQUATIONS 159 

Writing this in mathematical form 
320 = 168 

.\ i> = ~ of 168 = = 56 Ibs. 
o o 

Notice that on the left-hand side, to obtain the w above we have 
divided the ^w by 3. Notice also that the right-hand side has also 
been divided by 3. 

Thus in detail : = i. e., w = 56 Ibs. 
j o 

Hence we see that if in the given equation we find the unknown 
multiplied by a coefficient (as the 3 above), then we can isolate the 




3 ur = 168 
U/ = 56 

Fig. 20. 

unknown and still keep the equality, by dividing each side by this 
coefficient. This is usually stated as " dividing across by 3." 

Example 138. If h = height of water in feet in a vertical pipe, 
then p the pressure in pounds per square inch at the bottom is given 
by the equation 2-3 p = h. Find the pressure when the height is 25 ft. 

Substituting for h 2-3 p = 25 
T- 2 '3 2 *3/> _ 2 5 

^3 "23 
= 10-9 Ibs. per sq. in. 

Having obtained a solution, a check on the correctness of the 
working is always available by substituting in the given equation 
the calculated value for the unknown, and then simplifying each 
side. If correct, the two sides of the equation will be found equal ; 



i6o ARITHMETIC FOR ENGINEERS 

otherwise some wrong operation has been performed. Beginners 
should certainly check all their equations. For the earlier examples 
of each type in this chapter the check will be shown. 

To prove the truth of the result in Ex. 138, we substitute the 
value 10-9 for p in the equation. 

Thus 23/> = 25 

Substituting 2*3 X 10-9 = 25 

Simplifying 25-07 = 25 

which is quite satisfactory, as the slight difference is due to the value 
10-9 being calculated to only 3 significant figures. 

With a simple equation as above, the operation performed 
(dividing by 2-3) is easily seen. With more difficult types the 
operations may involve many steps. It is therefore advisable to 
state at the left-hand side what operation has been performed on 
each line as shown above. Then when returning to the equation 
for checking or reference, it is easy to pick out any operation. 
The steps can be very briefly stated : thus " -f- 2-3 " in the previous 
Example, No. 138, means " dividing both sides by 2-3 " and similarly 
" X 25 " would mean " multiplying both sides by 25." The reader 
can also use abbreviations such as " simpli." for " simplifying." 
Adopted right from the commencement of equations, the scheme 
becomes a habit and is of distinct advantage in more difficult work. 
The sign /. (therefore) is placed before the last step in the working 
to emphasize the conclusion arrived at. 

The following example shows that the method is just the same 
if the coefficient is a decimal. 

Example 139. Find the value of x from the equation "ix 6. 

Proof. 

Substituting 30 for x in given 
equation 

2 X 30 = 6 
6 =6 



2* = 6 

"2.X 6 



-2 ~ V 2 

. x = 30 



.'. Result is correct. 



With practice the number of steps may be reduced. Thus, in the 
above example, the step = may be omitted, the left-hand side 
being understood. The working would then appear thus : 

2* = 6 



SIMPLE EQUATIONS 161 

Exercises 40. On Equations requiring Division. 

Solve the following equations : 

1. 50 = 16-5. 2. lid = 134. 3. '75D = 16. 

4- 3-7* = 3-5. 5. -1055 = 11-05. 6. "004M = 17. 

7. 251* = 1-35. 8. 440 == 11-5. 9. R X 1*34 = 37. 

10. If la = V calculate the value of a when / -667 and V = 15. 

11. The formula Pv = pV refers to the expansion of gases. Find 
the value of v when P = 165, p = 20, and V = 13-4. 

12. The speed of a belt is given by the formula V = irDN where 
V = speed in ft. per min., D = dia. of pulley in ft., and N = revs, 
per min., and ir = 3-14. A planing machine belt is to run at 1500 ft. 
per min., over a pulley running at 240 revs, per min. What must be 
the diameter of the pulley ? 

13. The formula pr = ftrj * relates to steam boilers. Calculate 
the plate thickness / (inches) for a large marine boiler when r = 114, 
p is to be 210, /= 18000, and 17 is to be '9. Give the actual result and 
also the thickness to the next -fa". 

14. In a hydraulic press P = "78$d 2 p where P = total load in 
pounds, d = dia. of ram in inches, and p = pressure of water in pounds 
per square inch. Calculate the pressure p when the ram is 12" dia. 
and is to exert a total pressure of 150 tons (i.e., P = 150 X 2240 Ibs.). 

15. Find the value of h from the formula v 2 = 2gh, which relates 
to hydraulics; when v = 30 and g = 32. 

16. Using the formula in Ex. 13, calculate p the pressure allowable 
(pounds per square inch) in a compressed air receiver where r = 9", 
t = .25, n = -66, and / = 10,000. 

17. For a direct-acting steam pump we have the formula '75D 2 P = 
d*p. A certain pump has D = 8, P = 60, d = 3-5. Calculate p using 
these values. 

18. If V = 2 -47^2 calculate the value of d when V = 257 and t = 1*56. 

Equations requiring Multiplication. Sometimes the co- 
efficient multiplying the unknown appears as a vulgar fraction, 

I P 

e.g., - P, or in the form -. To isolate the unknown here we must 
o ? 

p 

multiply by the denominator, since x 5 = P, the 5's cancelling ; 
we have therefore to " multiply across by 5." 

p 
Example 140. Solve the equation =8-3. 



P = 8- 3 


Proof. 


5 3 


Substitute for P in equation 


X 5 -y = 5 X 8-3 


... 413 _ 8 . 3 


' P = 4 I- 5 


/. 8-3 =8-3 




Result is correct. 



* v is the Greek small letter "eta." 
M 



162 ARITHMETIC FOR ENGINEERS 

Example 141. Solve the equation -^- =13. 

- m - = 13 
30-5 

X30-5 '-^ =i3X3-5 

m = 396-5 

After the first few examples it is possible to dispense with part 
of the second step, and to set down as in the next example. 

Example 142. Find the value of / from the equation - = 2-31. 

^ - 2 ' 31 

X '43 ' = 2 *3l X -43 = -993- 

The above two sections may be summed up into one general 
rule : 

Either side of an equation may be multiplied, or divided, by any 
number provided that the other side is also multiplied, or divided, by the 
same number. 

Thus if A = B 
then *A = xB] 
and A B Y wnere x is an y number 



Exercises 41. On Equations requiring Multiplication. 

Solve the following equations : 



-3 



10. For an electric battery supplying current we have the formula 
c = - where c current, e = volts lost in cell, and r = resistance of 
cell. If c = -56, and r = 2-5, find e. 

11. The formula r = c refers to long columns. Calculate the value 
of / when c 160 and k =* I-O75. 



SIMPLE EQUATIONS 163 

y 

12. If = N, a formula relating to belts, calculate the value of 

trD 

V when IT = 3-14, D = 1-25, and N = 165. 

T? T? 

13. The formula ^ = ^- is employed when measuring electric 

K. 2 -TM 

resistance by a Wheatstone bridge. s If R 3 = 10, R 4 = 1000, and R 2 = 
247, calculate the value of R^ 

14. The formula c = - relates to the clearance volume in a 

Y ~~~" I 

gas engine. If c = 25, r = 6-5, calculate v. 

15. The formula - =~ applies to the expansion of gases. Calculate 
the value of v when P = 115, p = 20 and V = '25. 

16. In a transformer if E t = primary voltage, E 2 = secondary 
voltage, Sj = turns in primary coil, and S 2 = turns in secondary coil, 

C* o 

then-^ 1 - = ^. Find St if E, = 6000, E 3 = 500, and S 2 = 12. 
H, 2 b 2 

Combination of Multiplication and Division. Both 
operations are required in equations like the following. They 
may be combined or performed in one step, but first will be shown 
separated for clearness. 

Example 143. Solve the equation - d 13, giving the diameter 
of a circle whose circumference is 13*. 

22 , 
y^= 13 

I , 13 

-r 22 - d = - 

7 22 

X7 <*=X7 = 4-14. 



The first step is to divide across by 22 since this is a multiplying 
coefficient. Secondly, multiply by 7, since this is a dividing 
number. 

Combining the two operations we multiply by 7 and divide by 22, 

7 
i. e ., multiply by . 



Then 


d 


= 13 


Proof. 




7 




22 








V x 4*14 = 13 


X 


d 


= 13 x 2 7 2 


7 
.-. 13-01 == 13 






= 4'H 


.', Result is corre 



164 ARITHMETIC FOR ENGINEERS 

Example 144. When a gas expands under the application of heat, 

V T 
the formula ~ = 7 is of importance. Find the value of t if V = 2-5, 

v = 7-8, and T = 550 

Y I 

t; "* * 
Substituting- | 55? x 

x < fil- - 550 2 

X 7-8 2-5* = 550 x 7-8 ... 3 

< _ 550 X 7-8 _ 

* - ?5 - 

The double multiplication performed in the above example in 
passing from line I to line 3 can be usefully combined into a step called 
Cross Multiplication. Notice that in line 3 we have on the left- 
hand side 2-5 X t\ i.e., the top of the left-hand fraction (in line i) 
X the bottom of the right-hand fraction. Similarly on the right-hand 
side we have the top of the right-hand fraction x bottom of left- 
hand fraction. Thus two multiplications have been performed, as 

indicated here, o^XCT / an( * the two products equated. By 

means of this we are enabled to at once straighten out any equation 
in which the whole of each side is separated into numerator and 
denominator by a division line ; the unknown can then be readily 
found. Exercises on proportion are best treated when worked on 

A C 

the above lines, since A : B : : C : D can be stated as ^ = T^ 

15 D 

Example 145. The horse-power of a steam engine is given by the 
formula H.P. - O Q Q -where H.P. = horse-power, p = mean effective 

pressure in Ibs. per sq. ins., L = stroke in ft., a = area of piston in sq. 
ins., and N = number of strokes per min. Find the stroke L when 
H.P. = 50, a =s 143, p = 27 and N = 300. 

Substituting the given values in the formula, the following equation 
is obtained (reversed for ease of reading). 

27 X L x 143 X 3^ = o 
330 T __ 50 X 330 I Approximation. 



27 X 143 X 3 27X 143 X 3 



say 17' 



5X3 
3x1x3 



5. = about 2'. 






SIMPLE EQUATIONS 165 

In equations of this form it is best to perform all the algebraic 
operations before doing any actual simplifying. Then one expres- 
sion is obtained for the result, and is easily dealt with. Where a 
slide rule is used this is a decided advantage. 

Thus in the above example it is best to work in the form shown 
rather than to simplify some of the figures on the left-hand side 
first. 

Exercises 42. Multiplication and Division combined. 

1. For a direct current generator ----- = K where E = volts, C 

= current in amperes, and K = kilowatts. A generator to develop 
150 kilowatts is to work at 220 volts. What current will it give ? 

2. The formula r ^ refers to a screw-jack. If p = -5, * = 3-14, 
and r = 250, find the value of R. 

^A 

3. The formula K = - gives the capacity of a parallel plate con- 
denser. Calculate the value of A when K = 18, k = 3-5, TT = 3-14 
and t = -15, 

d 2 s 

4. A rough rule for petrol motors is H = where H = horse- 
power, d = dia. of cylinder in inches, and s = stroke in inches. A 
motor with a cylinder 3*25* dia. is to develop 3-5 H.P. Calculate its 
stroke. 

5. The formula d*l = -~- refers to dynamos. Calculate the value 

of c for a small multi-polar machine where d = 18, / = 26, K = 37-5, 
and N = 475. 

6. In connection with the strength of beams Z = ->*-. Calculate 
the value of b if Z = 24 and h 6. 

Lv 2 

7. The formula h =f- refers to pipes conveying water. In a 

certain test L = 12, v = 6*5, d = 4 and h -85. Calculate the value 
of/. 

8. For hydraulic pipes t QQ where / = thickness of pipe in 

inches, d == dia. in inches, and H = head of water in feet. A pipe is 
5" dia. and \" thick. Calculate the value of H. 

9. The pull of a lifting magnet is P = ^ Ibs. Calculate 

the value of A if P = 3490 and B = 16000. 

10. The formula H = -. refers to the magnetic field produced 

by a coil of wire. In a certain coil / = 20, T == 250 and H = 75. 
Calculate the value of C. 



166 ARITHMETIC FOR ENGINEERS 

PD 

11. The formula S = - --- relates to the strength of boilers when 

metric units are used. Calculate the value of P when S = 7-5, D 1-35, 
and / = 12. 

12. If H = horse-power obtained from a waterfall, Q = quantity 
of water in cubic feet per second, and h height of fall, then H = - 

where G = 62-4. Calculate the value of Q when H = 40 and h = 10-5. 

2 R 2 

13. The formula / --f P relates to flat circular plates loaded 

uniformly. If R = 18, t = -625, and / = 9000, calculate the value of p. 

2^N 

14. Find the value of N from the formula -^ = when * = 3-14 

DO w' T 

and <a = 35*6. 

2/2 

15. The formula e = ^ relates to the twisting of shafts. Calculate 

the value of /, if 9 = -35, / = 5-5, C = 5400, and d = '375. 

(In the following few examples the unknown is in the denominator. 
They are best solved by cross-multiplication,} 

18. The formula R = - relates to the resistance of an electric 
a 

conductor. In a pair of feeders supplying current to a number of 
lamps / = 21600, R = -286, and s oooooo66. Calculate the value 
of a. 

17* The formula r relates to a screwjack. Find p if r 200, 
R = 20, and * = 3' 14. 

P V P V 

18. The formula - L - 1 = - 2 2 relates to the mixture in a gas engine. 

T! r 2 s 

If P t = 15, Vj = 1-3, r t = 480, P 2 = 102, and V 2 = 25, calculate r 2 . 

19. The capacity of a plate condenser is given by the formula 
kA 

K = ,. Calculate the value of f if k = 3-6, K == 150, A = 650, 

47T( 

and TT = 3'i4- 

tt>L 2 

20. The formula d = -^r refers to overhead electric conductors. 

Calculate the value of T when d = 12, L = 13 20 and w '5. 

Equations requiring Subtraction. As an example consider 
the equation w + *5 =71. Let w represent the weight of a bag 
of cement, as before; then this equation is represented by Fig. 21. 
On the left-hand scale pan we have the bag of cement (weight 
unknown) together with a weight of 15 Ibs. All this is balanced 
by 71 Ibs. on the right-hand pan. To obtain the bag of cement by 
itself on the left, we must remove the 15 Ibs. from the left-hand 
pan, and to preserve the balance we must also remove 15 Ibs. from the 
right hand. 



SIMPLE EQUATIONS 



167 



Then, 10 + 15 71 

- 15 w + 15 15 = 71 15 

/. w = 56 Ibs., i. e., the sack weighs 56 Ibs. 

Hence when there is a known quantity added to the unknown, 
the latter is isolated by subtracting the known quantity from each 
side of the equation. 




Example 146. Find the value of r from the formula R + r = s 
when R = 199 and s = 256. 



Substituting given 199 + r == 256 

values, 

199 199 + r 199 = 256 199 

r =57 



Proof. 

*99 + 57 = 256 
/. Result is correct. 



Example 147. If w + w l = w. it find the value of w i when w = 21*7 
and w 2 = 22-63. 

Substituting, 21*7 + w l = 22*63 

21-7 21-7 + w l 21*7 = 22-63 2I *7 



Proof. 2 1 -7 + -93 = 22-63, /. Result is correct. 
In practice the second step may be shortened thus 

21-7 4- a/! == 22-63 
21*7 t^ = 22-63 217 

= -93 

Equations requiring Addition. Consider the equation 
w 10 == 46. If w again represents the weight of our bag of cement, 
this equation is seen in Fig. 22. On the left-hand scale pan is a 
bag of cement (weight w when full) from vrtiich 10 Ibs. of cement 



i68 



ARITHMETIC FOR ENGINEERS 



have been taken out. This partly filled bag is found to weigh 
46 Ibs. Now let us supply 10 Ibs. to the left-hand pan to make up 
the weight of a full sack ; then 10 Ibs. must be added to the right- 
hand pan to keep the balance. So that 

w 10 = 46 

+ 10 W 10 + 10 = 46 + 10 

/. w = 56 Ibs. = weight of sack. 

Hence when there is a known quantity subtracted from the un- 
known, the latter may be isolated by adding this known quantity 
to each side. 





lOlbs has been 
fa Ken owV of .Hn 
bag. (We9hr 
when full -It/) 




4-6 Ibs 



U/ -1O = 

UT * 

* 

Fig. 22. 



4-6 
4-6 
56 



Example 148. Solve the equation T 35 

T - 35 = 81 

+ 35 T - 35 + 35 = 81 + 35 
.*. T = 116 



= 8i. 

Proof. 

116 35 = 81 
/. Result is correct. 



Example 149. Find the value of T from the equation T 2 73 = 131. 



+ 273 



T - 273 

T - 273 + 273 

T 



131 + 273 
404 



Proof. 

404 - 273 = 131 
Result is correct. 



As before, the work may be shortened by combining steps 2 and 3. 
Thus, T 273 = 131 
+ 273 T = 131 + 273 = 404. 

The operations in the two previous sections may be summed 
up in the following rule : 

Any number may be added to or subtracted from either side of an 
equation provided that the same number is also added to or subtracted 
from the other side. 



SIMPLE EQUATIONS 169 

Thus if A = B 

then A + x = B + x). . , 

j A ^ r where x is any number 

and A - x & x) J 

The operations may be conveniently dealt with in a more me- 
chanical manner. Consider the shorter form of working in the four 
previous examples 
In Ex. 146, the 199 was + when on the left, and when on the right 

147, 217 + 

148, ,, 35 ,i ,, + ft 

149, 273 + 

It is noticeable that in each case the sign becomes changed. Hence : 
Any term (i. e., set of symbols or numbers between a pair of + and 
signs) in an equation may be written on the other side if its sign be 
changed. This is called "transposing." Thus in Ex. 148, the 35 is 
transposed, i. e. t it is written on the other side with an altered sign. 

Exercises 43. On Equations requiring Addition 
and Subtraction. 

Solve the equations 

1. x -f 5 = 12. 2. p -}- 147 = 96. 3. h -f -015 = 37. 
4. 275 -f L = 1236. 5. / -f 1200 = 6250. 

6. 11-5 + D = 23-85. 7. Y -f i = 7'7- 

8. n -f -333 = 1-667. 9. -004 -f C = -00415. 

10. The formula H = S -f #L refers to wet steam. Calculate the 
value of S when H = 1115, x = -895, and L = 889. 

11. The formula W = H -f Cr refers to the energy given out by a 
cell. If W = 2-7, f = -028 and C = '5, calculate the value of H. 

12. The formula D = Cs + k is used when measuring distances 
with a telescope. Find the value of k when D = 166-2, C = 100, and 
s = 1-65. 

Solve the equations 

13. m 5 = i'i. 14. x 21 = 27-85. 
15. H - 34 = 2-5. 16. Y - -15 = -16. 

17. y -006 = 'ooi. 18. D -05 = 1-13. 

19. P 14-7 = 21-3. 20. L -105 = 125-605. 

21. / -55 = ~ -ox. 

22. The formula u = v at is used in mechanics. Find the value 
of v when u = 25, a = 3-8, and t == 7. 

23. From the formula L -f- "]t = 1114, find L, the latent heat of 
steam if t = 250. 

Equations combining the Four Rules. The equation 
usually met with requires two or more of the foregoing operations. 



170 ARITHMETIC FOR ENGINEERS 

Example 150. Find the value of a from the equation v = u at, 
when v = 6-5, u = 33, and t = 3-5. 

Substituting values 6-5 = 33 3*5# 

Transposing 3-5 a = 33 6-5 

26-5 
-3'5 a = -^j- = 7-57 

Proof. Left-hand side = 33 3*5 x 7-57 

= 33 26-5 = 6-5 

/. Result is correct. 

Note. Before the value of a can be obtained the 3-5 a must be isolated, 
which is done by transposing. This also gives a plus sign to the term 
containing the a. 

Example 151. Solve the equation 27-6 -f '45 a = 13 

27-6 + -45* = 13 
27-6 -450 = 13 27*6 = 14-6 



Note. After transposing, the right-hand side, when evaluated, 
becomes minus. Then, dividing across, we have a minus divided by 
a plus quantity. Remembering the points in Chap. Ill, the result is 
minus. 

Example 152. If r is the internal resistance of a battery, the 
number of batteries connected in parallel, and R the external resistance, 
then the total resistance R! is given by the formula 



Find the value of r if Rj = 1-07, R = -45 and n = 4 
Substituting the given values 
1-07 = '45 + 



- -45 l'0 

X 4 '62 X 4 = r = 2-48 

Note. The term ~ is first isolated by subtracting the term 45. 
Then is converted into r by multiplying across by 4. 

Example 153. If F is a temperature on the Fahrenheit (English) 
Thermometer, and C the corresponding temperature on the Centigrade 

(French) Thermometer, then F = - C -f- 3 2 Find the Centigrade 
reading when F = 212. 




SIMPLE EQUATIONS 171 

Substituting 212 = - C -f 32 

32 212 32 = - C 

X - 180 X ^ = C 

9 9 

C = IPO 

Example 154. If p is the pitch (in inches) of a riveted joint, d the 
rivet diameter, and -n the efficiency of the joint, then 2-Z '-- = TJ. 
Find the value of /> when d = -75 and rj = -7. 



Substituting given values, 

X /> - -75 = '7 X p. 

Transposing /> -yp = -75 

Simplifying -3^ = -75 

^7 e 

-4- 3 p = -*- D = 2-5 inches 

Note. Since p is the only denominator we multiply through by p, 
the fact that there is also a p in the numerator making no difference. 

Exercises 44. On Equations combining the Four Rules. 

1. If F is the Fahrenheit temperature, and C the corresponding 
Centigrade temperature, then F = -C -f 32. Calculate the tempera- 
ture C when F = 86. 

2. If t is the thickness of a boiler flue and L the maximum un- 
supported length, then according to the Board of Trade rule : 
L i2o/ 12. Find the value of t when L = 3'-6". (Note. Work in 
inches.) 

3. The formula H = S -f #L refers to saturated steam. Calculate 
x, the " dryness fraction" when H = 1113, S = 338, and L = 876. 

4. The formula B = H -f 4*! refers to magnetism. Calculate the 
value of I when B = 18300, H = 22*5 and * = 3-14. 

5. From the formula v = u + at, important in mechanics, calculate 
a, if v 47, u = 5, and t = 28. 

6. The formula R -f = S relates to electric batteries in parallel. 
Calculate the value of r when R = 15, n = 10, S = 17-5. 

7. If 25 = 14-6, calculate the value of /. 

8. In connection with wire ropes we have the formula D = + yd. 

If D = i J and d = '064, find the value of n. 

9. When turning soft cast iron the formula v ~ 115 13000 may be 
used, where v = cutting speed in ft. per min. and a = area of cut in sq. 
ins. If v is to be 70 ft. per min., what will be the value of a ? 



I 7 2 ARITHMETIC FOR ENGINEERS 

10. The expression p A -f- ^ relates to thick cylinders. Calculate 
the value of B when p = 750, A = 700, and r = 6*5 . 

11. The formula c 2 = Shr 4/1 2 refers to a circular arc. Calculate 
the value of r when c 15 and h 4-7. 

12. The tapping size of a Whit worth bolt is given approximately 
by the formula d = -QD -05, where d is tapping size and D = out- 
side diameter. A certain bolt must be ia6* at the bottom of the 
thread (*. e., d = i-i6). Calculate the outside diameter of the bolt to 
the next largest sixteenth. 

13. The formula v* = w 2 -f 2as refers to moving bodies changing 
their speed. Calculate the value of a when v = 3500, u 3750 and 
s = 30. 

14. The latent heat of ammonia is L = 566 8/, where t Fahren- 
heit temperature. Calculate the temperature at which L = 502. 

15. Using the formula in Ex. 14, above, calculate t when L = 595. 

16. The formula L = 176 -27* gives the latent heat of sulphur 
dioxide, where / == temperature F. Calculate the value of / when 
L = 160. 

17. The equation nr wR = o gives the condition for the maximum 
current from a certain arrangement of cells. Calculate R when r = 2-5, 
n = 5, and m = 7. 

18. Using the formula in Ex. i, calculate the temperature C when 
F = 25. 

19. The formula ad = c refers to the relative cost of gas and 

B 

electricity as lighting agents, where a = cost of electricity in pence 
per B.O.T. unit, and b = cost of gas in pence, per 1000 cu. ft. If a 4, 
d = 3j, c = ij, and e = 2, find 6. 

20. The formula * == -oooiPD -f -i refers to steam boilers. Calcu- 
late the value of P when t = -625 and D == 42. 

21. The formula s = i -\ is given in connection with helical 

springs, to allow for reversing loads. Calculate the value of n when 
s = 3'7- __ 

22. The formula -. = a, relates to bodies in motion. If v = 
15.5, / = .75 and a = 76, calculate the value of u. 

- 4 

23. The formula ^ = n relates to the efficiency of heat engines. 

If n is to be 33 and t is 520, what must be the value of T ? 

j? v 

24. Calculate the value of E from the formula C = -, which 

refers to a dynamo supplying current, when V = 108, C = 215, and 
r = '028. 

25. Using the formula in Ex. 6, above, find n when S = 31-3, 
R = 30, and r = 6'5. 

26. The formula g = -- refers to a " Porter " governor. 

Calculate w when H = -33, N = 270, and W = 54. (ffin/. First 
evaluate the left-hand side.) 



SIMPLE EQUATIONS 173 

Equations with Several Terms. In some calculations 
equations are obtained with several terms on each side, with the 
unknown quantity appearing in more than one term. In these 
cases all the terms containing the unknown are collected to one 
side, and all those containing the known to the other side. On 
simplifying, the resulting equation is quite simple. 

Example 155. Solve the equation 3* = 255 4* 

3* = 255 - 4* 

Transposing, 3* -f 4* = 255 
Collecting, jx 255 

+ 7 ,-553-36.4 

Explanation. In this case the unknown x occurs on two sides of 
the equation, and hence either the 3* or the 4* must be transposed 
to the opposite side. By transposing the latter, the sign of the unknown 
is kept positive. After transposing, the " like terms " 3* and 4* are 
" collected " as in Chap. Ill, p. 134, reducing the equation to a simple 
form. 

Proof 3* = 255 4* 

.'. 3 X 36-4 = 255 4 X 36-4 
109-2 = 255 145-6 
109-2 = 109-4, and result is satisfactory. 

In this and other cases the two sides do not prove exactly equal. 
This is due to limiting our calculations to three significant figures 
but the agreement is sufficiently close to prove our method correct. 

Example 156. In an experiment to find the specific heat of copper 
the following equation was obtained : 430005 = 3960 -f 15105. Find 
the value of s. 

43000 5 = 3960 -f 1510 5 

Transposing, 43000 s 15105 = 3960 
Collecting, 41490 5 = 3960 



5 = - = .0955 



Example 157. Solve the equation 5 -f- R =: 15 3R. 

5 + R = 15 - 3 R 
Transposing, R + 3R 15 5 
Collecting, 4R = 10 

TD I0 

-5-4 R= 7 = 15 

Note. Two changes are necessary here to collect the two terms 
with the R on the left and the numbers upon the right. 



I 7 4 ARITHMETIC FOR ENGINEERS 

Example 158. The equation 300 -f 883* = 1148 + 1-4 occurred 
in a boiler test, " x " representing the dryness of the steam. Calculate 
the value of x. 

300 -f 883* = 1148 -f i4 
Transposing 883* = 1148 -f 1-4 300 

4-883 - = 84 



Example 159. Find the value of n from the formula e(m a) = 
a, when e = 1-33, m = o, and a = i. 

Substituting the given values 

i-33(o i) = i 
Simplifying 1-33 X i = i 

Simplifying 1-33 = n i 

Transposing I 1-33 = n 



After substituting, the bracket (o i) is reduced to i. 
The rest is simple if the multiplication of unlike signs be remembered. 

Equations containing Brackets. These may be solved with 
the help of the laws in Chap. Ill, p. 143, for removing brackets from 
expressions. After any necessary removal or insertion of brackets, 
the equation can usually be dealt with by some of the foregoing 
methods. Any bracket which contains figures only may be re- 
moved by simplifying these figures; removal by algebraic means 
is then unnecessary. 

Example 160. Solve the equation 65 4 s = 7(3 + *) 

65 - 45 = 7(3 + s) 

Removing brackets 65 45 = 21 + 75 
Transposing 65 21 =75 + 45 

Simplifying 44 = n 5 

44 

-n -" = i 

Explanation. First remove brackets on right-hand side, by multi- 
plying every term inside the bracket by the 7. Then transpose the 
known values to the left, and the unknowns to the right, thus keeping 
the sign of the latter plus. Then proceed as usual. 

Example 161. The following formula occurs in heat experiments : 
W(T /) = w(t /J. Find the value of / when w = 71-5, T = 95, 
W = 8-3, and /, = 15. 



SIMPLE EQUATIONS 175 

W(T - t) = w(t - /,) 

Substituting given values 8*3(95 /) = 71*5^ 15) 
Removing brackets 788-5 8-3* = 71 -$t 1072-5 

Transposing 788-5 + 1072-5 = 71 -5* -f- 8-3* 

Collecting 1861 = 79-8* 

-79-8 < 

Note. In removing brackets care must be taken to multiply all 
the inside quantities by the outside number. 

Example 162. The formula V = R __ gives the " velocity ratio " 

of a Weston pulley block, when R = radius of large pulley and r = 
radius of small pulley. Find the value of R when V = 25 and r = 2. 

Substituting given values 25 = ^ 

J-Y ~ 2 

X (R 2) 25(R 2) = 2R 

Removing brackets 25R 50 = 2R 

Transposing and simplifying 23R = 50 



Note. Multiply through by the denominator R 2, inserting 
brackets to ensure correct multiplication. 

Example 163. Find the value of a from the formula e = n ~~ ~ 

16 m a 

(used in toothed gearing), when e = - , n = 55, and m = o. 

Substituting given values, = _ J ZT 

Cross-multiplying i6a = 5(55 a) 

Removing brackets i6a = 275 50 

Transposing 275 = i6a 5a= na 

~ 2 75 
-f- ii a = -jj> = - 25 

Example 164. The equation 450 7-50; 7-6(45 -f w) = o refers 
to a problem on a loaded beam. Determine the value of w. 

45 ~ 7'5 W - 7' 6 (45 + ) = o 

Removing brackets 450 7-5^ 342 y-6w = o 
Collecting 108 15-1^ = 

Transposing- 108 = 15-1 w 

108 
f- 15-1 w = Y 5 ~i = Zlii 

Note. Change the sign when removing the bracket. 



176 ARITHMETIC FOR ENGINEERS 

Example 165. Solve the equation 15 (# i) 4(# - 3) = 

2(7*5 + *) 

I5(* i) - 4(x - 3) = 2(7-5 + *) 

Removing brackets 15* 15 4* -f 12 = 15 + 2* 
Transposing 15* 4* 2* =154-15 12 

Collecting 9# = 18 

4- 9 # =2 

Exercises 45. Equations requiring Removal or 
Insertion of Brackets. 

1. Solve the equation 5(3* 4) = 10. 

2. The equation n = -65(15 n) occurred in a calculation on a 
reinforced concrete beam. Find the value of n. 

3. Find the value of R from the equation 20 (R 2-5) = 2R, which 
refers to a Weston pulley block. 

4. Find the value of a in the equation 22(1-5 ~ a ) 3(4 a ) 
which occurred in a problem on epicyclic gearing. 

5. Find the value of a in the equation -85(1 a) = 2 a, 
referring to toothed gearing. 

6. The formula L -f T t = W(/ ^) occurs in connection with 
jet condensers. Calculate the value of / when L = 1015, T = 142, 
W = 28-5, and fj = 54. 

Solve the equations 

7. 3* 2(5 2*) = 24. 8. n 2(n + i) = 3 5. 

9. 3* - (* + 3) = i - 4(7 - 2/). 

10. The equation 8i(/ 3000) = 20-25(7 -f 3) occurred when 
finding the stress in the metal of a hydraulic cylinder. Find the value 
of the stress /. 

FA 

11. The formula = W(H -f A) relates to forces caused by shock. 

Calculate the value of A when F = 11200, W 56, and H = 18. 

12. Find the value of r from the equation 2(7 5) = 3 4- 5(2 f). 

13. The formula w(L -f- /) = W(T /) deals with a mixture of ice 
and water. Calculate the value of t when w = 28-5, W = 397, L = 80, 
and T = 18. 

14. The equation 130$ = 3(i9'5 c \ + 5 6 (7*5 ~ <0 occurred 
in a calculation to find the voltage drop in an electrical conductor 
supplied at both ends. Solve for c. 

15. The formula $G(a -f b) = h(a -f zb) relates to the strength of 
masonry dams. Calculate the value of a when G = 15*6, h = 33, and 
6- 15- 

C* I Q 

16. The formula J = m is used in connection with galvano- 
meter shunts. Calculate the value of S when G = 54 and m = 20. 

i* ^ 

17. The formula g = H , refers to an economises If g = -15, 
T = 250, and H 1185, calculate the value of /. 



SIMPLE EQUATIONS 177 

18. The formula e = - applies to epicyclic trains of gear 
wheels. Calculate the value of n when e = , a = 30 and m = o. 

19. Using the formula in Ex. 18, calculate the value of n when 
e ss -25, m = 5, and a = 1*4. 

Solve the following equations ; 

20- j^* = - -5- 21. =^ = 3-5- 

22. Calculate the value of/ from the equation {. -- = , which 

7 H / - 95 9 

relates to the strength of a hydraulic cylinder. 

23. The formula R = J ~ refers to clearance in a steam engine. 
If R = 2 53, and r = 2-67, calculate the value of C. 

24. The formula C = ~* r"~ relates to the electric current supplied 

R -f- nr r * 

by a number of cells connected in series. Calculate the value of n 
when C = 8, E = 1-08, R = 5, and Y = 1-23. (Give the calculated 
result and the result to the next whole number.) 

25. The formula -/-- = r- relates to the strength of reinforced 

d n t 

concrete beams. Find the value of n when d = n, m = 15, c = 600, 
and t = 16000. 



26. The formula w = ->- - .- relates to the weight of built-up 

girders. If w = 13-3, W = 95, / = 76, / = 6-5 and r = 14, calculate 
the value of the ratio c. 

Equations containing Brackets : Cases where Removal 
can be avoided. Any equation containing brackets can, of 
course, be dealt with by their removal as in the previous section. 
In certain cases, however, by retaining the brackets, the working 
is easier. This is true for those equations where the unknown 
only occurs once, that once being inside the bracket. The following 
examples will illustrate. 

Example 166. An experiment to determine the latent heat of 
steam (in metric units) produced the following equation : 5i-9(L -f- 60) 
= 1190 x 26. Calculate the value of L. 

5i'9(L -f- 60) = 1190 x 26 

T , f 1190 X 26 _ 

~ 51-9 L -f 60 = ~" 5I . 9 ~- = 596 

Transposing L = 5 96 60 = 536 

Explanation. Note that L only occurs once, and is inside the 
bracket. Dividing across by the coefficient 51-9 the bracketed quantity 
(L -f- 60) is isolated, and since there is a + s ig n outside, the bracket is 
not necessary. Continue as usual. 



178 ARITHMETIC FOR ENGINEERS 

For comparison, the same example is shown worked by removing 
the brackets in the ordinary way 

5i-9(L -f- 60) = 1190 x 26 
Remove brackets 5i'9L + 3114 = 30900 
Transpose 5i'9L = 30900 3114 

= 27786 (say 27790) 

A L -?gf -536 

Note that by the first method the working is simpler, and the 
operation of multiplying the 60 by the 51*9 is saved. 

Example 167. The Fahrenheit temperature F and its corresponding 
Centigrade temperature C, are connected by the equation C = -(F 32). 

A refrigerating machine keeps a cold store at a temperature of 8C; 
what is the equivalent F reading ? 

Substituting given value 22 = --(F 32) 

X* _".?..,_ 

Simplifying 39-6==F 32 

Transposing 39-6 -{- 3 2 F 

F = 7-6 

Note. Multiplying across by - isolates the bracketed quantity. 

Example 168. The formula wjL -f- (t 32)} = W(T t) occurs 
when rinding the latent heat of water in English units. Solve for L 
when w = '125, T = 54, W = i-i, and t = 37. 

Substituting values, -I25{L + (37 32)} = 1-1(54 37) 
Simplifying- *I25{L + 5} = i-i x 17 

T , 18-7 

-r 125 L + 5 = -^ = I49'5 

Transposing L == 149-5 5 = 144-5. 

Example 169. Find the value of $ from the formula 

W(T-*) = (w l + ws)(t-t l ) 
when, 
W = 25 w l = 68-3 w 15-5 T = 91 / = 36 *! = 16-3 



Substituting values 25(91 36) = (68-3 -f 15-5 s)($6 16-3) 

Simplifying 25 x 55 = (68-3 + 15-55) x 19-7 

-19.7 ^==68.3 + 15.5 5 

Simplifying and transposing : 69-8 68-3 = 15-55 

1-5 == i5-5 
-r 15-5 = ~ - '0968. 



SIMPLE EQUATIONS 179 

Exercises 46. Equations with Brackets where Removal 

is not needed. 

1. A formula for the piston speed of a petrol engine is 333 (r -f 2) = V, 
where r = stroke-ratio and V = speed in ft. per min. If V is to be 
1 200, calculate the value of r. 

2. The expression w(L -f t) W(T t) refers to a expression on 
the latent heat of water. Calculate the value of L when w = 120; 
W = 452, T = 30-5 and t = 7-5. 

FA 

3 The formula = W(H + A) relates to the strength of bodies 

under impact. Calculate the value of H when F = 12-6, A == -27, 

w = - 5 . 

4. Find the value of the Fahrenheit temperature F from the formula 
R = ^(F 32) when R = 57'5. (R is Reaumur temperature.) 

5. Using the formula in Ex. 4, above, calculate the value of F when 
R = - 10. 

6. The formula c 2 = h(D h) refers to circular arcs. Calculate the 
value of D when c = 3-78 and h = -77. 

7. In heat experiments we have the formula H = (W + ">)(T /). 
If T = 60, t = 23, W = 435, and H = 18650, find w. 

8. The formula = h(zr h) applies to the measurement of flat 

4 
circular arcs. Calculate the value of r when c = 3-5 and h = -43. 

9. The formula Z = h\K + ~J is connected with the strength of 

built-up beams. If Z = 176, h = 27, and a = 13-5, calculate the 
value of A. 

10. Using the formula given in Ex. 9, above, calculate the value 
of a when Z = 255, A = 7-5, and h = 33. 

11. The formula R = R (i + refers to the increase in resistance 
of an electric conductor with a rise of temperature. In a test on a 
sample of aluminium wire R = -327, R = -42 and t = 65. Calculate 
the value of o. 

12. The formula H = -2d 2 (r -f i) gives the horse-power of a petrol 
engine when d = cylinder diameter in inches, and r = stroke-ratio. 
Calculate the value of r when H = 5 and d = 3-3. 

13. The expression w(T t) = (W + e)(t ^) occurs in heat 
experiments. In a particular case w = 96-8, W = 278-5, T = 96, 
t = 35, / x = 15*5. Calculate the value of e. 

14. p = f(i gc 2 ) is Johnson's formula for columns and struts. 
Calculate the value of g when p = 4-73, / = 7, and c = 45. 

15. In a test to find the latent heat of water we have the statement 
w (t T) = u^L -f (T 32)}. Find the value of L when w 20, 
i 60, T 45, w l i. 



i8o ARITHMETIC FOR ENGINEERS 

16. The formula c = ~~J~T:> relates to electric batteries. If c = 2, 
E = 2-7, and r = '032, calculate the value of R. 

17. The formula p = ~ , . relates to the allowable pressure on 

^ 60 v + 20 r 

slides. If p is to be 300 what is the value of v ? 

18. The formula j-j 1 = -^ G T ' is used when measuring high 

resistances. In a particular test D l = 135, D 2 = 207, R = 100,000, 
RI = 30,000, G = 4150, and r = 7000. Calculate the value of x. 

Fractional Equations. Certain equations contain terms in 
the form of vulgar fractions, more or less complicated, in which 
the unknown may occur as part of the numerator, denominator, or 
of both. 

To explain the method let us solve the equation 

2 + "J ~~ 2~ 

Considering the left-hand side we have to add two terms which 
inform are exactly like vulgar fractions : the fact that they contain 
an unknown quantity need not create any difficulty, and in fact 
these terms can be treated exactly as fractions. It is best, how- 
ever, to work according to the method already treated in this 
chapter when clearing an equation of quantities below the line. 
Thus, if we multiply both sides of the equation by the product of 
all numbers below the line, in this case 2 X 5 X 2 = 20, we can, 
by cancelling, clear the equation of fractions. 

Thus- | + ~ = ^ 

10 4 10 

20* , 0X2* v6 X SQ 

X/}f* ' A \_ ~ V_ _ . ^ _ ~ A 

zu v. -p v ~ \j " 

io# + 8* = 3-6 X 10 
Collecting iSx = 36 



Proof. Substitute the value 2 for x in the given equation 

Then- \ + 2 - * 2 = 3 ; 6 

/. i -f -8 = 1-8, which is true, and 
* the result is correct. 



SIMPLE EQUATIONS 181 

The second step, with a little practice, may be omitted. Thus, 
when multiplying by 20, the process is carried out mentally in the 
following way : 

v 

First term, 20 times - 2 into 20 goes 10 times 



O V 

Second term, 20 times 5 into 20 goes 4 times. 



10 times x = IQX. Write IQX. 

5 into 20 goes 4 times. 

4 times 2% = Sx. Write Sx. 



Other side, 20 times 2 into 20 goes 10 times 
10 times 3-6 Write 10 x 3-6. 

In the examples worked, this multiplying step will be shown in 
detail, but will be enclosed in brackets, thus [ ], to indicate that 
it is for explanation only. 

It will be seen that in the next example we multiply by 12 in- 
stead of by the product of 6 and 4, although the latter would give 
the same result. Multiplying by a lower number that will still 
cancel the 6 and the 4, we get figures that are easier to work with. 
Obviously, then, we multiply by the L.C.M. of the numbers below 
the line. 

Example 170. Solve the equation ---- = n. 

7 a * 
6-4"" 

X L.C.M. of 6 and 4 - 12 [&-*!? _ &? = X .12"] 

L ^ \ i _J 

.'. 140 3 = ii x 12 

Collecting na =11x12 

. -B-^-,, 

Note. When multiplying across, the 1 1 must be included, although 
it has apparently no denominator. Its denominator is really i, as shown 
in the second step, but, of course, has no effect on the figures and so 
may be omitted in the working. When the term or terms on the other 
side are also fractional, then the L.C.M. taken must be that of all the 
denominators, as in the next example. 

Also in dealing with a term in which the numerator is composed 
of two parts, e.g., 25__5^ ft must be carefully remembered that the 
line of division performs the same office as a bracket, the expression 



182 ARITHMETIC FOR ENGINEERS 

0# tf 

merely being a shorter statement of -- +--, as used in ordinary 

vulgar fractions ; and when multiplying, both the quantities on top 
of the line must be multiplied. 

Example 171. Solve the equation ~ 3 "t-5 __ j 

3* + 5 ^1 , i 

7 23 

_ _ _ _ . r~ 6 21 14- 

X L.C.M. of 7, 2, ^ 



[b 21 14-1 

W X (3* -I- 5) _^* ,. VJ 
X "" * + *~J 



and 3 = 42 

.'. l8,Y -j- 30 = 2I# -f- 14 

Transposing 30 ~ 14 = 2ix i8x 
Simplifying 16 =3* 

^ 3 *^J& 

Note. After cancelling the left-hand side, remember that both 
the terms 3* and 5 are to be multiplied by the 6, giving i8x -f- 3. 
It may assist beginners if a bracket is put round such terms as the 
3* -f 5. Then, on multiplying, the form 6(3* + 5) is obtained, which 
shows more clearly that both the terms have to be multiplied. This 
becomes very important when minus signs occur; see Ex. 173. 

Example 172. Solve the equation -^-"^5 + 5 ^ p _j_ 5 ~ P^ 



) + X 5 _. Ig , 

/. i2/> + 10 + 30 = iBp -f 5 - p 
Transposing 10-1-30 5 = i8p p i2p 
Collecting 35 = $p 

-7-5 P^T. 

Explanation. The L.C.M. of the denominators being 18, we multiply 
throughout by 18. In the second step the compound numerators are 
shown closed in brackets to show that they must both be operated on. 
Remember to multiply the first p on the right-hand side. 

When the signs between the terms are minus, great care must 
be exercised, as a change of sign occurs. The following example 
will illustrate : 



SIMPLE EQUATIONS 183 

Example 173. Solve the equation - * 

J> - 3* + 3 ___* 

6 5 " 3 

X L.C.M. of 6, 5, $q X 5 W3* + 3) _ ^ X 

and 3 = 30 L ^ ^ ^ 

/. 25 18* 18 = 10* 

Transposing 25 18 = IO,Y + 18* 

Collecting 7 = 28* 

28 x J 8 - = J or '25 

Explanation. Only the second term on the left side calls for any 
comment. On multiplying across by the 30 this term becomes 
313* -r 3^ anc i on cancelling reduces to 6(3* + 3). On re- 
moving the bracket, as the sign outside is minus, the + sign is changed, 
and we have i8# 18, as shown in the working. 

An unknown in the denominator should create no difficulty; 
the equation is cleared of fractions in the same way. 

Example 174. If r and 5 are two electrical resistances connected 
in parallel, and R is the equivalent resistance, then these quantities 

are connected by the formula j = -p . Find the value of r when 

s 6 and R = 2. 

Substituting given values ?. -|_ JL = L 



X L.C.M. of 6, 2, and 

r = 6r 



/. 6 + r = 3f 

Transposing 6 = 37 r = 

^2 '=5=3 



Proof. 
i , i 



2 2 

Example 175. Find the value of m from the equation i = 






X L.C.M. = fm, |5W $m x = \m X-| 
L ^* ^J 

5tn 10 = 2m 

Transposing 5m 2m = 10 
$m = 10 

-r 3 = = 3'33 



i8 4 ARITHMETIC FOR ENGINEERS 

Exercises 47. On Fractional Equations. 

Solve the following equations ; 

A 2 . a , 
1. a -\ -- = 6-5 
3 ^ 5 D 

3 * ""- 1 - * ~ 2 = * ~~- 
234 

5. r - i = ^=-7 6. * - 2 + ** ~ i + *- 

^4 



9 . -- 
2 



2 





i n n 





2 3 




2 C 


" 


~ 5 ~ 




d 2 -f- 3<i 





2 5 


8. 


? 2 __ 30 




4 


10. 


A + 2 __ 223 




3 


19, 


JK _,_. 2 2 


A*?. 


# 3 



5 2 5 

V V 

13. The formula = C is used when measuring electrical 

resistances. Find R if V = 35, C = 7, and r = 590. 

14. Using the equation in Ex. 13, find V if R = 6-7, r 535, and 
C - 5-4- h a 2b 

15. Calculate the value of a from the formula g = . ? "^" ^ , 

referring to the strength of dams; when g = 7-3, h = 18, and 6 = 7. 

16. The formula ^ = ** """ * refers to the refraction of light 

at spherical surfaces. If v = 15, u = 20, and r 10, find the value of p. 

Equations requiring Square Root. Although quadratic 
equations are not taken in this book, there are some simple equations 
containing squares which may be solved without any further know- 
ledge. Such equations are those in which the square is the only 
power of the unknown which is present, as in the following examples. 
Briefly the method is to solve the equation as though # 2 (say) were 
the unknown quantity. Arriving at some numerical value for x 2 , 
the value of x is then easily obtained. The equations may be of 
any of the forms in the foregoing sections, and exactly the same 
methods will be adopted to determine the value of the square of 
the unknown. 

Example 176. The formula F = -00034 WRN 2 is used in connection 
with engine governors. Find N when W = 3, F = 69, and R = 6. 

Substituting values 69 = -00034 x 3 X '6 X N a 

-r -00034 x 1-8 N 2 = ^ = 112700 

Taking V N = ^112700"= 336 revs, per min. 

* a is a Greek small letter " theta." ( A is a Greek capital " delta." 
1 /u is a Greek small letter " mu." 



SIMPLE EQUATIONS 185 

Explanation. Treating N 2 as the unknown, the equation is divided 
across by the coefficient of N 2 , leaving N a alone. Taking the square 
root of the value of N 2 gives the value of N. 

Example 177. If h ft. is the "height " of an engine governor and N 
its speed in revs, per min., then h = -2^-. Find the speed when the 
height is 0-35 ft. 

Substituting -35 = ** 

X N 1 '35N 2 = 2938 

-35 N' 



Taking \/~~~ N = ^8400 = Qi-6 revs, per min. 

T> 

Example 178. The formula q A -- g~is used in connection with 

thick cylinders. Find the value of r when q 1120, A = 560, 
B = 5040. 



Substituting given values 1120 = 560 - 

Transposing - \ - 560 + 1120 = 560 

X r 1 54 56or 2 

3040 _ H . 9 

Taking ^^~ r = Vg = 3 

Note. Here it is advisable to transpose first, thereby reducing the 
size of the figures. 

Example 179. In connection with the strength of materials we 

- W 2 yr 

find the formula ^ - = W 2 . Find the value of m when E l = 13400, 
m "~ m i i 

E 2 = 14400. 

Substituting- _L_ 

X (w a i) m 2 = i-o74(m 2 i) 



Removing brackets w 2 = I-O74W 2 1-074 
Transposing i'74 = i'074m 2 m 2 = -074* 



1-074 
*O74 m 2 = '--*- = 14*5 

/^ . t J 



Taking \/~~ m = ^14-5 = 3-81 



186 ARITHMETIC FOR ENGINEERS 

Exorcises 48. On Equations requiring Square Root. 

c 2 

1. The formula s = relates to manilla ropes, where c = circum- 
ference in inches and s = safe load in tons. Calculate the value of c 
for a rope to support a load of -45 ton. 

2. The formula V = 247^/ 2 relates to the volume of an anchor ring 
of round section. If V = 18-7, and d 6, find t. 

3. The formula / = - refers to the stress caused by centrifugal 
force. If / = 200, w = -83, g = 32 and a = 2-4, find v. 

4. The formula H = -5 relates to engine governors. Calculate 

the value of n when H = "375. 

B 2 A 

5. The pull on an electric magnet is P = in Ibs. In a 

r to 11,180,000 

magnet where P is to be 600 the value of A is 7-2. Calculate the value of B. 

6. For a direct-acting steam pump we have the formula '75D 2 P = d*p. 
If p 150, P = no, and d = 8J, find the value of D. 

4)0? 

7. The formula / =* ~^- relates to the strength of a flat square plate. 

If f = 15000, p = 15, and a = 21, find the value of t. 

8. In the formula given in Ex. 7, calculate the value of a when 
/ = 2800, p = 150, and t = 1-125. 

9. The formula / = j a - refers to the strength of a flat circular 

plate. Calculate the value of R if /= 9000, p = 60, and / = '625. 

10. If a timber beam be loaded at the centre of the span, the breaking 

load in Ibs. is W = , , where b, d and / are the width, depth and span 

in inches, and a = a constant depending on the material. Find depth d 
for a beam 3" wide to support 6000 Ibs. on an 18" span, when a = 19200. 

e 2 L, 

11. The formula h = ~~ji relates to water pipes. Calculate g 

(galls, per min.) for a pipe 4" dia. (= d), and 450 ft. long ( = L) when 
the friction head h = 30 ft. 

t/ 2 

12. The formula f- h = H occurs in hydraulics. If h = 3*35, 

H = 3*53, and g = 32, calculate the value of v. 

13. The formula y = Hj H 2 relates to the expansion of steam 

in a turbine nozzle. If Hj = 1200, H 2 = 1090, g = 32-2, and J = 778, 
find the value of V. 

14. The formula q = A T refers to the strength of thick cylinders. 

If A = 7200, B = 64800, and q = 14,400, find r. 

15. The formula j i = x occurs in some problems in hydraulics. 
Calculate the value of c if x = '0628. 

16. The formula P = ^i- xJ^L is used in reference to the strength 

s o 

of flat stayed surfaces in boilers. Calculate / when P = 180, c = 125, 



SIMPLE EQUATIONS 187 

17. Calculate d t the diameter of cylinder (ins.) required for a petrol 
motor to develop a horse-power (H) of 7-5, using Henderson's formula 
H = -2d 2 (r -f- i). when r = 1*2. 

18. Work Ex. 17, using Poppe's formula H = *8i(d -yg) 2 , i. e. t 
find d when H = 7*5. 

Equations requiring Squaring. Often it is necessary to 
solve an equation containing the square root of the unknown 
quantity, as in the remaining examples in this chapter. These 
are easily solved by previous methods, with only one extra opera- 
tion. Thus, in the formula, V = 8-02 Vh, when V is known, the 
value of Vh can be easily found, and squaring gives the value of h, 
since Vh X "Vh = h. Sometimes the operation of squaring must 
be performed at an early stage in the working, and sometimes right 
at the end, but in all cases the whole of one side must be under the 
root sign before any squaring is performed. 

Example 180. If water falls from a height h ft. it acquires a velocity 
of V ft. per sec. as given by the formula V 8-02 Jh~ Find the 
height to give a velocity of 66 ft. per sec. 

Substituting 66 = 8-02 VJT 



Simplifying Vh =8-23 

Squaring h = (8-23)* = 677 ft. 

Explanation. Here the A//Tmust be isolated before h can be found. 
Hence the equation must be divided by the coefficient 8-02. Then 
squaring the value of V/Tgives the value of h. As a proof, the reader 
should substitute the result in the given formula, and see whether V 
comes to 66, as it should do. 

Example 181. If T is the time in sees, of the double swing of a 
pendulum and / is its length in ft., then T = 27r^/ where ir = 3-142. 
What length of pendulum is required to give a double swing of 2 sees. ? 



Substituting 




Simplifying V --/ = -3183 



32-2 
Squaring ~ 2 ~ = ('S^) 2 = -1013 

x 32-2 / = -1013 x 32-2 = 3-262 ft. 

say, 3 ft. 3i*. 



188 ARITHMETIC FOR ENGINEERS 

Example 182. In connection with water flowing in channels and 



pipes, we have the formula c \ where g = 32. By experiment 
we find that c = no. What is the corresponding value of /* ? 



Substituting 


110 = ^ 


/^ -5 . 

A* 




Squaring 
Simplifying 


HO 2 = 
I2IOO = 


2 X 32 




u 

6 4. 


X M 

-r I2IOO 


12 1 00 /A = 
p = 


_ 6 .4 _ 

I2IOO 


OO529 



Note. Squaring is the first operation that must be performed as 
the whole of the right hand side is under the root sign. 

Example 183. Find the value of Y from the formula h V 2 * 2 5 ~ y2 
(which relates to an engine governor) when h -833. 

Substituting -833 = \/2 T 25 r 2 

Squaring '694 = 2-25 r 2 
Transposing r 2 = 2-25 -694 = -556 

Taking V r = >/-~55 6 "= '7\ 6 - 

Note. This example involves both squaring and square root. To 
obtain Y we must know f 2 ; and to obtain this we must know the value 
of 2 '25 r 2 . Therefore the first operation must be to square each 
side. Then r 2 is obtained by transposition, and r follows from a square 
root. 

Exercises 49. On Equations requiring Squaring. 

1. The formula w = 1000 Vd refers to roller bearings for long bridges. 
Calculate the value of d if w = 2500. 

2. The formula d = i^VjTrelates to riveted work. Find the value 

of/if <*= 

3. Calculate the value of r from the Lanchester motor rating formula 



H = -4^V^when H = 5-5, and d = 3-75. 

4. If T = 2ir\/ calculate the value of g when T = 1-82, / = 2-73 

o 

and TT = 3*14. 

5. The formula B =\/- applies to curves on tramway tracks. 
If B = 6, and t = 1*375. calculate the value of R. 

6. Calculate the value of i from the formula v = cVmi, which 
relates to the flow of water, when v = 3-8, c = 95, and m = 1-7. 



SIMPLE EQUATIONS 189 

7. In the formula given in Ex. 6, calculate the value of w if v = 6, 
c = 120 and i = -00045. 

VTJ" 
-r- refers to reinforced concrete beams. 

Calculate the value of /* when B = 190,000, b = 10, and/ = 15. 

/~~2^sP 

9. The formula p \J r> ~~ relates to helical tooth gears. Find 

R if p = 2j, P = 5450, and k = 1075. 

/P" 

10. The formula T = V i refers to the strength of locomotive 

fireboxes. If T = 8, calculate the value of P. 

Vx~ 

11. In connection with slide valves we have the formula = = n. 

i- V* 
If n = 2-41 calculate the value of x. 

12. In connectionjwith the strength of thick cylinders we have the 
formula -^ = \/y^-?- Calculate / when D = 4, d = 3, and p = 1500. 

13. Calculate the value of p in the foregoing formula when D = 7, 
d = 4, and / == 12000. 

14. A formula connecting the cylinder diameter ^of a steam engine 

with the indicated horse-power is d 2O$\/ -TJJJ- Calculate L if 
d = 12, I.H.P. = 37, /> = 40, and N = 180. 

15. The formula A = h 2 \ -, -61 gives the area of a segment 

of a circle. Calculate D if A = 8-83 and h = 2-21. 

Exercises 49a. Miscellaneous Examples on Equations. 

1. For a compressed gas cylinder the formula p(R t) ft is 
applicable, where p = internal pressure, R = outside radius, / stress 
in the material, and t thickness. Calculate the necessary value of 
t for a case where p = 1*5, R 3-5, and/ = 8. 

2. The wave length of an oscillatory electric circuit is given by the 
formula X = ^\/LC where L = inductance, C = capacity, and k a 
constant. Calculate the value of C when k = 1884, L = 130 and A 
is to be 500. (X is the Greek small letter " lambda lf .) 

3. For the Pitot tube used for measuring the velocity of a fluid 
the fundamental formula is p = Jpf 2 . For a particular case p = 

W 

2490 h, p = 0-73, and v = -0268 . Substitute these values in the 

fundamental formula and reduce all the constants to give a new formula, 
giving h in terms of W and a. (p is the Greek small letter " rho ".) 

4. The inductance of a circular solenoid of circular section is given 
by the formula L = -0126 w 2 [R VR 2 fl 2 ], where n = number 
of turns of wire. Calculate the value of n required if L is to be 100, 
and R = 4*5 and a = 1-5. 

5. Referring to the formula in Ex. 4, calculate the value of a when 
L = 150, n =E 220 and R = 4*8. 



CHAPTER V 
TRANSPOSITION OF FORMULAE 

The General Solution. The process of solving an equation 
is not always confined to cases like those in Chap. IV, where a 
numerical answer is obtained. In many cases an equation is com- 
posed entirely of symbols, and the solution is a more or less com- 
plicated algebraic expression. Such an equation is called a literal 
equation, and the solution is a general solution, i. e. t one that will 
suit for all numerical values of the quantities involved. 

The applications of this are as follows : Consider a formula 
such as H = 1082 + 3/. Frequently several values of H are 
known, and it is required to find the corresponding values of /. 
If each value of H be substituted in turn in the given formula, and 
the resulting equations worked out as in Chap. IV, a considerable 
amount of time and space will be occupied in repeating a number 
of similar statements. This can be saved if, in calculating the 
various values of /, all the algebraic operations be performed before 
any substitution. Thus from the above formula it can be calculated 

that t = ^^ , and all that is necessary is to substitute each 

o 

value of H in turn, and to evaluate the right-hand side in each 
case (see Ex. 186, p. 192). Often in the process of finding the 
general solution some of the symbols and figures may cancel (see 
Ex. 187, p. 192), thus reducing the amount of arithmetic necessary 
in the final substitutions. 

Again, when studying problems in some branch of science, it 
is constantly necessary to find a formula by which a particular 
quantity can be calculated. An examination will show some con- 
nection between the various quantities involved, as, for example, in 
the case of a boiler, where we can determine the relation prl = ftlrj. 
If it is necessary to calculate / for any particular conditions a formula 

-far 
for t can be obtained from the above relation, i. e. t t = ~. The 

W 
190 



TRANSPOSITION OF FORMULA 191 

operation of finding this formula from the known conditions is 
purely an algebraic one, and is independent of the values involved. 

The name " transposition " will be given to the process of finding 
a new formula for some particular quantity contained in a given 
formula. In the majority of cases it will be found better to trans- 
pose the given formula before substituting numerical values. 

The Unknown Quantity. Considering an equation such as 
ap + r = /, we cannot merely say " solve this equation/' as there 
are four symbols whose values are unknown. Hence it is necessary 
to specify the particular symbol whose " value " is desired. Strictly 
speaking a " value " means a numerical value, but it is quite evident 
what is meant. Thus suppose a formula for p is required from 
the statement ap + * = f> we can express ourselves in several ways, 
e. g. : " solve for p in the equation " ; " find the value of p in the 
equation " ; " find a formula for p " ; or " transpose for p." 

Some difficulty may be experienced at first in distinguishing 
the unknown quantity from the remaining symbols, especially if, 
in the course of working, the unknown occupies different positions. 
To assist in this direction the unknown will be shown in heavy type 
throughout the working of most of the examples in each section ; 
the student, in his first few cases, can put a ring round the unknown 
quantity. 

The methods of working in this chapter are practically identical 
with those in Chap. IV if the unknown be regarded as a symbol, 
and all the other symbols as numbers, and treated accordingly : 
also the order of treatment is similar. For the first example of 
each type a numerical equation will be shown as an illustration : 
and proving these will readily show that the method is correct. 

Equations requiring Division. As with similar cases in 
Chap. IV, the equation must be divided across by the quantity 
which multiplies the unknown. 

Example 184. If P represents the pressure of water in Ibs. per 
sq. ft., and p the pressure in Ibs. per sq. in., then i^p = P. Find a 
formula for p. 



I44P 



* '44 



Similar Numerical Example. 



3* = 15 



+ 3 



e. In the numerical case one other step is possible, i. e. t the 
evaluation of . This, of course, cannot be done in the left-hand 
case until the value of P be known. 



192 ARITHMETIC FOR ENGINEERS 

Example 185. If E = primary voltage, e = secondary voltage, 
C = primary current, and c = secondary current in a transformer, then 
neglecting losses ec = EC. Transpose for c. 

ec = EC 
EC 

* * ~ T 

Example 186. If D = the diameter of a belt pulley in feet, N = 
revs, per min., and V = velocity of belt in ft. per min., then V = 3-I4DN. 
Find a formula for D, and calculate the values of D when V has the 
values 800, 1000 and 1200, N being 200. 

3-i 4 DN = V 

V 





-r 3'MN 




D 


(i) V 


= 800 .*. 


D = 


3-14 X 2^ 


(2) V 


= IOOO 


D = 


IOOO 
3*14 X 2OO 


(3) V 


= 1200 


D = 


I2OO 


3'14 X 2OO 



= 1-273 ft 

= 1-592 ft. 
= 1-91 ft. 

Note. It should be easily seen that the multiplier of the unknown 
D is 3-I4N. Hence the equation is divided throughout by this quantity. 

Example 187. The relation -d* = irdl occurs when finding the lift 

of a mushroom valve, d being the valve diameter and / its lift. Find 
a formula for /. What should be the lift of a 3" valve ? 

- - ;* 

+ '-"-Si 

Cancelling, / = 

4 

When d = 3" I = ^ 

4 

Explanation. Divide across by ird to get / on left-hand side by 
itself. The expression is then cancelled by -nd, leaving a very simple 
result. It is evidently better to calculate the lift from the final expres- 
sion than to substitute for and d in the given equation and then 
work out. 

Equations requiring Multiplication. Equations where 
the unknown is divided by a number, or symbol, require to be 
multiplied by this number or symbol to isolate the unknown. 



TRANSPOSITION OF FORMULAE 193 

TT 

Example 188. The formula = p connects the pressure p of 
water in Ibs. per sq. in. with the head H in ft. Find a formula for H. 

Similar Numerical Example. 



2-3 



X 2-3 H 



x 

2 = * 
X 2 # = 4X2 



Example 189. If Q = the quantity of electricity in coulombs, and 
/ the time in sees in which it flows, then C the current in amperes is 

given by C = y . Transpose for Q. 

Q -C 

7 ~ 
x * Q = a 

w 

Example 190. With an alternating electric current p = y^r where 

p = power factor, W = true power (watts), E = volts, and C = current. 
Find a formula for W, and find the values of W: (a) when E = no, 
C = 35, and p = *8i ; (6) when E 6000, C == 23, and p = *739 

--p 

EC " p 

X EC W = pEC 



(a) W = -81 X no X 35 = 3120 watts 

(b) W = *739 X 6000 x 23 = 102000 watts. 



Exercises 50. On Equations requiring (a) Division, 
(b) Multiplication. 

(a) DIVISION. 

1. Transpose for a in the formula F = Ma. 

2. The formula w = eCt relates to electrolysis. Transpose for e. 

3. In connection with the expansion of gases we have the formula 
Pt/ = pV. Transpose for P. 

4. In Exs. 40, No. 12, we have the formula V = wDN. Find a 
formula for N. 

5. Transpose for R 4 in the equation R^ == R 2 R 3 , which relates to 
the measurement of electrical resistances. 

6. In Exs. 40, No. 14, we have the formula P = f yS^d 2 p. Transpose 
for p. 

7. Transpose for / in the formula pr = ftrj t which relates to steam 
boilers. 

8. With reference to an alternating current the formula W = ECp 
occurs. Transpose for C. 

9. Solve the equation t; 2 = 2gh, for h, this occurring in hydraulics. 

10. In Exs. 40, No. 17, we have the formula *75D 2 P = d 2 p. Solve 
for P. 



194 ARITHMETIC FOR ENGINEERS 

11. Transpose for s in the formula H = ws (T *), which relates 
to problems on heat. 

12. Lewis's formula for the strength of wheel teeth is L = c X p X 
w X F. Find a formula for w. 

(b) MULTIPLICATION. 

T? 

13. If R = 7*- transpose for E. 

Ly 

M 

14. The statement / = 7 refers to stress in beams. Find a formula 

for Z by cross-multiplying. 

15. If Q is the actual discharge through a nozzle, and Q l is the 
theoretical discharge, then the " coefficient of discharge " c = ~ . Find 
a formula for Q. 

16. The formula *r r t^ refers to levers. Solve for E. 

W D 

17. The formula V = ~- relates to an electrically charged sphere. 

ftr 

Transpose for Q. 

18. Transpose for v in the formula c __ which relates to the 

clearance in gas-engine cylinders. 

A* TTr-^t f . ti TVX u j- No. of teeth _. , 

19. With reference to gear wheels: Pitch dia. = - ~j .- -, Find 

a statement for " No. of teeth." 

H W 

20. The equation = ^ occurs in connection with engine governors, 

and F = 'OO034WRN 2 . Substitute this value of F and transpose 
for H. 

Equations requiring Multiplication and Division. To 

obtain a formula for the unknown the equation must be multiplied 
by the quantities dividing the unknown, and divided by those 
multiplying it. The operation may be done separately if so desired, 
but very little practice should enable the student to combine the 
two operations. 

Example 191. If p = the number of pairs of poles in an alternator, 
n = the number of revs, per sec. and / = the frequency of the current, 



then p = /. Transpose for />. 



Similar Numerical Example 



pn 

2 - = 

2 *" 9 

X 2 



TRANSPOSITION OF FORMULA 195 

PD 
92. The formula S = - oc 

boilers, using metric units. Transpose for P. 



PD 
Example 192. The formula S = - occurs in connection with 



2t 

x D 



Example 193. If c*r = f -J R, find a formula for R. 

, (c\ 2 T> 

c*r = (- 2 )R 

c 2 
Simplifying cV = R 



Cancelling by c* /. R = 4^. 

e. The resulting formula is far easier to use than the original 
expression. 

Example 194. In an investigation concerning the arrangement of 
the cells in an electric battery the following equation appeared : 

^- = : and it was desired to find the value of p. Transpose for p. 
R 

E 2 p _ w 

" R 



Explanation. The unknown being p, the equation must be divided 
by E* and multiplied by ^Rr ; combining the two operations the equa- 



tion is multiplied by -g. R is then cancelled. When the result 

contains both numbers and symbols it is usual to place the numbers 
in front. 

Example 195. The Board of Trade formula for steel tube- plates 
in boilers is p = 2OO L """- ', p being working pressure, T thickness, 

W width of combustion chamber, D pitch of tubes, and d inside diameter 
of tubes. Transpose the formula for the thickness T. 

20oooT(D d) 

WD 
X WD 20oooT(D d) 

4- 2Oooo(D d) 




196 ARITHMETIC FOR ENGINEERS 



e. The bracketed quantity (D d) should be looked upon as 
a single symbol or number; there is no need for removing the bracket. 
The operations of multiplying and dividing are done separately in this 
example as the expression has rather a heavy appearance. 

In the following example the unknown is below the line, but 
exactly the same principles apply. 

Example 196. The capacity of a plate electric condenser is given 

k A 
by the formula K = ^. Transpose for /. 

,, __ kA Note. Clear equation of 

"~~ 4ir* fractional appearance by cross- 

Cross-multiplying, 4?r/K = A multiplying, which brings un- 

__ A known to left-hand side. Then 

~ r ^ ~~ 4rrK proceed as usual. 

Exercises 51. Multiplication and Division combined. 

1. The Board of Trade Rule for Morrison Furnaces is P 



where P is working pressure, T is thickness, D is diameter. Find a 
formula for T. 

EC 

2. In Exs. 42, No. i, the formula -- = K appears. Transpose 

forC. 

3. The horse-power taken by a D.C. motor is given by the formula 

EC 

-p = H, where E = volts, C = current in amperes, and H = horse- 

740 

power. Transpose for C. 

4. The expression W = 2-71* refers to the weight of air.* Solve 

for p. 

Hdf 

5. In Exs. 42, No. 8, we have the formula t = . Solve for d. 

6. In connection with the horse-power transmitted by belts the 

SV 
formula H = ---- occurs. Transpose for S. 

7. The formula Z = -^- occurs in connection with beams. Transpose 
for b. 

8. Solve for S in the formula H = j^ Q t which refers to the rating 



of petrol engines. 

9. Transpose for d in the formula W = C-^-, which refers to the 
weight of chimneys. 

* T is a small Greek letter "tau." 



TRANSPOSITION OF FORMULA 197 

10. In Exs. 42, No. 10, the formula H = . occurs. Transpose 

for CT. 

R L 

11. In a law applying to wedges we have p = -. Obtain a formula 

for P (i. e. t cross-multiply). 

12. The formula V = refers to volume of air. Solve for p. . 

2'7I/> 

kA. 

13. Find a formula for k from : K = ., which gives the capacity 

4** 
of a parallel plate condenser. 

Lv a 

14. Transpose for d in the formula h =* f~^- which refers to pipes 

conveying water. 

15. In Exs. 42, No. 5, the formula d*l ^,~. Transpose for c. 

16. Transpose for w in the formula W = w x 6 ~ *-. 

2 R 2 

17. In Exs. 42, No. 13, we have the formula/ = - -> p. Solve for p. 

18. Transpose for h in the formula H= ~^~jr which relates to 
the height of chimneys. 

tzT<* 

19. The formula C = r> i relates to electric batteries in series. 

Find a formula for E. 

20. Transpose for / in the equation = / 6 , which relates to 
beams. 

Equations requiring Addition and Subtraction. When 
an equation contains terms separated by + and signs, then 
these terms may be transposed as desired provided that the signs 
are changed. 

Example 197. If t = Fahrenheit temperature and T = absolute 
Fahrenheit temperature, then / + 460 = T. Transpose for /. 



t + 460 = T 
Transposing the 460, t = T 460 



Similar Numerical Example. 
x + 2 = 5 
.'. *= 5 - a 



Example 198. If p the pitch of a riveted joint, d = dia. of 
rivet, and w effective length between rivets, then p d w 
Transpose for p. 

p d w 
Transposing d p = w + d 



1 9 8 ARITHMETIC FOR ENGINEERS 

Example 199. The formula v = -f at occurs in connection with 
velocity calculations. Find a formula for u. 

u -f at v 
at u = v at 

Note. The term " at " is considered as one symbol and taken to 
the other side with a change of sign. 

p 
Example 200. If H h , transpose for h. 



+ h H = + A 

P P 



Explanation, The unknown has a sign, and if the H be taken 
across a formula would be found for h. But since transposing a 
term changes its sign, the unknown can be made -f by taking it to 

p 
the other side. Then transpose the term ^. 

Exercises 52. Equations involving Addition and 
Subtraction. 

1. Transpose for p in the formula P = p + 14-7. 

2. The formula H = S -f #L refers to wet steam. Transpose for S. 

3. Solve for / in the formula T = / -f 273. 

4. In the formula B = H -f 4*1 transpose for H. 

5. The formula M 2 = Mj + ^M refers to properties of sections. 
Transpose for Mj. 

6. Solve for R in the formula R -f - = S. 

n 

7. Transpose for c in the formula ad -- c (see Exs. 44, No. 19). 

> 

8. If E = the E.M.F. of a battery, e = volts lost internally and 
V = " potential difference," then E e V. Transpose for e. 

9. Transpose for H in the formula W = H -f Cr. 

10. The formula u = v at is used in mechanics. Transpose for v. 

V V 

11. Transpose for C in the formula ^ = C -- . 

12. Solve for A in the formula -y = A 4- ^-. 

n o 

Equations requiring the Four Rules. Where various 
combinations of multiplication, addition, etc., occur, then two 
main operations are necessary : 



TRANSPOSITION OF FORMULA 199 

1. Transpose as in the last section to obtain the value of the 
term containing the unknown. 

2. Divide or multiply this result by the quantity which 
respectively multiplies or divides the unknown, as in the first 
section. This may be seen by the working of a simple numerical 
example 

2X + I = 7 

_ ( transposing to find the value 

I 2X ' 7 """"" I i c 

' \ Of 2X 

_._ __ TJ _i f dividing by 2 to obtain the 

2 \ value of x 

The method will be exactly the same if the 2, the i, and the 7 
in the given equation be replaced by symbols, % being considered 
as the unknown throughout. 

Thus, let the 7 be replaced by a symbol, say r t 

Then 2x + i = r 

i 2x r i 

r i 

* /7 y __ ______ 

-r 2 X - - - 

Now let the i be replaced by a symbol, say d, 

Then zx + d = r 

d 2x = r d 

r-d 
-ra * = 

Finally replace the 2 by a symbol, say a, 

Then ax + d = r 

d ax =r - d 

r-d 
~ a x = ----- 



Example 201. If D = dia. of a Whitworth bolt, and d == dia. at 
the bottom of the screw thread, then the following relation is approxi- 
mately true : d -gD -05. Transpose for D. 



-f -05 
-r -9 




200 ARITHMETIC FOR ENGINEERS 

Example 202. The Fahrenheit temperature F, and the Centigrade 

temperature C, are connected by the formula F = -C -f- 32. Transpose 
for C. 5 

|C + 32 = F 

- 3* *C = F - 32 

X C 5(F -32) 

Explanation. The value of -C is obtained by transposing the 32. 

Now the equation has to be multiplied by 5 and divided by 9 to get C ; 
and the whole of the right-hand side must be so treated. Therefore a 

bracket is first placed round the F - 32 and then the - written outside 
to conform to this. 

Example 203. Let d = dia. of a punch, / = thickness of plate, and 
D = dia. of hole in the bolster ; then D d -}- ^. Transpose for t. 

, t _ D Note. After transposing the 

6 ~~ d, the whole must be multiplied by 

_. ^ D d 6; a bracket must therefore be 

~~ 6 ~~~ placed round the D d to indi- 

X 6 t = 6(D d) cate this. 

Example 204. The formula i + 4 7I "K = A* is used in connection with 
magnetism. Transpose for K. 

I -f 47rK = /i 
I 4?rK = JJL I 



Example 205. The formula c 2 = 8hr ^h 2 relates to circular arcs. 
Transpose for r. 

Shr 4/i 2 = c 2 
+ A Shr = c 2 + 4 ^ 



Exercises 53. On Equations combining the Four 

Rules. 

1. Transpose for / in the formula L = 120^ 12. 

2. The formula W = ^GaH + 7 was obtained from tests on a 
Diesel engine. Transpose for H. 

3. Find a formula for q when n = 1-035 + !$, a statement relating 
to the expansion of steam. 



TRANSPOSITION OF FORMULAE 201 

4. The formula K = 48 -f ^Y relates to the consumption of power 
in a textile factory. Transpose for Y. 

5. The formula / = i'54PD -f 2-6 relates to the strength of steam 
boilers. Transpose for P. 

6. The formula F = -R + 32 connects Fahrenheit and R6aumur 

4 
temperatures. Find a formula for R. 

7. Find a formula for n from the equation R + r = S. 

8. In Exs. 44, No. 9, the formula v 115 i3Ooa occurred. Find 
a formula for a. 

9. In the formula L = 566 8/, relating to the latent heat of 
ammonia, transpose for /. 

10. Transpose for / in the formula 25* =* 14-6. 

11. The formula D = CS -f K is used when measuring distances 
with a tacheometer. Transpose for C. 

12. Find the value of a from the formula v = u -f at. 

13. In Exs. 44, No. 17, we have the equation nr wR = o. Solve 
for r. 

14. Solve the equation in Ex. 13 above, for R. 

15. Find a formula for I from the equation B = H -f- 4* I, which is 
used in magnetism. 

16. Find a formula for a from the equation v a = w 2 -f- 2 as, which 
relates to velocity problems. 

17. Transpose for r in the formula / a = 6-$pr + i-8 a . 

18. Find a formula for n from the equation s = i H which 

relates to helical springs. 

19. Transpose for r in the formula R -f - = S, relating to electric 
batteries in parallel. 

20. The formula D = 1- yd occurs in connection with wire ropes. 

Transpose for n. 

TD 

21. Transpose for B in the formula p = A H ^ which relates to 
thick cylinders. 

22. In Exs. 44, No. 19, we have the formula ad = c. Solve for 6. 

c 

A 

23. The formula i y- = n, relates to epicyclic gearing. Transpose 

for A. 

