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ii4 ARITHMETIC FOR ENGINEERS
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ARITHMETIC FOR ENGINEERS
PLEASE SEND FOR DETAILED PROSPECTUSES Mathematics for Engineers. By w. N. ROSE,
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Metric System for Engineers. By CHARLES
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The Directly-Useful
D.U.
Technical Series
Founded by the late WILFRID J. LINEHAM, B.Sc., M.Inst.C.E. General Editor : JOHN L. BALE.
Arithmetic for Engineers
INCLUDING
SIMPLE ALGEBRA, MENSURATION, LOGARITHMS,
GRAPHS, TRIGONOMETRY, MfD
THE SLIDE RULE
WITH AN APPENDIX ON
VERNIERS AND MICROMETERS
BY
CHARLES B. CLAPHAM
B.Sc. (IIoNS.) RNG., JT.M.l.hlECH.El
Lecturer and Demonstrator in the Mechanical Engineering
Dept., Finsbury Technical College; Author of
" Metric System for Engineers"
FOURTH EDITION
LONDON CHAPMAN & HALL, LTD.
1925
FRINt'Fn IN GKK\T Kll'HAIlP Cl.AV & t-l) LiUNt.AV, Sri-J
BRITAIN I'.Y N^, LjMllH),
EDITORIAL NOTE
THE DIRECTLY-USEFUL TECHNICAL SERIES requires a few words by way of introduction. Technical books of the past have arranged themselves largely under two sections : the Theoretical and the Practical. Theoretical books have been written more for the train- ing of college students than for the supply of information to men in practice, and have been greatly filled with problems of an academic character. Practical books have often sought the other extreme, omitting the scientific basis upon which all good practice is built, whether discernible or not. The present series is intended to occupy a midway position. The information, the problems, and the exercises are to be of a directly useful character, but must at the same time be wedded to that proper amount of scientific ex- planation which alone will satisfy the inquiring mind. We shall thus appeal to all technical people throughout the land, either students or those in actual practice.
THE EDITOR.
AUTHOR'S PREFACE
THE following work is an endeavour to treat the elementary portions of what is usually called " Practical Mathematics " in a thorough and practical manner, suitable for elementary students of technical schools and for home study. Although a great many books on the same subject already exist, the Author has been unable to find one dealing with the necessary matter in sufficient detail and with a sufficient amount of engineering application to meet the needs of his own students. For private study, too, he is of the opinion that the existing text-books treat the elementary matter in a manner too cursory to really fulfil their object. The treatment in the following pages, therefore, has been developed from his own lecture notes and class instruction on the subject, and mimerous diagrams have been introduced to assist in making the work clear.
Many will think, no doubt, that the title " Arithmetic for Engineers " is not sufficiently comprehensive for the matter con- tained, but it was found impossible to frame a short original title to adequately describe the contents. An examination of the examples and exercises should show that the whole of the matter is directly useful ; all purely academic work such as Highest Common Factor, Recurring Decimals and the like being discarded. Wherever possible the examples are truly practical, i. e., are problems actually met with in the Drawing Office, Workshop and Laboratory, while the data are of correct dimensions. Such examples should stimulate the reader's interest in the mathematical work and show the applications of the principles to practice ; at the same time a little general engineering knowledge will be gained. Even at the risk of becoming verbose, all matter is treated at length, every principle being followed by worked qxamples. Where slight varia- tions of the problem may cause difficulty to the beginner, or where special precautions have to be observed on certain points, several illustrative examples are given. A case in point is that of the extraction of a square root on pp. 102 et seq. For the private student this is particularly necessary, as he is often unable to obtain guidance in working a new problem which presents features slightly different from those in the example worked to illustrate the principle.
