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ii4 ARITHMETIC FOR ENGINEERS
OSMANIA UNIVERSITY LIBRARY
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ARITHMETIC FOR ENGINEERS
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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.
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The DirectlyUseful
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 & tl)
LiUNt.AV, SriJ
BRITAIN I'.Y
N^, LjMllH),
EDITORIAL NOTE
THE DIRECTLYUSEFUL 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 textbooks 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 classroom 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 submultiples 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 Antilogarithms Finding antilogarithms : 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 rightangled triangle
Proportions of 45 and 6o3o rightangled 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 coordinates 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
antilogarithms 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 horsepower 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, horsepower, etc., the units are fixed by general
consent.
Multiples and Submultiples. 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 submultiples 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 horsepower, 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 " twentyfour 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) ,, " horsepower."
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 " threefourths " or " threequarters ") 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 midline 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 12inch rule, while thirtysecondths
and sixtyfourths are occasionally marked. In the monetary system
where the sovereign () is the unit twentieths and two hundred
andfortieths 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 threequarters
of an inch (commonly " two and threequarter 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.
1174 =2 with remainder 3.
' v = 1
(6)101712 = 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 numbers
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 12inch 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 bolthead 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 fortyeighths or into twentyfourths. 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*
82x2x2 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 34 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, $, 18
4 = 2X2
23 18 = ^ X 3 X 3
.'. L.C.M. = 2x2x3x3
36.
Explanation. The L.C.M. is quite simple. Taking the first frac
tion, 4 into 36 goes 9. Then 9 x 1=9, which is written above the
long line. Then 3 into 36 goes 12, and 12 x i = 12, which appears
as shown. Lastly, 18 into 36 goes 2, and 2x1 = 2. Addition of
the numerators completes the example.
Where the given quantities include whole numbers, then
the easier and safer method is to add the whole numbers inde
pendently of the fractions. Usually the whole numbers may be
VULGAR FRACTIONS 17
added mentally. It is unwise to convert the mixed numbers into
improper fractions, as unwieldy figures result, and the subsequent
labour is greatly increased. Where improper fractions are given
(an unlikely condition), they should be converted into mixed
numbers.
Example g. Find the value of3j44 + 7i + 6J.
Add the whole numbers mentally and write the result in front of
the addition of the fractions, thus
The expression then =
24
L.C.M. =8x3
= 2 4
20 f ill Ma Y often be done mentally.
Example 10. The dimensions of the spindle for a sluice valve are
shown in Fig. 4. Determine the overall (i. e., total) length of the
spindle.
Fig. 4. Spindle for a Sluice Valve.
Writing down the dimensions in order we have
= 18 4___ 10 t 8 JL_14
16
L.C.M. = 16
Examples like this are of frequent occurrence when dealing
with drawings, and with practice they can be done mentally. It
is then necessary to be thoroughly familiar with the ordinary
measurements of J, T \, etc., and to know at once the simpler con
versions such as f = { = ! etc  Sucn knowledge is obtained
by doing a considerable amount of mechanical drawing. Then
the above example would be done mentally as follows : ioj + f
is ioj, add 6 gives i6J, add J gives 17$, 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 crosshead 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 3j%
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 headstock. 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 + 444 + 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. 32 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" selfacting 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
129 \\
This step usually omitted.
Example 20. Find the value ofijxjf2g.
I*Xir2f
_ * x 7 x _5 _ 35
 2 X 8 X i  6^
4
Example 21. A large pipejoint 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 oilseparator 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 = JJ, 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 reread 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 cuitmqoff tool
' 1
Example 29. Castiron piston rings, fa" wide, are to be turned from
a rough casting. The width of the cuttingoff 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 + VMY + 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 cuttingoif 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
cuttingoif, 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
wroughtiron 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
hacksaw, from extruded bars each 6'6* long. (See e, Fig. 12.) If
the sawcut 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. Fortyeight 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 righthand 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 lefthand 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 submultiple ^ 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
Submultiples
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 815327 means 815 + T 3 ^ + T rr +
But, by the rules of Chap. I, T ^  T ^, and 
so that the number 815327 = 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 thirtynine 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, 62425 reads sixtytwo 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 1500 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, 0305, 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 31400
is only 314, 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 007854 = 7854.
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 3937. 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 3937, which = 393^
= m*. Converting to ^ths we have  3 ^, which = Stfp X 10,
so that 3937 is 10 times 39*37. The converse is, of course, equally
true. Thus 3937 is ^ of 3937. 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 footnote on p. 36.
38 ARITHMETIC FOR ENGINEERS
As examples : 25 X 10 =25 (the dot is not needed final]) 7 ).
25 ~io = 25
7854 x 100 = 7854
31416 x 1000 = 31416.
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 = 1760.
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 314 r 100 "0314, for we may consider 314
as 0314.
Also 0625 f 10 = 00625.
Exercises 8. On Movement of Decimal Point.
What is the result of
1. () 5'5 X 10. (b) 55 10. (c) 134 X ioo.
(d) 134 4 ioo. (e) 1536 x 1000. (/) 4536 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) 73 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 3937 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) 03; (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) 03 = 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.
Decmal
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) 2205; (b) 134; (c) 305.
4. (a) 964; (b) 3142; (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 12498" ; 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 25364978
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 64978, 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 horsepower 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 baseline 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) 1666; (c) 01734; (d) 7071; (e) 1001 ,
(/) 2 355; fe) i'57 8 ; W '995 r 
( a ) *33 as 3 is less than 5; (6) 17 as 66 is greater than 50 ;
(c) oi7as3 4 ,, 50; (d) 71 71,, 50;
(e) 10 as 01 05; (/) 240,, 55,, ,, 50;
(g) 16 as 708 is greater than 500; (h) io 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 27183; (d) 23026; (e) 1897;
(a) 3740; (b) 394.' ( c ) 272; (d) 230 (i.e.. 23);
(*) 190 (i.e.. 19); (/) 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
230. The last o has no use, so that our result is 23. 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 190, i. e., 19, or only 2 figures, which, however,
cannot be avoided.
Example 34. Express the number 31415928* correct to
(a) 5 significant figures ; (b) 4 significant figures ;
(c) 3 significant figures.
(a) 31416; (b) 3142; (c) 314
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 3937: 5280; 2205; 3045; 0365.
2. 29,850,000; 10,500,000; 0807; 1083; 454.
3. 00000066; 6065; 'OoniS; 3209; 15708.
4. 109; '0000087; 0000105; 6009; 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; 3048; 4343; 62425; 001118.
6. 14380; 1393; 32182; 6009; 37499.
Express the following numbers correct to 2 significant figures : 
7. 0807; ^282; 147; 0646; 537.
8. 2205; 6065; *i6o4 ; 29,850,000; '1145.
9. The Imperial Gallon contains 277274 cubic inches. Express this
(a) correct to 4 significant figures ; (b) correct to 3 significant figures.
10. The British pound weight contains 4535924 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. 271828.
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 253, 1257 an< ^ '^ 2
253
1257
O02
3849
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 = 3849.
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 648, 0312, and
2 '3 7. Find the value required.
648
0312
2*37
! ! The 2 may be neglected.
9162 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 2327
2327
1875
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 castiron 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) 1945, 1225; (&) 57' 6 3 5 6 '5 2 ; ( c ) 7'7 6 > 6 9'09
(a) 1945 (&) 57* 6 3 (c) 77 6
1225 5652 6909
720 sq. ins. in sq. in. 167 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 622 ft. tons from the sum of 3*4,
456 and 37'6 ft. tons. Find this value.
3 4 ft. tons
45* 6
Sum
866
622
Difference = 244 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
752
B
409
441
C
9'75
D
414
I I2
E
201
2 '73
F
199
469
G
878
Totals .
2950
2689
Deduct Total Fore Sights 2689
(*. e. t Col. 3)
2*61
Result : Place G is 261 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 233 ft., and then to subtract
283 ft. from this sum. Find the resulting measurement.
We have 119 + 233 283
= 352  283
= 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.
921 f 56 (25 f 179)
= 9266 429
= 4976, say 498.
Exercises 11. On Addition and Subtraction of Decimals.
1. Find the sum of 105, 6358 and 252.
2. Add together 6735, 8562, 10514 and 57.
3. Add together 1255, 67*825, 15625, 13125 and 1203125.
4. The total head against which a centrifugal pump has to work
is made up of three heads : loss of head in pump 55 ft., loss by friction
in the pipe 936 ft., and height through which water has to be lifted
2505 ft. Find the total head.
5. In a laboratory experiment the following weights (Ibs.) were
placed on a certain hook : 7, 29, 3, 3, 29, 29, 3. 3, 2, 2, 15, 15, ii,
2, 3, 15. Find the total weight on the hook.
6. The lengths of the lines in a traverse survey were as follows :
1363, 38, 697, 941, 3*48, 61 1, 966, 508, 684, 207 chains. Find
the total length of line surveyed.
7. In calculating the bending moment on a continuous girder the
expression 5 j 375 f 106 f 142 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
2507
i8'73
657
62
491
Mond Gas . .
1000
2300
I500
300
4900
Natural Gas
41
164
2 5
937
3MI
9. Find the difference between 6875 an( i 59'&5'
10. Find the result of 1253 6845 269.
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 23
(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 . . . 638203 grammes
,, ,, after ,, ... 645091 ,,
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 .... 14286 ,,
,, ,, ,, f cold water f steam condensed 14805 ,,
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 = 30035 in.
Minimum ,, ,, ,, 30000 ,,
Maximum ,, ,, shaft = 29965 ,,
Minimum ,, ,, ,, = 29930 ,,
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
1396
C
418
9\57
D
4'95
Pt.
B.S.
F.S.
A
2475
13
337
3*9 I
C
215
4") 9
D
621
E
213
533
F
215
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
10265
9'455
2119
C
3'44i
5'55
5210
D
460
33^5
076
E
661
9885
5285
A
449
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*75 142 + 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) 127 47 1555 + 123 175.
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 245 X 13:
V.AZ _ <7 45 _ 2 iff
* 45 * T<rff i<7i7
also 13 = if\ = {.
Hence we have ^g X  J = \\% = 3185.
Again notice that the digits in 3185 are the product of the digits
in 2 '45 and 13 ; also the total number of decimal places in the numbers
multiplied is 3 (245 has 2 decimal places and 13 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., 245 X 13.
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 13. 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
13 . . . . .1 decimal place
735
245 _
3185 . .... 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 245 and 1*3 is 3185.
Example 42. If i cubic foot of water weighs 625 Ibs. and cast
iron is 72 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 625 X 7*2 Ibs.
625 i decimal place
1250
4375
450'QO 2 decimal places.
The total number of decimal places in 625 and 72 being two, two digits
are pointed off from the right in the product 45000.
.'. 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 275 X 373 X 678.
Long hand working. Working when each result is
reduced to 3 sig. figs.
1670 1670
275 275
8 35 835
11690 11690
334 3340
373
say 4590 to 3 sig. figs.
137775 373
321475
17130025 1377
678
1712070
137040200
say
119910175 y
102780150
11614156950 I368
^ 119700
102600
11593800
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 295 by 138. 2. Multiply 235 by 172.
3. 057 139 4. 1595 .. 7 8 5
5. 314 ,, 196. 6. ,, 0087 ,, 054.
7. ,, 13400 ,, 0036. 8. ,, 191*5 ,> '0986.
9. 0046 ,, 212. 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 castiron rail
section (a figure used in connection with the strength of the
section) gave the figures 2 X 1*78 X 27 X 27. 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 625 X 026. Find the
required value.
16. The calculation for the weight per foot of a large wroughtiron
pipe produced 1025 X '7^5 X 34 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 steamdriven air compressor
52 ARITHMETIC FOR ENGINEERS
is to be 1*38 times the area of the air cylinder. Find the steamcyhnder
area when the aircylinder 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 TTTS ~ < _ IITB ^ __ 235
men n 7575 flro  r 5 100*
Expressed decimally the result is 2*35. Now it will be seen that*
the digits in 235 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 235.
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 8672 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.
8672 X 100 8672 (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 8672, it is placed after
the 3 in quotient.
A rough test is available: 8672 is nearly 9, and 271 is nearly 3 ; dividing
9 by 3 gives 3, which is in agreement with our result 32.
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 345 may, of course, be written as 34500. The following
example illustrates.
54 ARITHMETIC FOR ENGINEERS
Example 45. Divide 345 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 345 r 12 = 2875.
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 7563. The answer is then stated as 756.
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 2314. Hence two noughts must be added to
DECIMAL FRACTIONS 55
make up the three places, and the example appears as ^Jjijp, the
decimal point in the numerator standing after the second o.
1 938)23 1400 (11940
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 23.)
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 2195 DV 5
3. 4138 2275.
5. 1*897 .. 2135.
7. ,, 0076 ,, 016.
9. 621 ,, 31.
2. Divide 3273 by 7.
4. 135 i. 4'36.
6. ,, 158 ,, 0065.
8. 098 1157.
10. ,, 33000 117.
11. If 147 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 2205 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 = 2205 Ibs., and i cwt. = 112 Ibs., find how
many kilogrammes go to i cwt.
16. If 746 watts i horsepower, find how many horsepower =
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 455 minutes.
19. ioo ,, ,, 237
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 625 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)70(04375
(a) 4)100 64^
* 25 " " 4 6 8
' A ^ '4375:
(fc) 8)100 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)160(01415
47
45*
* 3\ 1 A = 314 to 3 sig. figs,
1 80 *" 
Example 50, In a certain experiment the following intervals of
time were read with a stopwatch
(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 = 17 mins.
(b) 2 min. 35 sec. = 2gJ min. = 2^ cancelling by 5
2^ = 2583 min. or 2*58 to 3 sig. figs.
(c) i min. 13 sec. = ijg min. = 1217 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. 45 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 587 . ,
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 75* 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 228 to 3 significant figures. Now supposing
our detailed working had been bungled, and a result of, say, 228 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 18
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 195 x 10
529 529 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
(195 x 10) X (529 x io X 10)
(31 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
~ 31 x 18 ~ 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 1239.
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 righthand 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 12
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 39 and write 4. Decimal point having
been moved one place to right, place a stoke below the line.
Bottom line. Convert 000328 to 328 and write 3. Decimal point
moved 4 places to right, place 4 strokes above the line.
Convert 1200 to 12 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(09898
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 314 X 6 b b J
Tens
_ 5
""
= about J = 2
.". an approx. result = 2 X 100 = 20
.'. Actual result must be 209.
\\\
Two strokes above.
/ Multiply by 10 X 10
= IOO.
Approximation
' = about 5
i X 15 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 15, as
with the 144 at (b) above, it is best to call it 15 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 714 X 60
" 4 rx~6"28 X 21
Approximating
Tens
1
.'. multiply by 10.
/. Result is approximately 2 X 10 = 20, and answer given is
reasonably correct.
(c) 141 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 horsepower (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 6009.
.'. 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 314 =
6400 X 322 J 5< " 32 X 144 X 6 55 '
K
5   = ' 494 '
< 3aa X 900
7. 2 x 7 8 5 x 625x140=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
325 X 12 " " " ^'
11 10 X IIP X_2640
33000 " " "
_ __
30 x 106 X 1056 " " " 7I4 *
?i2<_3i4_ x 15 X 24
  4 x  3l ., 214.
314 X 24 X 1000
 ~~~~
15 ? x goo x 322
* 9 : 67rx^7~x~ : 6 " " " 44 '
16 4 X 314 X 18 X_7\5 X_72_'5
342
17.  5 35 _. 5 ~~ ., 245.
12
7 x 0625 x 2240
6 X 232
18. zTT^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 righthand term under the
line.
tons () 6 '5 X 65 = 423
 
6*5 X 65
536 ~
.
+ ~6 <*> 536 X 536 = 287
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 .
719 3*6
4. The current to pass through an electromagnet was calculated
to be ^  %~ amperes. Complete the calculation.
5. The resistances in an electric circuit, which is to carry a current
of 45 amperes, are 35 ohms, '75 ohms, and 238 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 horsepower of the stream, which is
it. 48 X 625 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 candlepower.