E c 

24. Transpose for E in the equation c = . 

Equations with Brackets. Where removal is necessary, or 
desirable, the removing will generally reduce the example to an 
equation similar to those in the last section. When the unknown 
is multiplied by some symbol (as by c in the following example), 
then it is usually advisable to remove the brackets. 



202 ARITHMETIC FOR ENGINEERS 

Example 206. The formula p = /(i ec) is used in calculating the 
sizes of struts and stancheons. Transpose for e. 

p=f(i -ec) 

Removing brackets, p = / fee 

Transposing, fee = / p 

+> -v 

Where the unknown stands alone as a term in the bracket then 
it is usually better not to remove the brackets. By multiplying or 
dividing, the value of the bracket is obtained. The brackets can 
then be left out, without altering the value of the expression, and 
the example then usually appears as those in the last section. The 
following two examples illustrate : 



Example 207. If R is the temperature on the R6aumur ther- 
>meter, and 1 
formula for F. 



mometer, and F that on the Fahrenheit, then R = -(F 32). Find a 

x ' 



|(P - 32) = 



X-J 

4 



32 F = ?R + 32 

4 



Example 208. The following formula is met with in connection 
with plate-web girders Z = h\K -f >) Transpose for A. 



A +-= 

_ a . __ Z a 

~ 6 A " /i~~ 6 

Exercises 54. On Equations involving Use of Brackets. 

[Note. In many cases the removal of the brackets will not be 
necessary. See above.] 

1. Transpose for T in the formula H ws(T /), which relates to 
heat experiments. 

2. The formula P = C(A <r + wA,) refers to reinforced concrete 
columns. Transpose for A c . 

3. Transpose for D in the formula c 1 = h(D h), which relates to 
circular arcs. 

4. In Exs. 46, No. i. the formula 333(7 + 2) = V occurs. Transpose 
for r. 



TRANSPOSITION OF FORMULAE 203 

5. The formula V(R 2*5) = 2R refers to Weston pulley blocks. 
Find a formula for R. 

c t 

6. Transpose for r from the equation = h(2r h). 

7. Solve for w in the formula H = (W -f w)(T /), which refers 
to heat experiments. 



8. The formula C = -d(W -f w) relates to experiments on the 

" calorific value " of fuels. Transpose for w. 

9. Transpose for F in the temperature conversion formula : 

C = 5{F- 3 2). - 

10. In Ex. 22, No. 9, the formula S = 600 (r -f- i) occurs. Trans- 
pose for r. 



11. Lloyd's Rule for Fox Furnaces is P = 

P = working pressure, D = greatest diameter, and T = thickness. 
Transpose for T. 

(T _ T )S 

12. The formula H.P. = --- - refers to transmission of power 

by ropes. Transpose for T 1B 

13. In Ex. 46, No. n, we have the formula R = R (i + at). Find 
a formula for a. 

14. In Ex. 46, No. 12, we have the formula H = '2d*(r -f i). 
Transpose for r. 

15. Transpose for g in the formula p f(i gc*). 

Wh 

16. In the formula P = ^7^ , < transpose for d. 

b(d 4-1) 

17. Find a formula for a from the equation Z = h(^A +-f J. which 

refers to the strength of plate girders. 

18. Find a formula for t from the equation in Ex. 7 above. 

19. In connection with the " wire test " of Whit worth screw threads 

we have the equation -^r^-^r^ *4 6l 7- Solve for d. 

Gases requiring the Insertion of Brackets. When the 
unknown symbol occurs in more than one term in the given formula, 
then usually brackets will have to be inserted, the unknown being 
taken out as a common factor, as on p. 147. 

Example 209. The formula W = cA -f mca refers to a reinforced 
concrete column. Transpose for c. 

c\ -f mca = W 
Taking out common factor c t c(A -f ma) = W 

W 

+ (A + ma) C = AT^ 

Explanation. The unknown occurs in two terms, hence its value 
cannot be found directly. But if the common factor c (the unknown) 



204 ARITHMETIC FOR ENGINEERS 

be taken out as shown, the bracketed quantity may be regarded as a 
single symbol multiplying the unknown, and by which we can divide 
across. 

Example 210. Transpose for n in the equation d w = /. 

d a = nf 

-f n d = nf + n 

Inserting brackets, = n(f -f i) 



Note. Transpose unknown from the left so that all terms containing 
it are on the right. Take out the unknown as a common factor and 
divide across by the bracketed quantity. 

Example 211. If em ea = n a, transpose for a. 

em ea = n a 

Transpose, em n ea a 

Taking out common factor a, == a(e i) 

. N em n 

T- (e - i) T^HE" " * 

There are other equations, of fractional form, which usually 
involve the removal or the insertion of brackets. The unknown 
may be in the denominator or in the numerator, or in both, and 
the equation is usually first cleared of fractions by multiplying 
through by the denominator (or the L.C.M. of the denominators, 
if there are several of these). 

Example 212. When two electrical resistances, r ohms and s ohms, 
are connected in parallel, then the equivalent single-resistance R ohms 

is given by the equation g = | --- . Find a formula for R. 



Note. 



= _ 

R rs 

Cross-multiplying, rs = R(s + r) 



__ 

- 



.--, 

Explanation. The right-hand side of the equation is the addition of 
two vulgar fractions. Add these by the rules of Chap. I. The L.C.M. 

IS If 

of r and s can only be rs, and therefore = and ~ = . Cross- 

i rS S rS 

multiply, put brackets round s and r to take the place of the horizontal 
dividing line, and divide by (s -f *)> giving the formula for R. 



TRANSPOSITION OF FORMULAE 205 

Example 213. Transpose for n in the formula - = x. 



X (n i) a = x(n i) 

= xn x 
Transposing, x = xn /i 



"Example 214. Transpose for /> in the formula \. , = s, a formula 
in connection with compound stresses. 



p + 9 

X (p + q) P - q = s(p + q) 

=* sp + sq 

Transposing, p sp sq -f q 

Inserting brackets, p(i s) = q(s -f- i) 

- (I - S) p - *<*-+-!) 

v ; ^ i s 

Explanation. Multiply across by the denominator p + g, placing 
brackets round the p -\- q, since the whole of this is to be multiplied 
by s. Then remove the brackets algebraically, so that all the terms 
containing p can be brought to the left, and the others to the right. 
Then insert brackets and divide across by i s. 

Example 215. When a number of secondary cells are connected 

wE 

in series we have the formula C = ^ - - , where C = current in 

K -f- nr 

circuit, n = number of cells, E = electro-motive force of one cell, 
7 = internal resistance of one cell and R = external resistance. Find 
a formula for n. 

c _ "E 
- R + nr 

X (R -f nr) C(R -f- nr) = nE 
CR 4- Car = nE 

Cwr CR = /:E - C/j;- 

= 11 (E - O) 

-ME-0) iS> = " 

Exercises 55. Equations requiring Insertion of Brackets. 

1. Transpose for e in the formula em -f a ti + ea. 

2. Find a formula for A (delta) from the equation = W(H -f- a) 

which relates to the strength of bodies under shock. 

3. Transpose for n in the formula C(nr -f R) = nE. 



206 ARITHMETIC FOR ENGINEERS 

4. Find a formula for T from the equation T / = ??T. 

5. Transpose for d in the formula d -f k(t d) = N, which relates 
to plate condensers. 

6. Transpose for t in the expression w(L -f t) = W(T t), which 
relates to the latent heat of water. 

fid 

7. In connection with wire ropes the formula D = -- \- jd occurs. 

Solve for d. 

8. Transpose for S in the formula G -f- S = Sw, which relates to a 
galvanometer. 

9. Find a formula for n from the equation n k(d n). 

10. Transpose for t in the formula g(H t) = T t, which relates 
to economise rs. 

11. The formula S = --- - refers to the strain on a chimney 
due to wind pressure. Transpose for C. 

12. Transpose for w, in the formula v l = - ; - which relates to 

^ l l m -\- m { 

momentum (or quantity of motion). 

13. Find a formula for a if e - : a gear-wheel formula. 

m a & 

14. Transpose for a if g = - . a ^ 2 - 

b R 2 r a 

15. Solve for /> in the formula = j>a"H"~a which refers to the 

stress in thick pipes. (Hint. Cross-multiply.) 

16. The equation h e = \e occurs in a problem on hydraulics. 
Find the value of e. 

Equations requiring Square Root. Many cases occur where 
the unknown appears in an equation as a square, and the equation 
is solvable by using the methods of the true simple equation with 
the one addition of taking a square root. Generally the equation 
should be worked as though the square of the unknown were the 
unknown quantity. When a value is obtained for the square, then 
by taking the square root of each side of the equation the value 
of the actual unknown is obtained. In no case should the root 
be taken until the whole of one side of the equation is contained 
under the index. 

Example 216. If V is the velocity of the wind in miles per hour, 
and P the pressure per sq. ft. caused by it on a flat surface, then the 
following formula (found by experiment) connects P and V ; P = KV 1 . 
Find a formula for V. 

P= KV a 



Taking V 



TRANSPOSITION OF FORMULAE 207 

Explanation. Consider (V*) as the unknown, and divide the equa- 
tion by K. Then take square root of each side, obtaining a formula 
for V. The whole of the left-hand side must be written under the root 
sign. 

Example 217. If C is the circumference of a white manilla rope in 

C 2 

inches, and S the safe load in tons that it can carry, then S = - . 

* 30 

Transpose for C. 

~~ 30 Note. Multiply by 30 to find 

X 30 3oS = C a value of C 2 . Then take V~~ of 

Takine V~ VS - C wh le ' giving value f C ' By 

id , 3 __ , separating the 30 and S as shown, 

C = ^3 **> A/30 is reduced to a definite figure. 
= 5-48 VS' 

Example 218. In connection with rope-driving we have the formula 

wv 2 
f . Find the formula for v. 

*^ ~g* 

x & fa = v 1 

w w 

Taking V~~ \^ = v 

Example 219. If H is the " height " of an engine governor in feet, 
and N its speed in revolutions per minute, then H = -Cff-. Transpose 
for N. 



X N* HN* =-- 2938 

H N* - 

' W ~ 

Taking V- N 



= . 

VH VH 

Note. After taking the root, the expression is simplified by taking 
roots of both numerator and denominator. The numerator being a 
number, its root may be evaluated, as shown. 

Example 220. If a circular plate be supported all round its circum- 
ference and be uniformly loaded, we have the following formula : 



208 ARITHMETIC FOR ENGINEERS 

2 R f 

/ = -- -pp t when / = stress in plate, R = radius of plate, / = thickness, 
and p = pressure. Find a formula to give thickness t directly. 



x / <*/ = - R'/> 

- 

Taking V- t = 



.-. t = -816 R \/| 

e. In clearing of fractions here, it is only necessary to multiply 
by /*. Using the whole denominator 3* 2 would necessitate dividing by 
3/to find the value of t 2 , bringing the 3 back to its original place. The 
final expression is much simplified by taking separate roots in the last 
steps, as shown. 

Example 221. The formula ^ i = x relates to the flow of 
water through orifices. Transpose for C. 

T % 

C 2 

4-1 A * + J 

X C 1 i = C 2 (* + i) 

+ (* + u Fi~- c ' 



Taking 

"Explanation. As the left-hand side contains two separate terms, 
first transfer the i to the other side, so as to find the value of the term 
containing the unknown. Then proceed as usual, taking care to insert 
the brackets in the third step. 

Example 222. Transpose for x in the formula (x -f i) 1 = c. 

(X + !) = C _ 

Taking \f~~ x + I = Vc 

I x = Vc I 

Note. As the whole of the left-hand side containing the unknown 
is squared, the square root is taken first of all. 



TRANSPOSITION OF FORMULAE 209 

Example 223. The Board of Trade formula for flat surfaces with 
screwed stays (over 6" pitch) is P = ^ *' : where P = working 

o ~~" O 

pressure ; C = a constant ; t = thickness of plate in sixteenths of an 
inch; and S = surface supported in sq. ins. Find a formula for t. 

_ C(t + i) 
r S - 6 
X (S - 6) P(S - 6) = CU f- i) 1 

- C (S - 6) = (t + i) 

V _ 

r(S 6) == t -f- I 



Note. The first two steps can be combined by multiplying by 
- ~ - ; they are separated here, as the expression is rather heavy. 
Note in last line that the i must not be included under the \/ sign. 

Equations requiring Squaring. Certain cases, where the 
unknown appears under a square root sign, may be solved by 
considering the square root of the unknown as the unknown, 
solving for this by previous methods, and finally squaring to give 
the true unknown. The squaring must only be performed when 
the whole of one side of the equation is under the square root sign. 

Example 224. The formula w = loooVd relates to the rollers for 
expansion bearings of large bridge girders, d being the diameter in ins., 
and w the load in Ibs. per in. of length. Transpose for the diameter d. 

w = loooVd 

-f- 1000 = Vd 

1000 

/ 
Squaring 



1,000,000 

Explanation. Divide across by 1000, as the value of Vd is first 
required. Square each side to give value of d. The formula may be 
simplified by squaring the top and bottom separately. 

Example 225. If B = the greatest permissible wheel base in ft., 
R = radius of quickest curve in ft., and T = width of rail slot in ins , 

on an electric tramway track, then B = \ . Find a formula to 

P 



2io ARITHMETIC FOR ENGINEERS 

give the radius of the quickest curve for a given wheel base and width 
of slot (t. e. find a formula for R). 



2RT Note. As the whole of the 

Squaring B J = right-hand side is under a 

3 B 2 X 3 __ p root sign, squaring is the first 

X 2T ~~ 2T ~ ~~ operation. 



Example 226. With the usual pitch of 4" for the screwed stays in 
locomotive fire-boxes, T = \/ -- i, where T = thickness of plate in 

sixteenths of an inch, and P = working pressure in Ibs. per sq. in. 
Transpose for P. 



._. Note. The i is first 

-j- i T -f- i = \J transposed to get the value 

p 2 of the term containing the 

Squaring (T -f- i) 2 = unknown. 

X 2 2(T -f i) 2 = P 

Example 227. If T is the time in seconds of one swing of a pendu- 
lum and I is its length in feet, then T = 27r\/ , where g is a constant. 
Find a formula for /. 



T . // 

_ 2* ~ r= y ~ 

Squaring ^ - j == 

T a / 

Simplifying ^^~g 

x ^ p = l 

Example 228. The equation C \ relates to problems on 
water supply. Transpose for /i. 



Squaring C* 

X /i CV 



TRANSPOSITION OF FORMULA 211 

Example 229. Transpose for x in the formula C = \ ~!~ 

i -f- x 



. ! Note. Comparing with 

Squaring C = + Example 221, it will be 

X (i -}- x) C 2 (i -\- x) = i seen that the operations 

! are in exactly the reverse 

+ C ' i 4- x = C2 order> 

- i x = i - i 

Exercises 56. Equations involving Square Root and 

Squaring. 

1. Transpose for C in the equation -C 2 R == x. 

E 2 

2. If -j^- = x t find a formula for E. 

3. From the formula E = JIw 2 deduce a formula for . 

4. The formula A = -525 K a refers to egg-shaped sewers. Trans- 
pose for K. 

B 2 A 

5. Transpose for B in the formula P = - 6 -- , which relates 

r 11,180,000 

to the pull on an electric magnet. 

6. Solve for D in the equation -75D 2 P = d 2 p. 

fia z 

I. Transpose for a in the formula / l t which relates to the 

strength of a flat square plate. 

8. The formula W = y occurs in Exs. 48, No. 10. Transpose for d. 

w\J 

9. The formula T = -^j- relates to trolley wires. Solve for L. 

10. Solve for n in the formula H = ^~ t which relates to engine 

governors. 

II. Transpose for i in the formula in Ex. 7 above. 

12. If b = depth and h = length of winding space, and d dia. 
of wire outside insulation, and S = no. of turns, then in a field coil 

S = -, Transpose for d. 



13. The formula I = y refers to fly- wheels. Solve for w. 

V 2 

14. In Exs. 48, No. 13, the formula - -^ = Hj H 2 occurs. Trans- 

pose for V. 

v* 

15. Transpose for v in the formula - - -f h = H. 

16. Solve for r in the formula q = A -- a , which relates to the 
strength of thick cylinders. 



212 ARITHMETIC FOR ENGINEERS 

17. Transpose for d in the formula H = -2d 2 (r -f- i). 

18. Transpose for x in the equation (x -\- i) 2 = 8i/ 2 . 

19. Cooper's formula for the spacing of stiffeners in plate girders is 

/ =_ _ --- 5 ---- Solve for d. 
' 8 



72 

EQUATIONS INVOLVING SQUARING 

20. Solve for # in the equation VAT i = y. 

21. Transpose for m in the formula v = CV;;u, which relates to 
the flow of water. 

22. The formula ^ = 1*2 Vt relates to riveted work. Transpose 
for t. 

/R" 

23. Transpose for B in the formula / = \/ -.- which relates to 

reinforced concrete beams. 

24. Transpose for /* in the previous formula (No. 23). 

25. The Lanchester mptor rating formula is H = -^d 2 V7. Find a 
formula for r. _ 

26. Transpose for N in the formula D = p\, which refers to 

condenser plates. 

/D 

27. The formula d = \/ ^- relates to cone pulleys where ^ = small- 

est diameter, D = large diameter, and R = ratio of cone. Transpose 
for D. 

28. Solve for / in the formula x = \/^~ -', which refers to deflec- 
tion of beams. _ 

29. Transpose for d t in the formula y- = \ ~r 

80. The formula t = 2w\/-p refers to spring calculations. Trans- 
pose for F. 

31. If j = V ?-?. transpose for /. (Hint. Cross-multiply after 
squaring.) 

32. Transpose for p in the previous formula (Ex. 31). 

33. Transpose for C in the formula X = k^/LC, which relates to 
wireless telegraphy. 



34. The equation a VV 2 2 occurs in connection with right- 
angled triangles. Transpose for c. 

35. In the formula in Ex. 31 above, which relates to thick cylinders 
subject to internal pressure, D always equals d -f 2t, where t is the 
thickness of the cylinder. For a particular case d = io. Substitute 
for D and d in terms of t in the equation which is the result of Ex. 31, 
and simplify to give an equation like / = ap, where a is a number. 
(Hint. If D = d -f- 2t, and d = loZ, then D = I2t.) 



CHAPTER VI 
USE OF LOGARITHMS 

Introductory. Much arithmetical calculation can be consider- 
ably shortened and simplified by using logarithms, which are certain 
numbers calculated by mathematicians and entered into tables 
for reference. By their use, the lengthy and laborious operations 
of multiplication and division (which are of very frequent occurrence), 
are replaced by the simpler operations of addition and subtraction ; 
while, in higher branches of calculation, there are certain operations 
which would be practically impossible without logarithms. 

It is not possible, in this book, to explain fully the principles 
underlying logarithms and their uses. The reader must accept the 
statements made until a later stage, but he can, with the aid of 
Chap. Ill, learn the meaning of a logarithm. Briefly, a logarithm is 
an " index." It has already been seen on p. 94, that in statements 
such as io 2 = 100, io 3 = 1000, and io 4 = 10000, the numbers 2 t 
3, and 4 are called "indices," while the quantities io 2 , io 3 , and io 4 
are " powers of io." Now, considering the values of these powers 
as numbers, the indices are said to be logarithms of the numbers, 
while the number io, which appears in all the given statements, 
is called the base of the logarithms. Describing the indices 2, 3, 
and 4 in greater detail, they are " logarithms of the numbers 100, 
1000 and 10000, respectively, to the base io." As many statements 
similar to io 2 = 100 are possible, we can use symbols to give a 
more general form and write 

a* = N 

where N is the number, % is the logarithm and a is the base, and % 
is then " the logarithm of N to the base a." 

Any positive number may be chosen as base. Similarly, the 
symbols % and N may have a variety of values, but with any one base, 
each value of N has a corresponding value of x, or, every number has 
its own particular logarithm. A base having been decided upon, 
it is possible, with the aid of higher mathematics, to calculate to 

213 



214 ARITHMETIC FOR ENGINEERS 

any accuracy desired the logarithms of all numbers to this base, 
and so a " system " of logarithms is constructed. The only system 
considered in this book will be that in which 10 is the base, this being 
the one of greatest practical importance. To keep the size of the 
tables within reasonable limits, and to shorten the actual working, 
only a certain number of significant figures is considered. Thus we 
have four-figure logarithms, which are logarithms calculated to 
four decimal places for numbers containing only four significant 
figures. 

For the great majority of engineering calculations these are 
quite suitable. The actual logarithm table for four-figure loga- 
rithms is quite small, only occupying two pages. For more accurate 
calculation there are 7-figure logarithms which are logarithms given 
to 7 decimal places, having been calculated for numbers containing 
7 significant figures. The logarithm table in this case is about 
100 times larger than the four-figure table, and is consequently 
more tedious to handle, while the working is correspondingly 
longer. 

Of course, when it is desired to find the logarithm of a number 
which is given to, say, 5 or 6 significant figures, then the number must 
first be reduced to 4 significant figures if the 4-figure tables are to 
be used. 

A logarithm (or " log " as it is commonly called) is usually a 
mixed number, such as 3-4512, but it is convenient to consider it 
as consisting of two parts (a) the whole number, (b) the decimal 
fraction.* It is only the decimal portion that is actually found 
from the tables, the whole number being supplied from an examin- 
ation of the number. The whole number may be either + or , 
as will be shown, but the decimal part, as found from the tables, 
is always positive. 

Finding Logarithms. (1) Whole Number. The whole 
number of a logarithm is found by observing the following 
rules : 

1. When the number is 1 or over, then the whole number of the 
logarithm is 1 less than the number of figures (or digits) in front of the 
decimal point, and is plus. 

It has no connection with the actual figures in the number. Thus, 
in 235-9 tne whole number of the log is 2, because there are 3 figures 
in front of the decimal point and 3 1 = 2. 

* In mathematical language the whole number is called the " charac- 
teristic " and the decimal portion the " mantissa." 



USE OF LOGARITHMS 215 

Similarly 

13500 ... 5 figs, in front /. whole number is 4 

29,500,000. . 8 ,, ,, 7 

2240 . . . 4 ,, 3 

14*7' . . . 2 ,, ,, ,, ,, ,, ,, ,, i 

3-142 . . . i ,, ,, o 

(because i i = o). 

2. When the number is less than 1, then the whole number of the 
logarithm is 1 more than the number of noughts between the first figure 
and the decimal point, and is minus. Again, it does not depend upon 
the actual figures m the number. 

Thus in -0952 the whole number is 2 because there is one nought 
between the decimal point and the first figure 9, and 1 + 1 = 2. 

Similarly 

0807 i nought between dec. point and first figure 

.'. whole number is 2 
0000087 ... 5 noughts between dec. point and first figure 

.*. whole number is 6 
000341 .... 3 noughts between dec. point and first figure 

.*. whole number is 4 
.3937 o nought between dec. point and first figure 

/. whole number is i 
(because i more than o is i). 

When the whole number is + , the complete log is written in 
the ordinary way, e. g., 3-2396, since both the whole number and 
decimal are here positive. But since the whole number of the log 
is sometimes , while the decimal is always -f , then the sign 
cannot be written in front of the log in the ordinary way. Thus 
with the number -095, the whole number is 2, and the decimal 
part is + *9777' Now we cannot write 2-9777, because this 
would mean that the whole of the number 2-9777 was negative 
instead of only the 2. Hence, to show this distinction, the minus 
sign is written above the whole number. Thus the log of ^095 is 
written as 2-9777, meaning 2 + -9777. 

Finding Logarithms. (2) The Decimal Part. It must 
first be noted that the decimal part of a log is independent of the 
position of the decimal point in the number ; it depends only on 
the significant figures. Thus the decimal part of the logs of the 
numbers : 39370, 39*37> '03937 are all the same, being -5952 in each 
case. Hence when finding logs from the tables, only the significant 



2l6 



ARITHMETIC FOR ENGINEERS 



figures in the number need be considered, no decimal point appear* 
ing anywhere in a logarithm table. 

The logarithm table is arranged in the following way : the 
extreme left-hand vertical column contains the first two significant 
figures of the number, while the extreme top horizontal column con- 
tains the third and fourth figures. In the body of the table the four- 
figure numbers are the decimal parts of the logs of the first three 
significant figures. The one or two figures on the right of the table 
are " differences " which are to be added to the four-figure numbers, 
to account for the fourth and last significant figure. By using 
these " difference " columns, the table is made as small as possible. 
For convenience the table is split horizontally at about the figures 

DIAGRAM ILLUSTRATING LOGARITHM TABLE. 





3 r Slgnificanr figure of fhe number 


4- rh signif I'cahr- f 13. 


L 
O _ 


O 


i 


a 


3 


*r 


5 


6 


7 


8 


9 


2 3 


^V 5G 


799 


-f.o 




























I" 






















// 




i* 


-1 






















Dif 


t trcn 


C.C. 


O 


Decimal porh'ono of loga^ifhrns -for rhe 


Col urnns. 


1 


frsr 3 significanlr -fiaures. 






These -figures 


c 


Given I'o -4-decimal places. 






robe 


added To 


J_ 






I 














Those 


\n rhe 


cer\^6 


c 

sr> 






















fo accouvilr 


for 


tO 






i 














a 4^ 


Siq. f W 


ure 


(VJ 




























98 




























U 99 





























54 in the left-hand column, and the lower half is placed on the 
right-hand side on the opposite page. The above diagram should be 
compared with the complete table at the end of the book. 

It must be remembered that the figures in the left-hand and 
top columns are not numbers but only significant figures. Thus, 
on the first line in the left-hand column we have 10. This should 
not be read as " ten " but as " one nought," and similarly with the 
other numbers. 

The decimal part is found in the following way : Required the 
decimal of the log of 3937. The significant figures are 3937. Look 
down the extreme left-hand column until the figures 39 are reached 
Pass horizontally across the table until the column is reached which 
is headed by the 3rd significant figure 3. Here we find the figures 



USE OF LOGARITHMS 



217 



5944. Keeping this number marked with the finger or a pencil, 
pass across to the " difference " columns until the column headed 
by the 4th significant figure 7 is reached. Here we find a figure 8. 
These two numbers 5944 and 8 are shown in heavy type and 
underlined in the following extract from the logarithm table. 

EXTRACT FROM LOGARITHM TABLE. 



10 





1 


2 


3 


4 


5 


6 


7 


8 


9 


1 2 3 


456 


789 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


4 9 13 
4 8 12 


17 21 26 
1C 20 24 


30 34 38 
28 32 37 


38 
39 


5798 
5911 


5809 
5922 


5821 
5933 


5832 
C944 


5843 
5955 


5855 
5966 


5866 
5977 


5877 
5988 


5888 
5999 


5899 
6010 


123 
123 


567 
457 


8 9 10 
^ 9 10 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 | 6107 


6117 1 1 2 3 


4 5 6 | 8 9 10 



Then the " difference " 8 is added to the extreme right-hand 
figure of the four-figure number 5944. Thus 

Opposite 39 at side and under 3 at top .... 5944 
., ,, ,, 7 in difference column 8 

Sum 5952 

Then the decimal portion of the log of 39*37 is -5952. The addi- 
tion can with a little practice be done mentally. The whole number 
being i, the complete log is 1*5952. 

As another example take the log of 501 '9. Look down the left- 
hand column until the figures 50 are reached. Pass across to the 
column headed I, when the figures 6998 are reached. In the " differ- 
ence " column under 9 and the same horizontal line we find 8. 

Then opposite 50 at side and under i at top .... 6998 
,, 9 in difference column 8 

Sum 7006 

The whole number being 2 the logarithm of 501*9 = 27006. 

The logs of numbers containing only 2 significant figures, e.g., 
93 ; '067, etc., are found in the column headed o, i. e. t the one next 
to the extreme left-hand column, the " difference " column not 
being required in such cases. 

With those numbers containing only i significant figure, e. g. 
700; -005, etc., a nought should be added mentally to the single 
significant figure to make two figures, as a single significant figure 
is not found in the left-hand column of the table. Then in the case 
of the 700, look opposite 70, and in the case of the '005 look oppo- 
site 50. 



2i8 ARITHMETIC FOR ENGINEERS 

It will be noticed from the given logarithm table (or from 
the extract just given), that in the earlier part of the table there 
are, for each number in the first column, two lines in the " difference " 
column. This arrangement (the copyright of Messrs. Macmillan 
& Co., Ltd.) gives greater accuracy. The logarithm is looked out as 
explained, and the " difference " figure is found in the same line as 
the 3rd significant figure of the number. Thus, for log 1048, follow- 
ing the 4 along horizontally, 34 is found in the " difference " column. 
For log 1078, the " difference " is in the lower line and is 32. 

Example 230. Find the decimal parts of the logarithms of the 
following numbers : (a) 606-5, (&) 6080, (c) 3-009, (d) -001016. 

(a) Opposite 60 at side and under 6 at top we find . 7825 
On same line under 5 in " difference " column . 4 

Then decimal part is -7829 

(b) Opposite 60 at side and under 8 at top, we find . . 7839 
No 4th significant figure .'. Decimal is .... -7839 

(c) Opposite 30 at side and under o at top we find . 4771 
On same line under 9 in " difference " column . 13 

/. Decimal part is '47^4 

Note. The " difference " column has 2 figures which are added to 
the 3rd and 4th figures as shown. 

(d) Opposite 10 at side and under i at top we find . 0043 
On same line under 6 in " difference " column . 26 



/. Decimal part is -0069 

Note. In the " difference " figures the 6 of 26 goes under the 3. 

Example 231. Find the complete logarithms of the following num- 
bers : (a) 5728; (b) -0001652; (c) 5; (d) 7004; (e) -1183; (/) -00000066; 
(S) 44<>o. 

(a) i figure in front of decimal point /. whole number is o. 

Opposite 57 and under 2 we find 7574 

> 8 in " difference " column . 6 



.'. Decimal is '7580 

.'. Logarithm = 0-7580 

(b) 3 noughts between ist figure and decimal point /. whole number 
is - 4. 

Opposite 1 6 and under 5 we find 2175 

. i, . 2 in " difference " column . 5 

2180 
/. Logarithm = 4-2180. ' 



USE OF LOGARITHMS 



221 



Finding Antilogarithms. (1) Significant Figures. Only 
the decimal part of the log is considered. The antilogarithm table is 
arranged thus : The first two decimal places of the log are contained 
in the extreme left-hand column, while the third and fourth places 
are found in the extreme top horizontal line. The four figure num- 
bers in the body of the table are the significant figures for the loga- 
rithms up to the third decimal place. The numbers of i or 2 figures 
to the right of the table are " differences " which are to be added 
to the four-figure numbers to account for the fourth decimal place. 

DIAGRAM ILLUSTRATING ANTILOGARITHM TABLE. 





3 decwal place of H 


.e loqaK\^m 




y^Ho . , . 

+r decimal place 




o 


1 


e 


3 


A- 


s 


G 


~7 


Q 


s 


,-as 


4 5G 


7 8B 


.00 




























en 






















"Di{ 


-Ceren 


ce 


_c 




4 


Sign 


if ICC 


anlr - 


f'3 u 


res 


m Y\ 


1C. VI 


jwb< 


JK 


I I 

Columns. 


S 




-f-or Vhe, -fiv^f 3 decimal places in 


These ^iguves to 
be adkc\ecl Vo 


4 




rhe loganHim 












Vhoae 


i V-he 


cev\lTre 


i 
























fe accou^V- ^ov Q 


-8 

j_ 1 






















V<M 


ccima 


p\ace 



It will be seen that, in the full tables, the decimal point appears in 
front of each pair of figures in the left-hand column. This is to 
distinguish such numbers as -01 from -i, etc. The operation of 
rinding the significant figures is carried out in a similar manner to 
that for finding a logarithm. Thus, let us find the significant figures 
for the antilog of 27829. Look down the left-hand column until the 
first two decimal places -78 are found. Move across the table until 
the column headed by the third decimal place 2 is reached. Here 
we find the figures 6053. Keeping this marked with the finger, or 
a pencil, move across to the " difference " columns until the column 
headed by the 4th decimal place 9 is found. Here we find the figure 
13 which is to be added to the four-figure number previously found. 

Thus opposite 78 and under 2 we find 6053 

M > 9 in the" difference" column 13 

/. Significant figures in antilog are . . 6066 



222 



ARITHMETIC FOR ENGINEERS 



These numbers are shown in bold type and underlined in the 
following extract from the antilogarithm table : 

EXTRACT FROM ANTILOGARITHM TABLE. 








i 


2 


3 


4 


5 


6 


7 


8 


9 


123 


456 


789 


00 


1000 


1002 


1003 


1007 


1009 


1012 


1014 


1016 


1019 


1021 


001 


111 


222 


77 
78 


5888 
6026 


6902 
6039 


5916 
G053 


6929 
6067 


5913 
6081 


6957 
6095 


6970 
6109 


5984 
6124 


6998 
6138 


6012 
6152 


134 
134 


578 
678 


10 11 12 
10 11 13 


79 


6166 


6180 


6194 


6209 


6223 


6237 


6252 


6266 


6281 


6295 


134 


679 


10 11 13 



As another example, let us find the significant figures in the 
antilog of 3-2104. 

Looking opposite *2i in the left-hand column and 

under the column headed o we find .... 1622 

On the same line and under 4 in the " difference " 

column 2 



.'. Significant figures in antilog are . 



1624 



Example 232. Find the significant figures in the antilogarithms of 
the following: (a) 0-7103; (b) 3-2997; (c) 2*0019; (d) 1-4779. 

(a) Opposite -71 and under o we find .... 

,, 3 in " difference " column 

/. Significant figures in antilog are . 



5129 
4 



5J33 



(b) Opposite *29 and under g we find .... 

,, 7 in " difference " column 

/. Significant figures in antilog are . 



_ 3 
1994 



(c) Opposite -oo and under I we find 1002 

,, ,, 9 in " difference " column . 2 

/. Significant figures in antilog are . . 1004 

(d) Opposite '47 and under 7 we find 2999 

9 in "difference" column. 6 



/. Significant figures in antilog are 



35 



Finding Antilogarithms. (2) Placing the Decimal Point. 

Having obtained the significant figures in the antilog the decimal 
point is fixed according to the following rules : 

1. If the whole number of the logarithm is plus, then the number of 
figures before the decimal point must be one more than the whole number. 

Taking (c), Example 232, the log is 2-0019 and the significant 
figures are 1004. Then the antilog will be 100-4, *'* the whole 



USE OF LOGARITHMS 223 

number is 2 and therefore there will be three figures in front of the 
decimal point. 

2. If the whole number is minus, then the number of noughts between 
the first figure and the decimal point must be made 1 less than the whole 
number. 

Taking (b) Example 232, the log is 3-2997, and the significant 
figures are 1994. Then the antilog will be -001994, i.e., the whole 
number is 3, and therefore there will be two noughts between the first 
figure and the decimal point. 

Example 233. Find the antilogarithms of the following (i. e. t find the 
numbers whose logs are) : (a) -7103; (b) 1*4779; (c) 2-9786; (d) 5*3001. 

(a) Significant figures are 5133; see (a) Ex. 232. Whole number is 
o, .'. there must be i figure before the decimal point. 

/. Antilog is 5-133. 

(b) Significant figures arc 3005 ; see (d) Ex. 232. Whole number is Y, 
.*. there must be no noughts between the first figure and the decimal 

point. 

.*. Antilog is -3005. 

(c) Opposite -97 and under 8 we find 9506 

,, ,, ,, ,, 6 in " difference " column . 13 

.*. Significant figures are 95*9 

Whole number is 2, /. there must be i nought between the first figure 
and the decimal point. 

/. Antilog is -09519. 

(d) Opposite -30 and under o we find 1995 

,, ,, ,, ,, i in " difference " column . o 

.*. Significant figures arc 1995 

Whole number is 5, /. there must be 6 figures in front of the decimal 

point. 

/. Antilog is 199500. 

A mistake sometimes made when using logarithms is to look 
up the antilog table in place of the log table, and vice versd. To 
avoid this, it should be remembered that there is no decimal point 
anywhere in the logarithm table; the decimal point only appears in 
the left-hand column of the antilogarithm table. 

When working examples logarithmically, it is usual to abbreviate 
the words " the logarithm of " to " Log/' so that the statement 

Log -09519 = 2-9786 
means " the logarithm of -09519 = 2-9786." 



224 ARITHMETIC FOR ENGINEERS 

Exercises 58. On Finding Antilogarithms. 

THE SIGNIFICANT FIGURES 

Find from the antilogarithm table the significant figures in the 
antilogs of the following : 

1. 1-245, 2-659, 4-50, 3-301. 

2. 0-005, 3-0004, 0-227, 1*707. 

3. i'7495>_2'3989 L i'73 7. 2-1486. 

4. -9989, 5'6739, i'_8o55, 3-9499- 

5. 0-2699, 3*4928, 1-8407, 2-0002. 

6. 4*9599, 795 6 > 3'9O39, 1-9009. 

PLACING THE DECIMAL POINT 

7. to 12. Place the decimal point in the significant figures found 
from Questions i to 6 above, according to the whole number in each 
case. 

COMPLETE ANTILOGARITHMS. 

Find the antilogs of the following : 

13. 3-0095, 2-193, i'3 OI -3069. 

14. 4-3286, 1-3595. i-3i, -4972. 

15. 5'iooi L 3-35ii,_-5956, 6-0969. 
16 -7575, 1-2091, 6-2201, 7-4771. 

17. 3-3502, -9896, T-oo2i, 1-9795, 1-8951. 

18. i-oooi, 3-2375, 3-1115, '1009, 1-8751. 

Multiplication by Logarithms. This is effected according 
to the following rule : 

Find the logarithms of the numbers to be multiplied. Add them 
together, and then find the antilogarithm of the sum. 

Expressed in symbols 

If X = A X B, then log X = log A + log B. 

Note that the multiplication of numbers is performed by the 
addition of their logarithms. 

The following examples illustrate the best method of setting 
down this work, and should be closely followed. Care should be 
taken that the decimal points and the various decimal places are 
arranged in straight vertical columns, as emphasised for decimal 
addition in Chap II. 

Example 234. Find the result of 7-2 X 62-5 by logarithms. 
Log 7-2 = 0-8573 

Log 62 -5 = 1-7959 

Log of product = 2-6532 = Sum 
Product = Antilog = 450 



USE OF LOGARITHMS 225 

The addition of the decimal portions presents no difficulty, as 
these are always +, but when adding the whole numbers care must 
be taken that the algebraic sum is obtained, as both + and 
numbers will be met with. 

Example 235. Perform the calculation of Ex. 43, p. 49 (-785 X 
0039), by logarithms. 

Log -785 = 1-8949 

Log -0039 = 3-59II 

Log of product = 3*4860 = Sum 
Product = Antilog = -003062 

Note. When adding the logs, the carrying figure from the first 
decimal place is -f- i. Then we have to add -f- i, i, and 3, which 
gives 3 for the whole number. 

When several numbers have to be multiplied then all the logs 
may be added up and the antilog found only at the end. 

Example 236. Calculate by logarithms the value of 3-14 X 3-14 X 
13000 X -021 tons. 

Log 3-14 = 0-4969 

Log 3-14 = 0-4969 

Log 13000 = 4*ii39 
Log *o2i = 2-3222 

Log of product = 3-4299 

Antilog = 2691 
.*. Result =2691 tons. 

Note. Carrying figure from the first decimal place is -f- I- Then 
the sum of -f- i, 2 and 4-4= i -f- 4 = 3- 

If the student should work this example by the ordinary method 
the amount of labour saved will at once be seen. 

Example 237. The expression -667 X -475 X 2-023 x 4*093 occurred 

in connection with the strength of a channel section. Complete the 
working, using logarithms. 

Log -667 = 1-8241 

Log -475 = 1-6767 

Log 2-023 = 0-3060 

Log 4-093 = 0-6120 

Log of product = 0-4188 
Product = Antilog = 2-623 

Note. Carrying figure from first decimal place is -f 2. The sum 
of -f 2, I, and i is o. 
Q 



226 ARITHMETIC FOR ENGINEERS 

Exercises 59. On Multiplication by Logarithms. 

Work the following by logarithms : 

1. 3-56 x 21-95 2. -095 X 316. 3. -00032 X 60 X 13- 

4. 1-414 X -5 X -7071. 5. -7 X 384*5 X '93 X 12. 

For further practice work a number of the following exercises by 
logarithms : 

Chap. II : Exs. 12; Chap. Ill : Exs. 21, Nos. i to 6. 

Division by Logarithms. The division of one number by 
another, using logs, is carried out according to the following rule: 

Find the logarithm of each number. Subtract the logarithm of the 
divisor from the logarithm of the number to be divided. Find the 
antilogarithm of the difference. 

Expressed in symbols 

If x = -- then log X = log A - log B. 

JD 

Note that the division of numbers is performed by the subtraction 
of their logarithms. 

Example 238. Find the value of ^- (last piece of calculation in 

2-47 

Ex. 57, p. 65). 

Log 238 2-3766 

Log 2-4? = 0-3927 

Log of quotient 1-9839 = Difference 
Quotient = Antilog = 96-36 

The subtraction of the decimal part presents no difficulty as it 
is always positive. But since the whole numbers may be both 
positive and negative, the subtraction must be performed in the 
algebraic way. 

.AQ j- 

Example 239. Find the value of ~- by logarithms. 

i oo 

Log -695 = 1-8420 

Log 1 60 = 2-2041 

Log of quotient = 3*6379 = Difference 
Quotient = Antilog = -004344 

Note. When taking the 2 from the i remember the algebraic 
method. Changing the sign of the quantity subtracted, the 2 becomes 
2. Adding 2 to i gives 3. 



USE OF LOGARITHMS 227 

When the decimal part to be subtracted is larger than the other 
decimal part, then the scheme of " borrowing one " must be very 
carefully watched on account of the presence of both + and 
numbers that may occur. The following example gives a case. 



Example 240. Find the value of - (see Ex. 47, p. 55) by 

logarithms. 

Log *oii93 = 2-0766 

Log 2-3 = 0-3617 

Log of quotient 3-7149 
Quotient = Antilog = '005187. 

Note. Coming to the first decimal place the 3 cannot be taken 
from o. Therefore borrow i from the 2; then 3 from 10 is 7. Now 
borrowing i from the 2 is taking + I from 2 which therefore becomes 3. 
Now proceeding to the whole numbers o from 3 leaves 3 

Hence the difference is 3*7149 

The following alternative explanation of the above will apply to those 
who are accustomed to " paying back " after borrowing. After sub- 
tracting the 3 from the 10, consisting of the o and the borrowed i, 
then i must be " paid back " to the o in 0-3617, making -f i from 2. 
Working algebraically this gives 3 as before. 

Example 241. Find the value of the expression - -- : figures 

'00075^2 

relating to an engine governor. 

Log 67*8 = 1*8312 

Log -0007502 = 4-8752 

Log of quotient = 4-9560 
Quotient = Antilog = 936o 

Note. Here again a i has to be borrowed to enable the subtraction 
of the first decimal place to be performed. Then in the top line, when 
" paying back," i taken from i leaves o. Now 4 has to be taken from o. 
Changing sign and adding gives 4- 4 for result. 



Exercises 60. On Division by Logarithms. 

Work the following by logarithms ; 

1. 219-5 -r 3i' 8 3- 2. -00963 ~ -785. 

3. 29580 -4- 1-986. 4. -1352 4- 23-66 

5. -00058 -r -000415. 6. 2382 ~ 299-8. 

Work by logarithms Exercises 13, Chap. II. 



228 ARITHMETIC FOR ENGINEERS 

Compound Examples. Examples involving both multipli- 
cation and division are very easily and quickly treated by means 
of logarithms. The following examples show the most convenient 
method of laying out the work. It is usually best to find the log 
of the whole of the top line, and the log of the whole of the bottom 
line, and then subtract. 

Example 242. Find the value of the expression ?~? 

B.T.U. : figures relating to the heating value of a gas. 

Top line Bottom line 

Log 2250 = 3'35 22 Log 454 = 2-6571 

Log 17-7 = 1-2480 Log -146 = 1-1644 

Log of product = 4-6002 Log of product = 1-8215 

Log of bottom line = 1*8215 

Log of quotient = 2-7787 = Difference 
Quotient = Antilog = 600-8 B.T.U. 

Explanation. First obtain the log of the top line and the log of 
the bottom line. Now, unless it is desired to know the two products 
there is no need to look up the antilogs of the two logs obtained. The 
log of the bottom line may be brought under the log of the top line, 
and the log of the quotient obtained directly. 

The student is warned against the practice of looking up the 
various logarithms and working the example without stating to which 
numbers the logarithms refer. The time necessary for the writing of 
statements such as " log 454 " is not worth considering, and when 
completed in this manner the example is understandable by any 
one at any time, and can always be readily referred to. 

Example 243. Work Ex. 55, p. 64, by means of logarithms. 

7*4J*J7 8 5 _*L_3 '5__ *L3 :5_X _8op 

33000 X 3 

Top line Bottom line 

Log 74-4 = ^'8716 Log 33000 = 4*5185 

Log -785 = 1-8949 Log 3 = 0-4771 

Log 3'5 = '544i Log of pro duct = 4-9956 

Log 3'5 = *544i 6 * * D 

Log 800 = 2-9031 

Log of product = 5-7578 
Log bottom line = 4-9956 

Log of quotient = 0-7622 
Quotient = Antilog = 5784; Sa 7 5*7 8 



USE OF LOGARITHMS 229 

Comparing this with the " longhand " working, it will be seen 
that the saving of labour is considerable. 

Some simple mental work will often save labour with logarithms, 
as in the above example, where 33000 X 3 = 99000. By looking 
up the logarithm of 99000 the working out of a log is saved. 

Example 244. Find the value of the expression : 

^ ^ v -000341 X 4 X *45 

figures relating to an engine governor. 

Log 51-9 = 1-7152 Log -000341 = 4-5328 

Log bottom line = 4-7881 Log 4 = 0-6021 

Log of quotient = ^1 Log ' 45 = 7 ' 6532 

Log of product = 47881 
Quotient = Antilog == 84550 

Exercises 61. Compound Examples. 

(In order that the student may check his calculations at various 
steps, log top line, log bottom line, and results are given in the answers 
for Nos. i to 4.) 

Find the value of - 

1 7'i i X 29-38 2 L3^ < 7 X j6'^ 

' -093 x 11-36 " "^ooo^i" X~-957~ 

3 7 8> 5 X 113 X 2-167 x 194 ^ 87-5 



33000 * -000341 X 5-6 X -416 

' -00374 X 15250 X I -021 * 33000 X 10-5 

7. The bending moment on a girder was given by the figures 

62 XJ-4X28 complete the working. 
2240 x 8 i 5 

8. Find the value of the expression Ibs. per sq. 

2x3x1*5x1*5 

iii. which was obtained in calculating the stress in the teeth of a gear 
wheel. 

9. A calculation for the area of the conductors to supply a certain 
house with electric light produced the figures 1QO x i sc *' ms " 
Find the actual area required. 

.- ^ , , ,, . 9-87 X 30,000,000 X '021 ., , . , 

10. Evaluate the expression - ~ rr^ Ibs. which 

refers to the buckling load on a steel column. 

Work by logarithms Nos. i to n of Exs. 16, Chap. II, and Nos. 
12 to 20 of Exs. 21, Chap. III. 

Examples involving -f and . There are plenty of examples 
in which the operations of addition and subtraction are mixed 



230 ARITHMETIC FOR ENGINEERS 

up with those of multiplication and division. If it is desired to 
evaluate these with the aid of logs, then two things must be 
remembered 

1. That addition and subtraction must be performed in the 
ordinary manner, viz., without the use of logarithms; 

2. That the addition of two logarithms is connected with the 
multiplication of their numbers. 

Hence any addition or subtraction of logs must be confined to 
those portions of the expression involving multiplication or division 
only. Then the antilogs or results of these portions must be found, 
and any addition or subtraction desired must be performed upon 
these partial results. 

In this type of example it is best to keep the form of the expres- 
sion as far as possible, and to do the logarithmic work at the side 
of the sheet. The student should state carefully to what numbers 
each log refers ; it is very desirable in this class of example to be 
able to trace all the steps in the working. 

The following will illustrate : 

Example 245. Find the value of the expression 7 x -575 X 1-75 + 
9-85 X 475 X '2375, figures relating to the strength of a channel 
section bar. 

Log 7 = 0-8451 

7 X -575 X i'75 + 9*85 x '475 X -2375 Log -575 = 1-7597 

= ' ^ ' + ' rrTi ' Log i'75 =0-2430 

Log of product = 0-8478 

= 8-155 

1 Antilog = 7-044 

N.B. The brackets are merely used 
to show the necessary steps. 

Log 9-85 = 0-9934 

Log -475 = 1-6767 

Log -2375 = * '3 75 6 

Log of product = 0-0457 
Antilog = in i 

Explanation. The first part of the expression is evaluated by logs, 
care being taken that the antilog 7-044 is found and written down under 
the main expression. Similarly the second part of the expression is 
evaluated by logs, and the antilog i-in written next to the 7-044. 
Great care must be taken that these antilogs are found before the -j- 
sign is obeyed. Note also that the final addition produces the actual 
result without the use of log tables. 



USE OF LOGARITHMS 

Example 246. Evaluate the expression 

i + 
figures relating to the eccentric load on a column. 



_64 

6-5 X 5 
2-82 X 2-82 



231 
tons, 



64 


Log 6-5 
Log 5 

Log of product 
Log of lower 
product 

Antilog 


= 0-8129 
= 0-6990 


1+ 6 ' 5 x 5 - 


^ 2-82 x 2-82 

64 


= i'5"9 
= -9004 


I -f 4-088 


6115 


= 4-088 



Log 2-82 = 0*4502 

Log 2-82 = 0-4502 

Log of product 0-9004 



__ 

5-088 

12*58 tons 



Log 64 = 1-8062 

Log 5-088 = 0-7066 

Log quotient = 1.0996 
Antilog =s 12-58 



Explanation. First evaluate by logs the right-hand term in the 
bottom line, and insert its antilog (4-088) in the second line of the main 
working. Then add the i to the 4-088, when the expression becomes a 
simple division, logs being used. The final antilog in this case is the 
answer. 

Example 247. Find the value of the expression 
6'395 ^-626 X -626 10 x io\ 

~32~-2~ \ ~~~4 h 72~~ / 

figures relating to an experiment on the twisting of wires. 
Expression = 



Log '626 
Log -626 



= 1-7966 
= 1-7966 



6-395 X 8-431 
32-2 



Log of product = 1-5932 
Log 4 = 0-6021 



Log 6-395 = 0-8058 

Log 8-431 = 0-9259 

Log oi product = 1*7317 
Log 32-2 = 1-5079 



2-9911 0-2238 

1-674 Antilog = -09797 Antilog = 1-674 

say, -098 

Explanation. First evaluate the bracketed terms ; the second 
term can be found mentally, being ~ = 8J, i. e., 8-333. Then the 

two results must be added, and their sum operated upon by the two 
numbers outside the bracket, logs being used. 



232 ARITHMETIC FOR ENGINEERS 

Exercises 62. Examples involving + and 

Evaluate by logarithms 

1. 9'5 2 X -459 4- 2-35 -f- -958. 2. 2-7 -f 3'5 8 X 2-895 X 3*142. 

8 9'75 -8-25 6-875_X_? 17-25 X 2-564-2-894 X -015 

' 10-385 -f-5'5 : 75 " 6785 x -0125 

B . 17-5 X * -56.5 + i?>5 x .^ x ^^ x 4 . 5Q 

6. The expression 7 X -57 X 1-75 -f 9-85 x -475 X ~~ occurred 

in connection with finding the centre of gravity of a channel section. 
Complete the working. 

7. The weight of a proposed bridge truss is estimated to be 

60 X 125 X 10 , ~ . , ., . . 

J tons. Complete the working. 

1800 x 7-5 125 x 10 r 6 

Work by logarithms, Nos. 13 to 25 of Exs. 16, Chap. II. 

Powers by Logarithms. Logarithms afford a very simple 
and rapid method of raising numbers to all kinds of powers. Only 
whole number powers, such as squares and cubes, will be taken in 
this book. 

A number can be raised to a power by logarithms according to 
the following rule : Find the logarithm of the number and multiply it 
by the index denoting the power. This product is the logarithm of the 
result, and its antilogarithm is the power required. 

Expressed symbolically 

If X a a then log X = n X log a 

A slightly different method of laying out the work is advisable 
in these cases. 

Example 248. Find the value of (6-25) 8 

Log (6-25) 8 = 3 x log 6-25 
= 3 x 0-7959 
= 2-3877 
Antilog = 244-2 = 6-25' 

Note. Brackets used as in the 6-25 above are not a necessity, but 
may help to keep the statement more clear, especially in written 
work. 

These examples present no difficulty when the whole number 
of the log is positive, as the multiplication is easily performed. 
But when the whole number is negative, then the log really con- 
sists of two parts, one + and the other , and care must be taken 
in dealing with it. 



USE OF LOGARITHMS 233 



Example 249. Find the value of '596*. 

Log -596 = 2 X log -596 

= 2 X I-7752 



Antilog = -3551 = -596*. 

To explain : When carrying out the multiplication by 2 it must 
be remembered that, of the log, only the I is negative, while the 
7752 is positive. Multiplying -7752 X 2 we obtain 1^504, the 
5504 being the decimal portion (positive) of the new log., and the I 
(also positive) being a carrying figure. Now multiplying the I by 2 
gives 2". Adding the carried I to the "2 gives I. Hence the new 
log is 1*5504, and its antilog is the required number. 

Example 250. Find the value of '0365 2 

Log -0365* = 2 x log -0365 
= 2 X 2-5623 
= 3-1246 
Antilog = -001332 = -0365* 

Explanation. Multiplying the decimal portion of the log by 2, 
the carrying figure is i, which is, of course, positive. Then 2 x 2 is 4, 
and adding the carried 4~ i gives 3. 

Example 251. The area of a circle of diameter d" is given by the 
formula 7854^. Find the area of the section of a wire No. ooo S.W.G. 
(372* diameter). 

Area = '7854^* 

= -7854 x -372' 

Then log area = log '7854 -f 2 x log -372 
= 1-8951 + 2 x i-575 
= 1*8951 4- 1-1410 
= 1-0361 
/. Area = Antilog = -1086 sq. in. 

Note. In this case the log of the square must be obtained by 
multiplying by the 2 before adding to the log of '7854. 

Exercises 63. On Powers by Logarithms. 

Find the value of the following by logs : 
1. (3-142)". 2. (7854) 1 - 3. 2-5*. 4. -0875*. 

5. 27-852. 6. 7-325* X -875. 7. (8-75 x 5'26 5 ) 2 . 

Work by logarithms, Nos. 18 to 30 of Exs. 23, Chap. III. 

Roots by Logarithms. Similarly any root may be easily 
and quickly extracted by means of logarithms. Only square roots, 
cube roots, and the like will be considered in this book. 



234 ARITHMETIC FOR ENGINEERS 

To extract any root of a number : Find the logarithm of the 
number and divide it by the figure denoting the particular root. The 
quotient obtained is the logarithm of the result, and its antilogarithm is 
the root. 

Expressed in symbols 

lOfiT & 1 

If X = "*Ja then log X = ~- or - x log a 

Example 252. Extract the square root of 22-09. 
Log ^22-09 = --log 22-09 

i of 1-3441 

= -6721 
Antilog =5 4-7 = \/22-o9 

Explanation. In the case of square root 2 is the figure denoting 
the root. Hence log 22-09 has to be divided by 2; the actual division 
can be done mentally. 

Referring to the ordinary working of this example on p. 103, 
the saving in time and labour is seen at once to be considerable. 

Roots by Logarithms. Adjustment of the logarithm 
when its whole number is negative. As when dealing with 
powers, these examples present no difficulty while the whole 
number of the log is positive, but an adjustment has to be made 
when it is negative. Two ways of making this adjustment will be 
shown. As an example let us find the value of ty-2512. In this 
case 3 is the number denoting the root. 

Then log $^2512 == - log -2512 

3 

= - of 1-4000 

METHOD I 

Now at first it does not seem easy to divide a number having 
+ and parts, but the difficulty is got over by changing the 
number, so that it only possesses one sign. Thus 1*4000 means 
I + '4000, which if evaluated according to the algebraic idea 
equals -6000. 

Now '6000 can be divided by 3 quite easily, giving -2000, 
which is then the log of the result. But it must be remembered 
that the logs in the tables (the decimal parts) are always positive, 



USE OF LOGARITHMS 235 

so the logarithm -2000 must first be converted (without, of 
course, altering its value) so that it will have a positive decimal 
part. 

Now let us add + I and I to the logarithm *20OO. The 
value will remain the same, since the addition of + i i is the 
addition of o. Then we may write -2000 + i i, which, com- 
bining the first two numbers, = + '8000 i = T-8ooo, and the 
decimal part is now positive, whilst the log is of the ordinary 
form. 

The complete working would appear as follows : 

Log ^-2512 = - log '2512 

i , _ 
= - of i '4000 

= - of -6000 
3 

= *2OOO 

= T'Sooo 

^^2512 = Antilog = '6310 

The chief new points in the above method are 

1. Convert the log to a single negative number, which really 
consists of taking the decimal from the negative whole number and 
reducing the digit of the latter by I. 

2. Find the required fraction of this. 

3. Add + i and i mentally, which is actually done by taking 
the decimal from i and increasing the digit of the whole number 
by i. 

METHOD II 

An alternative method of " making the adjustment " is as 
follows : add a sufficient negative number to the whole number to 
make it exactly divisible by the divisor representing the root, 
i. e., in this case add 2 to the I, making it 3, which is then exactly 
divisible by the divisor 3. Now to the decimal part add the same 
number, only positive, in order to keep the same value for the log, 
i. e. t in this case add + 2 to the decimal (making it 2-4000), the + 2 
and the 2 neutralizing each other. Then the original logarithm 
7-4000 becomes 3 + 2-4000, in which the negative part is exactly 
divisible by the divisor 3. Dividing the 3 by 3 gives Y, and dividing 
+ 2-4000 by 3 gives + *8ooo. Hence our new log is "1-8000, in 
which the decimal is +, so that the log is of the ordinary form. 
If the whole number is already exactly divisible this method is 



236 ARITHMETIC FOR ENGINEERS 

shorter than Method I, as division can be directly proceeded with. 
Performed mentally, Method II is in all cases the quicker. The 
complete working in this case is as follows : 

Log ^-2512 = -log -2512 
= -of 1-4000 

= --(3 + 2-4000) 

= T -f- *8ooo 
= 1-8000 
^2512 = Antilog = -6310 

Note. Brackets are needed in the third line to denote that both 
the 3 and the 2-4 are to be divided by 3. 

Example 253. Evaluate ^-000247, figures which occurred in a 
problem on water supply. 
Working by Method I 

Log /v/ -000247 ~ -of log -000247 
= g of 4-3927 
= ^of - 3-6073 




Antilog = 

Explanation. Converting the logarithm 4-3927 into a number of 
single sign, the -f *39 2 7 is taken from the 4-0000, giving 3*6073. 
Dividing by 5, '7215 is obtained. To convert we add -f i and i. 
The i is written as I, and the -f i added to the -7215, giving 
-f -2785. The final log is then 1-2785. 

Working by Method II 

Log ^-060247 = - of log -000247 
= j-of 4'3927 

= j(5 -f 1-3927) 

= 1-2785 
Antilog = -1899 

Explanation. To make the whole number of the log 4*3927 
exactly divisible by 5, a i must be added, when it becomes 5. 
+ i must therefore be added to the -3927, making it 1-3927. Dividing 
5 by 5 gives I and dividing 1-3927 by 5 gives -2785, so that our final 
log is 1-2785, as before. 



USE OF LOGARITHMS 237 

Exercises 64. On Roots by Logarithms. 

Find the value of the following by logs ; 

1. ^5^5- 2. A/2I7-8. 3. 



4. V *oo 1 65. 5. ^97-8. 6. v'iTT^Tooo. 

7. ^19-06. 8. v'lTSrJ. 9. ^f-Ti2. 

10. v 77 ^. 11. ^-0565. 12. ^-0005573? 

13. ^-00043. 14. ^^0000087. 15. ^ib^oy. 

16. Rankine's rule for the thickness (ins.) of a masonry arch is 

zr. Calculate the thickness when radius r = 10-5, 

17. The formula ^-- gives the area of cross-section (sq. ft.) of a 



chimney H ft. high, when G = total grate area in sq. ft. of the boilers. 
Calculate the value of the expression when G = 30 and H = 70. 

18. The formula P = -,. refers to live rollers. Calculate the 

Vr 
value of P when Q 500, r = 2-5, and 5 = -0073. 

Vx 

19. The expression ~T>= relates to the motion of a steam engine 

slide valve. Find its value when x = -625. 

20. Calculate the diameter of a piston-rod from the formula 
D = 1*2 VdL, when d = 13 and L = '667. 

/d? 

21. The expression 1*73 V y- relates to the collapsing of short tubes 

under external pressure. Calculate its value when d 2*5 and / = -5. 

22. The greatest allowable wheel base on a tramway track of R ft. 

radius, where the width of rail groove is w" t is \/ ---- ft. Calculate 
this length when R = 70 ft. and w = 1-25". 

23. The expression \ occurred in a certain hydraulic problem. 
Find its value when g = 32, x = 6, and y i'2$. 

24. In calculating the sizes of motor andjdynamo field coils Esson's 

Formula may be used, which is C= V R . Calculate the value of 
C when t = 50, A = 1150, and R = 10. 

25. The current allowed in bare overhead electric conductors is 

/T\3f 

given as \/ R amps. Calculate this value when D = 9-45, / = 20, 

and R = '00000177. 

26. The " impedance " of an A.C. circuit is VR a ^"/T 2 L*. Cal- 
culate its value when R = 2*3, p = 546, and L '015. 



27. The formula P = ^VD + '625 '175 ins. gives the pitch of a 
U.S. standard thread where D = diameter of bolt in inches. Calculate 
P for a i i" bolt (i. e., D = 1-5). 



ARITHMETIC FOR ENGINEERS 

28. The expression \ ^-^ 4 - gives the diameter of wire re- 
quired for the field coils of a shunt-wound dynamo. Calculate the 
size of wire required when E =* no, D = 15, d = 10, and CT = nooo. 

29. The horse-power formula given by Messrs. Rolls Royce for 

petrol engines is H.P. = '2$(d J VS per cylinder, where d = cylinder 

diameter and S = stroke. Calculate the H.P. if d = 4-33 and S = 5-12. 

30. The " number of threads per inch " on bolts over i* diameter 

10 
having British standard fine threads is given by the expression "ygy 

where d ~ diameter of bolt in inches. Calculate the number of threads 
per inch on bolts of the following diameters : (a) 2", (b) i \". (Note. In 
each case give the actual answer, and also the nearest whole number of 
threads.) 

Further examples will be found in Exercises 24 (p. 109), Chap. III. 

Various Examples. We will close this chapter with some 
examples of a more difficult nature involving all the foregoing 
points. 

Example 254. The diameter of a turbine nozzle is given by the 
figures V -T-JTT inches. Complete the working. 

Log '1675 = 1-2240 

Log -785 = 1*8949 

Log of quotient 1-3291 

= '6709 (By Method I, p. 234) 
Log V quotient = *3354 (Dividing by 2) 

= 1-6646 
Antilog = -461 9*; say -46*. 

Example 255. In a problem on water supply the figures v/73-6 2 
ft. per sec. appeared. Find the required value. 

Log. 73** = ^ of log (73 -6 2 ) 

= * of (2 X log 73-6) 
= - of (2 X 1*8669) 

- I of 37338 
= -7468 
/. Antilog = 5-582, say, 5-58 ft. per sec> 



USE OF LOGARITHMS 239 

Example 256. It was calculated that under certain conditions the 

speed of an engine governor would be \J - R.P.M. 

000341 A 4 x -55 

Complete the calculation. 
Bottom line 



Log -000341 = 4-5328 
Log 4 = 0-6021 


Log 69-2 S=B 1-8401 
Log bottom line = 4*8753 


Log -55 = i-744 


Log of quotient = 4-9648 


Log of product = 4-8753 


Log V quotient = 2-4824 



Antilog = 303-7 R.P.M. ; say, 304. 

Example 257. The deflection of a beam loaded centrally is given 

WI 3 
by the formula - QT "f ins. Find the deflection of such a beam where 



W = 1 1 -5 tons, L = 24 ft., E = 12500 tons per sq. in., I = 725-7. 

Converting the 24 ft. into ins., L = 24 X 12 = 288 ins. Deflec- 

tion = OT ^ 1 == - * * 5 - - --- (substituting given values). 
48E1 48 X 12500 x 725-7 v b b ' 

Top line Bottom line 

Log (288 3 ) Log 48 = 1-6812 

= 3 X 2-4594 = 7-3782 Log 12500 = 4-0969 

Log 11-5 = 1-0607 Log 725-7 = 2-8607 

Log top line = 8-4389 Log of product = 8-6388 

Log top line = 8*4389 
Log bottom line = 8-6388 

Log of quotient == 1*8001 

Antilog = -6311 
Deflection = -631" 

Example 258. A number giving the inclination of a water supply 
pipe is given by the expression ( - -~^i~~=^Y. Complete the working. 

Bottom line 

Log Vi^8i7 Log 2-5 = 0-3979 

= J X 0-2593 = 0-1297 Log bottom line = 2-1711 

Log 1 10 = 2-0414 Lo g O f quotient = 1^268 

Log bottom line = 2-1711 Log (quotient) 2 = 4-4536 

Antilog = -0002843 

Required number = -000284, say. 



240 ARITHMETIC FOR ENGINEERS 

Example 259. Dr. Pole's formula, giving the quantity of gas flow- 
ing through a pipe in cu. ft. per hr., is I3$od 2 \/ j~, where d = diameter 

of pipe in ins., h = effective gas pressure in ins. of water, / = length of 
pipe in yds., s = specific gravity of gas.* Find the discharge when 
d = 6*, h = 1*6*, / = 1250 yds., and s = '425. 

Discharge = I35<>^ 2 V y cu. ft. per hr. 

=* I 35 X 6 2 V ---- ^ ----- (substituting the given figures) 
1250 X *4^5 

= 1350 X 36\/ -- - (doing a little mental calculation) 
1250 x *4^*5 

Log 1250 =3-0969 Log 9-6 =0-9823 

Log -425 = 1-6284 Log bottom line =2-7253 

Log of product = 2-7253 Log of quotient = 2-2570 

Log Vquotient = 1-1285 

Log of the sq. root 1-1285 
Log 36 = 1-5563 

Log 1350 = 3-1303 

Log of product = 3*8151 

Antilog =s 6533 

Discharge = 6533 cu. ft. per hr. 



2. The Board of Trade rule for safety-valve springs of square 

3 /~SD 
section is d = \/ - . Find d when S = 3350 and D = 4. 



Exercises 65. Various Examples. 

8 /T~ 
1. The formula d = \f ^j relates to shafts under torsion. If T 

558 and / = 9000, calculate the value of d. 
Board of Tr 

3 /~SD 
= \/ - . 

3. The diameter in inches of a pipe to transmit g gallons of water 

V^ 
per min. a distance of L ft. with a loss of head of h ft, is d = \/ - 

Calculate d if g = 150, L = 1470, and h = 25. 

4. The following expression relates to helical springs : \/ -~ 

v>A 

Calculate its value when w = 250, n = 12, r = 1*75, C = 12,000,000, 
and A= 1-5. 

5. The distance apart (in ft.) at which the bearings of shafting 
should be placed is $Vd 2 , when d = diameter of shaft in inches. Calcu- 
late the distance for a shaft 2J" diameter. 

* A figure connected with the weight of the gas. 



USE OF LOGARITHMS 241 

6. The formula K \/ -r relates to a certain kind of galvanometer. 

a 4 

If d = 117 and rf 4 = 219, calculate the value of K. 

7. The diameter of a propeller fan to deliver Q cu. ft. of air per 
sec. with a consumption of H horse-power is ^/'^? C)OII 5Q 3 . jf 
H = '85 and Q = 230, find the diameter. 

8. Find the value of the expression (-,-=-_-) , which relates to a 

\9oVi -57/ 
water-supply pifte. 

9. Using Dr. Pole's formula (Ex. 259), Q = I35od 2 \/ -.-, calculate 

the quantity of gas Q if d = 10, h^== 2-5, / = 2355, and s = -43. 

8 /3~2oRC* 

10. The formula D = \ ~~ relates to bare overhead con- 

ductors. Calculate D when R = 'OOOooiS, / = 20, and C = 270. 

11. With reference to steam temperature and pressure we have 

6 = V ^5 X io lg ~ 35 ' 16 * Flnd the VaIUG f ' 

12. With reference to the Venturi meter for measuring water we 



have the formula Q = _\. Calculate Q if C == -975, 

h = 6*3, #! = -785, a a = -136. 

i6W 
13. The equation A == x( v~^ (R 4 f 4 ) gives the deflection of a 

conical spiral spring. Calculate the deflection A when W 500, 
X = -15, G = 12 X io 6 , d = -192, R = -8, r = -3. 

14-. The true air speed v of an aeroplane flying with an indicated 
speed of V m.p.h. in air of density p is given by the expression 



v = V f^ - 



At an altitude of 10,000 ft., p = 910. Calculate the 



true air speed represented by an indicated speed of 100 m p.h. 

15. For a variable electrical condenser of the " square law " type, 
R = -v/4^ + r 2 . Calculate the value of R when a = 6, 6 = 3-142, 
* = '375- 

16. The pressure of the atmosphere, p inches of mercury, at an 
altitude of H feet above sea-level is given approximately by the formula 

2Q*Q2 

H = 62580 log . Calculate the altitude H corresponding to a 

pressure of 16 inches of mercury. (Hint. log =(log 29-92 

log p}, and when the two logarithms have been looked out they become 
ordinary numbers for the purpose of the subsequent calculation.) 

As so^n as the student, by sufficient practice, has made him- 
self reasonably proficient in the use of logarithms, he should use 
logarithms for the numerical work in subsequent calculation, and 
if possible, should take up the use of a slide rule. 
R 



CHAPTER VII 
MENSURATION 

LENGTHS AND AREAS 

IN this chapter it is assumed that the reader can use, or is learn- 
ing to use, the ordinary drawing instruments such as T-square, 
set-squares, and compasses ; also that he has some knowledge of 
the method of showing solid objects on paper by " plan, elevations, 
and sections." 

Measurement of Length. In the introductory chapter it is 
stated that the Yard is the British standard of length or Unit ; and 
that multiples and sub-multiples of the yard are employed for 
larger and smaller distances. The smallest sub-multiple unit 
employed is the Inch, which is ^ of the yard ; below this, fractions 
of an inch are used, in the vulgar form for general use, and in the 
decimal form for the finer measurements. The largest multiple in 
ordinary use is the Mile, which is 1760 yards. Between these 
extremes there are various other units as given in the table on p. 3. 

In addition, the following relations are very useful: 

36 inches = i yard 
1760 yards = i mile 
5280 feet = i mile 

For certain classes of work special units are employed. Thus, 
in nautical work the " nautical mile " is used for long distances, 
while the speeds of ships are stated in " knots." 

NAUTICAL MEASURE 

i nautical mile = 6080 feet = 1-15 ordinary miles 
i fathom = 6 feet = 2 yards 

[Depths of sea bottom are given in fathoms] 
i knot = i nautical mile per hour 

= 6080 feet per hour 
242 



MENSURATION 243 

In Land Surveying, measurement is made with a chain 22 yards 
or 66 feet long, called Gunter's Chain. For certain special work a 
chain 100 feet long is used. Unless otherwise stated, the " chain " 
is taken to mean the Gunter Chain. 

SURVEYORS' MEASURE 

100 links = I chain (Gunter) 
10 chains = i furlong 
80 chains = I mile 
[i link = 7-92 inches] 

Conversion or Reduction. When stating lengths or distances, 
it is customary to use two or more units for sizes above i ft. Thus 
on drawings we find such dimensions as 22'-6", 3 / -4j", or the 
length of a road may be given as 3 miles, 5 furlongs, and 35 yards. 
But for many calculations it is desirable that such a length should 
be stated with a single number, i. e., it should be in terms of one 
unit only. The 22'-6" would then be " converted " into (say) 
feet, becoming 22-5 ft., while the 3 / -4f /r might all be converted 
into inches, becoming 40-75". Again, a calculated dimension is 
usually obtained with a single unit, e. g., 4*833 ft. ; on a drawing 
this would be given as 4'-io", the decimal of a foot having been 
converted into inches. The amount of this " conversion " or 
" reduction " required in connection with length measurement is 
not very great. 

The method of working is best illustrated by examples. 

Reduction to Smaller Units. 

Example 260. Reduce the distance 4m. 3 fur. 25 po. 3 yds. to 
yards. 

m. fur. po. yds. 

4 3 25 3 Explanation. Reduce 4 miles to 

8 furlongs by multiplying by 8, since 

, i m. = 8 fur. ; while multiplying, add 

^ U " in the 3 fur.; similarly convert the 

1^ 35 fur. to poles by multiplying by 40, 

1425 poles adding in the 25 poles. The method 

5J of multiplying by 5^ to convert to 

~ JT" yards should be noted as being quick 

i and easy. First multiply the 1425 

_Z by 5 and add in the extra 3. Then 

Result = 7840$ yds. multiply by i and add the two 

" ~* results. 



244 ARITHMETIC FOR ENGINEERS 

Example 261. (a) Convert 5'-7j* into inches; (b) convert 4-74 ft. 
into feet and inches. 

( a ) 5'-7i* (&) 474 *t ..... only the decimal of a ft. 

12 12 is multiplied by 12. 

60 -f 7J = 67^ 8-88 ins. 

c'-" " ^ 4'- 8 " + * 88 of an inch 



/. Dimension = 4'-8f" 
Note. Conversion of this type is largely done mentally. 

Example 262. Convert 35 miles per hour to feet per second. 

i mile = 5280 ft. 

.'. 35 m. per hr. = 35 x 5280 ft. in i hr. 
i hr. = 60 mins. = 60 X 60 sees. = 3600 sees. 



.% 35 X 5280 ft. in i hr. = - ft m i gec> 

= 51-3 ft. per sec. 

Conversion to Larger Units. 

Example 263. A warship opens fire at a range of 12,000 yds. How 
far away is the target : (a) in miles and yards ; (b) in miles only ? 

(a) 1760)12000(6 m. 
10560 

1440 yds. Distance 6m. 1440 yds. 



(6) 1440 yds. = ~- m. = -818 m. 

/. Distance is also 6-818 m., say 6-82 m. 

Example 264. Reduce 16950 ft. to miles, etc. 



5650 yds. 

2 



11)11300 half yds. 
4,0) 102,7 + 3 half yds. 
8) 25 + 27 po. 
3 +i ^r. 
Result = 3 m. i fur. 27 po. i yd. i ft. 6 ins. 

Explanation. The feet are converted into yards by dividing by 3. 
As it is not easy to divide by 5} direct, the yards are multiplied by 2 
(i.e., converted to J yds.) and then divided by n. The 3 remainder 



MENSURATION 245 

is the 3 half yds., i.e.. i\ yds. or i yd. i ft. 6 ins. To divide by 40, point 
off the o in the 40 and the last figure in the 1027 as shown (i. e., dividing 
by 10). Then divide the 102 by 4 and place the remainder 2 before 
the 7. 

Addition and Subtraction in Length Units. It is occasion- 
ally necessary to add and subtract various lengths when given in 
two or more units. The method is exactly the same as when dealing 
with money. In addition, add up the figures of the smallest unit 
first; convert to the next higher unit and write the remainder. 
Repeat, carrying forward the converted figure from the previous 
column. 

Example 265. Find the sum of the following dimensions from the 
wheel base of a 4-coupled locomotive, which gives the overall length 
of the engine : 6'-io^, io'-2^, 8'-7", 4'-7i". 

6' loj" 

10 2j Note. Sum of inch column is 

8 7 27J = 2'-3j". Write 3^ carrying 

4 7J 2 to feet column. 



Subtraction is performed similarly. Occasionally a I from the 
feet column has to be " borrowed " and converted into inches in 
order to carry out a subtraction in the inch column. 

Example 266. A locomotive is 13'- if" from rail to top of chimney 
and 7 / -9 // from rail to centre line of boiler. Find the distance from 
centre line of boiler to top of chimney. 

I 3^" I I" In subtracting the inches it is 

7 "9 _ necessary to "borrow" a i from 

Difference = 5 '_-4JP the feet column. 

Exercises 66. On the Conversion, etc., of Length. 

1. Reduce 2 m. i fur. 20 po. 5 yds. to yards. 

2. Reduce 7 fur. 3 yds. to yards. 

3. Reduce i m. 28 po. 3 yds. to feet. 
Convert : 

4. 115-5 yds. to inches. 5. 27-85 ft. to inches. 
6. 5 yds. 25 ft. to inches. 7. 5'-6k" to inches. 

8. n'-ioj" to feet. 9. i'-9F to yards. 

10. 2785 ft. to a decimal of a mile. 

11. A machine shop requires the following lengths of a certain 
width belting : i4'-6", 28 / -o / ', 7'-4", IQ'-IO". What total length must 
be ordered ? 



246 



ARITHMETIC FOR ENGINEERS 



12. A plate girder requires the following lengths of plate for its 
flanges : i6'-o", 29'-6*, 45 / -4*, 45 '-4", 2C/-6", i6'-9*. Find the total 
length of plate required. 

13. In connection with boiler flues it was necessary to find the 
difference between 3'-8" and 2'-io". What is this difference ? 

14. The overall length of a lathe bed is 6 '-6". The fast head-stock 
occupies i'-2i" and the loose heads tock n". What is the greatest 
length available between the centres ? 

Simple Geometrical Terms. The well-known " Square " 
and " Circle " are examples of " plane geometrical figures/' i. e., flat 
figures which have something definite about them, such as straight 
sides, or sides made up of smooth curves ; such figures are con- 
stantly occurring in engineering work. Before proceeding to any 




Acute Angle Righr Angle ObKuse Angle 



& 1 1 1 1 M M M I I ' i ' 1 1 1 ' i i 1 1 1 1 1 ' i ' 1 1 i ^.C 

^^MHMMi^ai^^MMj lMMM . I I -* "~ 

Rule shor Rule fully open 

Fig. 23. Illustrating Various Angles. 

actual calculation some descriptions and explanation of terms are 
necessary. 