Stress has been laid, in Chap. II, onllimiting the number of figures to be given in a result, as class-room experience shows that many students will persist in using, and stating in results, far more figures than are necessary from the practical point of view, where
viii AUTHOR'S PREFACE
an accuracy of I per cent, is often as mfcch as is desired. The method of " approximating for a result " shown on pp. 58 et seq. is, it is believed, not generally known. In Chap. Ill positive and negative quantities, which are of importance in logarithms and higher* work, are given special attention, as they often present difficulty, to the beginner. The simple equation is treated very metho'dically in Chap. IV, with concrete illustrations of the earlier examples, and is followed immediately by a similar treatment of the literal simple equation, which to a beginner often presents great difficulty, even when he is well able to work a similar numerical example. The use, only, of logarithms is taken in Chap. VI ; it was hoped to add something of the theory, but space would not permit. The Mensura- tion in Cnaps. VII and VIII may appear rather extensive, but calls for little knowledge beyond the evaluation of a formula or the solution of a simple equation. This division of the subject was framed to be of use for reference in the Drawing Office, in which direction the tables on pp. 312, 313, 345 and 346 should prove useful. The " more exact " formula for the circumference of an ellipse on p. 260 (due to Boussinesq) is not often seen, but is very accurate. It is to be feared that many writers repeat the formula n\/2(a2 -f bz) as being more exact than fr(a + b), when really it is but little better and has only a very limited range of application. Chap. IX takes Graphs in an elementary degree, considerable atten- tion being paid to the details of setting out and finishing off, in which direction many students fail. The last chapter is devoted to the Slide Rule, with illustrations of readings and settings, and the method of instruction was tested while in manuscript form. The chapter was carefully worked through by a novice, who, finally, and without other instruction, could use the rule with ease and certainty : these pages should be helpful to the private student.
It is hoped that teachers of Practical Mathematics will be saved much research work by the numerous classified practical exercises throughout the book : answers are given to all these exercises.
To MR. W. J. LINEHAM, B.Sc., M.I.C.E., the Author tenders his sincere thanks for much kindness throughout his career, and also for generous assistance and useful criticism in the production of the book.
The Author's gratitude is also tendered to MR. JOHN L. BALE for his generous help with the book in all its stages, from the earliest research, throughout the manuscript and proof forms to the final production.
The notification of errors, clerical and otherwise, will be gratefully appreciated.
CHARLES B. CLAPHAM Goldsmiths' College,
New Cross, S.E.,
May, iQ/6.
PREFACE TO THE THIRD EDITION
IN preparing a third edition of "Arithmetic for Engineers" a chapter on Elementary Trigonometry has been added. It is hoped that this addition, by filling a gap in the original treatment, will render the book of greater value to Technical Schools, in many of which a first course of Practical Mathematics includes at least the elementary conceptions of this branch of the subject. The new chapter covers little more than the meaning of the important ratios sine, cosine, and tangent, but is believed to include sufficient to show how even this small amount can be of great utility to the practical man in drawing office and workshop. The treatment is similar to that followed throughout the rest of the book, viz., detailed explanations followed by really practical illustrative examples and exercises at every step.
The matter on Verniers and Micrometers, originally added as an Appendix to the Second Edition, has been retained as an Appendix, as it is rather in the nature of a special application.
It is a matter of deep regret to the Author that, by the death in 1919 of Mr. W. J. Lineham, founder of the D.U. Scries, he has been deprived of much valuable assistance and criticism.
The Author wishes to acknowledge with many thanks the assistance of the various readers who have kindly notified him as to errors in the first two editions, and he hopes that this useful assistance will continue in the future.
CHARLES B. CLAPHAM.
Goldsmiths' College, New Cross, S.E. October,
PREFACE TO THE FOURTH EDITION
THE opportunity has been taken to revise one or two sections and to add a short account of the finding of logarithms and anti- logarithms on the slide rule. About fifty new exercises have been added, covering a variety of details including some simple applications to wireless and aircraft.
The notification of errors is gratefully acknowledged; the Author will always be pleased to receive further information in this direction.
CHARLES B. CLAPJIAM.
Finsbury Technical College, Leonard St., E.C. 2.
September, ^925.
CONTENTS
PAGE
INTRODUCTORY i
Measurement and units — Multiples and sub-multiples — Abbrevia- tions, mathematical signs and terms — Tables of measurements.
CHAPTER I VULGAR FRACTIONS 5
Fractions generally — Forms of the vulgar fraction — Cancelling — Addition and subtraction of vulgar fractions — Multiplication and division of vulgar fractions — Compound examples — Brackets.
CHAPTER II DECIMAL FRACTIONS 35
Decimal notation — Movement of the decimal point — Conversion from decimal to vulgar fractions — Degree of accuracy and significant figures — Addition and subtraction of decimals — Multiplication and division of decimals — Conversion from vulgar to decimal fractions — Compound examples — Approximation for result — Averages — Percentages — Ratio — Proportion .