Each lamp takes 1*2 watts for every candlepower, (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 49 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 648 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 builtup 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 + 88 + 5*6) + 15 X
(78 f 67) 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 . 1645 , 1645 !V75 t^ t_ AI_ i
_J_^  ^ J>   2 ^  5^  2L2 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
"*" 282 x 287
21. Find the value of the expression 130 f '85 (6065 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(226 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 wroughtiron 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)4985
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 horsepower. 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 = 364 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
3062
946
31^3
3I80
3I90
326I
3282
942
3313
3322
3)2835 12)38588
945 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 152, i'5Q5
and 149. 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,
3i7. 3i7. 3i6, 33, 34'5. 32'4> 265, 200, 238, 240, 240, 245, 247,
245 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, 1115, 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 113% (<*) 028%.
(a) Shifting decimal point two places to left (to divide by 100),
30% = 3
(b) Similarly 375% = 375
(c) 113% = ' OI I3
(d) 028% = 00028
Example 62. Convert the following percentages into vulgar frac
tions : (a) 45%, (6) 12*%, (c) 33*% (d) 2625%, (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) 2625%= 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 = 75
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% J25
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 steelmakers' 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 193 Ibs. per foot.
Listed weight = 193 Ibs. per ft.
2. *>
2}% allowance = x 193
= 482 Ibs. per ft.
/. Greatest weight = 19*3 + '48 = 19*78 Ibs. per ft.
and least = 193 '48 = 1882 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 463
amperes, the true value being 45 amperes. Find the percentage
error (i. e. t express the error as a percentage of the true value).
Apparent value == 463
True value = 4*50
/. Error = 13 too high
.\ % error = ~ 3 X I0 = 289% too high,
45 _____
i. e. t for every 100 amperes indicated the error is 29 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.
= 605 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 = 5905%,
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 45, 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 335 Ibs. was found on
analysis to contain : Carbon 252 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 . . . . 252 Ibs. Total coal ... 335 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
252 X ioo
~ TaJ
234 x ioo
3'35 '
_ 2 3*4
752%
588%
3*97%
7'95%
3'35
197 X ioo
3'35
197
3'35
133 X ioo
3'35
13*3
3'35
266 x ioo
3'35
266
3*35
3'35
Total 9999%
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 9999%, 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 " Selfhardening " steel
gave the following results : Carbon 63%, chromium 104%, 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% 10000
Chromium . . 1*04,, 187
Manganese . . 05 ,,  
Silicon . . . 15 ,, 9813% Iron
'. Iron " by difference " = 9813
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 castiron bodies :
i y holes 45. per dozen holes ; i" holes 25. 6d. per dozen.
For caststeel 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 wroughtiron shaft will only transmit 70% of the h.p. that the
samesized steel shaft will transmit at the same speed. If the h.p.
of a certain steel shaft is 256, what is that of the samesized wrought
iron shaft ?
6. A square foot of wroughtiron 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 154, oxygen 181, nitrogen 1143.
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
323
146
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" = 15 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., jS  v5. 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 onethird 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 firstnamed 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 = 569, or, say, 57 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 steamdriven 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 horsepower.
Then, since i kw. = 134 h.p.
276 kw. =s 276 x 134 = 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
725%.
Ratios are used when speaking of the relative sizes of the cylinders
in a compound or multistage 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 tripleexpansion 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. = ^  242 : 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 323 : 242 : i,
meaning that if the H.P. area be i, then the I.P. area is 242 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. = ^ 1335 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 highspeed electriclighting engine; crank 4", connecting
rod i'S\
(b) A powerdriven aircompressor; 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 41.*
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 2475 sq. ft.
(b) Heating surface 1616 sq. ft. ; grate area 536 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 n6 tons per sq. in.
(b) Hard rolled bronze: Tensile 269 tons per sq. in.; shear 1606
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 248. Breaking stress 469.
(b) Lowmcor iron. Elastic stress 125. Breaking stress 22*1.
10. At atmospheric pressure *oi6 cu. ft. of water, when evaporated,
becomes 2637 cu. ft. of steam. Find the " relative volume " (i. e. t the
ratio of steam volume to water volume).
11. A twostage aircompressor 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 tripleexpansion 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 2945 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 143 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 25".
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 175 Ibs. per sq. in.
19. If a castiron 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 castiron 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 15. Similarly
the ratio of 12 to 8 is  1 / which also equals 15. 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 45 (i. e., 45 : 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 horsepower.
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 = 403 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 145 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 1:22: 46 : 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 45 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(174 f 25) f 30 x 65 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 0032"
thick, spaced uniformly oi" apart. The two supporting end plates
are 0104" 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 twinengined 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 . 145 Ib. Compass Type A . 45 Ib.
Pressure Head . . . 25 Ib. Compass Type B . . 67 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 . 125 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 = 663 ohms
then V = C X R
= 4'5 x 663 = 298 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 pev
\ \
 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 113)
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 = 117 and e = 105, find the volts necessary
for an arc 5 millimetres long with a current of 10 amperes.
Voltage required = a + bL +
39 + I0 . 5
495 + 642
= 55'92, say 559 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 horsepower (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)<3'23 118)
= 45 X 823 X 205
= 76 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 electroplating = 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 = 314, D = dia. of pulley in ft., and N = revs, per min.
of the pulley. Find the speed when D = 35 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 wroughtiron 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 = 314, 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 inchpounds 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 = 25.
92 ARITHMETIC FOR ENGINEERS
A h
12. The formula ~ occurs in connection with the strength of
A
beams. Find its value when A = 36, A r = 167, h = 55.
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.
prJJ
15. Find the value of the expression ~ t which relates to a centrifugal
pump, when g = 32, H = 125, V = 275, and v = 23.
A
16. The formula K = . refers to electrical condensers. Find K
4?r(
when k == 2'2, A = 975, TT = 314 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 = 225, L = 175, 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 = 314, 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 = 25, N 1250, and W = 53.
20. The B.H.P. of a threephase motor is given by the formula
1732^7? ^ 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 128 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 highspeed 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 i8aD f 1315^, where D = long axis, and d = short
axis. Find circumference when D = 25 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 = 125.
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 = 25, R = 57,
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 = 35 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 = 27, 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 crosssection 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 = 25, and
19. The weight of a girder before designing may be obtained from
the expression J^ i tons. For a particular plateweb girder w = 50,
r = 12, c = 1300, / = 45, / = 65. 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 = 27.
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 323 and Y = 155.
\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 = 28,
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 righthand 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 25.
(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) 25/> 2 = 25 X 25 2 = 25 X 625 = 15625.
(/) (2'5^) a = (2'5 X 2 5 ) 8 = 625* = 390625.
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 314, 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
293 tons per sq. in. _ ,, ,
23 K * Bottom line = 45125
say 45100.
P v 2
Example 86. The expression HjiH is very important in
hydraulics. Find its value when H = 15, P 2200, G = 625, v = 4,
and g = 32.
H h 7* H = 15 4 ^ 4 
Cr 2g ^ 625 2 X 32
= J 5 + 35^ + ^
= J 5 + 35'2 + 25 = 5045
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 = 241 000028(500 6o) a
= 241 000028 X 440 *
= 241 000028 X 193600
= 241 542
= 1868, say, 187 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 = 15.
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 = 176 Ibs.
20. Find the value of the expression A 2 & 2 , which relates to the
strength of a beam, when A 2 = 222 and h = 57.
21. Find the value of the expression J I o> 2 , which refers to flywheels,
when I = 472 and = 357.
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 33 and 5 = 3*74?
bh*
24. Find the value of the expression , which refers to the strength
of beams, when b = 75 and h = ii.
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 lowpressure 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 45 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 = 18.
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 = 402*.
30. Burls' formula for a purpose similar to that in Ex. 29 is
o8d s (i f *i5f) + 15 Ibs. Calculate the weight for a car engine when
d = 3 > i5 / " and r = 15.
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 3voltmeter 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 = 65 amperes, V = 218 volts,
V, == 68 volts, V 2 = 156 volts.
33* The formula J la  j refers to radialarm bogies on locomo
tives. Find the value (ft.) when a = 12 ft. and b = 675 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
225
i'5
36
6
196
H
4
2
49
7
225
15
6*25
25
64
8
256
16
9
3
81
9
289
17
1225
3'5
IOO
10
324
18
16
4
121
ii
361
19
2025
45
144
12
400
20
Square.
Number.
Square.
Number.
Square.
Number.
The squares are in the lefthand columns reading from the head
ings at the bottom; square roots are in the righthand 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 righthand 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 lefthand 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:
'2209' (47
16
609
609
Then, V2209 = 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 lefthand 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 righthand 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.
263
26*3
"789
1578
526
5261
69169
.'. \/692 5 = 263.
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 263. As a proof, 263 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 4000' (30659
606
6125
61309
4000
3636
36400
30625
577500
551781
== 3O66
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 (16867
26
328
3366
33727
184
156 Decimal point is placed when this
product is obtained.
2850 ^
_ ^2845 = 1687
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 = 1414 >/3~= 1732 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. 1156. 2. 24025. 3. 441. 4. 841.
5. 842724. 6. 11025. 7. 162409. 8. 11881.
9. 6084. 10. 0729. 11. 009025. 12. i 0201.
13. 00000961. 14. 25281. 15. 249001. 16. 102,400.
17. 6,760,000.
Work out the following square roots to 4 decimal places and state
the answer to 3 :
18. 227. 19. 261. 20. 2558. 21. 3682.
22. 7631. 23. 5837. 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 =* 3584 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
i73'oo' (1315
i
/28oo + 75
V 2800 750
4 ^2<r
23
73
4NAT 7 73
4 X I3I5 = 526*
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 sealevel
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 = 35* and r = 1*285.
30. Find the value of the expression A 'o6\/A, which relates
to chimney stacks, when A = 126.
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 * = 314. Calculate the time
when / = 326 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 / = 125 and r = 85.
37. The expression B f \/B 2 + T 2 occurs in shaft calculations.
Calculate its value when B = 455 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 1201 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 = 55, 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 pocketbooks 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^^.Wi
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^VpSP
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 directacting
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, takep. 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 325 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< 23 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 flywheels. 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 = 3142.
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 crosssection. If the ring is of circular section wire,
a = * d 2 if d is diameter of wire. If if = 314 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 sealevel 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 = 12. 7.1= 4, m = 6. 8. I = 19, m = 3.
9. / = 3, m = 8. 10. / 15, m = 14.
11. Add 4 3570 to 4 5160. 12. Add 4 2180 to 3560.
13. ,, "2250
15. ,, 0050
17. 4 i
19.  1505
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.
4 105.
16.
,, 2
.. + 305
046.
18.
,,43
,, 205.
61.
20.
,,42
,.  16 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. 25M, 3w, i5W, 7'5n.
40. 28u>, $w t 575^, 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 is9
ic (a
3(jie)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 $wl* to and add.
Sum of 8 and 5 is 3.
The " unit " wl* remains.
Change sign of ywl 2 to and add.
Sum of 2 and 7 is 5.
Change sign of
Sum of 4 and 3 is 7.
+ and add.
Change sign of 6wl* to + an d add.
Sum of 9 and + 6 = 3.
Exercises 32. On Subtraction of + and Quantities.
Find the value of the expression W w in the following cases :
1. W = n, 0; = 5.
4. W= i, w = 10.
7. w = 2,w = f 12.
2. W = 4, w == 7.
5. W = 3, w 2.
8. W = 4, w = * 3.
3. W = 7, w = 3.
6. W = 15, w = } 3.
9.W = 3,; = 3.
10.
Subtract 15
from 37.
11.
Subtract 1*2 from
 24.
12.
. '7
16.
13.
i'9
46.
14.
26
"3
15.
 218
 128.
16.
tt 313
.. 359
17.
229
116.
18.
5'35
156.
19.
358
~ *759.
20.
7'2
5*5
21.
  1*355 >,
2746.
22.
238
tt 205.
23.
x *357
3359
24.
3
2155.
25.
235
5
26.
ywl
i^wL
27.
I9^/"
7 rf.
28.
,, i2ad
2 i$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? = $a 2 , which
completes the terms in a 2 . Then take all the terms of another kind,
here numbers only, and repeat : 10 and f 10 = o. Thenf 3 alone
remains.
When a term has been added in, it is advisable to tick it off
before proceeding to another. This serves to show what terms
have been dealt with, and indicates when all terms are included.
This method is sometimes known as " collectingup like terms/'
Example 116. Collect up like terms in the expression 750 3/ +
5 /a _ t + 24 _ 3.5/2 + 4 . 7/<
V V v/ V V V V
75  3* + 5*  < + 24 ~ 35' 1 + 47* = 774 + '7< + 15**
Taking the numbers first, 750 + 24 = 774. Next taking terms in
t : 3* and t = 4*, then 4* and f 47* = f "jt. Now taking
terms in t 2 (quite distinct from those in t) : f 5^ and 35^ = f i$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. 275y 2 + 1457 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 .rp = 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 = 35, R = '6. 6. a = 49, R = ii.
7. a =  2, R = 1563. 8. a =  15, 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 = 15, / == 3, x = i'5
15. a = 15, /= 6, # = i.
Find the value of the expression  for the following conditions :
16. p = 10, q = 25. 17. p = 10, = 25.
18. /> = 10, q = 2*5. 19. = 10, q ~ 25.
20. p =  7'5. ? = 3 21. p =  75, ?   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 = 17, S = 327, R = 5; (2) a = 15, S =* 3, R = 75.
(1) a + S  R = 17 + 327  5 = 447*
(2) a + SR = 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 + 45 + (~
_ =
R  r  25  5
= 5 X{ 1515}^ J775 _ 8
 30  30 i.
Note. After substituting, the expression 5 + 45 + ( 25) = 95 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 = 125, S = 275, R = o. 10. a = 106, S = 35, 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. 375.
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 = 15.
22. n = 56, a = 28, 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 = 848, B = 212, r =45. 26. A = 848, B = 212, r = 5.
27. A = 36, B =  575, r = 45. 28. A = 36, B =  575, 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 = 75, b = 7, R = 20.
31. The formula v = u f at is important in questions on velocity.
Find v when u = 5, t = 35, 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 = 25. 33. v = 6, w = 22*5, * = 35.
34. v = 10, u = 5, * = 1*5. 35. v = 12, w = 15, t = 15.
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 == 37. 43. a = 15, b = 6, R = 135.
44. a = 15, b 6, R = 135.
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 + 74
= 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
9~43
= 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(63)
=8+5x65x3
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
= 154 =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 1484 terms inside by 7. " Collect up "
= 11144 *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 5f 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.
1434 [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 wormgeared
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. 1019 {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. in, 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. 1221 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 2pr. (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)
ab
a b
2 ab
 ab + b*
a* 2ab + b 2
Multiply by f a
axa==a 2 ; &xfa= 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
abb*
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 twoterm 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 flywheels, 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 flywheel, 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 i2/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 rightangled tri
angles we have the statement b 2 = a 2 c 2 . Find by the " expansion
method " the value of b 2 when a = 9625 and c = 3125. (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
" lefthand side " of the equation, meaning all written to the left
of the equals sign, and the " righthand side/' meaning all on the
right of the sign,
The allimportant 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 56lb. weights.
Then evidently if three sacks weigh three 56lb. weights, one
sack will weigh one 56lb. 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 lefthand side, to obtain the w above we have
divided the ^w by 3. Notice also that the righthand 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 23 p = h. Find the pressure when the height is 25 ft.
Substituting for h 23 p = 25
T 2 '3 2 *3/> _ 2 5
^3 "23
= 109 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 109 for p in the equation.
Thus 23/> = 25
Substituting 2*3 X 109 = 25
Simplifying 2507 = 25
which is quite satisfactory, as the slight difference is due to the value
109 being calculated to only 3 significant figures.
With a simple equation as above, the operation performed
(dividing by 23) is easily seen. With more difficult types the
operations may involve many steps. It is therefore advisable to
state at the lefthand 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 23 " in the previous
Example, No. 138, means " dividing both sides by 23 " 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 lefthand 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 = 165. 2. lid = 134. 3. '75D = 16.
4 37* = 35. 5. 1055 = 1105. 6. "004M = 17.
7. 251* = 135. 8. 440 == 115. 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 = 134.
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 = 314. 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 directacting steam pump we have the formula '75D 2 P =
d*p. A certain pump has D = 8, P = 60, d = 35. 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 =83.
P = 8 3
Proof.
5 3
Substitute for P in equation
X 5 y = 5 X 83
... 413 _ 8 . 3
' P = 4 I 5
/. 83 =83
Result is correct.
* v is the Greek small letter "eta."
M
162 ARITHMETIC FOR ENGINEERS
Example 141. Solve the equation ^ =13.
 m  = 13
305
X305 '^ =i3X35
m = 3965
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  = 231.