Angle is the geometrical name for " corner." More particu- 
larly, it is the opening between the two lines making a corner. 
Thus, when a f elding rule is opened the two legs enclose or include 
an angle, as shown in Fig. 23. At a, in this figure, the opening is 
small and the angle is said to be " acute " meaning sharp. At c 
the opening is very large and then the angle is said to be " obtuse," 
meaning blunt. At b, the opening is exactly halfway between the 
acute and obtuse positions, or the rule is just halfway between 
being quite shut and being full open. This angle is a Right Angle, 
or a square corner, and is the most important of all. In ordinary 
language we should say that the two legs of the rule are " square 
with each other." 



MENSURATION 



247 



It should be noticed that the angle between two lines is not 
the distance between them ; the distance is different at different 
places. Two lines may include an angle although they do not 
meet and make a proper corner, as at e, Fig. 23 ; here the two full 
black lines are inclined to each other and therefore enclose an angle. 
The appearance of a corner is obtained if each line be continued to 
the left until the two meet as shown dotted. In geometrical lan- 
guage this continuing of a line is called " producing," and the line 
is said to be " produced." To indicate any particular angle on a 
figure the corners are lettered. When only two lines meet at a corner 
a single letter is sufficient. Thus at d in Fig. 23, we have the angle 
D. When more than two lines meet at a point several letters are 
necessary. Thus, the angle BDC is the one between the lines BD 



Wate 



face.-a Hon 2.0* fa I li 




a Verhcal line Pairs of Perpendicular lines 





a 



Parallel Lir\es 
!Hg. 24. Illustrating Various T-irics. 

and CD. The angle ADB is that between the lines AD and BD ; 
while the angle ADC is the whole angle between the lines AD and 
CD, *. e., the sum of BDC and ADB. 

Parallel lines are those which are the same distance apart every- 
where ; they would never meet however far produced in each direc- 
tion. Parallel lines may be either straight or curved, as in Fig. 24, 
or like the rails on a railway or tramway track. 

A Vertical line is one which stands straight up, and is at right 
angles to the earth's surface, as with a plumb line (see Fig. 24). 

A Horizontal line is one which is quite level, like the surface of 
a pond, as shown in Fig. 24. It is at right angles to a vertical 
line. 

Lines which are at right angles to each other are said to be 
perpendicular to each other. Neither line need necessarily be vertical 



248 ARITHMETIC FOR ENGINEERS 

or horizontal, as shown in Fig. 24. The vertical and horizontal 
directions are of course perpendicular to each other. 

A figure is said to be symmetrical about a certain line when the 
portions on either side of that line are exactly the same shape, 
but reversed (see Fig. 24). 

A plane figure is one that is quite flat and has no thickness. 

SIMPLE PLANE FIGURES 
(Letters refer to Fig. 25) 

All figures having four straight sides are called Quadrilaterals. 
Forms shown from A to F are examples. 

When opposite pairs of sides in quadrilaterals are parallel the 
figure is called a Parallelogram, and the opposite sides are then 
equal. Examples are shown from A to D. 

A Rectangle is a parallelogram in which all angles are right 
angles. Strictly the term applies to both A and B, but it is cus- 
tomary to restrict it to the form in which adjacent sides are unequal 
as at A, which is sometimes known as an " oblong." When all 
four sides are equal as at B, the figure is called a Square. 

C. A Rhomboid is a parallelogram in which the angles are not 
right angles and in which adjacent sides are not equal. This is 
the general form of parallelogram. 

D. A Rhombus is a parallelogram in which all four sides are 
equal, but the angles are not right angles. When placed so that 
the longest dimension is vertical, we recognise the " diamond " or 
" lozenge " shape. 

E. A Trapezoid is a quadrilateral having only one pair of parallel 
sides. There are various forms as shown. 

F. A Trapezium is a quadrilateral in which all sides are unequal. 
A name which better describes this figure is Irregular Quadrilateral. 

G. A Triangle is a three-sided figure, and has therefore three 
angles. At a, all three sides are equal in length, and the triangle 
is said to be " equilateral "; the angles are then all equal. At b 
only two sides are equal and the triangle is " isosceles." At c and 
g all three sides are unequal. 

The Right-angled Triangle (shown at g) is one in which one angle 
is a right-angle. This triangle is the most important in higher 
work. 

H. A Hexagon is a six-sided figure, in which all sides are equal 
and all angles are equal ; the opposite sides are then parallel. This 
is the shape of the common nut and bolt head. 

J. An Octagon is an eight-sided figure, in which all sides are 



MENSURATION 



249 



A) Recrcmgle (B 




c. 




. . 

CL. Equilateral b. Isosceles g. 

5)^JTVi angles Angled 




^oT^t?e7*7 { 

' (U) SecVor / 
(H) Hexagon (J) Ocragon (R) Grde \^ Cairc^e y 





Chord \ 




o{ Circle, / 



Semicircle 




(P) Ellipse (Various Forms) 

Fig. 25. Simple Geometrical Figures of Importance to-the Engineer. 



250 ARITHMETIC FOR ENGINEERS 

equal and all angles are equal ; the opposite sides are then parallel. 
Chipping chisels are frequently made from " octagon bar steel/' 

The remaining straight-sided figures are of no impoitance here. 

K. The Circle is a figure with only one continuous outline which 
is always the same distance from some point called the centre. 
This distance is the " Radius " (plural, radii). The dimension right 
across from one side to the other when taken through the centre is 
the diameter of the circle, and is, of course, equal to twice the 
radius. The actual curved outline is called the Circumference. 

L. A Sector of a circle is a piece of a circle with three sides ; two 
sides being radii (and therefore straight) and the third side a piece of 
the circumference. This curved piece is called an arc, and is said 
to " subtend " the angle at the centre (i. e., it is opposite this 
angle). 

M. A Quadrant is a particular sector in which the angle between 
the two radii is a right angle. Since there are four such sectors 
in the whole circle a quadrant is a quarter of a circle. 

N. A Segment is a piece of a circle with only two sides ; one an 
arc and the other a straight line from side to side, which is called 
a Chord. When this chord passes through the centre of the circle 
the segment becomes exactly half a circle and is then called a 
Semicircle (shown at O). It should be noted that the distance along 
the arc is considerably longer than that along the chord. 

P. The Ellipse (or Oval) is a figure with one continuous outline 
and is longer one way than the other, being in fact a stretched 
or flattened circle. It is symmetrical about two perpendicular 
lines passing through its centre, the longer being called the " major 
axis," and the shorter the " minor axis." At X, the major axis 
is much longer than the minor axis. This is approximately the 
shape of the cross-section of the tube in a steam -pressure gauge. At 
Y the shape is a medium one, as in a boiler manhole. At Z the 
major and minor axes are very nearly equal ; when equal, our ellipse 
becomes a circle. Figures like the above drawn with circular arcs 
are not necessarily ellipses, as certain laws exist as to the shape 
of the curves. These laws cannot be dealt with until a much later 
stage. 

The majority of the shapes we deal with in engineering work 
are these regular figures or combinations of them, since the regular 
figures are easily produced by machine processes, e.g., the circle 
in the lathe, and rectangular shapes on the planer. Irregular shapes 
require either expensive machines to produce, or much hand labour, 
which is both costly and slow. 



MENSURATION 251 

Perimeter or Circumference. The actual distance round 
the outline (or the length of the boundary) of any figure is called 
its Perimeter or Circumference. With the straight -sided figures the 
calculation of the perimeter is an easy matter when the length of 
the sides are known. If all the sides are not known, but sufficient 
information is given to draw out the figure, then it should be drawn, 
and the sides actually measured. 

In all the straight-sided figures the perimeter is the sum of the 
lengths of all the sides, particular cases being formulated as follows : 

1. Rectangle : Let a = length of long side and b = length of 
short side. Then Perimeter = 2a -f 26 or 2 (a + b). 

2. Square : Let s = length of side. Perimeter = 4s. 

3. Equilateral Triangle : Let s = side. Perimeter = 3s. 

4. Hexagon : Let s = length of side. Perimeter = 6s. 

5. Octagon : Let s = length of side. Perimeter = 8s. 

Circumference of a Circle. Meaning of TT. The length of 
the circumference of a circle cannot be so easily reckoned as that 
of a straight -sided figure. Roughly the circumference is a little more 
than three times the diameter. If several circles be drawn and 
their diameters and circumferences measured in some way, then it 

.,11 t j XT. j. xt. A - Circumference . , . 

will be found that the ratio - ~-.- - , is always very nearly 

3-14. This figure or " constant " is a very important one, and 
occurs in a great many calculations. .In formulae it is usually 
denoted by the Greek letter TT (pi), its value not being known with 
absolute exactness, although it has been calculated to 700 decimal 
places ! Approximate values of varying accuracy may be used 
depending on the class of work in hand, so that for practical calcula- 
tions we can consider it to be known exactly. 

To 8 significant figures TT = 3-1415928 
,,4 = 3<1 ^ 2 

The latter is quite suitable for the great majority of engineering 
calculations. Then we have 

Circumference of a Circle = ir x diameter 

or in symbols, if c = circumference and d = diameter 

C = ird. 

Since the diameter of a circle is twice the radius, then if r = radius, 

d = 2? 

Then c = 2?rr, a useful form. 



252 



ARITHMETIC FOR ENGINEERS 



Occasionally IT is given in the form of a vulgar fraction ; the com- 
monest value being - 2 r 2 - *' g - 3r- The decimal equivalent of this 
is 3*142857 .... so that the difference is not great. Another, and 
very accurate, form is -J-JS, *' -, 3i\V * which the decimal equivalent 

is 3-i4*59 2 9' 

Determination of TT by Measurement. Students should 
find the value of ?r by actual measurement of the circumference of a 
circle. There are several methods possible, of which one follows : 

Draw a circle with a diameter of 3" or more and make a mark 
somewhere on the circle as at A, Fig. 26. Open a pair of dividers 





Stepping round Irhe circle With 'dividers 
Fig. 26. Determination of TT by Measurement. 

to some small even distance, say J", and starting from A, step care- 
fully round the circle as indicated. The points of the dividers 
must be very carefully placed on the circumference each time. 
Count the number of steps taken from the start until the mark A 
is again reached. The last step may be less than J", in which case 
it should be estimated as or of a step. If the step is small and 
the circle fairly large, the length along the arc from one point to 
the next is almost exactly equal to the length of the chord, i. e. t the 
straight distance between the points. 

Then if N be the 'number of steps and / the length of the step 
the circumference == N/. 

Thus, with a 3" dia. circle, and J* step, 37! steps were made 
round the circle. 

Then circumference = 37! x \* 



crcum. 

,. --- 

dia. 



9-438 



3-146" 



which is very near to the actual value of 3-142. 



MENSURATION 



253 



If (in a class) several students do this, each having a different 
sized circle, all the ratios should come in the neighbourhood of 
3-14, and the average value should be very near to 3-142. 

Examples involving Circumferences of Circles. 

Example 267. A steam engine cylinder is lagged with wood strips 
held in place by two brass bands as shown at a, Fig. 27. Determine 
what length of brass strip is necessary for each band. 




Views of | 

Hne coil.^ 
Fig. 27. Examples on Circumference of Circle. 

The portion round the cylinder is less than a full circumference by 
the \" between the ends. In addition to this there are the two pro- 
jecting pieces each \" long, for the bolt. 

Circum. of circle = vd 

= 3-142 x ii 34*5 6 * 

Add for two lugs each J" long, ij* and deduct \" piece for space, 
therefore add i*. 

/. Length required = 35 '56, say 35!" 

or, 2 / -ii$" 



254 ARITHMETIC FOR ENGINEERS 

Example 268. A spur wheel is to have 45 teeth of }* pitch. What 
will be the "pitch diameter" of the wheel (i.e., the diameter of the 
circle on which the pitch is measured see b t Fig. 27) ? 



Circum. = 45 x }" = 3375 

33'75 
o:> ' J 

3-142 



~. circum. 
Dia. = ------ = 



In certain cases, given the number of revolutions per unit of 
time, we need to know the surface speed of a revolving wheel, i. e. t 
the speed of a point on the circumference. Obviously in i revolu- 
tion a point on the circumference will travel a distance equal to 
the circumference, in 2 revolutions 2 X circumference, or for N 
revolutions N x circumference ; or 

Surface speed (ft. per min.) = Revolutions (per min.) 

x circumference (ft.) 

(The surface speed is also called the " circumferential " or the 
" peripheral " speed periphery being another name for circum- 
ference.) 

The following examples will illustrate : 

Example 269. The commutator of a dynamo is 19* dia., and 
makes 470 revolutions per minute. Find the surface speed in feet 
per minute. 

Surface speed = circum. x revs. 
Circum. of commutator = x = 4-975 ft. 

.'. Surface speed = 4-975 x 470 

= 2338 ft. per min. 

Note. In first line 19 is divided by 12 to bring inches to feet. 

Example 270. A milling cutter is 3" dia. and is to be used with a 
cutting speed of 50 ft. per min. At what speed (revs, per min.) must 
it run ? 

T-, . cutting speed 

Revs, per min. == -? - ~ " 

r circumference 

Cutting speed = 50 ft. per min. 50 x 12 ins. per min. 
Circum. of cutter = 3 x = 9-426" 

j-Q \X T ^ 

.". Revs, per min. = - : - = 63-6 revs, per min. 

Example 271. A copper expansion bend in a long length of steam 
pipe is shown at c, Fig. 27. The portion BCD is J of a complete 
circle, and the portions AB and DE are each j of a circle. Find 



MENSURATION 



255 



the total length of pipe from flange to flange necessary to make 
the bend. 

Portion BCD = J of circum. of circle 

= I X x 15 = 35'4* 

Portion AB = f of circum. of circle 
Portion DE = | 
/. AB 4- DE = | 

= | x IT x 5 = ii-S" 

Two straight portions 10" each = 20 -o" 



/. Total length required 
say, 5'-7i* 



67-2" 



Example 272. A close-coiled helical (or coil) spring is 3* average 
dia. (i.e., dia. measured at centre of coils) and contains 10 complete 
coils. What length of wire is necessary to make the spring, allowing 
an extra 8" for hooks at the ends ? (See d, Fig. 27.) 

Note. Considering one coil or turn it is seen from A, Fig. 27, that, 
like a turn of a screw thread, it is not a true circle, its length being a 
little more than the circumference of a circle, but the difference is so 
small in a closely coiled spring that for practical purposes it can be 
neglected. Then the length of i turn is calculated as for a true circle, 
and is multiplied by the number of coils. 

Length of i turn = ir x 3 = 9-42" 
/. Length of 10 turns = 9-42 X 10 = 94'2 /r 
Add for ends 8 

.*. Total length required per spring = 102-2" 
or, I 




Fig. 28. 



(Nor oil 
Fig. 29. 



Example 273. A resistance for an arc lamp is to be in the form 
of a close wound coil as in Fig. 28. The total length of wire is to be 
44 yds., and the coil is to be wound on a cylinder of 2j" dia. How 
many turns will there be ? If the wire is ^ dia,, what length of 
cylinder will the wire occupy ? 



256 



ARITHMETIC FOR ENGINEERS 



Circum. of 2%" circle 
Total length 



, X 2-5 = 7-85" 

44 yds. 

44 x 36 ins. = 1584* 

, . Total length 1584 
No. of turns = ^ TT- ?-? = o = 202 turns 

/. Length of coil = 202 X fo = 12*62 5" 



Example 274. A coil of leather belting (see Fig. 29) has an outside 
dia. of 10" and an inside dia. of 4", and there are approximately 12 
turns. What is the approximate length of belting in the coil ? 

Note. Any single turn of the coil is not a true circle but a turn of 
a spiral as shown at A, Fig. 29. For practical purposes its length is 
that of a circle. Since all turns are of different diameters it would 
be very tedious to calculate the length of each turn separately. There- 
fore we take the average or mean (or centre) turn and calculate on 
that. For every turn smaller than the mean diameter there is one 
larger than the mean. 

^_, ,. outside dia. -f inside dia. 
Mean dia. = 



. Mean circum. = w x 7 = 22* 
Total length in 12 turns = 12 x 22* = ~ 

22 ft. 



ft. 




Example 275. Fig. 30 shows a field coil for a 
dynamo, with dimensions. Find the length of the 
wire in the coil, which is required in calculating the 
resistance, etc., of the field windings. 

Considering I vertical layer of the wire. 



No. of turns in i layer = - 5 o == 
J -070 



', say 57 turns.* 
Considering the horizontal layers. 



No. of layers 



078 



= 25-6, say 25 layers. 



Fig. 30. 



Thus we have 25 layers with 57 turns in each 
layer. 

.*. Total number of turns in the coil = 25 x 57 = 1425 turns. 
Now the turns vary in diameter; the smallest being in the inner- 



* In examples of this nature it is best to discard any decimal of 
a turn rather than work to the nearest whole number ; it is preferable 
to have the calculated answer a little small. 



MENSURATION 257 

most layer and the largest in the outermost. Then we may calculate 
on the length of an average turn, 

Outside dia. = 12" Inside dia. = 8* 
/. Dia. of average turn = = 10* 

/. Length of average turn = *d = - ~ 2 ' 62 ft. 

/. Total length of wire in coil = Average turn x No. of turns 
= 2-62 X 1425 = 3730 ft. 

Example 276. An electrically driven hoist is to run at a speed 
of 200 ft. per min., and the winding drum is to revolve at 26 revs, per 
min. What must be the diameter of the drum ? 

Note. In 26 revs, the drum must wind up 200 ft. of rope ; hence 
we can find the length wound on in i rev., which gives the circum. of 
the drum. 

Circum. of drum = - - f - = 7-7 ft. 

26 ' ' 

| /. Dia. = ? = 2-45 ft. 
say, 2 '-5 * 

Exercises 67. On Circumferences of Circles. 

(Letters refer to Fig. 31.) 

1. The rings of a locomotive boiler, 4 /> -3 // dia., are butt-jointed 
as shown at A. Calculate the length of plate necessary for each ring. 

2. The rings in a large marine boiler are built up of three equal pieces, 
butt-jointed as shown for the single joint at A. If the boiler is i4 / -7j /r 
dia., find the length of the pieces. 

3. An angle ring for the top of a cylindrical tank is to be 3'-2^* 
dia. (measured on the centre line). What length of angle bar must 
be cut ? 

4. Find the length of plate required for a lap-riveted boiler ring 
6'-o / " dia., if the lap-joint requires an extra 4^ ins. 

5. A wrought-iron pulley is 3 / -o /!r dia. Find the length of plate 
required for the rim. There are 10 arms evenly spaced. Find the 
distance apart of the holes which must be drilled in the rim to receive 
the ends of the arms. 

6. A concrete culvert is reinforced with two sheets of expanded steel, 
as shown at B, with a lap of 8*. What is the width of each sheet ? 

7. In a steam engine cylinder lagged as shown at a, Fig. 27, the 
lagging is to be composed of wood strips ij" wide and is to be n* dia. 
as nearly as possible. Find the number of strips required per cylinder. 

8. An electric cable is to be " armoured " with a ring of steel wire, 
No. 8, S.W.G. (i. e., i6 // ) diameter, as at C. The diameter of the circle 
through the centre of the wires is to be 2j". How many wires are 
required ? (Give the nearest whole number.) 

S 



258 ARITHMETIC FOR ENGINEERS 

9. An alternator armature is 60" inside diameter and is to have 
60 slots, as at D. What will be the " pitch " of the slot (i. e., the 
distance from the edge of one slot to the same edge of the next slot) ? 
If the slot is i- 2" wide, what will be the width of the " tooth " ? 

10. A spur wheel is to have a pitch circle diameter of 3'-9i /r and 
a pitch of 2 1". How many teeth will it have ? (Note. Work in inches 
and give nearest whole number.) 

11. A boiler-maker's wheel tool for measuring curved lines on 
plates is exactly 2 ft. circumference. What is its diameter in inches ? 

12. A worm wheel in an electric lift drive is to have 43 teeth of 
if* pitch. What will be its pitch circle diameter? 

13. A spur wheel in a lathe back gear is to have 79 teeth, and 
6 teeth for every i" of diameter. Find its diameter; also calculate 
the pitch of the teeth. 

14. An emery wheel 18" dia. runs at 1200 revs, per min. Find 
its circumferential speed in feet per min. 

15. A wood-planing machine with a 5* dia. cutter is to run at 
2000 ft. per min. Determine the revs, per min. 

16. An electric motor has a belt pulley 6" dia., and runs at 550 revs, 
per min. Calculate the speed of the belt in feet per min. If this belt 
drives a machine shop shaft by a pulley 24" dia., find the revs, per 
min. of the shop shafting. 

17. Find the length of pipe required fronAflange to flange to make 
the expansion bend shown at E. 

18. A band saw is to run over two equal pulleys 3 ft. dia., their 
centres being 5 ft. apart as shown at G. Calculate the length of saw- 
blade required. 

19. At F is shown a bent copper steam pipe. What length is 
required from flange to flange ? 

20. A pipe hanger is required to the dimensions shown at H. What 
length of bar is necessary ? 

21. Dimensions are given at J of the curved portion of the splasher 
guard over the driving wheel of a locomotive. What length of plate 
is required ? 

22. Dimensions of a sheet metal guard over the back gear of a 
milling machine are given at K. Find what length of strip is 
required. 

23. Dimensions are given at L of a strap brake on a traction engine. 
Determine wlmt length of strap is required. 

24. A closely coiled spring for a coil clutch is shown at M. Find 
what length of bar is required, there being 4^ complete coils and 4* 
extra for the ends. 

25. An inductance coil, as shown at N, is required for a wireless 
installation. If it has 60 turns and is wound on a cylinder of 3^" dia., 
what length of wire is required ? 

26. A flat coiled spring in an indicator reducing gear is ij" inside 
dia. and 3* outside dia., as shown at O, and there are 22 coils. Find 
the length of strip in the spring. 

27. A volute spring for the buffer between engine and tender on a 
locomotive has n complete coils, and is 6" outside and 2" inside dia., 
as shown at P. Calculate the length of material required. (Note. 
This may be treated as a flat coil spring.) 



MENSURATION 




Fig. 31. --Exercises on Circumferences of Circles and Ellipses. 



260 ARITHMETIC FOR ENGINEERS 

28. A coil of leather belt as in Fig. 29 is 8* outside and 2j* inside 
dia., and has n coils. What length of belt does it contain ? 

29. A roll of sheet zinc is found to have 24 coils and is ft" outside 
and 3 J" inside dia. What length does it contain ? 

30. At Q is shown a " Gutermuth Valve/' used in certain pumps. 
The outside diameter is 4", and the inside i", and there are 4 complete 
coils. Find the length of strip required, allowing 4^ for the straight 
flap and \* for attaching to the central spindle. 

31. Calculate the length of wire in the field coil of an electric motor 
as in Fig. 30, the dimensions being outside dia. 8", inside dia. 5", length 
3*, winding depth i \" , and dia. of wire (outside insulation) 07". 

32. A cylinder is 8" dia. and the spring rings are required to be 
open }V when in the cylinder (see R). If the rings are turned 8J" dia. 
how much must be cut out ? 

Circumference of Ellipse. The exact formula for this is 
very complicated and far too cumbersome for practical use. Hence 
various approximate formulae have been proposed. Of these two 
are given, the first being the simpler and the second the more exact. 
The accuracy desired in the particular example should decide which 
formula is to be used. The more exact formula gives less than 
i% error if major axis is less than 5 times minor. 

Simpler form : Circumference of Ellipse = TT (a + b) where a = 
\ major axis and b = J minor axis (see Fig. 25). The error in 
using this formula is 2*7% when a = zb and as much as 10% when 
a~$b. 

Note. When our ellipse becomes a circle, a = b = r , then circum- 
ference = ir(r -f- r) = if X 2f = 2-nr as already known. 

More exact form : Circumference of Ellipse = 7r{l*5(a -f b) Vab} 

Example 277. An oval or elliptic chain link is shown at a, Fig. 32. 
Find the length of bar in the link. 

This will be worked by the more exact formula. The student 
should work by the simpler one to see the difference. 

a = J major axis = \ X 2 = i* 
b = i minor axis = J X ij = f* = '625* 
A Circum. of ellipse = ?r{ 1-5(0 + b) Vab\ 

= 7r{i-5(i + -625) - Vi x -625} 
= 7TJ2-44 - -79} 
= TT X i-,65 
/. Length of rod per link = 5-18" 

Example 278. Fig. 32 shows at c an elliptic gear wheel. The 
major axis is g" long and the minor axis 4-7" long, and there are 30 
teeth. What is the " pitch " (i.e., distance from centre of one tooth 
to centre of next) ? 



MENSURATION 

In this case the more exact form is desirable. 
* = i X 9 =4'5 



261 



6 = i X 47 = 2-35 
Circum. = ir 



.*. a 4- 6 = 6-85 

and Va6 = VZ-^ 



.'. Pitch = 



= ir{i-5(a + b) - Vrtfc} = 

== 3-I4 X 7-05 = 22-1" 

Circum. __ 22-1 
No. of teeth ~~ 30 " 
-737". say, f* 



2 -35 

= 3-25 
- 3-25! 




Elevahow 



ynp^i 



LX 1 




d 



FIG. 32. Examples on Circumference of Ellipse. 



Example 279. At d t Fig. 32, are given dimensions of an oval hand 
hole on a cylindrical tank, under a vacuum. For staunchness the pitch 
of the bolts must not exceed 6 times the bolt diameter. Using J*-bolts 
how many must be used to satisfy this rule ? (Note. It is preferable 
to use an even number of bolts. If a mixed number be obtained then 
use the next larger even number.) 

In this case the simpler formula may be used - 

Major axis = SJ" Minor axis 6J 
/. a = 4-25 ; b = 3-25 ; and a + b = 7-5 
/. Circum of ellipse = TT x 7-5 = 23'52 r ' 

Bolt dia. = i" = -625" 
.'. Largest allowable pitch = 6 x -625 



Least possible number of bolts = --^>- 
Next even number = 8 



3*75" 

=6-27 
use eight $"-bolts. 



The arrangement would then be as at b, Fig. 32. 

A circumference formula often used is TT V2(# 2 + b*), but this is 
almost as approximate as ir(a +b), and will in certain cases give 
an error of 7 %. 



262 ARITHMETIC FOR ENGINEERS 

Exercises 68. On Circumferences of Ellipses. 

(Letters refer to Fig. 31.) 

1 to 3. Find the circumferences of each of the following ellipses, 
using both the simple and the more exact formulae : 

1. Major axis 6". Minor axis i$". 

2. Major axis 15*. Minor axis 10*. 

3. Major axis 12". Minor axis 10*. 

4. An elliptic gear as at c, Fig. 32, has major axis 2* and minor 
axis i -4 1 4", and 38 teeth. Calculate the pitch. 

5. Calculate the length of rod for an elliptic chain link (see a, Fig. 32), 
if the centre-line dimensions are major axis 3^, minor axis 2 J". 

6. Mustard tins of elliptic section are to be made of dimensions 
shown at S. What length of sheet tin is required for the walls of each 
tin if an extra J* be allowed for the joint ? 

7. A boiler mud-hole is elliptical, similar to d, Fig. 32, and the 
pitch line of the rivets measures iS" x 13!". If the rivets are pitched 
2* apart how many rivets will be required ? (Give nearest even 
number.) 

8. At T is shown, with dimensions, a magnet coil, elliptical in 
section. If the wire is -vyS" dia., calculate the length of wire required 
for the coil. (Note. Work this exactly as Example 275, but use the 
circumference of the average ellipse,) 

9. A semi-elliptic arch is shown at V. How many 3* bricks are 
required for the inner ring ? 

10. The conveyor tunnel shown at U is semi-circular at the top 
and semi-elliptic at the bottom. Find how many 3" bricks are required 
to line the inside. 

11. Part of the cross section of a split concentric electric cable is 
shown at W. The outer ring of wires is elliptical, the wire being No. 19, 
S.W.G. (04 /x ) dia. Determine how many wires there are in this ring. 

12. At X are given dimensions of a y " Simplex " oval conduit for 
carrying electric wiring. It is made out of one strip of steel, rolled and 
brazed. Calculate the width of strip required. 

Measurement of Angles. Degrees. When measuring the 
size of an angle, two systems may be employed. The one in general 
use is that in which the unit is the degree. 

The degree is ^ part of a complete revolution. 

At X, Fig. 33, is shown a circle divided up into 36 equal parts, 
each division indicating 10 degrees. The divisions are numbered 
off in a left-handed or anti-clockwise direction, i. e. t in a direction 
opposite to that in which the hands of a clock move. The first 
division is split into 10 parts, so that each of the smallest divisions 
represents i degree. The unit is a very small one, as can be seen 
from Y, Fig. 33, which shows an angle of 5 degrees. When writing 
down angles, the word " degree " is replaced by a small o () at the 
right-hand top corner of the figure. Thus 90 reads as 90 degrees. 



MENSURATION 



263 



The right angle being J of a revolution, is J of 360, i. e., 90. 
Similarly two right angles, i. e., J a revolution, is 180, and three 
right angles 270. 

For finer measurements, such as are required in surveying work, 
the degree is sub-divided into minutes and seconds, as shown in 
the following table. 

ANGULAR MEASURE 

60 seconds (sees.) = i minute (min.) 
60 minutes = I degree ( or deg.) 

360 degrees = i revolution (rev.) 

The minute is indicated by a single tick, and the second by two 
ticks. Thus, 35 15' 28^would be read as 35 degrees, 15 minutes, 
28 seconds. 

The minute and second are far too small to be shown in a picture , 
and their use in instruments is only possible by means of special 
apparatus. Another system of angle measurement adopts as its 
unit the Radian (= 57-3), but this will not be dealt with here. 

The Protractor. 

To measure and set out angles in degrees a portable divided 
circle is used, known as a protractor. It may be either semi- 



45-0 




70 



Fig. 33. Measurement of Angles. 

circular (the most common form), circular or rectangular, and may 
be of celluloid, brass, wood, or ivory. Z, Fig. 33, shows a semi- 
circular one ; other forms are shown in instrument-makers' catalogues. 
They are usually marked in both directions of reading, i. e. t left and 



264 ARITHMETIC FOR ENGINEERS 

right handed, so as to be equally useful either way. A convenient 
size is 6" diam. ; below this, reading is limited to an accuracy of 
about i. 

In measuring angles the bottom line AB opposite o and 180 
is placed carefully along one line of the angle, so that the central 
mark C is at the point of the angle. Then the angle is read off on 
the scale, taking care to read in the correct direction, since the 
divisions are marked for both directions. Thus, the protractor in 
Fig. 33 is shown reading an angle of 63. Full degrees are obtained 
exactly from direct markings; parts of a degree may be estimated 
by eye, although half-degrees are sometimes indicated. 

In setting out an angle, the instrument is placed as for reading, 
and then the required angle is marked by a fine dot or tick at the 
proper figure on the scale. Removing the protractor this tick is 
joined to the point at which C was placed. 

Important Angles. Besides the right angle there are certain 
angles of frequent occurrence which deserve special mention. The 
angle of 45 or half a right angle, shown in Fig. 34, appears on the 
45 set-square, i. e. t the one with two equal sides. 






Fig. 34. Important Angles. 

The angle of 30, i. e., J of a right angle, and the angle of 60, 
i. e., f of a right angle, appear on the 30 - 60 set-square, and 
are also shown in Fig. 34. 

In any triangle the sum of the three angles is always 180. Thus 
if two of the angles are known and their sum is subtracted from 
180, the third angle is found. 

Addition and Subtraction of Angles are carried out in 
columns, in much the same manner that money is dealt with. 
These operations appear very considerably in surveying work 
where the measurement of angles is very important. 

Example 280. The angles of a four-sided paddock were measured, 
with the results given below. If the work has been exact then the 
sum of the angles should be 360. Find this total, and the error of 
measurement, if any. 



MENSURATION 



265 



Station 
(or corner). 


Measured Angle. 


A 


33 4*' 


B 


in 51' 


C 


49 "' 


D 


164 12' 



360 oo 

358 57 



Error = 



Total = 358 57' 

Note. The sum of the minutes column is 117 = i-57'. Carry 
i to degree column. Difference between total and 360 gives error. 
Remember that a i has been borrowed from the 360 when taking the 
57' away. 

Example 281. In a traverse survey, with a theodolite, of a five- 
sided figure, the angles in the following table were obtained. Find the 
total, which should be 1260 exactly. Find the error if any. 



Station 
(or corner). 


Angle. 


A 


277 15' 


00* 


B 


215 8' 


20" 


C 


229 29" 


20" 


D 


246 II' 


40* 


E 


291 52' 


40" 



Total 



Error 



1260 



57' oo" Note. Top line to be 

oo oo subtracted from bottom. 



3' oo" too small. 



Reduction of Angles. It is sometimes easier when calculat- 
ing, to use angles in degrees and decimals of a degree, rather than 
degrees, minutes, etc. Hence it is necessary to convert from one 
style to the other. 

Example 282. Convert 39*38 into degrees, minutes, and seconds. 

The degrees remain : decimal to be converted. 
38 X 60 = 22-8' 
8' x 60 = 48" 
.'. Angle = 39 22' 48" 

Example 283. Reduce the angle 73 19' 12" to degrees and decimals. 

12 I 

12* = ^- mins. = = *2' 
60 5 

/. 19' 12" .-= 19-2' 



Angle 



32 degs. 
73-32 



2b6 ARITHMETIC FOR ENGINEERS 

Exercises 69. Addition, Subtraction, and Reduction 
of Angles. 



1. The angles -measured in a traverse sur- 
vey are given in the table. Find the total of 
these measured angles. This should be exactly 
540; any difference is an error. Find the 
error if any. 



Station. Angle. 



A 
B 
C 
D 
E 



115 12' 

H9 3' 

110 o' 

121 n' 

74 54' 



2. A series of measured angles have been adjusted to account for 
slight errors in reading, etc. These angles are : 277 15' 36"; 
215 8' 56"; 229 29' 56"; 246 12' 16" and 291 53' 16". The total 
should be exactly 1260. Try if this is so. 

3. The three angles of a triangle should total up to 1 80. In a certain 
triangle two angles, when measured, were found to be 35 14' and 
98 47'. Find the third angle. 

Convert the following angles into degrees, minutes and seconds : 

4. 19-35. 5. 17-28. 6. 75'06. 7. 0785. 
Convert the following angles into degrees and decimals of a degree : 
8. 25 15'. 9. 30 18' 50". 10. 65 22' 30". 

11. n 5' 50". 12. 49 13' 49"- 

The Right-angled Triangle. In any right-angled triangle 
we have the following very important relation : 

Square of Hypotenuse = Sum of the Squares of the other two sides ; 

or in symbols, if I = hypotenuse (a word used for the side opposite 
the right angle, in a right-angled triangle, i. e., the longest side), 
and h and b the other two sides, then 

I 2 = h 2 -f b 2 

Thus, if the two sides of a triangle enclosing a right angle be made 
3" and 4" long, then on measurement the other side will be found 
to be 5" long exactly. 

Thus, 3 2 + 42 = 9 + 16 = 25 = 5 2 

This particular triangle is shown in Fig. 35, where the squares on 
each side are drawn in detail. It can be seen that the number of 
square inches in the big square is equal to the number in the other 
two squares. The student should draw several right-angled triangles 
at random and measure the sides of each; square all the values 
obtained for each triangle and add together the two smaller figures, 
which will be found to equal the third and largest square. 

From the above formula we obtain by simple transposing 
h 2 = I 2 b 2 and b 2 = I 2 h 2 . Hence, when dealing with right- 
angled triangles : if the length of two of the sides be known, the 
length of the third can always be calculated. 



MENSURATION 



267 



The fact that a right-angled triangle with shorter sides of 3" 
and 4" has a long side of 5" is useful for setting out a right angle 
by measurement only. A triangle is drawn with sides 3, 4, and 5 
long (in inches, feet, etc., or in chains for surveying work) thus 
producing a right angle between the 3 and 4 sides. 




Fig. 35- 




I^'- T -1 2 F 
End View 



Fig. 36. 



Example 284. The length of a governor arm is 10". If the radius 
is 7-5", what is the height ? See Fig. 37. 

Let / = length of arm, h = height, and r radius. 
h* = I* - r* 

= io 2 - 7'5 2 

= 100 - 56-25 = 4375 

/. h = V~^375 = 6-61* 

Example 285. A triangular air duct for use in a cooling tower is 
made up as shown in Fig. 36, which also gives dimensions. What will 
be the width of plate necessary for each side ? 

The portion on either side of the centre line forms a right-angled 
triangle, short side i'; vertical side 2'. 

Then I 2 = i 2 -f 2 a = 5 
/=V5~= 2-236' 

= 2 / -2| // , say 
Lap for riveting at top = i%" 



/. Plate must be 2 / -4f /r wide 



268 



ARITHMETIC FOR ENGINEERS 



Example 286. An open-coiled spring has an average diameter of 
2j", and has 8 complete turns or coils in a length of 16*. What total 
length of wire is necessary to form the spring ? (See Fig. 38.) 

In this case, unlike that taken on p. 255, the length of one 
complete turn is appreciably more than the circumference of the 
coil, since the forward movement is so large. Fig. 38 shows that 
a screw thread or helix is merely a straight inclined line wrapped 
round a cylinder. Thus, at (a) is shown a sheet of paper with an 
inclined line drawn upon it, resting flat against a cylinder. At 
(b) the paper is partly wrapped round and at (c) it is completely 




= >85 
Fig. 38. Open Coiled Helical Springs. 

wrapped round. The straight line now presents the appearance 
of a screw thread. If therefore we unfold one turn of a screw thread 
or helix we can find the length of the inclined or screw surface. 

In Fig. 38, the triangle PQR represents one turn of the centre 
line of our spring unwound. P is the starting point p, and Q is 
the finish of the complete turn as at q. Then QR is the distance 
advanced in one turn, i. e. t the " pitch," and PR is the distance round 
the cylinder from p to p again, i. e. t the circumference of the 
cylinder. Then PQR is a right-angled triangle and PQ, the actual 
length of the centre line, may be calculated. 

Circum. of cylinder = *d = IT x 2-5 = 7*85* 



Pitch = - 



Length 



No. of Coils 



16 
8 



SB ZZ = 2" 



MENSURATION 269 

Then, if / = length of one turn = PQ, 

12 _. 7> 3 5 a 4- 2* = 61-6 + 4 = 65-6 
/. / =s V65-6 = 8 i* =s length per single turn 
/. Length for 8 turns = 8-1x8 = 64-8" 
say, 65* or $'-5* 

Proportions of 45 and 60 30 Right-angled Triangles. 

With the aid of the foregoing relation we can examine certain useful 
proportions in the two right-angled triangles most often met with. 

The 45 Right-angled Triangle. 

See A, Fig. 39. Let the short side be I. 
i. e. t b = h = i (b = h is only true for a 45 right-angled triangle) 

Then / 2 - 6 2 + A 2 



. e., if short side is i, hypotenuse is 1*414 
or Hypotenuse = 1*414 x Short side. 




Fig. 39- 
From this we get 

Short side - Hypotenuse - -707 x Hypotenuse 
1.414 

If a square is divided into two parts by a diagonal each triangle 
is a 45 90 triangle. Hence the diagonal is 1*414 x side. 

The 60 30 Right-angled Triangle. See B, Fig. 39. 

In this case we cannot deduce the relations by use of the right- 
angled triangle only : some higher work is needed. But by proving 
certain proportions found by measurement they will be better 
remembered. If the student draws a 60 30 right-angled triangle 
carefully he will find on measurement that the shortest side is half 
the longest side (or hypotenuse). 

Thus if b = i, then / = 2 



270 



ARITHMETIC FOR ENGINEERS 



Then h* = I 2 - b* 

=4-1=3 

/. / = \/3 = 1732 as may be verified by measurement. 

Thus if shortest side = i, longest side = 2, and other or inter- 
mediate side = 1732 ; or 

Longest side = 2 x shortest side 
Intermediate side = 1*732 x shortest side 

From the above we also obtain 

Shortest side = '5 x longest side 
Intermediate side = 1732 x shortest 

= 1732 x (-5 x longest) 

= -866 of longest side 
which again may be verified by measurement. 

Exercises 70. On Right-angled and 60 - 30 Triangles. 

(Letters refer to Fig. 40.) 

1. A telegraph pole is strengthened by a wire rope stay as shown 
at A. Determine the length of the stay between the points of attach- 
ment. 

2. A crane jib is to operate at a radius of 18 ft., when the top is 
20 ft. above the bottom, as at B. What is the length of the jib ? 




Side 
Fig. 40. Exercises on Sides of Triangles. 



End View 



3 to 5. A crane jib is 27'-6 / " long. At what radius does the head lie 
when the jib makes the following angles with the horizontal ? (See C.) 
3. 60 4. 45. 5. 30. 

(i. e. t what are the distances A, B, and C.) ? 

6. At D are given dimensions of a North light roof truss. Calculate 
the length of the two rafters AB and AC. 

7. In the roof truss in Ex. 6, calculate the length of the strut AD, 
using the result of AB found in Ex. 6. 



MENSURATION 271 

8. At E are given dimensions of the conical portion of a cooling 
tower shell. Calculate the total width of plate required for the inclined 
portion. 

9. In connection with the flow of water in open channels of the 
iorm shown at M, Fig. 60, it is necessary to find the length of the inclined 
side AB. Find this length for the case shown. 

*10. A segmental arch has a span of 37 ft. and a radius of 22 ft. as 
shown at F. Calculate the " rise," i.e. the height AB. (Hint. First 
find the distance BO, and subtract from the radius.) 

11. In connection with alternating electric currents, we have the 
right-angled triangle shown at G. If resistance = -5 ohm and re- 
actance = *oi, calculate the value of the impedance. 

12. Referring to G, if impedance = -83, and resistance = '47, find 
the reactance. 

13. Bright hexagon-head screws are to be cut from round stock as 
indicated at H, and are required to be 1-3" across the flats of the head. 
What must be the least diameter of the round stock ? (Hint. The 
least size is that across the corners. The method is indicated by the 
small triangle.) 

14. Calculate the length of piping required for the evaporating coil 
shown at J, if the average diameter is 21". 

15. Find the length of hand-railing required for a spiral staircase, 
the diameter being 10 ft. and the pitch 9 ft. 6 ins., if there are 12 complete 
turns. 

16. An open coiled helical spring (see Fig. 38) has an average dia- 
meter of 4^" and has 15 complete turns in a length of 2 '-6*. Find 
the length of wire necessary to form the spring. 

17. A spiral shoot in a warehouse (similar to a spiral staircase) is 
made of steel plate edged on the outside with angle bar. The diameter 
is 6 ft. and the pitch 12 ft. Calculate the total length of angle bar 
required for n complete turns. 

Length of Arc of Circle. 

I. Given Radius and Angle. The length of the arc of a circle 
can be calculated exactly if its radius and the angle at its centre 
be known. If the ends of the arc are joined to the centre from 
which it was drawn a sector is formed. If the central angle is n 

then the sector is -y- of a complete circle. Since the whole cir- 



cumference of a circle is Zirr, the circumference of the sector is only 

-- of zirr. 
360 



Arc of Sector = -- x 2irr 



2 x 3-14 nr 

.__ _ *-* ' v/ /liv __ 

~p /\ fir ~iZ~r~~l^ 

360 57*3 

The length is in the same units as r. 



272 



ARITHMETIC FOR ENGINEERS 



Example 287. Fig. 41 shows an end view of a 5/5" corrugated 
sheet with dimensions (i. e. t it has 5 complete waves from A to B, each 
of 5" pitch). Find the width of flat sheet necessary to make this 
corrugated sheet. 




End View of 



Fig. 41. 5/5" Corrugated Sheet. 

Consider i half wave as shown at (a). 

It is a sector of a circle, radius 2 y and angle at centre 



Length of arc = 



rn 
57-3 



57*3 



= 2 . 6 * 

5 



In all there are 1 1 of these half waves, 6 on top of, and 5 below the 
centre line, as may be seen from the figure. 

.'. width of sheet required = n x 2-65 = 2g'2* 

say, 2 '-5J* 



2. Gi'v^n C/^or^ aw^ Height. In many cases the angle n is not 
known, the dimensions c (i. e., chord) and h (height) being given. 
The sector may be drawn and n and r measured, or the following 
approximate formula may be used, with an error of less than i%. 

T ^i_ f 8a c 

Length of arc = - 

where a = chord of half the arc == AC in Fig. 78, and c = chord 
of whole arc = AB. 

The value of a may be calculated from the right-angled triangle 

ACD. Thus a 2 = (|)* + h* and a - y -| + h. 

3. Given Chord and Radius (i. e., c and r). Height h must be 
calculated from the right-angled triangle OAD (Fig. 78), where 

r 2 = (-Y+ OD 2 or OD = Jr 2 - - and h then = r - OD. Then 
proceed as in 2. 



MENSURATION 



273 



Exercises 71. On Arc of Sector. 

(Letters refer to Fig. 42.) 

1. A curved channel-section bar in a vaulted roof is to be n'-6* 
radius and is to subtend an angle of 75, as shown at A. Find the 
length of bar required. 

2. A pipe bend is required to be of 39" radius, and to subtend an 
angle of 55. What length is required from flange to flange ? 




Fig. 42. Exercises on Arc of Sector. 

3. At B are given the arrangement and dimensions of a band- 
sawing machine. Calculate the length of the saw blade. 

4. Calculate the length of the bent pipe shown at C. 

5. At D are given dimensions of a bent copper pipe on a steam 
engine. Calculate its length from flange to flange. 

6. Calculate the length of the curved pipe shown at E. 

AREA 

Measurement of Area. The area of any figure is the amount 
of surface it contains. It is made up of two measurements com- 
monly known as length and breadth (or width), and consequently 




-- 1 poof te" - 
Fig. 43- 



Fig. 44. 



Fig. 45. 



its units are derived from those in the ordinary table of length 
measure. The Unit of area or surface measure is the Square Foot, 
i. e., the area of a square whose side is I Foot. The smallest sub- 
multiple in use is the Square Inch, i. e. t the area of a square of i* 

T 



274 ARITHMETIC FOR ENGINEERS 

side. Larger multiples include the square yard, the square pole, 
etc., as set out in the table on p. 4. The latter half of the table 
is only used in the measurement of land where very large areas 
have to be dealt with. The reason for the 144 in converting square 
inches to square feet is seen in Fig. 43. Here a square of i foot 
side is divided up into squares of i" side. Each side being 12" 
long, it is seen that the big square contains 12 rows of small squares, 
each row itself containing 12 small squares. 

.*. i sq. ft. = 12 x 12 = 144 sq. ins. 

It is also seen from this that the area of a square is found by " squar- 
ing " the side, or multiplying the length by the breadth. 

Also a square of i yard side has a side 3 ft. long, and therefore 
contains 3 x 3 = 9 sq. ft., 

i. e., i yd. = 3 ft. /. I sq. yd. = 3 2 = 9 sq. ft. 
Similarly i pole = 5| yds. .*. i sq. po. = 5j 2 = 30^ sq. yds. 

Area units = length units x length units = (length units) 2 or 

square units. 

The Surveyors' Measure was designed to enable areas to be 
easily calculated in acres. 

Since i chain == 22 yds., i sq. ch. = 22 2 = 484 sq. yds. 
.'. 10 sq. ch. = 4840 sq. yds. = 1 acre 

Now 1 acre = 4 roods 

= 4 x 40 sq. po. = 160 sq. po. 

= 160 x 3o sq. yds. = 4840 sq. yds. 



Reduction. The reduction from one unit to another used by 
the average engineer is practically confined to the use of the 144 
and the 9. 

Thus, to convert square inches to square feet divide by 144, and 
to convert square feet to square inches multiply by 144, 

and similarly in conversion with the square yard. 

Conversion among the other units is necessary when any land 
measurement occurs, the method being exactly similar to that of 
length. 

Example 288. The survey of a plot of land gives the area as 115*2 
square chains. Convert this into acres, etc. 



MENSURATION 275 

10 square chains = i acre 
. 115*2 sq. ch. = 11-52 acres 
_ 4 
2-08 rds. 



3-20 sq. po. 

3Q 
6-00 

| Of *2 ...... '05 

6-05 sq. yds. 
_ 9 

J45 sq. ft. 
Area = n ac. 2 rds. 3 sq. po. 6 sq. yds. 0*45 sq. ft* 

Example 289. A plot of building land measured with the 100 ft, 
chain gave an area of 82,728 sq. ft. State this in acres, etc. 

9)82728 sq. ft. 4o)3Q.3 sq. po. 

9192 sq. yds. 4) 7 rds. + 23 sq. po. 

4 i~ac. + 3 rds. 

121)36768(303 sq. po. 
363 
468 
363 

105 quarter sq. yds. over 

= 26J sq. yds. 
.'. Area = i ac. 3 rds. 23 sq. po. 26J sq. yds. 

Note. Instead of dividing by 30} as 30-25, the sq. yds. are multi- 
plied by 4, and then divided by 121, since 30! = - . By this means 

the decimal point is not introduced. The remainder being J sq. yds. 
is divided by 4 to bring to sq. yds. 

Areas of the Simple Figures. 

In calculating areas, dimensions must be in the same units, i. e., 
feet and feet, or inches and inches, etc. A length in feet must 
never be multiplied by a width, say, in inches. 

Square and Rectangle. 

Area = length x breadth 
Fig. 44 shows in detail the truth of the method. A rectangle is 



276 



ARITHMETIC FOR ENGINEERS 



9* by 4* (i. e., g" long by 4* wide), and evidently there are four rows 
of little squares, each row containing 9 sq. ins. 

.*. in all there are 4 x 9 = 36 sq. ins. 
i. e., length x breadth. 

A square is merely a rectangle with equal sides. 

Example 290. In a certain two-stroke petrol motor the exhaust 
port is in the form of three rectangular holes each f * x J" as shown in 
Fig. 45. Determine the area of the port. 

Area of i rectangular hole = I X J = fV == '1875 sq. in. 
/. Area of 3 rectangular holes = 3 X '1875 = -5625 sq. in. 

.". Exhaust port area = -562 sq. in. 

Example 291. A steam port in an engine cylinder is 16" wide, and 
the area required through the port (which is rectangular) is '165 sq. ft. 
What must be the length of the port ? 

Area = Length X width 
Area 



Length = 



Width 



' x M4 _ ... 
16 - Z 4 



/. Length of port = i \" (to the nearest -^*) 

Note. The area is given in sq. ft. and length in ins. For calculation 
both must be in the same units. Hence the 144 is introduced to convert 
sq. ft. to sq. ins. The result is in inches. 

Example 292. When calculating the discharge of water through 
channels it is necessary to find the " hydraulic mean depth " which 
is " area of cross section of channel ~ wetted perimeter." Find the 



Water Level 




z. ~i~- ~_ 1 _~ 






* - 


_ _ . fO . _ 




Cross SecVvov\ 



Fig. 46. 



Fig. 47- 



hydraulic mean depth for the channel shown in Fig. 46 when the water 
runs 3'-o" deep. 

Area of cross section = area of rectangle 5' X 3' = 15 sq. ft. Wetted 
perimeter is that portion of the channel perimeter touched by the water. 



MENSURATION 



277 



/. Wetted perimeter 
Hydraulic Mean Depth 



: 5 + 3 + 3 

Area 



ft 



Wetted perimeter 



When dealing with areas we frequently meet with expressions 
such as " Ibs. per square inch/' " amperes per square inch/' etc. 
meaning the number of Ibs. (or amperes, etc.) on every i square 
inch of area. Thus if the total load (number of Ibs.) on a piston 



is 5000 Ibs. and the area is 100 sq. ins., the pressure is = 50 jbs. 

100 

on i sq. in., or " 50 Ibs. per sq. in/' Similarly if a conductor is 
carrying 1000 amperes per sq. in., and has an area of -25 sq. in. 
the total current is 1000 x '25 = 250 amperes. 

Example 293. A valve chest cover is 20" X nj" (inside dimen- 
sions) as shown in Fig. 47, and the steam pressure is 100 Ibs. per sq. in. 
What is the total force tending to blow the cover off ? If the cover 
is held down by 12 bolts what is the force on each bolt ? 

Area of rectangle under steam pressure == 20 x nj 

= 225 sq. ins. 

At 100 Ibs. on each sq. in., total force on cover 
= 225 x 100 = 22500 Ibs. 



Then force on each bolt = = 1875 Ibs. 



x- 



The areas of other figures can be obtained by dividing the 
surface up into squares, by two 
perpendicular sets of parallel lines 
The figure should be drawn either 
full size or to scale upon " squared 
paper " (see p. 349) where the 
perpendicular lines are already 
provided. The complete squares 
and fractions of squares enclosed 
by the outline of the figure are 
then counted up, and any piece 
which is less than half a square is 
neglected, whilst any piece more 
than half a square is counted as 
one. Thus, taking the quarter of Fig. 48. Area by " Counting Squares/' 
a circle shown in Fig. 48 (only a 

quarter need be set down), there can be counted 71 complete squares ; 
and around the outline there are 8 pieces greater than half a square 





2 7 8 



ARITHMETIC FOR ENGINEERS 



The area of the quadrant is then 79 sq. ins., and the total area of 
the circle 4 x 79 = 316 sq. ins. The pieces less than half a square 
(which are to be neglected) are; shown black in Fig. 48 ; while those 
pieces greater than half a square are shown shaded. The correct 
area as obtained by the formula on p. 287, is 314-2 sq. ins., so that 
the result obtained from " counting squares " is very close. 

Any figure may be treated in the foregoing manner, but the 
method is laborious, and areas can usually be found more readily 
by formulae or simpler graphical methods, as will be shown. 

Rhomboid. This figure may be regarded as a distorted rect- 
angle. In Fig. 49, at (a) we have a rectangle divided into a number 
of narrow horizontal strips. If the lowest strip be fixed, and each 
of the others shifted a small distance to the right of the one just 
below it, then the rectangle will appear as at (b), i. e., as a rhomboid. 




Fig. 49. 

Now the number and size of the strips have not been altered in any 
way, and therefore the area of the rhomboid must equal the area of 
the rectangle. If the thickness of the strips be made very small thsn 
the slant sides of the rhomboid will become practically straight lines. 

Then, Area of Rhomboid = Area of Rectangle 

= Base x Perpendicular Height 

and this statement is true for all parallelograms. 

It is very important that the perpen- 
dicular height be used when calculating 
this area, and not the length of the slant 
side. If the height is not actually given, 
the figure should be drawn out and the 
perpendicular height measured. 

Example 294. The steam port for the 
piston valve in a high-speed engine has 10 
holes with dimensions as shown in Fig. 50. 
Fig. 50. Determine the total area through the port. 

Area of i hole = base X perpendicular height 

= -75* X 1-25" = -938 sq. ins. 
/. Area of 10 holes = 9-38 sq. ins. total port area, 




MENSURATION 



279 



Example 295. The steam port for a piston valve of a marine engine 
is to have an area of 97 sq. ins. There are 12 holes, a shown in Fig. 51. 
Calculate the height of the holes. 



Area per hole = -- = 8-083 sq. ins. 