CHAPTER III SYMBOLS AND THEIR USES . . , 86
Symbols and formulae — Signs of x , -j- , -f- and — : Brackets — Simple evaluation — Powers and indices — Square root — Evaluation including square root — Other roots — Powers and roots of fractional expressions — Laws of indices : Multiplication, division and powers — Substitution of symbols — Positive and negative quantities — Addition and subtraction of -|- and — quantities — Addition and subtraction of several terms — Multiplication and division of -f- and — quantities — Powers of minus quantities — Evaluation in- cluding positive and negative quantities — Removal and insertion of brackets — Taking out a common factor — Multiplication with expressions of two or more terms — Evaluating such expressions as (a ± 6)1.
xi
xii CONTENTS
CHAPTER IV
PAGE
SIMPLE EQUATIONS 157
Equations generally — Equations requiring division, multiphca* tion, and combination of the two — Equations requiring subtraction and addition — Equations combining the four rules — Equations with several terms — Equations with brackets — Fractional equations — Equations requiring square root — Equations requiring squaring.
CHAPTER V TRANSPOSITION OF FORMULAE 190
The general solution — Equations requiring division, multiplication, and combination of the two — Equations requiring addition and subtraction — Combination of all four rules — Equations with brackets — Equations requiring square root — Equations requiring squaring.
CHAPTER VI USE OF LOGARITHMS 213
Introductory — Finding logarithms : i. The whole number. 2. The decimal part — Anti-logarithms — Finding anti-logarithms : i. Sig- nificant figures. 2. Placing the decimal point — Multiplication and division by logarithms — Compound examples — Examples involving plus and minus — Powers and roots by logarithms — Various examples.
CHAPTER VII MENSURATION — LENGTHS AND AREAS 242
Measurement of length — Conversion and reduction — Addition and subtraction in length units — Simple geometrical terms — Simple plane figures — Perimeter or circumference — Circumference of a circle — ir and its determination by measurement — Examples involving circumference of circles — Circumference of ellipse — Measurement of angles — The protractor — Important angles — Addition and sub- traction of angles — Reduction of angles — The right-angled triangle — Proportions of 45° and 6o°-3o° right-angled triangles — Length of arc of circle.
AREA
Measurement of area — Reduction — Areas of the simple figures — Square and rectangle — Rhomboid — Triangle — Trapezium — Trape- zoid — Hexagon — Octagon — Circle — Determination of diameter of circle from area — Hollow circle or ring — Sector of circle — Area of fillet — Segment of circle — Ellipse — Area of irregular figures — Table of areas and circumferences of plane figures.
CONTENTS xiii
CHAPTER VIII
PAGE
MENSURATION (cont.) — VOLUMES AND SURFACE AREAS . .314
Volume — Conversion — Volume of regular solids — Prisms — Cylinder — Hpllow cylinder and tubes — Sphere — Segment of sphere — Pyra- mids— Frusta of cones and pyramids — Calculation of weights.
SURFACE AREAS
Cylinder — Sphere — Cone — Square pyramid — Frustum of square pyramid — Frustum of cone — Table of volumes and surface areas of solids.
CHAPTER IX
CURVES OR GRAPHS ........ 347
Curves and their uses — Rectangular co-ordinates — Squared paper — Method of plotting : I. Choice of scales. 2. Setting out scales. 3. Plotting the points. 4. Drawing in the curve. 5. Further information required on the sheet — Interpolation — Cases where origin is not required — The straight line — Plotting of negative values — The equation to a straight line — Obtaining the equation or law.
CHAPTER X THE SLIDE RULE 385
Description — Division of the scales — Method of reading — Advice as to holding the rule — Operations on the slide rule — Division — Multiplication — Combined multiplication and division — Ratios and percentages — Square root — Squaring — Reading logarithms and anti-logarithms from the rule.
CHAPTER XI TRIGONOMETRY 410
The trigonometric ratios — Ratios of the angleb 30°, 45°, and 60° — Ratios of the angles o° and 90° — Relation between sine and cosine of complementary angles — The table of trigonometric ratios — Use of the trigonometric ratios in formulae — Problems involving simple trigonometry — The reciprocal trigonometric ratios — Further points concerning the table of ratios — Important relationships between the ratios — Graphs of the important ratios.
ANSWERS 434
MATHEMATICAL TABLES ....... 454
APPENDIX 459
Verniers and Micrometers.
INDEX 488
ARITHMETIC FOR ENGINEERS
INTRODUCTORY
Measurement and Units. — From the practical standpoint our calculations are never concerned with numbers alone, but always with some properties of materials or machines : simple, such as length and weight ; or complicated, such as horse-power and electrical resistance. We have always, therefore, to deal with quantities which have been measured and which, besides a number, will have a name, such as feet, pounds, etc.