^  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 = 25, find e.
11. The formula r = c refers to long columns. Calculate the value
of / when c 160 and k =* IO75.
SIMPLE EQUATIONS 163
y
12. If = N, a formula relating to belts, calculate the value of
trD
V when IT = 314, D = 125, 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 = 65, 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 = 414.
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
.. 1301 == 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 = 25,
v = 78, and T = 550
Y I
t; "* *
Substituting  55? x
x < fil  550 2
X 78 25* = 550 x 78 ... 3
< _ 550 X 78 _
*  ?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 25 X t\ i.e., the top of the lefthand fraction (in line i)
X the bottom of the righthand fraction. Similarly on the righthand
side we have the top of the righthand 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 horsepower of a steam engine is given by the
formula H.P.  O Q Q where H.P. = horsepower, 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 lefthand 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 screwjack. If p = 5, * = 314,
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 = 35, TT = 314
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 35 H.P. Calculate its
stroke.
5. The formula d*l = ~ refers to dynamos. Calculate the value
of c for a small multipolar machine where d = 18, / = 26, K = 375,
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 = 75, D 135,
and / = 12.
12. If H = horsepower obtained from a waterfall, Q = quantity
of water in cubic feet per second, and h height of fall, then H = 
where G = 624. Calculate the value of Q when H = 40 and h = 105.
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 * = 314
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, / = 55, C = 5400, and d = '375.
(In the following few examples the unknown is in the denominator.
They are best solved by crossmultiplication,}
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 = 13, 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 = 36, 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 lefthand 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 righthand pan. To obtain the bag of cement by
itself on the left, we must remove the 15 Ibs. from the lefthand
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 = 2263.
Substituting, 21*7 + w l = 22*63
217 217 + w l 21*7 = 2263 2I *7
Proof. 2 1 7 + 93 = 2263, /. Result is correct.
In practice the second step may be shortened thus
217 4 a/! == 2263
21*7 t^ = 2263 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 lefthand 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 lefthand 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/)
46 Ibs
U/ 1O =
UT *
*
Fig. 22.
46
46
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. 115 + D = 2385. 7. Y f i = 7'7
8. n f 333 = 1667. 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 = 27, 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 = 1662, C = 100, and
s = 165.
Solve the equations
13. m 5 = i'i. 14. x 21 = 2785.
15. H  34 = 25. 16. Y  15 = 16.
17. y 006 = 'ooi. 18. D 05 = 113.
19. P 147 = 213. 20. L 105 = 125605.
21. / 55 = ~ ox.
22. The formula u = v at is used in mechanics. Find the value
of v when u = 25, a = 38, 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 = 65, u = 33, and t = 35.
Substituting values 65 = 33 3*5#
Transposing 35 a = 33 65
265
3'5 a = ^j = 757
Proof. Lefthand side = 33 3*5 x 757
= 33 265 = 65
/. Result is correct.
Note. Before the value of a can be obtained the 35 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 276 f '45 a = 13
276 + 45* = 13
276 450 = 13 27*6 = 146
Note. After transposing, the righthand 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 = 107, R = 45 and n = 4
Substituting the given values
107 = '45 +
 45 l'0
X 4 '62 X 4 = r = 248
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 2Z ' = 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 = 25 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 * = 314.
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 = 175.
7. If 25 = 146, 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 47.
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 = ii6). 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 = 25,
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 = 313,
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 lefthand 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 ,55336.4
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 364 = 255 4 X 364
1092 = 255 1456
1092 = 1094, 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
54 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 + 14 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 14 300
4883  = 84
Example 159. Find the value of n from the formula e(m a) =
a, when e = 133, m = o, and a = i.
Substituting the given values
i33(o i) = i
Simplifying 133 X i = i
Simplifying 133 = n i
Transposing I 133 = 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 righthand 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 = 715, T = 95,
W = 83, and /, = 15.
SIMPLE EQUATIONS 175
W(T  t) = w(t  /,)
Substituting given values 8*3(95 /) = 71*5^ 15)
Removing brackets 7885 83* = 71 $t 10725
Transposing 7885 + 10725 = 71 5* f 83*
Collecting 1861 = 798*
798 <
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 = ^
JY ~ 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
Crossmultiplying 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 450 750; 76(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 75^ 342 y6w = o
Collecting 108 151^ =
Transposing 108 = 151 w
108
f 151 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(75 + *)
Removing brackets 15* 15 4* f 12 = 15 + 2*
Transposing 15* 4* 2* =15415 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 25) = 2R, which
refers to a Weston pulley block.
4. Find the value of a in the equation 22(15 ~ 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 = 285, 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) = 2025(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 = 285, 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. =^ = 35
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 = 267, 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 = 108, R = 5, and Y = 123. (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 builtup
girders. If w = 133, W = 95, / = 76, / = 65 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 : 5i9(L f 60)
= 1190 x 26. Calculate the value of L.
5i'9(L f 60) = 1190 x 26
T , f 1190 X 26 _
~ 519 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 519 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
5i9(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 396==F 32
Transposing 396 { 3 2 F
F = 76
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 = ii, and t = 37.
Substituting values, I25{L + (37 32)} = 11(54 37)
Simplifying *I25{L + 5} = ii x 17
T , 187
r 125 L + 5 = ^ = I49'5
Transposing L == 1495 5 = 1445.
Example 169. Find the value of $ from the formula
W(T*) = (w l + ws)(tt l )
when,
W = 25 w l = 683 w 155 T = 91 / = 36 *! = 163
Substituting values 25(91 36) = (683 f 155 s)($6 163)
Simplifying 25 x 55 = (683 + 1555) x 197
19.7 ^==68.3 + 15.5 5
Simplifying and transposing : 698 683 = 1555
15 == i55
r 155 = ~  '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 = strokeratio 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 = 305 and t = 75.
FA
3 The formula = W(H + A) relates to the strength of bodies
under impact. Calculate the value of H when F = 126, 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 = 378 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 = 35 and h = 43.
9. The formula Z = h\K + ~J is connected with the strength of
builtup beams. If Z = 176, h = 27, and a = 135, calculate the
value of A.
10. Using the formula given in Ex. 9, above, calculate the value
of a when Z = 255, A = 75, 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 horsepower of a petrol
engine when d = cylinder diameter in inches, and r = strokeratio.
Calculate the value of r when H = 5 and d = 33.
13. The expression w(T t) = (W + e)(t ^) occurs in heat
experiments. In a particular case w = 968, W = 2785, 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 = 473, / = 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 = 27, 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 jj 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 lefthand 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* = 36 X 10
Collecting iSx = 36
Proof. Substitute the value 2 for x in the given equation
Then \ + 2  * 2 = 3 ; 6
/. i f 8 = 18, 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 36 Write 10 x 36.
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 *
64""
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 "t5 __ j
3* + 5 ^1 , i
7 23
_ _ _ _ . r~ 6 21 14
X L.C.M. of 7, 2, ^
[b 21 141
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 lefthand 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 10130 5 = i8p p i2p
Collecting 35 = $p
75 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 righthand 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 \  = 65
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 = 67, r 535, and
C  54 h a 2b
15. Calculate the value of a from the formula g = . ? "^" ^ ,
referring to the strength of dams; when g = 73, 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 18 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 035 ft.
Substituting 35 = **
X N 1 '35N 2 = 2938
35 N'
Taking \/~~~ N = ^8400 = Qi6 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 = io74(m 2 i)
Removing brackets w 2 = IO74W 2 1074
Transposing i'74 = i'074m 2 m 2 = 074*
1074
*O74 m 2 = '* = 14*5
/^ . t J
Taking \/~~ m = ^145 = 381
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 = 187, and d 6, find t.
3. The formula / =  refers to the stress caused by centrifugal
force. If / = 200, w = 83, g = 32 and a = 24, 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 72. Calculate the value of B.
6. For a directacting 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 = 1125.
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 = 322, 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 horsepower (H) of 75, 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 = 802 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 802 Jh~ Find the
height to give a velocity of 66 ft. per sec.
Substituting 66 = 802 VJT
Simplifying Vh =823
Squaring h = (823)* = 677 ft.
Explanation. Here the A//Tmust be isolated before h can be found.
Hence the equation must be divided by the coefficient 802. 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 = 3142.
What length of pendulum is required to give a double swing of 2 sees. ?
Substituting
Simplifying V / = 3183
322
Squaring ~ 2 ~ = ('S^) 2 = 1013
x 322 / = 1013 x 322 = 3262 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 = 225 r 2
Transposing r 2 = 225 694 = 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 = 55, and d = 375.
4. If T = 2ir\/ calculate the value of g when T = 182, / = 273
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 = 38, c = 95, and m = 17.
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 = 241 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 horsepower 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 = 883 and h = 221.
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 35, 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 = 073, 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 = 15.
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 righthand 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 lefthand
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 = 3I4DN.
Find a formula for D, and calculate the values of D when V has the
values 800, 1000 and 1200, N being 200.
3i 4 DN = V
V
r 3'MN
D
(i) V
= 800 .*.
D =
314 X 2^
(2) V
= IOOO
D =
IOOO
3*14 X 2OO
(3) V
= 1200
D =
I2OO
3'14 X 2OO
= 1273 ft
= 1592 ft.
= 191 ft.
Note. It should be easily seen that the multiplier of the unknown
D is 3I4N. 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 lefthand 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.
23
X 23 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 crossmultiplying.
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 gasengine 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
Crossmultiplying, 4?r/K = A multiplying, which brings un
__ A known to lefthand 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 horsepower 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 = 271* 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 horsepower 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 crossmultiply).
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 + 147.
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
rd
ra * =
Finally replace the 2 by a symbol, say a,
Then ax + d = r
d ax =r  d
rd
~ 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 righthand 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 = 1035 + !$, a statement relating
to the expansion of steam.
TRANSPOSITION OF FORMULAE 201
4. The formula K = 48 f ^Y relates to the consumption of power
in a textile factory. Transpose for Y.
5. The formula / = i'54PD f 26 relates to the strength of steam
boilers. Transpose for P.
6. The formula F = R + 32 connects Fahrenheit and R6aumur
4
temperatures. Find a formula for R.
7. Find a formula for n from the equation R + r = S.
8. In Exs. 44, No. 9, the formula v 115 i3Ooa occurred. Find
a formula for a.
9. In the formula L = 566 8/, relating to the latent heat of
ammonia, transpose for /.
10. Transpose for / in the formula 25* =* 146.
11. The formula D = CS f K is used when measuring distances
with a tacheometer. Transpose for C.
12. Find the value of a from the formula v = u f at.
13. In Exs. 44, No. 17, we have the equation nr wR = o. Solve
for r.
14. Solve the equation in Ex. 13 above, for R.
15. Find a formula for I from the equation B = H f 4* I, which is
used in magnetism.
16. Find a formula for a from the equation v a = w 2 f 2 as, which
relates to velocity problems.
17. Transpose for r in the formula / a = 6$pr + i8 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) =
XJ
4
32 F = ?R + 32
4
Example 208. The following formula is met with in connection
with plateweb 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 41)
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 singleresistance R ohms
is given by the equation g =   . Find a formula for R.
Note.
= _
R rs
Crossmultiplying, rs = R(s + r)
__

.,
Explanation. The righthand 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 = electromotive 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)
ME0) 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 gearwheel 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. Crossmultiply.)
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 lefthand 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.
= 548 VS'
Example 218. In connection with ropedriving 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
41 A * + J
X C 1 i = C 2 (* + i)
+ (* + u Fi~ c '
Taking
"Explanation. As the lefthand 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 lefthand 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 = righthand 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 fireboxes, 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 eggshaped 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. Crossmultiply 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 fourfigure 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 fourfigure loga
rithms is quite small, only occupying two pages. For more accurate
calculation there are 7figure 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 fourfigure 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 4figure tables are to
be used.
A logarithm (or " log " as it is commonly called) is usually a
mixed number, such as 34512, 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 2359 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
3142 . . . 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., 32396, 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 29777, because this
would mean that the whole of the number 29777 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 29777, 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 lefthand 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 fourfigure 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 4decimal 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 lefthand column, and the lower half is placed on the
righthand 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 lefthand and
top columns are not numbers but only significant figures. Thus,
on the first line in the lefthand 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 lefthand 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 righthand
figure of the fourfigure 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 lefthand 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 lefthand 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) 6065, (&) 6080, (c) 3009, (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 = 07580
(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 = 42180. '
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 lefthand 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 fourfigure 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
for Vhe, fiv^f 3 decimal places in
These ^iguves to
be adkc\ecl Vo
4
rhe loganHim
Vhoae
i Vhe
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 lefthand 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 lefthand 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 fourfigure 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 32104.
Looking opposite *2i in the lefthand 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) 07103; (b) 32997; (c) 2*0019; (d) 14779.
(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 20019 and the significant
figures are 1004. Then the antilog will be 1004, *'* 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 32997, 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) 29786; (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 5133.
(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 lefthand 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 = 29786
means " the logarithm of 09519 = 29786."
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. 1245, 2659, 450, 3301.
2. 0005, 30004, 0227, 1*707.
3. i'7495>_2'3989 L i'73 7. 21486.
4. 9989, 5'6739, i'_8o55, 39499
5. 02699, 3*4928, 18407, 20002.
6. 4*9599, 795 6 > 3'9O39, 19009.
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. 30095, 2193, i'3 OI 3069.
14. 43286, 13595. i3i, 4972.
15. 5'iooi L 335ii,_5956, 60969.
16 7575, 12091, 62201, 74771.
17. 33502, 9896, Too2i, 19795, 18951.
18. ioooi, 32375, 31115, '1009, 18751.
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 72 X 625 by logarithms.
Log 72 = 08573
Log 62 5 = 17959
Log of product = 26532 = 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 = 18949
Log 0039 = 359II
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 314 X 314 X
13000 X 021 tons.
Log 314 = 04969
Log 314 = 04969
Log 13000 = 4*ii39
Log *o2i = 23222
Log of product = 34299
Antilog = 2691
.*. Result =2691 tons.
Note. Carrying figure from the first decimal place is f I Then
the sum of f i, 2 and 44= 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 2023 x 4*093 occurred
in connection with the strength of a channel section. Complete the
working, using logarithms.
Log 667 = 18241
Log 475 = 16767
Log 2023 = 03060
Log 4093 = 06120
Log of product = 04188
Product = Antilog = 2623
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. 356 x 2195 2. 095 X 316. 3. 00032 X 60 X 13
4. 1414 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
247
Ex. 57, p. 65).
Log 238 23766
Log 24? = 03927
Log of quotient 19839 = Difference
Quotient = Antilog = 9636
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 = 18420
Log 1 60 = 22041
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 = 20766
Log 23 = 03617
Log of quotient 37149
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 03617, 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 = 48752
Log of quotient = 49560
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. 2195 r 3i' 8 3 2. 00963 ~ 785.
3. 29580 4 1986. 4. 1352 4 2366
5. 00058 r 000415. 6. 2382 ~ 2998.
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 = 26571
Log 177 = 12480 Log 146 = 11644
Log of product = 46002 Log of product = 18215
Log of bottom line = 1*8215
Log of quotient = 27787 = Difference
Quotient = Antilog = 6008 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 744 = ^'8716 Log 33000 = 4*5185
Log 785 = 18949 Log 3 = 04771
Log 3'5 = '544i Log of pro duct = 49956
Log 3'5 = *544i 6 * * D
Log 800 = 29031
Log of product = 57578
Log bottom line = 49956
Log of quotient = 07622
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 519 = 17152 Log 000341 = 45328
Log bottom line = 47881 Log 4 = 06021
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 2938 2 L3^ < 7 X j6'^
' 093 x 1136 " "^ooo^i" X~957~
3 7 8> 5 X 113 X 2167 x 194 ^ 875
33000 * 000341 X 56 X 416
' 00374 X 15250 X I 021 * 33000 X 105
7. The bending moment on a girder was given by the figures
62 XJ4X28 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.
. ^ , , ,, . 987 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 175 +
985 X 475 X '2375, figures relating to the strength of a channel
section bar.
Log 7 = 08451
7 X 575 X i'75 + 9*85 x '475 X 2375 Log 575 = 17597
= ' ^ ' + ' rrTi ' Log i'75 =02430
Log of product = 08478
= 8155
1 Antilog = 7044
N.B. The brackets are merely used
to show the necessary steps.