_ Area _ 8-083 
~~~ Length ~~ 3-25 

= 2-485, say 2^" 



Height 




Triangle. Consider any triangle, as ABC, Fig. 52. From two 
of the corners, say A and C, draw lines parallel to the other sides 
as AD and CD. Then the figure ABCD is a rhomboid and its area = 
Ih, where / = base and h = perpendicular 
height. It can be seen, and measurement 
will prove, that the added triangle ADC is 
exactly the same size and shape as the original 
one ABC. Hence the area of triangle ABC 
is half the area of the parallelogram ABCD. Fig. 52. 

.'. Area of triangle = \ area of parallelogram = J Ih 
= ^ Base x Perpendicular Height 

It does not matter which side of the triangle is called the base, 
provided that the corresponding perpendicular height is taken. 
Thus consider the triangle in Fig. 53 






Fig. 53- 



Fig. 54- 



Base. 


Perpendicular 
Height. 


Area. 


BC =/ = 
AC = /! = 
AB = / 2 = 


3'5 

2I 

3' 2 


h =1-89 

h t = 3-15 
A 2 = 2-07 


J(3'5 X 1-89) 

j(2-i x 3-15) 

J(3-2 x 2-07) 


= 3-307 
= 3-307 
= 3'307 



Whence the area is the same in each case. 



280 



ARITHMETIC FOR ENGINEERS 



Should the perpendicular height not be given, then the triangle 
should be drawn out, and the height measured. In a right-angled 
triangle if either perpendicular side be taken as base, then the other 
perpendicular side is the perpendicular height. (An additional rule 
for area of triangles is given in the table at the end of these sections 
on Area.) 

Example 296. The area of the plot of ground shown in Fig. 54 
is measured by dividing the shape into triangles, finding the areas of 
these and calculating approximately the surplus outside the triangles. 
If this surplus area is 18-6 square chains, find the total area of the 
plot. 

Triangle ABC. Area == J base X height 
= J X 16-7 X 7-4 
= 61-75 sq. ch. 
Triangle ADC. Area = X 16-7 x 5-1 

= 42-5 sq. ch. 
61-75 

Surplus area 18-6 



Total area = 122-85 sq. ch. = 12-285 



Example 297.- 



A cooling tower has two air ducts in the form of 
triangles as shown at a, Fig. 
55. What is the total cross 
sectional area in square feet ? 
It is desired to replace them 
by a single* triangular duct 
with base and height equal, 
which shall have the same 
area as the two combined. 
What must be the size of the 
new duct ? 




-l"2 ; 
Mew Arra*". 

Fig. 55. Air Ducts for Cooling Tower. 



Area of i duct = J base x height 

= J X 2 X 2 = 2 Sq. ft. 

/. Total area of two = 4 sq. ft. 

Let / = base of new triangle ; then since base is to equal perpen- 
dicular height, base also = /. 

/. Area of new triangle = J base x height 

= j/ X / - i/ 2 
/. V 2 = 4 
/. I 2 - 8 
/. / = V8~= 2-83' or 2'-io /r 

The new duct will then appear as at b, Fig. 55. 



MENSURATION 



281 



Trapezium. This may be divided up into two triangles by 
joining a pair of opposite corners with a straight line, measuring 
the bases and perpendicular heights of each, and then calculating 
the areas of the separate triangles, and finally adding the results 
together. Ex. 296, just dealt with, illustrates this. 

A briefer statement in symbols is as follows : Refer to Fig. 56. 

Take the diagonal /. 

Measure extreme height of figure perpendicular to this = h. 



Then, Area of Trapezium * 



This last applies to any quadrilateral, 
but h must always be taken perpen- 
dicularly to /. (The student should 
work the above taking diagonal L, 

Fig. 56.) 

Thus, given the diagonals of a 
square instead of the sides : 

Area = | diagonal x diagonal. 




Fig. 56. Area of Trapezium. 



Trapezoid. Consider the trapezoid ABCD in Fig. 57. Draw 
a line EF exactly half way between the two parallel lines. From 
points E and F draw lines perpendicular to the parallel lines as 
GH and KL. Continue the top line as shown dotted. Then it can 
be seen by both eye and measurement that the triangle AHE is 




Fig. 57- 



Fig. 58. 



exactly the same size and shape as triangle EGD. Similarly triangle 
FKC is the same size and shape as FLB and replaces it. Then 
the area of the rectangle HLKG is the same as the area of the 
trapezoid. 

But area of rectangle = length x height 

The length is the length EF, halfway between the top and bottom, 
which is thus the average between the top and bottom. 



282 ARITHMETIC FOR ENGINEERS 

The average length = b -^ m + to P = i 

2 2 

Then, Area of Trapezoid = Average length x height 

fa + b\ . 
= (__)xA 

Example 298. Find the hydraulic mean depth (i. e. t area 4- wetted 
perimeter, see Ex. 292) for the channel shown in Fig. 58. 

Average width = I -~~ - 8 ft. 

/. Area = average width x height 
= 8 X 3 = 24 sq. ft. 

Length of slant sides of trapczoid may be measured after carefully 
drawing out the section, or may be calculated with the aid of the right- 
angled triangle relation, thus 



Slant side BC = \/2Hh~3* 

= Vi 3 = 3-61 ft. 

Then, wetted perimeter = base + 2 slant sides 
= 6 + 2 X 3-61 
= 13-22 ft. 



/. Hydraulic mean depth = 



area 



wetted perimeter. 

-?*- - i-8 I5 



13-22 

Example 299. Fig. 59 gives dimensions of a positive conductor 
rail employed on an electrified railway. Find the area of the section 
in square inches, neglecting any small radii. 

Cross section is difference between large and small trapezoids. 
LARGE TRAPEZOID 

Mean width = ^-t-S* = 4-625 

/. Area = mean width x height 

= 4-625 x 4*625 = 21 4 sq. ins. 

SMALL TRAPEZOID 

- 1 i 
Mean width = 




Fig. 59- ' Arca = 3*375 X 3-25 = 10-96 sq. ins. 

/. Net area of section = 21*4 10-96 
E=- 10*44 sq. ins. 



MENSURATION 



283 



Exercises 72. On Areas of Triangles and Quadrilaterals. 

(Letters refer to Fig. 60.) 

1. The grate of an American locomotive is 9'-6 // long and 6 '-3 J" 
wide. Determine the grate area in square feet. 

2. The heating surface in a locomotive fire-box is as follows : 
Top, 5 '-4" long, 3 '-3" wide ; 2 sides each 5 '-6" long, 5 '-4" high ; 2 ends 
each 3 '-3* wide, 5 '-6" high. Calculate the total heating surface in 
square feet. 

3. When calculating the strength of masonry dams it is necessary 
to know the total pressure of the water on the face for a length of i ft. 
The total pressure on such a rectangular area is 62-4 HA IDS., where 
H = depth to the centre of the rectangle and A = the area of the 
rectangle. Find the total pressure when the total depth of water 
is 21 ft. 




Fig. 60. Exercises on Areas of Triangles and Quadrilaterals. 

4. A column base plate is 2'-3* square and rests on a York stone 
block. Allowing a pressure of 20 tons per sq. ft. on the block, what 
load can the column carry ? 

5. In a test to determine the bearing pressure permissible on a 
certain soil a platform having 4 legs each 8" square was loaded with 
31 cwts. Calculate the bearing pressure in Ibs. per sq. ft. 

6. In an Edwards' air pump the inlet ports consist of 16 rect- 
angular holes each 4!" X 4t* Find the total port area. 

7. At A is shown the cross section of the rim of a dynamo arma- 
ture. Calculate the net cross section (i. e. t the shaded area). 

8. A rectangular screening plate 3 '-2" by 2 '-2* has ij" square holes 
as shown at B, there being 18 rows of holes and each row containing 
12 holes. Determine the total area through the holes. What percent- 
age is this of the plate area ? 



284 ARITHMETIC FOR ENGINEERS 

9. In a tensile test on a piece of boiler plate the cross section of 
the specimen was if* x \". The breaking load was 25-8 tons. Calcu- 
late the breaking stress (i. e. load per sq. in.). 

10. In a tensile test on a piece of mild steel plate the following 
results were obtained: original section 1-753" wide, -64" thick; final 
section 1-472" wide, '482" thick. Calculate the reduction in area, 
and express it as a percentage of the original area. 

11. Calculate the area of the T-section shown at C, neglecting any 
radii at the corners. 

12. At D is shown the cross section of a compound girder built of 
plates and channels. Find the net area of the cross section shown 
shaded. (Note. Do not include space occupied by rivets.) 

13. When charging a certain make of lead accumulators, a current 
of 5 amperes per sq. ft. of positive pkite may be allowed. An accumu- 
lator has three positive plates each 9" x 7". Calculate the charging 
current in amperes. (Note. Each plate has two sides.) 

14. The rafters in a roof truss are 28 ft. long and are spaced 12 ft. 
apart. If the wind pressure is 25 Ibs. per sq. ft., what is the total load 
upon the roof between a pair of rafters, due to the wind ? 

15. An ammeter shunt for switchboard use has four manganin strips 
each 3* long and if" wide. Find the total amount of radiating surface. 
(Note. Each strip has two sides.) 

16. If the strips in Ex. 15 above are -02" thick, calculate the total 

cross section (i. e., width x thickness). The resistance is ohms 

where s specific resistance, / = length of strips, and a = total area. 
Calculate the resistance if s -000043. 

17. A parallel screen has 96 holes with dimensions as at E. Find 
the total area through the holes. 

18. Find the area through the holes forming the inlet port of an 
Edwards' air pump, which has 12 holes as shown at F. 

19. The steam ports in a pressure regulator on a turbine-driven 
centrifugal feed pump, are triangular, as shown at G. If there are 
6 holes in all, find the total port area. 

20. When designing girders to carry a wall over an opening we 
assume that the load on the girder consists only of the brickwork 
enclosed within an equilateral triangle as shown shaded at H. Calcu- 
late the area of the face of the wall supported by a i7'-6* span 
girder, and the total load (per half-brick thick), if i sq. ft. of brickwork 
(per half-brick thick) weighs 125 Ibs. 

21. At J are given dimensions of the " keystone section bar," 
employed as third rail by the Underground Electric Railway. Calculate 
the area of the section, neglecting the radii. (Hint. Deduct the 
triangles from the rectangle.) 

22. In connection with deflection of beams it is necessary to calcu- 
late the area of certain triangular diagrams. Find a statement for the 
area in the case shown at K. 

23. Expanded steel of diamond mesh, as shown at L, is used in an 
oil separator. Calculate the total area through the holes in a sheet 
6'-o" wide by 4 / -o" high, which contains 144 holes across and 32 holes 
vertically. 

24. A flat screening plate has 20 diamond-shaped holes, the long 
diagonal being 2" and the short one f ". Find the total area through 
the holes. 



MENSURATION 



285 



25. Find the area enclosed (in square chains) between the straight 
lines on the plan of a plot of ground, given in Fig. 61. (Any dimen- 
sions required beyond those given on the picture are to be measured 
on the scale of chains shown.) 



i -5 o i a 

Scale of Chains 




Fig. 61. Area of Triangles. 

26. A section of a supply channel is shown at M. Calculate the 
cross-sectional area. 

27. The " hydraulic mean depth " = area of section -f- wetted 
perimeter (see Ex. 292). Calculate this for the channel shown at M. 

28. The kathode in an electrolysis experiment is shown at N. It 
is immersed until the level of the liquid reaches the narrow shank. 
Determine the area in the liquid. Allowing a current of -09 amps, 
per sq. in., what total current may be applied ? 

29. At O are given dimensions of the section of the third rail employed 
on the Central London Railway. Determine the area of the cross 
section, neglecting any rounding of corners. (Note. The section is a 
rectangle minus a trapezoid.) 

30. The third rail section used on the Metropolitan District Railway 
is shown at P. Determine the area in square inches. (Note. For 
calculation divide the figure up as suggested by dotted lines ) 

31. At Q is shown the section of the conductor rail on the L.C.C 
conduit tramway system. 

Calculate the cross-sectional 
area. (The mode of division 
is suggested.) 

32. When calculating the 
volume of earth to be dealt 
with in constructing cuttings 
and embankments, sections 
occur such as in Fig. 62. In 
finding the area of these 
sections the sloping ground 




Fig. 62. 



line is replaced by a straight horizontal line, converting the section into 
a trapezoid of approximately the same area. Find in square feet the 
two adjusted areas (cross-hatched), shown dimensioned in Fig. 62. 

Hexagon. It is a property of the hexagon that the length of 
its side is the same as that of the radius of the circle through the 



286 



ARITHMETIC. FOR ENGINEERS 



points, so that a hexagon drawn in a circle of J" rad. has a J" side. 

If the opposite corners be joined by straight lines then all will 
pass through the centre of the outside circle as 
at O in Fig. 63. This will divide the hexagon 
into six equal triangles like OAB, which are 
equilateral, i. e., OA = AB = OB 

Then the area of the hexagon = 6 times area 
of triangle OAB. 

Let vS = side of hexagon 
then AC = J side = -5 S 

The triangle OAC has a right angle at C and a 
60 angle at A. 




Fig. 63. 
Then referring to p. 270, 

the intermediate side CO 



also if/ = 



1732 X short side 
- 1732 x -58 
= -866 S 
Area of triangle = base x height 

= JS x -866 S = -433 S a 
/. Area of hexagon = 6 x -433 S 2 
= 2-598 S* 
say, 2-6 S- 2 
distance across the flats 

Area = '868 / 2 



Octagon. Let S = side of octagon (see Fig. 64). 

AB = S and in an octagon = 45 
.'. AC = 707 S (see p. 269) 

Side of outer square = S + 1-414 S 

= 2-414 S 

/. Area of outer square = 5-827S 2 

Deduct for four corners as ABC = 2 squares of side 707 
= 2 x 707 x 707 = i-oo S 2 

Deducting gives 4-827 S* 
/. Area of octagon = 4*83 S 2 
also if f distance across the flats 

Area = *829/ 2 

Circle. The area of a circle may be examined by the aid of 
Fig. 65. At (a) a circle is divided up into 16 equal sectors by means 
of 8 diameters. These sectors are numbered from i to 16. If the 



MENSURATION 



287 



circle is cut along these diameters the sectors may be arranged as 
at (b), forming a parallelogram if the slight curvature at the ends 
of the sectors be neglected. The more sectors that the circle is 
divided into, the straighter will the top and bottom of this parallelo- 
gram become. It can be seen that the length of the parallelogram 
is only that occupied by sectors numbers I to 8 : i. e., half a 
circumference. 

Then, length of parallelogram = half circum. = ~*~ = irr 

Also height = length of sector = radius = Y 

Then, area of circle = area of the parallelogram 

= length x perpendicular height 
= Ti-r x r = Trr* 
or Area = IT X Square of Radius. 




Fig. 64. 



Fig. 65 Area of Circle. 



Another form which is often useful is in terms of the diameter, 
and may be obtained as above 



Half circum. of circle 



ird j j. d 

- and radius = - 

2 2 



A Tr TT, O 

/. Area = - - x - =-a 2 
224 

Taking TT = 3-142, *" - 7854 
.'. Area of circle = -7854 d 2 . 

Note. 3-14 and 785 are sufficiently accurate for most practical 
work. 

This may also be obtained from the first form, by substituting 

for r the value . 
2 

TU 2 / d \* " 

Thus 7T,* = *-/- = 



288 



ARITHMETIC FOR ENGINEERS 



Example 300. A water-pipe is 14* diameter. What is its cross- 
sectional area ? 

Radius = 7" 

/. Area = irr 2 = TT x 7 2 

= 3-14 X 49 = 154 s q- ins. 

Example 301. A 3/20 electric cable is composed of three wires each 
of No. 20, S.W.G. (i. e., -036" dia.) as shown in Fig. 66. Find the total 
area of the section and the current allowable at 1000 amperes per 

sq. in. 

Area of i strand = 



/. Total area of section 



At 1000 amps, per sq. in. 
Current allowable 




Fig. 66. 



= -7^5 x (-036)2* 

'7^5 X -001296 
= -001017 sq. in. 
= 3 X -001017 
= -003051 sq. in. 

003051 X 1000 
3-05 amps. 



[* If the student finds it of any assistance he can put a bracket round 
the quantity to be squared.] 

Example 302. A reinforced concrete beam is to have a total area 
of steel of 1-07 sq. ins. How many bars must be used if they are to 
be I" dia. ? 

Area of J* dia. circle = -785 X -8752 

-785 X -766 = -601 sq. in. 

. -vr t , total area i -07 

.. No. of bars = 7 r- = -,-*- = 1-78 

area of i bar -60 1 ' 

We must, therefore, use 2 bars. 



Example 303. A set of expansion bends 
for a main steam-pipe is arranged as in 
Fig. 67. The combined areas of the small 
bends is slightly greater than the area of 
the main steam-pipe. This pipe is 15" dia., 
and there are three small bends each 9* dia. 
Determine by how much the total area of the 
bends is greater than that of the main pipe. 
Also express this as a percentage of the main 
steam-pipe area. 




MENSURATION 
Area of main steam-pipe 

Area of i bend 
.'. Total area of bends 



289 



785 x 15* 

= '785 X 225 = 176-6 sq. ins. 

= -785^ 

= -785 X 9 a 

= -785 X 81 = 63-6 

= 3 X 63-6 = 190-8 



Excess area of bends = 190-8 176-6 = 14-2 sq. ins. 



.'. As a percentage of the main steam-pipe area this 

8% approx. 



14-2 X IPO 
176^6" 



Example 304. A cottered joint for pump rods is shown at a, Fig. 68, 
and an enlarged cross section through the cotter hole is given at 6, 




Cross sccVion 
of rod A Y\\Yd 

colrlrer hole-. 
Fig. 68. 

with dimensions. Determine the effective or net area (shown shaded) 
at this section. If the load on the rod is 7500 Ibs., what is the stress 
(i. e. t force per sq. in.) ? 

Area of ij" dia. circle = -785 x i*5 2 

= -785 X 2-25 = 1-766 sq. ins. 
Deduct plain rectangular hole i y x $$" = -656 

(neglect rounded ends) 

/. net area = I'lio sq. ins. 

^ load 7500 .. 

/. Stress = = = 6750 Ibs. per sq. in. 

area i-n -LL 1 _ 

Determination of Diameter of Circle from Area. When 
the area of a circle is known the diameter can be obtained by solving 
an equation. Either the radius or the diameter may be obtained 
directly 



Thus 



area 



,2 _ ?^ 



Taking V 



area 

7T 



u 



290 ARITHMETIC FOR ENGINEERS 

Hence to determine the radius of a circle, knowing its area, divide the 
area by *, and extract the square root of the result. 



Similarly 


5* 


= area 


i. e. t 


785^ 


= area 


-2-785 . 


'. rf 2 


area 
= ^85 


,/ 


J 


/area 



Taking V 

Hence to obtain the diameter of a circle, when its area is known, divide 
the area by '785, and extract the square root of the result. 

Example 305. The area of a steam-pipe is to be 12-3 sq. ins. What 
will be its inside diameter ? 

j8$d 2 = 12-3 



say, 4" dia. 

Example 306. The area of the low pressure cylinder of a com- 
pound condensing engine is to be 3 J times the area of the high-pressure 
cylinder, and is 30" dia. Find the diameter of the H.P. cylinder. 

Area of L.P. cylinder = *r 2 

= 3-14 X 15* 

= 3-14 X 225 = 706 sq. ins. 

Area of L.P. cylinder = 3^ x area of H.P. 

A f TT T- Area of L.P. 

.'. Area of H.P. = - 

3-25 

sq. ins. 
j *;> 

* d ^^ ^/ ~ ~~^ ~~ 
*7o ^ 

say 

The above may be extended to three or four cylinders when 
the ratio of the area of each cylinder to the H.P. (or L.P.) is 
known. 

Example 307. In an expansion bend of the form shown in Fig. 67, 
the main steam-pipe is 18" dia. There are to be four bends, and their 
combined area is to be about 8% greater than the steam-pipe area. 
Determine the diameter of the bends. 




MENSURATION 291 

Area of main steam-pipe = *r 2 

= 3'H X Q 1 

= 3-14 X 81 = 254 sq. ins. 

o 

Taking 8% of this = - X 254 = 20-32 sq. ins. 
i oo 

.*. Total area of bends = 254 -|- 20-3 = 274-3 sq. ins. 
There being four bends the area of each 



.% -785^* = 68-6 if d = dia. of branch 



say, 9f" dia. 

Example 308. The total force on a valve spindle is 8400 Ibs. and 
the stress in the spindle (i. e. t the force per sq. in. of area) is not to be 
more than 2500 Ibs. per sq. in. What must be the diameter of the 
spindle ? 

Area required in spindle = -4 = 3-36 sq. ins. 

2^00 



= 2-07* 
/. Spindle diameter = 2^* 

Exercises 73. On Areas of Circles, 

(Letters refer to Fig. 69.) 

1. The outlet from the turbine wheel in a hydro-electric instal- 
lation is circular and 18 ft. dia. Determine the cross-sectional area. 

2. Find the cross-sectional area of a wire the diameter of which is 
109". 

3. A tray in a contact feed water heater has three rows of holes f * 
dia., there being 56 in each row. Determine the total area through the 
holes. 

4. At W, Fig. 31, is shown the arrangement of a split concentric 
electric cable. Determine the cross-sectional area in the central 
portion (which contains 18 wires, No. 16 S.W.G., i. e. t 064"' dia.) ; in 
the outer ring (which contains 34 wires, No. 19, S.W.G., i. e., -04" dia.) ; 
and the total area. 

5. A concrete column is reinforced with 20 steel rods J* dia. Find 
the total area of steel. 

6. A motor piston is 3^" dia. Find the total force upon it if the 
pressure is 300 Ibs. per sq. in. 

7. Calculate the area through the boiler flue shown at A. 



ARITHMETIC FOR ENGINEERS 

8. Calculate the area through the discharge flume for a water 
turbine, as shown at B. 

9. A wire rope consists of six strands, each having 24 wires of No. 15, 
S.W.G. (-072" dia.). Calculate the total cross-sectional area of the 
rope. Assuming that every square inch requires a breaking load of 
100 tons, what is the breaking load of this rope ? 

10. A i\" eye-bolt is 942" dia. at the bottom of the screw thread 
(i. e. t at the weakest place). Find its cross-sectional area, and allowing 
a stress of 9000 Ibs. per sq. in., calculate the load it will carry, in tons. 

11. A locomotive safety valve has two valves each 2%" dia., held 
down by a central spring as shown at C. Find the total force on the 
Spring when the steam is just blowing off at 140 Ibs. per sq. in. 




Fig. 69 Exercises on Areas of Circles and Hollow Circles. 

12. A two-stage air compressor has a low-pressure cylinder lo" dia. 
and a high-pressure cylinder 6J" dia. Find the ratio of piston area 
of low pressure to high pressure. 

13. At D is shown an armature slot with 4 round conductors 'i6* dia. 
Owing to waste space between wires and thickness of insulation, the 
slot is not completely filled with conductors. Find the " space factor," 
which is " area of conductors area of slot." 

14. A steam-collecting pipe in a boiler is 7" dia., and is to be per- 
forated with rectangular slots 2* x J". How many such slots are 
required so that the total area through the holes is ij times the pipe 
area ? 

15. A boiler manhole is 16* dia. and the vsteam pressure is 60 Ibs. 
per sq. in. The cover is held on by sixteen J"-bolts. Find the stress 
in the bolts (i. e. t the force per sq. in. of bolt area) if the diameter of 
a !"-bolt at the bottom of the thread is -733*. 

16. A perforated screening plate is made with holes at ij" pitch 
as at E. Determine the area of perforation per square foot of plate. 
If the area of perforation is to be not more than two-fifths of the whole 
plate area, find if this rule is satisfied here. 



MENSURATION 293 

17. The low-pressure exhaust port in a Willans' central valve 
engine is composed of six holes in a hollow tube, as shown dimensioned 
at F. Determine the total area through the port. 

18. At G is shown the front of an ash-pit in a Lancashire boiler. 
In a boiler test it was necessary to know the area through the ash-pit. 
Find the required area. (Note. The area is very nearly a semicircle 
less a rectangle.) 

19. H gives dimensions of the Field tube in a Niclausse boiler. 
Find the ratio of the outer area to the inner area (the two areas are 
indicated by shade lines). 

20. It is calculated that a reinforced concrete beam requires a 
total area of steel of 1*56 sq. ins. If round rods y dia. are to be used, 
how many will be required ? (Take the nearest whole number.) 

DIAMETER FROM AREA. 

21. An exhaust main is to take the exhaust steam from three engines. 
Two of the exhaust pipes are 5$" dia. and the third is 2%" dia. Calculate 
Ihe diameter of the exhaust main, the area of which is to equal the 
total area of the other three exhaust pipes. 

22. A 64/38 flexible wire for electric work is made up of 64 strands 
of No. 38 gauge (i. e., -006" dia.). Find what diameter of solid wire is 
equivalent in cross-section to this flexible wire. 

23. A copper conductor, No. 6/0, S.W.G. (-464' dia.) is to be re- 
placed by an aluminium conductor. If the area of the latter is to be 
1-7 times the area of the former (being a worse conductor), what must 
be the diameter of the aluminium conductor ? 

24. A hydraulic riveter is required to give a total force of 100 tons 
and the pressure water is supplied at 1500 Ibs. per sq. in. Find the 
area and the diameter of the ram. (Hint. Bring total force to Ibs.) 

25. A Venturi meter in a pipe 30" dia. is to have a throat area 
one-eighth the pipe area. What will be the diameter of the throat ? 

26. The exhaust pipe of a petrol motor is of the cross-section shown 
at J. The circular part is to be ij" radius and the whole area required 
is 2-4 sq. ins. What must be the depth of the rectangular piece ? 
(Hint. Area of half-circle subtracted from given area gives area of 
rectangular piece.) 

27. In a hydraulic accumulator pump the piston is 4" dia. and the 
piston rod area is required to be one-half of the piston area. Calculate 
the diameter of the piston rod. 

28. The Board of Trade Rule for safety valves is A = p where 

A = area of safety valve in sq. ins., G = total grate area in sq. ft., and 
p = working pressure -f 15 Ibs. A Lancashire boiler to work at 120 Ibs. 
per sq. in. has two grates each 6 ft. long and 2 / -6 // wide. Calculate 
the diameter of the safety valve. 

29. In an expansion bend of the form shown in Fig. 67, the main 
steam-pipe is 12%" dia. The total area of the bends is to be 10% more 
than the steam-pipe area. Calculate the diameter of each branch if 
there are to be three branch pipes of equal area. 

30. The areas of the cylinders in a triple expansion engine are in 
the ratios of i : 2-8 : 7*1. If the low-pressure cylinder diameter is 7'-4*, 
what will be the diameters of the high and intermediate pressure 
cylinders ? (Hint. Area of L. P. cylinder is 7-1 times H. P. area. Work 
as in Example 306. Also area of I. P. cylinder is 2-8 times H. P. area.) 




294 ARITHMETIC FOR ENGINEERS 

Hollow Circle or Ring (also called an Annulus). The area 
of a ring, i. e. t the area enclosed between two circles, one inside 

the other (as in Fig. 70) is evidently 
the larger area minus the smaller. 
If the diameters are known, then 
by calculating each area separately, 
and subtracting, the value can be 
found. But it is usually more 
convenient to have a single formula 

_. for this area, thus 

Fig. 70. 

Area of large circle = ?rR 2 if R = rad. of large circle 
Area of small circle = Ti-r 2 ,, r = small 
Then Area of Ring = large circle small circle 

= TrR 2 - Trr* 

= ir(R 2 r 2 ) taking out the common factor IT. 

By using this form only one multiplication by ?r has to be 
made. 

A similar form can be obtained in terms of the diameters, thus 

Area of large circle = 785D 2 if D = dia. of large circle 
Area of small circle = 785^ d = small 
Then Area of Ring = 7850* - 785^ 

= *785(D 2 d 2 ) taking out the common 
factor 785. 

This may also be stated as 785(0 + d)(D d), as dealt with in 
Chap. III. 

The above also applies if the inside circle is not in the centic 
of the outside one (i. e., is eccentric) provided that it is completely 
inside. (See a, Fig. 70.) 

Example 309. A hollow shaft is 15* outside diameter and n" inside 
diameter. Find the area of its cross-section. 

Area = 785(02 - d*) 
= 785(152 - Hi) 

= 7 8 5( 22 5 - I21 ) 

= 785 X 104 = 81-6 sq. ins. 

Working on the radius form 

R = j'$" and r 5-5" 
Then area = 3-14(7-52 - 5-52) 

= 3-14(56-25 - 30*25) 

= 3-14 X 26 = 81-6 sq. ins. 



MENSURATION 



295 



Example 310. A thrust bearing in a 5-ton testing machine has 
4 collars each 2" inside and 2J" outside diameter as in Fig. 71. Find 
the bearing pressure per sq. in. when the full load of 5 tons is on the 
machine. 

Area of i collar = 785(02 d 2 ) 



= 785(7-56 - 4) 

= 7 8 5 X 3-56 = 2795 sq. ins. 

/. Total area of 4 collars = 4 x 2795 

= 11-18 sq. ins. 

i- iu 5 X 2240 

/. Bearing pressure in Ibs. per sq. in. = J --- g -- 

= 1002 Ibs. per sq. in. 

Example 311. In a double-beat valve of a steam engine (Fig. 72) 
the upper valve is 7" dia. and the lower one 5^" dia. If the steam pres- 
sure between the discs is 120 Ibs. and the weight of the valve and its 




Fig. 71. 



Fig. 72. 



Section on CD 
Fig. 73- 



Practical Examples involving Area of Hollow Circle. 

fitting is 25 Ibs., what force must be exerted by a spring (not shown) 
which holds the valve down on its seat ? 

Note. The top disc being the larger, and pressure being inside, 
there is a force acting upwards trying to lift the valve. This is 
partly neutralized by the weight of the valve, etc., which acts 
downwards. 



296 ARITHMETIC FOR ENGINEERS 

Effective area = area of upper valve minus area of lower 
= -785 X 7 2 - -785 X 5-5 2 
= '785(7 2 ~ 5'5 2 ) 
= '7 8 5(49 ~ 30-25) 
= -785 x 18-75 = 14-7 sq. ins. 
Then upward force of steam 120 x 14*7 

= 1764 Ibs. 
Then final upward force = force on spring 

steam force minus weight of valve 
= 1764 25 = 1739 Ibs. 

Example 312. Bolts subjected to heavy shocks are sometimes 
made with a hole drilled up the unscrewed portion as in Fig. 73, so 
that the area of the bolt at any section as AB is equal to the area at 
the bottom of the thread. A particular bolt in a connecting-rod end 
is 3i* dia., and the diameter at the bottom of the thread is 3*34". What 
must be the diameter of the central hole ? 

Area at bottom of screw thread = -785 X 3*34 a 

= -785 X 11-15 8-75 sq. ins. 
Then net area as at AB = -785(02 d 2 ) 



= -785(12-25 - d*) 
and this is to equal 8-75 sq. ins. 
/. .785(12-25 -*) =8-75 
-.785 /. 12-25-* =^ 

Transpose .*. 12-25 11-15 = d* 

d* =1-1 
Take V~~ /. d = Via = 1-05*, say, i^'dia. 

Note. This was the actual size of hole used in the particular bolt 
mentioned. 

The area of a hollow circle may also be obtained in anothei 
way which is sometimes preferable. Suppose we imagine the ring 
to be made of a long thin rectangle bent round into a circle, the 
length of the rectangle being the average circumference of the circle, 
and the width, of course, the width of the ring. 

Then area = length of rectangle X width 

Now length of rectangle = average circumference of ring 

= TT x average diameter 
and width of rectangle = width of ring 

= i difference of diameters 



MENSURATION 297 

Then area = TT x average diameter x difference of diameters. 
If this be put into symbols and evaluated we obtain a proof of its 
accuracy. 

Average diameter = 

2 

Width = |(D - d) 

Area = TT x D + d x J(D - d) 



=^ (D* - *) J P- '53- 

The foregoing method by average diameter, etc., is somewhat 
easier when the ring is very thin, i. e., the inside and outside diameters 
are nearly equal. Then it is usually easier to find an average 
diameter, and multiply by the thickness, than to find the difference 
of the two squares. The method can only be applied to cases of 
concentric circles. 

Example 313. Find the cross-sectional area of the metal in the lap- 
welded wrought-iron pipe shown in Fig. 74. 

Inside diameter is 15*, and thickness is J". Then 
outside diameter is 15^, and average diameter = 15^". 

.'. Average circum. = v x 15 J 

and area = * x 15-25 x -25 = n-gSsq.ins. 
say, 12 sq. ins. 

As a comparison the student should work this Fig. 

example by the other method. 



Exercises 74. On Areas of Hollow Circles. 

(Letters refer to Fig. 69.) 

1. A hollow shaft has an outside diameter of 6J* and an inside 
diameter of 3^. Find its cross-sectional area. 

2. A hollow cast-iron column is 10" inside diameter and 12 out- 
side. Find the area of its cross-section, and the load it can carry at 
3 -5 tons per sq. in. 

3. In a bolt such as is shown in Fig. 73, the bolt diameter is 2j* 
and the diameter at the bottom of the thread is 2i8 / ". What size of 
hole is required ? 




298 ARITHMETIC FOR ENGINEERS 

4. A chimney stack is 8 ft. inside diameter. If the effect of 
friction, etc., be considered equivalent to the loss of a layer of air 2* 
thick round the inner surface, as shown by the plain area at K, calculate 
this area in square feet. (Hint. Use " average diameter " method.) 

5. A wrought-steel riveted pipe is 16" internal diameter and is J* 
thick. Find its cross-sectional area. (Use " average diameter " method.) 

6. Calculate the area of the section of a column shown at L. 

7. At M is shown the plan view of the grid in a flexible disc valve. 
Calculate the area through the holes. (Hint. Deduct the area occupied 
by the bars from the area of the annulus.) 

8. The plunger in a hydraulic press consists of a hollow rod moving 
over a screw as shown at N, the area producing the pressure being that 
of the annular area (the plain area in end view). If the area required 
is 2-75 sq. ins. and the inside diameter is 2*, what must be the outside 
diameter ? 

9. The arrangement of a compound air compressor with differential 
pistons is shown at O. The low-pressure piston is the large piston 
at the top and the high-pressure piston is the annular area between 
the large and small pistons. Calculate the L. P. and H. P. piston areas 
and find the ratio of L. P. area to II. P. area. 

10. In a hydro-electric plant the total weight of the turbine wheel, 
generator rotor, etc., is 550,000 Ibs., and is to be supported by oil 
pressure acting on an annular ring 6'-o" outside and 2 / -i /r inside 
diameters. What pressure of oil (Ibs. per sq. in.) is necessary ? 

Sector of Circle. Suppose a sector of a circle contains an 
angle of 60. Since a full circle contains 360, then the sector is 

--, i. e., J of a full circle. Hence its area is J that of the circle, 



t. ., far 2 . 

2V 

Similarly for any other angle, say 27. Then this sector is -~ 

of a circle and its area is ~ x ?rf 2 . 

360 

Hence, if n be the angle of a sector in degrees, 
as in Fig. 75, 

Area of sector = 

t5DU 

Thus, when n == 90, i. e., the angle is a right angle, 

QO 

Area = --> - x ?ry 2 = far 2 , as is already known, since a sector of 
90 angle, i. e., a quadrant, is a quarter of a circle. 

Splitting the formula into factors we get -^- x ^ ; the first 

57'3 6-28 

Jy 

factor length of arc (/), as on p. 271, whence area = , since ir 
cancels twice into 6*28. 




MENSURATION 



299 



Example 314. A rotary valve of a petrol engine has a port shaped 
as in Fig. 76, which also gives dimensions. Determine the area through 
the port, in sq. ins. 

Notice here that the port is a sector of a hollow circle. The same 
principle applies as laid out above. 



R = if* = 1-625 J ' = T| = '9375 
Then area of complete ring = TT (1-625* *9375 2 ) 
.-. Area of Port = ^(1-625* - -9375 2 ) 

= g( 2 -6 4 - -8 7 9) 

^ x * x _L7_6 
' 8 




Fig. 76. 



Area of Fillet. The fillet or quarter-circle is very common 
in the corners of various shapes in engineering work, and it is occa- 
sionally necessary to find the area of the shaded 

portion shown in Fig. 77, when the radius is a _A _o 

big one. In most cases the rounding of corners, 
etc., may be neglected. 

Let r = radius of the fillet 
Then area of the square OABC = r 2 




and area of quarter-circle = J x Ti-r 2 = 785^ 

/. Area of fillet = area of shaded portion 

= the square the quarter- circle 



= r 2 (i 785) taking out the coefficient r 2 
-215r 2 (or approximately \r 2 ) 

Segment of a Circle. Referring to Fig. 78, it can be seen 
that the area of the segment ACB can be obtained by taking the 
area of the triangle OAB from the area of the sector OACB 

Then, if r = radius of arc ; C = length of chord 

h = height of segment ; n = angle at centre in degrees 



300 



ARITHMETIC FOR ENGINEERS 



Then, area of sector 
Height of triangle OAB = r - h 




Fig. 78. Area of Segment. 



/. Area of triangle OAB = (r h) ii. 

Taking (ii) from (i) gives area of segment. 

To use these formulae the segment must be drawn out to scale and 
any necessary dimensions not given must be measured off. Thus, 
if two of the quantities C, h and r are known, the segment can be 
drawn, and the remaining dimension and n measured. 

The area of a segment can be calculated directly by the use 
of higher work, but several approximate formulae can be used. 
Thus, given diameter (D) of complete circle, and height (h) of seg- 
ment, which are the two measurements usually known, 



Area of segment = ^/ 



^ - -608 



or given length of chord (C) and height of segment (h\ 



Area of segment 



,^ 
3 *- 



For segments greater than a semicircle find the area of the 
remaining portion or segment of the circle, and subtract this from 
the area of the whole circle. 

When the segment is a flat one, i. e., h is less than JC, then the 

o 

area is given approximately by the simple formula -Ch t the error 

being less than i%. The error increases as the segment becomes 
less flat, and is 4-5% when h = JC. 



MENSURATION 301 

Example 315. In calculating the number of stays necessary for a 
boiler head, as in Fig. 78, which gives dimensions, it is required to find 
the area (shaded) which they will support. Find this area. 

The area is a segment of a circle 

D = 5 '-5* = 5-42'; h = 2'-2j" - 2-21'; /. A 2 = 4-88 
Area of segment = -A 2 \/-v -608 

-4 X 4*8 \/|;g- -608 

t= i x 4-88VT-842 

= 4 X 4*88 X 1-358 
= 8-83 sq. ft. 

Ellipse. The area of an ellipse cannot be examined in as 
simple a manner as the area of a circle. The following formula is 
obtained by the use of higher mathematics : 

Area of Ellipse = trab 

where a = J major axis and b ~ \ minor axis (see p. 249, Fig. 25). 
This is very similar to the formula for the area of a circle. Thus, 
when an ellipse becomes a circle a = b = r . 

Then area, = irdb 

= Trrr = Trf*, the well-known formula. 

Example 316. The mud-hole door of a boiler is elliptical, and is 
15* x n*. What is the total force acting on it when the steam pressure 
is 60 Ibs. per sq. in. ? 

a = J major axis = -~ = 7-5* 

, t .11 

6 = 4 minor axis = = 5-5 

Then, area = irab 

= * X 7-5 X 5-5 = 129-5 sq. ins. 
.*. Total force at 60 Ibs. per sq. in. = 60 x 129-5 
= 7770 Ibs. 

Example 317. Fig. 79 shows the standard proportions of egg- 
shaped sewers for drainage. The lower portion is practically half an 
ellipse. Find an expression for the total area if k = the depth. 



3 o2 ARITHMETIC FOR ENGINEERS 

Area of ellipse = trab .'. Area of half ellipse = ab 



a = \ major axis = -k 



... , 2k k 

b | minor axis = J x - - = - 



.'. Area of half ellipse = ~ X - k X ~ 





^ 


Area 


of half circle 


_ _/o$- 

2 " 

785 / 2 ^\* 

~ 2 x vi ; 

785 X 4& 2 


*w 

\ * 


y^?^^L 
THAI/CIRC?! 

/^/^/ixx/^/X^VV/ 


v 
^ 

'1 


Here d = 

.*. Area of half circle 
> 

M. 

1 . 


I Jf. 
J^ 

*e 


^ftsfyor^dx/a 

.HALF ELLIPSE^ 

y|l, J 
|i? 
^i^L- 


2x9 
= T'y/if;^ 2 


T^i or nrt 





/. Total area = half circle + half ellip' e 




Exercises 75. On Areas of Sectors and Segments 
of Circles and Ellipses. 

(Letters refer to Fig. 80.) 

SECTOR OF A CIRCLE 

1. Find the area in sq. ins. of a sector of a circle 6J" radius, if the 
angle is 55. 

2. The sector of a circle has a radius of I'-if" and the arc sub- 
tends an angle of 215. Find the area of the sector in sq. ft. 

3. Calculate the area of the port of a rotary valve shaped as in 
Fig. 76, if the large radius is i-^", the short radius }%*, and the angle 
45. (Hint. This is a sector of a hollow circle.) 

4. The throttle regulator in a petrol engine is shown at A. Neglect- 
ing the rounded corners, find the total area through the ports, in sq. ins. 

SEGMENT OF A CIRCLE 
5 to 7. B shows the segment of a circle with necessary dimensions. 

5. Calculate the area exactly by subtracting the triangle from the 
sector. 



MENSURATION 



303 



6. Calculate the area using the approximate formula - 



-608. 



Find the difference between this result and the true one (as found from 
No. 5), and express it as a percentage of the true area. 

A 3 2. 

7. Calculate the area, using the approximate formula -~ -\ Ch, 

and find the percentage error, as in Ex. 6. 

8. At C is given all necessary dimensions of a segment where /* is 

Q 

less than . Calculate the area exactly by subtracting the triangle from 

4 
the sector. 



9. Calculate the area using the approximate formula -Ch. 



Find 



how much this differs from the true result of Ex. 8, and express this 
difference as a percentage of the true area. 




Fig. 80. Exercises on Areas of Segments, etc. 

10. At D is shown a circular culvert running one-third full. Calculate 
the area occupied by the water in sq. ft. 

11. Calculate the wetted perimeter, and find the hydraulic mean 
depth, which is Area -f- Wetted Perimeter, for the above case. (Hint. 
Obtain chord by drawing; or by a right-angled triangle with a vertical 
to the centre.) 

12. Calculate (a) the area of the cross-section; (b) the wetted 
perimeter; and (c] the hydraulic mean depth (area ~ wetted peri- 
meter) for the circular channel in Ex. 10, above, when running two-thirds 
full (i. e., 3 ft. deep as at D). 

13. A furnace Hue is shown at F. Calculate the area of the cross- 
section. 

14. T$ie plan of a bridge pier is shown at G, and the pier consists 
of two segments. Find its cross-sectional area. 

15. At H is given dimensions of the filling culvert to a dry dock. 
Calculate the area of the section. 

16. Particulars of the steam inlet to an oil separator are given at E. 
Calculate the cross-sectional area enclosed between the two arcs at 
section BB. The inlet pipe being 6* dia., find how many times the area 
at BB is greater than the pipe area. 

17. Repeat the calculation in Ex. 16 above, for section A A. 



304 



ARITHMETIC FOR ENGINEERS 



ELLIPSE 

18. Find the area of an ellipse whose major axis is 14!" and minor 
axis ioj*. 

19. In finding the capacity of an elliptic petrol tank for the back 
of a motor-car the area of the cross-section was required. Find this 
area if the two axes were i'-i" and 8J". 

20. Find the cross-sectional area of an elliptical magnet core 
18" X 9*. 

21. Find the cross-sectional area of an elliptical funnel 13* X 10*. 

22. The egg-shaped culvert used for drainage is very nearly a half 
ellipse in the lower two-thirds usually occupied by the water. Calculate 
the area of the water section in the culvert shown at J. 

23. A manhole door in the low-pressure end of a turbine casing is 
elliptical, 16* x 12". What is the total force on the door in Ibs., if 
the net pressure of the atmosphere is 14-4 Ibs. per sq. in. ? 

[Further examples involving areas of ellipses will be found in 
Exercises 80 on " The Volumes of Various Prisms."] 

Area of Irregular Figures. The area of a figure with an 
irregular outline cannot be obtained by the simple application of a 
formula; but can be found by drawing and measurement on the 
actual figure. 

C 



a 



The fwo -figures have equal areas* 
Fig. 81. Area of Irregular Figure. 

At a, Fig. 81, is shown a figure with one irregular side BC; the 
sides AB and CD are parallel and at right angles to the side AD. 
The figure has been divided into four strips of equal width by the 
parallel lines EF, GH, and JK. Taking any one of these strips, 
say No. 3, it can be seen that the side HK, although strictly curved] 
is very near indeed to a straight line. Suppose we assume for the 
present that it is a straight line : then the figure GJKH is a trapezoid 

and area of GJKH = area of a trapezoid 

average height x width 



MENSURATION 305 

where A 3 = the height at the centre of the width and / = length 
of figure. Assuming the side HK as straight, then the height at 
the centre is the average height. If h lt A 2 > ^3 and & 4 are the heights 
at the centres of the 4 strips, then the areas of the various strips are 

Strip No. i h 1 x - 

4 

* **;[ 

/ 

1 A, x - 



4 A 4 x -, the width of all the 

4 

strips being -. 

Then the total area is very nearly 

i I , i I , , I , , I 
*i4+*4+*3- 4 +*44 

. . , , /A, + ^7 + Ao + A A XT' J. j.1 r I T 

which = / x ( - 1 - - 2 8 - J - 4 j taking out the common factor /. 

Note that the bracketed factor is the average of the four central heights. 
At b, Fig. 8 1, is shown a rectangle of the same length as the given 
figure, with height equal to this calculated average height. The 
area of the given irregular figure is, then, the area of this rectangle. 
Now if the length is divided into a larger number of strips, i. e. t the 
width of each strip is made less, then the curved side of each strip 
is practically a straight line, and its area for all practical purposes 
is given exactly by " width x central height " : therefore the area 
of the whole figure is given by length x average of these central 
heights. The heights of a curve aie usually called " ordinates," and 
thus these heights at the centre of strips have been called " mid- 
ordinates " and the foregoing method of finding irregular areas is 
called the Mid-Ordiuate Method. This method may be summed 
up in the following rule : 

Divide the figure up into a number of narrow strips and measure the 
height at the centre of each strip. Find the average of these heights, and 
multiply by the length of the figure, measured perpendicularly to the 
heights. 

The number of strips into which the figure should be divided 
depends upon the state of the curve; when fairly stiaight a few 
would suffice and when very irregular many should be used. For 
x 



306 



ARITHMETIC FOR ENGINEERS 



general purposes 10 strips is a very convenient number to adopt, 
as the average is then very easily obtained from the total by merely 
shifting the decimal point. Using the mid-ordinate method, areas 
can be measured with an accuracy of within I %. 

Method of Dividing-up. Where the length is an even amount, 
as 6" or 8", and the number of strips is 10 or 20 or some round 
number, then the widths of the strips can be set off with an ordinary 
rule divided into tenths. But with uneven lengths a mechanical 
method of dividing-up is necessary, and that shown in Fig. 82 is 
both rapid and successful. The two end lines of the figure are 
produced for some distance. If 10 strips be required, a rule is laid 
diagonally across the figure and adjusted so that the two end lines 






\ 


6 


\ 

\ 


1 




N 




V 




1 


V 






j 




x i I^T^-v*^ 




1 


1 

1 


! I4l 1 ! m ! 


A^^of^X^jij 


fguire vvKew cliNftded op. \! ! j 


TV\e "wid-ord inches" ort * >v. 


shewn as ^yll \iv<.s, 



Fig. 82. Method of Dividing for Mid-Ordinate Rule. 

just enclose a distance easily divisible by 10 : in this case 5", i. e. t 
10 half inches. Then a line is drawn along the edges of the rule, 
and the 10 spaces marked off together with the centre of each 
space. Removing the rule, lines are drawn through these centre 
points parallel to the end lines as indicated at b, Fig. 82. The 
drawing and setting off should be done very carefully with a finely 
pointed pencil. It is not really necessary to show the actual strips, 
as only the centre or mid-ordmates are required to be measured. 

In many cases, the figure has no parallel end lines, and perhaps 
no straight base line, being completely irregular. In such cases 
two parallel lines are drawn one at each extreme of the figure as 
shown in the following examples ; the " length " is then the per- 
pendicular distance between these lines, and the mid-ordinates are 
drawn parallel to these end lines. 



MENSURATION 



307 



Example 318. In order to find the volume of material held in an 
elevator bucket it is necessary to find the area of its end (shown in 
Fig. 83). Find this area in sq. ins, 




Diagram is 
abouf JrFull Sije. 

Fig. 83. Use of Mid-Ordinate Rule. 

The top has been taken as the base or length, and 10 mid-ordinates 
set up perpendicular to this by the method described. The measured 
lengths are shown. 

Sum = 47-4 
/. Mean height = 4-74 

Area = Mean height X length 

= 4-74 X 12-5 = 59-25 sq. ins. 

Example 319. In calculating the volume of water impounded in a 
reservoir it was necessary to find the area of a number of irregular 




Scale of Chains. 



jo eo 30 -40 
Fig. 84 




Atmospheric Line. 
Fig. 85. 



Use of Mid-Ordinate Rule. 

figures. Find the area in square chains and acres of one of these 

shown in Fig. 84. (Measurements to be made on the scale attached.) 

The figure has no straight side, so that any over-all dimension may 



3 o8 ARITHMETIC FOR ENGINEERS 

be taken as the length. Two parallel lines have been drawn touching 
the extreme ends, and on measurement they are found to be 68 chains 
apart. Ten mid-ordinates have been drawn, and their lengths are 
shown on the figure. 

Then sum of 10 mid-ordinates = 295-3 
.'. Mean height 29-53 chains 

Area = length X mean height 

= 68 x 29-53 = 2008 sq. chains 
or 200-8 acres. 



Example 320. Fig. 85 shows an indicator diagram from a non- 
condensing steam engine. Find the average height. If i" of height 
represents 24 Ibs. per sq. in., find the average, or mean effective pres- 
sure. (Note. Only the outline and the atmospheric line appear on 
the diagram as given.) 

In this case there is no straight base, but the "atmospheric" line 
must be the length as pressures are measured vertically. Two lines 
are set up perpendicular to the atmospheric line and touching the 
extreme ends of the figure. Ten mid-ordinates are drawn and measured 
with the results shown, and the average obtained. 

Average or mean ordinate = -718" 



i* of ordinate represents 24 Ibs. per sq. in. 
.*. Mean pressure = 24 x -718 

= 17*2 Ibs. per sq. in. 

Note. When dealing with the indicator diagram it is usually only 
the mean height that is required and not the area. 

Exercises 76. On Area of Irregular Figures. 

[Note. In many cases the figures to the following exercises are 
shown much smaller than they would be drawn in practice; all draw- 
ing and measurement should therefore be done with the utmost care. 
For ease of working the student is advised to make a careful tracing 
with a fine pencil of the figures and to work on the tracing. Where 
drawings arc not full size a scale of inches, etc., is given and all measure- 
ments should be made on this scale.] 

1. Fig. 86 shows at A the cross-section of a six-pole dynamo 
magnet frame. Find the area of the section. 

2. C, Fig. 86, shows the cross-section of an aqueduct for water 
supply. Find the area below the water level in sq. ft. 

3. The water line section (half only) of a reinforced concrete 
barge is given at A, Fig. 87. Find the area of the whole section in 
sq. ft. 

4. Find the cross-sectional area of the magnet frame shown at B, 
Fig. 86. 



MENSURATION 



309 




Scale of Inches 
for A a*d B 




Fig. 86. Exercises on Area of Irregular Figures. 



ARITHMETIC FOR ENGINEERS 



5. Fig. 86 at F shows the development of one blade of a screw 
propeller. Find its area in sq. ft. 

6. The curve at D, Fig. 86, was obtained from the tensile test of 
a steel bar. The area beneath the curve is the work done in breaking 
the bar, in inch tons. Find the work done. (Hint. Measure vertical 
distances on scale of tons. Then " height in tons X length in ins." 
gives work in inch tons.) 

7. B, Fig. 87, shows a " contour " line obtained when surveying 
in connection with a proposed reservoir. Find the enclosed area in 
square chains. 

8. Find the cross-sectional area of the cast-iron beam section, half 
of which is shown at E, Fig. 86. (Hint. The quickest method is to 



101234 fe 6 10 12 

ERffiffiiffi 

Scale o{ Peer 




4-0 



50 

zE 



Scale of c.hcun-5 
Fig. 87. Area of Irregular Figures. 






I MILE 



calculate as a rectangle the area of the half web from top to bottom, 
as shown by dotted lines. Then only the projecting portions of 
the flanges need be averaged. The bottom flange projections are 
best done by vertical mid-ordinates, and those of the top flange by 
horizontal ordinates. Multiplying by 2 will give the area of the whole 
beam.) 

Questions 9 to 12 give examples of indicator diagrams. In each 
case the average height and the mean effective pressure are to be 
found. 

9. A, Fig. 88. Diagram from the H.P. air cylinder of a two-stage 
air compressor, i* = 50 Ibs. per sq. in. 



MENSURATION 311 

10. B, Fig. 88. Full load diagram from scavenging pump of a 
two-cycle Marine Diesel Engine, i" = 10 Ibs. per sq. in. 

11. C, Fig. 88. Gas Engine Diagram, i" = 243 Ibs. per sq. in. 

12. D, Fig. 88. Diagram from four-cycle Diesel Engine, i" = 300 
Ibs. per sq. in. 




Aj .Diagram from HP Cylinder* of 

Compound Air* Compressor* 



Une 




Fig. 88. Indicator Diagrams. 

TABLE. For convenience of reference, all the formulae, etc., 
relating to areas and circumferences, mentioned in the previous 
pages, have been collected into the following table : 



312 ARITHMETIC FOR ENGINEERS 

TABLE OF AREAS AND CIRCUMFERENCES OF PLANE FIGURES. 



Title 



Figure. 



Circumference or 
Perimeter. 



Rectangle 



Square 



Rhomboid . 



Triangle . . 



Equilateral 
triangle . 



Hexagon . 



Octagon . 



Trapezoid 



Irregular 
quadrilate- 
ral or tra- 
pezium . . 



Circle . . . 










b) 



Ib 



Ih 



s= perimeter 



ah 

2 



6s or 3'46/ 



8s or 3'3 2 / 



Vs(s- a )(s-b)(s-c) 



4335* 



2-65* or -866/* 



4-835* or 829/2 





Sum of all four 
sides. 



ltd or 2irr 



Divide into two tri- 
angles by either diagonal. 
Find area of each tri- 
angle and add. 

_ Ih 

Or area = -- 



MENSURATION 3 ! 3 

TABLE OF AREAS AND CIRCUMFERENCES OF PLANE FIGURES (continued). 



Title. 


Figure. 


Circumference or 
Perimeter. 


Area. 


Hollow circle 
(annul us) . 

Hollow circle 
(eccentric) 

Sector of 
circle . . 

Sector of hol- 
low circle . 

Fillet . . . 

Segment of 
circle . . 

Ellipse . . 

Irregular 
figures . . 




i rn - 

' 5T3 

ir(a + b) approx 
or ?r{i-5(a + 6) - 
more nearly 

Step round 
curved por- 
tions in small 
steps, with 
dividers ; add 
in any straight 
pieces. 


^(D-P)= s - 7 854(D*~d) 
or ,r(R - r z ) 
or TT x mean dia. x thick- 
ness 

7854 (!>-<*) 
or 7r(R 2 -f 2 ) 

ir)7' a lv 
l6o" Or T 

7M7(R*-r J ) 




_.., 


360 
2i5f 2 or approx. \v* 


'^N \ c X^ 


Area sector triangle 
Various approx. for- 
mulae on p. 300. 

- Vab} rrab 

Divide into narrow 
strips ; measure their 
mid-ordinates. Then 
Area = aver, mid-ordi- 
nate x length / 


' v 


ff TfV 


L^13 


1, -j 



CHAPTER VIII 



MENSURATION- continued 
VOLUMES AND SURFACE AREAS 

Volume is the amount of space occupied by a body, or is its 
bulkiness, and includes measurement in three main directions, 
usually known as length, breadth, and depth or thickness. The 
unit of volume is the cubic foot (cu. ft.), which is the volume of a 
solid body, with six faces, each face being a square of i ft. side, 
as shown in Fig. 89. This solid is called a cube. 





Fig. 89. The Cube. 



Fig. 90. Eolation between the Cubic 
Inch and Cubic Foot. 



For smaller measurements the cubic inch (cu. in.) is used, which 
is the volume of a cube of i" side. The relation between the cubic 
inch and the cubic foot may be obtained by reference to Fig. 90. 
Here the big cube represents a cube of i ft. side, and is therefore 
a cubic foot. Since i ft. = 12 ins. then the cube may be sawn 
into 12 slabs, each i" thick, by cutting along the dotted lines. The 
cube, when sliced, is shown in Fig. 90, the slabs being slightly 
separated from one another. Now each slab is i in. thick and 
i ft. side, i. ., 12 ins. each side, and may therefore be divided up 

3M 



MENSURATION 315 

into a number of little cubes each i in. side as indicated on the 
foremost slab. Evidently the first slab contains 12 x 12 == 144 
cubes of I in. side, i. e., 144 cu. ins. There being 12 slabs in all 
each of which may be divided like the first one, the whole cubic 
foot is obviously 144 x 12 cu. ins., i. e., 1728 cu. ins. 

Then i cu. ft. == 12 x 12 X 12 cu. ins. = 1728 cu. ins. 
Similarly for the cubic yard 

I yard = 3 feet 
.'. i cubic yard =3x3x3= 27 cubic feet 

These three units are the only ones of their type employed, and 
our table of volume measure is then 

Smallest unit = i cubic inch (cu. in.) 
1728 cubic inches = i cubic foot (cu. ft.) 
27 cubic feet = i cubic yard (cu. yd.) 

The only conversions which are required may be summed up in the 
following : 

Cubic inches = cubic feet x 1728 

Cubic feet = cubic inches -f- 1728 

Cubic feet = cubic yards x 27 

Cubic yards = cubic feet -f- 27 

There is another unit often employed when dealing with the capacities 
of various vessels, etc., and which relates to the weight of water. 
This is the gallon (volume of 10 Ibs. of water).* 

* i gallon (gall.) = 277-4 cubic inches 

= '1605 cubic feet 
or 6*24 gallons = i cubic foot * 

Occasionally when dealing with the storage of grain, etc., the 
following are required : 

8 gallons = i bushel 
8 bushels = i quarter. 

The weight of an article depends on its volume, and it is very 
often necessary to find the weight, knowing the volume and the 
weight per cubic foot or per cubic inch of the material of which it 
is made. This merely requires a simple multiplication. 

* This is when measured at 62 F; the weight of a cubic foot of 
water depends upon the temperature, and becomes less as the tempera- 
ture rises. The greatest weight is 62-425 Ibs., at 39 F. 



316 ARITHMETIC FOR ENGINEERS 

Conversion. The amount of conversion and reduction wanted 
is very small. The following examples should suffice : 

Example 321. The volume of an engine cylinder is 1830 cu. ins. 
Convert this into cubic feet. 

Volume of cylinder = 1830 cu. ins. 

= ~ 8 -^cu. ft. = i-o6cu. ft. 
1720 ___. 

Example 322. A pipe is discharging -28 cu. ft. of water per second. 
Find the discharge in gallons per minute. 

28 cu. ft. per sec. = -28 x 6-24 galls, per sec. 

= '28 X 6-24 X 60 galls, per min. 
= 105 galls, per min. 

Example 323. A pump is rated at 4800 gallons per hour. How 
many cubic feet of water can it discharge per second ? 

6-24 galls. = i cu. ft. 

/. 4800 galls, per hr. = \ - cu. ft. per hr. 
0*24 

4800 ,. 

_. 3 _-. cu. ft. per sec. 
6-24 X 3600 r 

= '214 cu. ft. per sec. 

Exercises 77. On Volume Conversion. 

1. The volume of a steam engine cylinder is 12-56 cu. ft. The 
clearance volume is 8% of this. Find the clearance vojume in cubic 
inches. 

2. A storage tank is to hold 5000 gallons of water. What must be 
its volume in cubic feet ? 

3. A pump for an experimental hydraulic plant is required to 
deliver 1*5 cu. ft. of water per second. What must it deliver in gallons 
per hour ? 

4. A feed pump delivers 160 cu. ins. of water per stroke, and 
makes 25 strokes per minute. How many gallons of water does it 
deliver per minute ? 

5. A " Navvy " or steam shovel had an average output of 63 cu. yds. 
per hour. What is this in cubic feet per minute ? 

6. If a casting has a volume of 3250 cu. ins. and i cu. ft. of the 
metal of which it is cast weighs 390 Ibs., find the weight of the casting. 

Volumes of Regular Solids. 

Prisms. The great majority of solid bodies met with in engi- 
neering work are either regular solids, or are made up of a number 
of regular solids. A " regular solid " is one which has some definite 



MENSURATION 



317 



geometrical shape. There are two main classes : " Prisms " and 
" Pyramids.' 1 A prism is a solid with two parallel ends, both of 
the same shape and size, joined by a number of sides which are 
plane rectangles. The ends are generally perpendicular to the 
sides. The shape of the end may be that of any of the regular 
geometrical figures, e. g., square, hexagonal, circular, etc., and the 
prism is named according to the shape of the end. Thus Fig. 92 
shows a rectangular prism. The " circular prism " is given the 
special name of cylinder. A square prism in which the length of 
the side is the same as the side of the square end is called a cube ; 
every face is then a square, so that any pair of opposite faces may 
be called the ends. A rectangular prism, i. e., one whose ends are 
rectangles, is sometimes known as a cuboid. 

Any prism may be regarded as the volume swept out by a 
plane figure when moved perpendicularly to its plane. Thus in 



r^c, 



*, 





Fig. 91. 



Fig. 92. Volume of Rectangular 
Prism or Cuboid. 



Fig. 91, the square ABCD is moved in the direction of the arrow 
to the position marked A 1 B 1 C 1 D 1 and in doing so sweeps out a 
square prism. 

Volume of a Rectangular Prism. Fig. 92 shows a cuboid 
whose end is 2" X 3", and whose length is 5". Now let the solid 
be cut into five pieces each i" long as indicated by the faint lines- 
Taking the end section, this may be cut up as indicated (Fig. 92) 
into a number of small cubes each of I* side; these small cubes 
are then cubic inches. Evidently in the end section there are 
2x3=6 cu. ins. 

Now every other section is like this one and therefore contains 
6 cu. ins. There being five sections in all, the total number of 
cu. ins. in the solid is 6 x 5 = 30. Thus the volume of the prism 



ARITHMETIC FOR ENGINEERS 



= 2 x 3 x 5 = 30 cu. ins. Supposing we call the 5* measurement 
the " length/' the 3" measurement the " breadth/ 1 and the 2* 
measurement the " thickness " : then evidently 

Volume of rectangular prism = length x breadth x depth 
All three dimensions must be in the same one unit, i. e., they must 
all be in inches, or all in feet, etc. If the measurements are given 
in various units, say one in feet and the others in inches, then all 
must be converted to some one unit, say feet or inches. Again, if 
some of the dimensions are in two or more units, e.g., length = 
3'-2", then before any multiplying is carried out this dimension 
must be converted into inches (38") or feet (3-167'). See Ex. 326. 

Example 324. A rectangular tank for measuring purposes is 5 ft. 
long and 2 ft. broad. Find how many gallons of water it holds when 
the water is 4 ft. deep. See Fig. 93. 

(All measurements are in feet, and may therefore be multiplied up 
directly, giving the result in cubic feet.) 




Fig. 93- 



Fig. 94- 



Volume of water in tank = 5X2X4 
= 40 cu. ft. 

i cu. ft. = 6-24 gallons 
/. Gallons in tank = 40 x 6-24 

= 249-6, say, 250 gallons. 

Example 325. In some milling tests on cast iron, a block 8*' wide 
was milled with a cut '32* deep, the feed being 13^ per min. (See Fig. 
94.) Determine the volume of metal removed per minute. 

In i min. the cutter will pass over a length of i3j". Then the 
metal removed is a rectangular prism 13^ X 8* x -32". 

.'. Volume removed == 13-25 x 8 x -32 
s= 33*92 cu. ins. 



MENSURATION 



319 



Example 326. A concrete foundation for an engine is to be g'-6* 
long, 3 / -9 /r wide, and I'-G" deep. Find what volume of concrete wil) 
be required. 

(All measurements must be converted to some one unit : in this 
case feet will be the most suitable.) 

Length g'-6" = 9-5 ft. Breadth 3 '-g* = 375 ft. Depth i '-6" = 1-5 ft. 

.'. Volume required = 9-5 X 3-75 X 1-5 
= 53-4 cu. ft. 



Exercises 78. On Volume of Rectangular Prisms. 

1. What volume of concrete will be required for an engine foundation 
15' long, 5 / -6 // wide, and 4 '-6" deep ? 

2. A railway wagon is y'-o" wide, 4 '-6" deep, and i3'-o /r long. 
What weight of coal will it carry when just full, if i cu. ft. of coal 
weighs i cwt. ? 




ig- 95- Exercises on Volumes of Prisms. 



3. A storage tank is 40 ft. long and 12 ft. wide. If the water be 
8'-6" deep, how many gallons of water does the tank hold ? 

4. A storage battery contains 60 cells each 12" X 15". How many 
gallons of acid will be required to fill the cells to a depth of 10"? 

5. A hod used for lifting bricks by crane is shown at A, Fig. 95. 
Taking i cu. ft. of brick as weighing 112 Ibs., calculate the weight to 
be lifted. (Neglect weight of hod.) 

6. Calculate the weight of a railway sleeper, 9 ft. long, 10* wide 
and 5" deep, if i cu. ft. of the wood weighs 52 Ibs. 

7. A rectangular acid tank is 2'-o" long, i / -3 // wide, and i'-io* deep. 
Calculate the number of gallons of sulphuric acid it will hold when 
filled to within 2" of the top. 

8. A steel ingot containing n cu. ft. of metal is rolled into a plate 
i* thick and 9 ft. wide. Find the approximate length of the plate. 
(Give the answer to the nearest foot.) 



320 ARITHMETIC FOR ENGINEERS 

9. The cross-sectional area of the Standard Bull-Head Rail used 
on British railways is 8-81 sq. ins. If i cu. ft. of the steel weighs 
490 Ibs., calculate the " weight per yard " of the rail (i. e., the weight 
of a piece i yd. long). 

10. A rectangular bin is required to have a volume of 150 cu. ft. 
The length is to be g ft., and the height is limited to 4 '-3*. What 
width must it be made ? 

11. An overhead storage tank, 13 ft. square, is required to hold 
5000 gallons of water. How deep must the water be ? 

12. A small transformer core is shown at B, Fig. 95. In a test it 
was necessary to know the volume of metal in the core. Find this 
volume. 

13. The cross-section of a reinforced concrete foot-bridge is shown 
at C, Fig. 95. The total length is 42 / -7 // . If i cu. ft. of concrete be 
taken as 120 Ibs., calculate the weight of the concrete. 

14. A rectangular channel is required to discharge 10 cu. ft. of 
water per second. If the water travels 4 ft. in i sec., what must be 
the area of the cross-section of the channel ? If the channel be made 
2 '-3* wide, how deep will the water run ? 

Volume of Other Prisms. When the shape of the end is 
other than rectangular we cannot take length x breadth x -depth 
to give volume. Notice that in the rectangle in Fig. 92, 3" x 2" = 
6* = area of the end, i. e., breadth x thickness. 

So that we can also say volume = area of end x length. 

Now whatever the shape of the end, provided that it be one of 
the regular shapes, we can calculate its area. Hence we have for 
any prism 

Volume of prism = area of end x length 

The same units must be used throughout, as emphasised in the 
previous section. 

Taking some particular cases 

Hexagonal prism : Let s = side of hexagon, / = length 

Area of end = 2 -6s 2 
/. Volume = 2'6s 2 / 

Cylinder : d = diameter, h = length, r = radius 

Area of end = - d 2 or nr 2 
4 

/. Volume = T d 2 /* or iri^h 

The remaining cases are given in the table at the end of the chapter, 
as they are all of the same nature. 

Example 327. A steam engine cylinder is 13^ dia. and the stroke 
is 1 6". Determine the " working volume " in cubic feet (i. e. t the 
volume swept out by the piston). 



MENSURATION 321 

This is a case of a cylinder 13* dia. and 16* long. 

Area of end irr 2 and r = 6 y 

= IT X 6*5 2 = 133 sq. ins. 

= I33 sq. ft. 
144 

Length = 16" = i'-4* = 1-33 ft. 

.'. Volume = area of end X length = I3J.2LJL33 cu ft 

144 

= 1-23 cu. ft. 

Example 328. A boiler feed pump is 5* dia. and has an 8" stroke. 
Find the discharge in gallons per minute when making 30 strokes per 
minute. 

Area of cylinder = d 2 = -785 x s 2 = 19-6 sq. ins. 
4 

19-6 ,, 
= * sq. ft. 
144 ^ 

Stroke 8" = - ft. = -667 ft. 
/. Volume per stroke = - x -667 = -0908 cu. ft. 

At 30 strokes per min., volume per min. -0908 X 30 cu. ft. per min. 

i cu. ft. 6*24 galls. 
/. Galls, per min. = -0908 x 30 X 6-24 
= 17 galls, per miii. 

In many cases it is necessary to invert the operations when 
dealing with volume, i. e., knowing the volume and some other 
dimensions to find, say, the length of the prism. 

Since volume = area of end x length 

., . .. volume 

then length = _ --= 

area of end 

j i j volume 

and also area of end = . - 

length 

With the particular case of the cuboid 

volume = length x breadth x depth 

. -r> -ui volume _, 

/. Breadth = f pr A , . , . Ltc. 

length x thickness 

Example 329. A tank is s'-o" long, and 3 '-6* wide. How deep 
will the water be when the tank contains 120 galls. ? 

Volume of water =120 galls. 

= g^j cu. ft. = 19-23 cu. ft. 
Then 19-23 = 5 X 3-5 X depth 
/. Depth = ~ 1 ^- = i-i ft. 



322 



ARITHMETIC FOR ENGINEERS 



Example 330. A cylindrical receiver for an air compressor is to 
have a capacity of 25 cu. ft. What diameter must it have if its length 
is 5'-o"? 

Volume = 

:. -785^ x 5 - 25 



^6-37 = 2 '5 2 ft. 

= 2 ft. 6J ins. 




Fig. 96. 



Fig. 97. 



Fig. 98. 



Example 331. In making bullets, lead wire is produced by squeez- 
ing molten lead from a cylindrical chamber through a hole in the top, 
as shown in Fig. 96. If the chamber is 4* dia. and 8" long, what length 
of J" dia. wire can be produced ? 

The 4" dia. cylinder, 8" long, is made into a long J* dia. cylinder, 
whose length is to be found. 

Volume of lead in chamber = d*h 

4 

= -785 x 4* X 8 = 100-5 cu - ms - 

Cross-sectional area of wire = -785 x -25* = -0491 sq. in. 
.', 100-5 = -0491 x length in inches 

.% Length = 2047" or 56-9 yds. 

Hollow Cylinders and Tubes. Take the hollow cylinder in 
Fig. 97, and let it be required to find the volume of the solid portion. 
Area of the end = 785 (D 2 - d 2 ) 
Then volume = area of end x height 
- -785 (D 2 - d z )h 

Alternatively, the volume of the solid " walls " of the tube is evi- 
dently the volume of the big cylinder of diameter D, less the volume 
of the small cylinder of diameter d. 



MENSURATION 



323 



Then volume of large cylinder = 7850% 
and volume of small cylinder = 785^ 

/. Volume of tube = 7850** - 785**** 
= -785h(D 8 - d 2 ) 

taking out the common factor 785/1 

Example 332. Fig. 99 shows the sump or reservoir of a circular 
cooling tower. How many gallons of water does it hold when the 
water is 3 '-6" deep ? 



Oevafiorv 




Fig. 99- 

Volume of water = 785(02 - d*)h 

= 785(22* 7-25 2 ) x 3'5 cu. ft. 
(reducing all measurements to feet) 

= 7 8 5 X 431-4 X 3*5 
= 1185 cu. ft. 

.*. Gallons of water = 1185 x 6-24 = 7396 

Example 333. In making lead pipes by extrusion, what would be 
the least volume of metal required to produce 60 ft. of pipe y inside 
diameter and }$" outside diameter ? What weight of metal would be 
required if the lead weighed -41 Ib. per cu. in. ? 

This is the case of a very long hollow cylinder. 

Area of end = 785(02 d 2 ) 

= 785('68 7 5 a - -5') 
= 785 x -2228 
= -1749 sq. in. 
Length = 60 ft. = 720 ins. 

/. Volume of metal = area of end x height 

= -1749 X 720 = 125-9 cu. ins. 

say, 126 cu. ins. 

/. Weight of metal at '41 Ib. per cu. in. = 126 x *4 J =51-66 Ibs. 

say, 52 Ibs. 



324 ARITHMETIC FOR ENGINEERS 

Exercises 79. On Volumes of Cylinders. 

1. A cylindrical gas-holder is 150 ft. dia. and 115 ft. high. How 
many cubic feet of gas does it contain ? 

2. An oxygen cylinder is 5" dia. and 3'-o* long. It is filled with 
gas under pressure so that every cubic foot of the cylinder contains 
1 20 cu. ft. of gas. How much oxygen does the cylinder hold ? 

3. The H.P. cylinder of a compound Corliss engine is 15" dia. and 
has a 36" stroke. Calculate the " working volume " in cubic feet (i. e., 
the volume swept out by the piston). 

4. A separator drain tank is i'~(>" dia. and 2 / -o* long. Calculate 
how many gallons of water it will hold, (i cu. ft. = 6-24 galls.) 

5. The governor ball on a high speed engine is a cast-iron cylinder 
3" dia. and 2 y thick. Calculate its weight if i cu. in. of cast iron weighs 
26 Ib. 

6. A three-throw force pump (i. e., one with three cylinders) has 
cylinders 14!" dia. and 24* stroke, and makes 100 r.p.m. Calculate 
the gallons of water it delivers in i min. (Hint. In i rev., pump 
delivers the volume of the 3 cylinders.) 

7. The clearance volume in a gas-engine cylinder is 209 cu. ins. 
It is desired to reduce this volume to 165 cu. ins. by attaching a plate 
g dia. to the back of the piston. What thickness of plate is required ? 

8. In a cement testing machine the load is applied by running 
water into a tall cylindrical tank. In a certain machine the greatest 
load required -962 cu. ft. of water to be in the tank. If the length of 
the tank is limited to 3'-9*, what must be its diameter ? 

9. The cylinder of an air compressor is required to have a working 
volume of 5 cu. ft. If the diameter is to be 20", what must be the 
stroke ? 

10. The cross-section of the discharge flume from a hydro-electric 
installation is given at E, Fig. 95. Calculate the cubic feet of water 
discharged per second if the water leaves at 4 ft. per sec. 

11. Determine the volume of concrete (cubic feet) required to build 
the culvert whose section is given at F, Fig. 95, if it is 650 yds. long. 

12. A condenser tube plate of rolled Muntz metal is 3 '-6* dia. and 
i Y thick. Calculate its weight if the metal weighs -31 Ibs. per cu. in. 

13. Dimensions of a Lancashire boiler are given at G, Fig. 95. How 
many gallons of water will be required to fill the boiler for its hydraulic 
test ? (Hint. The water volume is the large cylinder less the two flue 
cylinders.) 

14. A Diesel engine cylinder is 19^" dia. and has a 37* stroke. When 
at the top of its stroke the piston is 2^ away from the cylinder cover. 
The resulting space enclosed is the " clearance volume." Express this 
volume as a percentage of the " working volume " (i. e., the volume 
swept out by piston) and also as a percentage of the total volume. 

15. In a test on a pump driven by producer gas the quantity of gas 
used was measured by noting the height fallen by the gas-holder top 
during the run. The holder was 20 ft. dia. and dropped 2 / -7j /r in the 
test. How many cubic feet of gas were used ? 

16. H, Fig. 95, gives dimensions of a link for a crane chain. Calcu- 
late the weight in pounds of one link, if wrought iron weighs -28 Ib. per 
cu. in., and also the weight per fathom (i. e., number of Ibs. for 6 ft.). 
(Hint. When finding the number of links in the 6 ft., the length of 



MENSURATION 



325 



Work to the nearest complete 



one link is the inside dimension 2* 
link.) 

17. A lead hammer head is shown at J, Fig. 95. If the metal 
weighs -41 Ib. per cu in., calculate the weight of the head. 

HOLLOW CYLINDERS 

18. A hollow steel shaft has an outside diameter of 7* and an inside 
diameter of 3j*. Calculate the weight of a 12 ft. length if the steel 
weighs 491 Ibs. per cu. ft. 

19. Calculate the weight per foot of a cast-iron pipe 9J* inside dia., 
and J" thick, if cast iron weighs 460 Ibs. per cu. ft. (Hint. Use 
" average diameter " method.) 

20. A circular brick chimney is TOO ft. high, 8 ft. average dia., and 
i$y average thickness. If i cu. ft. of brickwork weighs 120 Ibs., 
estimate the weight of the chimney. 

21. An elevated storage tank is shown at D, Fig. 95. Calculate the 
gallons of water it will hold. 

22. The body of a gas-storage cylinder is made by " cold drawing " 
from a flat plate. If the body is produced from an annular plate as 
indicated at F, Fig. 102, and the volume of the annular plate is equal 
to the volume of the cylinder body, calculate the thickness of plate 
required for dimensions shown. 

We may also have prisms whose ends are segments of circles 
or combinations of any of the regular figures, or even irregular 
shapes. In all these cases find the area of the end and multiply 
by the length. See Exercises 80. 

-3-6 




Fig. 100. Exercises on Volume of Prisms and Spheres. 



Exercises 80. On Volumes of Various other Prisms. 

(Letters refer to Fig. 100.) 

1. A sandbin built against a wall is shown in end view at A. Cal- 
culate the cubic yards of sand it contains when filled to the dotted 
line. Also find the total weight of the sand if i cu. ft. weighs 105 Ibs. 

2. Taper fire-bricks for use in the furnace arches of ""mechanical 
stokers are shown at B. Calculate the weight per 1000 bricks if i cu. ft. 
of fire-brick weighs 137 Ibs. 



326 



ARITHMETIC FOR ENGINEERS 



3. A swimming bath is shown at C. Find how many gallons of 
water are required to fill it. 

4. Find (a) the weight of a foot of hexagonal bar y across the 
flats and (b) the weight of a foot of hexagonal bar of \" side, both to be 
used for making bolts and both of steel weighing -28 Ib. per cu. in. 

5. An octagonal bar J* across flats is to be used for chisels. Find 
its weight per foot if the steel weighs 489-3 Ibs. per cu. ft. 

6. At D, Fig. 1 02, is shown a V-block for shop use. Find its weight 
if of metal weighing 489 Ibs. per cu. ft. 

7. Calculate the capacity in gallons of an elliptic milk-can 10" X 6* 
and 15" deep, if filled to within i J" of the top. 




Fig. 101. Voiume of Prisms with Irregular Ends. 



8. Calculate the capacity in gallons of the elliptical petrol tank for 
the back of a motor-car, shown at B, Fig. 102. 

9. The dimensions of the cross-section of a f" " Simplex " oval 
conduit similar to that at X, Fig. 31, are as follows : Mean axes -95* 
and -42*, and thickness '042*. Find the weight of a 100 ft. length if 
steel weighs -28 Ib. per cu. in. (Hint. This is a hollow elliptical 
cylinder, and the " average " sizes of the section are given.) 

10. Calculate the capacity in gallons of the petrol tank for a car, 
as shown at D. (Hint. As the corners have a large radius, allow for 