In some of the examples to follow, numbers alone will be used when the calculations are only for the explanation of a certain mathematical process. A measured quantity is described as being so many times larger than some particular " standard " of refer- ence. This " standard " is called a unit, a word denoting the number I, and which for each particular kind of measurement is given a distinctive name. Thus the unit in the table of weights is the pound, and in the table of length the yard, and so on. These units are fixed by law in the case of the more common measure- ments, such as length, weight, etc., and official samples are pre- served in Government offices. With more special measurements, such as velocity, horse-power, etc., the units are fixed by general consent.
Multiples and Sub-multiples. — It is very seldom that we find any particular unit suitable for all measurements of a certain kind. Thus the yard would be very inconvenient when measuring long distances on the earth's surface, for our measurements would come to millions of yards. Such numbers, besides being difficult to imagine, would render calculation very laborious. Therefore to make our measurements of long distances consist of a smaller number we employ a larger unit which contains an exact number of standard units. Such " large " units are called multiples of u
2 ARITHMETIC FOR ENGINEERS
the standard. The mile (1760 yards) and the ton (2240^ Ibs.) are examples. On the other hand, when measuring very small lengths (such as the thickness of sheet metal) the yard would be too large a unit, and therefore small units called sub-multiples are adopted. Common examples are the inch, of which 36 make I yard, and the ounce, of which 16 make i pound. Taking the more special measure- ments used by the engineer, the number of multiples and sub- multiples is small, while in some cases, as with horse-power, only the standard unit is employed.
Abbreviations, Mathematical Signs and Terms. — In order to save time and space, and to make our calculations more easily read, it is customary to use certain signs for phrases which are constantly occurring in mathematical work. These must be committed to memory. The more frequent ones are given below; others will be introduced as required.
= stands for " equals " or " is equal to " ; + ,, „ " plus/' meaning " added to "; — ,, „ "minus," meaning "subtract or take away"; x ,, ,, " multiplied by "; -+• „ „ " divided by"; „ " therefore."
Thus the statement 3 — 2 + 5 = 6 would be read as " three minus two plus five equals six," meaning that if 2 is subtracted from 3, and 5 added to what is left, the result equals 6.
Very often a division is not stated with the sign -r, but in the form -28^, meaning 24 divided by 8. This form is similar to the sign ~, the dot representing the numbers, and is then often read as " twenty-four over eight."
The following words have mathematical meanings —
Sum is the result of an addition. Difference is the result of a subtraction. Product ,, „ „ „ multiplication. Quotient ,, ,, ,, ,, division. Dividend is a number to be divided.
An expression is any mathematical statement containing numbers, signs, etc.
If several numbers be multiplied together they are said to be factors of the product.
INTRODUCTORY 3
Thus since 3 x 4 X 5 = 60, the numbers 3, 4 and 5 are " factors of 60." Any number may be split up into factors by knowledge of the multiplication tables, but it is seldom necessary to find the factors of large numbers.
Several different factors may be found of the same number. Thus besides the above factors of 60, we may also have 20 and 3, 30 and 2, 2, 10 and 3, etc. This is because some of these factors can themselves be split up into factors. Thus 30 x 2 can be written as 6 x 5 X 2, and so on.
The simplest factors of all are those which cannot themselves be split up, e. g., 2, 3, 5, 7, n, 13, 17, 19, etc. These are called "prime factors."
If a number contains another number exactly (/. e.t can be divided exactly by that number), then it must contain the prime factors of that number. Thus 36 contains 12, i. e., 36 can be divided exactly by 12; then 36 contains the prime factors of 12, i.e., 2, 2 and 3, as may be proved by simple division.
Also the following abbreviations will be met with —
i. e. meaning " that is " ;
e. g. ,, " for example " ;
viz. „ "namely";
approx. „ "approximately" or "very
nearly " ;
revs, per min. or r.p.m. ,, " revolutions per minute " ; m.p.h. ,, " miles per hour " ;
h.p. (H.P. also used) ,, " horse-power."
Previous Knowledge. — In the following pages it is assumed that the reader is acquainted with the four simple rules of arithmetic — addition, subtraction, multiplication, and division when applied to whole numbers ; and with the simple kinds of measurements, such as money, length, weight, time, etc.
The tables of these measurements are given below for reference —
BRITISH TABLE OF LENGTH
12 inches (ins.) — i foot ;
3 feet (ft.) =i yard;
5j yards (yds.) = i rod, pole, or perch;
40 poles (po.) = i furlong;
8 furlongs (fur.) =• i mile.