Log 985 = 09934
Log 475 = 16767
Log 2375 = * '3 75 6
Log of product = 00457
Antilog = in i
Explanation. The first part of the expression is evaluated by logs,
care being taken that the antilog 7044 is found and written down under
the main expression. Similarly the second part of the expression is
evaluated by logs, and the antilog iin written next to the 7044.
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
65 X 5
282 X 282
231
tons,
64
Log 65
Log 5
Log of product
Log of lower
product
Antilog
= 08129
= 06990
1+ 6 ' 5 x 5 
^ 282 x 282
64
= i'5"9
= 9004
I f 4088
6115
= 4088
Log 282 = 0*4502
Log 282 = 04502
Log of product 09004
__
5088
12*58 tons
Log 64 = 18062
Log 5088 = 07066
Log quotient = 1.0996
Antilog =s 1258
Explanation. First evaluate by logs the righthand term in the
bottom line, and insert its antilog (4088) in the second line of the main
working. Then add the i to the 4088, 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
= 17966
= 17966
6395 X 8431
322
Log of product = 15932
Log 4 = 06021
Log 6395 = 08058
Log 8431 = 09259
Log oi product = 1*7317
Log 322 = 15079
29911 02238
1674 Antilog = 09797 Antilog = 1674
say, 098
Explanation. First evaluate the bracketed terms ; the second
term can be found mentally, being ~ = 8J, i. e., 8333. 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 4 235 f 958. 2. 27 f 3'5 8 X 2895 X 3*142.
8 9'75 825 6875_X_? 1725 X 25642894 X 015
' 10385 f5'5 : 75 " 6785 x 0125
B . 175 X * 56.5 + i?>5 x .^ x ^^ x 4 . 5Q
6. The expression 7 X 57 X 175 f 985 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 75 125 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 (625) 8
Log (625) 8 = 3 x log 625
= 3 x 07959
= 23877
Antilog = 2442 = 625'
Note. Brackets used as in the 625 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 I7752
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 25623
= 31246
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
= 18951 + 2 x i575
= 1*8951 4 11410
= 10361
/. 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. (3142)". 2. (7854) 1  3. 25*. 4. 0875*.
5. 27852. 6. 7325* X 875. 7. (875 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 2209.
Log ^2209 = log 2209
i of 13441
= 6721
Antilog =5 47 = \/22o9
Explanation. In the case of square root 2 is the figure denoting
the root. Hence log 2209 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 ty2512. In this
case 3 is the number denoting the root.
Then log $^2512 ==  log 2512
3
=  of 14000
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 = T8ooo, 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 24000), the + 2
and the 2 neutralizing each other. Then the original logarithm
74000 becomes 3 + 24000, in which the negative part is exactly
divisible by the divisor 3. Dividing the 3 by 3 gives Y, and dividing
+ 24000 by 3 gives + *8ooo. Hence our new log is "18000, 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 14000
= (3 + 24000)
= T f *8ooo
= 18000
^2512 = Antilog = 6310
Note. Brackets are needed in the third line to denote that both
the 3 and the 24 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 43927
= ^of  36073
Antilog =
Explanation. Converting the logarithm 43927 into a number of
single sign, the f *39 2 7 is taken from the 40000, 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 12785.
Working by Method II
Log ^060247 =  of log 000247
= jof 4'3927
= j(5 f 13927)
= 12785
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 13927. Dividing
5 by 5 gives I and dividing 13927 by 5 gives 2785, so that our final
log is 12785, 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/2I78. 3.
4. V *oo 1 65. 5. ^978. 6. v'iTT^Tooo.
7. ^1906. 8. v'lTSrJ. 9. ^fTi2.
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 = 105,
17. The formula ^ gives the area of crosssection (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 = 25, 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 pistonrod 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 = 125".
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 = 945, / = 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 = 15).
ARITHMETIC FOR ENGINEERS
28. The expression \ ^^ 4  gives the diameter of wire re
quired for the field coils of a shuntwound dynamo. Calculate the
size of wire required when E =* no, D = 15, d = 10, and CT = nooo.
29. The horsepower 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 = 433 and S = 512.
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 TJTT inches. Complete the working.
Log '1675 = 12240
Log 785 = 1*8949
Log of quotient 13291
= '6709 (By Method I, p. 234)
Log V quotient = *3354 (Dividing by 2)
= 16646
Antilog = 461 9*; say 46*.
Example 255. In a problem on water supply the figures v/736 2
ft. per sec. appeared. Find the required value.
Log. 73** = ^ of log (73 6 2 )
= * of (2 X log 736)
=  of (2 X 1*8669)
 I of 37338
= 7468
/. Antilog = 5582, say, 558 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 = 45328
Log 4 = 06021
Log 692 S=B 18401
Log bottom line = 4*8753
Log 55 = i744
Log of quotient = 49648
Log of product = 48753
Log V quotient = 24824
Antilog = 3037 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 = 7257.
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 7257 v b b '
Top line Bottom line
Log (288 3 ) Log 48 = 16812
= 3 X 24594 = 73782 Log 12500 = 40969
Log 115 = 10607 Log 7257 = 28607
Log top line = 84389 Log of product = 86388
Log top line = 8*4389
Log bottom line = 86388
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 25 = 03979
= J X 02593 = 01297 Log bottom line = 21711
Log 1 10 = 20414 Lo g O f quotient = 1^268
Log bottom line = 21711 Log (quotient) 2 = 44536
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 =30969 Log 96 =09823
Log 425 = 16284 Log bottom line =27253
Log of product = 27253 Log of quotient = 22570
Log Vquotient = 11285
Log of the sq. root 11285
Log 36 = 15563
Log 1350 = 31303
Log of product = 3*8151
Antilog =s 6533
Discharge = 6533 cu. ft. per hr.
2. The Board of Trade rule for safetyvalve 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= 15.
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 horsepower 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/
watersupply pifte.
9. Using Dr. Pole's formula (Ex. 259), Q = I35od 2 \/ ., calculate
the quantity of gas Q if d = 10, h^== 25, / = 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 = 3142,
* = '375
16. The pressure of the atmosphere, p inches of mercury, at an
altitude of H feet above sealevel 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 2992
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 Tsquare,
setsquares, 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 submultiples of the yard are employed for
larger and smaller distances. The smallest submultiple 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 = 115 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 = 792 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 225 ft., while the 3 / 4f /r might all be converted
into inches, becoming 4075". 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 474 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^ 888 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>
= 513 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 6818 m., say 682 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 4coupled 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. 1155 yds. to inches. 5. 2785 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 headstock
occupies i'2i" and the loose heads tock n". What is the greatest
length available between the centres ?
Simple Geometrical Terms. The wellknown " 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 Tirics.
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 threesided 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 Rightangled Triangle (shown at g) is one in which one angle
is a rightangle. This triangle is the most important in higher
work.
H. A Hexagon is a sixsided 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 eightsided 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 tothe 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 straightsided 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 crosssection 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 straightsided 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
314. 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 = 31415928
,,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 JJS, *' , 3i\V * which the decimal equivalent
is 3i4*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.
9438
3146"
which is very near to the actual value of 3142.
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
314, and the average value should be very near to 3142.
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
= 3142 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
3142
~. 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 = 4975 ft.
.'. Surface speed = 4975 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 = 9426"
jQ \X T ^
.". Revs, per min. =  :  = 636 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 = iiS"
Two straight portions 10" each = 20 o"
/. Total length required
say, 5'7i*
672"
Example 272. A closecoiled 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 = 942"
/. Length of 10 turns = 942 X 10 = 94'2 /r
Add for ends 8
.*. Total length required per spring = 1022"
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 25 = 785"
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
= 256, 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
= 262 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  = 77 ft.
26 ' '
 /. Dia. = ? = 245 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 buttjointed
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,
buttjointed 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 lapriveted boiler ring
6'o / " dia., if the lapjoint requires an extra 4^ ins.
5. A wroughtiron 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 boilermaker'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 woodplaning 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 = 2nr 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{ 15(0 + b) Vab\
= 7r{i5(i + 625)  Vi x 625}
= 7TJ244  79}
= TT X i,65
/. Length of rod per link = 518"
Example 278. Fig. 32 shows at c an elliptic gear wheel. The
major axis is g" long and the minor axis 47" 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 = 235
Circum. = ir
.*. a 4 6 = 685
and Va6 = VZ^
.'. Pitch =
= ir{i5(a + b)  Vrtfc} =
== 3I4 X 705 = 221"
Circum. __ 221
No. of teeth ~~ 30 "
737". say, f*
2 35
= 325
 325!
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 = 425 ; b = 325 ; and a + b = 75
/. Circum of ellipse = TT x 75 = 23'52 r '
Bolt dia. = i" = 625"
.'. Largest allowable pitch = 6 x 625
Least possible number of bolts = ^>
Next even number = 8
3*75"
=627
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 centreline 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 mudhole 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 semielliptic arch is shown at V. How many 3* bricks are
required for the inner ring ?
10. The conveyor tunnel shown at U is semicircular at the top
and semielliptic 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 lefthanded or anticlockwise 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
righthand 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 subdivided 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 (= 573), 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
450
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 instrumentmakers' 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 halfdegrees 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 setsquare, 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 setsquare, 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 foursided 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 = i57'. 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 = 228'
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" .= 192'
Angle
32 degs.
7332
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. 1935. 5. 1728. 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 Rightangled Triangle. In any rightangled 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 rightangled 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 rightangled 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 rightangled 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 75", 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  5625 = 4375
/. h = V~^375 = 661*
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 rightangled
triangle, short side i'; vertical side 2'.
Then I 2 = i 2 f 2 a = 5
/=V5~= 2236'
= 2 / 2 // , say
Lap for riveting at top = i%"
/. Plate must be 2 / 4f /r wide
268
ARITHMETIC FOR ENGINEERS
Example 286. An opencoiled 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 rightangled triangle and PQ, the actual
length of the centre line, may be calculated.
Circum. of cylinder = *d = IT x 25 = 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* = 616 + 4 = 656
/. / =s V656 = 8 i* =s length per single turn
/. Length for 8 turns = 81x8 = 648"
say, 65* or $'5*
Proportions of 45 and 60 30 Rightangled Triangles.
With the aid of the foregoing relation we can examine certain useful
proportions in the two rightangled triangles most often met with.
The 45 Rightangled 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 rightangled 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 Rightangled 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 rightangled 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*
=41=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 Rightangled 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
rightangled 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 hexagonhead screws are to be cut from round stock as
indicated at H, and are required to be 13" 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 handrailing 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 314 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
573
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 265 = 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 rightangled 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 rightangled 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 channelsection 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. = 1152 acres
_ 4
208 rds.
320 sq. po.
3Q
600
 Of *2 ...... '05
605 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 3025, 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 twostroke 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 3142 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 highspeed 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 125" = 938 sq. ins.
/. Area of 10 holes = 938 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 =  = 8083 sq. ins.
_ Area _ 8083
~~~ Length ~~ 325
= 2485, 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 =189
h t = 315
A 2 = 207
J(3'5 X 189)
j(2i x 315)
J(32 x 207)
= 3307
= 3307
= 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 rightangled
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 186 square chains, find the total area of the
plot.
Triangle ABC. Area == J base X height
= J X 167 X 74
= 6175 sq. ch.
Triangle ADC. Area = X 167 x 51
= 425 sq. ch.
6175
Surplus area 186
Total area = 12285 sq. ch. = 12285
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~= 283' 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 = 361 ft.
Then, wetted perimeter = base + 2 slant sides
= 6 + 2 X 361
= 1322 ft.
/. Hydraulic mean depth =
area
wetted perimeter.
?*  i8 I5
1322
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 = ^tS* = 4625
/. Area = mean width x height
= 4625 x 4*625 = 21 4 sq. ins.
SMALL TRAPEZOID
 1 i
Mean width =
Fig. 59 ' Arca = 3*375 X 325 = 1096 sq. ins.
/. Net area of section = 21*4 1096
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 firebox 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 624 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 258 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 1753" wide, 64" thick; final
section 1472" wide, '482" thick. Calculate the reduction in area,
and express it as a percentage of the original area.
11. Calculate the area of the Tsection 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 turbinedriven
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 halfbrick thick), if i sq. ft. of brickwork
(per halfbrick 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 diamondshaped 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
crosssectional 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 crosssectional
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 (crosshatched), 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
= 2598 S*
say, 26 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 + 1414 S
= 2414 S
/. Area of outer square = 5827S 2
Deduct for four corners as ABC = 2 squares of side 707
= 2 x 707 x 707 = ioo S 2
Deducting gives 4827 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
= Tir 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 = 3142, *"  7854
.'. Area of circle = 7854 d 2 .
Note. 314 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 waterpipe is 14* diameter. What is its cross
sectional area ?
Radius = 7"
/. Area = irr 2 = TT x 7 2
= 314 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
305 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 107 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 = ,* = 178
area of i bar 60 1 '
We must, therefore, use 2 bars.
Example 303. A set of expansion bends
for a main steampipe is arranged as in
Fig. 67. The combined areas of the small
bends is slightly greater than the area of
the main steampipe. 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
steampipe area.
MENSURATION
Area of main steampipe
Area of i bend
.'. Total area of bends
289
785 x 15*
= '785 X 225 = 1766 sq. ins.
= 785^
= 785 X 9 a
= 785 X 81 = 636
= 3 X 636 = 1908
Excess area of bends = 1908 1766 = 142 sq. ins.
.'. As a percentage of the main steampipe area this
8% approx.
142 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 225 = 1766 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 in 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
2785 .
'. 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 steampipe is to be 123 sq. ins. What
will be its inside diameter ?
j8$d 2 = 123
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 highpressure
cylinder, and is 30" dia. Find the diameter of the H.P. cylinder.
Area of L.P. cylinder = *r 2
= 314 X 15*
= 314 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. = 
325
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 steampipe is 18" dia. There are to be four bends, and their
combined area is to be about 8% greater than the steampipe area.
Determine the diameter of the bends.
MENSURATION 291
Area of main steampipe = *r 2
= 3'H X Q 1
= 314 X 81 = 254 sq. ins.
o
Taking 8% of this =  X 254 = 2032 sq. ins.
i oo
.*. Total area of bends = 254  203 = 2743 sq. ins.
There being four bends the area of each
.% 785^* = 686 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 = 336 sq. ins.
2^00
= 207*
/. Spindle diameter = 2^*
Exercises 73. On Areas of Circles,
(Letters refer to Fig. 69.)
1. The outlet from the turbine wheel in a hydroelectric instal
lation is circular and 18 ft. dia. Determine the crosssectional area.
2. Find the crosssectional 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 crosssectional 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 crosssectional 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\" eyebolt is 942" dia. at the bottom of the screw thread
(i. e. t at the weakest place). Find its crosssectional 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 twostage air compressor has a lowpressure cylinder lo" dia.
and a highpressure 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 steamcollecting 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 twofifths of the whole
plate area, find if this rule is satisfied here.
MENSURATION 293
17. The lowpressure 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 ashpit in a Lancashire boiler.
In a boiler test it was necessary to know the area through the ashpit.
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 crosssection 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
17 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
oneeighth the pipe area. What will be the diameter of the throat ?
26. The exhaust pipe of a petrol motor is of the crosssection shown
at J. The circular part is to be ij" radius and the whole area required
is 24 sq. ins. What must be the depth of the rectangular piece ?
(Hint. Area of halfcircle 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 onehalf 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
steampipe is 12%" dia. The total area of the bends is to be 10% more
than the steampipe 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 : 28 : 7*1. If the lowpressure cylinder diameter is 7'4*,
what will be the diameters of the high and intermediate pressure
cylinders ? (Hint. Area of L. P. cylinder is 71 times H. P. area. Work
as in Example 306. Also area of I. P. cylinder is 28 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 = Tir 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 crosssection.
Area = 785(02  d*)
= 785(152  Hi)
= 7 8 5( 22 5  I21 )
= 785 X 104 = 816 sq. ins.
Working on the radius form
R = j'$" and r 55"
Then area = 314(752  552)
= 314(5625  30*25)
= 314 X 26 = 816 sq. ins.
MENSURATION
295
Example 310. A thrust bearing in a 5ton 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(756  4)
= 7 8 5 X 356 = 2795 sq. ins.
/. Total area of 4 collars = 4 x 2795
= 1118 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 doublebeat 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 55 2
= '785(7 2 ~ 5'5 2 )
= '7 8 5(49 ~ 3025)
= 785 x 1875 = 147 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 connectingrod 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 1115 875 sq. ins.