them by deducting the " fillets " when finding the area, see p. 299.) 

11. At H are given dimensions of a locomotive wheel balance 
weight. Calculate the weight if of steel weighing -28 Ib. per cu. in. 

12. F shows the cross-section of an arched reinforced concrete floor. 
If the length is 18 ft., find the total volume of concrete between two 



MENSURATION 327 

girders. If the concrete weighs 120 Ibs. per cu. ft., what is the weight 
of the floor between the two girders ? (Hint. In finding the cross- 
sectional area subtract the segment from the rectangle.) 

13. A section of chisel-steel bar is shown at E. Calculate the 
weight per foot if i cu. in. of cast steel weighs -28 Ibs. Find the cost of 
8 ft. at 6d. per Ib. 

14. J gives dimensions of a crescent-shaped balance weight for the 
driving wheel of a locomotive. Calculate the weight if i cu. in. of steel 
weighs -28 Ibs. 

15. An oil-storage tank partly filled with oil is shown at G. Find 
the number of gallons it contains. 

16. Dimensions of a hollow concrete slab for the fire-proof con- 
struction of flooring are given at L. Calculate the volume of concrete 
required for one slab and also its weight if i cu. ft. of concrete weighs 
120 Ibs. (Hint. In finding the area of the end of the prism deduct 
the ellipse and the pieces put of the corners from the sum of the segment 
and the rectangle below it.) 

17. When designing dams and retaining walls it is necessary to 
know the weight of i ft. length of the wall. B, Fig. 101, gives the section 
of a certain dam. Calculate the weight of i ft. if the masonry weighs 
156 Ibs. per cu. ft. (Hint. Use horizontal mid-ordinates.) 

18. A, Fig. 10 1, shows the section of a British Standard Bull-Head 
Rail. Find the weight per yard in Ibs. (Hint. Calculate the area of 
the web between a pair of lines indicated ; work by vertical mid-ordinates 
on the two flanges.) i cu. ft. of the steel weighs 490 Ibs. 

Volume of Sphere. The volume of a sphere or round ball 
cannot be examined in any simple way. At this stage, then, the 
following rule must be accepted. 

4 
Volume of sphere = ^TrR 3 where R = radius of sphere 

or in terms of diameter R = 



- ~ - - or -5236D* 

3^ 2 6 

By a similar deduction to that for the hollow cylinder (p. 323), 

4 
Volume of hollow sphere = ^-(R 3 r 3 ) 

o 

Example 334. A spherical float for an automatic valve is 5" dia. 
How much water will it displace when half submerged ? The upward 
force that the water exerts on a floating body is equal to the weight 
of water displaced. What will be the force in this case if i cu. in. of 
water = -0361 Ib. ? 



328 ARITHMETIC FOR ENGINEERS 

irD 3 

Volume of sphere = ~^- 

ir X 5 3 IT X 125 , 

= ~ = - 7- - = 65-5 cu. ms. 
.'. Volume of water displaced when half submerged 

= 4 X 6 5'5 = 32*75 cu - ins* 
Upward force = weight of displaced water 
= 3275 X -0361 = 1-18 Ibs. 

Example 335. A cauldron for melting down fat is cylindrical, with 
a hemispherical bottom, as shown in Fig. 98 ; determine the capacity 
of the tank. 

The volume required is composed of a cylinder 6'-o* dia. and 4'-o* 
long, and half a sphere of 3'-o" rad. 

Volume of cylinder = -rrr z h 

= Tf X 3 2 X 4 = 113 cu. ft. 
Volume of hemisphere == \ volume of sphere 
= J X rr 8 

= f X ir X 3 s 

= 27r X 3 2 = 2ir X 9 = 56-5 CU. ft. 

/. Capacity of tank = 113 + 56-5 = 169*5 cu - ft - 

Volume of segment of sphere = ? -c"( 3r2 + A 8 ) 

where r = radius of the base (not radius of the sphere) 
h = height of segment. (See Table on p. 346.) 

Example 336. An evaporating pan is segmental and is 6 ft. dia. 
and 2'-9" deep. Calculate its capacity in cubic feet. 

Radius of base = 3 ft. ; height = 2*73 ft. 
Volume - *($r* + h 2 ) 

= '-^p- 5 ( 27 + 7-56) 

= -XJ5Xj4 J 6 = . 



Exercises 81. On Volume of Sphere, etc. 

1. Calculate the weight of a cast-iron governor ball 2}" dia., weighing 
26 Ib. per cubic inch. 

2. A lead melting pot is hemispherical and is f dia. What volume 
of lead can it hold ? How many vice clamps, each requiring 6J cu. ins., 
can be cast from one charge of the above pot ? 

3. An air vessel for a pump is shown at M, Fig. 100. Calculate 
the air volume in cubic feet when the average position of the water is 
as shown. 



MENSURATION 



329 



4. A storage tank has. hemispherical ends as shown at K, Fig. 100. 
Calculate the quantity of water held when half full (in gallons). 

5. An elevated storage tank, with a hemispherical base as shown 
at N, Fig. 100, is supported on a trestle tower. Calculate the gallons 
of water it will hold, 

6. An evaporating pan is segmental and is 7'-6 /r dia. and 3 ft. deep. 
Calculate its capacity in cubic feet. 

7. Fig. 102 shows at E a rivet with a segmental head. Calculate 
the weight of 100 such rivets if made of steel weighing 490 Ibs. per 
cu. ft. 

8. Shrapnel bullets are of lead, spherical and of J" dia. Calculate 
the weight per 1000 if i cu. in. of lead weighs -41 Ib. 

9. At C, Fig. 102, is shown a balance weight used on a testing 
machine. Find its weight if of cast-iron weighing 450 Ibs. per cu. ft. 



End View Side Elevation Q 




Plan - Top* 



Plan 



Fig. 102. Exercises on Volumes. 



10. Find the weight of a hollow sphere of 3" inside and 3^* out- 
side radius which was for part of an experimental boiler. (Take the 
weight of the material as -26 Ib. per cu. in.) 

Pyramids. A pyramid is a solid having a base of any geo- 
metrical shape, from which the sides taper uniformly to a point 
or apex. As with prisms, pyramids are designated by the shape 
of their ends, in this case called the base. Thus a, Fig. 103, shows 
a square pyramid. The circular pyramid b, Fig. 103, is given the 
special name of cone. From the engineering standpoint the cone 
and square pyramid are the most important solids of this class. 

The volume of a pyramid cannot be examined in any simple 
manner, and must therefore be accepted as truth until a later 
stage. It is, however, one-third of the volume of the circumscribed 
prism, this latter being a prism whose end is the same shape and 



330 



ARITHMETIC FOR ENGINEERS 



size as the pyramid, and whose height is -the same. Thus at a, 
Fig. 103, the full lines show a square pyramid and the chain- 
dotted lines indicate the circumscribing prism. 

Since the volume of a prism is " area of base x height " 

Volume of any pyramid = area of base x height 

Taking the two important cases : with the square pyramid 

Let s = side of square base ; h = height of pyramid 

Then volume of circumscribing prism = s 2 h 
and Volume of square pyramid = Js 2 /i 

Similarly for a cone. The circumscribing prism is a cylinder of 
radius r and height h, as at b, Fig. 103. 

Volume of circumscribing cylinder = rrr 2 h 
/. Volume of cone 




Fig. 103. Cone and Pyramid. 

It must be carefully observed that by the " height " is meant the 
" height perpendicular to the base," and not the length of the sloping 
side. 

Example 337. A wooden cone for demonstration purposes is &" dia. 
and 1 8" high. Determine its volume and also its weight, if made in 
pine weighing -0318 Ib. per cu. in. 



Volume of cone = 



= J X 4 2 X 18 

~ ?L* : l6 _ x l8 = 

3 " 



301-4 cu. ins. 



/. Weight = 301-4 X -0318 = 9-6 Ibs. 

Example 338. It is desired to cut down the cone mentioned in the 
previous example, into a square pyramid as indicated on the left of 
Fig. 103, keeping the diameter of the cone as the diagonal of the square 
base. What will be the volume of the square pyramid ? 

It is not necessary here to find the side of the square. Referring to 




MENSURATION 331 

p. 281 area of square = half product of diagonals. In this case the 
diagonals are 8*. 

Area of base = J product of diagonals 

= -J X 8 X 8 = 32 sq. ins. 
Volume = J base X height 

= i X 32 X 18 = 192 cu. ins. 

From the formulae given either the height or the size of the base 
can be determined, when the volume, etc., are known. 

Example 339. A conical measure as in Fig. 104 is required to hold 
one pint (34-66 cu. ins.) when filled to a depth of 5 J*. 
What must be the diameter at the surface of the 
liquid ? 

Volume of cone = J-n-f 2 /* 

34-66 = 3 ' 14r> X 3' 5 

,2 = 3 X 34'66 
3'i4 X 5-5 
/. r = V6-03 = 2-455* 

.'. Diameter at surface = 2 X 2-455 4-91* 

Ing. 104. 

Frusta of Cone and Pyramid. If a cone or pyramid is cut 
into two parts by a plane parallel to the base, the lower portion 
is termed a Frustum (plural " frusta "). Thus Fig. 105 shows a 
frustum of a square pyramid. Evidently the volume of the frustum 
= Volume of large cone minus volume of small cone cut off. (See 
Fig. 105.) The usual dimensions known will be the sizes of the 
small and large ends and the height of the frustum, i. e., the 
perpendicular distance between the ends. If the height of the 
complete cone is known (or is obtained by drawing out and 
measuring) then the volume of the frustum can be obtained as 
indicated above. 

But it may often be inconvenient to find the height of the 
complete cone, and in any case the finding of two volumes is a 
long job. A formula for the volume is then useful. 

Volume of any frustum ^(A + a + A/A'a) 

6 

where h = height of frustum 

A = area of large end, and 
a area of small end 

(This formula cannot be proved here, as it involves a considerable 
amount of algebraical work, including a quadratic equation.) 



332 ARITHMETIC FOR ENGINEERS 

The formula is applicable to any frustum, whatever the shape 
of the end. As the most important cases are those of the square 
pyramid and cone we will derive a particular formula for each. 

FRUSTUM OF SQUARE PYRAMID. Let S = side of base 

s = side of top 
Then A = area of base = S 2 
and a = area of top = s 2 

' Vo f lume of frustu !) = *(A + a + VKJ) 
of square pyramid/ 3 

*( S + S 2 + VsV) substituting for A 
3 ' and a 

- |(S* + s 2 + Ss) 

FRUSTUM OF CONE. Let R = radius of base 

r = radius of top 

Then A = area of base = ?rR 2 
a = area of top = irr 2 



+ ^ + , Rr) educing the square 
out the common 



If diameters be used instead of radii 
the volume - (D 2 + d 2 + 

The reader is cautioned against calculating the volume of a 
frustum by the expression " height x average of the two end 
areas/ 1 This approximate formula can only be used in cases of 
slight taper and where the height of the frustum is a very small 
fraction of the height of the complete pyramid. Large errors may 
be introduced by the indiscriminate use of this formula; and ex- 
amination for each case will take as long as working by the exact 
formula. 

Example 340. Fig. 106 gives dimensions of a coal bunker, con- 
sisting of a square prism and the frustum of a square pyramid. Find 



MENSURATION 



333 



the capacity of the bunker in cubic feet, and also the weight cf coal it 
holds, if i cu. ft. of coal = cwt. 



Volume of the square prism = 10-5 X 10*5 x 7*5 

= 826-9 cu. ft. 

Volume of lower portion 1 __ ^/ 
(frustum of sq. pyramid) J ~~~ 3 * 



_t 2 i c x 
"*" s ' ' 



3 

= 1-667(110-2 -f 2-25 + 15-75) 
= 1-667 X 128-2 = 213-7 cu - ft - 

Total capacity of bunker = 826-9 + 213-7 s=s 1040-6 cu. ft. 
Weight of coal held == 1040-6 X J cwts. = 520-3 cwts. 

Capacity (coal level with j _ saVj 520 cwts or 26 tons 
top of bunker) J ._ 



Pyramid 
cuiroff 



THS 

FRUSTUM. 



Fig. 105. 






Fig. 106. 



Fig. 107. 



Example 341. An ash receiver in a suction ash plant, with a 16" 
dia. hole for discharging, is cylindrical, with a taper bottom, as shown 
with dimensions in Fig. 107. Find its capacity. 

Volume of cylindrical portion = d*h 

= -785 x 12* x 18 
= -785 X 144 X 18 
= 2035 cu. ft. 
Conical frustum, radius of top = 6 ft. 

radius of bottom = 8" = -667 ft. 

/. Volume = ~(R* + r* + Rr) 



X 9>25 (6* 



+ -667 2 + 6 X -667) 

+ '445 + 4) 

392 cu. ft. 



/. Total capacity"! . 
of receiver J 



IT X 9*25 X 40-44 

3 
2035 + 392 = 2427 cu. ft. 



334 ARITHMETIC FOR ENGINEERS 

Example 342. Fig. 108 gives dimensions of a gas-holder tank. 
Determine how many gallons of water are required to fill it to a depth 
of 35 ft. 




cular* \n plan 

Fig. 108. Fig. 109. 

The volume of water here is the volume of a cylinder, minus the 
volume of the conical frustum in the centre. 

Volume of cylindrical portion = irR 2 h 

= IT X 6i 2 X 35 = 408,900 cu. ft. 

Frustum to be deducted 

Radius of large end = 55' 

Radius of small end = lo'-S" = 10-67 ft. 

Height = 25-5' 



.'. Volume = ~(R 2 + r z + Rr) 



+ 55 X 10-67) 




/. Volume of water = 408900 99420 



.*. Gallons of water = 309480 X 6-24 
= 1,932,000 galls. 

Exercises 82. On Volumes of Pyramids, Cones, and 

Frusta. 

1. Find the volume of a square pyramid if the side of the base 
is 5* and the perpendicular height is 12^''. Find what fraction this 
volume is of the volume of a square prism, of the same height and same 
base. 

2. A square pyramid is to have a volume of 160 cu. ins. and a 
base of 7^ side. What will be its height ? 

3. A cone is nj" high and $%" in diameter at the base. Find its 
volume. What relation can you find between this volume and the 
volume of a cylinder of the same height and on the same base (5^ dia.) ? 

4. A, Fig. no, shows a supply bucket (for holding sand) as used in 
a special testing machine. Find its capacity in cubic inches. The 



MENSURATION 



335 



sand is allowed to run into a cylindrical tin 10* dia. If this tin is to 
hold all the contents of the supply bucket, what must be its least depth ? 

5. A conical part of a measuring glass, as shown in Fig. 104, is 
to have a length of 4" and is to hold half a pint. What will be its 
diameter at the surface of the liquid (i pint = 34-66 cu. ins.) ? 

6. A bucket for a coal conveyor consists of a frustum of a square 
pyramid, and is 26" square at the top, 18" square at the bottom, and 
12" deep. Find how many tons of coal 20 such buckets will carry, if 
the coal weighs cwt. per cu. ft. (Hint. Reduce all dimensions to 
feet.) 

7. Calculate the capacity in cubic feet of a coal bunker such as 
is shown in Fig. 106, with the following dimensions. Upper rectangular 
portion 12 ft. square, 10 ft. deep. Lower tapered portion i'-2* square 
at the bottom opening and 9 ft. deep. If i cu. ft. of loose coal weighs 
56 Ibs., what weight in tons does the bunker hold ? 

8. At A, Fig. 102, is shown a group of grain silos or bins filled 
with wheat. Find the total capacity in cubic feet and also the weight 
of the wheat if i cu. ft. weighs 48 Ibs. 




Fig. no. Exercises on Volumes of Cones and Frusta. 

9. B, Fig. no, gives dimensions of a gas-holder tank. Calculate 
the quantity of water it contains in gallons. 

10. A milk churn is 12* dia. at the top and 18" dia. at the bottom, 
and s'-6" high. How many gallons of milk will it hold ? (The churn 
is a frustum of a cone.) 

11. The approximate dimensions of a steel- works ladle are given at 
C, Fig. no. Calculate the weight of steel (in tons) it holds at 490 Ibs. 
per cu. ft. 

12. A surge tank at a hydro-electric station is shown at D, Fig. no. 
Calculate the gallons of water it contains when the water reaches the 
high- water level. 

The Calculation of Weights. In the drawing office the 
calculation of the weights of proposed machine parts is of common 
occurrence. In all cases the volume of the object must first be 
found, and then be multiplied by the " density " of the material 
used. This density is the weight of a unit volume of material. 
The usual forms are " Ibs. per cubic inch " and " Ibs. per cubic 
foot." Thus the density of cast iron is -26 Ib. per cu. in. 



336 



ARITHMETIC FOR ENGINEERS 



Since 1728 cu. ins. equal I cu. ft., then 

Ibs. per cubic foot = Ibs. per cubic inch x 1728 
The densities of the many materials employed vary considerably 
with the composition, method of manufacture, etc., and will be 
found in engineering pocket-books. The following densities of the 
more common materials should be memorised : 



Material. 


Weight in Ibs. 
per cu. -in. 


Weight in Ibs. 
per cu. ft. 


Cast iron 


26 


A CQ 


Wrought iron .... 
Mild steel 


-2 7 3 
284 


480 
4QO 


Brass and gun-metal . . 
Lead 


307 
*4I 


530 
70S 


Water 


O^6l 


62*4 









The majority of engineering structures or parts are combina- 
tions of the simple solids already taken, so that the total volume 
can usually be obtained by dividing the object up into its component 
parts, and applying the formulae already given. Certain shapes 
appear, however, which do not lend themselves, either easily or at 
all, to the application of a formula. Such figures, by a " give-and- 
take " process, can usually be reduced to a form permitting of easy 




Fig. in . Calculation of Weights. 



calculation. With practice the process of " give-and-take " can 
become very reasonably accurate. When calculating weights, an 
exact answer is seldom required, as the weight of the actual article 
is affected by so many conditions not considered in our calculations, 
such as the blow-holes in casting, lack of uniformity in material, 
and the like. 

Space does not permit of a number of examples being taken 
here, but Fig. in will illustrate the principle. At X is shown a 



MENSURATION 337 

forging for an engine crank, which, for calculation, divides naturally 
into the two bosses and the crank arm. The bosses are easily dealt 
with, but when finding the volume of the crank arm the area of 
the figure ABCFED is required, which is not easily found by formula. 
A straight line is then drawn across each curved end as at LM and 
PN, so that at each end the areas b, cut off, balance the areas a, 
put on. The balance is obtained by eye only and its accuracy is 
dependent upon judgment and experience. The area of the irregular 
figure is thus replaced by that of the trapezoid LMNP, which can 
be dealt with by the formula given on p. 282. 

Similarly with the bracket casting at Y, Fig. in, the complex 
figure ABC, forming the back b, can be replaced by a rectangle 
AEDC if the line ED be drawn so that the area m is judged to 
balance the area n. 

Fillets, if small in comparison with the rest of the object, can 
usually be neglected, but when large, as in the rib a, at Y, should 
be treated by the formula on p. 299. 

Surface Areas. 

In many cases we are concerned with the area of the surface 
of some of the solids previously mentioned. In the case of the 
cuboid and square prism the calculation of the area of the surface 
is simple, given the necessary dimensions. An example will there- 
fore be sufficient ; frequently only part of the area is required. 

Example 343. The inside of a lift cage is to be lined with sheet 
zinc, on top and bottom, sides and back. The dimensions are given 
in Fig. 109. Find how many square feet of zinc are required. 

Back 5'-o* X 2'-o"; area = 2X5 = 10 sq. ft. 

Two sides each 3'-6" x 2 / -o /r ; area = 2x3-5X2 =14 sq.ft. 
Top and bottom each $'-Q" X 3 '-6"; area = 2 X 5 X 3*5 = 35 sq. ft. 

/. Total area required = 59 sq. ft. 

Cylinder. In all prisms the surface can be divided into two 
parts : the lateral surface, which is the surface of the sides ; and the 
two ends. Suppose that a cylinder is covered with paper. Then 
if it be slit along a straight line lengthwise and opened out, as in 
Fig. 112, the area of the rectangular sheet is evidently the area of 
the curved or lateral surface of the cylinder. Now the length of 
the sheet is the length of the cylinder, and the width of the sheet 
is the circumference. 
z 



333 



ARITHMETIC FOR ENGINEERS 



Then if d = diameter of cylinder and / = length 

Area of curved surface j = j h circumference 
or lateral surface / 

=== i X TTw 

= irdl or 2-n-r/ 

All measurements must, of course, be in the same units. 
The two ends each have in area ?rf 2 

.'. Area of ends 2 vr 2 
/. Total surface area = lateral -h ends 

= 2/r/(7 -f r) taking out the common factor 
In many cases, however, only the lateral surface is required. 



TTd 





\\ E.ad 

V<v/V 



Fig. 112. Illustrating Surface of Cylinders. 



Example 344. The smoke tubes in a locomotive boiler are shown 
in Fig. 112. Find the heating surface of i tube (i. e., the lateral surface). 
If the boiler has 220 such tubes in all, calculate the total heating surface 
of the tubes. 

Lateral surface is that of a cylinder i J"dia. and n ft. (=132") long. 
.*. Heating surface of i tube = irdl 

= 3-14 X 1-75 X 132 sq. ins. 

= 3'!4_ * --*2>-X - 13 - 2 = 5-036 sq. ft. 
/. Total heating surfaces of) 



tubes 



j 



= 5-036 X 220 

= 1108 sq. ft. 



Example 345. A surface condenser is to have 12000 sq. ft. of 
cooling surface (lateral). How many tubes J" dia. and 14 ft. long will 
be required ? 

Lateral surface of i tube = irdl 

= 3-14 X -75 X 14 X 



3 sq. ft. 



12 
= 2-75 sq. ft. 

-a- , . . , , Total surface 

.'. Number of tubes required = -^^ 7 - T- 

Surface of i 

12000 



2-75 



= 4368 



MENSURATION 



339 



Example 346. The field coil of an electric motor is 4* dia. and 
4J" long. Determine the watts radiated per sq. in. of lateral surface 
when the voltage is 100 and the current is -48 ampere. 

(Note. Watts = volts x amperes.) 

Surface of coil = irdl 

= 3*i4 X 4 X 4-5 sq. ins. 
= 56-5 sq. ins. 
Watts radiated = volts x amperes 

= 100 x -48 = 48 watts 

/. Watts radiated per sq. in. = - = -85 

Example 347. A dynamo core is 24" long, and it is to have at least 
2000 sq. ins. of curved surface for radiation. Determine a suitable 
diameter. 

Surface = itdl 
/. 2000 = 3-14 X d X 24 

, 2OOO .. - - , ,. 

whence a = = 26*54 , say, 26^ dia. 

Surface of Sphere. This cannot be examined in any simple 
manner, and the following formula must, therefore, be accepted 
until a later stage. 

Surface of sphere = 47rR 2 where R = radius of sphere. 

Example 3470. What is the surface of a sphere $\" dia. ? 

Radius = ij" = i'f5* 
Surface = 4irR a 

= 4* X (i75) a = 38*5 s q- ins - 

Curved surface of segment of sphere = 2?rR/i 

where R = radius of sphere ; h = height of segment 

In cases where h is not given the following formula can be 

ised where r = rad. of base of segment. 

Curved surface = 2;rR(R - 1/R 2 - r*) 

Example 348. Find the area of the dished end of a 
vater-tube boiler drum, as shown in Fig. 113. 

Surface = 2irRA 

= 2ir x 3 X ~ s q- f t. 

= 7-85 sq. ft. 

Fig. 113, 

Example 349. The dished ends of boiler drums are flanged from a 
ircular plate as indicated in Fig. 114. Assuming that the thickness 




340 



ARITHMETIC FOR ENGINEERS 



of the plate and the volume of the metal are not altered, find what 
diameter of plate is required to produce the end shown in Fig. 114. 

Thickness of plate is very small in comparison with other dimen- 
sions, and so we may calculate the volume of metal as " surface area x 
thickness. " If volume and thickness are to be the same in each case, 
surface areas must also be equal. 

.'. Surface of circular plate = surface of dished end 

(i face only) = surface of segment of sphere -J- 

surface of cylindrical flange. 

Surface of segment = 2ir"Rh =2 X 3-14 X 30 X 6= 1130 sq. ins. 
Surface of cylinder = irdl = 3*14 X 36 X 5 = 565 ,, 

.'. Total surface of dished end = 1695 > 



r 




Surface of circular plate = 

)2 = J695 

4 / J 695 (- 
V -785 = 4 6 '5 

i. e. a plate 3 '-TO!* dia. is required. 



D 



Fig. 114. 



In cases of " cold drawing " of tubes, etc., 
where the thickness alters, the volumes may be 
equated instead of surface areas. 



Surface of Cone. Make a mark on the base of a cone, and 
lay the cone on its side with this mark to the ground, as at a, Fig. 
115. Roll the cone quite freely until the mark goes just once 

round, and arrives at the 
ground again as at c. It will 
be found that a sector OAB 
has been marked out, which is 
the curved surface of the cone. 
The radius of the sector is 
evidently the " slant height " 

Fig. ns-Surfacc of Cone. J * he C , ne > L <' the **&* l 

of the sloping side. Also the 

length of the arc of the sector is evidently the circumference of 
the base of the cone, i. e., -nd if d = diameter of cone base. 

Then, Curved surface of cone = Area of Sector 

= J arc x radius (see p. 313) 
= l*d X I 




I may be obtained by drawing, or by calculation of a right-angled 



MENSURATION 



341 



triangle, as indicated in Ex. 350. The total surface is evidently 
the curved surface plus the base. 

i. e. } vrl -f 7rr z = */(/ -f- r) 

Example 350. Find the curved surface of a cone 8* dia. and 10" high. 

The slant height is required and will be calculated here. It is the 
hypotenuse of a right-angled triangle whose perpendicular sides are 
12" and 4". 

/. Slant height / = Vio 2 +"4* 

= Vii6 = 10-77* 
/. Curved surface = -nvl 

= if X 4 X 10-77 = 135-3 sq. ins. 

Surface of Square Pyramid. The lateral surface consists of 
four triangles. Let s = side of base and I = slant 
height, i. e. t the distance from point to centre of 
side of base, and not the slant edge; the distinction 
is shown at Fig. 116. 

The area of one triangle = %sl 
.'. Lateral surface of pyramid = 4 x %sl 

-2s/ 

The base area = s 2 
.'. Total surface = 2sl + s 2 
= s(2/ -{- s) 

Example 351. Find the area of the lateral surface of a square 
pyramid, if perpendicular height is 24" and side of base 9", 

Slant height ^242 + 9 2 ~ 
= A/576 -f- 81 
= Vds 7 = 25-64" 
/. Area of lateral surface = zsl 

= 2 X 9 X 25'64 
461 sq. ins. 

Surface of Frustum of Square Pyramid. The lateral sur- 
face consists of four faces each of which is a trapezoid. Let S = side 
of base, s = side of top, and / = slant height as shown 
at Fig. 117. 
Then- 
Area of one trapezoid = - -- x / (as on p. 282) 




Lateral surface 



2/(S 




The total surface easily follows, being : lateral surface + S* + s 8 . 



342 



ARITHMETIC FOR ENGINEERS 



Where /, the slant height, is not given it may be obtained by 
drawing, or by calculation, using the right-angled triangle property. 

Example 352. Find how many square feet of plating are required 
to make the tapering portion, shown in 
Fig. 1 1 8, of a coal bunker. 

Slant height = VJ*"+ 4-25*" 




TV-ue -shape* o^ a Pace. 
Fig. Ii8. 



= V43-o6 = 6-56' 
/. Lateral surface = 2/(S + s) 

= 2 X 6-56(10-5 + 2) 
= 2 X 6-56 X 12-5 
= 164 sq. ft. 



Surface of Frustum of Cone. This is somewhat similar to 
the square pyramid. 

Lateral surface 



where 



2 

= TT/(R + r) 
/ = slant height as before 



Example 353. Fig. 119 gives dimensions of a " Galloway " tube, 
placed in the flue of a boiler to provide extra heating surface. Determine 
the heating surface (lateral) in the case shown. 

The heating surface is the lateral or curved 
surface of the frustum. 

Slant height = 




+ (5-375 3 r i25) a 
~ 



Curved surface^ 
of frustum j " 



= ^789-1 = 28- 



-09 



X 28-1 



IT X 17 X 28-1 

_ 



= 750 sq. ins. 
or, 5-2 sq. ft. 



Exercises 83. On Surface Areas. 

(Letters refer to Fig. 120.) 

1. Calculate the total heating surface (= tubes + firebox) of a 
large American locomotive having 446 tubes, each 2j* dia. and 24 ft. 
long, and a firebox of 353 sq. ft. heating surface. 

2. An English locomotive boiler has 157 tubes 2J-* dia. and 24 
tubes 5J" dia., all being n / -6" long. Calculate the total heating 
surface in sq. ft. (i.e., lateral surface). 



MENSURATION 



343 



3. A surface condenser has 2680 tubes f* outside dia., and i4'-2 y 
long. Calculate the cooling surface in sq. ft. If the condenser handles 
52,000 Ibs. of steam per hour, calculate the weight of steam condensed 
per sq. ft. of surface. 

4. In tests on the adhesion of round steel bars embedded in con- 
crete, a bar ij" dia. and 3 ft. long was pulled length-wise from the 
concrete with a force of 18 tons. Calculate the adhesive stress (i.e., 
the force in Ibs. per sq. in. of lateral surface). 

5. In a test to determine the loss of heat from a bare steam pipe 
the pipe was 6" outside diameter and 3 '-6* long. In i hour, 2210 units 
of heat were lost. Calculate the loss per sq. ft. 

6. A Leyden jar is 4" mean dia. and is coated with tinfoil for a 
depth of 5" up the sides and on the base as shown at A. Calculate the 
area of the coating. (One side only.) 




Fig. 120. Exercises on Surface Areas. 



7. If the bevel on a mushroom valve be neglected, and the valve 
considered as merely a circular plate closing the pipe, as shown at B, 
the* area through which steam passes is the surface of the cylinder as 
shown. When the valve is full open this area should equal the area of 
the pipe. Find what must be the " lift " of a 6" dia. valve for full 
opening. 

8. Refer to Ex. 7 above : if d = dia. of valve, and / = lift, find 
what the value of / must be for full opening. (Hint. Equate formula 
for area of pipe to formula for surface of cylinder and solve for /.) 

9. A tank for use in boiler tests is to be 2'-6* dia. and 3 '-6* high 
with an open top. Calculate its weight if i sq. ft. of the sheet metal 
of which it is made weighs 2-86 Ibs. 

10. A float for a carburettor is shown at C. Calculate its weight 
if of sheet brass weighing -87 Ib. per sq. ft. (Hint. There are four 
surfaces to consider; the outside and inside lateral surfaces, and the 
two annular ends.) 

11. A dynamo armature is io|* dia. and 23" long, and the power 
lost by radiation is 680 watts. Calculate the watts lost per square inch. 



344 ARITHMETIC FOR ENGINEERS 

12. The pole piece in a two-pole dynamo embraces an angle of 120 
at the centre as shown at F. The armature is 15* dia. and 28* long. 
Calculate the area of the lateral surface of the armature opposite the 
pole piece. (This area is wanted when calculating the field winding, 
etc.) 

13. The area beneath the edge of the inlet tube in the oil separator 
(see E, Fig. 80) is to equal i times the area of the steam pipe. Further 
dimensions are given at D (Fig. 120). Calculate the height h if the 
steam-pipe is 6" dia. 

14. In a fat melting tank as shown in Fig. 98, the space between 
the inner and outer vessels is a steam jacket. If the metal is y thick 
calculate the heating surface in square feet. (Hint. The heating 
surface is the outside surface of the inner vessel.) 

15. A pressed sheet metal dome is shown at E. Calculate its 
weight if the metal weighs 2-6 Ibs. per sq. ft. 

16. A dished boiler end plate, as shown in Fig. 114, has the following 
dimensions : dia. 3 '-6", mean rad. of spherical part 3 7 -o", height of 
segment 6*, length of flange 5*. Calculate the diameter of the flat 
plate required to make it, assuming that the thickness does not alter 
during flanging. 

17. At G is shown the combustion chamber in a vertical boiler. 
Calculate the heating surface (i. e., lateral surface of cylinder plus area 
of spherical end; neglect the entering flue and the tube plate). 

18. A taper flue in a Lancashire boiler is 2'-8" dia. at one end and 
3'-3" dia. at the other, and is 3'-3* long. Calculate its heating surface 
(lateral surface). 

19. The furnace of a vertical boiler is shown at H. Calculate the 
heating surface (i. e., lateral surface of conical frustum plus curved 
surface of segment of sphere. Neglect the two openings). 

20. Calculate the weight of the plating necessary for the coal bunker 
in Fig. 1 06, if the plate of which it is made weighs 20 Ibs. per sq. ft. 

21. At J is shown an iron ring of circular section (called an " Anchor 
Ring " see end of Volume Table, p. 346) wound with a coil of wire. 
Calculate the watts radiated per square inch of surface if the watts 
wasted in a test were 10-5. Also find the volume of the iron ring within 
the coil. 

22. In the " Brinell " hardness test for metals, a hard steel ball is 
pressed into the test piece under a certain load, and the diameter* of 
the spherical depression produced is measured. Then the " hardness 
number " is " Load 4- Curved Surface of Depression." In a certain 
test the load was 3000 kilogrammes and the diameter of the depression 
was 3-8 millimetres, the steel ball being 10 millimetres dia. Calculate 
(a) the curved area of the depression ; (6) the hardness number. 

23. In a certain aeroplane instrument a pivoted tube is kept head 
to wind by a trailing conical tube of sheet metal. This conical tube 
is the frustum of a cone ; diameter at large end 4*, at small end f *, 
length on centre line yj". It is necessary to know its weight for the 
purpose of balancing the instrument. Calculate the weight if the 
metal weighs -53 Ib. per sq. ft. 



MENSURATION 344 

Exercises 83a. Miscellaneous Examples on 
Mensuration. 

(Letters refer to Fig. I2oa.) 

1. The plates of a variable electrical condenser are semicircular, 
2j* diameter. If the portion round the axis for a radius of J* be 
ineffective, find the effective area of the two sides of a single plate. 

2. A flywheel casting has a rim of rectangular section, 6 '-2" outside 
and 5 '-4" inside diameter, 8" broad. The hub is 7" dia. by 9* long with 
a 3" dia. cored hole. There are 6 arms connecting rim to hub, of 
elliptical cross section <\\" x 2%". Calculate the weight in cast iron. 

3. A frame aerial consists of a square pancake coil as indicated at 
A. If there are 14 complete turns, spaced \" apart, the side of the 
outside square being 4 ft., estimate the total length of wire required. 
(Hint. 14 turns include 13 spaces of J" wide. Calculate the average 
length of a turn.) 

4. At B are given dimensions of a cast iron segment used for lining 
an underground railway tunnel. Calculate its weight. (Hint. The 
cylindrical part is a sector of a tube; use average diameter method. 
The side flanges can be treated similarly and the end flanges are rect- 
angular. The material at corners must not be included twice.) 

5. A conical spring for the overhead valves of an aero engine is 
shown at C. Estimate the total length of wire required per spring, 
allowing ij coils extra to the working coils to allow for the flattened 
ends. (Hint. Find the length of a turn of average diameter.) 

6. A hardness test consists in pressing a hard steel cone into the 
material under a certain load and measuring the diameter of the conical 
depression so produced. The apex angle of the cone is 90 so that the 
depth of the conical depression, measured along the axis of the cone, is 
one-half of its diameter. The " hardness number " is then " Load 
Curved Surface of Depression." Calculate (i) the curved surface of 
the depression, (2) the hardness number, when a load of 1000 kilo- 
grammes produces a depression 3 millimetres diameter. 

7. A special milling cutter is required having a diameter over the 
teeth of 4!". The teeth are to be cut with a 45 angular cutter as 
indicated at D. The width of the gap is required to be \" and the 
width of the top of the tooth is to be about O5 // . Determine how many 
teeth must be cut and the actual width of the top of the tooth. (Note. 
An exact number of teeth must be cut.) 

8. Big end bolts for marine engine connecting rods are frequently 
reduced in area in the shank as a provision against shock. One way 
of doing this is as shown at E, where three semicircular flutes are 
milled in the shank. If the diameter of such a bolt is 3" and the diameter 
at the bottom of the threads is 2- 634", find what diameter of flute is 
needed to make the area at the section AB equal that at the section 
CD. (Hint. The area at AB is that of a circle less the area of three 
flutes, and each of the latter may be taken as a complete semicircle.) 

9. Earthenware draw-in conduits for carrying electric cables under- 
ground are made of octagonal cross section with a circular hole. If the 
octagon be 4" across the 'flats and the hole 3" diameter, estimate the 
weight of a section 15" long, the material weighing 150 Ibs. per cubic foot. 

10. A motorcycle petrol tank is a cylinder 5" diameter and 8" long, 
with conical ends, the length of cone, measured on the axis, being 3 // . 
Estimate the capacity of the tank in gallons, when quite full. 



344B 



ARITHMETIC FOR ENGINEERS 



11. A cast iron balance weight is spherical, with a hole cored 
through the centre as shown at F. If the ball is 10", and the hole 
5", diameter, determine the weight of the balance weight. (Hint. 
From the volume of the sphere subtract the volume of the cylinder 
forming the hole and also the volume of the two segments of the 
sphere cut off beyond the ends of the hole. The length of the hole 
must first be calculated from the right-angled triangle OAB.) 




BiiiicfefRl 

-< r i * * *- / 

E) O B L ~ K 




Fig. I2oa. Diagram for Exercises 83a on Mensuration. 



12. A number of weights for testing purposes are required of the 
form shown at G, to be of brass and to weigh i Ib. each. If the out- 
side diameter be 3f // , the central hole be \" diameter, the thickness 
be fy" t determine whether a slot about $" wide will give enough 
volume to permit of adjustment to the exact weight, i.e., the weight 
when calculated from the tentative dimensions should be a little more 
than i Ib. (Note. For practical purposes, the area of the top Of the 
weight may be obtained by subtracting the area of the rectangle shown 
crossed from that of the hollow circle.) 

13. An aeroplane petrol tank is shown at H, with dimensions. The 
shape of the end may be considered as consisting of two semi-ellipses. 
Calculate the capacity in gallons when quite full. 



MENSURATION 345 

TABLE OF VOLUMES AND SURFACE AREAS OF SOLIDS. 



Title. 



Figure. 



Any prism . 



Rectangular 
pi ism or 
cuboid . . 



Cube . . . 



Square prism 



Hexagonal 
prism . 



Octagonal 
prism . 



Cylinder . . 



Hollow cylin- 
der ... 



Elliptical 
prism . 



Sphere . . 



Hollow 
sphere . . 










Volume. 



Area of base 



X height 
Ibh 



S 2 / 



2-6S 2 / 
or -866/ 2 / 



4-83S 2 / 
or 'S29/ 2 / 



or 7854^7* 



,r(R2 - 



irabh 



-?rR 3 or - 
or -52360* 



Surface Area. 



Circumference of base 

x height 



Whole area = 6S 



Lateral surface = 4S/ 
Ends = 2S 2 



Lateral = 6S/ or 3-46^ 
(For ends see Table in 
Chap. VII.) 



Lateral = 8S/ or 



Lateral = 2irrh 
Two ends = 2irr* 
Whole area = 2-rry(h -f- r) 

Outer lateral^ _ 
surface J~" 2?rK/l 

Inner lateral) , 

surface j=2i 



Lateral 

= irA{i-5(fl-f 6)-V 
or it(a+b)k 

(less accurate) 



346 ARITHMETIC FOR ENGINEERS 

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 of 
any pyra- 
mid . . . 



Frustum of 
square 
pyramid . 



Frustum of 
cone. . . 




^o 

or -5236/1(3)-'+^) 




area of base 

X height 



JS'A 



/f=height of frus- 
tum 

A=area of large end 
B= area of small end 



B+VAB) 



Anchor ring . 




Curved surface = 2irRA 

or 2irR(R-V r JR>~^ 2 )" 
where R=rad. of sphere 



Lateral = J circum. of 
base X slant height 



Lateral = 2$/ 



Lateral = 



Lateral=J mean circum. 
X slant height 



Lateral = 2/(S + 5) 
(/ == slant height) 



Lateral = r/(R + ') 
(/ = slant height) 



Round section 



Square section 



DS 2 



CHAPTER IX 



CURVES OR GRAPHS 

[All the worked examples should be drawn out by the student.] 

Curves and their Uses. After a machine has been built, 
especially if it is one of a new type, it is usual to test its perform- 
ance in order to see if it satisfies the conditions for which it was 
designed, to determine any defects, or to obtain information for 
later designs. Thus, if the machine is an electric motor, it is neces- 
sary to find its " efficiency 11 (i.e., what fraction of the electric 
energy supplied, is given out usefully) at various loads. The motor 
is worked at a number of different loads (" brake horse powers ") 
varying from nothing to rather more than the rated horse power, 
and certain measurements are made at each load from which the 
horse power and the efficiency are calculated. Thus two sets of 
figures are obtained, and for each horse power there is one particular 
efficiency. Such a set of figures is given below, calculated from 
the results of tests on a direct current motor rated at 90 H.P., 
and each efficiency is written beneath its particular horse power. 



BH.P. . . . 


10 


20 


42 


60 


74 


O2 


IO2 


III 








T* 




/"T 








Efficiency . . 


60 


"74 


83 


86 


89 


QO 


905 


905 



Looking at these figures alone does not supply much useful 
information. Thus we cannot say what is the efficiency at horse 
powers other than those in the table ; again, although it can be seen 
that the efficiency increases as the B.H.P. increases, it is not an 
easy matter to see how it increases, i. e>, whether faster or slower 
than the H.P. 

For these reasons the results of the test are shown graphically 
by drawing a "graph" or " curve " as shown in Fig. 123 et seq. 
The two quantities are then said to be correlated by the graph, and 
from it the efficiency at any H.P. between the lowest and the highest 

347 



348 ARITHMETIC FOR ENGINEERS 

stated can be found, and from the shape of the curve we can say 
how the efficiency changes. 

Besides being useful for testing purposes, curves are of great 
value in science and calculations. Many equations may be easily 
solved by their aid, and in some cases a graph is the only means 
of solving an equation. Again, when designing machines it is 
frequently necessary to calculate from a certain formula, and when 
the class of work demands its frequent use it is a great advantage 
to draw a curve for the formula ; then when any value is required 
it is merely read from the curve, no calculation being necessary. 
In this case a graph is often called a chart. 

Since a graph gives to the eye a much better understanding of 
the changes in a varying quantity than a table of figures, it is 
customary with quantities constantly changing to have graphs 
drawn automatically. Thus " recording ammeters " and " record- 
ing voltmeters " in electric stations show for any period of the 
day how much energy has been given out, and how the generators 
have been regulated. Waterworks are also provided with " re- 
corders " which register the quantity of water being delivered at 
any time. The height of the barometer and the temperature, 
together with many other physical measurements, are automatically 
recorded in observatories. A further advantage in these cases is 
that, being automatic, practically no attention is required, whereas 
if figures were read at certain times there is a danger of certain 
readings being omitted or wrongly read. The common indicator 
diagram is an example of an automatically drawn graph giving the 
pressure in an engine cylinder at any point of the stroke. 

Rectangular Co-ordinates. The principle of the graph is 
as follows. Each pair of figures is represented by the position of a 
point upon a sheet of paper, the distance of the point being measured 
from two fixed lines at right angles to each other, such as the left- 
hand edge and the bottom of the sheet. These two lines are " lines 
of reference " and are known as the axes ; the point at which they 
cross is called the origin. For each set of figures a " scale " has to 
be arranged in order that a quantity like horse power can be shown 
by a distance. Taking our case of the electric motor we might say 
that i" should represent 20 H.P., and that i" should represent 
0'2 efficiency. The method of choosing these scales will be dealt 
with later. 

Let us now take one particular pair of figures as an example, 
say the pair H.P. = 60, efficiency = -86. Then if every 20 H.P. 
is to be shown by i", 60 H.P. will be shown by 3*. Also if every 



CURVES OR GRAPHS 



349 



2 efficiency is to be shown by I*, -86 efficiency will be shown bv 

OS- J 

4-3", since = 4-3. Thus this pair of figures will be shown by a 



These two dimen- 



= GO YP 



point 3" from one axis and 4-3" from the other, 
sions are known as the rectangular co- 
ordinates of the point, and the position 
of the point on the paper is shown in 
Fig. 121. Continuing this process for 
each point, they can be " plotted " in 
the same manner as the one taken, and 
are shown as dots in Fig. 121. Evidently 
they are points on a regular curve. 

Either quantity may be measured 
horizontally or vertically, but certain 
arrangements are kept to in various 
kinds of work in order to obtain uniform 
plottings. In the particular case just 
taken the H.P. would be measured 
horizontally and the efficiency verti- 
cally, as shown. It is then said that 
to a base of H.P.," the 
horizontally. 

Squared Paper. It would obviously be very inconvenient if 
the rectangular co-ordinates of each point had to be calculated and 



/ORIGIN 

HORIZONTAL 



AXI 



Fig. 121. Rectangular 
Co-ordinates. 

the efficiency is plotted 
base " quantity being the one plotted 



A i SQUARES B T-SQUARES C I mm/ SQUARES. 

Every 5 TH line heavy Every IO TH line heavy Every I O line heavy 
Fig. 122. Types of Square Paper. 

measured, as shown above. This is entirely avoided by employing 
squared paper. This paper is ruled all over both horizontally and 
vertically, with equi- distant parallel lines, thus dividing the surface 
up into a number of small squares. Several varieties may be 
obtained, but the two most useful ones are those where the lines 



350 ARITHMETIC FOR ENGINEERS 

are -j 1 ^" apart, and i millimetre apart. The lines are printed in 
blue or gray, these colours being easily seen without tiring the eye, 
but at the same time being fainter than black, and therefore not 
so prominent as to interfere with the appearance of the " curve " 
when drawn in black. Usually every fifth or tenth line is printed 
heavier than the others so that numbers of 10, 15, 20, etc., can be 
quickly and easily counted. Fig. 122 shows three different types 
of paper : 

A. lines T V apart, every 5th line heavy 

B. ,, loth 

C. i millimetre apart, every loth line heavy 

Type A will generally be found most useful, although type C is in 
considerable use. Squared paper can be obtained in various sized 
sheets or in exercise books, at slightly more than the cost of ordinary 
paper. Its use in exercise books is not recommended, as the fold 
in the book renders the drawing of the curves rather difficult ; it 
is easier to draw on a flat sheet of paper and to stick this in the book 
finally. 

The most common size of sheet is " foolscap" (13" x 8"), and the 
examples taken here will be arranged, so far as possible, to suit 
such a sheet. Allowing a small margin all round, the largest work- 
ing size is about 12" x 7*. The curves shown will not appear 
so distinctly as on the actual paper, as the division lines are of 
the same colour as the curves. Also, since the plottings cannot 
be shown full size, all the -^G* lines cannot be drawn. The first 
example (Fig. 123) is shown half size in order that the style of work 
may be clearly seen. The heavy lines i* apart are shown, together 
with every other T V line, while the heavy " lines are indicated by 
the short lines standing out from the axes. The remaining exam- 
ples are shown about one-fourth full size ; only the heavy lines 
\* apart being drawn. 

Method of Plotting. The method of plotting a curve from a 
table of given values, and the various points to be observed, will 
now be dealt with in the order in which the work should be done. 
The example will be the curve of efficiency and B.H.P., of which 
the figures are given on p. 347. The B.H.P. will be plotted horizon- 
tally and the efficiency vertically. The finished curve is shown half 
size in Fig. 123, and should be referred to continually in reading 
the following description. In all cases a student should endeavour 
to produce a neat and complete job ; a graph always looking more 



CURVES OR GRAPHS 351 

important and workmanlike when finished with neatly printed titles, 
figures, etc. 

1. Choice of Scales. This is decided by the condition that 
where possible the curve should fill the sheet ; but it is very im- 
portant that the scales should be such that the decimal system can 
be used with ease. Thus i" (i. e., 10 squares) should equal I, or 
10, or 20, or 50 units, etc., and should not be used to represent 
3 or 6 or 8 units. This condition should be obeyed even if it requires 
drawing a smaller curve or using a larger sheet of paper. 

Taking our particular case, the maximum value of the H.P. 
to be plotted is in, let us say no. Taking i* to represent 10 H.P. 
we require IT/, which is just within the length of our working page 
and therefore quite satisfactory. Then, scale of H.P. is 1" = 10 H.P. 
For the efficiency, the maximum value is -905, say I. Space will 
not permit of using 10" to represent I, so that we must employ 
5" = i, i. e., i" = -2. There is nothing between which will give a 
decimal system. Then, scale of efficiency is 1" = *2. 

2. Setting-out Scales. The axes should now be drawn in ink, 
each upon one of the heavy lines, and not less than \" from the edge 
of the paper. This margin is necessary for titles, etc. If the curve 
is not likely to fill the sheet with a convenient scale, the margin 
may be increased to say ij". The lines may be fairly thick, and 
may, with advantage, be put in with the ruling pen and Indian ink. 
Both horizontal and vertical scales should be marked as follows : 
Taking the horizontal scale, first mark the origin o, and then mark 
every second heavy line (i. e., every i") as 10, 20, 30 and so on to 
no. The figures should be printed, and placed centrally under 
their particular heavy line, a little beneath the axis. Parallel and 
near to the axis the title of the scale should be neatly printed, 
describing the kind of measurement, and the unit in which it is 
measured (see Fig. 123, et seq.). Taking the vertical scale, the origin 
is numbered o, and then each inch is marked -I, -2, -3, and so on 
to l-o, as shown. The title in this case is merely " efficiency." 

3. Plotting the Points. The points should first be plotted in 
pencil as follows. From the table of figures on p. 347, the first pair 
of figures is H.P. 10, efficiency -6. Taking the base figure first, 
i. e. t the 10 H.P., look along the H.P. scale for the line marked 10. 
Follow this line upwards with the pencil until, glancing at the 
efficiency scale on the left-hand side, the heavy line marked -6 is 
reached. Make a small dot at the crossing point of the two heavy 
lines, and draw a small ring round it in order that it may easily 
be found again. 

A A 



352 



ARITHMETIC FOR ENGINEERS 




tf. 

UJ 



o 
CL 



fn 

03 



u 
(f) 

tt 
O 
X 



P< 

a 

ctf 
X 

w 



&o 

E 



UJ 



CD 



CURVES OR GRAPHS 



353 



The next pair of figures is H.P. = 20, efficiency = 74. Find 
the heavy line at 20 on the base scale and follow upwards with 
the pencil until the heavy line between *6 and -8 on the efficiency 
scale is reached, i. e., 7. Then since the distance between -7 and 
8 is divided into five parts, each of the small squares represents 
2, and standing from the 7 the light lines read 72, 74, 76, and 78 
(this is shown full size at A, Fig. 124). The required point is then 
at the intersection of the second of these light lines (i. e., 74) and 
the vertical heavy line at 20. The third pair of figures is H.P. 42, 
efficiency -83. Along the base, 42 is at the second fine line past 
the 40 (since i" = 10 H.P., then from 40 to 50 the lines read 41, 
42, 43, and so on to 50). Following this upwards with the pencil, 
the figure -83 is required on the efficiency scale. Now the first fine 



8- 



6 







Fig. 124. Estimation of Intermediate Readings. 



line after the -8 is -82, and the next -84. Hence -83 is halfway be- 
tween these, and a dot is made here on the 42 vertical line, as shown 
full size at B, Fig. 124. The remaining points are similar until the 
one H.P. 102, efficiency -905, is reached. The position of -905 is 
found as indicated at C, Fig. 124. From -9 to -92 is blank on the 
squared paper, and halfway between these points is -91. Now '905 
is halfway between -9 and -91 ; hence -905 lies one-quarter of a small 
square above -9 as shown. When marking the actual dot the middle 
point between the -9 and the -92 lines should be estimated and 
marked ; then the middle point between this and the *g line gives 
the point required, as shown twice full size at D, Fig. 124. Although 
this process is lengthy to describe it is quite easy to perform. A 
little practice is required in estimating the intermediate points, but 
if care be taken it is quite possible to estimate one-fourth or one-fifth 
of a ^j" division. 

Estimation of intermediate figures such as the above is only 
possible with a decimal scale. Thus if I* (*'. e. t 10 squares) *2, 



354 ARITHMETIC FOR ENGINEERS 

then one small square (i. e.> ^*) represents *O2 and a half of it *oi. 
Thus in this case any second decimal place can be accurately and 
quickly found. (With an odd scale such as ij", i. e. t 15 squares = '2, 
one small square would be '0133, and it would be practically 
impossible to find correctly any particular decimal figure that was 
not marked.) 

When all the points have been plotted in pencil, their general 
appearance should be noticed. If they appear to lie on a fairly 
smooth curve, then the points may be marked in ink, but if any 
oi them appear to be very much out of line with the others the 
plotting of these particular ones should be checked to see if a blunder 
has been made. When correct, the points should be marked in ink. 
This applies to all cases of tests and experiments. Only in a very 
few cases is it allowable to draw the graph without the points. 
The commonest method of marking the point finally is by a small 
fine dot and a ring round it, ^ dia. or less, which may easily be 
put in by hand. When two or more curves appear on one sheet, 
and the points are liable to become mixed, it is desirable to use 
different methods of marking the points (three of which are shown 
in Fig. 135). Different curves on the same sheet can also be shown 
by differently coloured inks. 

4. Drawing-in the Curve. The curve should now be sketched in 
freehand with pencil. Sometimes all the points will be found to lie 
exactly on the curve as in Fig. 129 (top curve), where the values are 
calculated from a formula. But in plotting the results of experimental 
work the points seldom lie quite regularly on any curve. Instrument 
errors, errors in instrument reading and other circumstances, all 
affect the values to be plotted. The points will follow some general 
direction, and when sketching in the curve, it is this general direction 
which must be followed. The points must not be connected by a 
wavy line, but the curve sketched so as to lie among the points. As 
a general guide it may be taken that for every point on one side of 
the curve there should be one on the other side, and about the same 
distance from it. Where three or four points appear to be nicely 
in the path of the curve then, of course, the curve should be drawn 
right through them. In cases where the points do not follow an 
exact curve there may be a difference of opinion as to the best line 
to put among the points. Where answers are required to such 
examples they cannot be taken as being absolutely correct. Differ- 
ences may be found in the third (and perhaps the second) significant 
figures. With a very few exceptions all plottings of natural or 
scientific observations will lie on a " smooth " curve, i. e., one 



CURVES OR GRAPHS 355 

without any sudden changes in direction. The most common 
exceptions are the temperature and the barometer readings. 

For the final drawing of the curve some mechanical aid is desir- 
able, such as an adjustable curve or, better still, a French curve. 
These are templates made usually in pear-wood and containing a 
large variety of curves; they can be purchased for a few pence. 
Several types may be seen in the instrument-maker's catalogue, 
but one which the author has found very useful is shown in Fig. 
125. For graphs it is desirable that the curve should have a long 
portion, such as AB, only very slightly curved near the end. A 
portion of the French curve agreeing well with, say, half of the 
sketched curve should now be found, and the French curve moved 
about until its edge exactly " picks up " some portion of the sketched 
curve. Then carefully ink in this piece of trie curve with a fairly 
fine line, using the ruling pen. 
Next move the French curve along 
and find another portion suitable 
for another piece of the curve 
and ink this in, taking care to 
make a smooth and continuous 
joint. Repeat until the curve is 
finished. Finally clean the pencil- Fig I25 ._ Frenc h Curve, 

ling off, and the finished curve 

will remain. With practice the French curve will be found quite 
easy to use. 

In all cases where any reading of values is required from the 
final curve, the line should be made fairly fine, but where only 
the general shape of the curve is required a fairly heavy line is 
advisable, as it stands out much better. Examples of this will be 
found in illustrations in technical books. For clearness the curves 
shown here are drawn thicker than is necessary. 

5. Further Information Required on the Sheet. This depends upon 
the style of work and the use to which the curve is to be put, 
but in all cases the title of the curve should be written on the sheet. 
In our case the title is " Efficiency Curve for 90 H.P. Electric 
Motor." For a drawing office job there would be some order 
number and a drawing number, with the name of the firm for whom 
the tests were made, the place of test and particulars of the type 
of motor, e. g. t " Shunt Wound, 500 volts at 400 revs, per min." 

When two or more curves appear on one sheet, a title must be 
placed on each curve, as shown in Fig. 128, to distinguish between 
them, and to show which scale applies to each curve, for frequently 




356 



ARITHMETIC FOR ENGINEERS 



two vertical scales will be needed when two curves are drawn. In 
cases like this, the second curve is plotted quite separately from the 
first one. Any such information which applies to a particular 
curve should be neatly printed parallel to the curve. 

Interpolation. This is the process of finding from the curve 
values between those given in the table. Thus supposing it is 
necessary to know the efficiency of the motor mentioned on p. 347, 
at 50 H.P. and at 85 H.P., the required values are " interpolated " 
in the following way. Taking the 50 H.P., glance along the H.P. 
scale (Fig. 123) until this value is reached. Follow the 50 H.P. line 
upwards until the point is reached where the curve crosses this 
vertical line. Now glance horizontally to the left and read off this 
position on the efficiency scale. If the intersection comes upon one 
of the ^ lines the reading is easy, but if it is between two -$* lines, 
then the exact reading must be estimated as when setting out the 
points on p. 353 (see Fig. 124). In the case we are considering, the 
50 H.P. line intersects the curve on the second T V line above -8, 
the reading, therefore, being '84. Thus we have interpolated the 
efficiency at 50 H.P. to be -84. 

In the second case, at 85 H.P., the intersection of the curve 
and the vertical line is about one-fourth of a small square below 
the -9 line, i. e., '005 below the '9. The reading is therefore -895. 

Example 354. The following table gives the safe working load 
in Ibs. allowable on a certain ball thrust bearing 2" dia., at various 
speeds, from 50 to 1000 revs, per min. Plot a curve of load upon a 
base of speed, and interpolate a suitable working load at speeds of 
130 revs, per min., and 750 revs, per min. 



Speed, r.p.m. . . . 


50 


IOO 


200 


3OO 


400 


6OO 


800 


IOOO 


Safe load, Ibs. . . . 


1 100 


780 


550 


440 


390 


3 IO 


275 


240 



Refer to Fig. 126. 

Scales. Base. Max. speed = 1000. Taking i" = 100 r.p.m., the 
base will require to be 10*, which is suitable. 

Vertical. Max. load = noo Ibs. Taking i" = 200 Ibs., this scale 
will require to be 5^ long, which is suitable. 

Plotting. This should present no difficulty. The points will all be 
found to lie exactly on the curve. 

Interpolation. At 130 r.p.m. the curve reads 680 Ibs. ; at 750 r.p.m. 
the curve reads 280 Ibs. - 



CURVES OR GRAPHS 



357 




O tOO ZOO 300 4-OO 5OO 6OO 7OO &OO 9OO IQOO 

SPEED IN R. P. M. 



O 5 10 15 eg 30 35 AO 




28o 



'2TO 







Vacuunr^ Curve . H 
















< 












Turb\ne C 




de 




nq 








^1 


^N 




















uPlanf 
















^5 


< 






































^ 


^> 







































* 


1 


k 




































^'v, 








































s ^ 


^ 




































cj 


^ 











































I2OOO 
10000 14000 



eeooo 

LOAD IN LBS OF STEAM PER 



Fig. 126. Curves to Examples 354, 355, and 356 



358 



ARITHMETIC FOR ENGINEERS 



Example 355. An experimental dynamo could be arranged " series " 
or " shunt " wound. Tests were made with each kind of winding to 
determine the " characteristic " curves, measurements of voltage and 
current being taken. The results are given in the tables below. Plot 
the curves (voltage on base of current) for the two cases, upon the same 
sheet, using the same scales. 

I. SERIES WOUND. 



Current, amps. . 


o 


9'5 


14-25 


17 
70 


20 


22 


25-^5 


26-5 


Voltage . . . 


8 


44 


62 


76 


79 


77'5 


72-5 



II. SHUNT WOUND. 



Current, amps. . 





8-75 
103 


17*5 

101 


29-25 


34*75 


4i 
94 


Voltage 


I0 5 


99 


96 



Find from the curves the following information 

(a) The greatest voltage for the series machine. 

(b) The current at this voltage for the series machine. 

(c) The voltage of the shunt machine for the current in (b). 

Refer to Fig. 126. 

Scales. Since both curves are to be plotted to the same scale and 
on the same sheet, the scales must be suitable for the largest values in 
either table. 

Base. The greatest current is 41 amps, (say 40) in the shunt-wound 
case. Taking i* = 5 amps., a base of 8* is required. 

Vertical. The greatest voltage is 105, in the shunt- wound case. 
A suitable scale is i" = 20 volts, giving a height of 5*. 

Plotting. As the curves are quite separate, and the points of each 
very far apart, there is no need here to use two types of point. On the 
horizontal scale figures such as 29-25 can be exactly obtained as ^" 
5 amps.; hence -25 is shown by half of a small square. 

On the vertical scale fa* = 2 volts, hence to show -5 for the last two 
points of the series case, one-fourth of a small square must be estimated. 

The Curves. In the series case all the points seem to lie well on the 
curve. In the shunt case the curve passes slightly above the third point 
and slightly below the fourth, going right through the others. Remember 
that the two curves must be labelled " series " and " shunt." 

Interpolation. (a) On the series curve the highest point reached 
(i.e., the greatest voltage) is 80 volts. 

(b) The current at this max. point is 23^ amps. 

(c) Following the vertical at 23$ amps, up to the shunt curve, the 
voltage is exactly joo. 



CURVES OR GRAPHS 



359 



Exercises 84. Cases Requiring the Origin. 

[Note. Answers can only be given where values have to be inter- 
polated or equations found. A suitable scale for some of the exercises 
will be found in the answers.] 

1. Some values of wind pressure upon a sloping roof are given 
in the following table. Plot a curve showing wind pressure on a base 
of angle, and read off the pressure for the following angles : (a) 38, 
(&) 52. 



Angle of roof, degrees 





10 


20 


30 


40 


45 


50 


55 


60 


Wind pressure, Ibs. 
per sq. ft. 





12 


23 


33 


42 


45 


47*5 


49'5 


50-5 



2. Plot a curve of temperature upon time from the following 
figures, which were obtained in a test on the heating of a small electric 
motor field-coil. 



Time (mins.) 


o 


2 


4 


6 

30 


IO 


12 


18 


24 


30 


35 

47 


40 


Temp. C . 


21-5 


23-5 


26 


35 


36-75 


40 


43 


45 


47 



3. The following figures give the voltage drop across the carbon 
brushes of a direct-current motor, at varying currents. Plot a curve 
showing drop on a base of current. Read off the voltage drop for the 
following currents : (a) 20 amperes, (b) 40 amperes. 



Drop, volts. . . 


35 


65 


88 


i'3 


i'5 


i-75 


1-8 


1-85 


Current, amps. 


4 


Q 


13*5 


21-5 


27-5 


37'5 


42 


47'5 



4. The weights of a particular exhaust valve are given below, for 
various sizes. Plot a curve showing weight on a base of bore, and 
estimate the probable weight of the following sized valves : (a) 8", 
(b) 16", (c) 22". 



Bore of valve in ins. . 


6 


IO 


12 


15 


18 

2OOO 


2O 


24 


Weight in Ibs 


400 


690 


840 


1300 


23OO 


35 



5. The following table gives the prices of certain expansion joints 
for steam-pipes. Plot a curve showing price on a base of diameter, 
and estimate the price of the following sizes : (a) 2%", (b) 3^", (c) 7". 



Diameter of pipe, ins. . 


2 


3 


4 


6 


8 


IO 


Price in shillings . . . 


46 


60 


68 


85 


124 


i8 5 



ARITHMETIC FOR ENGINEERS 

6. The following figures were obtained in tests on a dynamo 
arranged as shunt, series, or compound wound. Plot a chart showing 
volts on a base of current for each machine, using the same scale for 
each. 

SERIES WOUND. SHUNT WOUND. COMPOUND. 



Voltage. 


Current in 
Amperes. 


16 


O 

19 


25 
36 


35 
61 


45 


85 



Voltage. 


Current in 
Amperes. 


83 


O 


80 


19-5 


77 


34*5 


75 


46-5 


70 


71 



Voltage. 


Current in 
Amperes. 


81 


O 


84 


13 


85 


23'5 


86 


38 


86 


48 



7. The following figures give the resistance to motion of an engine 
and train, in Ibs. for every ton of total weight. Plot a curve showing 
train resistance on a base of speed. With the aid of the curve find the 
total resistance of a train whose complete weight is 285 tons and whose 
speed is 35 miles per hour. 



Speed, miles per hr. . 


o 

12 


3 

7'5 


5 

6-5 


10 


12-5 


15 


17-5 


20 

77 


25 


30 
6-8 


40 


50 


70 
19-7 


Resistance, Ibs. per ton 


S-4 


5-26 


5-3 


5*32 


6 


9 


n-8 



8. The following figures give some results of crushing tests of 
mixtures of sand and Portland cement. Plot a curve of strength on 
base of proportion. Interpolate : (a) the probable strength of a mix- 
ture of 4J to i ; (b) the proportion required to give a strength of 1500 
Ibs. per sq. in. 



Proportion of sand 
to i cement 


i 


2 


3 


4 


5 


7 


8 


10 


Crushing strength 
in Ibs. per sq. in. 


6600 


6000 


5000 


2900 


2050 


1250 


1050 


650 



9. The following figures give the least sizes of safety valve per- 
mitted by the B.O.T. requirements for steam boilers. Construct a 
chart showing area of valve on a base of boiler pressure. Find from 
the chart the area required for pressures of 50 Ibs., 180 Ibs., and 90 Ibs. 
per sq. in. 



Boiler pressure, Ibs. per sq. 
ins. 


15 


30 


45 


60 


80 


TOO 


120 


I 4 


1 60 


200 


Area of valve in sq. ins. per 
sq. ft. of grate area 


1-25 


833 


625 


5 


'394 


326 


"277 


241 


214 


174 



10. The breaking strength of tempered steel wire of various dia- 
meters is given in the following table, the diameters being given by 
their wire gauge number. Plot a curve of strength upon a base of 
gauge number. 



CURVES OR GRAPHS 



361 



Diameter, S.W.G. 





2 


4 


6 


8 


9 


12 


16 


20 


Breaking strength 
in Ibs. 


18460 


13400 


9470 


6490 


45oo 


3650 


I90O 


720 


830 



Find from the curve the sizes of wire with the following breaking 
strengths: (a) 15,000 Ibs.; (6) 5000 Ibs.; (c) 350 Ibs. (Note. As 
only whole numbers exist in the wire gauge the next smaller whole 
number must be given in the answer.) 

Cases where the Origin is not Required. In the examples 
shown up to the present the figures to be plotted have included 
values from o upwards, and each scale, therefore, when marked off, 
has been stated at o, i. e., we have plotted from the origin. In the 
majority of cases this is necessary or at least advisable, but it is 
sometimes advisable to omit the 
origin on the sheet, as seen in the 
example which follows. Here in 
the case of the load the figures vary 
from 10,000 to 25,000, i. e. t there 
are no figures from o to 10,000. 
Similarly the vacuum varies from 
2 7*35 to 28-65, i. e., there are no 
figures from o to 27. Now if this 
curve were plotted from the origin 



4: 



O 5000 (0000 15000 20000 25000 
Load . Lbs o Steam berlruf 



Fig. 127. 



the result would be as shown in Fig. 127, the curve lying in 
a very small portion of the paper and leaving the majority of 
the sheet blank. In addition to this, the plotting of the second 
decimal place given in the vacuum figures would be practically 
impossible with the small scale which would have to be adopted. 
Therefore those portions of the scales in which there are no figures 
to plot are omitted ; that portion of the sheet lying between the 
chain dotted lines is stretched out to fill our foolscap sheet. The 
load scale will then start at 10,000 and the vacuum scale at 27. 

Example 356. The following table gives the vacuum produced by 
the condensing plant of a steam-turbine installation. Plot a curve 
with the load as base. 



Load in Ibs. of steam 
per hour 


IOOOO 


15000 


17500 


2OOOO 


22500 


25OOO 


Vacuum. Inches of 
mercury 


28-65 


28.3 


28 


27-9 


27-65 


27-35 



The guarantee condition was 28* at the full load of 18,000 Ibs. Find 
from the curve the actual vacuum at 18,000 Ibs. 



362 



ARITHMETIC FOR ENGINEERS 



Refer to Fig. 126. 

Scales. The origin will not be used on either scale. 

Base Scale. Start, say, at 10,000. Then the scale must accom- 
modate from 10,000 to 25,000, i. e. t a distance of 15,000. A base of 
15* would be very suitable, giving i* = 1000 Ibs., but is too large. 
The only convenient scale then is i* = 2000 Ibs., giving a base of i\" . 

Vertical Scale. Start, say, at 27 and go to 29, i. e., a difference of 2. 
Taking 2" for i, the total height will be 5", which is suitable. The 
scale is then y = -2 ; and since y is divided into five small squares, 
every small square is -04, and the second decimal place (*oi) can be 
plotted if a quarter small square be estimated. 

Plotting. The points follow a very flat curve, which might even be 
taken as straight, except that the general direction of the points suggests 
a curve ; none of the points appears to be exactly on the curve. 

Interpolation. At 18,000 Ibs. the curve reads exactly 28", so that 
the guarantee condition is satisfied. 

Example 357. Experiments were made upon three different types 
of electric glow lamp to determine their resistance at various voltages. 
The results are given in the tables below. Plot upon the same sheet, 
and to the same scales, a curve of resistance with voltage as base for 
each lamp. State, from an inspection of the curves, how the resistance 
varies with increase of voltage, in each case. 



I. Carbon 
Lamp. 



Voltage . 


75 


80 


84-5 


90 


93'5 


96 


Resistance 
in ohms 


937 


93 


91-9 


90 


88-6 


88 



II. Tungsten 
Lamp. 


Voltage 

Resistance 
in ohms 


70 


75 


80 
267 


85 


90 


95 
290 


98 


IOO 


251 


259 


274 


283 


297 


300 



III. Tantalum 
Lamp. 


Voltnge 


73-5 


8 4 


90 


94 


96 


IOO 


Resistance 
in ohms 


392 


400 


428 


443 


453 


459 



Refer to Fig. 128. 

Scales. The lowest voltage tabulated is 70 and the highest 100, 
t . e. t only 30 difference. The origin of the voltage scale, therefore, will 
not be shown. The lowest resistance is 88 and the highest 459, giving 



CURVES OR GRAPHS 



363 



500 Resistance -Vo\t-gq& 



a considerable difference. In this case, then, the origin may be shown. 
Thus, the resistance scale will start at o, and the voltage scale at 70. 

For this example it will be found advisable to use the foolscap sheet 
of squared paper with its long side vertical, as this arrangement enables 
better scales to be used. 

Base Scale. Requires from 
70 to 100, i.e., 30 difference. 
Taking i" to 5 volts, a base of 
6" is required, which is suitable 
for the short side of the paper. 

Vertical Scale.- Required to 
take 460. Allowing 50 ohms 
per inch, a total height of 9" is * 
required and suits the long side ^ 
of the paper. O 

Plotting. In the case of the 
tantalum and tungsten lamps 
the plotting is not difficult. As 
i* = 50 ohms, every small 
square is 5 ohms, and to plot to 
i ohm, one-fifth of a small 
square must be estimated. For 
the carbon lamp the decimals 
will have to be neglected. As 
one-fifth of a small square is the 
smallest that can be reasonably 
estimated, then i ohm is the 
least that can be plotted. 

Therefore 93-7 must be called 94, and so on. Only one style of point 
need be used, as the curves are some distance apart. 

Deduction. In the tantalum and tungsten lamps the resistance 
evidently increases with increase of voltage. In the carbon lamps the 
resistance decreases slightly with an increase of voltage. 




75 80 95 90 

VOLTAGE 



\oo 



Fig. 128. Curves to Example 357. 



Exercises 85. Origin not Required. 

1. Plot a curve of Indicated Horse Power on a base of speed from the 
following figures, the result of a trial of a Diesel-engined vessel. Inter- 
polate the H.P. required at 10 knots. 



Speed in knots . 


77 


9-0 


9-65 


10-8 


117 


I.H.P 


800 


1 200 


1440 


22OO 


2600 



2. In an experiment to determine the effect of salt in water upon 
the boiling-point, the following figures were obtained. Plot a curve 
showing boiling-point upon a base of salt percentage. Estimate (a) the 



364 



ARITHMETIC FOR ENGINEERS 



boiling-point if there is 20% in solution; (b) the percentage of salt 
required to produce a boiling-point of 107 C. 



Percentage of 
salt in solution 


o 


3'2 


6-3 


9'4 


12*2 


17-9 


22'9 


277 


3i'9 


Boiling - point 
inC. 


IOO 


ioo'5 


IOI*2 


IO2 


102-5 


I04-5 


106 


107-8 


108-8 



3. The following figures give the horse power taken by the circu- 
lating and feed pumps of a small condensing steam-engine at various 
speeds. Plot, on a base of speed, a curve for each pump, using the 
same scales. 



Speed. Revs, per min. . 


40 


45 


5 


55 


60 


65 


H.P. of circulating pump 


10-5 


!5'5 


18 


27 


4i-5 


47 


H.P. of feed-pump . . 


4 


5'5 


6'5 


7'5 


9 


15 



4. The table below gives the specific resistance of the acid in accu- 
mulator batteries for different specific gravities. Plot a curve showing 
resistance on a base of specific gravity. Read off the specific gravity 
at which the resistance is least and also the resistance at that point. 



Specific gravity . 


1-04 


i -06 


1-09 


i-i 
i'5 


I-I5 


1-2 


1-25 


i'3 
1-7 


J '35 


i'4 
2-9 


Specific resistance 


2'5 


2 


1-9 


i'3 


i'3 


i'4 


2-15 



5. Friction tests on a certain bearing gave the following results, 
the speed being constant. Plot a curve with friction on a base of 
temperatures. Read off the friction when the temperature is 85. 



Temperature F. . . 


60 


70 


80 


90 


IOO 


no 


I2O 


Coefficient of friction 


008 


008 


0063 


0052 


0045 


0039 


00 35 



6. The following figures are from a turbo-electric plant, the load 
being constant while the vacuum varies. Plot a curve of steam con- 
sumption on a base of vacuum. 



Vacuum in ins. of 
mercury 


o 


14 


i5'3 


19-3 


22 


25'3 


26-6 


Steam consumption, 
Ibs. per K.W. hour 


34-3 


31-2 


29H 


27-4 


25'6 


24 


23-2 



CURVES OR GRAPHS 



365 



Note. As the measurement of large quantities of steam is trouble- 
some, the points will appear rather irregular. Judgment only can 
settle the best curve through the points.) 

7. The voltages obtained from an accumulator with varying strength 
of acids are given below. Plot a curve of voltage on a base of specific 
gravity, and interpolate the voltage when the specific gravity is i-i. 



Specific gravity 
of acid 


i 


Voltage . . . 


i*5 



I -006 


1-015 


1-025 


1-038 
1-88 


i -08 


*'*5 


1-2 


1-25 


1-7 


1-8 


1-85 


1-94 


2-0 


2-03 


2-O7 



8. The weight of i cu. ft. of water at various temperatures is given 
below. Plot a curve showing weight on a base of temperature. Inter- 
polate the weight of i cu. ft. at temperatures of (a) 180; (b) 320. 



Temperature F. . . . 


32 


39 


46 
62-42 


80 


IOO 


I2O 


140 


Weight of i cu. ft., Ibs. 


62-42 


62-425 


62-23 


62-02 


61-71 


61-38 



1 6O 


200 


220 


26O 


3OO 


4OO 


61 


60-08 


59-58 


58-46 


57-26 


53-63 



9. The following table gives the safe currents that may be carried 
by the usual sizes of iron wires employed for starting resistances. Plot 
a curve of current on a base of diameter. What sized wires will be 