ARITHMETIC FOR ENGINEERS
WEIGHT (Avoirdupois) 16 drams = I ounce (oz.) ;
16 ounces = i pound ;
28 pounds (Ibs.) = I quarter ; 4 quarters (qrs.) = I hundredweight (cwt.) 20 cwt. = i ton.
also 112 Ibs. = i cwt.;
and 2240 Ibs. = I ton.
SQUARE MEASURE
144 square inches (sq. ins.) = I square foot ;
9 square feet (sq. ft.) = i square yard ;
30 J square yards (sq. yds.) = i square pole ;
40 square poles = i rood ;
4 roods = i acre ;
640 acres = i square mile.
CUBIC MEASURE
1728 cubic inches (cti. ins.) = i cubic foot ; 27 cubic feet (cu. ft.) = I cubic yard.
MEASURE OF CAPACITY
4 gills = i pint ;
2 pints = i quart ;
4 quarts = i gallon ;
2 gallons = i peck ;
4 pecks = i bushel ;
8 bushels = i quarter.
CHAPTER I VULGAR FRACTIONS
Fractions generally. — As the quantities with which we have to deal seldom contain an exact number of units, we have to con- sider how to deal with parts, or fractions, of a unit. Since I is the smallest whole number in our written figures, and we have to deal with quantities less than I, it is evident that our ordinary system of notation cannot be employed to represent the magnitude of such quantities.
Two systems of representation are in general use, both employ- ing the following idea : the unit is divided into a certain number of equal parts, and any particular fraction is said to be equal to so many of those parts. The number of parts into which the unit is divided may be chosen within wide limits, but in practice only the most convenient are employed.
In the Vulgar System this number may be any convenient whole number whatever, for example : 8, 13, 100, 346, etc.
In the Decimal System the number is always i with one or more noughts after it, i. e.t 10, or 100, or 1000, etc.
The decimal system is thus a special case of the vulgar system, but appears rather different in practice on account of the different method of statement adopted. In Great Britain the vulgar system is employed chiefly in connection with money and the everyday use of the ordinary weights and measures. The decimal system is certainly more valuable for scientific purposes and for the more accurate measurements demanded in the workshop, while on the Continent it has been adapted, with few exceptions, to all measurements.
Vulgar Fractions. — In the vulgar system the following nota- tion is adopted : below a short horizontal line is placed the number which shows exactly into how many equal parts the unit has been divided; above the line is placed the number which indicates to how many of these parts our fraction is equivalent. Thus in the case of £ (read as " three-fourths " or " three-quarters ") the unit
5
ARITHMETIC FOR ENGINEERS
has been divided into 4 equal parts, and 3 of them taken. This is shown graphically at a, Fig. i, where the length AE represents our unit (say i foot, for example) divided into 4 equal parts, AB, BC, CD, and DE. The fraction f is then represented by the length AD.
|
"< |
2_ i ^ |
• • • > |
|
|
n JL , |
4- e ' > |
||
|
A |
B |
c |
D E |
|
1 — ' |
_ A- |
||
|
^ |
1 UNIT ' |
a
.15
I
UNIT
|
E rf |
||
|
A IB |
c |
D |
|
, , i | . |
1 1 |
I ' |
|
D £ £ $ 1 £ 3 UNITS |
Fig. i.— Illustrating Vulgar Fractions.
The number underneath the line, which indicates the size of the equal parts, is called the Denominator ; while that above the line, which shows the number of equal parts taken, is called the Numerator. Thus in the above example, 3 is the numerator and 4 the denominator.
VULGAR FRACTIONS 7
Similarly b and c, Fig. i, indicate respectively the fractions ^ (four fifteenths) and JJ (thirteen twenty -thirds), in which 4 and 13 are numerators and 15 and 23 are denominators. We shall fre- quently refer to the numerator and denominator as the " top " and " bottom " of the fraction.
From the foregoing, the line between the numerator and de- nominator is merely a means of distinguishing between them. On p. 2 it is stated that this mid-line indicates the division of the upper number by the lower one. This is quite in accordance with our fractional notation, as may be seen from d, Fig. i. Here the distance AE, which represents 3 units, is shown divided into 4 equal parts on the top side of the heavy line. Beneath the line each of the 3 units is shown divided into 4 equal parts, and it is easily seen that AB = f of i unit. Therefore the fraction J unit is the result of dividing 3 units into 4 equal parts, and hence we may consider any vulgar fraction as indicating the result of dividing the numerator by the denominator.