Then net area as at AB = 785(02 d 2 )
= 785(1225  d*)
and this is to equal 875 sq. ins.
/. .785(1225 *) =875
.785 /. 1225* =^
Transpose .*. 1225 1115 = d*
d* =11
Take V~~ /. d = Via = 105*, 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 crosssectional area of the metal in the lap
welded wroughtiron 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 1525 x 25 = ngSsq.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 crosssectional area.
2. A hollow castiron column is 10" inside diameter and 12 out
side. Find the area of its crosssection, 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 wroughtsteel riveted pipe is 16" internal diameter and is J*
thick. Find its crosssectional 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 275 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 lowpressure piston is the large piston
at the top and the highpressure 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 hydroelectric 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
57'3 628
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* = 1625 J ' = T = '9375
Then area of complete ring = TT (1625* *9375 2 )
.. Area of Port = ^(1625*  9375 2 )
= g( 2 6 4  8 7 9)
^ x * x _L7_6
' 8
Fig. 76.
Area of Fillet. The fillet or quartercircle 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 quartercircle = J x Tir 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 45% 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* = 542'; h = 2'2j"  221'; /. A 2 = 488
Area of segment = A 2 \/v 608
4 X 4*8 \/;g 608
t= i x 488VT842
= 4 X 4*88 X 1358
= 883 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 wellknown formula.
Example 316. The mudhole 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 = ~ = 75*
, t .11
6 = 4 minor axis = = 55
Then, area = irab
= * X 75 X 55 = 1295 sq. ins.
.*. Total force at 60 Ibs. per sq. in. = 60 x 1295
= 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^
yl, 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 onethird 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 rightangled triangle with a vertical
to the centre.)
12. Calculate (a) the area of the crosssection; (b) the wetted
perimeter; and (c] the hydraulic mean depth (area ~ wetted peri
meter) for the circular channel in Ex. 10, above, when running twothirds
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 crosssectional 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 crosssectional 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 motorcar the area of the crosssection was required. Find this
area if the two axes were i'i" and 8J".
20. Find the crosssectional area of an elliptical magnet core
18" X 9*.
21. Find the crosssectional area of an elliptical funnel 13* X 10*.
22. The eggshaped culvert used for drainage is very nearly a half
ellipse in the lower twothirds usually occupied by the water. Calculate
the area of the water section in the culvert shown at J.
23. A manhole door in the lowpressure 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 144 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 MidOrdiuate 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 midordinate method, areas
can be measured with an accuracy of within I %.
Method of Dividingup. 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 dividingup 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 "widord inches" ort * >v.
shewn as ^yll \iv<.s,
Fig. 82. Method of Dividing for MidOrdinate 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 midordmates 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 midordinates 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 MidOrdinate Rule.
The top has been taken as the base or length, and 10 midordinates
set up perpendicular to this by the method described. The measured
lengths are shown.
Sum = 474
/. Mean height = 474
Area = Mean height X length
= 474 X 125 = 5925 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 MidOrdinate 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 overall 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 midordinates have been drawn, and their lengths are
shown on the figure.
Then sum of 10 midordinates = 2953
.'. Mean height 2953 chains
Area = length X mean height
= 68 x 2953 = 2008 sq. chains
or 2008 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 midordinates 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 crosssection of a sixpole dynamo
magnet frame. Find the area of the section.
2. C, Fig. 86, shows the crosssection 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 crosssectional 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 crosssectional area of the castiron 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
40
50
zE
Scale of c.hcun5
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 midordinates, 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 twostage
air compressor, i* = 50 Ibs. per sq. in.
MENSURATION 311
10. B, Fig. 88. Full load diagram from scavenging pump of a
twocycle 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 fourcycle 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 )(sb)(sc)
4335*
265* or 866/*
4835* 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{i5(a + 6) 
more nearly
Step round
curved por
tions in small
steps, with
dividers ; add
in any straight
pieces.
^(DP)= 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
midordinates. Then
Area = aver, midordi
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.) = 2774 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 62425 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. = io6cu. 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 624 galls, per sec.
= '28 X 624 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 ?
624 galls. = i cu. ft.
/. 4800 galls, per hr. = \  cu. ft. per hr.
0*24
4800 ,.
_. 3 _. cu. ft. per sec.
624 X 3600 r
= '214 cu. ft. per sec.
Exercises 77. On Volume Conversion.
1. The volume of a steam engine cylinder is 1256 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 (3167'). 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. = 624 gallons
/. Gallons in tank = 40 x 624
= 2496, 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 == 1325 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" = 95 ft. Breadth 3 'g* = 375 ft. Depth i '6" = 15 ft.
.'. Volume required = 95 X 375 X 15
= 534 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 crosssectional area of the Standard BullHead Rail used
on British railways is 881 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 crosssection of a reinforced concrete footbridge 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 crosssection 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* = 133 ft.
.'. Volume = area of end X length = I3J.2LJL33 cu ft
144
= 123 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 = 196 sq. ins.
4
196 ,,
= * 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 624
= 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. = 1923 cu. ft.
Then 1923 = 5 X 35 X depth
/. Depth = ~ 1 ^ = ii 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
^637 = 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 = 1005 cu  ms 
Crosssectional area of wire = 785 x 25* = 0491 sq. in.
.', 1005 = 0491 x length in inches
.% Length = 2047" or 569 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* 725 2 ) x 3'5 cu. ft.
(reducing all measurements to feet)
= 7 8 5 X 4314 X 3*5
= 1185 cu. ft.
.*. Gallons of water = 1185 x 624 = 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 = 1259 cu. ins.
say, 126 cu. ins.
/. Weight of metal at '41 Ib. per cu. in. = 126 x *4 J =5166 Ibs.
say, 52 Ibs.
324 ARITHMETIC FOR ENGINEERS
Exercises 79. On Volumes of Cylinders.
1. A cylindrical gasholder 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. = 624 galls.)
5. The governor ball on a high speed engine is a castiron cylinder
3" dia. and 2 y thick. Calculate its weight if i cu. in. of cast iron weighs
26 Ib.
6. A threethrow 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 gasengine 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 crosssection of the discharge flume from a hydroelectric
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 gasholder 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 castiron 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 gasstorage 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.
36
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 firebricks for use in the furnace arches of ""mechanical
stokers are shown at B. Calculate the weight per 1000 bricks if i cu. ft.
of firebrick 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 4893 Ibs. per cu. ft.
6. At D, Fig. 1 02, is shown a Vblock for shop use. Find its weight
if of metal weighing 489 Ibs. per cu. ft.
7. Calculate the capacity in gallons of an elliptic milkcan 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 motorcar, shown at B, Fig. 102.
9. The dimensions of the crosssection 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 crosssection 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 chiselsteel 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 crescentshaped balance weight for the
driving wheel of a locomotive. Calculate the weight if i cu. in. of steel
weighs 28 Ibs.
15. An oilstorage 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 fireproof 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 midordinates.)
18. A, Fig. 10 1, shows the section of a British Standard BullHead
Rail. Find the weight per yard in Ibs. (Hint. Calculate the area of
the web between a pair of lines indicated ; work by vertical midordinates
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  = 655 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 = 118 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 = 565 CU. ft.
/. Capacity of tank = 113 + 565 = 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 + 756)
= XJ5Xj4 J 6 = .
Exercises 81. On Volume of Sphere, etc.
1. Calculate the weight of a castiron 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 castiron 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, onethird 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 "
3014 cu. ins.
/. Weight = 3014 X 0318 = 96 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 (3466 cu. ins.) when filled to a depth of 5 J*.
What must be the diameter at the surface of the
liquid ?
Volume of cone = Jnf 2 /*
3466 = 3 ' 14r> X 3' 5
,2 = 3 X 34'66
3'i4 X 55
/. r = V603 = 2455*
.'. Diameter at surface = 2 X 2455 491*
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 = 105 X 10*5 x 7*5
= 8269 cu. ft.
Volume of lower portion 1 __ ^/
(frustum of sq. pyramid) J ~~~ 3 *
_t 2 i c x
"*" s ' '
3
= 1667(1102 f 225 + 1575)
= 1667 X 1282 = 2137 cu  ft 
Total capacity of bunker = 8269 + 2137 s=s 10406 cu. ft.
Weight of coal held == 10406 X J cwts. = 5203 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 4044
3
2035 + 392 = 2427 cu. ft.
334 ARITHMETIC FOR ENGINEERS
Example 342. Fig. 108 gives dimensions of a gasholder 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" = 1067 ft.
Height = 255'
.'. Volume = ~(R 2 + r z + Rr)
+ 55 X 1067)
/. Volume of water = 408900 99420
.*. Gallons of water = 309480 X 624
= 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 = 3466 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 gasholder 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 hydroelectric 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 pocketbooks. 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 gunmetal . .
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 " giveand
take " process, can usually be reduced to a form permitting of easy
Fig. in . Calculation of Weights.
calculation. With practice the process of " giveandtake " 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 blowholes 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 = 2x35X2 =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 2nr/
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
= 314 X 175 X 132 sq. ins.
= 3'!4_ * *2>X  13  2 = 5036 sq. ft.
/. Total heating surfaces of)
tubes
j
= 5036 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
= 314 X 75 X 14 X
3 sq. ft.
12
= 275 sq. ft.
a , . . , , Total surface
.'. Number of tubes required = ^^ 7  T
Surface of i
12000
275
= 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 45 sq. ins.
= 565 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 = 314 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
vatertube boiler drum, as shown in Fig. 113.
Surface = 2irRA
= 2ir x 3 X ~ s q f t.
= 785 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 314 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. nsSurfacc 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 rightangled
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 rightangled triangle whose perpendicular sides are
12" and 4".
/. Slant height / = Vio 2 +"4*
= Vii6 = 1077*
/. Curved surface = nvl
= if X 4 X 1077 = 1353 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 = 2564"
/. 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 rightangled 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*"+ 425*"
TVue shape* o^ a Pace.
Fig. Ii8.
= V43o6 = 656'
/. Lateral surface = 2/(S + s)
= 2 X 656(105 + 2)
= 2 X 656 X 125
= 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 =
+ (5375 3 r i25) a
~
Curved surface^
of frustum j "
= ^7891 = 28
09
X 281
IT X 17 X 281
_
= 750 sq. ins.
or, 52 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 lengthwise 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 286 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 twopole 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
steampipe 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 26 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 105. 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 38 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
onehalf 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 drawin 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 rightangled 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 $" wide will give enough
volume to permit of adjustment to the exact weight, i.e., the weight
when calculated from the tentative dimensions should be a little more
than i Ib. (Note. For practical purposes, the area of the top Of the
weight may be obtained by subtracting the area of the rectangle shown
crossed from that of the hollow circle.)
13. An aeroplane petrol tank is shown at H, with dimensions. The
shape of the end may be considered as consisting of two semiellipses.
Calculate the capacity in gallons when quite full.
MENSURATION 345
TABLE OF VOLUMES AND SURFACE AREAS OF SOLIDS.
Title.
Figure.
Any prism .
Rectangular
pi ism or
cuboid . .
Cube . . .
Square prism
Hexagonal
prism .
Octagonal
prism .
Cylinder . .
Hollow cylin
der ...
Elliptical
prism .
Sphere . .
Hollow
sphere . .
Volume.
Area of base
X height
Ibh
S 2 /
26S 2 /
or 866/ 2 /
483S 2 /
or 'S29/ 2 /
or 7854^7*
,r(R2 
irabh
?rR 3 or 
or 52360*
Surface Area.
Circumference of base
x height
Whole area = 6S
Lateral surface = 4S/
Ends = 2S 2
Lateral = 6S/ or 346^
(For ends see Table in
Chap. VII.)
Lateral = 8S/ or
Lateral = 2irrh
Two ends = 2irr*
Whole area = 2rry(h f r)
Outer lateral^ _
surface J~" 2?rK/l
Inner lateral) ,
surface j=2i
Lateral
= irA{i5(flf 6)V
or it(a+b)k
(less accurate)
346 ARITHMETIC FOR ENGINEERS
TABLES OF VOLUMES AND SURFACE AREAS OF SOLIDS (continued).
Title.
Figure.
Volume.
Surface Area.
Segment of
sphere . .
Zone of
sphere . .
Any pyramid
Square pyra
mid . . .
Cone
Frustum of
any pyra
mid . . .
Frustum of
square
pyramid .
Frustum of
cone. . .
^o
or 5236/1(3)'+^)
area of base
X height
JS'A
/f=height of frus
tum
A=area of large end
B= area of small end
B+VAB)
Anchor ring .
Curved surface = 2irRA
or 2irR(RV r JR>~^ 2 )"
where R=rad. of sphere
Lateral = J circum. of
base X slant height
Lateral = 2$/
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 Coordinates. 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
43", since = 43. Thus this pair of figures will be shown by a
These two dimen
= GO YP
point 3" from one axis and 43" 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 coordinates of each point had to be calculated and
/ORIGIN
HORIZONTAL
AXI
Fig. 121. Rectangular
Coordinates.
the efficiency is plotted
base " quantity being the one plotted
A i SQUARES B TSQUARES 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 onefourth 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. Settingout 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 lo, 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 lefthand 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 onequarter 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 onefourth or onefifth
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. Drawingin 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 pearwood and containing a
large variety of curves; they can be purchased for a few pence.
Several types may be seen in the instrumentmaker'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 onefourth 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 4OO 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
1425
17
70
20
22
25^5
265
Voltage . . .
8
44
62
76
79
77'5
725
II. SHUNT WOUND.
Current, amps. .
875
103
17*5
101
2925
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 shuntwound
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 2925 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, onefourth 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
505
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 fieldcoil.
Time (mins.)
o
2
4
6
30
IO
12
18
24
30
35
47
40
Temp. C .
215
235
26
35
3675
40
43
45
47
3. The following figures give the voltage drop across the carbon
brushes of a directcurrent 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
i75
18
185
Current, amps.
4
Q
13*5
215
275
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 steampipes. 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
195
77
34*5
75
465
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
65
10
125
15
175
20
77
25
30
68
40
50
70
197
Resistance, Ibs. per ton
S4
526
53
5*32
6
9
n8
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
125
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 2865, 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 steamturbine 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
2865
28.3
28
279
2765
2735
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
845
90
93'5
96
Resistance
in ohms
937
93
919
90
886
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
735
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\tgq&
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, onefifth of a small
square must be estimated. For
the carbon lamp the decimals
will have to be neglected. As
onefifth of a small square is the
smallest that can be reasonably
estimated, then i ohm is the
least that can be plotted.
Therefore 937 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 Dieselengined vessel. Inter
polate the H.P. required at 10 knots.
Speed in knots .
77
90
965
108
117
I.H.P
800
1 200
1440
22OO
2600
2. In an experiment to determine the effect of salt in water upon
the boilingpoint, the following figures were obtained. Plot a curve
showing boilingpoint upon a base of salt percentage. Estimate (a) the
364
ARITHMETIC FOR ENGINEERS
boilingpoint if there is 20% in solution; (b) the percentage of salt
required to produce a boilingpoint of 107 C.
Percentage of
salt in solution
o
3'2
63
9'4
12*2
179
22'9
277
3i'9
Boiling  point
inC.
IOO
ioo'5
IOI*2
IO2
1025
I045
106
1078
1088
3. The following figures give the horse power taken by the circu
lating and feed pumps of a small condensing steamengine 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
105
!5'5
18
27
4i5
47
H.P. of feedpump . .
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 .
104
i 06
109
ii
i'5
II5
12
125
i'3
17
J '35
i'4
29
Specific resistance
2'5
2
19
i'3
i'3
i'4
215
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 turboelectric 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
193
22
25'3
266
Steam consumption,
Ibs. per K.W. hour
343
312
29H
274
25'6
24
232
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 ii.
Specific gravity
of acid
i
Voltage . . .
i*5
I 006
1015
1025
1038
188
i 08
*'*5
12
125
17
18
185
194
20
203
2O7
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
6242
80
IOO
I2O
140
Weight of i cu. ft., Ibs.
6242
62425
6223
6202
6171
6138
1 6O
200
220
26O
3OO
4OO
61
6008
5958
5846
5726
5363
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
n5
67
45
10. A certain watersupply 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 setsquare 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, onefifth 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
forhLoad 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 wormwheel lifting machine as
determined by experiment. Plot a curve with load as base.
Load, Ibs.
60
107
167
22
265
32
36*7
42
465
515
Effort, Ibs. .