required to carry the following currents : (a) 10 amperes ; (b) 25 
amperes ? (Note. Take the next smaller whole number in each case.) 



Diameter in S.W.G. . 


12 


14 


16 


18 


20 


22 


Safe current in amps. 


5i 


30 


20 


n-5 


6-7 


4-5 



10. A certain water-supply problem was solved by plotting, the 
figures being as follows. Plot two curves to the same scales upon a 
base of h, (i) for i; (2) for I. Read off very carefully the value of h 
at which the curves cross each other. 



h 


80 


90 


IOO 


no 


1 20 


13 


i 


2343 


238 


242 


-245 


-2485 


252 


I 


2826 


2706 


.258 


2444 


2291 


2116 



366 



ARITHMETIC FOR ENGINEERS 



The Straight Line. The name " curve " is applied to any 
graph, even though the points may lie upon an exact straight line. 
Probably none of the various curves met with in mathematics are 
more important than the straight line. In all cases where the 
relation or equation connecting two quantities is to be found from 
experimental results the straight line must be finally used. 

In drawing a straight line through the plotted points a ruler 
or set-square may be laid down and a trial line drawn in pencil. 
It will be found easier, however, to space the line evenly among the 
points, by using a piece of cotton or thread. Stretch the cotton 
between thumb and forefinger of each hand. Laying this taut 
thread through the points, it can be seen at once how they are 
balanced on either side. When satisfactory it should be noticed 
if the thread passes through any two of the plotted points a fair 
distance apart, or any prominent intersection of the squares, and 
by such guides the line can be drawn. 

Example 358. The Fahrenheit and Centigrade temperatures are 
connected by the equation F = 9 C + 32. Calculate values of F for 

the following values of C : o, 40, 80, 120, 150. Construct a " tempera- 
ture conversion chart " to convert between Fahrenheit and Centigrade 
readings, and to read from o C. to 150 C. 

The calculation is best carried out in tabular form thus, the values 
to be plotted being found in the first and last columns 



C. 


-?c + sa . 




F. 





~ X + 3 2 


= o -f 32 


32 


40 


5 


= 72+32 


IO4 


80 


9 X 80 , 

___. + 32 


= M4 + 32 


176 


I2O 


9 X 120 


= 216 + 32 


248 


5 


150 


9 X 150 


= 270 + 32 


302 


5 "^ 



Refer to Fig. 129. 

Scales. Origin will be required. 

Base. Required to take 150. Allowing i* = 20, a length of jy is 
required, and is suitable. 

Vertical. Required to take 302, say 300. Allowing i* 50, a 
height of 6" is required. 



CURVES OR GRAPHS 



367 



Plotting. On the base scale the points are easily found. On the 
vertical scale, since i" = 50 every small square = 5. Hence to plot 
to i, one-fifth of a small square must be estimated. All the points 
will be found to lie exactly on a straight line. 



TEMPERATURE FAHRENHEIT 

> J5 8 8 g S 




T< 


am 


pera^ure Conversion 






s 








^q 


_g_rjt 














/ 


y 
































X 


X 
































>X 


































*\ 
































X 


x* 






























r 


iX 
































x 


xl 
































X 


/ 






























> 


^ 
































X 


X 
































x 


































S 









































































eo to eo QO 100 eo 1^0 i>o 
TEMPERATURE CENTIGRADE 



6 

5 

CD 

z 3 

o 

. 

LLJ 
o 




. 


Lf 


forh-Load Curve $or Wovm Gear 






J 






































































^x 






































X 


^ 






































x 


H 






































x 


U^ 









































xjj 


f 




































> 








































X 


X 






































^ 


X 






































2r" 


X 








































X 






































X 


J 








































x 













































O 5 to J 5 ao e5 30 35 40 ^5 50 

Lo AD IN L BS 
Fig. 129. Curves to Examples 358 and 359. 

Example 359. The following table gives the corresponding values 
of effort and load in a small worm and worm-wheel lifting machine as 
determined by experiment. Plot a curve with load as base. 



Load, Ibs. 


6-0 


10-7 


16-7 


22 


26-5 


32 


36*7 


42 


46-5 


51-5 


Effort, Ibs. . 


9 


1-42 


2-12 


2-6 


3'3 


3-72 


4-32 


4-8 


5'53 


6-0 



B B 



368 ARITHMETIC FOR ENGINEERS 

Refer to Fig. 129. 

Scales. The origin will be required. 

Base. Required to take 51-5, say 50. Taking i* 5 Ibs., a base 
of lo" will be needed, and is suitable. 

Vertical. Required to take 6 Ibs. Taking i" = i lb., a height of 6* 
will be needed, and is suitable. 

Plotting. Since i* = 5 Ibs. on the base scale every small square 
-5 lb., and the first decimal place can be plotted by estimating 
one-fifth of a small square. 

On the vertical scale i" = 10 Ibs. and each small square = I lb. 
As we cannot with accuracy estimate finer than one-fifth of a small 
square, one-fifth lb. (i. e. t *2 lb.) is the smallest amount we can plot. 
The second decimal place must therefore be neglected. 

The Curve. It will be seen that the points generally follow a 
straight line. A trial with thread as described shows that the 
line will be suitably balanced if drawn through the two extreme 
points. 

Exercises on plotting of straight lines will be found in Exercises 
87, when finding laws. 

The Plotting of Negative Values. In the previous examples 
only positive values have been taken for each of the two quantities 
plotted, as required by the majority of curves met with. There 
are, however, certain cases where the values to be plotted are nega- 
tive, or perhaps both positive and negative. These are accommo- 
dated as follows : Instead of drawing our axes at the side and 
bottom of the sheet of paper, they are placed with the origin in 
the centre of the sheet and drawn right across, thus dividing the 
surface into four quadrants as shown in Fig. 130. Then on the 
vertical scale all 4- values are plotted from the origin upwards, and 
all values downwards. On the horizontal scale all -f values are 
plotted from the origin to the right, and all values to the left. 

Let us call the quantity plotted along the base x, and the quantity 
plotted vertically y. Then the four quadrants will take the various 
signs as follows : 

Right-hand top quadrant x -f and y + 
Left-hand top x y + 

Left-hand bottom x y 
Right-hand bottom x -f y 

Taken in this order, i. e., starting from the right-hand top quadrant, 
and going round " against the clock " the quadrants are numbered 



CURVES OR GRAPHS 



369 



for reference as 1st, 2nd, 3rd, and 4th (Fig. 130). The first quadrant 
where both quantities are positive is the one used up to the present. 
Evidently if our table of figures contains no negative values of any 
kind, only the right-hand top 
quadrant is needed, and is then 
arranged to fill the paper. The 
remaining three quadrants are more 
used in higher mathematics. 

If necessary to specify any one 
point, it may be done by writing 
down the two co-ordinates with 
their signs, stating the x or base 
value first. Thus a point whose 
distance along the base is + 3, or 
3 to the right, and whose vertical 
distance is 2, or 2 below the 
origin, would be stated as (3, 2). 

The following example is merely to . ~ . , . , 

_ b . r J Fig. 130. Complete Axes for 

illustrate the above. Plotting. 





1-4- 




2 ND QuAD. 


+ 3- 


. ("QUAD. 


oc 




. oc -t- 


y + 




y + 




-*- 




-3 -2 - 
1 1 1 




S \ i - - i 


1 1 1 

x - 


-i - 


. oc -f 


y " 




y - 




-z- 






-3 - 




3 R QuAD. 




4- QUAD. 


-4- 





Example 360. Plot the following points and letter each one as 
stated : A (3, 4) ; B (- 2, 3-5); C (- 1-5, - 3) ; D (1-5, - 5-5), the 
figures being inches. As a check on the plotting, join up the points 
in the order given, and measure the four sides of the quadrilateral so 
formed, viz. the lines AB, BC, CD, DA. The sheet of paper is to be 
used with the long side vertical. 

Refer to Fig. 131 
The axes must first be drawn in the centre of the sheet. 

Plotting. Point A. Co-ordinates (3, 4). Plus sign understood. 
Therefore the point is in the ist quadrant, 3 horizontally and 4 
vertically. 

Point B. Co-ordinates (2, 3-5). The horizontal distance is 
negative and the vertical distance positive. Thus the point lies in 
the 2nd quadrant, 2 horizontally to the left and 3-5 vertically 
upwards. 

Point C. Co-ordinates (1*5, 3). Since both are negative the 
point is in the 3rd quadrant, 1*5 to the left and 3 downwards. 

Point D. Co-ordinates (1*5, 5*5)- The horizontal distance 
being positive and the vertical negative, the point is in the 4th quadrant 
i -5 to the right and 5-5 downwards. 

In Fig. 131 the lines to be measured as a check are shown by a 



370 



ARITHMETIC FOR ENGINEERS 



chain-dotted line, and are measured as follows : AB = 5*02 ", BC = 6*52* 
CD 3-9% DA = 9-62". 

The following example requires the axes to be in the centre 
of the sheet, since the figures to be plotted require equal space in 
all four quadrants : 

Example 361. The following figures are taken from the results of 
an experiment on a sample of cast steel, to determine its curve of 




Fig. 131. Curve to 
Example 360. 



Fig. 132.- Curve to Example 361. 



magnetisation (known as the " B-H " curve), which indicates by its 
shape, etc., the suitability of the material for magnet cores, etc., in 
electrical work, (a) Plot the curve, with the magnetising force H as 
base ; (b) read off the values of B where the curve crosses the vertical 
axis, and, neglecting the minus sign, take the average of these two 
values (known as the residual magnetism) ; (c) read off the values of 
H, where the curve crosses the horizontal axis, and, neglecting the 
minus sign, take the average of these two values (known as the coercive 
force). 



CURVES OR GRAPHS 



H. Magnetising force in 
ampere turns per cm. 


94 


67-5 


47 


27 


14 

IOIOO 


B. Flux density in lines 
per sq. cm. 


I35 


12600 


12000 


III30 



11 


3 


4 


-7'5 


-10-5 


B 


8520 


5^30 


3900 


960 



-10-5 


-14 
1920 


-18-5 


-28 


-37'5 


960 


4840 


84OO 


IOIOO 



H 


-49 


-64 


-90 


-6 3 


37 


B 


IIOOO 


I2OOO 


I3OOO 


12700 


i i 800 



-3 



10400 9000 



H 


+ 3 


4-6-5 


+ 9 


+ 13 


+ 17 


B 


-7300 


5100 


2100 


+ 775 


+ 455 



+ 22 
6800 



30 48 



8700 10700 



Refer to Fig. 132 

Scales. Base. The greatest positive value is 94, and the greatest 
negative value 90. Then the total value to be accommodated is 
184, say 1 80. Allow i" = 20, then 9* is required. This is rather 
wider than our foolscap sheet, but the scale is so convenient that it is 
advisable to use a larger sheet. 

Vertical. The greatest positive value is 13,050, and the greatest 
negative one is 13,000. Then the total height required is about 26,000. 
Allow i" = 2000, then 13" will be required. This is again longer than 
our sheet, but the scale is very convenient and very suitable for the 
larger sheet. 

Plotting. The first six points lie wholly in the first quadrant, both 
values being positive. The next three have base value , and vertical 
still 4-, i. e., they lie in the second quadrant. The following n points 
have both values negative, and thus lie in the third quadrant. The 
next three points have base values + an d vertical values , and there- 
fore lie in the fourth quadrant. Finally, the remaining points lie in 
the first quadrant again, and complete the plotting. The curve lies 
practically through all the points, and the two branches should be 
joined to the extreme point at each end, making a " closed " curve. 

Interpolation. (b) The curve crosses the vertical axis in two places : 
at + 7600 and at 8300. 



372 



ARITHMETIC FOR ENGINEERS 



Neglecting the sign the average = 



__ 7600 -f 8300 __ 



= 795 



.*. Residual magnetism = 7950 lines per sq. cm. 
(c) The curve crosses the horizontal axis in two places : at -f- 12 
and at n. 

Neglecting the sign the average = = 11-5 

.*. Coercive Force = 11-5 amp. turns per cm. 

Example 362. The following values were calculated when solving 
a certain difficult equation 



Value of t 



Value of y 



362 


362-5 


363 


363-5 


0033 


0028 


0013 


-0012 



364 
- -0045 



Plot the curve with the values of t as base, and read off accurately 
the value of / where the curve crosses the base line. 

Refer to Fig. 133 

Scales. The origin will not be required on the base, as there are no 
figures up to 362, and the highest is only 364. Hence we start at, say, 
361, which will keep the curve away from the vertical axis. Then 
from 361 to 364 is 3. Allowing i" = -5, a base of 6" is required, and 
is suitable. 



OO4 i 



u. 
O 

(f) 

UJ 

3 







^ .004 



Fig. 133. Curve to Example 362. 

Vertically the largest + value is -0033 and the largest value is 
0045, giving a total depth of '0078, say -008. Allowing \" -ooi, a 
total depth of 4* is required. This is suitable, although by no means 
filling up our sheet. As only -}- values occur in the base scale, the 
second and third quadrants are not required. The vertical axis may 
then be placed towards the left-hand edge of the sheet, with the 
horizontal axis about halfway up as shown in Fig. 133. 

Plotting. As the scale is a very open one, the values can be plotted 
very exactly. 

Interpolation. The curve crosses the horizontal axis at 363-27. 



CURVES OR GRAPHS 



373 



Exercises 86. Plotting of Negative Values, etc. 

Plot the following points, and as a check measure (in inches) the 
distances named. (Take a scale of y = i on both axes.) 

1. A (5, - i), B(-i, 5 ), C(2,i). Measure AB, BC, and CA. 

2. A(- 4,3-5), B(- 4 , - 3-5), C(- 5 , - i), D(- 5 ,i). Measure AC 
and BD. 

3. In solving a difficult equation in connection with the steam 
engine the following figures were obtained. Plot y on a base of r t and 
find the value of r at which the curve cuts the horizontal axis. 



Values of r 


5 


6 


7 


8 


9 


10 


Values of y 


.67 


45 


23 


- -03 


~ -30 


- '59 



4. The following figures were calculated when solving a difficult 
hydraulic equation by plotting. Plot the figures (with d as base) and 
read off the value of d at which the curve cuts the horizontal axis. 



Values of d . 


1-2 


i-4 


i-3 


,6 


i-7 


Values of y . 


1-142 


-39 


US 


705 


1-385 



5. In connection with the flow of water in pipes it was necessary 
to solve a certain equation by plotting. Plot the figures given below 
(with e as base) and read off the value of where the curve cuts the 
horizontal axis. 



Angle e in degrees 


298 


303-5 


309 


315 


321 


Values of y . . 


2-173 


912 


-29 


- 1-435 


2-499 



(Note. The origin will not be required on the e scale.) 
6. The following figures are for a B-H curve of a sample of armature 
iron. Only the half curve above the horizontal axis is given. Plot 
the half curve and read off the intersection on the vertical axis. 



B 


o 


1000 


2500 


5OOO 


6000 


7500 


7000 


6000 


5000 


3000 


1000 


o 


KL 


i-55 


1-7 


2 O 


2 6 


3'i 


4-0 


2-3 


65 


--25 


I'2 


-1-55 


-1-7 



7. Plot the following figures with values of x as base. Interpolate 
the following points 

(a) The values of y where the curve cuts the vertical axis. 

(b) The values of x where the curve cuts the horizontal axis. 

(c) The smallest value of x. 



Values of x 


3 


i'3 


-*4 


-1-4 


-17 


-1-77 


-i'5 


-'75 


*3 


2-25 


Values of y 


1-7 


1*2 


'5 


-15 


e 


-1-25 


-1-67 


2*2 


-2-73 


-3'4 



374 



ARITHMETIC FOR ENGINEERS 



8. Plot the following figures with x as base, and read off the three 
values of x where the curve cuts the horizontal axis. 



Values of x . 


-4 


-3'5 




-3 




2'5 


2 


__ i 


-4 




I 


Values of y . 


-8 


5-62 




14 


I 


8 


18 


10 


1-94 




-5'5 






























I 




i* 


8 




'3 


3 


3'8 




4'i 






- 18 




2 


5'2 





26-3 


22 


-6-1 




3'4 



9. Below are given some figures of the pressure at various 
temperatures of the two refrigerating agents in common use (sulphur 
dioxide, SO; and Ammonia, NH 3 ). Plot upon the same sheet and. to 
the same scales, a curve showing pressure on a base of temperature for 
each agent. 



Tern 


p.F. 


-40 


-30 


20 


10 





20 


40 


60 


70 


tn CT 


S0 2 


3-i 


4'4 


6 


8 


10-4 


17*3 


27-3 


4-5 


50 


in CO 


NH 3 


9-2 


13 


I7-5 


23 


29-5 


47 


73 


107 


128 



The Equation to a Straight Line. In many cases after an 
experiment has been made, and the values plotted, it is necessary 
to find out some equation or formula connecting the two quantities 
plotted. This enables us to understand the relationship better, 
and reduces the result of the experiment to a simple statement 
which any one can use without having to obtain a graph. This 
equation is spoken of as the law of the curve or the equation to the 
curve. Only the case of the straight line will be taken here, but 
equations can be obtained to any curve. 

Consider the straight line curve AD in Fig. 134. Draw a hori- 
zontal line Ad through the point A, where the curve cuts the vertical 
axis. Then any ordinate consists of two parts, a constant piece 
and a variable piece. Thus take any ordinate as 6B. It is evi- 
dently equal to bn + nB* Now it can be seen that the part bn 
is the same all along the curve, as the line Ad is parallel to the 
base. Let us call this the 1st constant. The portion above the 



* The expression bn here does not mean b x n, but only the line 
from b to n in Fig. 134. 



CURVES OR GRAPHS 



375 




line Ad, e. g., nB, evidently changes with the position of the ordinate, 
i.e., with the base value Ob. Considering this variable piece only, 
it can be seen that the triangles such as ABn are always the same 
shape ; in fact triangle ACp is only triangle ABn drawn to a bigger 
scale. These are known as similar 
triangles. Now, it is a principle in 
geometry that with triangles such as 
these " the ratio of the two perpen- 
dicular sides of each triangle is always 
the same/' Thus if Bn is half of An t 
Cp will be half of Ap. (This the 
student should verify by measure- ^ c 

ment.) Thus the ratio " vertical side l **' 3 *' 

-f- horizontal side" will be the same in any triangle, e.g., ABn, 
ACp, etc. In triangle ABn, the vertical side is Bn and the hori- 
zontal side An, and in triangle AC^>, the vertical side is Cp and the 
horizontal side Ap. 

Then ^ = -~, and so on for any other ordinates. 
An Ap 

Now the vertical side = variable piece of ordinate, and the hori- 
zontal side = base value. Then we have a ratio which is constant 
for our curve and which we can call the 2nd constant, i. e., 
variable 



base value 

/. variable piece = 2nd constant x base value 
Now, from some lines back, 

Any ordinate = ist constant + variable piece 

= ist constant + 2nd constant x base value 

To put this into a shorter and more mathematical form 

Let y represent any ordinate ; and % the corresponding base value 
Let a represent the ist constant ; and b the 2nd constant 

Then writing these symbols in place of the words in our previous 

statement, we have 

y = a -f bx 

This is the form of the equation to any straight line. The constants 
a and b, of course, vary for different cases, but are constant for any 
one curve. The signs of a and b may be + or , depending on 
the position of the line. 

The following examples are intended to show how the values 
and signs of the constants a and b affect the line. When plotting 



ARITHMETIC FOR ENGINEERS 

straight lines from equations such as the following there is no need 
to calculate and plot many points. It can be seen from Ex. 363 
that such calculated points lie exactly on the line. To fix the 
position of a straight line only two points are required, the line 
being merely drawn through them with a straight-edge. But then 
any blunder in calculation would not be shown by the plotting. 
Thus it is necessary to plot three points, say two extreme ones and 
another approximately in the centre of the line. Then all three 
should lie exactly on the straight line, otherwise there is an error. 

Example 363. Plot on the same sheet the curves showing the three 
following equations, between the limits x = 2 and x = + 2. 
(a) y = i + 1-5*; (b) y = 1-5*; (c) y = 2 + 1-5*. 
Calculations : 



X 


I + i'5v 


ay 


2 


I + -3 


=== 2 


+ '5 


i 4- -75 


= i'75 


-f 2 


i +3 


= 4 



X 


i-5r = y 




x 


-2 + 1-5* 


**y 


_ 2 


3 




_ 2 


-2 + - 3 


_ 5 


+ 5 


75 




+ '5 


- 2 + -75 


- I-2 5 


+ 2 


3 




+ 2 


-2+3 


I 



The curves are shown plotted in Fig. 135. 

The following points should be noted from the above plottings : 
When the sign of the constant "a" is +, the line cuts the vertical axis 
above the origin. See curve (a). 

When the constant a is o (i. e., when there is no constant a), the 
line passes through the origin. See curve (b). 

When the sign of "a " is , the line cuts the vertical axis below the 
origin. See curve (c). 

Naturally the distance from the origin is given by the value 
of the constant a. 

Example 364. Plot on the same sheet the curves showing the 
following equations between the limits x = 2 and x = + 2, 
(a) y = 2 + 3*; (b) y = 2 + -5*; (c) y = 2 2X. 

Calculations : 



x 


3* + 2 


-y 


2 


6 + 2 


4 


5 


I '5 + 2 


3'5 


2 


6 + 2 


8 



# 


5* +2 


= y 


2 


-1+2 


i 


5 


25 + 2 


2-25 


2 


I +2 


3 



The curves are shown plotted in Fig. 136. 



# 


2 2* 


= y 


2 


2 - ( - 4) 


6 


5 


2 1 


I 


2 


2-4 


2 



CURVES OR GRAPHS 



377 



The following points should be noted from these plottings. 
When the sign of the constant b is +, the line slopes from the left- 
hand bottom corner to the right-hand top corner. This can be 
called a " positive " slope. When the sign is , the line slopes from 
the left-hand top to the right-hand bottom corner. This can be 
called a " negative " slope. 

The greater the value of the constant b, the greater is the amount 
of slope. Should the constant have for its value o, then the slope 





Fig. 135. Curves to Example 363. Fig. 136. Curves to Example 364 

is o i. e., the line is horizontal and its equation is then y = a, the 
a being the intercept on the y axis. The value of the constant b 
is called the slope of the line, and is the ratio " variable piece -f- base 
value/' as already stated. Thus, taking the ordinate at -f 2, on 
curve (a) in Fig. 136, the variable piece is 6, and the base value 2. 



Then, constant b = 



3, which, of course, appears in the given 



equation. 

To obtain the Equation to a Straight Line. This easily 
follows from the previous section. In all cases the curve must be 



5 - 
4- 



?3 



3- 




378 ARITHMETIC FOR ENGINEERS 

plotted from the origin, if the law is to be found. If necessary, 
the line should be continued to cut the vertical axis. 

The reading at which the line cuts the vertical axis is the value 
of the constant "a." The sign must, of course, be included when 

reading. Thus in Fig. 137 the line cuts 
the vertical axis at 1-57. 

To find constant b glance along the 
curve near its extremities, and find a 
point somewhere near each end, where 
the co-ordinates are definite figures 
(i. e., the reading does not have to be 
estimated). Thus in Fig. 137, at point A 
the line has the co-ordinates (-5, 2) and 
at B (4, 5). From these two points 
v . . _ draw horizontal and vertical lines to 

' cut, as shown, at the point C. Note 

Fi S' X 37- the length of the vertical BC, on the 

vertical scale ; in this case 3 (i. e., 5 2) : also the length of the 
horizontal line AC on the base scale ; in this case 3-5 (i. e. t 4 -5). 

Then, constant b = r . T ,-, and the sign is determined by 
horizontal & J 

the direction of the slope. 

o 

For the line in Fig. 137, b = --- = '857, and the sign is -f from 

the direction of the slope (see p. 377). 
Then the equation is 

y = i-57 + -857* 

Any two points can be taken in estimating b, and it is therefore 
best to choose points of definite and easy co-ordinates, by which 
means it is often possible to see the value of b at a glance. The 
method of checking the equation is given on p. 379. 

Obviously the whole variable piece of the ordinate at 4 (viz. 
Bp) and the whole base value mp (i. e., 4) might have been taken; 
but as the point m seldom cuts the axis at some round figure, an 
estimation has to be made, which is longer and more liable to error. 
Since the triangles ABC and mBp are similar, the ratio " vertical 
side -f- horizontal side " is the same in each case. Thus the vertical, 
Bp = 3*43, and the horizontal mp = 4, therefore the constant 

b ~ = -857 as before. 

4 

Before plotting the straight line to determine the equation from 
any experiment, it must be decided which quantity is desired in 



CURVES OR GRAPHS 



379 



terms of the other, or in other words which is to be the " y " quantity, 
and which the " x." The quantity required in terms of the other 
is plotted vertically. Thus supposing figures were given of H.P. 
and steam used in an engineering test and it was desired to find the 
equation for steam in terms of H.P., then steam would be plotted 
vertically (i. e. t the " y " quantity). In the examples given in this 
book it will be stated which quantity is required in terms of the 
other. In any case the equation can be transposed. 

Example 365. The following figures were obtained from a test on a 
small quadruple expansion steam engine. 



B.II.P 


4Q 


60 


QI 


QQ 


116 


1^7 


I-dS 


















Steam used, Ibs. per hr . 


960 


1190 


1440 


J 55 


175 


2OOO 


2090 



De (ermine the equation giving steam used in terms of the H.P., 
in the form y = a -f- bx. Check the equation by calculating from it 
the steam used 'per hour, when the B.H.P. is 36. 

Refer to Fig. 138. 

As steam is required in terms of B.H.P., plot steam vertically and 
B.H.P. horizontally. The scales and plotting should present no diffi- 
culty, and the points will be found to follow a straight line very well. 
The Equation. Let H = B.H.P. and W = Ibs. of steam per hour. 

Then, W = a + 6H, where a and b are constants 

Producing the line to cut the vertical axis, the intersection is at -f- 370, 
which is the value of a. 

Two points have been selected near the ends of the line where the 
co-ordinates have definite values. These are : (124, 1840) and (20, 600). 
From these points vertical and horizontal lines have been drawn as 
shown. 

The length of the vertical line is 1840 600 = 1240 
The length of the horizontal line is 124 20 = 104 

vertical _ 1240 
104 



Then constant b 



11-91 



horizontal 

and the sign is -f- since the slope is to the right and upwards, 
the equation required is 

W = 370 -f 11-9H 



Then 



Check. 

when H = 36, 



W = 370 -f- 11-911 
= 370 -f- n-9 X 36 
= 37 + 4 28 = 7 ( )8 



For H = 36 the curve gives W 
reasonably correct. 



800 barely, so our equation is 



ARITHMETIC FOR ENGINEERS 



Example 366. The following table gives the diameters of the 
largest step on the cone pulley of some large lathes. Plot the curve, 
and assuming a straight line, find its equation, giving diameter of step 
in terms of height of centre. 



Height of centres, ins. . 


12 


18 


24 


30 


3<> 
34 


42 


66 


Dia. of largest step, ins. . 


22 


26 


30 


3i 


40 


5 




O 4O 6O 8O IOO |2O 14O IGO 

BRAKE HORSE POWER H 



Q 50 

z 
30 

eo 



Cone PULLEY Pioms - 



Larhes' 



MG 



*&-- 



r 




i i 



i i i i 



O IO 20 3O 40 50 60 70 

HEIGHT OF CENTRES IN INS H 
Fig. 138. Curves to Examples 365 and 366. 

Refer to Fig. 138. 

To give diameter in terms of height of centre, diameter must be 
plotted vertically and height horizontally. 

The scale and plotting are shown in Fig. 138. The points lie rather 
irregularly, but a straight line can be drawn to balance. 

Equation. Let H = height of centre in ins., and D = diameter 
of largest step in ins. 

Then, D = a + 6H, where a and b are constants 



CURVES OR GRAPHS 



38i 



The line cuts the vertical axis at -f 16, which is then the value 
of a. 

This intersection is a good point to use for finding 6; (54, 44) is 
taken at the other end. Then, the length of the vertical line is 
44 1 6 = 28, and of the horizontal line 54 o = 54. 

Then, constant b = i-^^^i = = -519, say, -52 
horizontal 54 D J J 

The slope being from left to right and upwards, the sign is -f 
Therefore the equation is, 

D = 16 + -52H 




^ O 100" ZOO 300" 4OO 50O 60O 700 80O 90O 1000 

^ TEMPERATURE FAHRENHEIT t 

Fig. 139. Curve to Example 367. 

Example 367. The following figures are the results of experiments 
on the yield point strength of mild steel at various temperatures. Plot 
the curve, and, assuming a straight line law, find the equation giving 
yield point strength in terms of the temperature. 



Yield point strength, 
tons per sq. in. 


16 


I5'2 


13 


I3'5 


ia-8 


1 1'3 


10-7 


9'6 


87 


7'7 


6-3 


5'5 


Temperature F . . . 


o 


66 


210 


300 


390 


480 


570 


60 


750 


840 


930 


IO2O 



Refer to Fig. 139. 

Strength is required in terms of temperature, therefore strength 
will be plotted vertically. 

A larger sheet of paper is used to accommodate a better scale. 

The points lie very irregularly, but the line shown is about the best 
that can be put in. 



ARITHMETIC FOR ENGINEERS 



Equation. Let W = yield point stress and / = temp. F. 
Then, W = a + bt t where a and b are constants. 

The line cuts the vertical axis at + 16, which is thus the value of 
the constant a. 

This point is convenient for finding the slope; the other point 
taken is (noo, 6). Then, the horizontal distance = uoo, and the 
vertical distance = 16 6 = 10. 

/- , , , vertical 10 

Constant , . - -,--.- = = -ooooo. sav, 'oooi. 

horizontal noo ^"b^y> *<*y, y 

In this case the slope is downwards from left to right, and therefore 
the constant b is negative. 
Then the equation is 

W 16 - -0091 1 



Exercises 87. On Straight Lines and Finding Laws. 

1. The following figures were taken when stretching a helical 
spring. Plot a curve of extension on a base of load, and find the 
equation giving extension (e) in terms of load (w). 



Load, Ibs. . . . 


o 




2 


4 


7 


9 
1-8 


Extension, ins. 


4 


8 


!*4 



II 



2. The following table gives the prices of some geared pulley blocks. 
(a) Assuming a straight line law, find an equation giving the price (P) 
in terms of the capacity (W) ; (b) using the equation obtained, calculate 
the price of a block having a capacity of 5 tons ; (c) read off the price 
of a 5-ton block from the curve. 



Capacity in tons . 


i 


I 


2 


4 


6 
I5'25 


8 


Price, . . . . 


4 


5 


7^5 


10-5 


18 



3. The following figures give the voltages necessary to cause an 
electric spark between two needle-points in air, when d millimetres 
apart. Plot a curve and find the equation giving voltage (V) in terms 
of distance (d). 



Distance apart in 
millimetres 


6 


10 


14 


18 


22 


26 


3 


Voltage 


5000 


7500 


IOOOO 


12500 


I5OOO 


17600 


20100 



4. In some experiments on the breaking strength of rolled copper 
at various temperatures the following results were obtained. Plot the 



CURVES OR GRAPHS 



383 



curve, and, assuming a straight line law, find the equation giving 
strength (/) in terms of temperature (/). 



Temperature, F 


60 


2IO 


^OO 


4IO 


^oo 


600 
















Breaking strength, tons per sq. in. . 


17-8 


17-4 


16-4 


16 


15 


14-2 



5. Taking a number of lathes of somewhat similar designs, the 
width of the cone pulley steps is given in the following table. Plot 
the figures, and, assuming a straight line law, find the equation giving 
width (w) in terms of height of centres (H). 



Height of centres, ins. . 


6 


8J 


9 


io| 


12 


15 


18 


Width of step, ins. . 


2* 


3t 


3J 


3J 


4l 


5i 


6 



6. The weights of some surface condensers are given in the follow- 
ing table, with the cooling surface. Plot a curve of weight on a base 
of cooling surface, and interpolate an approximate weight for con- 
'densers having (a) 5000 sq. ft. ; (b) 2600 sq. ft. Find also an equation 
giving weight (W) in terms of square feet of cooling surface (S). 



Cooling surface, sq. ft. 


400 


IOOO 


2OOO 


3OOO 


4OOO 


6OOO 


Weight in Ibs. . . . 


3040 


6310 


II4OO 


I7IOO 


2IIOO 


32700 



7. The following figures were obtained in the tensile test of a piece 
of mild steel bar within the elastic limit. Plot a curve of load upon a 
base of extension, and find the " slope " of the line. Note. The curve 
must pass through the origin. 



Load in tons . 


1-76 


3-52 


5-28 


7-04 


Extension in ins. . . 


0024 


0047 


0075 


0102 



8. A test of a small oil engine gave the following results 



BH.P 


I'O 


2'I 


3*o 


4.*2 












Oil used per hr. in Ibs. . 


1-07 


2-16 


2-85 


3-91 



4-40 



5-3 



4-90 



Plot total oil used, to a base of horse power. Calculate also the oil 
used in Ibs. per H.P. hour, and plot these figures upon the same base. 
Interpolate the total oil used and the oil used per H.P. hour at 5 H.P. 
c c 



ARITHMETIC FOR ENGINEERS 



9. A test on the resistance of a metal filament lamp gave the follow- 
ing values - 



Voltage in volts 


30 


40 


5 


60 


70 


80 


Current in amps. . 


2 


23 


24 


26 


27 


'3 



Calculate for each pair of values the resistance in ohms (i. e. t voltage 
-f- current), and plot a curve showing resistance on a base of voltage. 
Find the equation giving resistance (R) in terms of voltage (V). 

10. In a certain experiment the following figures were obtained. 
Find the equation giving L in terms of /. 



Values of / . 


6025 


81 


1*017 

818 


1-243 


1-6 


Values of L . . 


1-023 


933 


746 


'55 



11. The following table gives the latent heat of ammonia at various 
temperatures. Find an equation giving latent heat (L) in terms of the 
temperature F. (t). 



Temperature F. 


- 30 


10 


20 


40 


80 


90 


Latent heat .... 


589 


573'5 


55 


534 


5 02 '3 


494-2 



12. The prices of some centrifugal pumps are given in the following 
able. Find an equation giving price (P) in terms of size (D). 



Size of pump . 



Price, 



44 



7 


8 


9 


10 


12 

102 


5^ 


62 


70 


86 



13. In connection with aircraft engine radiators the relation between, 
the boiling point of water and the altitude is important. The follow- 
ing table gives the boiling point at various altitudes indicated by the 
oidinary altimeter. 



Altitude in feet 





5,000 


10,000 


15,000 


20,000 


25,000 


30,000 


Boiling Point 
degrees Centi- 
grade 


100 


94'9 


90-0 


85-25 


80-6 


7 c-i 


7*-75 



Find the law giving boiling point (T) in terms of altitude (H). 
Assume a straight line passing through the first point. 



CHAPTER X 
THE SLIDE RULE 

[The student should study this chapter with a slide rule at hand.] 

Introductory. The slide rule is a mechanical aid to calcula- 
tion ; a " labour-saving " device by means of which many arith- 
metical calculations can be performed with very little expenditure 
of time and energy. Its principle is logarithmic, and although the 
slide rule may be used without any knowledge of logarithms, the 
student is advised to study the latter first, as he will then appreciate 
much better the possibilities of his instrument. 

Some students appear very loath co learn the method of em- 
ploying this most useful instrument, even when they possess one. 
This is partly due to the fact that at first their calculations with 
the slide rule are more troublesome and perhaps take longer than 
when worked in the ordinary manner. This difficulty must be 
expected in the early stages of yising a strange machine and only^ 
practice will surmount it. The student should endeavour to use 
his slide rule upon every possible occasion, when its use will gradually 
become second nature to him. To employ the slide rule quickly 
and easily it is essential that he should be able to " approximate" 
with ease, and the material on p. 61 should first be thoroughly 
grasped. 

Description. The slide rule is made in three varieties, the 
straight, the circular, and the spiral. The straight type is the only 
one treated here, as it is by far the most common. The circular 
type is more easily carried in the pocket, but is hardly as accurate. 
The spiral form is very bulky, and very expensive, and is therefore 
beyond the reach of the average student. The straight type is 
made in various sizes, varying from about 5* to 40' long. For all 
general purposes the " lo-inch " slide rule is by far the most suit- 
able, and is the one most commonly met with. An example of 
the 10" rule is shown in Fig. 140, half full size. It consists of a 
frame or stock F, grooved centrally to receive the slide S. Both 
these parts are of mahogany, and the surfaces carrying the various 

385 



386 



ARITHMETIC FOR ENGINEERS 





OJ 

-J 
UJ 



t-ro 






a 



J2" 



I 

in 



bO 

S 



.- 



THE SLIDE RULE 387 

scales are faced with white celluloid. The central portion of the 
stock is slotted lengthwise by the thin slit /, and a piece of thin 
sheet steel is placed at the back. This gives strength and elasticity, 
permitting the stock to be closed in, so as to grip the slide better, 
or opened out to give greater freedom to the slide. New rules are 
usually rather stiff in action, but with use they become more free, 
so it is unwise to open out the stock, as this will tend to looseness 
in later years. A stiff rule can be eased by the application of a 
little blacklead to the grooves and rubbing edges. When through 
use the slides become slack, the stock should be closed in, so that 
the slide cannot be shaken from any set position. This is done 
by removing the slide, and wrapping a handkerchief round the 
frame. Then holding it with both hands near the end, a steady 
grip should be applied, which brings the edges of the grooves slightly 
together. Then repeat further along, until the whole length is 
travelled. On replacing the slide it will be found appreciably 
tighter. 

A light aluminium frame, called a cursor or runner, and carrying 
a piece of thin glass embraces the whole width of the rule, sliding 
in grooves upon the edges. A thin steel spring P, of the " carriage " 
type, in the top of the frame prevents any shake. A fine black line 
is engraved vertically in the centre of the glass, and passes across 
the 4 scales carried on the face of the rule. These scales are usually 
lettered at the ends as ABCD, starting at the top, and A and B 
are identical, A being on the stock and B on the slide. Also the 
scales C and D are identical, C being carried on the slide and D 
on the stock (see Fig. 140). Frequently these four are spoken of 
as two only, A and B being called the " top scale " and C and D 
the " bottom scale/' These scales occupy a length of 10* (the 
exact length is often 25 cms., but the rule is still known as a 10* 
rule). In the later designs there is sufficient space beyond each 
end of the scales to permit of the cursor line being placed right 
over the end division. 

Division of the Scales. The " bottom scale " is divided into 
9 spaces, the 10 lines starting from the extreme left being numbered 

i, 2, 3 8, 9, i. These are the ist significant figures of the 

numbers used. The lengths of the 9 spaces decrease steadily from 
left to right, and in consequence of this the subdivision is not 
uniform. From I to 2 the space is divided into 10 parts, and each 
of these into a further 10 parts. These also decrease steadily from 
left to right as seen at A, Fig. 141. The subdivision from I to 2 
gives the 2nd significant figures (of numbers between i and 2), 



388 



ARITHMETIC FOR ENGINEERS 



and the smallest divisions are the 3rd significant figures. By 
estimation the 4th can be obtained, but it is not very reliable. 

From 2 to 4 each main division is split into 10 parts, and each 
of these into 5 parts (see B, Fig. 141). The first 10 parts give the 
2nd significant figure, while the 5 parts only give exactly the 
3rd significant figures 2, 4, 6 and 8. By estimation the inter- 
mediate ones are obtained ; an attempt at the 4th figure may be 
made, but is difficult. From 4 to the end each main division is 
also divided into 10 parts (see C, Fig. 141), but each of these divisions 
is only split into 2 parts. The 2nd significant figure is marked, and 

K <3*3 c.ise. 

Qillll|iill|lili|im| 



TtTTTTTimti 



Sub-dlvis\on -fvom 



5 
I 




Sub-di vision -from 2. to 




6 7 8 

Sub -division -frow A- Yo End. 

Fig. 141. Subdivision of, and Readings on Scales C and D. 

also the 5 in the 3rd significant figures. By estimating one-fifth of 
the smallest division the 3rd significant figure can be obtained, 
but no more. The accuracy of the 10* slide rule is thus limited to 
the 3rd or 4th significant figure, which is quite sufficient, however, 
for ordinary engineering work. 

The " top scale " consists of two halves, end on, and exactly 
the same, each half being similar to the " bottom scale." The 
subdivision is not as fine, since the lengths are not as long. This 
is dealt with further on p. 405 (see Fig. 149). 

The back of the slide carries three scales, only one of which will 
be considered in this book. 

Usually the edges of the body carry a scale of inches and tenths 



THE SLIDE RULE 389 

on one edge and a scale of centimetres and millimetres on the other ; 
these are very convenient for drawing, etc. ; they are not shown in 
Fig. 140. 

Many varieties of slide rules are made, with special scales for 
extracting various roots, areas of circles, resistances of electric 
conductors, etc., but unless one is doing a considerable amount of 
such special work, the extra scales do not warrant the expense. 

Method of Reading. As with logarithms, only the significant 
figures are dealt with on the face of the slide rule, and when finding 
any reading on the scales A to D only the significant figures of the 
number are noticed. The ist significant figure is one of the main 
printed numbers I to 9. Great care must be taken at first that the 
printed numbers from between 1 and 2 are not mistaken for these 1st 
significant figures. The 2nd significant figure is one of the next 
subdivision (which from i to 2 are numbered, but nowhere else). 
For ease of reading the fifth line of these subdivisions is rather 
longer than the rest. The 3rd significant figure may be actually 
marked (as between i and 2) or may have to be estimated. Accurate 
estimation can only be obtained by practice, preferably under 
careful supervision. The cursor is particularly valuable when 
estimating, as its line is fine and extends right across the scales. 
In all cases the beginner should be exact in his settings and read- 
ings ; speed in using the slide rule is only attained by practice. 

In Fig. 141 are shown various sample readings on the bottom 
scale, only the significant figures being given. In reading these 
values the digits should be read separately and not as a complete 
number ; thus 21 is read as " two one/' and not " twenty-one " ; 
336 as " three three six/' etc. At a is shown 21; at b 336; at c 
152 ; and at d 785, all of which require no estimation, the third 
figure being marked. At e is shown 285 ; the first two figures 28 
are easily found. Then from 8 to 9 the subdivisions read 2, 4, 6, 8 
and the 5 is halfway between the 4 and the 6. At /is shown 317, 
the last figure being halfway between the third and fourth sub- 
division lines. At g is shown 578; the 57 is easily found. The 
next short line is 575 and the following long one 58. Then, as 578 is 
3 more than 575, three-fifths of the next division must be estimated. 
Similarly 832 is shown at h. An estimate of the fourth figure for 
numbers between i and 2 is given at k, which shows 1343. 

Advice as to Holding the Rule. Fineness of adjustment is 
essential, and the rule should be held so as to obtain this. It is 
usually convenient to hold the left hand end lightly in the left 
hand. Placing the right hand underneath the rule the cursor may 



390 



ARITHMETIC FOR ENGINEERS 



be moved with the forefinger and thumb on the top and bottom 
respectively. This is preferable to using the right hand on top, 
as no shadow obscures the cursor. The slide may be moved from 
its closed position by either forefinger. When once out it may 
be manipulated by finger and thumb. A very fine adjustment 
may be obtained by placing the thumb and forefinger holding the 
slide against the end of the frame as indicated in Fig. 142. A 
slight bending of the thumb and forefinger, while keeping them 
pressed against the end of the frame, will now produce a very fine 
motion of the slide. This will be found particularly useful when 
using a new rule, which is rather stiff. If the slide is manipulated 
without supporting the thumb and forefinger against the frame, 
it will continually be overshooting the mark. 




Fig. 142. Holding Slide Rule to obtain Fine Adjustment. 

Operations on the Slide Rule. The rule can be used for 
multiplication, division, finding of various roots, finding logs and 
antilogs, and for other operations not treated here. It must be 
noted that the slide rule cannot perform addition and subtraction. 
By far the most common operation to be performed is that of 
combined multiplication and division. 

All multiplication and division is to be performed on the bottom 
scale. 

In every case the operation is started and finished on the frame 
or D scale. In the following descriptions the word " cursor " will 
frequently be used to denote the line on the cursor. 

Division. To divide one number (say, 7-5) by another (say, 3) 
proceed as follows : 

1. Place the cursor exactly over the number to be divided (in 
this case 7-5) on the frame or D scale. 

2. Move the slide to bring the divisor (in this case 3) on the 
slide or C scale exactlv under the cursor. 



THE SLIDE RULE 391 

3. Shift the cursor to the end of the slide, setting it carefully 
over the extreme line. 

4. Read under the cursor on the frame or D scale. This reading 
gives the significant figures in the quotient, in this case 25. 

5. In approximating for a result settle the position of the decimal 
point and place it in the significant figures read. 

The approximation in this case is very simple. Thus ^ is 

evidently a little more than 2 ; then our result is 2-5. 

This order of operations must be strictly kept. At first the 
examples taken will be those not requiring any estimation of a 
reading. The setting of the rule is shown in the figures referred to 
for each case, the actual readings of the results being indicated. 



.Divi'sor on Slide x 
/ |A| 




\N? to be divided 
on O 5ce*le on D Scale o** D *=>cal<= 

Fig. 143. Division by Slide Rule. 

The student should perform each step in the order given in the 
following examples, comparing his various settings with the diagrams. 
The method of carrying out the operation of division is shown in 
Fig. 143. 

Example 368. Work the following divisions with the slide rule. 

(a) 5*; (b) *2; (c) *3 ; (d) 33* 
v ' 4 v ' -46' v ' 3-12' v ' 51-5 

Refer to Fig. 144. 

In the following examples the same letters are used in the figures 
as in the explanation. 

(a) 1. Place cursor at 56 on D scale. 2. Shift slide to right, placing 
4 under the cursor. 8. Shift cursor to left-hand end of slide. 4. Read 
off 14 on D scale (the cursor will appear over a printed figure 4. Re- 
member that this is 14, being between i and 2). Approximating, ^~ 
will be something over 12. Then our result is 14. 

(b) 1. Place cursor at 23 on D scale. 2. Shift slide to left, placing 
46 under the cursor. 3. Shift cursor to right-hand end of slide. 

4. Read off 5 on D scale. Approximating, ;? == ^ = about 5. So 
our result is 5. 



392 



ARITHMETIC FOR ENGINEERS 




Ic 


1 j 






J.ll.l,, 


III 


DM III III! 


mi iiimni]iiii|inijiiii 


mill 










{ 


1 1 ' 

e 3 ^ 


r 


TResulr 



'Cursor 

1. Set- cursor ar 56 onD , M , , ^ . , _. 
o e i. >i /- j 3. Under Ion C.read 14- on D. 

2. Set" -4- on C u^der cursor 

SETTING SLIDE (a) READING RESULT 



in] 



JOE 




Resuir. 

I. Scf cursor at- S3 onD ^,i.-^ ir- r^ 

o e w /\ /- /- ^ 3. Under lonC.read 5 on D. 

2. Ser- 46 on C under cursor .^ ' 



SETTING SLIDE 



READING RESULT 



8 



I 






C 


1 ^ 




,',11,1,. 










1TTI 


T| 


| 


\\ 


Jill 


II 


| lllMIIM 


mi 


11 ll IT! 


I III! 


Illll 


II 


u 


III 


III! 1 






t : 


b 





Resulh 



'Cursor 

. SeV cursoral- 85 on D 
2 Ser 312 on Cundar cursor 3. Under I on O y read 66 on D. 

SETTING SLIDE READING RESULT 

891 



I ll 



t 



' 



Cunsor I Result 

I Ser cursor a Ir 33 on D 
a ser 5^5 on C under cursor .oUnder I on C ; read G45 onD 

SETTING SLID.E READING RESULT. 

Fig. 144. Settings for the Divisions of Example 368 



THE SLIDE RULE 393 

(c) 1. Place cursor at 83 on the D scale. 2. Shift slide to right, plac- 
ing 312 under the cursor. 3. Shift cursor to left-hand end of slide. 

4. Read off 266 on D scale. Approximating, ~ = say, , i. e. t 

something over 200. Then our result is 266. 

(d) 1. Place cursor at 332 on D scale. 2. Shift slide to left, placing 
515 under cursor. 3. Shift cursor to right-hand end of slide. 4. Read 

645 under cursor on D scale. Approximating, - = about - or 

5i'5 500 

-, = '6. Then our result must be '645. 

When the student can work exercises like the foregoing with 
accuracy, he may attempt those requiring estimation of readings 
such as the following : 

Example 369. Perform the following divisions with the slide rule. 

gf 

Refer to Fig. 145, which indicates concisely the operations. 

(a) 1. Place cursor halfway between the graduations denoting 
1 17 and 1 1 8. 2. Shift slide to left, placing 5 under the cursor. 3. Shift 
cursor to right-hand end of slide. 4. On the D scale the line will be 
found midway between the lines denoting 234 and 236. The reading 

is then 235. Approximating, ^5 is evidently about 2 ; then our 
result is 2-35. 

(6) 1. Place cursor halfway between the graduations showing 
344 and 346. 2. Shift slide to right, placing 12 under the cursor. 
3. Shift cursor to left-hand end of slide. 4. On the D scale the line 
will be found one-fourth of a small division to the left of the gradu- 
ation 288 (see enlarged view in Fig. 145). The previous graduation 
reads 286. Then halfway is 287 and our reading is 2875. Approxi- 
mating, 2 A3 js about , i. e., 3. Then- our result must be 2-875. 

(c) 1. Set cursor at either extreme end of the D scale, the right- 
hand for preference. 2. Shift slide (to the right) placing 746 under 
the cursor. The graduation 745 is marked. Then one-fifth of the 
distance between 745 and 750 must be estimated (see enlarged view in 
Fig. 145). 3. Shift cursor to (left) end of slide. 4. Read 134 exactly. 

Approximating, ^- is evidently just over i. Then our result is 1-34. 

(d) 1. Set cursor between the graduations of 605 and 610, estimating 
three-fifths of the smallest division to the right of 605, giving 608 
(see enlarged view in Fig. 145). 2. Shift slide to the right, placing 528 
beneath the cursor. The graduations 525 and 530 are marked and 
three-fifths of the small division must be estimated beyond the 525. 



394 



ARITHMETIC FOR ENGINEERS 




Cursov 
Scr cursor ah 1175 on D. 
5* on C under cursor. 

SETTING SLIDE 




3. Sef cursor at" I on C. 



r. Read 2.55 under cursor on D 

READING RESULT 

o Q7C 

Enlarged View -for 1 
Khe esttwiatfon in J> -i t ; 

esel i ess 

287 



I 7 t 5 I 2t 

||j; Enlarged Vtew for hhe 



Result- T Cursor 

.Se,r cursor a V- 34 5 on U. 
2. Sc^ 12 on C under cursor 
3-Str cursor at" Ion C 
4. Read 875 under cursor on D 



Enlarged View -for Hne, 




in d 60OJ liUJJGlO 
I 605 ' 



D 



tun 



Cursor* ' Result 

I. SeV cursor a r 1 on D. 3 . Under I on C, read lb^ on D. 
Se* 74G on C onder cursor 

SETTING SUDE (C ) READING RESULT 




D 



l 



iiiiliiii nil ini|iiii 



Resulr: 

3. SeV-cursoroK 1 onC. 
. Ser 58on C under cursor ^* ^^^^ ^51 under cursor on D. 

SETTING SHOE (S) READING 

Fig. 145. Settings for the Division of Example 369 



THE SLIDE RULE 395 

8. Shift cursor to left-hand end of slide. 4. The line will be found 
a shade to the right of graduation 1150, say 1151, this being our 
reading. Approximating, ~ is just over i. Then our result is 1*151. 

The principle of the Slide Rule cannot be fully explained until 
the reader understands the theory of logarithms, but some idea of 
the principle may be given. It has been seen in Chap. VI, that to 
divide one number (X) by another number (Y) we subtract log Y 
from log X, and obtain the log of the result. Now the bottom scale 
of the slide rule is really a logarithm table. The distance of any 
graduation from the left-hand end represents the logarithm (decimal 
part only) of the reading at that graduation. Thus the distance 
from i to 2 represents the log of 2 (i. e. t -301), the distance from 
i to 55 represents the decimal part of log 55 (i. c. t 744)> and so on - 









& _ E i H 



LI Lo. o 'N; JM o^g 

^ L 9 f Divisor I Lob.x-f LoaYl 
r ~ 



.of.N9 



Fig. 146. Illustrating Principle of Slide Rule. 

Let us consider the case 9-4-2, and refer to a, Fig. 146. The 
figure 9 is first found on the D scale; and distance LM is log 9. 
Now the slide is moved to the right to bring 2 on the C scale at 
point M; and distance NM is log 2. Then evidently distance 
LN = LM - NM = log 9 log 2. Now at N on the D scale we 
read off 45 (read as four five). Then distance LN is log 45 and 
therefore 9 log 2 = log 45 ; or since subtraction of logs means 

division of numbers ^ = 45 (four five) as we know. 

Exercises 88. On Division. 

Perform the following divisions with the Slide Rule 

1. 24 -T- 2. 2. 55 ~ 25. 3. 6-8 -r -81 

4. -392 - -67. 5. -194 ~ 6-3. 6. -153 -^ 15. 

7. 96-5 -109. 8. 2-28 ~ -0252. 9. -384 -i- 25-6. 

10. 308 -^ 725. 11. 223 4- 425. 12. 27-5 4- i-i 

13. -083 - 3-03. 14. 4475 -r- 384- 15. 397 + 8 4'5- 

16. 5-28 20. 17. 417 -- 182. 18. -682 -r -745. 

19. 13*55 -r '0209. 20. 1012 -f- 9-17. 



396 ARITHMETIC FOR ENGINEERS 

Multiplication. To multiply one number by another, for 
example, 3*5 by 2 '2, proceed as follows : 

1. Place the cursor exactly over the number to be multiplied 
(in this case 3'5) on the frame or D scale. 

2. Move the slide to the right (or left as may be required by 
operation 3) and place the end line exactly under the cursor. 

3. Move the cursor until it is exactly over the multiplier (in 
this case 2-2) on the slide or C scale. 

4. Read under the cursor, on the stock or D scale, the significant 
figures in the answer (in this case 77). 

5. By approximating for a result fix the position of the decimal 
point, and place it in the significant figures read. 

The approximation here is simple; 3-5 x 2-2 is about 3x2, 
i. e., 6. Hence our answer must be 7-7. 

The moving of the slide to the right or left in the second opera- 
tion is necessary in order that the multiplier may appear within 



on Slide 



Jl 



r '< 


^ lei 


l?l ^^"^ 


i pi ?! 


01 _ 


^|0| 


l !/^ 


H ^-^1 Io| 


CN9 \ro be muth plied ^ Pvoduct ow 
oy> D Scqle 


D Scale ^ 


1 N? ro be- 
multiplied on D Scale 



Fig. 147. Multiplication by Slide Rule. 

the limits of the stock or D scale. Thus taking 3-5 X 4, if the 
slide be moved to the right, its end being at 3-5 on the D scale, 
then when searching for the 4 on the slide or C scale, this will be 
found outside the stock altogether. Then the slide must be changed 
to the left-hand side, its right-hand end being opposite 3-5 on the 
D scale, when the 4 on the slide will be found opposite 14 on the 
D scale. When first using the instrument the student will be 
unable to say to which side the slide should be moved, but with 
practice he will recognise that a result is going to be smaller or 
greater than 10, and will move the slide naturally to the right or 
left respectively. The method is illustrated by Fig. 147. 

Example 370. Work the following multiplications on the slide 
rule : (a) 32 X 2-5; (b) 785 X 4; (c) 1-34 X 48-5; (d) -14 X 80. 

When the student has worked Exercises 88, on Division, he should 
be able to follow this example without a figure showing details of 
setting. 

(a) 1. Set cursor to 32 on D scale. 2. Move slide to right and 
place the left-hand end under the cursor. 8. Move cursor to right 



THE SLIDE RULE 397 

and place over 25 on the slide or C scale. 4. Read off 8 on the D scale. 
Approximating, 32 X 2*5 will be something more than 64, hence our 
result is 80. 

(b) 1. Set cursor at 785 on the D scale. 2. Move slide to left and 
place the right-hand end under the cursor. 3. Move cursor to left 
and place over 4 on the slide or C scale. 4. Read off 314 on the D 
scale. Approximating, -785 X 4 will be about 3, hence our result 
is 3-14- 

(c) 1. Set cursor at 134 on D scale. 2. Move slide to right and 
set left-hand end under the cursor. 3. Move cursor to right and place 
over 485 on slide. 4. Read off 65 on the D scale. Approximating, 
1-34 X 48-5 will be rather more than 48-5. Hence our result is 65. 

(d) 1. Set cursor at 14 on the D scale. 2. Move slide to left and 
place right-hand end under the cursor. 3. Move cursor to left and 
place over 8 on slide or C scale. 4. Read off 112 on D scale. (This 
may appear to be 12, but it should be remembered that the end of the 
D scale is i, the next i to the right being n.) Approximating, -14 X 80 
is rather more than one-tenth of 80, or 8. Hence our result is 11-2. 

The above example requires only the graduations marked. The 
following will need estimation, and also introduces " continued multipli- 
cation," i. e. t the multiplication of three (or more) numbers. 

Example 371. Work the following multiplication on the slide 
rule : (a) 2-71 X 3-2; (b) 68-7 X 16-7 X -366; (c) 5-33 x -946; 
(d) -349 x 20-3 X 8-35. Refer to Fig. 148, which indicates concisely 
the operations. 

(a) 1. Place cursor halfway between graduations 270 and 272, 
thus giving 271. 2. Shift slide to the right and place left-hand end 
under the cursor. 3. Move cursor to right and place over the gradua- 
tion 32 on the slide or C scale. 4. Read on the D scale ; the cursor 
line appears to be a little less than halfway between 865 and 870 ; 
hence read as 8670 (see enlarged view in Fig. 148). Approximating, 
2-71 x 3*2 is about 9; hence result is 8-670. 

(6) 1. Place cursor between graduations 685 and 690, estimating 
two-fifths of the small division beyond the 5, thus giving 687. 2. Move 
slide to left, and place right-hand end under the cursor. 3. Move 
cursor to left and place over the graduation 167 on the slide or C scale. 
[Cursor should now read a shade less than 115 : in practice this result 
is not read ; it is given here as a check.] 4. Move slide to right, placing 
i under the cursor. 5. Move cursor to right, placing it over 366 on 
slide. 6. Read 42 on D scale under cursor. Approximating, 68-7 X 
i67 X '366 is, say, 60 x 20 x 1 or 400. Hence result is 420. 

(c) 1. Place cursor between graduations 530 and 535, estimating 
three-fifths of the small division beyond the 530, thus giving 533 (see 
enlarged view in Fig. 148). 2. Move slide to the left and place the 



398 



ARITHMETIC FOR ENGINEERS 




867 
Y 



! si ] 

Enlarged View -for rV\e 



esVtVnairforu 



^Cursor Result 

I. Set- cursor aV- 271 onD. 3. SeV- cursor at- 
Z SeV- \ on C ur,der cursor 3 e on D 

4- Read 8G7unde,rcursoronU. 



SETTING SLIDE 




READING RESULT 



4- 



4- 



Cursor ' Cursor scune, pos\r\ovi 

I.Seir cursor at- 687 3. Ser cursor 4. Sef \ o n C 

on D _ . arlG7onC urder cursor 

ow C urider x-^ 



cursor 
1 ST 5ETTIINOOF SUlOE 



Enlarged View/ for ? 
cslrioiaVvon in C m y w 

ew for 



Resu\v- 
5 Seycursora.r 

>. ReaJ <f onD 
undev cursor 
OP SHOE 



533 



ii!j|T 

in 1 J , 
550 540 
535 



3. SeV' cursor or. ^^VG on C - 




Result: 

lonC 5. Sex cursor a V* 
oia *-' ox\r ^lO5on C. unde,r cursor 85 S ovi c 

2. Ser \onC under x-^ e Read5Qe.on 

cursor. \S/ D under cursor 

1 ST SETTING orSuoe ND SETTING OF SUDE. 

Fig. 148. Settings for the Multiplications of Example 371. 



THE SLIDE RULE 399 

right-hand end under the cursor. 3. Move cursor to the left and place 
it between graduations 945 and 950 on the C scale, estimating one 
fifth of the small division beyond the 945, thus finding 946. 4. Read 
on the D scale ; the line appears a shade to the left of graduation 505, 
say 504 (see enlarged view in Fig. 148). Approximating, 5-33 X -946 
is practically 5-33 X I, so that our result must be 5-04. 

(d) 1. Place cursor halfway between graduations 348 and 350, 
thus giving 349. 2. Move slide to the right, placing the left-hand end 
under the cursor. 3. Move cursor to the right, placing it midway be- 
tween graduations 202 and 204 on the C scale, thus giving 203. [Cursor 
should now read a shade less than 71, as a check.] 4. Shift slide to 
left, placing the end under cursor. 5. Shift cursor to left, placing it 
over 835 on C scale. 6. Read 592 on D scale (see enlarged view in 
Fig. 148). Approximating, -349 X 20-3 X 8-35 is about J x 20 x 8 
= about 50. Hence result is 59-2. 

It is known from Chap. VI that the multiplication of two numbers 
by logarithms requires the addition of their logs. Now, when the 
slide is moved to the right and the left-hand end placed at (say) 
3 on the D scale (see 6, Fig. 146) the distance LF is log 3. When 
the cursor is moved to (say) 25 on the C scale, the distance FH is 
log 25 (decimal part only). Then evidently the total distance 
LH = LF + FH = log 3 + log 25. Now at H on the D scale (i. e. t 
counting from L) we read 75. Then distance LH = log 75. There- 
fore log 3 -f- log 25 = log 75 ; or since addition of logs means 
multiplication of numbers 3 x 25 = 75 as we know. 

Exercises 89. On Multiplication. 

Perform the following multiplications by slide rule 
1. 22 X 3'5- 2. -54 X 4. 3. 7fc\x -26. 

4. 33-8 x 2-1. 5. 202 x 4-95- 6. -643 x -49. 

7. 1-99 X 98. 8. 6-1 x -141. 9. -0312 x -173. 

10. 4-25 X 2-66. 11. -205 X 2-78. 12. 34-9 x 47. 

13. 2-83 x 38. 14. 10-25 x 70. 15. -4375 X *7 2 - 

16. 50-3 X 3*7. 17. -774 X 119- 18. 4-08 x 2-29. 

19. -866 X 37'3- 20. 120-5 x 8-775. 21. -285 x 3-12 X 42. 

22. 32 X 4 8 X 86. 23. 79-2 x -53 X 1-5. 

24. 115 X 1-2 X 5-175- 25. 345 X -638 x -205 X 223. 

Combined Multiplication and Division. It is in calcu- 
lations of this nature that the slide rule is particularly valuable. 



Consider an expression such as ~-rrv77~^r^ "- Now ky tne aid 

35 x 144 x 3-14 

of the previous section it would be possible to multiply up the top 
DD 



400 ARITHMETIC FOR ENGINEERS 

line, giving 2,100,000. Similarly the bottom line can be multiplied 
up, giving 15,830. Then the division -~ could be performed 

as in the previous section but one, giving 132-5. But a quicker way 
of working, in which fewer settings of the slide are necessary, is 
to start from the first figure in the top line, divide by the first on 
the bottom, multiply by the second on the top, divide by the next 
on the bottom, and so on, zigzagging from top to bottom. It will 
be noticed from the detailed method about to be given for the 
above expression, that after the first setting on the D scale, the 
slide and cursor are moved alternately, and all setting is done on 
the slide or C scale. The final reading is, of course, taken on the D 
scale, and in this case comes under the end of the slide. 

The intermediate results are not read, but to enable the student 
to check his setting at each step when learning, these intermediate 
results will be shown in square brackets to distinguish them from 
the necessary readings. 

1. Set the cursor on the D scale, to the first number in the 
top line (in this case 53). 

2. Set the slide with the first number in the bottom line under 
the cursor (in this case 35). 

[Now the left-hand end of the slide should be opposite 1513 
on the D scale.] 

3. Move the cursor, placing it on the C scale over the second 
number in the top line (in this case 33). 

[The cursor should read 499 on the D scale.] 

4. Move the slide placing the second number in the bottom 
line (in this case 144) on the C scale under the cursor. 

[The left-hand end should now read 347.] 

5. Move the cursor to the next number in the top line reading 
on the slide or C scale. 

Repeat this zigzag process until all the numbers have been used, 
when the answer is read off on the D scale, either under the end of 
the slide (see Ex. 375) or under the cursor (see Ex. 373). Finally 
approximate for the result, and fix the position of the decimal 
point. 

Example 3 7*.-Find the value of ii 

1. Set cursor at 133 on D scale. 2. Move slide to left, placing 
35 on C scale under cursor. [Right-hand end of slide should now read 
38 on D scale.] 3. Move cursor to 45 on slide or C scale. [Cursor should 
now read 171 on D scale.] 4. Shift slide to right, placing 1728 on C scale. 



THE SLIDE RULE 401 

under the cursor. [Right-hand end should now read 99 on D scale.] 
5. Move cursor, placing over 19 on C scale. [Cursor should now read 
1877 on D scale.] 6. Move slide to left, placing 4 under the cursor. 
7. Shift cursor to right-hand end of slide, and read off 47 on D scale. 

Approximating, we have ,, J ~\, -Alii 

F1 5 ' 3 x % X \ \\\\ 

= A or .^ * Approx. answer 

3 = -33 X 10 = 3-3 

/. Result = 4-7. 



Example 373. Find the value of > - = . 

r D/0 13-6 X 26 

1. Set cursor at 214 on D scale. 2. Move slide to right, placing 136 
under cursor. [Left end of slide should now read 1572.] 3. Move 
cursor to left, placing it over 125 on C scale. [Cursor should now read 
1967 on D scale.] 4. Move slide to left, placing 26 on C scale under the 
cursor. [Right-hand end should now read 756 on D scale.] 5. Move 
cursor to right, placing it at 3 on C scale. 6. As this completes the 
expression, read the result under the present position of the cursor on 
the D scale, i. e., 2268. 

Approximating, for result - -rJJ- 

1 X % \\\ 

= 3 /. Approx. answer 

= 3 X 100 = 300 
/. Result = 226-8. 

Sometimes when moving the cursor to find a certain number 
in the top line, it will not be found on that portion of the slide 
within the stock. Then an extra shift is necessary. The following 
example will illustrate : 

Example 374. Evaluate ~j *^' 

1. Set cursor at 27 on D scale. 2. Move slide to left, placing 7 under 
the cursor. [Right-hand end should now read 386.] On moving cursor 
to pick up 236 on slide, this reading will be found beyond the left-hand 
end of the stock. Then proceed thus : 3. Shift cursor to right-hand 
end of slide. 4. Move slide from left to right, placing left-hand end 
under the cursor (which is at 386). 5. Move cursor to right, placing 
it over 236 on slide. [The cursor should now read 91 on the D scale.] 
6. Move slide to left, bringing 593 under the cursor. 7. Bring cursor 
to left-hand end of slide, and read off 1536 on D scale. 

Approximating, we have 5-rr-| 

.'. Approx. answer 



sav -IA 
7' y> * = -14 x 10 = 1-4 



Result 1*536. 



402 ARITHMETIC FOR ENGINEERS 

When the bottom line contains one number more than the 
top line, then an extra shift has to be made, as shown in the 
following : 4 



Example 375. Calculate I2 '32<-4_52 

^ 0/J 25-4 x 3'35 X 14-6 



1. Set cursor at 123 on the D scale. 2. Move slide to the left, 
placing 254 on the C scale, under the cursor. [Right hand end of 
slide should now read 484 on the D scale.] 3. Move cursor to 45 on 
the slide. [Cursor should now be over 218 on the D scale.] 4. Move 
slide to the right, placing 335 on the C scale under cursor. [Right-hand 
end of slide should now be opposite 65 on the D scale.] 

This end position is the result of the working of the first four 
numbers, counting in our zigzag order ; the value has now to be divided 
by 14-6. 5. Move cursor to right-hand end of slide. 6. Shift slide to 
the right, placing 146 on the C scale under the cursor. 7. Move cursor 
to left-hand end of slide and read off 445 on the D scale. 

2 

1 X ^ 111 

Approximating, we have ^ o~ i TT 

% X 5 X A llf 

2 __ fifi .*. Approx. answer 

ra 3 ~~ = 66 X 10 = 6*6 

.*. Result = 4*45. 
Example 376. Evaluate the expression - ~ 5 2 ^ 



785 X -73 X -00546 

1. Set cursor at 525 on D scale. 2. Move slide to left, placing 785 
under cursor. [Right-hand end of slide should now be over 669 on 
D scale.] 3. Shift cursor to 8 on slide. [Cursor should now be over 
535 on D scale.] 4. Move slide, placing 73 under cursor. [Right- 
hand end should now read 733 on D scale.] 5. Move cursor to right- 
hand end of slide. 6. Move slide to right, placing 546 under the 
cursor. 7. Move cursor to left-hand end of slide, and read off 134 on 
D scale. 

5 x & 
Approximating, we have -- 



8 X $ X Jf 

^ ^ /. Approx. answer 

8=3 ? ^ " = '125 x 100000 

/. Result =* 13420. = 12500 

The student will find many examples for practice in the worked 

examples of Chaps. II, III, etc. These afford valuable practice, as 
there is a check on any errors. 



THE SLIDE RULE 403 

Exercises 90. On Combined Multiplication and Division. 

[In the answers to some of these exercises intermediate as well as 
final results are given to give a more ready check on the working.] 

1. i-oi x -98 -f- -102. 2. 11-75 x '7 1 "i- 81 

3 35:5 XJ575 4 4'35 X 25 

32-4 " i'7 X 12-3 

5 195 X 3 H 6 -975 X 10-6 

" 13 X 1-2 * 2*82 X '805 

7 2Q-4 X 6-9 X 2-34 g '0735 X 47 X 109 

25-6 X '146 X 76 * 9*55 X 13-6 X 2-96 

9 5L'75_X ^52 2< 5 10 -325 X 5'2 X 6-6 

15-3 X 7'35 * 8> 4 X 134 

11 7*05 X 7-7 12 114-5 X -0183 

"' 23-2 "' -216 

13 ___ -303 _ 14 >Q 95 _ 

1-775 X 1-16 ' -243 X 9'6 



15. - ir - 7r 16. 

785 X 14-65 16-55 

17 _ 515 __ 18 _ -9375 X i -7 
00034 x 7-6 x -4375 " ' 6 x -753 x 1-27 

19 23-8 X -94 X 114 2Q 11-05 Xj64 

2-13 X 17-1 * 1-04 X -918 

Ratios and Percentages. Special Case. A ratio or per- 
centage is easily worked on the slide rule as a division. When, 
however, several numbers have to be expressed as a fraction or 
percentage of some one number, the amount of time and labour 
necessary can be reduced by working the example as a multiplica- 
tion instead of a division. For example, let it be required to express 
each of the numbers 256, 137, 373, 400, 65, and 78 as a decimal of 
435. Now 256 as a decimal of 435 is 256 435, and similarly 
for the others, 435 being the denominator in each case. Then the 
ordinary operation of division would be as follows : 

Place cursor at each number in turn on D scale 

Set slide with denominator 435 at this position of the cursor 

Read under end of slide on D scale 

Thus, the slide has to be shifted each time. 

But ~5- = 256 x -i- and similarly for each of the other numbers. 
435 435 

Thus we have, decimal required = each number x . Now 

435 435 

is a constant quantity, and if the end of the rule be placed at this 
reading on the D scale, all the multiplying will be done without 



404 ARITHMETIC FOR ENGINEERS 

shifting the slide, as the cursor will be placed at various positions 
on the slide or C scale and the result read on D. It is not necessary 

to know the value of ; the position of the end of the slide is 

found automatically if 435 on the C scale be placed at i or 10 on 
the D scale. After that only the cursor need be moved, but some- 
times the slide may have to be shifted once. 

The numbers to be expressed as fractions or percentages will 
generally be somewhat alike in size, and thus once the decimal 
point is fixed by approximating for one number, the others will 
follow naturally. 

The following example will illustrate the method : 

Example 377. Express each of the following numbers as a decimal 
of 435 : () 2 5 6 ; ( b ) 137; (c) 73i (d) 4; M 65; (/) 378. 

Move slide to right, placing 435 on the C scale at the extreme end 

of the D scale. Then the left-hand end of slide gives the figures of - 

435 

Now numbers greater than 435 will be found beyond the end of the 
D scale, and cannot be taken on this setting, but will require the slide 
to project to the left. Then divide the numbers into two groups, one 
set being all below 435 and the remainder above. Indicate one set 
by, say, a tick. 

Taking the numbers below 435 in turn, i. e., a, b, d, and /. Then for 
(a) set the cursor at 256 on the C scale and read 588 on the D scale. 
The result is evidently about J. Then 256 is -588 of 435. 

For (b) move cursor to 137 on C scale and read 315 on D, Result 

is -315- 

For (d) set cursor at 4 on C scale and read 92 on D. Result is -92. 
For (/) set cursor at 378 on C and read 869 on D. Result is -869. 

(Note that the slide has not been moved the whole time, but it must 
now be moved to the left, with 435 on C scale at i on D scale.) 

For (c) set cursor at 73 on C scale and read 1677 on D. Result is 

-1677. 
For (e) set cursor at 65 on C scale and read 1493 on D. Result is 

'1493- 

Exercises 91. On Percentages. 

Work by slide rule, Exercises 18, Chap. II. 

Square Root. The actual operation upon the slide rule when 
extracting square root is very simple indeed, but difficulty may be 
experienced at first in fixing the position of the decimal point. 



THE SLIDE RULE 405 

The slide is not used at all, only the extreme top or A scale, and the 
extreme bottom or D scale being employed. It will be seen in 
Fig. 149 that the top scale is divided into two halves or scales each 
exactly the same, and similar to the bottom scale, except that they 
are not so finely subdivided, since the space is more limited. The 
division and method of reading these scales will be readily understood 
from the diagram. The left-hand half of the top scale will be called 
the first half, and the right- hand the second half. 
To extract the square root proceed as follows : 

1. Start from the decimal point and mark off the digits in pairs, to 
the left only if a mixed number, and to the right if wholly a decimal. 

This is similar to the pointing off, when finding a square root in 
the ordinary way, on p. 102. 

2. K the extreme left-hand period contains one digit use the 1st half 
of the top scale ; it it contains two digits use the 2nd half. In the case 
of a number wholly a decimal, any noughts immediately after 
the decimal point are not considered as digits, as they are not 
significant figures. 

3. Place the cursor on the A scale, at the number, in the particulai 
half decided by (2) above, and read under the cursor on the D scale. 
This gives the significant figures in the square root. 

4. Finally fix the decimal point in the result according to the 
following : 

(i) When the number is a mixed number. There will be as many 
figures in the whole number of the square root as there are periods 
in the number. 

(ii) When the number is a decimal. There will be as many 
noughts immediately after the decimal point in the square root 
as there are complete periods of noughts in the number. 

It is this setting of the decimal point which is found trouble- 
some, and to better understand the above rules the following two 
examples should be worked through carefully : the first deals 
with the case when the number is a mixed one, and the second when 
the number is wholly a decimal. In any case a rough squaring 
of the root will give a check on the results. Should the student 
find that the graduations of the B and C scales on the slide confuse 
him, when reading from the A to the D scale, he should pull the 
slide out -some distance so as to leave clear the region in which 
he is working. 

Example 378. Extract the square roots of the following numbers, 
with a slide rule: (a) 2250; (b) 259; (c) 44100; (d) 53-7; (e) 3-3. 



4O6 



ARFTHMETIC FOR ENGINEERS 




THE SLIDE RULE 407 

(a) Mark off in pairs from the decimal point thus : '22 '50. The 
left-hand period contains two figures, and therefore the second half of 
the upper scale must be used. Place cursor over 225 on right of A scale 
and read 474 on D scale (see A, Fig. 149). As there are two periods 
in 7 22'5O, there will be two figures in the whole number of our result. 
Thus required square root is 47-4. 

(b) Mark off in pairs, giving 2 '59. Left-hand period has only 