9
142
212
26
3'3
372
432
48
5'53
60
B B
368 ARITHMETIC FOR ENGINEERS
Refer to Fig. 129.
Scales. The origin will be required.
Base. Required to take 515, 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
onefifth 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 onefifth of a small
square, onefifth 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 :
Righthand top quadrant x f and y +
Lefthand top x y +
Lefthand bottom x y
Righthand bottom x f y
Taken in this order, i. e., starting from the righthand 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 righthand 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 coordinates 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.
14
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, 35); C ( 15,  3) ; D (15,  55), 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. Coordinates (3, 4). Plus sign understood.
Therefore the point is in the ist quadrant, 3 horizontally and 4
vertically.
Point B. Coordinates (2, 35). The horizontal distance is
negative and the vertical distance positive. Thus the point lies in
the 2nd quadrant, 2 horizontally to the left and 35 vertically
upwards.
Point C. Coordinates (1*5, 3). Since both are negative the
point is in the 3rd quadrant, 1*5 to the left and 3 downwards.
Point D. Coordinates (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 55 downwards.
In Fig. 131 the lines to be measured as a check are shown by a
370
ARITHMETIC FOR ENGINEERS
chaindotted line, and are measured as follows : AB = 5*02 ", BC = 6*52*
CD 39% DA = 962".
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 " BH " 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
675
47
27
14
IOIOO
B. Flux density in lines
per sq. cm.
I35
12600
12000
III30
11
3
4
7'5
105
B
8520
5^30
3900
960
105
14
1920
185
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
465
+ 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 = = 115
.*. Coercive Force = 115 amp. turns per cm.
Example 362. The following values were calculated when solving
a certain difficult equation
Value of t
Value of y
362
3625
363
3635
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 lefthand 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 36327.
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( 4,35), B( 4 ,  35), 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 .
12
i4
i3
,6
i7
Values of y .
1142
39
US
705
1385
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
3035
309
315
321
Values of y . .
2173
912
29
 1435
2499
(Note. The origin will not be required on the e scale.)
6. The following figures are for a BH 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
i55
17
2 O
2 6
3'i
40
23
65
25
I'2
155
17
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
14
17
177
i'5
'75
*3
225
Values of y
17
1*2
'5
15
e
125
167
2*2
273
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
562
14
I
8
18
10
194
5'5
I
i*
8
'3
3
3'8
4'i
 18
2
5'2
263
22
61
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
3i
4'4
6
8
104
17*3
273
45
50
in CO
NH 3
92
13
I75
23
295
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 straightedge. 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 + 15*; (b) y = 15*; (c) y = 2 + 15*.
Calculations :
X
I + i'5v
ay
2
I + 3
=== 2
+ '5
i 4 75
= i'75
f 2
i +3
= 4
X
i5r = y
x
2 + 15*
**y
_ 2
3
_ 2
2 +  3
_ 5
+ 5
75
+ '5
 2 + 75
 I2 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
225
2
I +2
3
The curves are shown plotted in Fig. 136.
#
2 2*
= y
2
2  (  4)
6
5
2 1
I
2
24
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 righthand top corner. This can be
called a " positive " slope. When the sign is , the line slopes from
the lefthand top to the righthand 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 157.
To find constant b glance along the
curve near its extremities, and find a
point somewhere near each end, where
the coordinates are definite figures
(i. e., the reading does not have to be
estimated). Thus in Fig. 137, at point A
the line has the coordinates (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 35 (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 = i57 + 857*
Any two points can be taken in estimating b, and it is therefore
best to choose points of definite and easy coordinates, 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
IdS
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
coordinates 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
1191
horizontal
and the sign is f since the slope is to the right and upwards,
the equation required is
W = 370 f 119H
Then
Check.
when H = 36,
W = 370 f 11911
= 370 f n9 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
ia8
1 1'3
107
9'6
87
7'7
63
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
18
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 5ton block from the curve.
Capacity in tons .
i
I
2
4
6
I5'25
8
Price, . . . .
4
5
7^5
105
18
3. The following figures give the voltages necessary to cause an
electric spark between two needlepoints 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. .
178
174
164
16
15
142
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 .
176
352
528
704
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. .
107
216
285
391
440
53
490
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
1243
16
Values of L . .
1023
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
4942
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
900
8525
806
7 ci
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 " laboursaving " 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 " loinch " 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
tro
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.
Qilllliillliliim
TtTTTTTimti
Subdlvis\on fvom
5
I
Subdi 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 onefifth 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 " twentyone " ;
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, threefifths 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, 75) by another (say, 3)
proceed as follows :
1. Place the cursor exactly over the number to be divided (in
this case 75) 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 25.
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 ' 312' v ' 515
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 lefthand 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 righthand 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]iiiiinijiiii
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 lefthand 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 righthand 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 righthand 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 235.
(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 lefthand end of slide. 4. On the D scale the line
will be found onefourth 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 2875.
(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 onefifth 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 134.
(d) 1. Set cursor between the graduations of 605 and 610, estimating
threefifths 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
threefifths 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
3Str 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 iniiiii
Resulr:
3. SeVcursoroK 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 lefthand 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 lefthand 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.xf LoaYl
r ~
.of.N9
Fig. 146. Illustrating Principle of Slide Rule.
Let us consider the case 942, 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. 68 r 81
4. 392  67. 5. 194 ~ 63. 6. 153 ^ 15.
7. 965 109. 8. 228 ~ 0252. 9. 384 i 256.
10. 308 ^ 725. 11. 223 4 425. 12. 275 4 ii
13. 083  303. 14. 4475 r 384 15. 397 + 8 4'5
16. 528 20. 17. 417  182. 18. 682 r 745.
19. 13*55 r '0209. 20. 1012 f 917.
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 22) 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; 35 x 22 is about 3x2,
i. e., 6. Hence our answer must be 77.
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 35 X 4, if the
slide be moved to the right, its end being at 35 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 lefthand side, its righthand end being opposite 35 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 25; (b) 785 X 4; (c) 134 X 485; (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 lefthand 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 righthand 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 314
(c) 1. Set cursor at 134 on D scale. 2. Move slide to right and
set lefthand end under the cursor. 3. Move cursor to right and place
over 485 on slide. 4. Read off 65 on the D scale. Approximating,
134 X 485 will be rather more than 485. Hence our result is 65.
(d) 1. Set cursor at 14 on the D scale. 2. Move slide to left and
place righthand 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 onetenth of 80, or 8. Hence our result is 112.
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) 271 X 32; (b) 687 X 167 X 366; (c) 533 x 946;
(d) 349 x 203 X 835. 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 lefthand 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,
271 x 3*2 is about 9; hence result is 8670.
(6) 1. Place cursor between graduations 685 and 690, estimating
twofifths of the small division beyond the 5, thus giving 687. 2. Move
slide to left, and place righthand 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, 687 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
threefifths 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!jT
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
righthand 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, 533 X 946
is practically 533 X I, so that our result must be 504.
(d) 1. Place cursor halfway between graduations 348 and 350,
thus giving 349. 2. Move slide to the right, placing the lefthand 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 203 X 835 is about J x 20 x 8
= about 50. Hence result is 592.
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 lefthand 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. 338 x 21. 5. 202 x 495 6. 643 x 49.
7. 199 X 98. 8. 61 x 141. 9. 0312 x 173.
10. 425 X 266. 11. 205 X 278. 12. 349 x 47.
13. 283 x 38. 14. 1025 x 70. 15. 4375 X *7 2 
16. 503 X 3*7. 17. 774 X 119 18. 408 x 229.
19. 866 X 37'3 20. 1205 x 8775. 21. 285 x 312 X 42.
22. 32 X 4 8 X 86. 23. 792 x 53 X 15.
24. 115 X 12 X 5175 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 314
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 1325. 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 lefthand 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 lefthand 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. [Righthand 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. [Righthand 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 righthand 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 = 33
/. Result = 47.
Example 373. Find the value of >  = .
r D/0 136 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. [Righthand 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 = 2268.
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. [Righthand end should now read 386.] On moving cursor
to pick up 236 on slide, this reading will be found beyond the lefthand
end of the stock. Then proceed thus : 3. Shift cursor to righthand
end of slide. 4. Move slide from left to right, placing lefthand 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 lefthand end of slide, and read off 1536 on D scale.
Approximating, we have 5rr
.'. Approx. answer
sav IA
7' y> * = 14 x 10 = 14
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 254 x 3'35 X 146
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. [Righthand
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 146. 5. Move cursor to righthand end of slide. 6. Shift slide to
the right, placing 146 on the C scale under the cursor. 7. Move cursor
to lefthand 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. [Righthand 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 lefthand 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. ioi x 98 f 102. 2. 1175 x '7 1 "i 81
3 35:5 XJ575 4 4'35 X 25
324 " i'7 X 123
5 195 X 3 H 6 975 X 106
" 13 X 12 * 2*82 X '805
7 2Q4 X 69 X 234 g '0735 X 47 X 109
256 X '146 X 76 * 9*55 X 136 X 296
9 5L'75_X ^52 2< 5 10 325 X 5'2 X 66
153 X 7'35 * 8> 4 X 134
11 7*05 X 77 12 1145 X 0183
"' 232 "' 216
13 ___ 303 _ 14 >Q 95 _
1775 X 116 ' 243 X 9'6
15.  ir  7r 16.
785 X 1465 1655
17 _ 515 __ 18 _ 9375 X i 7
00034 x 76 x 4375 " ' 6 x 753 x 127
19 238 X 94 X 114 2Q 1105 Xj64
213 X 171 * 104 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 lefthand 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 lefthand 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 lefthand 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) 537; (e) 33.
4O6
ARFTHMETIC FOR ENGINEERS
THE SLIDE RULE 407
(a) Mark off in pairs from the decimal point thus : '22 '50. The
lefthand 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 474.
(b) Mark off in pairs, giving 2 '59. Lefthand 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 i6i.
(c) Point off, giving 4 / 4i / oo, and as lefthand 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 '537. Lefthand 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 732.
(e) Point off, giving '33. Lefthand 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 '33, hence there will be one figure in the whole number
of the result. Required square root is then 1815.
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 Lefthand 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. Lefthand 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^. Lefthand 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 lefthand 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. 155. 4. 213.
5. 985. 6. 737. 7. 0505. 8. 0034.
9. 117. 10. 307.
Extract the following square roots by Slide Rule. In each case
check your result by multiplication
11. ^250. 12. ^73' 13. V3Q. 14.
15. Vio6. 16. ^5^25. 17. ^187. 18.
19. WI75. 20. V307. 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) 237*; (b) 3I6 2 ; (c) 663 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 562.
(6) Place cursor at 316 on the bottom scale and read i (i. e.> the
central figure) on the top scale. Now 316* 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 3I6 2 , we multiply 31*6 x 316 on the C and D scales,
reading off 999. The actual result to four significant figures being
9986, 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 subdivisions 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 subdivisions.
To find a logarithm (decimal part), e. g., log 26.
1. Move the slide to the right, placing its lefthand 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 Ushaped 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 = 0415.
To find an antilog. (significant figures) e.g., antilog 1658.
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 lefthand end of the slide the figures 455, which
are the significant figures in antilog 1658. The decimal point
must be placed in the usual way. Then antilog 1658 = 455.
CHAPTER XI
TRIGONOMETRY
The Trigonometric Ratios. In the Mensuration chapter the
rightangled 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 rightangled 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' = 0847
'
T 4. 1 ATYIT 2'08"
In triangle ADE, =7> = 0<8 49
T . . . ATV , FG 288" Q
In triangle AFG, ^ = ^77 = 0847
allowing for small errors in drawing and measurement, the value
of this ratio is evidently constant.
If another set of rightangled triangles be taken, as at Y, having
a different common acute angle, a similar result will be found;
the ratio . ^ 5 ^ is 0839 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, jg = ~n = 0529 and this ratio also will be found
to be constant for all rightangled 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 rightangled
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 *~~ = rp 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 rightangled 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 rightangled 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 rightangled 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 rightangled 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 =^=1782
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
00872 and 00875 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 09962.
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 00872, while the value of the sine
( = c v ) is very nearly i, being 00062. 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 rightangled 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 rightangled 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 lefthand column and read downwards, while the angles
from 45 to 90 appear in the extreme righthand 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 lefthand 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 = 0866 and tan 30 =
05774
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 righthand 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
lefthand 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 = 01736
Tan 55. Refer to righthand column for angle and the heading
" Tangent " at the foot. Opposite 55 and above " Tangent "
is found the value 14281
Cos 37. Use lefthand column of angles and top headings. Opposite
37 and under " Cosine " is found the value . . . 7896
Tan 5. Use lefthand column and top heading. Opposite 5 and
under top heading of " Tangent " find .... 0875
Cos 63. Use righthand 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 lefthand column ; but if the name of
the ratio is at the foot of the column, then the angle must be read
from the extreme righthand 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 lefthand 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 righthand 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 lefthand column of angles is then referred to and gives the
value ........... 44
Example 383. Find the angles whose tangents are (a) 18807,
(b) 1051, and also (c) the angle whose cosine is 5592.
(a) The tangent 18807 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 lefthand 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 righthand 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) 26051; (7) 10724; (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 righthand 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 = 105,
V = 500 and 9 = 32, find the power developed.
Power = AV cos 9
Substituting given values
= 105 X 500 x cos 32
= 105 x 500 x 8480
From Table,
cos 32 = 8480
= 4452 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 (o9).
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 = 505 and q = 105, 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 = 1908,
n = 1667, an d Y = 225.
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 directacting engine is given by
the expression vYsinGj ) where V velocity of crank pin,
6 crank angle, and n = ratio of connecting rod length to crank
length. If V = 79, = 38, and n 45, 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 = 125, 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 = 405, 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 = 25, d = 95, and / = 225.
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 rightangled 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 rightangled 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 anglemeasuring instrument) is set up
AD o
Fig. 154. Diagrams for Examples 388392.
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 rightangled 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 = 18008', 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 12 = 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 = 2083" EG = 2051"
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 15 = 513"
Also Q^ = cos 20
= '9397 /. OD = 9397 X 15 = 1409".
Now OB = OA + AB
= i'5 f 175 = 3*25"
and BD = OB  OD
325 1409 = 1841".
Now in the rightangled triangle BCD
BC 2 == BD 2 f CD 2 (refer p. 266)
BC 2 = i8 4 i 2 f 5132
= 3'3_88_+ 263  3651
BC = "X/3'65 1 = 1911
Extension of spring = BC AB
= 1911 175 = '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 71,"
indicating that, when gliding, a horizontal distance of 71 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 halfway
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 rightangled 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 interrelated
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 " coratio " 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 rightangled 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 TT5
cos A AB AB
14 . . . BC AB
inverting divisor and multiplying = ^ x rr%
BC
and cancelling AB = vp
which relation is already known as the tangent of A.
sin A
cos A
= tan A.
Again referring to the same rightangled 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 rightangled 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
10
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. ;
coaldust 20^, say 20^ Ibs. ; (b) Old sand 790^9, say 791 Ibs. ; new
sand 263^, say 263 Ibs. ; coaldust 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) 295 kw. ; (b) 231 kw. ; (c) 305 kw.
5. (a) 270,000,000 ; (6) 500,000 ; (c) 300
6. (a) 426; (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; 322; 601; 3750
7. 'O8i ; 28; 15; 065; 54 8. 22; 610; 016; 30,000,000; n
9. (a) 2773 cu. ins. ; (b) 277 cu. ins.
10. (a) 45359 grammes; (6) 454 grammes
11. (a) 14286; (b) 1429; (c) 14 12. (a) 27183; (b) 2718; (c) 27
Exercises 11
1. 9928 2. 26381 3. 22004375
4. 3496 ft. 5. 36 Ibs. 6. 6705 chains
7. 1123 8. Dowson 4442 ; Mond 36 ; natural gas 9575
9. 89 10. 2995 11 () Brass 18; (b) steel ioi
12. 6888 grammes 13. (a) 11637 grammes; (b) 519 grammes
14. (a) 1888; (b) 2878 15. 0035"; 0105" 16. 2066 17. 10995
jColumn R = 7'4o6j Error = . OQ2 g = . OQ2
\ F = 7404]
20. 3*83 21. 623 tons 22. 3
Exercises 12
1. 4071, i. e., 407 2. 4042, i. e., 404 3. 7923, t. ., 792
4. 12*52075, i. e. t 125 5. 61544, . 0., 615 6. 0004698, t. e. t 00047
7. 4824, t. e. t 482 8. 188819, *'.<?., 189 9. 09752, i. e. t 0975
10. 214445, i. e. t 21400 11. 3105 m.p.h., say 31 m.p.h.
12. Boat's speed = 121 m.p.h. /. boat is slightly faster
13. 2595 14. 239 Ibs. 15. 975 16. 274 Ibs.
17. 2713, say 2710 cu. ins. 18. 0783 in.
19. 515 Ibs. per sq. in. 20. 4777, say 478 Ibs. per sq. in.
21. 156 sq. ins. 22. 2816, say 282 ohms
Exercises 13
1. 439 2. 4676 3. 1819 4. 3096 5. 08885
6. 2431 7. 475 8. 00847.^, 9. 2003. 10. 282
11. (a) 1 1 2 atmospheres; (b) 612 atmospheres 12. 651; 6405; 62
13. 454 14. 2205 Ibs. 15. 508 16. 134
17. 485 sq. ft. 18. 22 Ibs. per min. 19. 422 Ibs. permin.
20. 311 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 16246" dia. ; " Not go in " end = 16262* dia.
Exercises 15
1. Yes 2. Yes 3. No; should be 315
4. No; should be 155 5. No; should be 0494 6. Yes
7. Yes 8. Yes 9. 478 10. 64
11. 880 12. 714 13. 214 14. 272
15. 44000 16. 34700 17. 0245 18. 705
Exercises 16
1. 34 ft. 2. 253 3. 255 4. 199 amps.
5. (a) 663 ohms. ; (b) 298 volts 6. 203 7. 655 h.p.
8. 619", say I" 9. (a) 11520 watts; (b) 1047, 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. 916 14. 3236 tons
15. 963 16. 4975 17. 337
18. 859 tons 19. 889 tons 20. 213 tons
21. 544. 22. 88 Ibs. per sq. in. 23. 1505, say 15 Ibs.
24. 536 Ibs. per sq. in. 25. 581 ft. tons 26. 1600 Ibs.
Exercises 17
1. 1505 2. 579 3. 0278 ohms
4. 03599"', say 036* 5. 396 6. 279 Ibs. per. min.
7. 1002, say 100 Ibs. per sq. in. 8. 11045, sa Y II0 volts
9. 2211, say 221 amps. 10. 5925, 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. 408 Ibs.
7. (a) '477 cu. ft. ; (b) 644 cu. ft. 8. Max. 2958 Ibs. ; min. 2842 Ibs.
9. 3'3 tons 10. 25%. 11. 141% 12. 112%
13. (0)3i7'5 gaUs. P er min.; (6) 583% above
14. Copper 762%; tin 238% 15. Copper 941%; tin 588%
16. Copper 65% ; tin 5% ; lead 30%
17. Copper sulphate 833% ; sulphuric acid 833% ; water 8333%
18. Water 13*3%; carbon dioxide 178%; nitrogen 63%; oxygen
9%
19. Carbon dioxide 104%; oxygen 1225%; nitrogen 774%
20. 99205% 21. 98136% 22. 93'97 2 % 23. 99'377%
24. 204 and 196 r.p.m. 25. 1827 an< ^ J 77*3 r.p.m.
26. 875 and 825 r.p.m. 27. 206%
28. Water 97*6% ; hypo 122% ; sugar of lead 122%
43 8 ANSWERS TO EXERCISES
Exercises 19
1. (a) 489 to i ; (b) 203 to i 2. (a) 5 to i ; (b) 59 to i
3. 639:1 4. 862:1 5. (a) 407:1
6. (a) 3*2:1; (6) 323:1 7. (a) 291:1
8. (a) 829 : i ; (b) 597 : i 9 (a) 529 : i
(fc) 1127: i
(b) 301:1
566 : i
10. 1648 11. (a) 441:1; (b) 268:1 12. 632:238:
13. 899:4:1735:1 14 (a) 148:1', (b) 133:1 15. 412
16. 12:1 17. 21 : i 18. (a) 4; (b) 246
19. (a) 158; (b) 48; (c) 542; (d) 563; (e) 605; (/) 709
Exercises 20
1. 42 Ibs. 2. 4i galls.
3. 502 sq. ins. 4. 163 Ibs.; increase = 18 Ibs.
5. H.P. 1594", say 16* dia.; I.P. 266", say 26 1* dia.
6. H.P. 7543 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. 375 cwts.
Exercises 20a. Refer p. 453.
Exercises 21
1. 868 volts
2. 1328 grammes
3. (a) 104 ; (6)
1596 4. 1540 ft. per min.
5. 153,400 Ibs.
6. 710 7. 90 inchlbs. 8. 288 sq. ins.
9. 247
10. 211 tonsft. 11. 48 12. 255
13. 131
14. 75" 15. 632 16. 1139
17. 28700
18. 297 H.P. 19. 315 B. H.P. 20.647
Exercises 22
1. i*
2. 622* 3. 11855 4. 883
5. H433
6. (a) 989 ft. per min.; (b) 216 ft. per min.
7. 652'
8. D = 1492", H = 4264", w = 1447*, h = 1421'
9. 1350 f t. P er
min. 10. 437 11. 59 12. 28
13. 185 F.
14. 384 amps. 15. 00833 16. 456
17. 467
18. 868 19. 341 tons
20. 802'
21. 3787 22. 504 H.P.
23. 2 33
24. 298 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) 225
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) 45; (ii)
75; (iii) 225
; (iv)
225
14.
W i; (ii) '5
; (i")
4* (iv}
* V /
i
15.
(i) 24; (ii) 54; (iii) 108;
(iv) 72
16.
(i) 432; (ii)
3<>;
(iii) 432; (iv)
432
17.
(i) 4000; (ii)
1000
; (iii) 30,000 ;
(iv)
0003
18.
491 sq. ins.
19.
112,640 Ibs.
20.
7213
21.
30,070
22.
102 tons
23.
339 H
.P.
24.
3'327
25.
2316
ft.
26.
438 H.P.
27.
871 H
.P.
28.
338
29.
7
68
Ibs.
30.
456 Ibs.
31.
195
32.
887 watts
33.
4
i ft.
439
X n
Ibs
Exercises 24
1. 34 2. 155 3. 21 4. 29 5. 918 6. 105
7. 403 8. 109 9. 78 10. 27 11. 095 12. i oi
13. 0031 14. 159 15. 499 16. 320 17. 2600 18. 476
19. 162 20. 50
6 21. 607 22. 276 23. 242 24. 301
25. 356
26. 758 nautical miles
27. 746 ft. per sec.
28. 674 ins.
29. 556 H.P.
30. 124
31. 835, i. e., &"
32. 1645", * e > *\
33. 159
34. 1998 sees.
35. 707 amps.
36. 917
37. 1 1 9
38. 212 ft.
89. 125
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. 802 Vh 7. 443^
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
lT 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. 247Dd 2 25. 26. ^ 27. ^B 21
4 ** 5
Exercises 31
1. 8 2. 3 3. 15 4. io 5. 17
6. 19 7. '2 8. 1*6 9. ii 10. 155
11. 8730 12. 1380 13. 10160 14. 6700 15. 1045
16. 1695 17. 954 18. 95 19 2115 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.  55*1. 40.  8670;
Exercises 32
1. 6 2. ii 3. 4 4. ii 5. i 6. 18
7. 14 8. i 9. o 10. 22 11. 36
12. 23 13. 27 14. 29 15. 9 16. 672
17.  3'45 18  6 '9i 19.  1117 20.  17 21. 4101
22.  33 23.  2002 24. 845 25. 735 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. 21 6. 5*39
7. 3126 8. 975 9. 24 10. 24 11. 24 12. 16
13. 10 14. 675 15. 9 16. 4 17. 4 18. 4
19. 4 20. 25 21. 1875
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. 636
11. 9*44
C 12. i778C
13.  i88 9 C.
14. o
15. 5 F.
16. 32 F.
17.  355 F.
18.  i
19. 5
20. 3
21. 167
22.  737
23. 175
24. 1804
25. 1895
26. 1696
27 644
28. 5'5
29. 423
30. 20 7
31.  2
32. 52
33.  471
34.  10
35. 18
36. 6025
37. 138
38. 2225
39. 1218
40.  3
41.  675
42. 1175
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. 1756 27* 28. s + 48/> + 9506
5
29. 1082 j 3* 30. 164* *5#* 31. i oi 8 + 0000306*
32. 7502 000504* 33. 145006 + 62o6oH + 7757O
34. 1076 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.
Rr)
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. 225^ 6^ + 4
439 2
18. 9>w 2 + 7'2w#> + i44/> 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. 82875
34 
Exercises 40
1. 33 2. 1117 3. 2133 4. 94 6 5. 1052 6. 4250
7. 005378 8. 2614 9. 2761 10. 225 11. 1624
12. 199, say 2 ft. 13. 1478", i. e. t i%" 14. 2972 Ibs. per sq. in.
15. 1406 16. 183 Ibs. per sq. in. 17. 235 18. 4275
Exercises 41
1. 30 2. 498 8. 132 4. 25 5. 1098 6. 1035
7. 22*05 8. 1962 9. 05775 10. 14 volts 11. 172
12. 648 13. 247 14. 1375 15. 1437 16. 144
ANSWERS TO EXERCISES 443
Exercises 42
1. 682 amps. 2. 199 3. 969 4. 398*, say 4* 5. 106,700
6. 4 7. 00671 8. 1250 ft. 9. 1524 10. 4762
11 T 3'33 12. 336 cu. ft. per sec. 13. 1627 14. 340
15. 644 16. 0498 17. 628 18. 628 19. 1242 20. 9075
Exercises 43
1.
7
2.
813
3.
36985
4.
961
5.
55
6.
1235
7.
67
8.
1334
9.
00015
10.
3I93
11.
2686
12.
12
13.
61
14.
685
15.
365
16.
3i
17.
005
18,
118
19.
36
20.
12571
21.
54
22.
516
23.
939
Exercises 44
1. 30 C.
2. 45" 3. 885 4. 1455 & i'5
6. 25
7. 208 8. 214 9. 0346 sq. ins.
10. 61260
11. 833 12. i34'. say if* 13.  30210
14. 80 F.
15.  3625 F. 16. 5925 F. 17. 179
18.  389 C.
19. 31 pence 20. 125 21. 405 22. 415
23. 776
24. 114 25. 5 26. 75.
Exercises 45
1. 2 2. 591 8. 278 4. 1105 5. 622 6. 914
7. 2 8. 75 9. 4 10. 5000 11. 182
12. 329 13. 114 14. 1015 15. 209 16. 284
17. 85 18. 14 19.  3 20. 3 21. 4
22. 3393 23. 0343 24. 417, i. e., 42 25. 396 26. 1333
Exercises 46
1. 1604 2. 791 3. 313 4. 161 5. 95* F  6. 1933
7. 69 8. 378 9. 427 10. 136 11. 00438 12. 1296
13. 243 14. 000158 15. 137 16. 1318 17. 2167 18. 52,840
Exercises 47
1. 75 2. 6 3. ii 4. 75 5. 2 6. 4
7. 2 8. 3 9. 286 10. 4 11. 0909
12. 6 18. 504 14 357 15. 2525 16. 15
Exercises 48
1. 3675 2. 112 3. 136 4. z475 5. 30,520
6. ii5 7. 664 8. 4*86 9. 9*375 10. i37 say i*
11. 223 galls, per min. 12. 3'39 13. 2348 14. 3
16. 97 16. 557 17 4'*3' 18 3*4*
444 ANSWERS TO EXERCISES
Exercises 49
1. 625 2. 458 3. 956 * 325 5  39'3
6. 000941 7. 556 8. 844 9. 1675 10. 162
11. 499 12. 5357 13. 6090 14. 15 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 '75D
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 2 126
?
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 147 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 ,
Lf 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
25V 6 c2 , * 7 H _w 8 9 W
*"+'* '!' .+! .2.. H ^
ia RRO 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 ' ma *' 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. 33502, 30523, 04972, 00249
18. 18949, 17190, 26567, 02355 19. 14624, 57634, 79069, 05228
20. 28910, 26898, 18495, 3'477 2 > 3*2375
21. 37226, 04343, 02385, 3699, 14048
22. 45328, 01962, 41440, 04343, 10026
23. 03623, 17582, 20249, 47756, 37839
24. 00004, l^S 2 , 25912, 11031, 15076
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. 1758, 4560, 31620, 2000 8. 1012, 1001, 1687, 5093
9. '5616, 02505, 5379, 1408 10. 9*974, 0000472, 639, 8910
11. 1862, 3111, 6929, 01 12. 0009118, 6246, 008015,7960
13. 001022, 156, 20, 2027 14. 21310, 2289, 2, 3142
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, 1262, 7501
Exercises 59
1. 7812 2. 3002 3. 2496 4. 4999 6. 3004
Exercises 60
1. 6896 2. 01227 3. 14890 4. 005715 5. 1397 6. 7945
Exercises 61
1. log top line = 23200 2. log top line = 16444
log bottom line = 00239 log bottom line = 46885
ResuR = 1977 Result = 90350
3. log top line =65717 4. log top line = 19420
log bottom line = 4*5185 log bottom line = 49001
Result = 1131 Result = no, too
5. 01171 6. 00001535 7. 2326
8. 2800 Ibs. per sq. in. 9. 09228 sq. in. 10. 1617 Ibs.
ANbWJbLKb 1U fcXttKCIblib 447
Exercises 62
1. 6822
2. 35'27 3. 7709 4. 5212
6. 277' 1
6. 8093 7. 6122 tons
Exercises 63
1. 9872
2. 6169 3. 9761 4. 00005861
5. 775 6
6. 343'9 7. 2122
Exercises 64
1. 7'5 l6 2.
1476 3. 4433 4. 04062 5. 4607 6. 1043
7. 2089
8. 2823 9. 1049 10. 9895 11. 2377
12. 2235
13. 144 14. 02057 15. 2158 16. 1347 ms 
17. 5343 sq.ft.
18. 2309 19. 3773 20. 3534 21. 9668
22. 737 **.
23. 2147 24. 4024 25. 5459 amps.
26. 8507
27. 1748 in. 28. 559 29. 8*299, say 83, H. P.
30 (a) 6485, i.
e., 6 threads ; (b) 8698, i. e., 9 threads.
Exercises 65
1. 3142 2. 1068 3. 1646*, say ij* 4. 4889
5. 921 ft. 6. 8549 7. 3582 ft, 8. 001076
9. 2i,2oocu. ft.perhr. 10. 128 11. 35141 12. 271
13. 131 14. 1 158 m. p. h. 15. 8692 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. 665 ins,
8. 11896 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. 764"
12. 24 ins.
13. 1317* dia.; 524* pitch 14. 5650 ft. per min.
15. 1528 r. p.m.
16. 864 ft. per min. ; 1375 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 1178*
; more exact 1296* 2. 39
27'; 3968*
3. 3454"; 346*
4. 142* 5. 833" 6
. 9" 7. 24
8. 5792 ft.
9. 97 10. 116 11
. 35 12. 178*
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. 2525
9. 30314 10. 65375 11. 110972 12. 4923
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. 596 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. 596 sq.ft. 2. 11175 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. ; 341% 9. 295 tons per sq. in.
10. 412 sq. ins.; 368% 11. 2485 sq. ins. 12. 451 sq. ins.
13. 131 amps. 14. 8400 Ibs. 15. 33 sq. ins.
16. !! sq. ins.; '001173 ohms. 17. 60 sq. ins. 18. 99 sq. ins.
19. 295 sq. ins. 20. 1326 sq. ft.; 16,570 Ibs. 21. 838 sq. ins.
22.  23. 1080 sq. ins. 24. 15 sq. ins.
25. 1934 sq. chains 26. 325 sq. ft. 27. 192
28. 1225 sq. ins. (both sides taken : edges neglected) ; ii amps.
29. 8435 sq. ins. 30. 9835 sq. ins. 31. 2183 sq. ins.
32. Embankment 389 sq. ft. ; cutting 554 sq. ft.
Exercises 73r
1. 254 sq. ft. 2. 00933 sq. in. 3. 1855 sq. ins.
4. Centre 0579 sq. in. ; Outer 0427 sq. in. ; Total 1006 sq. in.
5. 12 sq. ins. 6. 2486 Ibs. 7. 1103 s q ft 8. 158 sq. ft.