one figure, therefore first half of upper scale must be used. Place cursor 
over 259 on left of A scale, and read 161 on D scale (see B, Fig. 149). 
As there are two periods in 2'59, there will be two figures in the whole 
number of our result. Thus required square root is i6-i. 

(c) Point off, giving 4 / 4i / oo, and as left-hand period has only 
one figure, use first half of upper scale. Under 441 on left of A scale 
we read 21 on D scale (see C, Fig. 149). As there are three periods in 
4 / 4i'oo, then three figures must exist in the result. As two only exist, 
a nought must be added and the required square root is 210. 

(d) Point off, giving '53-7. Left-hand period contains two figures, 
therefore second half of scale A is used. Under 537 on right of A scale 
we read 732 on D scale (see D, Fig. 149). There is one period in the 
whole number of '53 -7, so there will be one figure in the whole number 
of the result. The required root is then 7-32. 

(e) Point off, giving '3-3. Left-hand period contains only one figure, 
therefore first half of scale A is used. Under 33 on left of scale A, 
read 1815 on D scale (see E, Fig. 149). There is only one period in the 
whole number of '3-3, hence there will be one figure in the whole number 
of the result. Required square root is then 1-815. 

Example 379. Extract the square root of the following numbers, 
with the slide rule : (a) '565; (b) -043; (c) -0085; (d) -00012. 

(a) Pointing off in pairs to the right from the decimal point gives 
56'5- Left-hand period contains two figures, hence second half of scale A 
is used. Place cursor over 565 on right of A scale, and read 751 on 
D scale (see F, Fig. 149). In the number -56' 5 there are no noughts 
following the decimal point. Hence there will be no noughts after the 
decimal point of the result. Thus the required square root will be 

75*1 

(b) Point off, giving -O4'3. Left-hand period contains one figure (the 
nought not being significant), hence first half of scale A is used. Under 
43 on left of scale A read 207 on D scale (see G, Fig. 149). A period 
consists of two digits, so there are no complete periods of noughts in 
04 X 3. Thus there will be no noughts after the decimal point in the 
result. The required square root is then -207. 

(c) Point off, giving -oo^. Left-hand period, with significant 
figures, contains two figures, hence second half of scale A is to be used. 
Under 85 on right of A scale read 921 on D scale (see H, Fig. 149). 



4 o8 ARITHMETIC FOR ENGINEERS 

There are two noughts after the point in the number -0085 (t. e. t one 
complete period of noughts). Hence there will be one nought following 
the point in the result. The required root is therefore -0921. 

(d) Point off, giving 'Oo'oi'2. The left-hand period, containing a 
significant figure, has one figure. Hence first half of scale A is used. 
Under 12 on left of scale A read 1095 (see K, Fig. 149). In the number 
oo'oi'2 there is only one complete period of noughts following the 
decimal point, so there will be one nought following the point in the 
result. The required square root is therefore '01095. 

Exercises 92. On Square Root. 

Find by slide rule the square roots of the following numbers, and 
say in each case which half of the top scale you would use 

1. 2600. 2. 44500. 3. 1-55. 4. 213. 

5. 98-5. 6. 737. 7. -0505. 8. -0034. 

9. 11-7. 10. -307. 

Extract the following square roots by Slide Rule. In each case 
check your result by multiplication 

11. -^250. 12. ^73' 13. V-3Q. 14. 

15. Vio6. 16. ^5^25. 17. ^-187. 18. 

19. W-I75. 20. V30-7. 21. ^289. 22. 

Squaring. This being the reverse of extracting square root 
the operations are reversed, but are much less troublesome, as 
there is no need to consider which half of the top scale A, must 
be used. As before, only the A and D scales are needed, and the 
procedure is as follows : 

1. Place the cursor over the D scale, at the number to be squared. 

2. Read on the A scale the significant figures in the result. 

3. Fix the position of the decimal point by approximating for a 
result. 

Example 380. Find by aid of the slide rule the value of the 
following: (a) 2-37*; (b) 3I-6 2 ; (c) 66-3 2 . 

(a) Place cursor at 237 on the bottom or D scale, and read 562 
on the top or A scale. Now 237 2 must be rather more than 2 2 or 4. 
Hence our result must be 5-62. 

(6) Place cursor at 316 on the bottom scale and read i (i. e.> the 
central figure) on the top scale. Now 31-6* must be rather more than 
30 2 , which is 900. Then the result must be 1000. 

(c) Place cursor at 663 on the bottom scale and read 44 on top scale. 
Now 663 2 must be more than 6o 2 , which is 3600. Hence the result must 
be 4400. 



THE SLIDE RULE 409 

In some cases, when the result appears in the crowded portion 
of the top scale (as from about 7 to 10) it is difficult to estimate 
a third figure reasonably correctly. In these cases, if an accurate 
answer is desired, it is better to work the square as a multiplication 
example on the bottom scale, where finer estimation is possible. 
Thus taking 3I-6 2 , we multiply 31*6 x 31-6 on the C and D scales, 
reading off 999. The actual result to four significant figures being 
998-6, this method gives a more accurate result than the former. 

Exercises 93. On Squaring. 

Square Examples i to 10 in Exercises 92. 

Further practice may now be obtained by working more com- 
plicated exercises such as those in Exercises 64 and 65. 

Reading Logarithms and Antilogarithms from the Rule. 

Of the three scales on the back of the slide one is a scale of equal 
parts ; it is generally the middle scale and is labelled L. This is 
used in conjunction with the A scale as a logarithm table, but the 
accuracy is limited to 3 decimal places. It must be noticed that the 
L scale reads from right to left, and is always used with the slide 
projecting to the right. The main divisions, which are numbered 
(the right hand end graduation is O), give the first decimal place in 
the logarithm, the next sub-divisions give the second decimal 
place, while the smallest graduations give every even third place, 
the odd figures in the third place being estimated as at the centres 
of the smallest sub-divisions. 

To find a logarithm (decimal part), e. g., log 2-6. 

1. Move the slide to the right, placing its left-hand end at the 
significant figures of the number, i. e., at 26, on the A scale. 

2. Turn the rule over (not end for end) and read the decimal 
of the log on the L scale opposite the appropriate small mark in 
the U-shaped gap at the right. The figures 415 should be found. 
Then the decimal part of log 2*6 is -415, and the whole number 
must be chosen in the usual way. Then log 2*6 = 0-415. 

To find an antilog. (significant figures) e.g., antilog 1-658. 
Reverse the order of operations in the previous paragraph. 

1. Turn the rule over, moving the slide to the right, and set 
the decimal of the log., i.e., 658, to the small mark in the gap. 

2. Turn the rule back to its normal position, and read on the 
A scale under the left-hand end of the slide the figures 455, which 
are the significant figures in antilog 1-658. The decimal point 
must be placed in the usual way. Then antilog 1-658 = 45-5. 



CHAPTER XI 

TRIGONOMETRY 

The Trigonometric Ratios. In the Mensuration chapter the 
right-angled triangle is referred to as being of considerable importance 
it is, in fact, the basis of a large branch of mathematics known as 
Trigonometry, which is of supreme importance in all branches of 
engineering and physical science. 

At X, Fig. 150, are shown several right-angled triangles of 
various sizes, but all having the same acute angle A. The measured 
lengths of the various sides are shown. If now for every triangle, 
such as ABC, we calculate the value of the ratio 

vertical side (such as BC) 
inclined side (such as AB) 

we find the following results 

In triangle ABC, ?= ^1' = 0-847 

' 



T 4. 1 ATYIT 2'08" 

In triangle ADE, =-7> = 0<8 49 



T . . . ATV , FG 2-88" Q 
In triangle AFG, ^ = -^77- = 0-847 

allowing for small errors in drawing and measurement, the value 
of this ratio is evidently constant. 

If another set of right-angled triangles be taken, as at Y, having 
a different common acute angle, a similar result will be found; 

the ratio . ^ -5 ^- is 0-839 for all three triangles. This ratio 



has evidently a different value for different angles, but is always 
the same for the same angle whatever be the size of the triangle. 
A xt. x- horizontal side , . . A , A , 

Again, the ratio ~{^ifHedl>ide~ ma ^ calculated; thus at 

410 



TRIGONOMETRY 



411 



X in Fig. 150, -j-g- = ~-n = 0-529 and this ratio also will be found 

to be constant for all right-angled triangles having the same acute 
angle. These ratios are instances of what are called the trigono- 
metric ratios of the angle at A. 

Referring now to (Z) Fig. 150, the sides of the right-angled 




Fig. 150. Right- Angled Triangles. 

triangle are given more general names with reference to their 
positions regarding the angle A ; thus, 

The side BC, lying opposite to the angle A, is the opposite side. 

The side AC, the shorter side enclosing the angle A, is the 
adjacent side. 

The side AB, the longest side, is already known as the 
hypotenuse. 

Of the various ratios that may be taken among the three sides, 
the three following are the most important : 



4 i2 ARITHMETIC FOR ENGINEERS 

^t . opposite side BC .. , ,. 

The ratio *~~ = -r-p called the sine. 

hypotenuse AB 

_ t .. adjacent side AC n , ., 

The ratio ~ . = -r-^ called the cosine. 

hypotenuse AB 

TI. j.* opposite side BC , ,, , . 

The ratio -- 7-^- = -r-~ called the tangent. 

adjacent side AC 

For convenience in writing in formulae, etc., these names are 
abbreviated into the following forms : 

sin A for sine of the angle A. 
cos A for cosine of the angle A. 
tan A for tangent of the angle A. 

The values of these ratios for all angles could evidently be 
obtained by careful drawing and measurement and subsequent 
division, a right-angled triangle being formed for every angle 
dealt with by drawing a line perpendicular to either side to cut 
the other side. Such a method would be liable to errors of drawing 
and measurement which for certain angles would be of serious 
proportions, and in all cases would be sufficiently large to limit 
the accuracy of the resulting quotients to a greater extent than 
would be permissible for many applications. It is possible, how- 
ever, to calculate the values of the ratios, to any desired degree of 
accuracy, by the methods of advanced mathematics, and the values 
so obtained are recorded in various tables for all angles at intervals 
of every second. Such tables are naturally very large, and various 
abridged forms are obtainable. For many practical problems it 
is sufficient if the tables give the values of the ratios for every 
single degree of angle; in this case the tables can be arranged to 
occupy a single book page and are reproduced on p. 458. 



Exercises 94. On the Trigonometric Ratios. 

1. Make a sketch of a right-angled triangle ABC, with the right 
angle at B. Which trigonometric ratios of the angle A will be found 
by evaluating the following ratios : 

/ % BC /TV AB / \ BC 

(ft) AB (&) AC (C} AC 

and what ratios of angle C are represented by 
/x7X AB / v BC 

(d) EC" W AC" 



TRIGONOMETRY 



413 



2. A right-angled triangle PQR is shown fully dimensioned in 
Fig. 151. Calculate the values of the follow- 
ing ratios for the angles P and R : 

(a) tan P, (b) cos R, (c) sin P, (d) cos P, 
(e) tan R, (/) sin R. 

3. Draw very carefully angles of 35, 68, 
20, 47 and 73, and from actual measure- 
ment calculate the values of the following 

Fl " i e i 

trigonometric ratios : & ' 

(a) sin 20, (b) tan 35, (c) tan 47, (d) cos 68, (e) sin 73, (/) cos 35. 

Ratios of the Angles 30, 45, 60. These angles are of fre- 
quent occurrence and their ratios will therefore be considered before 
examining the complete table. The three angles are shown in 






866 



I 70 7 ( 




Fig. 152. Trigonometric Ratios of Certain Important Angles. 

Fig. 152, the hypotenuse of the right-angled triangle being taken 
as i in each case. 

45 60 



30 
sin 30 = -5 = .5 

cos 30 = ^=-866 



sin 45 =-^^=-7071 



cos 45 = 



tan 45 = 



-7071 



sin 60 = 



866 



-=866 



cos 60 = - 5 - = -5 
tan 60 =^=1-782 



It should be carefully noted that sin 60 is not twice sin 30. 
In fact, no trigonometric ratio increases in the same proportion 
as the angle. This point is again referred to on p. 432. 

Ratios of the Angles and 90. At X, Fig. 153, is shown 
a very acute angle, viz. one of 5, across which has been drawn a 
line BC, perpendicular to the side AC. The values of the ratios 
" sine/' " cosine/' and " tangent " for the angle A are expressed, 
of course, as 




414 ARITHMETIC FOR ENGINEERS 

. A BC A AC , . BC 

sin A = AB cos A = AB tan A = ^ 

It should be noticed (a) that the length of BC is small compared 
with the length of AB, and also with the length of AC, and (b) that 
the length of AC is slightly less than, but nearly equal to, the 

length of AB. The values of the 
ratios sin A and tan A will then be 
small since the numerator is in both 
cases much smaller than the de- 
nominator ; actually the values are 
0-0872 and 0-0875 respectively. In 
the case of the cosine, the numera- 
tor and denominator are very nearly 
equal, although the latter is the 
larger, so that the ratio is very 
nearly i ; actually it is 0-9962. 

Now imagine that the angle A 
becomes steadily smaller owing to 
the line AB swinging about the point 
A towards the side AC. Then BC becomes still smaller, until when 
AB and AC are coincident, i.e the angle is zero, BC = o. At the 
same time, the length of AC increases and gets nearer and nearer 
to the length of AB until, when the angle A is zero, AB and AC 
are equal in length and are, in fact, identical lines. With an angle 
of o, therefore, the length of the " opposite " side of the right- 
angled triangle is o and the length of the " adjacent " side is 
equal to the length of the " hypotenuse/' The values of the 
three trigonometric ratios are then 

sin = PP site side ^ = o 

hypotenuse hypotenuse 

since the result of dividing nothing by any number is still nothing. 
Similarly, 

ten O o = opposite side = _^_p = Q 

adjacent side adjacent side 
But, 

A0 adjacent side . . ,, , , , , . 

cos = -y^ = 1, since the top and bottom quantities 

are equal. 

An angle of 85 is shown at Y, Fig. 153, lettered in the same way 
as the preceding angle of 5. In this case it is the " adjacent " 



TRIGONOMETRY 415 

side AC which is small, while the " opposite " side BC is nearly 

equal to the hypotenuse. The value of the cosine ( = , -, ) 

^ - L \ hypotenuse / 

is then small, being 0-0872, while the value of the sine 

( = c v ) is very nearly i, being 0-0062. The value of the 

V hypotenuse/ y y ' b ^ 

tangent is comparatively large, since it is the result of dividing a 
long length (opposite side) by a short length (the adjacent side) ; 
the actual value is ii'43. 

If now this angle be increased until it ultimately becomes 90, 
then the adjacent side BC diminishes to zero and the opposite side 
becomes equal to, and coincident with, the hypotenuse. The 
values of the three ratios then become 

o opposite side . . ,, , , , , , .... 

sin 90 = ~ -= 1, since the top and bottom quantities 

hypotenuse , r ^ 

are equal. 



c an adjacent side 
cos 90 = , . - = 



0. 



hypotenuse hypotenuse 

opposite side opposite side 

tan 00 = y , . , - = -*-*- . 

^ adjacent side o 

Now it is impossible to say what value is obtained by dividing 
any number by nothing. The result is larger than any number that 
we can think of and is termed infinity ; it is usually denoted by 
the symbol oo . 

Then tan 90 = oo . 

Relation between Sine and Cosine of Complementary 
Angles. A very important relation exists between the sines and 
cosines of any two angles which together total 90. Such angles 
are termed " complementary angles " and each is said to be the 
" complement " of the other. Thus 30 is the complement of 60, 
20 is the complement of 70, 45 is the complement of itself and 
so on, the sum of the two angles being 90 in every case. Thus 
the complement of any angle is the amount by which the angle 
differs from 90. 

Referring back to the right-angled triangle at (Z), Fig. 150, 
the values of the ratios sine and cosine of the angle A have already 
been expressed in terms of the sides AB, AC, BC. It is also possible 
to express the values of the same ratios for the angle B in terms 
of the same sides. The general names affixed to the sides in the 
figure will need changing over when referring to the angle B; 

EE 



416 ARITHMETIC FOR ENGINEERS 

thus although AB will still be the hypotenuse, AC will become 
the " opposite " side and BC will be the " adjacent " side. 

. ... . . , (opposite) AC 
-Ihen sin B will be given by ^__ = ^ 

T-, -11 t. i. (adjacent) BC 

and cos B will be given by ;- = AB' 



AC 
But the ratio ^ is already known as cos A. 

A. ij 

T>/" 

and the ratio -pg is already known as sin A. 
AJb> 

Thus, in the triangle ABC of (Z), Fig. 150, 

T>/- 

sin A = xg = cos B. 

AC 

and cos A = ^5 = sin B. 

Now the three angles of any triangle always total 180 (refer 
p. 264). Then, since the angle C is a right angle, i.e. = 90, the 
remaining two angles A and B must together be equal to 90, 
i.e. the angles A and B are " complements " of each other, 
B = 90 A and A = 90 B. 

Then, substituting (90 A) in place of B in the relations above, 

sin A = cos (90 A) 
cos A = sin (90 A) 

or in general terms, the sine and cosine of an angle are equal to the 
cosine and sine, respectively, of its complement. 

Thus sin 30 = cos 60 sin 20 = cos 70 

cos 30 = sin 60 sin 45 = cos 45 

and so on. 

The last equation is easily seen to be true on referring to any 
45 right-angled triangle, in which the " adjacent " and " opposite " 
sides are equal. 

The Table of Trigonometric Ratios, The abridged table 
giving the values, to 4 decimal places, of the various ratios for angles 
from o to 90 at intervals of one degree appears on p. 458. By 
making use of the relation discussed in the preceding section, it 
is possible to tabulate the values of both sine and cosine with only 
91 values for the 91 angles involved, in place of the 182 values that 
would be necessary were the table arranged in a purely straight- 



TRIGONOMETRY 417 

forward manner. The method of arrangement will be easily 
understood after a little use of the table. 

The table contains 10 columns, of which the two outer columns 
contain the values of the angles in degrees, while the four central 
columns contain the values of the sines, cosines and tangents of 
the angles. From o to 45 the values of the angles appear in 
the extreme left-hand column and read downwards, while the angles 
from 45 to 90 appear in the extreme right-hand column and read 
upwards. 

From to 45 the values of the sine, cosine and tangent are 
given in the columns so headed at the top of the table. Thus in 
the column headed " Sine " at the top, the figure -5000 is found 
opposite the angle 30 in the extreme left-hand column ; then 
sin 30 = 0*5. On the same horizontal line, in the columns 
headed " Cosine " and " Tangent " at the top, are found the values 
8660 and *5774 respectively ; then cos 30 = 0-866 and tan 30 = 
0-5774- 

From 45 to 90 the values of the various ratios are given in 
the appropriate columns headed at the foot of the table. Thus 
the value of sin 75 is found by taking the value opposite 75 on 
the extreme right, but in the column headed " Sine " at the foot of 
the table, viz. -9659. 

Example 381. Read the values of the following ratios from the 
table : sin 81, sin 10, tan 55, cos 37, tan 5, cos 63. 

Sin 81. This angle is over 45, and therefore the right-hand column 
of angles will be referred to and the value taken from the column 
headed " Sine " at the foot of the table. Opposite 81 on the 
extreme right and above " Sine " at the foot is found the value 
9877. Then sin 81 = -9877 

Sin 10. Being smaller than 45, the angle will be found in extreme 
left-hand column, and the column headed " Sine " at the top is 
used. Opposite 10 on the extreme left and under " Sine " at 
the top is found -1736. Then sin 10 = 0-1736 

Tan 55. Refer to right-hand column for angle and the heading 
" Tangent " at the foot. Opposite 55 and above " Tangent " 
is found the value 1-4281 

Cos 37. Use left-hand column of angles and top headings. Opposite 
37 and under " Cosine " is found the value . . . -7896 

Tan 5. Use left-hand column and top heading. Opposite 5 and 
under top heading of " Tangent " find .... -0875 

Cos 63. Use right-hand column and bottom titles. Opposite 63 
and above "Cosine," find -4540 



418 ARITHMETIC FOR ENGINEERS 

Exorcises 95. On reading Trigonometric Ratios from the 

Table. 

Find, from the table on p. 458, the values of the following ratios : 
(i) cos 15, (2) tan 71, (3) sin 18, (4) sin 44, (5) sin 46, 
(6) tan 28, (7) tan 89, (8) cos 46, (9) cos 77, (10) cos 4. 

It is frequently necessary to determine the angle for which 
one of the trigonometric ratios has a known or given value, i.e. 
the table has to be used in the reverse order. It must be remembered 
that in each case two columns are devoted to the values of each 
of the ratios tabulated, one headed at the top and the other at 
the bottom of the table. Then, if the name of the ratio is at the 
top of the column "in which the given value is found, the angle must 
be read from the extreme left-hand column ; but if the name of 
the ratio is at the foot of the column, then the angle must be read 
from the extreme right-hand column. 

In cases where the exact value given is not found in the table 
it will be sufficient for the purposes of this book if the nearest value 
is referred to. 

Example 382. Find from the table the angles whose sines have the 
following values : 
(a) -2079; (b) -9063; (c) -7071; (d) -6947. 

(a) An examination of the two " Sine " columns shows that the value 

2079 occurs in the column whose title is at the top. The angle 
must then be read from the left-hand column and is . .12 

(b) The sine -9063 occurs in the column marked " Sine " at the bottom. 

The angle must then be read from the right-hand column and 
is 6 

(c) The sine -7071 is found in both columns of sines and 45 is read 

in both angle columns. 

(d) The sine -6947 is found in the column marked " Sine " at the top. 
The left-hand column of angles is then referred to and gives the 
value ........... 44 

Example 383. Find the angles whose tangents are (a) 1-8807, 
(b) -1051, and also (c) the angle whose cosine is -5592. 

(a) The tangent 1-8807 occurs in the column with title " Tangent " 

at the foot. The corresponding angle is then read from the right- 
hand column and is 62 

(b) The tangent -1051 is found in the column labelled " Tangent " at 

the top. The left-hand column is then referred to and gives the 
value 6 



TRIGONOMETRY 419 

(c) The cosine -5592 occurs in the column entitled " Cosine " at the 
foot of the table. The right-hand column of angles is then referred 
to and gives the value ........ 56 

Exercises 96. On reading the Trigonometric Table 

(continued). 
Find the angles whose cosines are : 

(I) -4384; (2) -9703: (3) -77?i; (4) -0349- 
Find the angles whose tangents are : 

(5) -1405; (6) 2-6051; (7) 1-0724; (8) -6009. 
Find the angles whose sines have the following values : 

(9) -3090; (10) -7547; (n) -5150: (12) -9511- 

Use of the Trigonometric Ratios in Formulae. The 

various ratios are frequently met with in engineering formulae, and 
in a great many such cases no more knowledge of trigonometry is 
required beyond the ability to read values from the tables, and once 
the numerical values have been substituted the expressions require 
only ordinary evaluation. 

Certain points of writing must be emphasized, however, to 
ensure that the ordinary algebraic rules of expression are still 
complied with. 

Thus, if a trigonometric ratio has to be raised to any power, 
the index must not be written at the right-hand top corner of the 
abbreviated expression for the ratio, unless brackets be used. 
Thus sin A 2 does not represent the square of " the sine of angle 
A " ; reading this expression in strict accordance with the rules 
of symbolic expression explained in Chapter III, it expresses the 
sine of the " square of angle A." The error is rectified if the 
expression be written as (sin A) 2 , but this form is never used, and 
the general practice is to write the index immediately after the 
abbreviated name of the ratio, thus sin 2 A. This is read as " sine 
squared A " and indicates that the sine of angle A is to be squared. 
Emphasizing by a numerical example, suppose it necessary to 
evaluate the " square of sin 8." Now sin 8 = -1392, and we 
then write 

sin 2 8 = (-I392) 2 == -0183. 

If the form sin 8 2 had been used, then, in strict accordance with 
principles, it would follow that 

sin 8 2 = sin 64 = -8988, 
which is widely different from what was intended. 



420 ARITHMETIC FOR ENGINEERS 

When the angle involved in the expression of a ratio is com- 
pound as, for example, a sum or difference of two angles, then the 
compound form must be enclosed in brackets. For example, the 
sine of the angle produced, say, by adding angles denoted by P 
and Q must be expressed as sin (P + Q) ; the form sin P + Q 
has not the meaning intended. 

In cases where any ratio of an angle is to be multiplied by a 
factor, then the factor should always be written in front of the 
ratio name. Thus " twice the sine of angle A " should be written 
as 2 sin A and not as sin A X 2 ; the latter form is ambiguous, 
as it may also mean " the sine of twice the angle A." If it is 
desired to express the latter idea the appropriate form of expression 
is sin 2 A. It has already been stated, and it will be further 
explained later, that no ratio increases in proportion to the angle, 
so that the expressions 2 sin A and sin 2 A indicate quite different 
values, as may easily be seen by giving any value to A, and evaluat- 
ing the two expressions. 

Expressions of the form cos are also met with and should be 

written very carefully to avoid any possible confusion with the 

- cos A , . , i A 

form which means - cos A. 

2 2 

Example 384. The power developed, in watts, by an alternating 
electric current is given by the expression AV cos 9, where A = current 
in amperes, V = voltage, and 9 = the " angle of lag." If A = 1-05, 
V = 500 and 9 = 32, find the power developed. 

Power = AV cos 9 
Substituting given values 



= 1-05 X 500 x cos 32 
= 1-05 x 500 x -8480 



From Table, 
cos 32 = -8480 



= 445-2 watts. 
Example 385. The efficiency of a screw and nut, where 6 is the angle 

of the thread and 9 is the "angle of friction," is equal to , -^- . 

^ tan (o-|-9). 

Calculate the efficiency for a case where 6 5 and 9 = 8. 

_- . tanO 

Efficiency = an (0 + 9) - 

o , , ., j_- , tan 5 tan 5 

Substituting given values- = ^ ( ~ + ~ 8} = g^ 

Taking values from table = ^-~ = '37 S 9> say -379. 



Example 386. When choosing a stock gear wheel cutter for milling 
spiral wheel teeth, the cutter is chosen by reference to the number of 



TRIGONOMETRY 421 

teeth in an " equivalent spur wheel/' The number of teeth in this 

N 
" equivalent spur wheel " is j- t where N = number of teeth required 

in spiral wheel and a = angle of the spiral teeth. If N = 12 and a = 
45, find the value of the given expression, taking the nearest whole 
number. 

Number of teeth in " equivalent spur wheel " 

= N 
~~ cos 3 a 

Substituting given values = ^ = , I2 ... 
6 b cos 3 45 (-707I) 3 

= ; 3 * ~r = 33'94. *. 34 teeth. 

Example 387. The following equation relates to the combination 
of stresses in a piece of material : 



cos 20 = --. 

P + 3 

If p = 5-05 and q = 1-05, find the value of the angle 9. 



Substituting- 



COS 20 = '6558. 

Referring to the table, the value -6558 does not appear, but the 
nearest cosine value is -6561, which is cos 49, 

20 = 49 



Exercises 97. On Evaluation involving Trigonometric 

Ratios. 

1. When a ray of light is bent in passing through a piece of glass, 
the value of the expression r is termed the " refractive index " of 

the glass, where i denotes the " angle of incidence " and r denotes 
the " angle of refraction." Calculate the value of the refractive index 
for the case where i = 31 and r = 20. 

2. The expression wh ~x"^ 9 relates to earth pressure on retain- 
ing walls. Find its value when w 125, h s* 20, and 9 = 22. 

3. The effective helix angle of an aeroplane propeller at various 
radii is given by the expression tan A = - where A = helix 



422 ARITHMETIC FOR ENGINEERS 

angle, V = translational velocity of the propeller, n = speed of 
rotation, r = radius. Determine the value of A when V = 190-8, 
n = 16-67, an d Y = 2-25. 



2.t 

4. The formula tan 26 = ___ relates to " principal stresses." If 
P 4'3' <7 = 3*63, and t -63, find the value of the angle 0. 

5. The depth of a wall foundation is given by the expression 

W /i sin cp\ 2 



W /i sin 9\ 2 
~B>w \i -f- sin 97 



Find the value of the expression when W = 4500, B = 2, w = 125. 
and 9 = 21. 

6. The velocity of the piston in a direct-acting engine is given by 
the expression vYsinG-j-- ) where V velocity of crank pin, 

6 crank angle, and n = ratio of connecting rod length to crank 
length. If V = 7-9, = 38, and n 4-5, find the value of the 
expression. 

7. When measuring distances by a tacheometer (a surveyor's 
instrument) the expression Cs cos 2 -f- k cos occurs. If C = 100, 
s = i -i 8, k = 1-25, and = 28, find the value of the expression. 

8. In connection with Question 7, above, the expression sin 26 + 

k sin 6 also occurs. Using the same values as in Question 7, find the 
value of this expression. 

P 

9. The formula ^ tan (p a) refers to an inclined plane. If 

P = 4-05, W = 58, and a 7, find the value of the angle p. 

10. The angle of a conical plug is given by the expression 

tan = 7 where D and d are the large and small diameters 

2 21 

respectively, and / is the axial length between them. Find the angle 
of a plug in which D = 2-5, d = -95, and / = 2-25. 

Problems involving Simple Trigonometry. Problems 
frequently occur, particularly in drawing office and tool room, which 
require for their solution the use of the ^principles discussed in the 
earlier portions of this chapter. As with most examples of problem 
nature, it is impossible to give a rigid method by which such prob- 
lems may be solved, and the process must be assimilated by working 
through as many illustrative examples as possible. 

In all cases a sketch showing the conditions of the problem 
must be made, in which right-angled triangles should be sought 
for and, if necessary, introduced by the addition of suitable lines. 
A knowledge of geometry is of great assistance in this connection. 
From the various right-angled triangles then available, some 



TRIGONOMETRY 



423 



trigonometric ratio of some angle can usually be expressed in terms 
of one known side and one unknown side, whence the value of 
the latter may be calculated. A few illustrative examples are 
appended. 

Example 388. The height of a factory chimney stack is to be 
found. A theodolite (i.e. an angle-measuring instrument) is set up 




AD o 

Fig. 154. Diagrams for Examples 388-392. 



at a point 200 ft. away from the base of the stack on level ground, and 
the top of the chimney is sighted, as indicated at (A), Fig. 154. The 
" angle of elevation " is found to be 42. Calculate the height of the 
chimney if the theodolite stands 4 ft. high. 

A right-angled triangle is evidently formed by the centre line of 
the stack (vertical), the line of sight, and the horizontal through the 
theodolite centre. In this triangle PQR, the side QR and the angle R 
are known, and the length of PQ is required. 



424 ARITHMETIC FOR ENGINEERS 

Then since ,~ = tan R 

G R 

PO 

- tan 42 = -9004. 

200 ^ * ^ 

PQ = -9004 x 200 = 180-08', say 1 80 ft. 

The height of the stack is evidently 4 ft. greater than this, since 
the point Q is 4 ft. above ground. Therefore the total height of the 
stack = 184 ft. 

Example 389. The proposed dimensions of a small ball bearing 
are given at (B), Fig. 154. Assuming the balls to be evenly spaced, 
calculate the clearance between two adjacent balls. (Note. This 
clearance must be measured along the straight line joining the centres 
of the two adjacent balls.) 

A sketch showing two adjacent balls to a larger scale is shown at 
(C), Fig. 154, in which the ball centres are lettered A and B respectively, 
and the centre of the inner race is lettered O. 

Then on the assumption that the 9 balls are evenly spaced, the 
angle BOA is * of a complete circle 

i.e. angle BOA = 3-^- = 40. 

The triangle AOB is " isosceles/' i.e. side OA side OB. Then, 
if a line OC be drawn perpendicular to AB, this line OC will divide 
both the angle BOA and the side AB into equal parts. 

angle COA = a (say) = 4!L 2 o. 
Now in the triangle AOC, 

AC 
^ 

and OA = radius of inner race -f- radius of ball 



AC 

-- = s i n 20 = -3420 

219 ^7 

whence AC = -3420 x -219" = -0749". 

Now DC = AC ball radius 

= -0749 -069 = '0059" 
and DC is half the actual clearance, 

/. Clearance = 2 X '0059 = -on 8" 

say -012" to the nearest thousandth. 

Example 390. A gauge was required of the form and dimensions 
shown at (D), Fig. 154. The method of working was to first make a 
complete isosceles triangle AEB as shown at (E), Fig. 154, the sides 



TRIGONOMETRY 425 

AB, AE, and BE, and also (as a check) the height, EG, being measured 
by micrometer. The top triangle EDG was then cut off until the 
required height of the remaining piece was obtained. 

Calculate the dimensions AB, AE (which equals BE), and EG, to 
the nearest thousandth of an inch. 

If the dotted line DH be drawn perpendicular to the base AB, then 
the triangle ADH is similar to the triangle AEG, and the angle ADH = 
10, i.e. half the apex angle, since the triangle AEB is symmetrical 
about the centre line EG. 

Then, in triangle ADH 

as 

.'. . = 'i7 6 3- .*. AH = -1763 x 1-2 = -2116". 

Then the half width at the bottom, 

= AG = AH + HG 

= -2116 -f- -15 = -3616". 

Then the whole width AB = -3616 x 2 = -7232" 

say > 723 // . 
Now in the triangle AEG, 





... '^=,736 
Also, in the triangle AEG, 

' 



The required dimensions are then 

AB = -723" AE = 2-083" EG = 2-051" 

Example 391. The movement of a rocking arm in a certain instru- 
ment is opposed by the extension of a helical spring, arranged as 
indicated at (F), Fig. 154. Determine the extension of the spring 
corresponding to a maximum rotation of the arm of 20. 

A skeleton diagram of the mechanism is shown at (G), Fig. 154, 
where OA is the initial position of the arm and OC is the deflected 
position, whilst AB is the unstretched, and BC the stretched, positions 
of the spring. The extension of the spring will be the difference 
between the lengths BC and AB. 



426 ARITHMETIC FOR ENGINEERS 

The line CD is drawn perpendicular to the line BO. Then from 
the triangle OCD the lengths of CD and OD may be found. 

CD 





Thus 



CD 

5 = -3420 /. CD = -342 X 1-5 = -513"- 

Also Q^ = cos 20 

= '9397 /. OD = -9397 X 1-5 = 1-409". 

Now OB = OA + AB 

= i'5 -f- 1-75 = 3*25" 
and BD = OB - OD 

3-25 1-409 = 1-841". 
Now in the right-angled triangle BCD 

BC 2 == BD 2 -f CD 2 (refer p. 266) 
BC 2 = i-8 4 i 2 -f -5132 

= 3'3_88_+ -263 - 3-651 
BC = "X/3'65 1 = 1-911 
Extension of spring = BC AB 

= 1-911 1-75 = 'i6i"._ 

Example 392. The arrangement of a ship's steering gear is shown 
at (H), Fig. 154. The turning moment on the tiller is the product 
of the force represented by AB and the arm OA. Find an expression 
for this turning moment in terms of the applied force F, represented 
by AC, the distance d and the angle 0. 

From the information given 

Turning moment = force AB x length OA . . . . (i) 
It is then necessary to find (a) value of AB in terms of F and 6 from 
the triangle ABC, and (b) the value of OA from the triangle OAD. 

Now in triangle ABC 



AC ft 

= cos8 



And in triangle OAD, 
OD 



and 



.... (3) 

Then substituting in (i) the values found for AB and OA in (2) 
and (3) respectively, 

^ . F d Fd 

Turning moment = ~ x n = - - r^. 
& cos cos cos 2 



TRIGONOMETRY 427 

Exercises 98. On Simple Trigonometrical Problems. 

[Letters, thus (E) refer to Fig. 155.] 

1. The " gliding angle " of an aeroplane is given as " I in 7-1," 
indicating that, when gliding, a horizontal distance of 7-1 units is 
travelled for a loss in height of i unit, as indicated at (A). What 
is the angle of the gliding path with the horizontal, i.e. angle ? 

2. The maximum gradient on the Pilatus mountain railway is 
given as 48%, i.e. a rise of 48 units for a distance of 100 units along 
the slope. What is the angle of slope to the horizontal, i.e. angle a 
at (B). 

3. At (C), OE represents the effective radius of a steam engine 
eccentric. The " angle of advance " is arranged so that the projection 
of OE on the centre line, i.e. OP, equals " lap -f lead." What must 
be the value of if the effective radius is 3j", the lead is to be J" and 
the lap is if"? 

4. The Whitworth standard screw thread has an included angle 
between the side slopes of 55, while the crest and root are so rounded 
that the sharp points are removed for a depth of of the pitch p, as 
shown at (D). Calculate, in terms of /?, the value of the full depth d 
and also the actual depth D. (Note. It will be sufficiently accurate 
if the value of any trigonometric ratio of 27 ^ be taken as half-way 
between the values for 27 and 28.) 

5. The governor mechanism of a centrifugal type tachometer is 
shown at (E). When running at tliQ slowest and fastest speeds 
respectively, the governor runs " solid " in the positions shown at 
(F) and (G). The angles and 9 for these positions are of importance 
with regard to the spring control. Determine the values of and 9 
from the dimensions given in the figure (E). The necessary right- 
angled triangles are indicated in figures (F) and (G). 

6. The diagram at (H) indicates the skeleton arrangement of an 
Izod impact testing machine; OA is the initial position of a swinging 
hammer which, after breaking the specimen situated at C, rises to a 
position OB. The energy absorbed by the specimen is equal to W x 
distance DE, where W is the weight of the hammer. Find an expression 
for this energy in terms of the quantities W and L and the angles 
a and (3. 

7. A plug gauge is to be made of the form and dimensions shown 
at (J) ; the circular portion, if completed, passes through the corner 
A of the square ABCD. Calculate the diameter to which the plug 
must be ground before the shaded portions are cut away. [Hint. 
Referring to the sketch at (K), the radius of the plug = OA = OE = 
EF OF, and OF may be expressed in terms of OA by reference to 
the triangle OAF.] 

8. A dovetail gap in a form gauge is to be tested by the fit of a 
turned disc as shown at (L). Calculate the necessary diameter of disc 
for the dimensions shown. [Hint. The triangle essential for the 
solution is shown at (M) as OAB. Determine the value of r by 
reference to the angle OAB and the side AB.] 

9. When measuring the thickness of spur wheel teeth by a gear 
tooth vernier caliper, as indicated at (N), the chordal thickness ADB 
and the height CD are measured. These dimensions differ from the 



428 



ARITHMETIC FOR ENGINEERS 




.^."oi*. 



C)> H/h 

Lap ^Leabl 




Fig. 155. Diagrams for Exercises 98. 



TRIGONOMETRY 429 

thickness and height determined from the circular pitch and need to 
be specially computed when the number of teeth in the gear is small 
and the pitch is coarse. 

If R = pitch radius of wheel and N = number of teeth in the 
wheel (i.e. number of pitches), determine expressions for (a) the chordal 
thickness ADB, and (b) the additional height DE. [Hint. Work from 
the triangle AOD.] 

10. At (P) is shown the indicating portion of a certain gyroscopic 
instrument in which the tilting of an arm A is multiplied by the 
quadrant Q and the pinion P, the spindle of the latter carrying a 
pointer N. The essential dimensions of the multiplying gear being 
as shown, calculate the angle that the pointer will move if the arm A 
tilts 10 from the mean position shown; the ratio of the quadrant 
and pinion gear is 4 to i. [Hint. The skeleton diagram at (Q) is a 
view of the mechanism looking in the direction of the arrow Y ; x is 
the vertical movement of the tilting arm in the plane of the 
quadrant Q. The diagram at (R) is a view looking along the arrow 
Z; the angle a turned through by the quadrant is determined by the 
value of x.} 

The Reciprocal Trigonometric Ratios. The ratios already 
considered, viz. sine, cosine and tangent, are by far the most 
important trigonometric ratios, but there are a few others which, 
although of lesser importance, are occasionally useful. 

Referring to the right-angled triangle at (Z), Fig. 150, the three 
chief ratios of the angle A have been defined thus : 

opposite side BC . adjacent side AC opposite side BC 

hypotenuse AB ~ Sin A hypotenuse "T5 cos A; adjacent side "AC "" tan A - 

Now if ratios be taken between the same pairs of sides, but in 
the reverse order, another three trigonometric ratios are obtained, 
which are named thus : 

hypotenuse AB A . hypotenuse AB 

opposite - BC - Cosecant A ; y adjacent - ^C - Secant A ; 

adjacent AC 





These names are abbreviated in a similar manner as in the 
case of the other ratios, thus : 

cosec A for cosecant of angle A. 
sec A for secant of angle A. 
cot A for cotangent of angle A. 

It should be noted that each of the new ratios is derived from 
exactly the same two sides as those defining the more important 
ratios above, but the pairs of sides occur in the reverse order, 
i.e. the numerator and denominator are interchanged in every case. 

Now, when a ratio between any two quantities is inverted, 



430 ARITHMETIC FOR ENGINEERS 

the two forms are said to be " reciprocals " of each other (refer 
P- 79)- 

Therefore 

Sine and Cosecant are reciprocal ratios, 
Cosine and Secant are reciprocal ratios, 
Tangent and Cotangent are reciprocal ratios ; 

or, expressing these statements in equation form 

I 1,1 

cosec = . ; sec = - - ; cot = - 
sin cos tan 

i i A i 

sin = ; cos = ; tan = -. 

cosec sec cot 

The difference in the values of these ratios, for angles between 
o and 90 is that, whereas the sine and cosine are never greater 
than i, the cosecant and secant are never less than i. The cotan- 
gent is like the tangent in so far as its value varies from o to infinity. 

It has already been shown that sine and cosine are inter-related 
for complementary angles, 

i.e. sin = cos (90 0) and cos = sin (90 0). 

Among the six ratios now known, similar relations can be 
shown to exist, viz., 

sec 6 cosec (90 6) and cosec = sec (90 0), 
cot 6 = tan (90 6) and tan 6 = cot (90 0) ; 

or in more general terms, each ratio of an angle has the same value 
as, and is identical with, the " co-ratio " of its complement. 

Further points concerning the Table of Ratios. A few 
other points relating to the table on p. 458 may now be examined. 

It will be noted that the two central columns headed " Tangent " 
at one end are also headed " Co- tan gent " at the other end. The 
reason for this is that the angles in the extreme outer columns on 
any horizontal line are complements of each other, and the tangent 
of an angle and the cotangent of its complement are equal. The 
arrangement of the tan and cot columns is thus similar to the 
arrangement of the sin and cos columns. 

There are two other columns at each side of the table which 
have not yet been referred to. The " Radians " column merely 
gives the values of the angles in the extreme outer, or " Degrees/' 
columns, expressed in terms of the " radian," a special unit, already 
referred to on p. 263, which is equal to 57*3. This method of 



TRIGONOMETRY 431 

measurement is of considerable importance in more advanced work, 
but need not be further considered at this stage. 

The third and eighth columns headed " Chord " give the 
lengths of the chord of a circle [refer p. 250 and (N) Fig. 25] whose 
radius is I unit for various angles subtended at the centre. Referring 
to (C), Fig. 154, the line AB is a chord of the circle through the ball 
centres. Now the ratio of the length of this chord to the radius 

of its circle, i.e. ^-.- in the figure, is always constant for the same 

angle subtended at the centre of the circle, and the " Chord " 
columns in the table give the values of this ratio for the various 
angles from o to 90. The values in these columns are related 
to those in the sine columns ; they are, in fact, the values of " twice 
the sine of half the angle/' 

Thus referring to (C), Fig. 154, 

the chord AB = AC X 2, and AC = OA sin a. 
AB = 2 X OA sin a. 

Now the angle subtended by (i.e. opposite to) the chord AB 

A 

is the angle AOB, which we may denote, say, by ; then a = . 
AB = 2 OA sin - 

2 

AB 6 

r^ = " chord " ratio = 2 sin - 
OA 2 

which is twice the sine of half the angle. 

The chord columns are sometimes useful for setting out angles 
without using a protractor. 

Important Relationships between the Ratios. Much of 
the practical application of trigonometry is dependent upon a 
knowledge of certain very important relationships that exist be- 
tween the various ratios other than those already referred to. It is 
not possible, nor desirable, to discuss these at this stage, but there 
are two very important instances which may be noted and 
remembered for use when the subject is further pursued. 

Referring to the standard right-angled triangle at (Z), Fig. 150, 
it is known that 

. A BC A AC . A BC 

sin A = 



Now let the sine of angle A be divided by its cosine, when both 
ratios are expressed as above. 

FF 



432 ARITHMETIC FOR ENGINEERS 

Then 

sin A * . A BC . AC 

r = Sin A -r- COS A = -r-,5 -7- -T-T5 

cos A AB AB 

14 . . . BC AB 

inverting divisor and multiplying = -^- x -rr% 

BC 

and cancelling AB = -v-p 

which relation is already known as the tangent of A. 

sin A 



cos A 



= tan A. 



Again referring to the same right-angled triangle and the same 
three ratios, let the sine and cosine be each squared and their 
squares added together. 

TU A BC ' 2 A BC2 

Then since sin A = - /. sin 2 A = 



A AC , A AC 2 

and cos A = ^g /. cos 2 A = 

2 A . * A BC 2 AC 2 
.. sin 2 A + cos- A = 



_ BC 2 + AC 2 
"" AB 2 ' 

Now in any right-angled triangle the sum of the squares on the 
two sides enclosing the right angle is equal to the square on the 
hypotenuse (refer p. 266) so that BC 2 + AC 2 may be replaced by 
AB 2 . 

2 A i 2 A BC 2 + AC 2 AB 2 
/. sin 2 A + cos 2 A = 



XB2 
sin 2 A + cos 2 A = 1. 

From this relation sin 2 A = i cos 2 A 
so that sin A = Vi cos 2 A 

also cos 2 A = i sin 2 A 

so that cos A = Vi ->sin 2 A. 

Graphs of the Important Ratios. It has already been stated 
that no ratio increases in proportion to its angle; thus sin 60 is 
not equal to twice sin 30, and so on. This general statement 
can easily be verified by reference to the tables, but a much better 
conception of the way in which the various ratios change as the 
angle is changed can be obtained by plotting a curve for each ratio 
to be examined. 

If the values 'of the ratios sine, cosine and tangent be taken 
from the table for angles at intervals of, say, every 5 or 10, and 



TRIGONOMETRY 



433 



these values be plotted vertically on squared paper with the corre- 
sponding angles plotted horizontally, then the resulting points 
will be found to lie on a perfectly smooth curve in every case. The 
shapes of these curves are characteristic and are reproduced to 
a small scale, for angles from o to 90, in Fig. 156. None of the 
curves contains any straight portion, but the direction is every- 
where different. 

The following features should be noted : 

Sine Curve. The curve is convex upwards, increasing from 
zero at o to i at 90. The greatest slope is at o, and the slope 
steadily diminishes until at 90 the curve has become horizontal. 



- 


- 


- 




- 


_ 


^ 






in 
o 






















x^ 





" 
- 1 

p 


- 










- 













- 






^s 


N 
















^ 






V 


^ 




















^ 














- 






s 
















^ 


/ 


-- 




._. 




i- ., 






V 


. 














/ 




















\ 












> 
' 














4 














\ 








, 






c 


urve of 
Sines 








Curve of 




V 








A 










Cosines 






Ny 




> 




































N 


f 





































L 

f 

1-0 

8 
(, 
4 
2 

n 










t 


7 








/ 


1 















































/ 






























/ 






















J 


t 




Curve of 
Tangents 
















-- 


f 















o 20 

ANGLE 



60 80 9 
DEGREES 



20 40 
ANGLE- 



eo 90 o 20 

AMGI 



AO 60 80 90 
t - DEGREES 



Fig. 156. Sine, Cosine and Tangent Curves from o to 90. 

Cosine Curve. The curve is convex upwards but decreases 
from i at o to zero at 90, the general slope being downwards. 
The cosine curve is identical in shape with the sine curve but is 
reversed left for right. 

Tangent Curve. The curve is concave upwards and increases 
from zero at o to i at 45 after which it rises at a very rapid rate, 
becoming practically vertical in the neighbourhood of 90. The 
actual value at 90 cannot be plotted as it is infinitely far up in the 
vertical direction. 

In this book only angles between o and 90 have been con- 
sidered, but if the subject be followed up then angles of greater 
than 90 must receive consideration. It will then be found that 
the various trigonometric ratios can possess negative as well as 
positive values, and that the sine and cosine curves become a 
series of ripples evenly spaced with regard to the horizontal axis. 
The study of this sine curve is of the greatest importance in 
engineering and all branches of science. 



ANSWERS TO EXERCISES 



Exercises 1 

1. Proper fractions : , f|, ; Improper fractions : \^, ty 

2. tai; - .. . 13- 9i 

3. Improper fractions : !,; Mixed numbers : 5Jf, 30$, i^ 

4. Proper fraction : ^ ; Improper fractions : ^, f ; Mixed numbers : 

*i. 7A 

5. f, 1, H> . 6. v, ?i. it. ?S. ?? 7. ^, v. . . W 
8. W> *F> . H 1 * > W. #' HI* 1 10- W. W, Yi a , W 

11. it, 3t. 2?, if 2? 12. ijV. 9t, 6t, 3}, 9j 

13. 3iV 9i 2^-, I, 16 14. i^, 4^, 41J-, 12$, 9& 

15. 6, 3^J. 2^, ii}f, 22^5 16. 1 1 A, 6i8, 3r^, 3iV 3 

Exercises 2 

1. f, *.}.*,! 2. l-}},i. ^ } 3. *.,*. |,t 

4. A, i }J, i? 5. i, j, i i, jj 6. 8#, }M> A, * 

7. 2j, 4 , i J. if, ij 8. 31, 5, 2^ 6J, if 9. 5 &, 6fc ioj, io|, i^ 
10. lo^, 4 f 3 | f i,V 11. 15$. 88, 5 f, 3 J 12. 550, ijj, I3& 

Exercises 3 

1. (a) i"; (b) r ; W if 2. (a) |"; (6) }}" ; (c)&" 3. () i|; (b) Jj 

4. () T&; (*) } 5. (a) ft; (b) ft 6. (a) gf ; (6) J 

7. () JI : (b) U 8. (a) ft ; (6) 9. (a) JJ ; (6) 
10. (a) f J ; (b) ft 11. 24 12. 12 13. 72 14. 144 
15. 60 16. 42 17. 144 18. 165 19. 300 20. 153 

21. 102 22. 165 23. 750 24. 4200 25. 180 26. 306 
27. JJ 28. f 29. M 80. i^ 31. ft 32. i 3 
33. if J 34. 7^ 35. 7^ 86. L Va 37. 3 ~3" 88. 8J" 
89. 3 / -iJ // 40. 8ft'' 41- 9iV' 42. i6J x/ 

Exercises 4 

1. /' 2. A" 8. A" larger 4. J 5. 6. A 7. J 

8. ifj 9. i aj 10. ^ 11. 2 A 12. | 13. i/ a 14. 2 {g 
15. 4f 16. m 17. 18. 6}i 19. ft* 20. f 21. 3 A 

22. 2ff 23. A 24. ff 25. i 26. 2j 27. 8p 28. 2^ 
29. 6Jf 30. ijg 31. A 32. ^ 33. i^ 34. 3*" 35. i^ 
86.11} 37.12^ 88. A 89. i A 40. o 41.2" 42.20}^ 
48. 3i" 44. 7i" 45. (a) ij|"; (ft) A" 

434 



ANSWERS TO EXERCISES 435 

Exercises 5 

1. 2 2. 2ft 3. A 4. 3 | 5. * 6. ft 7. ft 

8. i 9. i? 10. M 11. 4 J 12. 6 13. 9j 14. T fc, 

15. A 16. iA 17. 3 J 18. 864 19. 2| 20. ^ 21. ^ 

22. A" 23- A" 24. iV 25. A" 26. 9* r.p.m. 27. T fo" 
28. (a) 1T foj"; (6) i^W 29. 6080 ft. 30. 12 miles. 31. noo yards 

32. 338 cu. yards 

Exercises 6 

1. 4 2. 2% 3. f 4. 12 5. A 6. f 7. 2 

8. 6 9. ^ 10. f 11. 3^" 12. ioo 13. 14" 14. 13}" 
15. 22" 16. 8" 17. 216 Ibs. per inch 18. 248 Ibs. per inch 
19. 256 Ibs. per inch 20. 624 Ibs. per in. 21. -fa 22. T J a 

23. A 24. A 25. * 26. f 27. i 28. A 29. ^ 

30. i 81. & 82. f 33. A 34. | 35. A 36. J 

Exercises 7 

1. ,V 2. i& 3. 3l 4. 3 J 5. i6J 6. ft 7. \\ 8. ij 

9. iA 10. 6f 11. A 12. iA 13. iJJ 14. | 15. ^ 16. JJ 
17. if 18. 6}| 19. 10 20. 2^ 21. logj 22. 8 
23. 45 screws per bar; 39 bars. 24. 50$", say 51" 25. 41!" 
26. 14 27. 138 28. 19!", say 20" 29. 163 ft. ; 4 lengths 

80. Cement 3 cwts. ; coke breeze 12 cwts. 

31. (a) Lead 90 Ibs., tin 60 Ibs.; (6) Tin ioo Ibs., lead 25 Ibs.; 
bismuth 25 Ibs. 32. Nitre 45 Ibs., ; charcoal 9 Ibs. ; sulphur 6 Ibs. 

33, (a) Tin, 74 Ibs. n ozs. ; bismuth, 37 Ibs. 5 ozs. 

(6) Lead, n Ibs. 3 ozs. ; tin, 44 Ibs. 13 ozs. ; bismuth, 56 Ibs. 

(c) Lead, 35 Ibs.; tin, 21 Ibs.; bismuth, 56 Ibs. 

84. (a) Old sand 122^, say 122 Ibs.; new sand 81^, say 8iJ Ibs. ; 
coal-dust 20^, say 20^ Ibs. ; (b) Old sand 790^9, say 791 Ibs. ; new 
sand 263^, say 263 Ibs. ; coal-dust 65^, say 66 Ibs. 

35. Lead 22^ Ibs., antimony 5 Ibs., bismuth 2^ Ibs. 

36. Cement y\ tons, sand 22j tons, aggregate 45 tons. 

37. 3 if ; 50' - 91", say 51 ft. 38. A 39. & 40. 3 7i 3 o 
42. fg 43. 27 

Exercises 8 

1. (*) 55; (&) -55; (*) 134; (<*) -0134; (e) 453600; (/) -4536 

2. (a) i'ii8; (b) -087; (c) -00000066; (6?) 350,000,000 

3. (a) 35oo watts ; (b) 7300 watts ; (c) 80 watts 

4. (a) 29-5 kw. ; (b) -231 kw. ; (c) 305 kw. 

5. (a) 270,000,000 ; (6) 500,000 ; (c) 300 

6. (a) 4-26; (b) -055 

Exercises 9 

1. W i: W A: W W 2. (a) ^; (6) I; ( C ) |^ 

8. (a) 2^; (6) iJJ; (c) ^ 4. (a) Jtt; ( 

5. (a) JJ; (6) SJ 



436 ANSWERS TO EXERCISES 

Exercises 10 

1. 3937; 528; 2205; 3045; 365 2. 2985; 105; 807; 1083; 454 

3. 66; 6065; 1118; 3209; 15708 4. 109; 87; 105; 6009; 2 

5. 454; 30*5; -434; 62 '4; -00112 6. 14400; 1390; 32-2; 601; 3750 
7. 'O8i ; -28; 15; -065; 5-4 8. 2-2; 610; -016; 30,000,000; -n 
9. (a) 277-3 cu. ins. ; (b) 277 cu. ins. 

10. (a) 453-59 grammes; (6) 454 grammes 

11. (a) -14286; (b) -1429; (c) -14 12. (a) 2-7183; (b) 2-718; (c) 2-7 

Exercises 11 

1. 99-28 2. 263-81 3. 220-04375 

4. 349-6 ft. 5. 36 Ibs. 6. 67-05 chains 

7. 11-23 8. Dowson 44-42 ; Mond 36 ; natural gas 95-75 

9. 8-9 10. 29-95 11- () Brass 1-8; (b) steel i-oi 

12. -6888 grammes 13. (a) 1163-7 grammes; (b) 51-9 grammes 
14. (a) -1888; (b) -2878 15. -0035"; -0105" 16. 20-66 17. 10-995 

jColumn R = 7'4o6j Error = . OQ2 g = . OQ2 

\ F = 7-404] 

20. 3*83 21. 6-23 tons 22. 3 

Exercises 12 

1. 40-71, i. e., 40-7 2. 404-2, i. e., 404 3. -7923, t. ., -792 

4. 12*52075, i. e. t 12-5 5. 61-544, . 0., 61-5 6. -0004698, t. e. t -00047 

7. 48-24, t. e. t 48-2 8. 18-8819, *'.<?., 18-9 9. -09752, i. e. t -0975 

10. 21444-5, i. e. t 21400 11. 31-05 m.p.h., say 31 m.p.h. 

12. Boat's speed = 12-1 m.p.h. /. boat is slightly faster 

13. 25-95 14. 23-9 Ibs. 15. 97-5 16. 27-4 Ibs. 
17. 2713, say 2710 cu. ins. 18. -0783 in. 

19. 515 Ibs. per sq. in. 20. 477-7, say 478 Ibs. per sq. in. 

21. 156 sq. ins. 22. 28-16, say 28-2 ohms 

Exercises 13 

1. 43-9 2. 4-676 3. 18-19 4. 3096 5. -08885 

6. 24-31 7. -475 8. -00847.-^, 9. 2003. 10. 282 

11. (a) 1 1 -2 atmospheres; (b) 6-12 atmospheres 12. -651; -6405; -62 
13. 4-54 14. 2-205 Ibs. 15. 50-8 16. 1-34 

17. 4-85 sq. ft. 18. 22 Ibs. per min. 19. 42-2 Ibs. permin. 

20. 31-1 Ibs. per min. 21. -0133 ohms, per ft. 22. -016 cu. ft. 

Exercises 14 

1. -625 2. -09375 3. -5625 4. -9048, say -905 

6- "171875 exact, say -172 6. -152 7. -45 

8. -555 9. -417 10. 3'i4 2 11- '3 min. 

12. -567 min. 13. -183 min. 14. -9 min. 15. -217 min. 



ANSWERS TO EXERCISES 437 

16. -0416 ins. 17. -0556 ins. 18. -in ins. 

19. -222 ins. 20. -286 ins. 21. -0909 ins. 

22. -0714 ins. 23. -0526 ins. 24. -0357 ms - 

25. " Go in " end 1-6246" dia. ; " Not go in " end = 1-6262* dia. 

Exercises 15 

1. Yes 2. Yes 3. No; should be 31-5 

4. No; should be -155 5. No; should be -0494 6. Yes 

7. Yes 8. Yes 9. -478 10. 64 

11. 880 12. 71-4 13. 21-4 14. 272 
15. 44000 16. 34700 17. -0245 18. -705 

Exercises 16 

1. 34 ft. 2. 2-53 3. 2-55 4. -199 amps. 

5. (a) 6-63 ohms. ; (b) 29-8 volts 6. 20-3 7. 655 h.p. 

8. -619", say I" 9. (a) 11520 watts; (b) 104-7, sa Y 10 5 amps. 

10. 24,600,000 Ibs. per sq. in. 11. 11,620,000 Ibs. per sq. in. 

12. 604 B.T.U. 13. 9-16 14. 32-36 tons 

15. 963 16. 4975- 17. 337 

18. 8-59 tons 19. 8-89 tons 20. 21-3 tons 

21. 544. 22. 88 Ibs. per sq. in. 23. 15-05, say 15 Ibs. 

24. 53-6 Ibs. per sq. in. 25. 581 ft. tons 26. 1600 Ibs. 

Exercises 17 

1. 1-505 2. -579 3. -0278 ohms 

4. 03599"', say -036* 5. -396 6. 27-9 Ibs. per. min. 

7. 100-2, say 100 Ibs. per sq. in. 8. 110-45, sa Y II0 volts 

9. 221-1, say 221 amps. 10. 592-5, say 593 cu. ft. per min. 

11. '462 mm. 12. '794. 

Exercises 18 

1. ij* holes, 6s. per doz. ; i" holes, 35. gd. per doz. 2. 2 los. 
3. 295. 3^. 4. (a) 4 4$.; (b) 3 35. 5. 179 h.p. 6. 40-8 Ibs. 
7. (a) '477 cu. ft. ; (b) 6-44 cu. ft. 8. Max. 2958 Ibs. ; min. 2842 Ibs. 
9. 3'3 tons 10. 2-5%. 11. 14-1% 12. 1-12% 

13. (0)3i7'5 gaUs. P er min.; (6) 5-83% above 

14. Copper 76-2%; tin 23-8% 15. Copper 94-1%; tin 5-88% 

16. Copper 65% ; tin 5% ; lead 30% 

17. Copper sulphate 8-33% ; sulphuric acid 8-33% ; water 83-33% 

18. Water 13*3%; carbon dioxide 17-8%; nitrogen 63%; oxygen 

9% 

19. Carbon dioxide 10-4%; oxygen 12-25%; nitrogen 77-4% 

20. 99-205% 21. 98-136% 22. 93'97 2 % 23. 99'377% 
24. 204 and 196 r.p.m. 25. 182-7 an< ^ J 77*3 r.p.m. 

26. 87-5 and 82-5 r.p.m. 27. 20-6% 

28. Water 97*6% ; hypo 1-22% ; sugar of lead 1-22% 



43 8 ANSWERS TO EXERCISES 

Exercises 19 

1. (a) 4-89 to i ; (b) 2-03 to i 2. (a) 5 to i ; (b) 5-9 to i 



3. -639:1 4. -862:1 5. (a) 40-7:1 

6. (a) 3*2:1; (6) 3-23:1 7. (a) 29-1:1 

8. (a) -829 : i ; (b) -597 : i 9- (a) -529 : i 



(fc) 112-7: i 
(b) 30-1:1 
-566 : i 



10. 1648 11. (a) 4-41:1; (b) 2-68:1 12. 6-32:2-38: 

13. 8-99:4:1735:1 14- (a) 14-8:1', (b) 13-3:1 15. -412 
16. 12:1 17. 21 : i 18. (a) 4; (b) 2-46 

19. (a) 1-58; (b) 4-8; (c) 5-42; (d) 5-63; (e) 6-05; (/) 7-09 

Exercises 20 

1. 42 Ibs. 2. 4i galls. 

3. 50-2 sq. ins. 4. 16-3 Ibs.; increase = 1-8 Ibs. 

5. H.P. 15-94", say 16* dia.; I.P. 26-6", say 26 1* dia. 

6. H.P. 754-3 sq. in.; ist I.P. 1660 sq. ins.; 2nd I.P.347O sq. ins. 

7. 3520 sq. ins. 8. 135 galls. 9. 175 galls. 10. 3-75 cwts. 





Exercises 20a. Refer p. 453. 




Exercises 21 


1. 86-8 volts 


2. 13-28 grammes 


3. (a) 104 ; (6) 


1596 4. 1540 ft. per min. 


5. 153,400 Ibs. 


6. 710 7. 90 inch-lbs. 8. 28-8 sq. ins. 


9. 24-7 


10. 211 tons-ft. 11. 48 12. 2-55 


13. 13-1 


14. -75" 15. -632 16. 11-39 


17. 28700 


18. 29-7 H.P. 19. 3-15 B. H.P. 20.6-47 




Exercises 22 


1. i* 


2. -622* 3. 1185-5 4. 883 


5. H43-3 


6. (a) 98-9 ft. per min.; (b) 21-6 ft. per min. 


7. 6-52' 


8. D = 1-492", H = -4264", w = -1447*, h = -1421' 


9. 1350 f t. P er 


min. 10. -437 11. -59 12. 28 


13. 185 F. 


14. 38-4 amps. 15. -00833 16. 45-6 


17. 4-67 


18. 8-68 19. 3-41 tons 


20. -802' 


21. 378-7 22. 5-04 H.P. 


23. 2 33 


24. 2-98 tons 25. 1762 Ibs. per hr 




Exercises 23 


1. 4 X 4 X 4; 


10 x 10 ; 3x3x3x3; -2 x -2 x -2; -5 x -5; 


X 10 X 10 X 10 X 10 


2. r X r; D x 


DxDxD\aXaXa\ hxhxhxhxh 



3. (i) 9; (ii) 6 4. (i) 64; (ii) 12 5. (i) 3; (ii) 2-25 

6. (i) 81; (ii) 12 7. (i) 10; (ii) 32 

8. (i) c x c X r ; (ii) c x r X r ; (iii) c x c X r X r ; (iv) c x r X c 
X r; (v) Results of (iii) and (iv) are equal 

9. (i) 2 x M x M ; (ii) 3 x M x M ; (iii) * X M X 2 x M ; 
(iv) 3 x M X 3 X M 



ANSWERS TO EXERCISES 





10. 


(i) a x a x 


a x r 


; (ii) a 


X r x 


r X 


r; (iii) a 


X a 


X 


r > 


; r > 


(iv) aXaXaXrXr 




11. 


(i) 3 X n X 


n X p 


; (ii) 3 


X n X 


px 


Pi (iii) 3 


X n ; 


x 


n x 


p > 




12. 


(i) 3 X n x 


p X n x p; 


(ii) 3 


X n 


X 3 X n 


x p; 




(iii) 


3 - 


X 


P 


X 3 X n X p 






















13. 


(i) 4-5; (ii) 


75; (iii) 2-25 


; (iv) 


2-25 














14. 


W i; (ii) '5 


; (i") 


4* (iv} 
* V / 


i 
















15. 


(i) 24; (ii) 54; (iii) 108; 


(iv) 72 




16. 


(i) 43-2; (ii) 


3<>; 


(iii) 4-32; (iv) 


4-32 




17. 


(i) 4000; (ii) 


1000 


; (iii) 30,000 ; 


(iv) 


0003 












18. 


4-91 sq. ins. 


19. 


112,640 Ibs. 


20. 


72-13 


21. 


30,070 




22. 


102 tons 


23. 


3-39 H 


.P. 


24. 


3'327 


25. 


2316 


ft. 




26. 


438 H.P. 


27. 


8-71 H 


.P. 


28. 


33-8 


29. 


7 


68 


Ibs. 




30. 


4-56 Ibs. 


31. 


195 




32. 


887 watts 


33. 


4 


i ft. 



439 



X n 



Ibs 



Exercises 24 



1. 3-4 2. 15-5 3. 2-1 4. 29 5. 9-18 6. 10-5 
7. 40-3 8. 109 9. -78 10. -27 11. -095 12. i -oi 
13. -0031 14. 15-9 15. 49-9 16. 320 17. 2600 18. 4-76 


19. 1-62 20. 50- 


6 21. 60-7 22. 27-6 23. 24-2 24. -301 


25. -356 


26. 7-58 nautical miles 


27. 74-6 ft. per sec. 


28. 6-74 ins. 


29. 5-56 H.P. 


30. 12-4 


31. 8-35, i. e., &" 


32. 1-645", * e -> *\ 


33. 15-9 


34. 1-998 sees. 


35. 70-7 amps. 


36. -917 


37. 1 1 -9 


38. 2-12 ft. 


89. 1-25 



8 . 



243 



. 

C 2 

9. JL 10. 

* 



15. 



gr* 
16. 



Exercises 25 



12. 



11. 

4*2 27 



ir 

13. 



17. 



18. 



19. 



14. ~~ 

900 3.375 

/3 
20. .85 7 . 



4. 
8. 



Exercises 26 

2. -8I7VRT 3. .-ggg^ 

5. -346 Vr 6. 8-02 Vh 7. 44-3^ 

9. -274D 10. 224 VH 11. 

13, -0763^ 14. -707 Vp - i 



440 ANSWERS TO EXERCISES 

Exercises 27 

1. a 5 2. / 3 3. N> 4. d 6 5. r* 6. a' 

7. H 8. io 9. io 7 10. w*r 11. px* 12. 6 

13. D 14. R 4 15. io 4 16. io 17. / 3 T 18. R 3 E 

19. * 2 M 3 20. p*d* 21. * 5 ;y 2 z> 



Exercises 28 

1. m 2. S 4 3. t 4. R 5. io 4 6. io 2 a 7. an 
8. *r 9. />d 3 10. fo> 11. io 8 R 12. io 6 

Exercises 29 

1. a 12 2. a 9 3. io 8 4. N* 5. D* 6. R 12 

7. io 8 8. io 12 9. R 10. m 10 11. r 4 12. c*d* 

13. c e <* 8 14. c 6 ^ 6 15. a 4 /) 8 16. <* 3 9 17. lo 1 ' 18. w 12 /? 4 

Exercises 30 

l-T 2 ' 2M ' 8 ' M ' *; 5 ' 8 6 '5 

7. I 8. i6n 9. (a) C 'R; (6) | 10. ^ 11. ^ 

12. 785^ 13. ^ 1*. 6' 15. ^ 16. gj- 17. g 

18. 5Jr 19. ^^ 20. 21. ^^ 2 22. -52360* 23.- 

2K / a 4 64 

24. 2-47Dd 2 25. 26. ^ 27. ^B 21 

4 ** 5 

Exercises 31 

1. 8 2. 3 3. 15 4. io 5. 17 

6. 1-9 7. '2 8. 1*6 9. i-i 10. 1-55 

11. -8730 12. -1380 13. 1-0160 14. -6700 15. 1-045 

16. 1-695 17. -954 18. -95 19- 2-115 20. -35 

21. 2 22. 3 23. - 6 24. - 3 25. o 

26. 7 27. i 28. 8 7 29. 3 30. 7 
31. o 32. 4# 2 ;y 33. r 2 34. ^pm 35. 2a 2 c 
36. i8u>/ 37. 8/> 2 . 38. 3^ 39. - 5-5*1. 40. - 8-670; 

Exercises 32 

1. 6 2. ii 3. 4 4. ii 5. i 6. 18 

7. 14 8. i 9. o 10. 2-2 11. 3-6 

12. 2-3 13. 2-7 14. 2-9 15. -9 16. 6-72 

17. - 3'45 18- - 6 '9i 19. - 1-117 20. - 17 21. 4-101 

22. - -33 23. - 2-002 24. -845 25. 7-35 26. ywl 

27. i2r/ 28. 25ad* 29. 2C 2 /> 30. yey 31. 



ANSWERS TO EXERCISES 441 

Exercises 33 

1. /> a zpq 2<? 2 2. 4#f 2 iar 4^ 

3. w -f 2/ / 2 / a 4. 4m 3 w 2 w 

5. a 3 6 3 6. 3 + q* 

7. 2# 8 x*y + 5)> 2 3# 3 ;y 3 # 4 

8. 3m/ 18 + 5/ 2 4m 2 9. 2 2r 2 10. / 4 12** 
11. 2* -55* 12. -25y 2 - -55y + 7 13. 7n 2 + 12 
14. iia + d 15. |c c 2 16. \$x* + # 10 
17. Id - 3* + 3** 18. |n 19. 36^ - ib*d - 36 20. 9* 4/ 

Exercises 34 

1. 20 2. 20 3. 20 4. 20 5. 2-1 6. 5*39 

7. 3-126 8. -975 9. 24 10. 24 11. 24 12. 16 
13. 10 14. 6-75 15. 9 16. 4 17. 4 18. 4 

19. 4 20. 2-5 21. 1-875 

Exercises 35 
1. 4 2. - 8 3. 16 4. 32 5. 64 6. - 4 7. 8 8. 16 

9. 32 10. 16 11. 64 12. 64 13. 6 4 r* 14. 641* 

Exercises 36 



1. 7 ii 2. 45i 


3. 436 4. 461 


5. 283 


6. 268 7. 273 


8. 32 9. 4 


10. 6-36 


11. 9*44 


C 12. i7-78C 


13. - i8-8 9 C. 


14. o 


15. 5 F. 


16. 32 F. 


17. - 35-5 F. 


18. - i 


19. 5 


20. 3 


21. 1-67 


22. - -737 


23. 175 


24. 180-4 


25. 18-95 


26. 16-96 


27- 6-44 


28. 5'5 


29. 4-23 


30. 20 7 


31. - 2 


32. 5-2 


33. - 4-71 


34. - 10 


35. 18 


36. 60-25 


37. 13-8 


38. 22-25 


39. 1218 


40. - 3 


41. - -675 


42. 1-175 


43. -65 


44. - -25 



Exercises 37 

1. M 2. C + M 3. x + 3 4. $p 2 5. 56 4 

6. 22 ^d 7. 3^ 2 w 8. i a 9. \r $r z 10. 7H + 8 
11. raA 12. 2 3 2/ 13. 2i> 2 v l 14. 3^ 7 

15. -133 -000279^ 16. 145006 + 62o6oH 17. 99 ~? 



18. PR - W 19. la + \b 20. - R 21. 

*2 * 

22. ^ + * 23. C2 ^" I 24. tC 25. -997 +-0002* 



26. 561 + 27. 175-6 -27* 28. s + -48/> + 950-6 

5 

29. 1082 -j- 3* 30. 1-64* *5#* 31. i -oi 8 + -0000306* 

32. -7502 -000504* 33. 145006 + 62o6oH + 7757O 

34. 10-76 -01* + -0048T 



442 ANSWERS TO EXERCISES 

Exercises 38 

1. a(i - e) 2. 

5. K(I - ) 6. 

9. 4 A(2f - h) 10. 
13. *( 5 - ) 

16. Wy(i - *j + ) 17. fc(a + 6) 18. 

21. 



R-r) 


3. c(A + ma) 


4. M(I + ^) 


"1 + 1} 


7. 2M& + 1 2 ) 


8. a(a + 26) 


*-?) 


11. *(i - L) 


12 ^f / 3 + / 3 



22. - / - ~ 23. *(* + a) + c(* + ) = (* + a)(* + e) 

24. , '5- ' 4 25. J^HB. 26. \ (a + 
Exercises 39 

1. 2AT* 5AT 3 2. X* + 3AT 2 28 3. 2/ 2 -f /AT # 2 

4. 2a 2 3ar 2r* 5. 2s 2 + PS 3p 2 6. 6 4 ft 3 66 2 

7. a 3 + & 3 8. a 3 - 6 3 9. 4^ 2 

10. Ar 2 - 4 #y + 4 >/ 2 11. QW 2 - i2n/> + 4/> 2 12. /> 2 



13. / 2 + 2^/ + ^ 2 Ar 2 14. + py + y 2 15. * 2 

16. ' _^ + 62 17. 2-25^ -6^ + 4 

439 2 

18. 9>w 2 + 7'2w#> + i-44/> 2 19- ^g + \* l + /2 

20. 4 w 2 2W + -25 21. 4 J 2 22. n 2 - 9 23. - x* 

x* v 2 Z 2 i 

24. f 2 - * 25. *V - ^ 26 ' 4 - F 

27. / 3 - 3& 1 - 2^ 3 28. 396 29. 4475 

30. 771 31. 855 32. 52 33. 82-875 

34 - 



Exercises 40 

1. 3-3 2. 11-17 3. 21-33 4. -94 6 5. 105-2 6. 4250 
7. -005378 8. -2614 9. 27-61 10. 22-5 11. 1-624 

12. 1-99, say 2 ft. 13. 1-478", i. e. t i%" 14. 2972 Ibs. per sq. in. 

15. 14-06 16. 183 Ibs. per sq. in. 17. 235 18. 42-75 

Exercises 41 

1. 30 2. 4-98 8. 13-2 4. 2-5 5. 109-8 6. 103-5 

7. 22*05 8. 19-62 9. -05775 10. 1-4 volts 11. 172 
12. 648 13. 2-47 14. 137-5 15. 1-437 16. 144 



ANSWERS TO EXERCISES 443 

Exercises 42 

1. 682 amps. 2. 19-9 3. 9-69 4. 3-98*, say 4* 5. 106,700 

6. 4 7. -00671 8. 1250 ft. 9. 152-4 10. 4-762 

11- T 3'33 12. 33-6 cu. ft. per sec. 13. 16-27 14. 340 

15. 64-4 16. -0498 17. -628 18. 628 19. 1-242 20. 9075 

Exercises 43 



1. 


7 


2. 


81-3 


3. 


36-985 


4. 


961 


5. 


55 


6. 


12-35 


7. 


6-7 


8. 


1-334 


9. 


00015 


10. 


3I9-3 


11. 


2-686 


12. 


1-2 


13. 


6-1 


14. 


6-85 


15. 


36-5 


16. 


3i 


17. 


005 


18, 


1-18 


19. 


36 


20. 


125-71 


21. 


54 


22. 


51-6 


23. 


939 







Exercises 44 



1. 30 C. 


2. -45" 3. -885 4. 1455 & i'5 


6. 25 


7. 20-8 8. 214 9. -0346 sq. ins. 


10. 61260 


11. 8-33 12. i-34'. say if* 13. - 30210 


14. 80 F. 


15. - 36-25 F. 16. 59-25 F. 17. 1-79 


18. - 3-89 C. 


19. 31 pence 20. 125 21. 405 22. 41-5 


23. 776 


24. 114 25. 5 26. 7-5. 



Exercises 45 

1. 2 2. 5-91 8. 2-78 4. 1-105 5. -622 6. 91-4 

7. 2 8. -75 9. 4 10. 5000 11. -182 

12. 3-29 13. 11-4 14. 10-15 15. 20-9 16. 2-84 
17. 85 18. 14 19. - 3 20. 3 21. 4 

22. 3393 23. -0343 24. 41-7, i. e., 42 25. 3-96 26. 1333 

Exercises 46 

1. 1-604 2. 79-1 3. 3-13 4. 161 5. 9-5* F - 6. 19-33 

7. 69 8. 3-78 9. 4-27 10. 1-36 11. -00438 12. 1-296 

13. 24-3 14. -000158 15. 137 16. 1-318 17. 2-167 18. 52,840 

Exercises 47 

1. 7-5 2. -6 3. ii 4. -75 5. 2 6. 4 

7. 2 8. 3 9. -286 10. -4 11. -0909 

12. 6 18. 5-04 14- 35-7 15. 25-25 16. 1-5 

Exercises 48 

1. 3-675 2. 1-12 3. 136 4. z-475 5. 30,520 

6. ii-5 7. -664 8. 4*86 9. 9*375 10. i-37 say i|* 

11. 223 galls, per min. 12. 3'39 13. 2348 14. 3 

16. -97 16. 5-57 17- 4'*3' 18- 3*4* 



444 ANSWERS TO EXERCISES 

Exercises 49 

1. 6-25 2. -458 3. -956 * 32-5 5 - 39'3 

6. -000941 7. 5-56 8. 84-4 9. 1-675 10. 162 

11. -499 12. 5357 13. 6090 14. 1-5 15. 5'4i 

Exercises 49a. Refer p. 453. 
Exercises 50 



1 F 2 w 3 ^ V 4 V 
1 M 2> 8> f *' irD 

7 . 8.W 9> *1 1Q j 

/T; E/> 2g '75-D- 
M dW 

13. CR 14. Y 15 - c Qi 16 - ~D~ 

19. Pitch dia. x Real Pitch ! 

Exercises 51 

i _ 2 . I00 ^ a 746H 


5. 
11. 
17. 

4, 


? 6 - ~8 P ^ 

H L 
o;(T If) " c/?F 

Vkr 18. c(r - i) 

1 or 2941 


WT c I2SOO< 

0. TT 


i35 


E 


E 




271 H 


330ooH 
o. y 


7 - F 


g I2 7 2 ?_ H 


9. 


m 10. 5L 

cA 2 1-26 


? 


40 T 


^o 4**K 


14. 


j v IK 


2 7iV 


13. A 


A K 


, 9 66W__ 


iT -^ 


1R. _ H ^ , 


1Q 


C(R -f wr) 0/> $wl 



Exercises 52 

1. P 14-7 2. H - *L 3. T 273 4. B - 4*! 

6. M 2 mM 6. S - 7. - ad 8. E V 



ts 



9. W Cr 10. u + a/ 11.^+- 12. g 

Exercises 53 , 





L-f 12 




W - 7 




n 





1*035 


4 


K- 48 


. 


120 




362 








i 




45 




5. 


1 I 2 ' 6 


6. 


J(F - 32) 


7. 


S 


~ 


R 


8. 


1300 


V 


9. 


566 -L 


10. 


50^ 29'2. 


11. 


D 





K 


12. 


v u 




8 




S 




t 


4 O 


mR 


44. 


Wf 


1K 


B 





H 


IB 


V* U 


2 


13. 


n 


IV. 


tn 






4 ' 


K* 




2S 





17 . - J_il<.- 18. 150(5 - i) 19. n(S - R) 20. ^^D - 7 < 
21. >^_ A) 22. (ad + c) 28. L(i n) 24. cr + e or c(r + 



ANSWERS TO EXERCISES 445 

Exercises 54 

* + ' 2- e-- A - 3 - + * *-3B- 2 

2-5V 6 c2 , * 7 H _w 8 9- W 

*"+'* '!-' .+! .2.. H ^| 



ia R-RO 14 H 

1O. ~ > , ~ I 4 *. , 1 I.O. 



fc 



* 



/ *7 'W 

47 fi( ^ A\ 1R T 1Q 

V^T / W 4- o; 

Exercises 55 

1 n ~ a 2 2\VH CR / 

l ' m-a *' Y~~W a ' E - Cr *' i -1 

- N ^/ WT u/L - I3D G 

5. r" 6 TTT f ~~~ , o. 

i k W 4 w n -\- 91 m 

9. _M. 10. S H ^T n. ?S + W6 12 . m( V .- 

I 4- ^ I <*"- - 



i a p '^ ~ '* 1J, ^v- 6 '* 36) 4* "\^ ' ) 4 a 2 j. 

lo - "- ** ^^ /. !*> T52^a~~~ !*> r** 




34 4 Vt 3 a 2 35. /=, 

Exercises 57 

1. i, 2, o, i, I 2. 2, 4, o, , 7 3. i, 3, 5> o 

4. 2, 3, I, 3, 2 5. i, i, 6, 7, i 6. 2, 3, 4, o, 2 



446 ANSWERS TO EXERCISES 

7. '3096, '8299, -9809, '0334, -4900 

8. '7505, '8388, '0969, -9031, -0043 

9. -5999, -9395* M362, -3010, -6021 

10. '3164, -2919, "8912, -8006 11. '6998, '8483, '5112, -6056 

12. "9647, '6430, '6900, '7481, "4972 

13. -8012, -1498, -4453* '54*4' >2 9i5 

14. '6618, -8362, '9763, -9972, -0931 

15. '5598, '1844, -7901, '9003, '0038 

16. -2980, -0124, '0924, '9763, -0013 17. 3-3502, 3-0523, 0-4972, 0-0249 
18. 1-8949, 1-7190, 2-6567, 0-2355 19. 1-4624, 5-7634, 7-9069, 0-5228 

20. 2-8910, 2-6898, 1-8495, 3'477 2 > 3*2375 

21. 3-7226, 0-4343, 0-2385, 3-699, 1-4048 

22. 4-5328, 0-1962, 4-1440, 0-4343, 1-0026 

23. 0-3623, 1-7582, 2-0249, 4-7756, 3-7839 

24. 0-0004, l^S 2 , 2-5912, 1-1031, 1-5076 

Exercises 58 

1. 1758, 4560, 3162, 2000 2. 1012, ioor, 1687, 5093 

3. 5616, 2505, 5379, 1408 4. 9974, 4720, 6390, 8910 

5. 1862, 3111, 6929, 1000 6. 9118, 6246, 8015, 7960 

7. 17-58, 456-0, 31620, 2000 8. 1-012, 1001, 1-687, 50-93 

9. '5616, -02505, -5379, 140-8 10. 9*974, -0000472, -639, 8910 

11. 1-862, 3111, -6929, -01 12. -0009118, 6-246, -008015,79-60 
13. -001022, 156, 20, 2-027 14. 21310, 22-89, -2, 3-142 

15. -00001259, 2245, 3'94 T > 1,250,000 

16. 5*722, -1618, -00000166, 30,000,000 

17. 2240, 9'7 6 3> '1005* 95'39, '7 8 54 

18. 10, 1728, -001293, 1-262, -7501 

Exercises 59 
1. 78-12 2. 30-02 3. -2496 4. -4999 6. 3004 

Exercises 60 
1. 6-896 2. -01227 3. 14890 4. -005715 5. 1-397 6. 7-945 

Exercises 61 

1. log top line = 2-3200 2. log top line = 1-6444 

log bottom line = 0-0239 log bottom line = 4-6885 

ResuR = 197-7 Result = 90350 

3. log top line =6-5717 4. log top line = 1-9420 

log bottom line = 4*5185 log bottom line = 4-9001 

Result = 113-1 Result = no, too 

5. -01171 6. -00001535 7. 2-326 

8. 2800 Ibs. per sq. in. 9. -09228 sq. in. 10. 1617 Ibs. 





ANbWJbLKb 1U fcXttKCIblib 447 




Exercises 62 


1. 6-822 


2. 35'27 3. 7-709 4. 521-2 


6. 277' 1 


6. 8-093 7. 6-122 tons 




Exercises 63 


1. 9-872 


2. -6169 3. 97-61 4. -00005861 


5. 775- 6 


6. 343'9 7. 2122 




Exercises 64 


1. 7'5 l6 2. 


14-76 3. -4433 4. -04062 5. 4-607 6. 104-3 


7. 2-089 


8. 2-823 9. 1-049 10. -9895 11. -2377 


12. -2235 


13. -144 14. -02057 15. 2-158 16. 13-47 ms - 


17. 5-343 sq.ft. 


18. 2-309 19. 3-773 20. 3-534 21. 9-668 


22. 7-37 **. 


23. 21-47 24. 4-024 25. 5459 amps. 


26. 8-507 


27. -1748 in. 28. 55-9 29. 8*299, say 8-3, H. P. 


30 (a) 6-485, i. 


e., 6 threads ; (b) 8-698, i. e., 9 threads. 



Exercises 65 



1. 3-142 2. 1-068 3. 1-646*, say ij* 4. -4889 
5. 9-21 ft. 6. -8549 7. 3-582 ft, 8. -001076 
9. 2i,2oocu. ft.perhr. 10. 1-28 11. 35-141 12. 2-71 
13. 1-31 14. 1 15-8 m. p. h. 15. 8-692 16. 17,020 ft. 




Exercises 66 




1. 3855 yds. 


1543 yds. 3. 5751 ft. 


4. 4158 ins. 


5. 334' 2 ins - 


6. 210 ins. 7. 66-5 ins, 


8. 11-896 ft. 


9. -601 yd. 


10. -5274 mile 11. 69'-8* 


12. i8 3 '-2* 




13. 10* 14. 4'-4|" 






Exercises 67 




1. I3'-4F 


2. i5'"3iT 3. IQ'-I* 


4. i9'-2i" 


5. 9'-5r; "iV 


6. 9'-5" 7. 28 


8. 44 


9. 3 0I 4"; i '94* 


10. 52 11. 7-64" 


12. 24 ins. 


13. 13-17* dia.; -524* pitch 14. 5650 ft. per min. 


15. 1528 r. p.m. 


16. 864 ft. per min. ; 137-5 r.p.m. 17- 4''7J" 


18. ig'-s" 


19. 7'-3i" 


20. 4'-5i* 21. n / -2* 


22. 2 '-7* 


23. 5 '-7" 


24. 9'-2* 25. 55 ft. 


26. i2 / -ni* 


27. n'-6i" 


28. i 5 '-ir 29. 34'-7* 


30. 3 X -3J* 




31. 1500 ft. 32. jjj* 






Exercises 68 




1. Simple 11-78* 


; more exact 12-96* 2. 39- 


27'; 39-68* 


3. 34-54"; 34-6* 


4. -142* 5. 8-33" 6 


. 9" 7. 24 


8. 5792 ft. 


9. 97 10. 116 11 


. 35 12. 1-78* 


G G 







448 ANSWERS TO EXERCISES 

Exercises 69 

1. 20' too high 2. Yes 3. 45 59' * 19 ' 

5. 17 16' 48" 6. 75 3' 36" 7. 47' 6" 8. 25-25 

9. 30-314 10. 65-375 11. 11-0972 12. 49-23 

Exercises 70 

1. 52'-2 1* 2. 26'-n" 3. i3'-9* 4, i9'-5i* 

5. 23'-9j" 6. AB = i6'; AC = 27'-8J" 7. i3'-ioj" 

8. 7'-!* 9. 5-96 ft. 10. lo'-ij* 11. -5001 

12. -684 13. ij* 14. 34'-7 /r 15. 393'-" 

16. i6'-iol* 17. 245'-ioJ" 

Exercises 71 

1. i5'-o|* 2. 37i" 3. 22'-5J" 4. 3'-8* 

5. 4'-ni" 6. 2 '-3 1" 

Exercises 72 

1. 59-6 sq.ft. 2. 111-75 sq.ft. 3 - I 37 6 lbs - 

4. 101 J tons 5. 1950 Ibs. per sq. ft. 6. 3*4*5 s q- ms 

7. 179 sq, ins. 8. 338 sq. ins. ; 34-1% 9. 29-5 tons per sq. in. 

10. -412 sq. ins.; 36-8% 11. 2-485 sq. ins. 12. 45-1 sq. ins. 

13. 13-1 amps. 14. 8400 Ibs. 15. 33 sq. ins. 
16. !! sq. ins.; '001173 ohms. 17. 60 sq. ins. 18. 99 sq. ins. 
19. 2-95 sq. ins. 20. 132-6 sq. ft.; 16,570 Ibs. 21. 8-38 sq. ins. 

22. -- 23. 1080 sq. ins. 24. 15 sq. ins. 

25. 19-34 sq. chains 26. 32-5 sq. ft. 27. 1-92 

28. 12-25 sq. ins. (both sides taken : edges neglected) ; i-i amps. 

29. 8-435 sq. ins. 30. 9-835 sq. ins. 31. 2-183 sq. ins. 
32. Embankment 389 sq. ft. ; cutting 554 sq. ft. 

Exercises 73r 

1. 254 sq. ft. 2. -00933 sq. in. 3. 18-55 sq. ins. 

4. Centre -0579 sq. in. ; Outer -0427 sq. in. ; Total -1006 sq. in. 

5. 12 sq. ins. 6. 2486 Ibs. 7. 11-03 s q- ft 8. 158 sq. ft. 

9. -586 sq. in. ; 58-6 tons 10. -697 sq. in. ; 2-8 tons 

11. i372lbs. 12. 2-37:1 13. -312 14. 32 15. 1 786 Ibs. per sq. in. 
16. 50-25 sq. ins.; Yes 17. 8-44 sq. ins. 18. 311 sq. ins. 

19. 3:1 20. 5 21. 8f 22. -048" 23. -605" 

24. 149-2 sq. ins.; 13!" dia. 25. io* 26. -182*, say ^ 6 " 

27. 2-83", say 2}' 28. 3F 29. 7-58", say 7 f* 

30. H.P. = 2'-9*; I.P. - 4 '-7i* 



1. 23 sq. ins. 
3. 1-23", say ij'dia. 
6. 27-2 sq. ins. 
9. L.P. = 176*7 sq. ins. 


2. 44-1 sq. ins.; 
4. 4*1 sq. ft. 
7. I9'i sq. ins. 
; H.P. = 63-7 sq. ins.; 


154-5 tons 
5. 12-75 sc 
8. 2}' 
L.P. : H.P. 



ANSWERS TO EXERCISES 449 

Exercises 74 



ins. 

= 2-77:1 
10. 154 lbs. per sq. in. 

Exercises 75 

1. 20-25 sq. ins. 2. 2-33 sq. ft. 3. -417 sq. in. 

4. 1-035 sq. ins. 5. 2-46 sq. ft. 6. 2-456sq. ft.; -163% low 

7. 2-451 sq. ft.; -366% low 8. 1-48 sq. ft. 

9. 1-42 sq. ft.; 4-05 % 10. 4-63 sq. ft. 11. 5-52 ft.; '838 

12. (a) n-27sq.ft.; (b) 8-61 ft. ; (c) 1-31 13. 12-07 sq. ft. 

14. 1371 sq. ft. 15. 50-93 sq. ft. 16. 60-9 sq. ins.; 2-155 times 
17. 44 sq. ins.; 1-56 times 18. 121 -7 sq. ins. 19. 86-8 sq. ins. 
20. 127 sq. ins. 21. 102 sq. ins. 22. 7-06 sq. ft. 23. 2170 lbs. 

Exercises 76 

! 95*3 s q- ms 2. 46-8 sq. ft. 3. 1124 sq. ft. 

4. 245-5 sc l* ms - 5. 28-86 sq. ft. 6. 26-1 inch tons 

7. 1910 sq. chns. 8. 7-38 sq. ins. 9. -96"; 48 lbs. per sq. in. 

10. 552"; 5-52 lbs. per sq. in. 11. -385"; 93-5 lbs. per sq. in. 

12. 329*; 98-7 lbs. per sq. in. 

Exercises 77 

1. 1736 cu. ins. 2. 801 cu. ft. 3. 33,700 gals, per hr. 

4. 14-48 galls, per min. 5. 28-35 cu - ft- P 61 m in. 

6. 734 lbs., or 6 cwts. 2 qrs. 6 lbs. 

Exercises 78 

1. 371 cu. ft. 2. 204-7 cwts., say 10 tons 5 cwts. 

3. 25,450 galls. 4. 390 galls. 5. 31-5 cwts. 

6. 162-5 lbs. 7. 26 galls. 8. 15 ft. 
9. 90 lbs per yd. 10. 3'-n" 11. 4 '-9* 

12. 206-3 cu. ins. 13. 12 tons 14 cwts. 14. 2-5 sq. ft.; 13!" 

Exercises 79 

1. 2,032,000 cu. ft. 2. 49 cu. ft. 3. 3-68 cu. ft. 

4. 22 galls. 5. 4-59 lbs. 6. 4440 galls, per min. 