9. 586 sq. in. ; 586 tons 10. 697 sq. in. ; 28 tons
11. i372lbs. 12. 237:1 13. 312 14. 32 15. 1 786 Ibs. per sq. in.
16. 5025 sq. ins.; Yes 17. 844 sq. ins. 18. 311 sq. ins.
19. 3:1 20. 5 21. 8f 22. 048" 23. 605"
24. 1492 sq. ins.; 13!" dia. 25. io* 26. 182*, say ^ 6 "
27. 283", say 2}' 28. 3F 29. 758", say 7 f*
30. H.P. = 2'9*; I.P.  4 '7i*
1. 23 sq. ins.
3. 123", say ij'dia.
6. 272 sq. ins.
9. L.P. = 176*7 sq. ins.
2. 441 sq. ins.;
4. 4*1 sq. ft.
7. I9'i sq. ins.
; H.P. = 637 sq. ins.;
1545 tons
5. 1275 sc
8. 2}'
L.P. : H.P.
ANSWERS TO EXERCISES 449
Exercises 74
ins.
= 277:1
10. 154 lbs. per sq. in.
Exercises 75
1. 2025 sq. ins. 2. 233 sq. ft. 3. 417 sq. in.
4. 1035 sq. ins. 5. 246 sq. ft. 6. 2456sq. ft.; 163% low
7. 2451 sq. ft.; 366% low 8. 148 sq. ft.
9. 142 sq. ft.; 405 % 10. 463 sq. ft. 11. 552 ft.; '838
12. (a) n27sq.ft.; (b) 861 ft. ; (c) 131 13. 1207 sq. ft.
14. 1371 sq. ft. 15. 5093 sq. ft. 16. 609 sq. ins.; 2155 times
17. 44 sq. ins.; 156 times 18. 121 7 sq. ins. 19. 868 sq. ins.
20. 127 sq. ins. 21. 102 sq. ins. 22. 706 sq. ft. 23. 2170 lbs.
Exercises 76
! 95*3 s q ms 2. 468 sq. ft. 3. 1124 sq. ft.
4. 2455 sc l* ms  5. 2886 sq. ft. 6. 261 inch tons
7. 1910 sq. chns. 8. 738 sq. ins. 9. 96"; 48 lbs. per sq. in.
10. 552"; 552 lbs. per sq. in. 11. 385"; 935 lbs. per sq. in.
12. 329*; 987 lbs. per sq. in.
Exercises 77
1. 1736 cu. ins. 2. 801 cu. ft. 3. 33,700 gals, per hr.
4. 1448 galls, per min. 5. 2835 cu  ft P 61 m in.
6. 734 lbs., or 6 cwts. 2 qrs. 6 lbs.
Exercises 78
1. 371 cu. ft. 2. 2047 cwts., say 10 tons 5 cwts.
3. 25,450 galls. 4. 390 galls. 5. 315 cwts.
6. 1625 lbs. 7. 26 galls. 8. 15 ft.
9. 90 lbs per yd. 10. 3'n" 11. 4 '9*
12. 2063 cu. ins. 13. 12 tons 14 cwts. 14. 25 sq. ft.; 13!"
Exercises 79
1. 2,032,000 cu. ft. 2. 49 cu. ft. 3. 368 cu. ft.
4. 22 galls. 5. 459 lbs. 6. 4440 galls, per min.
7. * 8. 685", say 7* dia. 9. 2295', sav 2 '4*
10. 3200 cu. ft. per sec. 11. 8250 cu. ft. 12. 482*5 lbs.
13. 5490 galls. 14. 675% of working volume; 632% of total volume
15. 825 cu. ft. 16. 927 lbs. per link; 3335 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) i64lbs.; (6) 2i81bs. 5. 158 Ibs. 6. 297 Ibs.
7. 2*29 galls. 8. 1286 galls. 9. 316 Ibs. 10. 189 galls.
11. '104 Ibs. 12. 62 cu. ft.; 7440 Ibs. 13. i95lbs.; 8s. 6d.
14. 23olbs. 15. 2285 galls. 16. 475 cu. ft.; 57 Ibs.
17. 404,000 Ibs. 18. 886 Ibs. per yd.
Exercises 81
1. 284 Ibs. 2. 898 cu. ins.; 138, i.e., 13 8. 2045 cu  ft
4. 867 galls. 5. 58,200 galls. 6. 805 cu. ft.
7. 378 Ibs. 8. 268 Ibs. 9. 128 Ibs. 10. 8 Ibs.
Exercises 82
1. 104*2 cu. ins. ; J 2. 854" 3. 996 cu. ins.; cone is one third
of cylinder volume 4. 615 cu. ins.; 782*, say 8" 5. 4*07"
6. i ton 14 cwts. 7. 1918 cu. ft. ; 4795, 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.; 695 Ibs. per sq. ft.
4. 317 Ibs. per sq. in. 6. 402 units per sq. ft. 6. 754 sq. ins.
7. ij" 8.  9. 926 Ibs. 10. 15 Ib.
4
11. 897 watt per sq. in. 12. 440 sq. ins. 13. 237", say 2\"
14. 134*6 sq. ft. 15. 408 Ibs. 16. 439 ft., say 4 ' 4 J"
17. 43'3 sq. ft. 18. 303 sq. ft. 19. 393 sq. ft.
20. 9530 Ibs. 21. 568 watt per sq. in. ; 241 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) 403 Ibs. per sq. ft.;
(6) 485 Ibs. per sq. ft.
2. Base i* = 5 mins.; vertical i" = io/
3. Base i* = 5 amps.; vert. J" = *2 volt; (a) 122 volts; (b) 178
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) 63 to i
9 ( a ) '57 so fin'; (&) '195 sq. in. ; (c) 36 sq. in.
10. (a) 14, i.e., No. i S.W.G.; (b) 74, i.e., No. 7 S.W.G. ; (c) 186,
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) 1051
C.; (b) 26%
4. Base from i, i* = 05 ; vert. from i, \* =* 2 ; (a) 1175 S P B T * I
(b) 128 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) 566 Ibs.
9. (a) 186, i.e., No. 18 S.W.G.; (b) 149, i.e., No. 14 S.W.G.
10. 1095
Exercises 86
1. AB = 8*5, BC = 5, CA = 36. 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 = 64 3. 79 4. 1475 6. 308 6. 5350
7. (a) 7 and 26; (b) 12; (c) 184 8. yj; 25 ; 4
Exercises 87
1. e = 2W 2. (a) P = 31 f i'94 w ; (&) 12 165.; (c) 12 i6s.
3. V = 1200 + 630*2 4. /= 185 '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 = 464 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 = 131  '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. 22 3. 84 4. 585
5. 0308 6. 0102 7. 885 8. 905
9. 015 10. 425 11. 525 12. 25
13. 0274 14. 11*65 15. 4'7 16. 264
17. 229 18. 916 19. 648 20. 1104
Exercises 89
1. 77 2. 216 3. i8'2 4. 71
5. 1000 6. 316 7. 195 8. *86
9. 0054 10. ix3 11. 57 12. 1640
13. 1075 14. 7175 15. 315 16. 186
17. 921 18. 9'34 19. 323 20. 1058
21. 37'35 22. 132,100 23. 63 24. 714 25. 10070
452
ANSWERS TO EXERCISES
1. 9'7
5. 425
9. 58
13. 1471
17. 101,700
Exercises 90
2. 103
6. 456
10. 00991
14. 0407
18. 275
3.
7.
63
116
11. 234
15. 468
19. 70
Exercises 91
See Answers to Exercises 18.
4. 52
8. 98
12. 97
16. 1076
20. 7"4i
Exercises 92
1. 2nd : 51
5. 2nd : 9925
9. 2nd : 3*42
13. 624
17. 432
1. 6,760,000
5. 9700
8. 00001156
2. ist : 211
3. ist : 1245
4. ist : 146
6. ist : 2715
7. ist : 225
8. 2nd : 0583
10. 2nd : 554
11. 1581
12. 27
14. 95*9
15. 103
16. 2292
18. 738
19. 1475
20. 554
21. 17
22. 9125
Exercises 93
2. 1,981,000,000
6. 543,000
9. 1369
3. 24
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) 1282; (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. 1505
6. 5'7i5.
2. 29042.
7. 572900.
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. 4952.
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. 1099". 9. (a) ADB = 2R sin ~ ; (b) DE = R (i cos
10. a = 37 approx. ; angle turned by pointer = 148 approx.
7. 1171".
90 \
N/
ANSWERS TO EXERCISES 453
Exercises 99
1. (A) 156"; (B)3'92"; (C)oo 9 "
2. (D) 2731" ; (E) 0987" ; (F) 1069" ; (G) 028 ; (H) 4398" ;
(J) ooi"; (K) 2051"; (L) 3*549"; (M) 0625"
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) *&tife2i%ins.; (E) if + T * ff = i& ins. ; (F) ft + ij ff
= if ins.
10. (G) 299*; (H) 17416; (K) 26134 11. (L) 4 10'; (M) 18 40'
12. (N) 233 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. 282855, say 2829
3. (a) 5.51 lb. persq. ft.; (6) 1654 lb  per H.P.
4. 3762" 5. 923% 6. 64 to i
7. (a) 10 lb. 5  oz.; (b) 362 % 8. 2774 lb.
Exercises 49a
1. 553 2. 000542 3. ^P^ 2 4. 175 5. 152
Exercises 83a
1. 471 sq. ins. 2. 2727 lb. 3. 208'  10" 4. 503 lb.
5. 383", say 38^" 6. (i) 9995 sq. mm. ; (2) 1000
7. 26 teeth; 059" 8. 1509", say ij" 9. 806 lb.
10. 708 gallon 11. 884 lb.
12. No; calculated weight is practically i lb. 13. 1786 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
6480
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
Gotangent.
Cosine
CO
1
1414
15708
90
1
2
3
0175
0349
0524
0698
017
035
052
070
0175
0349
0523
0698
0175
0349
0524
0699
572900
286363
190811
143007
9998
9994
9986
9976
1402
1389
1377
1364
15533
16359
15184
15010
89
88
87
86
5
0873
087
0872
0875
114301
9962
1351
14835
85
6
7
8
9
1047
1222
1396
1571
105
122
140
157
1045
1219
1392
1564
1051
1228
1405
1584
05144
81443
71154
63138
9945
9925
9903
9877
1338
1325
1312
1299
14661
14486
14312
14137
84
83
82
81
10
1745
174
1736
1763
56713
9848
1286
13963
80
11
12
13
14
1920
2094
2269
2443
192
209
226
244
1908
2079
2250
2419
1944
2126
2309
2493
51446
47046
43315
40108
9816
9781
9744
9703
1272
1259
1245
1231
13788
13614
1 3439
13265
79
78
77
76
15
2618
261
2588
2679
37321
9659
1218
13090
75
16
17
18
19
2793
2907
3142
3316
278
296
313
330
2756
2924
3090
3256
2867
3057
3249
3443
34874
32709
3 0777
29042
9613
9563
9511
9455
1204
1190
1176
1161
12915
12741
12566
12392
74
73
72
71
20
3491
347
3420
3640
27175
9397
1147
12217
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
22460
9336
9272
9205
9135
1133
1118
1104
1089
12043
11868
11694
11619
69
68
87
66
25
4*563
433
4226
4663
21445
9063
1075
11345
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
18040
8988
8910
8829
8746
1060
1045
1030
1015
11170
10996
10821
10647
64
63
62
61
30
6236
518
5000
5774
17321
8660
1000
10472
60
31
32
33
34
5411
5585
5760
5934
534
551
568
585
5150
5299
5446
5592
6009
6249
6494
6745
16643
16003
15399
14826
8572
8480
8387
8290
985
970
954
939
10297
10123
9948
9774
59
58
57
56
35
6109
601
5736
7002
14281
8192
923
9599
55
36
37
H
6283
6458
6632
6807
618
635
651
668
6878
6018
6157
6293
7265
7536
7813
8098
13764
13270
12799
12349
8090
7986
7880
7771
908
892
877
861
9425
9250
9076
8901
54
53
52
51
40
6981
684
6428
8391
11918
7660
845
8727
50
41
42
43
44
7156
7330
7505
7G79
700
717
733
749
6561
6G91
6820
6947
8693
9004
9325
9657
11504
11106
10724
10355
7547
7431
7314
7193
829
813
797
781
8552
8378
8203
8029
49
48
47
46
45
7854
765
7071
10000
10000
7071
765
7854
45
Cosine
Ootangent
Tangent
Sine
Chord
Radians
Degrees
Angl
e
APPENDIX
VERNIERS AND MICROMETERS
CONTENTS
PAGE
VERNIERS 460
Introductory Vernier Calipers The OneThousandth 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 TenThousandth
Micrometer Zero Setting of TenThousandth 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 drawingoffice 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 overall
length is that of 9 main scale divisions, as shown in the fullsize
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 23"), 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. onehundredth of an
inch. If now the moving jaw be moved to the right until line i
Clamping Screw.
Coincidence"^ A V fo /Coincidence r ^ /^Comodencev
^.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'iii^ WBBI^** * ""^^i
\ Vernier
reading 23
\Vernier
reading 231
\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 24
\ 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
onehundredths 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 *" ParlView <
Vernier scales see Enlaced Views BacKFacc,:
Fig ISQ.arA. of JawPointe.closed. ^ IN ft OUT I
Fig. 158. Vernier Caliper, reading ^ v inch. (About halfsize.)
Thus at (D), Fig. 157, the vernier zero cuts off 23 on the main
scale, and coincidence occurs at 2 on the vernier scale. Then the
full reading is 23" + 02" = 232".
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 Englishreading 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 setscrew 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 OneThousandth 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 onethousandth, 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 IOOO
Coincidence V
3V4 5 6 7 8 9
O 5 10 15 20 25
Vernier reading 1353
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 13 and two small main scale
divisions, i. e. 13 + 05 . . . . . . . = 135
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 08 and one small main scale division,
i. e. 08 + 025 ^ = 0825
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 1575
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 onethousandth 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 22 and three small main scale
divisions, i. e. 22 } 075 .......= 2275
Coincidence on vernier is at 13, i. e. 1 JJ  . . . = 013
.'. Complete reading = 2288
At (B).
Vernier zero cuts off 45 and two small divisions, i. e.
4'5 + '5 = 4'55
Vernier coincidence is at 19, i. e. y^fo . . . . = 019
.'. Complete reading = 4569
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 18 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 18 f 075 .......
there being no thousandths to add.
1875
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 sixtyfourths. 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 sixtyfourths 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 rmies 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 halfdegree) 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
halfdegree ......... 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 wiih
Hie zero.
If coincidence, were at C,
rhen reading would toe !$.
If coincidence were. a+ fc,
reading would be l*33
Coincidence ^^ r
O l Q' /. Reading is l3Zb.
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 halfway between these two readings, e. g. if
coincidence at a gave 132" and coincidence at b gave i'33" then
the actual reading could be taken as halfway between 132 and
i'33> * e I325".
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 Jinch
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 OneInch Micrometer, one inch being the largest
dimension that it will measure.
The Ushaped 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 Englishreading instruments is invariably a single
Vthread 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 truingup 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 oneinch, 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 onethousandth, 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 tenthousandths, or
the reading may be stated to the nearest thousandth. The exact
reading of tenthousandths is treated later.
If the tenthousandths 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 th,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 halfway between
two thimble graduations, the halfthousandth, '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 tenthousandths.
This method is shown when dealing with the tenthousandth
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 halfinches; 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 TenThousandth Micrometer. For fine work demand
ing measurement to a greater accuracy than onethousandth of an
inch, the micrometer caliper* is fitted with a vernier scale by
which tenthousandths 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 tenthousandths (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 LJUJU4UI Coincidence
765432 M I 098765*
Reading IG74 Reading '3837 Reac*\ng 7239
JKS432IO \
Reading A441 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 tenthousandth, 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 tenthousandth 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 tenthousandth. 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 TenThousandth Micrometer. For reading
to a tenthousandth 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 tenthousandth, and frequently, therefore, the tool is left
reading one or two tenthousandths 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 . "RatchetStop 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
tenthousandths. 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
" Collectingup " 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
Coordinates, 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, 9496
, 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, 115118
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
Midordinatc 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 coordinates, 348
Reduction of angles, 265
Removal of brackets, 141
Rhomboid, 248
, Area of, 278
Rhombus, 248
Rightangle, 246
Rightangled 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
Straightline 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
, Rightangled, 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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