7. * 8. 6-85", say 7* dia. 9. 2-295', sav 2 '-4* 
10. 3200 cu. ft. per sec. 11. 8250 cu. ft. 12. 482*5 lbs. 

13. 5490 galls. 14. 6-75% of working volume; 6-32% of total volume 

15. 825 cu. ft. 16. -927 lbs. per link; 33-35 lbs. per fathom 
17. 13*75 Mbs. 18. 1180 lbs. 19. 36*2 lbs. per ft. 
20. 151*5 tons 21. 10,240 galls. 22. -2275*, sa X I" 



450 ANSWERS TO EXERCISES 

Exercises 80 

1. 3'33 cu. yds.; 9450 Ibs. 2. 9130 Ibs. 3. 63,300 galls. 

4. (a) i-64lbs.; (6) 2-i81bs. 5. 1-58 Ibs. 6. 2-97 Ibs. 

7. 2*29 galls. 8. 12-86 galls. 9. 31-6 Ibs. 10. 18-9 galls. 
11. '104 Ibs. 12. 62 cu. ft.; 7440 Ibs. 13. i-95lbs.; 8s. 6d. 

14. 23olbs. 15. 22-85 galls. 16. -475 cu. ft.; 57 Ibs. 

17. 404,000 Ibs. 18. 886 Ibs. per yd. 

Exercises 81 

1. 2-84 Ibs. 2. 89-8 cu. ins.; 13-8, i.e., 13 8. 2-045 cu - ft 

4. 86-7 galls. 5. 58,200 galls. 6. 80-5 cu. ft. 

7. 37-8 Ibs. 8. 26-8 Ibs. 9. 12-8 Ibs. 10. 8 Ibs. 

Exercises 82 

1. 104*2 cu. ins. ; J 2. 8-54" 3. 99-6 cu. ins.; cone is one third 
of cylinder volume 4. 615 cu. ins.; 7-82*, say 8" 5. 4*07" 

6. i ton 14 cwts. 7. 1918 cu. ft. ; 47-95, say 48 tons 

8. 13720 cu. ft.; 659,000 Ibs. or 294 tons 9. 417,000 galls. 

10. 27*1 galls. 11. 22*9 tons. 12. 4,090,000 galls. 

Exercises 83 

1. 6653 sq. ft. 2. 1440 sq.ft. 3. 7480 sq.ft.; 6-95 Ibs. per sq. ft. 
4. 317 Ibs. per sq. in. 6. 402 units per sq. ft. 6. 75-4 sq. ins. 

7. ij" 8. - 9. 92-6 Ibs. 10. -15 Ib. 

4 

11. -897 watt per sq. in. 12. 440 sq. ins. 13. 2-37", say 2\" 
14. 134*6 sq. ft. 15. 40-8 Ibs. 16. 4-39 ft., say 4 '- 4 J" 
17. 43'3 sq. ft. 18. 30-3 sq. ft. 19. 39-3 sq. ft. 

20. 9530 Ibs. 21. -568 watt per sq. in. ; 2-41 cu. ins. 

22. (a) 1 1- 8 sq. millimetres; (b) 254. 23. -206 Ib. 

Exercises 83a. Refer p. 453 
Exercises 84 

1. Base i* = 10; vertical i* =* 10 Ibs.; (a) 40-3 Ibs. per sq. ft.; 
(6) 48-5 Ibs. per sq. ft. 

2. Base i* = 5 mins.; vertical i" = io/ 

3. Base i* = 5 amps.; vert. J" = *2 volt; (a) 1-22 volts; (b) 1-78 
volts 

4. (a) 525 Ibs.; (6) 1500 Ibs.; (c) 2920 Ibs. 

5. (a) 545. ; (b) 625. ; (c) 1055. 

6. Base i* = 10 amps.; vertical I* = 20 volts 

7. Base i* = 10 m.p.h. ; vertical Y = 2 Ibs.; 2220 Ibs. 

8. (a) 2400 Ibs. per sq. in. ; (6) 6-3 to i 

9 ( a ) '57 so fin'; (&) '195 sq. in. ; (c) -36 sq. in. 
10. (a) 1-4, i.e., No. i S.W.G.; (b) 7-4, i.e., No. 7 S.W.G. ; (c) 18-6, 
i.*., No. 18 S.W.G. 



ANSWERS TO EXERCISES 451 

Exercises 85 

1. Base from 7, i* = 5 knot; vert. from o, = 200 H.P. ; 1660 
H.P. 

2. Base from o, J" = 2%; vert. from 100, i* = 2; (a) 105-1 
C.; (b) 26% 

4. Base from i, i* = -05 ; vert. from i, \* =* 2 ; (a) 1-175 S P- B T * I 
(b) 1-28 resistance 5. '0057 

7. Base from i, = -02; vert. from 1*4, i* = i ; 1*96 volts 

8. Base from o, " = 20; vert from 53, i* = 2lbs.; (a) 60*55 
IDS.; (b) 56-6 Ibs. 

9. (a) 18-6, i.e., No. 18 S.W.G.; (b) 14-9, i.e., No. 14 S.W.G. 
10. 109-5 

Exercises 86 

1. AB = 8*5, BC = 5, CA = 3-6. Note. These measurements 
should be made on the squared paper and not with a rule, as the 
squares are often not exact to size. 

2. AC = BD = 6-4 3. 7-9 4. 1-475 6. 308 6. 5350 
7. (a) -7 and -2-6; (b) -1-2; (c) -1-84 8. yj; -25 ; 4 

Exercises 87 



1. e = -2W 2. (a) P = 3-1 -f i'94 w ; (&) 12 165.; (c) 12 i6s. 

3. V = 1200 + 630*2 4. /= 18-5 '00684* 5. w = -65 -f '305H 

6. (a) 27,200 Ibs.; (b) 14,600 Ibs. ; W = 900 -f- 5*228 7. 710 

8. Total oil = 4-64 Ibs. ; oil per H.P. hour = -93 Ib. 

9. R = 63 -f 2 -72V ; the last point is disregarded ; it is far from 
the general direction of the others and indicates a blunder when 
testing. 10. L = 1-31 - '471* 

11. L t= 565 '795*. Note. On the vertical (latent heat) scale plot 
from 490, taking i" = 20. 12. P = 9'56D 13 

13. T = 100 -00096711. 

Exercises 88 

1. 12 2. 2-2 3. 8-4 4. -585 

5. -0308 6. -0102 7. 885 8. 90-5 

9. -015 10. -425 11. -525 12. 25 

13. -0274 14. 11*65 15. 4'7 16. -264 

17. 2-29 18. -916 19. 648 20. 110-4 

Exercises 89 

1. 77 2. 2-16 3. i8'2 4. 71 

5. 1000 6. -316 7. 195 8. *86 

9. -0054 10. ix-3 11. -57 12. 1640 

13. 107-5 14. 717-5 15. -315 16. 186 

17. 92-1 18. 9'34 19. 32-3 20. 1058 

21. 37'35 22. 132,100 23. 63 24. 714 25. 10070 



452 



ANSWERS TO EXERCISES 



1. 9'7 

5. 42-5 

9. 5-8 
13. -1471 
17. 101,700 



Exercises 90 



2. -103 

6. 4-56 

10. -00991 

14. -0407 

18. -275 



3. 
7. 



63 
1-16 
11. 2-34 
15. -468 
19. 70 



Exercises 91 

See Answers to Exercises 18. 



4. 5-2 

8. -98 

12. 9-7 

16. -1076 

20. 7"4i 



Exercises 92 



1. 2nd : 51 
5. 2nd : 9-925 
9. 2nd : 3*42 

13. -624 

17. -432 



1. 6,760,000 

5. 9700 

8. -00001156 



2. ist : 211 


3. ist : 1-245 


4. ist : 14-6 


6. ist : 27-15 


7. ist : -225 


8. 2nd : -0583 


10. 2nd : -554 


11. 15-81 


12. 2-7 


14. 95*9 


15. 10-3 


16. 2-292 


18. 7-38 


19. 1-475 


20. 5-54 



21. 17 



22. -9125 



Exercises 93 

2. 1,981,000,000 
6. 543,000 
9. 136-9 



3. 2-4 



4. 45,400 

7. -00255 

10. -09425 



Exercises 94 

1. (a) tan A; (b) cos A; (c) sin A; (d) tanC; (e) cos C. 

2. (a) -78; (b) -6158; (c) -6158; (d) -7894; (e) 1-282; (f) -7894. 

3. (a) -342; (b) -7002; (c) 1*0724; (d) -3746; (e) -9653; (f) '8192. 



1. -9659. 
6. -5317- 



1. 64. 
7. 47. 



1. 1-505- 
6. 5'7i5. 



2. 2-9042. 
7. 57-2900. 

2. 14. 
8. 31. 



2. 1138. 
7. 93*08. 



1. 8. 

5. e = 28' 



Exercises 95 
3. -3090. 
8. -6947- 

Exercises 96 / 
3. 39. 4. 88. 

9. 1 8. 10. 49. 

Exercises 97 
3. 39. 
8. 49-52. 

Exercises 98 
3. 35- 



4. -6947. 
9. "2250. 

5. 8. 
11. 31 

4. 31. 
9. 42. 

4. d = 



5. -7193- 
10. -9976. 



6. 69. 
12. 72. 



5. 4*017. 
10. 38. 



2. 29. 
9 = 63. 6. WL (cos j8 cos a.) 

8. 1-099". 9. (a) ADB = 2R sin ~- ; (b) DE = R (i cos 

10. a = 37 approx. ; angle turned by pointer = 148 approx. 



7. 1-171". 
90 \ 
N/- 



ANSWERS TO EXERCISES 453 

Exercises 99 

1. (A) 1-56"; (B)3'92"; (C)o-o 9 " 

2. (D) 2-731" ; (E) 0-987" ; (F) 1-069" ; (G) -028- ; (H) 4-398" ; 

(J) o-oi"; (K) 2-051"; (L) 3*549"; (M) 0-625" 

Exercises 100 

1. fa in. 2. -001" 3. -002" 4. -ooi" 

5. 6 minutes 6. 3 minutes 7. 20 seconds 

8. (A) i^ ins. ; (B) 3 f -f & = 3 J ins. ; (C) 2 J- -f ^ = 2^- ins. 

9- (D) *&tife-2i%ins.; (E) if + T * ff = i& ins. ; (F) ft + ij ff 

= -if ins. 
10. (G) 29-9*; (H) 17-416; (K) 26-134 11. (L) 4 10'; (M) 18 40' 

12. (N) 23-3 -23 1 8'; (O) 5'8 =5 4S' 

13. (P) 10 15'; (Q) 45 39' 14. (R) 2 18'; (S) 42 43' 

Exercises 101 

1. (A) -455"; (B) -284"; (C) -I75"; (D) - 3 7"; (E) '213-; (F) -027" 

2. (G) -3692",;.*. ^CX)"; (H) -1875"; (J) 5938' / ,t.e. -594" 



Exercises 102 

(A) -0813-; (B) -3148"; (C) -3512" 



Exercises 20a 

1. 5'734 2. 2828-55, say 2829 

3. (a) 5.51 lb. persq. ft.; (6) 16-54 lb - per H.P. 

4. 3-762" 5. 9-23% 6. 6-4 to i 
7. (a) 10 lb. 5 | oz.; (b) 3-62 % 8. 27-74 lb. 

Exercises 49a 
1. -553 2. -000542 3. ^P-^ 2 4. 175 5. 1-52 

Exercises 83a 
1. 4-71 sq. ins. 2. 2727 lb. 3. 208' - 10" 4. 503 lb. 

5. 38-3", say 38^" 6. (i) -9995 sq. mm. ; (2) 1000 

7. 26 teeth; -059" 8. 1-509", say ij" 9. 8-06 lb. 

10. -708 gallon 11. 88-4 lb. 

12. No; calculated weight is practically i lb. 13. 17-86 gallons. 



454 



MATHEMATICAL TABLES 

LOGARITHMS 



























































10* 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


9 13 
8 12 


17 21 26 
16 20 24 


SO 34 38 
28 32 37 


11 


0414 


046? 


1 0492 


0531 


0569 


0607 


0645 


0682 


0719 


0766 


8 12 
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 


7 11 

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 


7 10 
7 10 


13 16 20 
12 16 19 


23 26 30 
22 25 29 


14 


1461 


1492 


1623 


1653 


1584 


1614 


1644 


1673 


1703 


1732 


6 9 
6 9 


12 15 18 
12 15 17 


21 24 28 
20 23 26 


15 


1761 


1790 


1818 


1847 


1876 


1903 


1931 


1959 


1987 


2014 


6 9 
6 8 


11 14 17 
11 14 16 


20 23 26 
19 22 25 


16 


2041 


2068 


2096 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


5 8 
5 8 


11 14 16 
10 13 15 


19 22 24 
18 21 23 


17 


2304 


2330 


2365 


2380 


2405 


2430 


2455 


2480 


2504 


2629 


358 
267 


10 13 15 
10 12 15 


18 20 23 
17 19 22 


18 


2553 


2677 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2766 


267 
267 


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 


16 18 20 
15 17 19 


20 


3010 


3032 


,3054 


S075 


3096 


3118 


3139 


3160 


3181 


3201 


2 4 6 


8 11 13 


15 17 19 


21 
22 
23 
24 


3222 
3424 
3617 
3802 


3243 
3444 
3636 
3820 


3263 
3464 
3655 
3838 


3284 
2483 
3674 
3856 


3304 
S502 
3692 
3874 


3324 
3522 
3711 
3892 


3345 
3541 
3729 
3909 


3365 
3560 
3747 
3927 


3385 
3579 
3760 
3945 


3404 
3598 
3784 
3962 


246 
246 
246 
245 


8 10 12 
8 10 12 
7 9 11 
7 9 11 


14 16 18 
14 15 17 
13 16 17 
12 14 16 


25 


3979 


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 
4G09 
4767 


235 

236 
235 


7 8 10 
689 
689 


11 i3 15 
11 13 14 
11 12 14 
10 12 13 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 






10 11 13 


31 
32 
83 
84 


4914 
6051 
5185 
5315 


4928 
5065 
6198 
5328 


4942 
5079 
5211 
5340 


4955 
5092 
5224 
5353 


4969 
6105 
6237 
5366 


4983 
6119 
5250 
5378 


4997 
6132 
6263 
5391 


601X. 
5145 
6276 
5403 


5024 
6169 
6239 
6416 


6038 
6172 
6302 
6428 


134 
134 
134 
134 


6 7 8 
678 
668 
568 


10 11 12 
9 11 12 
9 10 12 
9 10 11 


25 


5441 


5453 


5465 


6478 


6490 


6602 


6614 


6527 


6639 


6551 


124 


567 


9 10 11 


36 
37 

88 


5563 
5682 


5575 
5G94 


6587 
5705 


5599 
6717 


6611 
6729 


5623 
6740 


5635 
6752 


6647 
6763 


6658 
6716 


6670 
5786 


124 
123 


567 
667 


8 10 11 
8 9 10 


39 


5911 


5622 


6933 


6944 


5955 


6966 


6977 


6988 


5999 


6010 


123 


467 


8 9 10 


4.0 


6021 


603 1 


6042 


6053 


6064 


6076 


6085 


6096 


6107 


6117 


123 


466 


8 9 10 


41 
12 
43 
44 


6128 
6232 
6835 
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 
6613 


6222 
6325 
6425 
6522 


123 
123 
123 
123 


456 
456 
466 
456 


789 
789 
789 
789 


45 


6532 


6542 


6551 


C561 


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 


6656 
6749 
6839 
6928 


66f>5 
6758 
6848 
6937 


6676 
6767 
6857 
6946 


6684 
6776 
6866 
6955 


6693 
6785 
6875 
6964 


6702 
6794 
6884 
6972 


6712 
6803 
6893 
6981 


123 
123 
123 
133 


456 
465 
445 
445 


778 
678 
678 
678 


50 


6990 


6999 


7007 


7016 


7024 


7031 


7043 


7050 


7059 


7087 


123 


345 


678 



MATHEMATICAL TABLES 
LOGARITHMS 



455 



























































51 


7076 


7084 






Tiin 


7118 




7141 


7"! 4 










52 
53 
54 


7160 
7243 
7324 


7168 
7251 
7332 


7177 
7259 
7340 


7185 
7267 
7348 


7193 
7275 
7356 


7202 
7284 
7364 


7210 
7292 
7372 


7218 
7300 
7380 


7226 
7308 
7388 


7236 
7316 
7396 


122 
122 


346 


677 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 






567 


56 


7482 


7490 






7613 


7520 


7528 


7636 


7543 


7561 








57 
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 


345 


567 


60 


7782 


7789 


7796 


780S 


7810 


7818 


7825 


7832 


7839 


7846 


112 


344 


666 


61 
62 
63 


7853 
7924 


7860 
7931 
8000 


7868 
7938 


7875 
7945 


7882 
7952 


7889 
7959 
8028 


7896 
7966 


7903 
7973 
8041 


7910 
7980 
8048 


7917 
7987 


112 
112 


244 
334 


666 
566 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


112 


334 


556 


AK 
























































66 
67 
68 


8195 
8261 


8202 
8267 


8209 
8274 


8215 
8280 


8222 
8287 


8228 
8293 
8357 


8235 
8299 
8363 


8241 

8306 
8370 


8248 
8312 
8376 


8254 
8319 
8382 


112 
112 


334 

334 


566 
656 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


112 


234 


456 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8600 


8606 








71 
72 
73 
74 


8513 
8573 
8633 
8692 


8519 
8579 
8639 
8698 


8525 

8585 
8645 
6704 


8531 
8591 
8651 
8710 


8537 
8597 
8657 
8716 


8543 
8603 
8663 
8722 


8549 
8609 
8669 
8727 


8555 
8616 
8675 
8733 


8561 
8621 
8681 
8739 


8567 
8627 
8686 
8745 


112 
112 
112 
112 


234 
234 
234 
234 


465 
466 
465 
466 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


112 


233 


465 


76 
77 
78 
79 


8808 
8865 
8921 
8976 


8814 
8871 
8927 
8982 


8820 
8876 
8932 
8987 


8825 
8882 
8938 
8993 


8831 
8887 
8943 
8998 


8837 
8893 
8949 
9004 


8842 
8899 
8954 
9009 


8848 
8904 
89GO 
9015 


8854 
8910 
8965 
9020 


8859 
8915 
8971 
9025 


112 
112 
112 
112 


233 
233 
233 
233 


465 
446 
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 
9238 
9289 


112 
112 
112 
112 


233 
233 
233 
233 


445 
446 
446 
446 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 




233 


445 


86 
87 
88 
89 


9345 
9395 
9445 
9494 


9350 
9400 
9450 
9499 


9365 
9405 
9455 
9504 


9360 
9410 
9460 
9509 


9365 
9415 
9465 
9513 


9370 
9420 
9469 
9518 


9375 
9425 
9474 
9523 


9380 
9430 
9479 
9528 


9385 
9435 
9484 
9533 


9390 
944O 
9489 
953S 


112 

Oil 
Oil 
Oil 


233 
223 
223 
223 


446 
344 
344 
344 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


Oil 


223 


344 


91 
92 
93 
94 


9590 
9638 
9685 
9731 


9595 
9643 
9689 
9736 


9600 
9647 
9694 
9741 


9605 
9652 
9699 
9745 


9609 
9657 
9703 
9750 


9614 
9661 
9708 
9754 


9619 
96G6 
9713 
9759 


9624 
9671 
9717 
9763 


9628 
9675 
9722 
9768 


9633 
9680 
9727 
9773 


Oil 
Oil 
Oil 
Oil 


223 
223 
223 
223 


344 
344 
344 
344 


95 


8777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


Oil 


223 


344 


98 
97 
98 


9823 
9868 
9912 


9827 
9872 
9917 


9832 
9877 
9921 


9836 
98S1 
9926 


9841 
9886 
9930 


9845 
989Q 
9934 


9850 
9894 
9939 


9864 
9899 
9943 
9987 


9859 
9903 
9948 


9863 
9908 
9952 


oil 

Oil 
Oil 


223 
223 
223 


344 
344 
844 































456 



MATHEMATICAL TABLES 

AN TILOG ABITHMS 








1 


2 


s 


4 


5 


6 


7 


8 


9 


123 


456 


789 


oc 


1000 


100S 


1005 


100 


J009 


1012 


1014 


1016 


1019 


1021 


1 


111 


231 


o 
o 
o 




1023 
1047 
1072 
1096 


1026 
1050 
1074 
1099 


1028 
1052 
1076 
1102 


103 
1064 
107 
1104 


1033 
1067 
1081 
1107 


1035 
1069 
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 


232 
232 
222 
233 


Off 


1122 


1125 


1127 


1130 


1133 


1136 


1138 


1140 


1143 


1146 








06 
07 
08 
09 


1148 
1176 
1202 
1230 


1161 

1178 
1205 
1233 


1153 
1180 
1208 
1236 


115 
118 
121 
133 


1169 
1186 
1213 
1242 


1161 
1189 
1216 
1246 


1164 
119 

121 
124 


1167 
1194 
1222 
liJOO 


1169 
1197 
1225 
1263 


1172 
1199 
1227 
1266 


1 1 
Oil 
Oil 
Oil 


1 1 2 
1 1 2 
1 1 2 
112 


222 
222 
223 
223 


10 


1269 


1262 


1265 


12C8 


1271 


1274 


127 


1279 


1282 


1285 


1 1 


112 


2 3 3 


11 
12 
13 
14 


3288 
IS 18 
1349 
1380 


1291 
1321 
1352 
1384 


1294 
1324 
1356 
1387 


1297 
1327 
1338 
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 


1 1 
Oil 
1 1 
1 1 


122 
1 2 2 
122 
122 


223 
12 2 3 
233 
233 

5TT" 


15 


1413 


1416 


1419 


1422 


1426 


1429 


1432 


1435 


1439 


1442 


Oil 


122 


16 
17 
18 
19 

80 


1445 
1479 
1514 
1549 


1449 
1483 
1517 
1552 


1452 
1486 
1521 
1566 


1455 
1489 
1624 
1660 


1469 
1493 
1628 
1563 


1462 
1496 
1531 
1567 


1466 
1600 
1635 
1670 


1469 
1603 
1538 
1574 


1472 
1607 
1542 
1578 


1476 
1510 
1645 
1581 


Oil 


1 2 2 


2 3 S 


Oil 
1 1 


122 
122 


233 
333 


1585 


1589 


1692 


1596 


1600 


1603 


1607 


1611 


1614 


1618 


1 1 


1 2 2 


3 3 3 


21 
22 
23 

24 

25 

26 
27 
28 
89 


1622 
1G60 
698 
738 


1626 
1663 
1702 
1742 


1629 
1667 
1706 
1746 


1633 
1671 
1710 
1750 


1637 
1676 
1714 
754 


1641 
1679 
1718 
1758 


1644 
1683 
1722 
1762 


1648 
1687 
1726 
1766 


1652 
1690 
1730 
1770 


1656 
1694 
1734 
1774 


Oil 
Oil 
Oil 
1 1 


222 


833 


221' 
322 


334 
334 


778 


1782 


1786 


1791 


795 


1799 


1803 


1807 


1811 


1816 


Oil 


222 


334 


820 
8G2 
905 
950 


1824 
1866 
1910 
1954 


1828 
1871 
1914 
1959 


1832 
1875 
1919 
1963 


837 

879 
923 
9U8 


1841 
1884 
1928 
197S 


1845 
1888 
1932 
1977 


849 
892 
936 
982 


1854 
1897 
1941 
1986 


185S 
1901 
1915 
1991 


Oil 
Oil 
Oil 
Oil 


223 
223 
223 
223 


334 
334 
344 
3 4 4 


so 


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 
21X>3 


2061 
109 

168 
208 


2066 
2113 
2163 
2213 


2070 
2118 
2168 
2218 


2075 
123 
173 
223 


2080 
2128 
2178 
2228 

3 


2084 
2133 
2183 
2234 


Oil 
Oil 
Oil 
112 


223 
223 
223 
233 


344 
344 
344 
445 

445 


35 


2239 


2244 


2249 


2254 


259 


2265 


2270 


276 


2280 


2286 


1 1 2 


3 3 3 


36 
37 
38 
39 


22$1 
2344 
2399 
2455 


2296 
2350 
2404 
2460 


2301 
2365 
2410 
2466 


2307 
23GO 
2415 
2472 


312 
366 
421 
477 


2317 
2371 
2421 
2483 


2323 
2377 
2432 
2489 


328 

382 
438 
495 


2333 

2388 
2443 
2500 


2339 
2393 
2449 
2506 


112 
112 

112 
112 


235 
233 


446 
446 


233 


456 


40 


2512 


2518 


2523 


2529 


535 


2541 


2547 


553 


2559 


2564 


112 


234 


456 


41 
42 
43 
44 


2570 
2630 
2G92 
2754 


2576 
2636 
2G98 
2761 


2582 1 
2642 
2704 
2767 


2588 
2649 
2710 
2773 


594 
655 
716 
780 


2600 
2661 
2723 
2786 


2606 
2667 
2729 
2703 


612 
673 
735 
799 


2618 
2679 
2742 
2805 


2G24 
2685 
2748 
2812 


112 
112 
112 
112 


234 
234 
334 
334 


455 

456 
456 
456 

556 


45 


2818 


2825 


2831 


2838 


844 


2851 


2858 


864 


2871 


2877 


112 


334 


46 
47 
48 
49 


2884 
2951 
8020 
3090 


2891 
2958 
3027 
8097 


2897 
2965 
3034 ! 
8105 


2904 
2972 
3041 
8113 


911 
979 
3048 
119 


2917 
2985 
3055 
3126 


2924 
2992 
3062 
S1S3 


931 
999 
8069 
141 


2938 
3006 
3076 
8143 


2944 
3013 
3083 
3155 


112 
112 
113 
112 


3 S * 
334 
344 
S 4 4 


6 6 
656 
666 
666 



MATHEMATICAL TABLES 
ANTILOQARITHMS 



457 








1 


2 


3 


4 


6 


6 


7 


8 


9 


1 2 3 


456 


789 


50 


3162 


8170 


S177 


8184 


3192 


3199 


8206 


3214 


3221 


3228 


112 


844 


667 


6 
52 
53 
54 


3236 
3311 
3388 
3467 


3243 
3319 
3396 
3475 


8251 
8327 
3404 
8483 


8258 
3334 
3412 
3491 


3266 
3342 
3420 
3499 


3273 
3360 
3428 
3508 


8281 
3357 
3436 
3516 


3289 
3365 
3443 
8524 


3296 
3373 
3451 
3632 


3304 
3381 
3459 
3540 








122 
122 
122 


346 
346 
346 


567 
667 
667 


55 


3548 


3556 


3565 


3673 


3581 


35S9 


3597 


3606 


3614 


8622 


122 


346 


677 


56 

57 
58 
69 


3631 
3716 
3802 
3890 


3639 
3724 
3811 
3899 


3648 
3733 
3819 
3908 


3656 
3741 

5828 
8917 


3664 
3750 
3837 
3926 


3673 
3758 
3846 
3936 


3681 
3767 
3855 
3946 


3690 
8776 
3864 
8954 


3698 
3784 
3873 
3963 


3707 
3793 
3883 
3972 








123 


346 


678 


128 


455 


678 


60 


3981 


3990 


8999 


4009 


4018 


4027 


4036 


4046 


4055 


4064 


123 


456 


678 


61 
62 
63 
64 


4074 
4169 
4266 
4365 


4083 
4178 
4276 
4375 


4093 
4188 
4585 
4385 


4102 
4198 
4295 
4395 


4111 

4207 
4305 
4406 


4121 
4217 
4316 
4416 


4130 
4227 
4325 
442G 


4140 
4236 
4335 
4430 


4150 
4246 
4345 
4446 


4159 
4256 
4355 
4457 


123 
123 
123 
123 


4 6 
456 
456 
456 


789 
789 
789 
789 


65 

66 
67 
68 
69 

-70 


4467 


4477 


4487 


4498 


4608 


4619 


4529 


4539 


4550 


4560 


123 


456 


789 


4571 
4677 
4786 
4898 


4581 
4688 
4797 
4909 


4592 
4699 
4808 
4920 


4603 
4710 
4819 
4932 


4613 
4721 
4831 
4943 


4624 
4732 
4842 
4955 


4634 
4742 
4853 
4966 


4645 
4753 

4864 
4977 


4656 
4764 
4875 
4989 


4667 
4775 

4887 
6000 


1 2 3 
123 
1 2 3 
123 


466 
457 
467 
667 


7 9 10 
8 9 10 
8 9 10 
8 9 10 


6012 


5023 


5035 


5047 


6058 


6070 


6082 


6093 


6105 


6117 


1 2 4 | 6 6 7 


8 9 11 


71 
72 
73 

74 


6129 
6248 
6370 
6495 


6140 
6260 
5383 
5608 


6152 

6272 
6395 
6621 


6164 
5284 
5408 
5534 


6176 
6297 
6420 
5646 


6188 
6309 
6433 
6559 


5200 
6321 
5445 
6672 


6212 
6333 
6468 
6685 


6224 
5346 
6470 
6598 


5236 
5358 

5483 
6610 


124 
124 
134 
134 


567 
667 
568 
568 


8 10 11 
9 10 11 
9 10 11 
9 10 12 


75 


6623 


6636 


6649 


5662 


6675 


6689 


6702 


6715 


5728 


5741 


134 


578 


9 10 12 


76 

77 
78 
79 


6764 
6888 
6026 
6166 


6768 
5902 
6039 
6180 


6781 
6916 
6053 
6194 


6794 
6929 
6067 
6209 


5808 
5943 
6081 
6223 


6821 
6957 
6095 
6237 


5834 
6970 
6109 
6252 


6848 
6984 
6124 
6266 


5861 
5998 
6138 
6281 


6875 
6012 
6152 
6295 


134 
1 3 4 
134 
134 


678 
678 
678 
679 


9 11 12 
10 11 12 
10 11 13 
10 11 13 


80 


6310 


6324 


6339 


6363 


6368 


6383 


6397 


6412 


6427 


6442 


1 3 4 


679 


10 12 13 


81 
82 
83 
84 


6457 
6607 
6761 
6918 


6471 
6622 
6776 
6934 


64-80 
6637 
6792 
6960 


6501 
6653 
6808 
6966 


6616 
6668 
6823 
6982 


6631 
6683 
6839 
6998 


6546 
6699 
6855 
7016 


6661 
6714 
6871 
7031 


6577 
6730 
6887 
7047 


6592 
6745 
6902 
7063 


236 
235 
235 
235 


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 


2 8 5 | 7 8 10 


12 13 15 


36 
87 
88 
80 


7244 
7413 
7686 
7762 


7261 
7430 
7603 
7780 


7278 
7447 
7621 
7798 


7295 
7464 
7638 
7816 


7311 
7482 
7656 
7834 


7328 
7499 
7674 
7862 


7345 
7516 
7691 
7870 


7362 
7534 
7709 
7889 


7379 
7551 
7727 
7907 


7896 
7568 
7745 

7925 


235 
235 
245 
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 


90 


7943 


7962 


7980 


7998 


8017 


8035 


8054 


8072 


8091 


8110 


246 


7 9 11 


13 15 17 


91 
92 
93 
94 


8128 
8318 
8511 
8710 


8147 
8337 
8531 
8730 


8166 
8356 
8551 
8750 


8185 
8376 
8670 
8770 


8204 
8395 
8590 
8790 


8222 
8414 
8610 
8810 


8241 
8433 
8630 
8831 


8260 
8463 
8660 
8851 


8279 
8472 
8670 
8872 


8299 
8492 
8690 
8892 


246 
246 
246 
246 


8 9 11 
8 10 12 
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 


96 
97 
98 
99 


9120 
9333 
9650 
772 


9141 
9364 
9672 
9796 


9162 
9376 
9594 1 
9817 


9183 
9397 
9616 
9640 


9204 
9419 
9638 
9868 


9226 
9441 
9661 
9886 


924? 
9463 
968) 
9906 


9268 
9484 
3705 
JW31 


9290 
9606 
9727 
9954 


9311 
9628 
9760 
9977 


246 
247 
247 
267 


8 11 IS 
9 11 13 
9 11 13 
9 11 14 


15 17 19 
15 17 2O 
16 18 20 
16 18 20 



458 



MATHEMATICAL TABLES 
TRIGONOMETRIC RATIOS. 





Ajigle. 


















De- 
grees. 


Radians. 


Chord. 


Sine. 


Tangent 


Go-tangent. 


Cosine 























CO 


1 


1-414 


1-5708 


90 


1 
2 
3 


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 
1-6359 
1-5184 
1-5010 


89 
88 
87 
86 


5 


0873 


087 


0872 


0875 


11-4301 


9962 


1-351 


1-4835 


85 


6 
7 
8 
9 


1047 
1222 
1396 
1571 


105 
122 
140 
157 


1045 
1219 
1392 
1564 


1051 
1228 
1405 
1584 


0-5144 
8-1443 
7-1154 
6-3138 


9945 
9925 
9903 
-9877 


1-338 
1-325 
1-312 
1-299 


1-4661 
1-4486 
1-4312 
1-4137 


84 
83 
82 
81 


10 


1745 


174 


1736 


1763 


5-6713 


9848 


1-286 


1-3963 


80 


11 
12 
13 
14 


1920 
2094 
2269 
2443 


192 
209 
226 
244 


1908 
2079 
2250 
2419 


1944 
2126 
2309 
2493 


5-1446 
4-7046 
4-3315 
4-0108 


9816 
9781 
9744 
9703 


1-272 
1-259 
1-245 
1-231 


1-3788 
1-3614 
1 3439 
1-3265 


79 

78 
77 
76 


15 


2618 


261 


2588 


2679 


3-7321 


9659 


1-218 


1-3090 


75 


16 
17 
18 
19 


2793 
2907 
3142 
3316 


278 
296 
313 

330 


2756 
2924 
3090 
3256 


2867 
3057 
3249 
3443 


3-4874 
3-2709 
3 0777 
2-9042 


9613 
9563 
9511 
9455 


1-204 
1-190 
1-176 
1-161 


1-2915 
1-2741 
1-2566 
1-2392 


74 
73 
72 
71 


20 


3491 


347 


3420 


3640 


2-7175 


9397 


1-147 


1-2217 


70 


21 
22 
23 
24 


3665 
3840 
1014 
4189 


364 
382 
399 
416 


3584 
3746 
3907 
4067 


3839 
4040 
4245 
4452 


2 6051 
2 4751 
23559 
2-2460 


9336 

9272 
9205 
9135 


1-133 
1-118 

1-104 
1-089 


1-2043 
1-1868 
1-1694 
1-1619 


69 
68 
87 
66 


25 


4*563 


433 


4226 


4663 


2-1445 


9063 


1-075 


1-1345 


65 


26 
27 
28 
29 


4538 
4712 

4887 
5061 


450 
467 
484 
501 


4384 
4540 
4G95 
4818 


4877 
5095 
5317 
5543 


20503 
1 9626 
1 8807 
1-8040 


8988 
8910 
8829 
8746 


1-060 
1-045 
1030 
1-015 


1-1170 
1-0996 
1-0821 
1-0647 


64 
63 
62 
61 


30 


6236 


518 


5000 


5774 


1-7321 


8660 


1-000 


1-0472 


60 


31 
32 
33 
34 


5411 
5585 
5760 
5934 


534 
551 
568 
585 


5150 
5299 
5446 
5592 


6009 
6249 
6494 
6745 


1-6643 
16003 
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 

H 


6283 
6458 
6632 
6807 


618 
635 
651 
668 


6878 
6018 
6157 
6293 


7265 
7536 
7813 
8098 


1-3764 
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 


50 


41 
42 
43 
44 


7156 
7330 
7505 
7G79 


700 
717 
733 
749 


6561 
6G91 
6820 
6947 


8693 
9004 
9325 
9657 


1-1504 
1-1106 
1-0724 
1-0355 


7547 
7431 
7314 
7193 


829 
813 
797 
781 


8552 
8378 
8203 
8029 


49 
48 
47 
46 


45 


7854 


765 


7071 


1-0000 


1-0000 


7071 


765 


7854 


45 








Cosine 


Oo-tangent 


Tangent 


Sine 


Chord 


Radians 


Degrees 


















Angl 


e 



APPENDIX 
VERNIERS AND MICROMETERS 



CONTENTS 

PAGE 

VERNIERS 460 

Introductory Vernier Calipers The One-Thousandth Scale The 
Back of the Caliper Least Count Various other Vernier Scales 
Lack of Coincidence. 

MICROMETERS 473 

Micrometer Wheel and Screw Micrometer Caliper Zero Adjustment 
To Use the Caliper Various Micrometer Tools Ten-Thousandth 
Micrometer Zero Setting of Ten-Thousandth Micrometer The 
Ratchet Stop. 



Answers to Examples will be found on page 453. 



APPENDIX 

VERNIERS AND MICROMETERS 

[This Appendix should preferably be studied with the various tools 

at hand.] 

VERNIERS 

Introductory. Mention has been made on p. 263 of a special 
device being necessary to read fine angular measurements such as 
the minute and second. Similarly, when length measurements 
correct to T ^ inch, and finer, are required, a finely divided scale 
will be found very difficult, or even impossible, to use. The 
smallest division obtainable on a steel rule is T ^ inch, and the 
lines are then so close together that a magnifying glass becomes 
necessary to take a reading. Such fine measurements, both linear 
and angular, are rendered readily possible by the use of an openly 
divided scale of ordinary pattern, in conjunction with an auxiliary 
scale called a vernier. By suitable arrangement this auxiliary scale 
can be made to read to almost any desired degree of accuracy in 
a comparatively easy manner. 

Vernier Calipers. The most common engineering application 
of the vernier is in the vernier caliper, a tool used in workshop 
and drawing-office for the accurate measurement of moderate 
lengths. Such tools are made to read in various scales, and for 
purposes of explanation the simple type shown at (A), Fig. 157, 
will be taken, in which the main scale S is divided into inches 
and tenths, and the vernier scale V is arringed to read hundred ths. 
Although not a common marking on vernier calipers, this provides 
the easiest introduction to the vernier principle. 

The vernier scale V consists of 10 equal divisions, whose over-all 
length is that of 9 main scale divisions, as shown in the full-size 
view at (B), Fig. 157. These divisions are numbered o to 10, 
starting from that nearest the zero on the main scale. Each 
vernier division is then T V of 9 main scale divisions, i. c. T V of 

460 



APPENDIX 



461 



9" = -09*. With the tool set as shown at (A) and (B) the dis- 
tance between the jaws is an exact number of tenths of an inch 
(actually 2-3"), and the two end lines on the vernier are in line 
(or "coincide") with lines on the main scale. The main scale 
division a being -i" and the vernier division b being -09", the 
distance c is evidently -i -09 = -01", i. e. one-hundredth of an 
inch. If now the moving jaw be moved to the right until line i 



Clamping Screw. 




Coincidence"^ A V fo /Coincidence r ^ /^Comodence-v 

^.D. 1 . 1 ' * I'/i? *' *T?i 1 .' .'.'.' J. 1 . 1 .) ^irhlr 

iHsi!"! 1111 \ / T 11 "" V /' M 

or c 5 ^J \ o J ht 5 I0 ) 5 

^^^i^'^""^'^**^^^ 1 **^^^^^/ ' ^^_- M ^^^*^^^i'ii-i^ WBBI^** * ""^^i 



\ Vernier 
reading 2-3 



\Vernier 

reading 2-31 



\Vernier 

reading 2'32. 



(2 /Coincidence 3 



I 

n 



^ Com* 1 



/ wiri^i wtc riv.^, vr Q v^^iri- * 

iM^i 1 1 1 .iTm^ if 1 1 I.JN iii i 

/| MI i 1 1 1 1 1 |v jj 1 1 1 1 1 1 1 1 1 1 ; 

|0 5 IOJ (O 5 |0 



Coincidence 2 



Vernier 

reading 2-4 



\ Vernier /^\ Vernier 

.^^ ,,,^ ^ , ~* readmgO*45 ^~/ reading 128 

Fig. 157. Simple Vernier Caliper, reading T $ v inch, with Sample Settings. 



on the vernier coincides with the nearest line on the main scale, 
the jaws will have opened -oi", and the new position of the vernier 
will be as at (C). Then the distance d is similarly *i" -09" = -01", 
and if the moving jaw be again moved to the right until line 2 on 
the vernier coincides with its nearest main scale division, as at (D) t 
a further opening of -01" will have taken place; the total distance 
between the jaws will then be 2*32" . This gradual movement of 
the moving jaw to the right can be continued, and the coincidence 



462 



APPENDIX 



of each successive line on the vernier will give the number of one- 
hundredths of an inch that the jaws are open beyond 2*3". When 
line 10 on the vernier is in coincidence, a total movement of 10 
one-hundredths or -i* will have been made from the first position, 
which is further indicated by line o on the vernier coinciding with 
2*4" on the main scale, as at (E). Then it will be seen that, at 
any particular setting, the number of the line on the vernier scale 
coinciding with any line on the main scale indicates the number 
of hundredths which the jaws are open beyond the number of 
exact tenths indicated by the zero of the vernier scale. 
To take a complete reading then, proceed thus : 

1. Read the inches and complete tenths up to the zero mark on the 
vernier. 

2. Note the line on the vernier which coincides with a line on the 
main scale. 

3. Add the figure noted in (2) as hundredths to the reading taken in (1). 



VemierScale 
reading 7ooo M5 - 




For derails of Mam and "** W 1 *" Parl-View < 

Vernier scales see Enlaced Views BacKFacc,: 

Fig ISQ.arA. of JawPointe.closed. -^ IN ft OUT I 

Fig. 158. Vernier Caliper, reading ^ v inch. (About half-size.) 

Thus at (D), Fig. 157, the vernier zero cuts off 2-3 on the main 
scale, and coincidence occurs at 2 on the vernier scale. Then the 
full reading is 2-3" + -02" = 2-32". 

With a very little practice this addition can be done mentally. 
As a further illustration, two other settings are shown at (F) and 
(G), Fig. 157, which should be easily understood. Exercises on 
reading this vernier are given in Exercises 94, No. I. 

The most widely used type of English-reading vernier caliper 
is shown in Fig. 158. The fixed jaw F is solid with the blade B, 
on which is marked the main scale of inches and tenths, each 
tenth being subdivided into 4 parts, i. e. fortieths. The moving 
jaw M carries the vernier scale, and can be clamped in any position 



APPENDIX 463 

by the set-screw P and the key K. An auxiliary runner R, which 
can also be clamped by screw Q, carries a milled edged nut N, 
between lugs, which is threaded on the adjusting screw S. The 
ends of the jaws are shaped down and rounded so that inside 
diameters can be measured. The smallest hole which can be 
gauged is determined, of course, by the width, w, across the 
points when the jaws are shut. This is usually either -25" or -3", 
according to the size of the tool ; this value, which may be found 
from the maker's catalogue, must be added to the reading taken 
when measuring inside diameters. 

To use the caliper, both screws P and Q are slackened, when 
the moving jaw M and the runner R may be pushed, together, 
into nearly the required position. Screw Q is then tightened, 
securing runner R to the blade. The nut N is now rotated to 
advance or withdraw the moving jaw until the object to be 
measured is just held, without pressure, between the jaws. Screw P 
is then tightened to secure the moving jaw from accidental move- 
ment until the reading of the vernier is taken. This fine adjust- 
ment is essential when readings are to be made to -ooi inch, although 
not necessary when only -01 is required. 

The One-Thousandth Scale.- The main scale is divided into 
tenths, and each tenth is subdivided into four parts, so that the 
smallest main scale division is J of ! = '025 inch or ^. The 
vernier scale has 25 divisions, whose total length is that of 24 
main scale divisions, see (A), Fig. 159, and which are numbered 
at every five from o to 25. Each vernier division is then ^ s of 24 
main scale divisions, i. e. 

JLof 24 = 24 - =-024, 
25 40 1000 ^ 

which is -ooi inch shorter than the main scale division of -025. 
The smallest reading is then '001 inch or one-thousandth, and is 
the amount added by successive coincidences of the vernier lines 
with the main scale. The complete reading of this vernier is then 
made as follows : 

1. Read the number of inches and complete tenths up to the vernier 
zero. 

2. Read off the number of small main scale divisions between the last 
complete tenth and the vernier zero. In each tenth the small divisions, 
in order, are (i) *025, (2) *050, (3) *075. Add the appropriate decimal 
to the reading in (1). 

ii H 



4 6 4 



APPENDIX 



3. Note the vernier graduation in coincidence with any line on the 
main scale and add this number of thousandths to the result of the 
addition in (2). 

Sample readings are shown in Fig. 159, at (B), (C), and (D), 
the diagrams being twice full size. 



[1 ^Coincidence 

1/2345 6/7 8 9 




|0 5 10 15 20 25 

Vernier reading I-OOO 




Coincidence V 

3V4 5 6 7 8 9 




O 5 10 15 20 25 

Vernier reading 1-353 




J .Coincidence 1 yCoincvdenoe^ J 

1/2 3 4 5 61 (5 6 7 8 9\ 

I/ I I I I I I I I ^ 




O 5 10 15 20 25 

i*To' N - 

Vernier reading O834- 
Fig. 159. Sample Settings of 




123 



O 5 10 15 20 25 
i^o IN - 

Vernier reading 1*595" 

inch Vernier. (About twice full size.) 



In Fig. 159:- 

At(B). 

The vernier zero cuts off 1-3 and two small main scale 
divisions, i. e. 1-3 + -05 . . . . . . . = 1-35 

Line 3 on the vernier coincides with a line on the main 
scale indicating an extra ^0%^ = '003 

Adding, the complete reading = 1*353 

At (C). 

Vernier zero cuts off 0-8 and one small main scale division, 
i. e. 0-8 + -025 ^ = 0-825 

Line 9 on the vernier coincides with a line on the main 
scale, indicating an extra Y^Q- = *oog 

Adding, the complete reading -834 



APPENDIX 465 

At (D). 

Vernier zero cuts off 1*5 and three small divisions, i. e. 

5 + -075 1-575 

Coincidence on vernier is at 20, i. e. 1 5g {5 . . . = -020 



.*. Complete reading = i\595 

Examination of the scales on an actual caliper will show that 
very careful inspection is necessary to distinguish the exact vernier 
line at which coincidence occurs. The lines immediately on either 
side of the coinciding line are only one-thousandth of an inch 
away from the nearest main scale graduations, but in opposite 
directions. This fact should be closely observed in the further 
sample settings of Fig. 160 (which are shown full size), as it assists 
considerably in identifying the coinciding line. 

In Fig. 1 60 : 

At (A). 

Vernier zero cuts off 2-2 and three small main scale 
divisions, i. e. 2-2 -}- -075 .......= 2-275 

Coincidence on vernier is at 13, i. e. 1 JJ - . . . = -013 



.'. Complete reading = 2-288 

At (B). 

Vernier zero cuts off 4-5 and two small divisions, i. e. 

4'5 + '5 = 4'55 

Vernier coincidence is at 19, i. e. y^fo . . . . = -019 



.'. Complete reading = 4-569 

At (C). 

Vernier zero cuts off no tenths and one small division, i. e. 

o + -025 . = -025 

Vernier coincidence is at n, i. e. iJJ<y . . . . = -on 



/. Complete reading = -036 

At (D). 

Vernier zero cuts off -3 and is very nearly in line with 
the first small division, but not quite. This is proved by 
coincidence being observed at 24 on vernier. Then vernier 
zero cuts off -3 and no small main scale divisions, i. e. -3 + -ooo = -300 

Vernier coincidence is at 24, i. e. yffoy . . . . = -024 

.'. Complete reading =a '324 



466 



APPENDIX 



At (E). 

Vernier zero cuts off 1-8 and appears to be in line with 
the third small division beyond this. It must not be assumed 
that this is actually the case unless both the vernier lines o 
and 25 are in coincidence. Inspection shows that both end 
lines of the vernier are in coincidence, and the reading is 
therefore 1-8 -f- -075 ....... 

there being no thousandths to add. 



1-875 



At (F). 

No tenths and no small divisions are cut off by the vernier 
zero, and coincidence is found at 21. The complete reading 
is then 



021 



I \ 

2 Coincidence-^ 3 ) /Coincidence-^ 5 ' 



345678/9 i 123 




r D ^v 5 1015202 

B; ioSo IN - 




1WVF ^ WV^^ f/ .~^~ ^ 

Vernier reading 2*288 Vernier reading 4*569 Vernier reading O036 




E) II I I I II II I I I II fc 
^J |0 5 1(^1520251 |0 5 10^520251 \T^ 

J5o?7 IN - ioo5 IN ' iooo-' N - 

Vernier reading 3*024 Vernier reading 1*875 Vernier reading 0*021 

Fig. i Go. Sample Settings of T5 ff(r inch Vernier. (Full size.) 

The examples at (D) and (E) above are important and should be 
carefully studied. 

Whenever the vernier zero appears to coincide with the '025, 
05 or -075, the other end should be carefully examined, as a 
coincidence may possibly occur at 24. It must not be assumed 
that there are no thousandths to add unless both end lines are in 
coincidence. 

After considerable practice in reading has been obtained the 
addition of the main scale reading and the vernier reading may 
be made mentally, but unless the user feels absolutely sure 
of his mental addition the separate readings should be written 
down. 



APPENDIX 



467 



Exercises 99. On Reading Verniers r J inch and 

TOW inch 

1. Read the settings of the ^ inch vernier shown in Fig. 161, 
(A) to (C). 

2. Read the settings of the rdro inch vernier cahper shown m 
Fig. 1 6 r. (D) to (M). 




Fig. 161. 'Exercises on reading Verniers. A to C, , ] 9 inch ; D to M, T7J V inch. 

(About full size.) 

The Back of the Galiper. The back of the blade is usually 
fully divided into inches and sixty-fourths. A vernier is not fitted 
to this scale, but measurements are made directly with the aid of 
two lines marked " in " and " out," as shown in Fig. 158. When 
the jaws are shut the line marked "out" will read o on the scale, 
while the line marked "in" Vill read off the distance w (see 
description of caliper on p. 463). The "out" line is used for 
outside measurements and the "in" line for inside measurements 
on the sixty-fourths scale, and no addition of the width w is then 
necessary. 

Least Count. To obtain the smallest reading possible with 
any given vernier from first principles, as done on pp. 460, 463, 



468 APPENDIX 

is rather lengthy, but by means of a little algebra a very simple 
formula can be deduced for the least count, as the smallest possible 
reading is sometimes called. 

Let v = length of i vernier division, 

s = length of i main scale division, 
and N = number of divisions on vernier. 

With very few exceptions the N vernier divisions are made 
equal in length to one less their number of main scale divisions. 
Then (N i) main scale divisions will be the same length as N 
vernier divisions, i. e. 

Nv = (N i)s 



Least Count " = scale division vernier division 
= s v 



= s 



-i s, substituting from (i) 



_ / __ N i\, taking out the common factor 
~ S V TT~; (refer p. 147) 
/N N + *\, combining the bracketed 
~~ S \ N / quantities 

= s X ~ or X S T , since N N = o 

N N 

i. e. the least count is the length of the smallest main scale division 
divided by the number of divisions on the vernier, provided that the 
N vernier divisions be made equal to N 1 main scale divisions. 
The least counts for the verniers already described will now be 
deduced from the formula : 

i. Main scale divided into tenths. Vernier with 10 divisions equal 
to 9 main scale divisions. 

Then s = = -i* N =* 10 



/. least count = ~ = = oi > ". 
N* 10 

2. Main scale division = - = -025^. Vernier with 25 divisions 

40 
equal to 24 main scale divisions. 

Then s = -025" 

/. least count = i = 

N 25 



APPENDIX 



469 



3. As a further example we will take a marking, occasionally found 
on vernier calipers, in which the main scale is divided into sixteenths, 
and the vernier has 8 divisions equal to 7 main scale divisions. 



Then 



= inch 
10 



/. least count = 1 = _. -i- 8 
N 16 



1 X * =-1. inch. 
16 8 128 



Various other Vernier Scales. Further examples of linear 
verniers of less importance will be found in Exercises 100. 

As an example of an angular vernier, that usually found on a 



Reac*mg^538' 




Scale aboof 6 r-mies -I "^ -full size. 

Fig. 162. Sample Settings of Box Sextant Vernier, reading to i minute. 
Box Sextant * will be taken. The instrument is arranged to be 
very light and portable; th3 scale is therefore of small radius, is 
finely divided on silver, and is read through a small magnifying 
glass carried on the instrument. 

The main scale is divided in degrees and half -degrees, and the 
vernier has 30 divisions equal in total length to 29 small scale 
divisions, see (A), Fig. 162. 

Then s == - degree = 30' (30 minutes) and N = 30, 



, least count = ~ = ^ 
30 



s 

_. 

N 



minute. 



* A surveying instrument for measuring, in the field, the angle 
between two objects viewed from a third point. 



470 APPENDIX 

The reading is carried out as follows : 

1. Read the degrees (and if one, the half-degree) cut off by the vernier 
zero. 

2. Note the vernier coincidence. 

3. Add the reading of (2), as minutes, to the reading of (1). 

Two sample settings are shown in Fig. 162 at (B) and (C). 

At (B). 

Vernier zero is beyond 37 but has not reached the next 
half-degree ......... 37 oo' 

Vernier coincidence is at 14 . . . . . . 14' 



Then complete reading = 37 14' 
At (C). 

Vernier zero lies between 45 and 46 . 45 30' 

Vernier coincidence is at 8 . . . . . 8' 



Then complete reading = 45 38' 



[verniev is used wii-h 

Hie zero. 




If coincidence, were at- C, 
r-hen reading would toe !$. 
If coincidence were. a+ fc, 
reading would be l*33 




Coincidence ^^ r 

O l- Q' /. Reading is l-3Zb. 

Scale -Three rimes Full Size. Scale - Full Size. 

Fig. i(>3. (A) Sample Setting of Abncy Clinometer Vernier. (B) Reading 
, :.,~u y ern i er w jth no coincidence. 



With sufficient practice in reading, the addition may be done 
mentally and the result entered up at once. 

The last example will be another angular vernier, that com- 
monly found on the Abney Clinometer.* Here the main scale is 
divided to show single degrees, while the vernier has 6 divisions 
whose total length is equal to 5 main scale divisions. 

* A surveying instrument for measuring angles in a vertical plane, 
such as the slope of the ground. 



Then 



c 60' 

.*. least count = XT = - 

N 6 



APPENDIX 

i = 60' and N = 6 

10 minutes. 



471 



The reading of this vernier is very simple, the successive vernier 
coincidences indicating 10, 20, 30, 40, 50 minutes beyond the 
complete degrees. A sample setting is shown at (A), Fig. 163, 
and should be easily understood. 









I " ' I ' "1 



Ml I I I 



JL1 



i I 



8 O 



8 



IN. Vernier. Mam Scale Division = ^ IN. 




=^5 IN. Vernier Mam Scale Division = 7 IN - 




002" Vernier 

igures 

Fig. 164 (part}. Exercises on Reading Linear Verniers. 



MamScale Divisiom= -O5' 
on vermer denote hundredfhs. 



Lack of Coincidence. With coarse reading verniers, such as 
the T J^ inch, it may happen that no coincidence can be detected, 
two vernier lines lying between two main scale lines, as at (B), 
Fig. 163. This indicates that the position is intermediate between 
the settings having coincidences at a and b respectively. It may 
usually be read as half-way between these two readings, e. g. if 
coincidence at a gave 1-32" and coincidence at b gave i'33" then 
the actual reading could be taken as half-way between 1-32 and 
i'33> * e- I-325". 



472 



APPENDIX 




IO Minute Vernier Mom Scale Division = I 



9 5 10 



:o (N) 



H 



30 











( 


) 5 tO 

[ . - - 


















1 






























1 





i( 


o 



6 Mmore,orl Vernier. Mam Scale Division = 




50 60 

() 

3Mrnure Vernier Mam Scale Division = 1 



O 5 10 15 20 25 30 



O 5 10 15 20 25 30 




I Minute Vernier Mam Scale Divis\on 

Fig. 164' (contd.) Exercises on Reading Angular Verniers. 



APPENDIX 



473 



Exercises 100. On Various Verniers 

Determine the " least count " of each of the following verniers (all 
actual examples) : 

1. Smallest main scale division = J inch. Vernier has 8 divisions 
equal to 7 main scale divisions. 

2. Main scale divided into J s inch, and 40 divisions on vernier = 39 
divisions on main scale. 

3. Main scale reads '05*. Vernier has 25 divisions equal to 24 
scale divisions. 

4. Smallest main scale division = -02", and 19 main scale divisions 
equal 20 vernier divisions. 

5. Main scale reads i and vernier has 10 divisions equal to 9 scale 
divisions. (Give result in minutes.) 

6. Smallest main scale division = i, and 20 divisions on vernier 
= 19 divisions on main scale. (Give result in minutes.) 

7. Smallest main scale division = J degree. Vernier has 60 divisions 
equal to 59 main scale divisions. (Give result in seconds.) 

Read the settings of the various verniers shown in Fig. 164: 
8. (A) to (C) Least count ^ inch 



9- (D) (F) 

10. (G), (H) and (K) 

11. (L) and (M) 
". (N) (O) 

13- (P) -. (Q) 
14. (R) (S) 



002 

10 minutes 
6 
3 



Micrometers. 

Micrometer Wheel and Screw. A method of fine measure- 
ment of considerable importance is that of the micrometer wheel 



-Po infer 





Section 
Fig. 165. Milling Machine Top Slide. 



End View 
-h nancHe remo 



and screw thread. It is employed on the better types of milling 
machine tables, slide rests of lathes, etc., where machining opera- 
tions are required to a fine degree of accuracy, and it is also the 
basis of fine measuring machines. 

Referring to Fig. 165, which shows a section and an end view 
of the top slide on a milling machine slide rest, A is the table 
supporting the work, B is the feed screw (usually of square thread 
form), C is the nut fixed to the remainder of the slide rest, while 



474 APPENDIX 

D is a large diameter disc or wheel, graduated on its outer edge, 
and called the micrometer wheel, and which is rigidly attached 
to the screw. A pointer, or a line prominently marked on the 
end of the table, serves to indicate the reading on the disc. 

Now the pitch (or better, "lead")'of the screw is the distance 
that the table will advance for one complete revolution of the screw 
and disc. If the screw has 10 single threads per inch, then the 
pitch is ! inch, and for one complete revolution of the screw the 
table would advance (or withdraw) ! inch. If half a revolution 
be made, the advance would be of ! = -05 ; if $ of a revolution 
be made, the movement would be T V of -i = oi inch, and so on. 
If the circumference of the micrometer disc be divided into 100 
equal parts, then a rotation of the screw through one of these 
parts would move the table ^^ of ! = -001 inch. The pitch of 
the screw and the number of divisions round the micrometer disc 
may be varied, but they are usually arranged so that the smallest 
advance registered is -ooi inch. Thus on a certain small tool- 
maker's lathe, the micrometer disc, which was about J-inch 
diameter, was divided into 50 parts; the pitch of the screw was 
g 1 ^ inch, so that each small division was equivalent to ^j of -^ 
= TT>W = 'O^ 1 inch. 

Then by simply counting the number of graduations rotated 
past the pointer, any desired advance of the table, in thousandths 
of an inch, may be obtained. There is no indication, except 
counting, of the complete revolutions made, but usually this causes 
no difficulty. 

The accuracy of the method is entirely dependent on the 
accuracy with which the screw is made, as regards pitch, and 
how the disc is divided. The latter can usually be done very 
exactly, while feed screws can be cut with a total error in pitch 
of not more than -005 inch in 10 inches, which is quite suitable for 
ordinary manufacturing. 

With this attachment to a milling machine table it is possible 
to graduate a scale, after working out the lengths of the divisions 
in thousandths of an inch. 

In use it is essential that the pressure between the screw and 
its nut be kept in the same direction all the time, i. e. any back- 
lash must be eliminated. This is ensured by approaching all 
settings with the same direction of rotation. If the disc is overrun 
for any particular setting, then it must be rotated well back to 
withdraw the table, and then again rotated in the original direction, 
with greater care, until the desired position is attained. 



APPENDIX 



475 



The Micrometer Caliper. The chief application of the fore- 
going principle is in the Micrometer Caliper, the most important 
workshop measuring tool. Fig. 166 shows the commonest form, 
which is adapted for outside or " male " dimensions, and which 
is spoken of as a One-Inch Micrometer, one inch being the largest 
dimension that it will measure. 

The U-shaped body or bow A carries a fixed anvil B. The 
opposite side carries a hollow sleeve C, the end of which is tapped 
and split and on its outer surface is threaded with a coned thread. 
A nut D on this thread enables the sleeve to be adjusted to grip 



Knurled-*. 




<2/ 



fixecl*anv// Adjustable Sle.e.ve. 

Fig. 1 66. Micrometer Caliper. 



the screw E with any desired degree of tightness, so that backlash 
may be eliminated. The screw E is the micrometer or gauge 
screw, and in English-reading instruments is invariably a single 
V-thread screw, right hand, of 40 threads per inch, i. e. pitch 
= $ inch = -025. Very great care is taken to make this screw 
accurate, and the total error in pitch on a i" length is usually 
less than 'oooi", *. e. one ten -thousandth. The inner end ter- 
minates in the anvil F, while the outer end is attached to the 
thimble G, which just passes over the sleeve C. The anvil faces 
should be ground and lapped quite flat and smooth, and should 
be perfectly parallel for all positions of the screw. 



476 



APPENDIX 



The sleeve C carries an axial scale H divided into tenths and 
fortieths, each tenth being numbered o, I, 2, etc., from left to 
right. One rotation of the thimble causes an axial movement of 
025", *'. e. the distance between two successive marks on the sleeve. 
This scale H registers the number of complete revolutions of the 
screw. The circumference of the thimble, on its bevelled edge, 
is divided into 25 equal parts, figured at every 5, as shown. A 




Fig. 167. Sample Settings of Micrometer Caliper. 

rotation of one of these parts, therefore, produces an axial movement 
of the anvil F of -^ of *O25 = -ooi inch. 

At (A), Fig. 167, the micrometer is shown with the anvils closed 
and reading o. Referring to (B), Fig. 167, an opening of -2" 
between the anvils has been made, i. e. the thimble and screw have 
been rotated 8 times (-025 X 8 = *2) from the closed position of (A), 
Fig. 167. If now a further rotation in the same direction be made 
until graduation 15 on the thimble coincides with the axial line 



APPENDIX 477 

on the sleeve, as shown at (C), Fig. 167, an additional opening of 
15 thousandths, i. e. -015, will have been made. The distance 
between the anvils is now -2* + -015" = '215". 

Outside micrometer calipers can also be obtained with larger 
bows to read up to 2", 3", 4^, etc., as desired, but in all sizes the 
travel of the gauge screw is invariably one inch. Then the dis- 
tance between the anvils when the micrometer is reading zero is 
i" less than the " size " of the tool, i. e. is i", 2", 3", etc., respec- 
tively. This may be called the " base " of the caliper and must 
be added to any reading taken, since the thimble and scale only 
indicafe the decimal part of the measurement. 

Zero Adjustment. When the anvils of a i" micrometer 
caliper are just closed, without pressure, after the faces have been 
wiped clean and dry, the edge of the thimble should be opposite 
the first line on the axial scale, marked o, and the line o on the 
thimble scale should coincide with the axial line on the sleeve; 
as at (A), Fig. 167. An adjustment is provided to obtain this 
condition when the tool is first assembled, and to maintain it after 
wear and the subsequent truing-up of the anvil faces. 

The mode of adjustment varies with the maker, but usually 
consists in either (i) making the fixed anvil B adjustable axially, 
or (2) making the sleeve C adjustable around its axis. 

In method (i) the " fixed " anvil is a good push fit in the bow, 
and can be pushed out by a screw L ; a locking screw M serves to 
lock the anvil when set. To adjust the tool, the locking screw M 
is slackened and the thimble set to read zero. The " fixed " 
anvil B is then set up to touch the anvil F by the adjusting 
screw L, and is finally locked in its position by screw M. The 
adjustment should then be tested by opening and closing the anvils, 
as it is possible to set the faces together with too much pressure 
by screw L. 

In method (2) the sleeve C is secured to the bow A by friction 
only, and can be rotated by the special spanner S, provided with 
the tool. To adjust, the anvils are closed, without pressure, and 
the sleeve C is then rotated until the axial line of the scale H 
coincides with the o on the thimble. The adjustment should then 
be tested to check that the anvils were not closed with undue 
pressure. 

With sizes larger than one-inch, a gauge is provided whose 
length is accurately made equal to the " base " of the tool. This 
is inserted between the anvils when setting to zero. 

It is important, when using any micrometer for the first time, 



478 APPENDIX 

to examine for any zero error and to adjust, if necessary. If the 
tool is in continual use then periodic examinations should be 
made. 

To use the Caliper. The thimble is rotated to open the 
anvils to pass over the work, and is then rotated in the reverse 
direction until the object is nipped, without pressure, between the 
anvils. 

To take the reading : 

1. Read off the number of complete tenths exposed by the edge o! the 
thimble. 

Call these "large divisions." 

2. Read off the last small division exposed beyond the last tenth. 

These small divisions, in order, read 

(i) *025, (2) '050, (3) *075. Call the particular decimal " small 
divisions.' 1 

3. Note the number of the graduation on the thimble which coincides 
with the axial line on the sleeve. 

Call this the " thimble divisions " and add, as thousandths, to 
the readings of (1) and (2). The thimble reading will always be 
between 'ooo and -025. 

Thus at (B), Fig. 167 : 

1. Two complete tenths are exposed; 

.*. Large divisions = *2 

2. No small divisions are visible beyond the last tenth; 

.*. Small divisions = -ooo 

3. Graduation 15 on thimble coincides with the axial 

line, i. e. ^^ ; .'. Thimble divisions = '015 



Adding, complete reading = -215 

The method of reading is closely allied to that for the one- 
thousandth vernier caliper. It is really simpler, as no vernier 
coincidence has to be detected. The micrometer caliper has 
advantages over the vernier type; It is easily capable of reading 
finer than one-thousandth, it is quicker and easier to handle, and, 
owing to its shape, is more sensitive for fine measurements. The 
vernier caliper, however, with its long thin jaws, can be used for 
measuring dimensions which are inaccessible to the micrometer. 

Some further sample settings are shown in Fig. 168, (A) 
to (E). 



APPENDIX 



479 



At (A). 

1. Two complete tenths are exposed by thimble. 

/. Large divisions = -2 

2. Two small divisions are exposed beyond the last tenth. 

/. Small divisions = -050 

3. Graduation 5 on thimble coincides with the axial line. 

.'. Thimble divisions = -005 

/. Complete reading = '255 

At (B). 

Pour complete tenths are exposed by thimble. 

* .*. Large divisions = -4 

2. One small division is exposed beyond the last tenth. 

.*. Small divisions = -025 

3. Graduation 10 on thimble coincides with the axial line. 

/. Thimble divisions -oio 



/. Complete reading = -435 




R eadmg 
Fig. 1 68. Sample Settings of Micrometer Caliper y^ inch. (Full size.) 

At (C). 

1. Five complete tenths are exposed. /. Large divisions = -5 

2. Three small divisions are exposed in addition. 

.*. Small divisions = -075 

3. Graduation 23 on thimble coincides with axial line. 

.'. Thimble divisions = -023 

/. Complete reading == -598 



At (D). 

1. Large divisions . 

2. Small divisions . 

3. Coincidence at 17. 

1 1 



. = -2 

. = -025 

/. Thimble divisions = -017 
.'. Complete reading -242 



480 APPENDIX 

At (E). 

1. Large divisions . . . * . . . . = -9 

2. Small divisions ........= -ooo 

3. Coincidence at 2. .*. Thimble divisions = -002 



/. Complete reading = -902 

In all the settings shown up to the present a graduation on 
the thimble has been in exact coincidence with the axial line on 
the sleeve. This will not always occur, and readings may have 
to be taken of such settings as appear in Fig. 169, where the axial 
line lies between two adjacent thimble graduations. In these cases 
the dimension does not consist of an exact number of thousandths, 
but contains a fraction of a thousandth. 

For ordinary manufacturing purposes it is seldom necessary to 
read anything smaller than a thousandth. The extra fraction of 
a thousandth may, however, be estimated as ten-thousandths, or 
the reading may be stated to the nearest thousandth. The exact 
reading of ten-thousandths is treated later. 

If the ten-thousandths are to be estimated, then the thimble 
reading is taken as the graduation immediately before the axial 
line on the sleeve; and the estimated tenths of a thimble division 
between this last graduation and the axial line are added to the 
reading in the fourth decimal place. 

If the reading is required to the nearest thousandth, then the 
thimble graduation which is nearer to the axial line is read. 
Examples are shown in Fig. 169. 

In Fig. 169 : 
At (A). 

1. Large divisions . . . . . . . = *2 

2. Small divisions . . . . . . . . = -ooo 

3. Axial line lies between 17 and 18 on thimble. 

.'. Thimble divisions = -017 

From 17 on the thimble to the axiaMine is estimated as -7 
of a thimble graduation . . . . . . . = -0007 



/. Complete reading = -2177 

To the nearest thousandth this is *2i8, as is shown by the graduation 
1 8 being the nearer to the axial line. 



APPENDIX 



481 



At (B). 

1. Large divisions . . . . . . . -7 

2. Small divisions . . . . . . . . = -05 

3. Axial line lies between thimble graduations 8 and 9. 

, .'. Thimble divisions = -008 

From 8 on thimble to axial line is about -3 of a thimble 
division ..........= ^0003 



/. Complete reading = "7583 

To the nearest thousandth this is ^758^ as is shown by graduation 8 
being th nearer to the axial line. 



- - f^l * r ^f 

I 9 I *ET 1 19! ??t? < ??? 10 

rtiitiiitf^N I L iiiiiiiiiiiiiiiiiiiiliiiliiiii r" -^^N 

>-pJ l> sU- 

/T*\ <-"s/V7/tf os ^/CT\ E&rimafie. as -3 07^ 



-7 of t-h,mblc 
vi sion 

'2177. 
Po nearest- 




'7583. 
758 To nearest- 



Fig. 169. Estimating Ten- thousandths on Micrometer Caliper. 

^/ (C). 

1. Large divisions . . . , . . . . = -o 

2. Small divisions . . . . . . . . -025 

3. Thimble divisions (axial line between 21 and 22) . . = -021 
The additional fraction of a thimble graduation is 

estimated as -2 . . . . . . . . . = -0002 



/. Complete reading = -0462 
To the nearest thousandth this is -046. 

When the axial line appears to be exactly half-way between 
two thimble graduations, the half-thousandth, '0005, is read, and 
the dimension cannot be stated to a nearest thousandth. 

The writing down of the component decimals of the reading 
should not be discarded until very considerable practice has been 
obtained. Even then mistakes of mental addition are possible, 
and it is safer to retain the writing down, reducing it, however, 
to two lines, the first containing the " large " and " small " divisions, 
and the second the thousandths and (if any) the ten-thousandths. 
This method is shown when dealing with the ten-thousandth 
micrometer. 



482 



APPENDIX 



Exercises 101. On Reading the Micrometer 



inch). 



1. Read the settings of the micrometer caliper shown in Fig. 170, 
(A) to (F). 

2. Read the settings shown in Fig. 170, (G) to (J), giving the 
readings including the estimated ten- thousandth and also to the 
nearest thousandth. 

Various Micrometer Tools. The measuring portion of the 
micrometer caliper, i. e. the sleeve, screw, and thimble, can be 
obtained separately, and is then known as a micrometer head. It 
can be attached to machines and tools where fine adjustment is 




Fig. 170. Exercises on Reading Micrometer, y^ inch. 

required, being gripped by the portion a of sleeve. See (A), 
Fig. 171. 

For measuring inside dimensions the inside micrometer caliper, 
shown at (B), Fig. 171, is employed. The measuring portion is 
the same as on the outside variety, and to enable a wide range 
of dimensions to be dealt with, extension rods of various lengths 
are supplied. 

The depths of holes and recessed parts are accurately measured 
by the micrometer depth gauge, shown at (C), Fig. 171. The gauge 
screw has usually a total movement of half an inch, while the 
measuring rod is graduated by angular grooves in half-inches; by 
setting the rod to the nearest suitable groove, a variety of depths* 
can be gauged. 



APPENDIX 



483 



There are many other applications of the micrometer which 
may be consulted at length in the manufacturers' catalogues. 

The Ten-Thousandth Micrometer. For fine work demand- 
ing measurement to a greater accuracy than one-thousandth of an 
inch, the micrometer caliper* is fitted with a vernier scale by 
which ten-thousandths may be read. 

Referring to Fig. 172 at (A), the sleeve is marked on the back 
with a vernier scale, the lines of which run the whole length of 
the sleeve, so that the vernier scale and the bevelled edge of the 
thimble; are in contact for any position of the screw. The vernier 
has 10 divisions, numbered o to 10, as shown, their total length 
round the sleeve being equal to 9 divisions on the thimble. This 




Fig. 171 . Various Micrometer Tools. 



is an example of the simple vernier described on p. 460, and the 
least count is then -^ of the amount indicated by one division on 
the thimble, i. e. $ of 'OOi = -oooi inch. 

To read this caliper, proceed as described on p. 478, reading up 
to the thimble graduation before the axial line on the sleeve. This 
gives the number of complete thousandths in the dimension. Then 
note which line on the vernier is in coincidence with a line on the thimble, 
and add this figure as ten-thousandths (4th decimal place) to the first 
reading. 

The caliper should be held with the axis vertical when reading 
the vernier, as otherwise the curvature of the vernier surface may 
cause a false coincidence to be read. 

Sample settings are shown in Fig. 172 at (B) to (G). 



4 8 4 



APPENDIX 



In Fig. 172 : 
At (B). 

1. Main scale reading, i. e. " large " and " small " divisions =a 150 

2. Axial line is between 17 and 18 on thimble . . = -017 

3. Vernier coincidence at 4 . .' . . . . = -0004 



/. Complete reading = '1674 



Coincidence LJ-UJ-U4-UI Coincidence 
765432 M I 098765-* 




Reading IG74 Reading '3837 Reac*\ng -7239 




JKS432IO \ 

Reading A4-4-1 Readimg '59*36 Reading -I960 

Fig. 172. Sample Settings of Micrometer Caliper, reading to T -yl^ inch. 



At (C). 

I Main scale reading .... 

2. Thimble division before axial line = 8 

3. Vernier coincidence at 7 . 



= '375 
= -008 
= -0007 



/. Complete reading = '3837 



APPENDIX 485 

At (D). 

1. Main scale reading .....,.= -700 

2. Thimble reading ....... =s -023 

3. Vernier reading = -0009 



.*. Complete reading =* 7239 

With further practice the thimble and vernier readings may 
be combined when read, the thimble reading consisting always of 
second and third places of decimals, and the vernier reading always 
the fourth place. 

Should a thimble graduation appear to coincide with the axial 
line on the sleeve when reading to a ten-thousandth, the reading 
must be made with caution. The vernier should be carefully 
examined for a possible coincidence near the ends, and no reading 
should be taken as consisting of an exact number of thousandths 
unless both the o and the 10 lines of the vernier are in coincidence. 

Examples are shown in Fig. 172 at (E) to (G). 

In Fig. 172 : 
At (E). 

1. Main scale reading .......= -425 

2. Thimble graduation 19 appears very nearly in line 
with the axial line, but examination of the vernier shows a 
coincidence at i. Then thimble and vernier = '019 -f- -oooi = -0191 

/. Complete reading = -4441 
At (F). 

1. Main scale reading .......= -575 

but appears very nearly to be -6, the line o on the thimble 

being just over the axial line. 

2. The vernier reading is then -024 and the vernier 
coincidence is seen to be at 8 . . . . . = -0248 

.'. Complete reading = -5998 
At (G). 

1. Main scale reading ....... ss '175 

2. Thimble reading is at 21 and vernier coincidence is at 

o and 10 ......... a* -0210 



.'. Complete reading = -1960 

Exercises 102. On Reading the Micrometer ( 1?y j ou inch). 

In Fig. 173 are shown settings of a micrometer caliper, together 
with a view of the vernier in each case. Take the reading for each 
of these settings from (A) to (C). 



486 



APPENDIX 



When using a ten-thousandth micrometer certain practical points 
need careful attention. Under no circumstances may the screw 
be jammed on to the job with any pressure. A light, almost 
imperceptible, pressure is sufficient, which is judged by the " feel " 
and can only be gained by considerable practice. 

It is frequently convenient to hold the bow of the instrument 
in one hand between the palm and the third and fourth fingers, 
the thimble being rotated by the thumb and first finger; the 
caliper should never be held very long, as the warmth of the hand 
is sufficient to cause an expansion of one ten-thousandth. It is 
a better plan, wherever possible, to hold the tool in a small oench 
stand, and to take the work to the instrument. 

Lastly, a line reading tool should never be used on rough 





Fig. 173. Exercises on reading Micrometer, 



inch. 



surfaces, or for any purpose where fine reading is not really neces- 
sary, as wear becomes of very considerable importance in such 
a tool. 

Zero Setting of Ten-Thousandth Micrometer. For reading 
to a ten-thousandth the micrometer cannot be said to read zero 
unless the vernier lines o and 10 are in coincidence, in addition to 
the thimble reading being o. Zero must be read then, in three 
places, i. e. on the main scale, on the thimble and on the vernier. 
It is a rather tedious process to set a micrometer to zero correct 
to one ten-thousandth, and frequently, therefore, the tool is left 
reading one or two ten-thousandths large or small, the exact amount 
being subtracted or added, respectively, to the readings taken. 
The method is, of course, always open to the objection that the 
correction may be forgotten. 

The Ratchet Stop. The ratchet stop shown in Fig. 174 was 
introduced to eliminate any error due to gripping the job with 
varying degrees of pressure. It contains a ratchet and pawl, and 



APPENDIX 487 

if the screw be always rotated by the small knurled end when 
closing on to the job, an approximately constant pressure is used 
on all occasions. If more than a certain pressure be applied the 
ratchet slips over the pawl without forcing the measuring screw 
round. Upon withdrawal, i.*e. backward rotation, the ratchet 
engages positively with the pawl and cannot slip. 




Fig. i 74 . "Ratchet-Stop for Micrometer Calipcr. 

The " feel " of the job between the anvils is, however, a more 
sensitive and better criterion for fine work. The ratchet, even 
when slipping, can exert an appreciable pressure, and by con- 
tinued or sudden rotation can jam the anvils on to the job, thereby 
causing the thimble to overrun its true reading by one or two 
ten-thousandths. It is, however, of advantage for taking measure- 
ments rapidly, and also where more than one person use the same 
micrometer. 



The tise of verniers and micrometers graduated in metric units is 
treated in " Metric System for Engineers " (Chapman & Hall). 



INDEX 



ABBREVIATIONS, 2 
Accuracy, Degree of, 40 
Addition of angles, 264 

of decimal fractions, 43 

of -f- and quantities, 123 

of vulgar fractions, 12 

Anchor ring, Surface area of, 346 

, Volume of, 346 

Artgle, 246 

Angles, Addition and subtraction of, 
264 

, Measurement of, 262 

, Reduction of, 265 

Annulus, Area of, 294 
Antiloganthms, 220, 221 
Approximation for result, 58 
Arc of circle, Length of, 271 
Area, 273 
Area of annulus, 294 

of circle, 286 

of ellipse, 301 

of fillet, 299 

of hexagon, 285 

of irregular figures, 304 el seq. 

of octagon, 286 

of rhomboid, 278 

of sector of circle, 298 

of segment of circle, 299 

of square and rectangle, 275 

of surface (see " Surface Area ") 

of trapezoid and trapezium, 281 

of triangle, 279 

, Reduction of, 274 

, Table of, 4 

, Table of, of plane figures, 312- 

313 
Averages, 67 



Brackets, 29 

. Insertion of, 146 

, Removal of, 141 et seq. 



Calculation of weights, 335 
Cancelling, 9 
Capacity, Table of, 4 
Chain, 243 
Charts, 348 



Chord of circle, 250, 431 
Circle, 250, 271, 298 

, Area of, 286 

, Hollow, area of, 294 

Circumference, 251 

of circle, 251 

of ellipse, 260 

, Table of, of plane figures, 312- 

313 

" Collecting-up " like terms, 134 
Common factors. 146, 147 
Complementary angles, 415 
Cone, 329 

, Surface of, 340 

, Surface of frustum of, 342 

, Volume of, 330 

, Volume of frustum of, 332 

Conversion of decimal fractions to 
vulgar, 38 

of units, 243, 274, 315, 316 

of vulgar fractions to decimal, 

56 

Co-ordinates, Rectangular, 348 
Cosecant, 429 
Cosine, 412 

curve, 432 

Cotangent, 429 
Cross multiplication, 164 
Cubic measure, 4 
Cuboid, 317, 318 
Curves, 347 

without origin, 361 

Cylinder, Definition of, 317 

, Surface of, 337 

, Volume of, 320 



Decimal equivalents, 39 

fractions, 36 

, conversion to vulgar, 38 

, operations on, 48, 52 

, notation, 35 

point, 36, 37 

Degree of accuracy, 40 
Denominator, 6 
Density, 335 

, Table of, 336 

Diameter of circle from area, 289 
Division of decimal fractions, 52 

oi + and quantities, 135 

of vulgar fractions, 25 



488 



INDEX 



489 



Ellipse, 250 

, Area of, 301 

, Circumference of, 260 

Equation to straight line graph, 374, 

377 

Equations, 157 

, simple (see " Simple equa- 
tions ") 

Evaluation of formulae, 89, 97, 108, 

139 

Expansion of (a &)*, 151 



Factors, 2 

, Common, 146, 147 

Fillet, Area of, 299 
Formulae, 86 

, Transposition of, 190 et seq. 

Fractional equations, 180 
Fractions, Decimal, 5, 36 

, Vulgar, various forms of, 7 

French curves, 355 
Frustum, 331 

of cone, Surface of, 342 

, Volume of, 332 

Frustum of square pyramid, Surface 

of, 341 

, Volume of, 332 

Frustum, Volume of any, 331 



Gallon, 315 

Geometrical terms, 246 
Graphs, 347 

H 

Hexagon, 248 

, Area of, 285 

Hexagonal prism, Volume, of, 320 
Hollow circle, Area of, 294 

cylinder, Volume of, 322 

Hypotenuse, 266, 269 



Indicator diagram, Mean height of, 

308 
Indices, 94-96 

, Laws of, 115, 117, 1 1 8 

Insertion of brackets, 146 

Interpolation, 356 

Irregular figures, Area of, 304 



Lateral surface, 337 
Laws of curves. 374 et seq. 
of indices, 115-118 



Least common multiple (L.C.M.), 14 
Length, Addition and subtraction of, 

245 

, British table of, 3 

, Conversion of, 243 

Logarithms, 213 et seq. 

, compound examples, 228, 232, 

238 

, Division by, 226 

, Evaluating powers by, 232 

, Multiplication by, 224 

, Roots by, 233 

, finding of, on slide rule, 409 

M 

Mathematical signs, 2 

terms, 2 

Mean value, 68 

Mid-ordinatc method of finding area, 

305 
Minus quantities (see "Negative 

Quantities ") 
Multiplication of decimal fractions, 48 

of + and quantities, 135 

of vulgar fractions, 22 



N 

Nautical measure, 242 
Negative quantities, 121 

. .operations on, 123, 127 

135, 138 

values, Plotting of, 368 

Numerator, 6 



Octagon, 248 

, Area of, 286 

Origin of curve axes, 348 



Parallelogram, 243 
Percentage, 71 
Perimeter, 251 
v (Pi), meaning of, 251 

, measurement of, 252 

Plane figures, 248 
Plotting, 350 

of negative values, 368 et seq. 

Powers, 94 

of fractional expressions, 112 

of negative quantities, 138 

of 10, 96 

Prime factors, 3 

Prism, Definition of, 317 

, Volume of, 320, 345 

Proportion. 82 



49 



INDEX 



Protractor, 263 

Pyramid, Definition of, 329 

, Square, surface cf, 341 

-, surface of frustum of, 341 

-, volume of, 330 

-, volume of frustum of, 332 
Pyramid, Volume of any, 330, 346 

Q 

Quadrilateral, 248 
Area of, 281 



R 

Ratio, 78 

Reciprocal, 79 

Reciprocal Trigonometric Ratios, 

429 

Rectangle, 248, 275 
Rectangular co-ordinates, 348 
Reduction of angles, 265 
Removal of brackets, 141 
Rhomboid, 248 

, Area of, 278 

Rhombus, 248 
Right-angle, 246 
Right-angled triangle, 248 

, Properties of sides of, 266 

Roots, no 

of fractional expressions, 113 

, Square, 100 

Rule of signs, 136 



Scales for curves, 348, 351 

Secant, 429 

Sector of circle, 250 

, Area of 298 

Segment of circle, 250 

, Area of, 299 

Segment of sphere, Surface of, 339 

, Volume of, 328 

Significant figures, 40 
Signs, Rule of, 136 
Simple equations, 158 

, Operations on, 158 et seq. 

Sine, 412 

curve, 432 

Slide rule, Description of, 385 

, Division of scales of, 387 

, Method of reading, 389 

, Operations on, 390 et seq. 

Sphere, hollow, Volume of, 327 

, Surface of, 339 

, Volume of, 327 

Square, 248, 275 
Squared paper, 349 
ffcjuare measure, 4 



Square pyramid, 329 

, Frustum of, 331 

, Surface of, 341 

, Volume of, 330 

Square root, 100 

, Table of, 101 

Straight-line graph, 366 

, Equation to, 374, 377 

Subtraction of angles, 264 

of decimal fractions, 43 

of -f- and quantities, 127 

of vulgar fractions, 18 

Surface area of any frustum, 346 

pyramid, 346 

Surface area of cone, 340 
of cylinder, 337 



of frustum of cone, 342 

- of frustum of square pyra- 
mid, 341 

of segment of sphere, 339 

of sphere, 339 

of square pyramid, 341 

, Table of, 345, 346 

Surveyors' measure, 243 
Symbols, 86 

, Substitution for, in formulae, 

118 



Table of area of plane figures, 312, 

313 
of circumferences of plane 

figures, 312, 313 
. of decimal equivalents, 39 

of densities, 336 

of squares and square roots, 101 

of surface area of solids, 345, 346 

of volume of solids, 345, 346 

Tables of measures, 3 
Tangent, 412 

curve, 432 

Transposition of formulae, igoetseq. 
Trapezium, 248 

, Area of, 281 

Trapezoid, 248 

, Area of, 281 

Triangles, 248 

, Area of, 279 

, Right-angled, properties of 

sides of, 266 

1 1 properties of 45 and 

60* 30, 269 
Trigonometric Ratios, 429 

, Reciprocal, 429 

Tubes, Volume of, 322 

U 

Units, I 

, Conversion of, 243 274 315, 

316 
Unknown quantity, The, 157, 191 



INDEX 491 

V Volume of solids, Table of, 345, 346 

of sphere. 327 

Volume, Conversion of, 316 of square pyramid, 3 

measure, Table of, 4 Vulgar fractions, 5 

of any frustum, 331, 346 1 operations on, 12 et seq. 

of any pyramid, 330, 346 

of cone, 330 

of cuboid, 318 w 

of cylinder, 320 

of frustum of cone, 332 Weight, Avoirdupois, Table of, 4 

of frustum of square pyramid, Weights, Calculation of, 335 

332 

of hexagonal prism, 320 

of hollow cylinder, 322 2 

of prisms, 320, 345 

of segment of sphere, 328 Zone of sphere, Volume of, 346 



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