Such of the simpler denominators as quarters (or fourths), eighths, and sixteenths, will be familiar to most readers as being the usual divisions on the common 12-inch rule, while thirty-secondths and sixty-fourths are occasionally marked. In the monetary system — where the sovereign (£) is the unit — twentieths and two hundred- and-fortieths are employed, but are better known by the names of shillings (s.) and pence (d.) respectively.
Other peculiar denominators such as those at b and c, Fig. i, are not often met with in practice.
Forms of the Vulgar Fraction. — A true fraction is less than a whole unit, and therefore its numerator must always be less than its denominator. When this is so the fraction is called a proper fraction. Thus f , f , i£, T*ff, are proper fractions.
When a measurement consists of an addition of whole units to a fraction of a unit, as, for instance, two inches and three-quarters of an inch (commonly " two and three-quarter inches "), it could be expressed as 2" + f ". It is customary, however, to omit the + sign and write the whole number immediately in front of the fraction, thus, 2§". Care should be taken, when writing in this form, that the figure representing the whole number is written larger than the figures composing the fraction ; otherwise the 2 may be confused with the 3 or the 4, and the quantity read as *-£ or TJ\. Numbers written in the form 2| are termed mixed numbers.
In the course of calculation we sometimes arrive at fractions in which the numerator is greater than the denominator, a con-
8 ARITHMETIC FOR ENGINEERS
dition which cannot denote a true fraction, since we have more parts than are contained in our unit. This type of quantity, which is known as an improper fraction, is only a statement of a mixed number in fractional form, and may be treated in all calculations as a proper fraction. Thus -1/, |£, ±^±, etc., are improper fractions. To convert a mixed number into an improper fraction proceed thus : — Multiply the whole number by the denominator and add the numerator. The result of this is the numerator of the improper fraction. The denominator remains the same. The method con- sists in finding the total number of equal parts contained in the mixed number and stating this in fractional form. Thus in (a), Ex. I, below, the denominator is 4, or each equal part is a quarter. We therefore convert the 2 into quarters, and as I unit = 4 quarters, 2 units = 8 quarters. The total number of quarters is then 8 + 3 ~ n, and therefore 2j = n quarters, i. e., -1/-.
Example i. — Convert into improper fractions (a) 2j; (b) n^
(a) Whole number X Denominator
2x4=8 add numerator = 3
- ,, Sum =11
• • 2f = V
This may be stated more mathematically thus,
2| = * (2_^3) ± 3 = 8±3 ^ ii 4 4 4~
(b) uA = (ri x l6) + 7 = I76±7_ - ~3 V ' ie 16 16 ~~ i(7~
With a little practice the operation may be performed mentally.
The reverse operation of converting an improper fraction into a mixed number is, naturally, the opposite of the foregoing, thus : _ Divide the numerator by the denominator, giving the whole number. The remainder of the division is the numerator of the true fractional piece.
Example 2. — Convert into mixed numbers (a) ^ ; (b) ^. (a) Numerator ~ denominator = whole number.
11-7-4 =2 with remainder 3.
•'• v = 1
(6)101-7-12 = 8 with 5 remainder.
* The use of the bracket signs ( ) is given later. Here it is adopted to show definitely that the multiplication of 2 and 4 must be done before adding the 3.
VULGAR FRACTIONS 9
This also may be done mentally after some practice. A special case occurs when the remainder of the division is o : that is, when the denominator divides into the numerator an exact number of times ; then the improper fraction is equal to an exact whole number, e- g-> IF = 2 and 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. 3°i 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. 2A> 3A-
7. ioj, i5j, nf, i9i 73V 8. loft, 1 1 4. i7f. 2Ii H?-
9. 14?, I5H. 1 7ft, 38J|. 10. I9», 4ift. i6ft, 3ift-
Convert the following improper fractions into mixed n-umbers —
11- f i I- V. V- 12- }i, ¥. S4ft, 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 0 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. fii8°- 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 -j3^. Either the eighths must be changed into sixteenths, or the sixteenths into eighths. Now it is seldom, if ever, convenient to convert any fraction into one of smaller denominator (i. e. larger equal parts). To illustrate this let us convert the sixteenths into eighths.
Then, since ^ = \, Y\ — ii eighths, and we may therefore write—
1 j_ _3 ___ i , i* _ 2* 8 "*" 16 ~~ 8 : "*" 8^ ~ 8 '
a form which, with any but the simplest numbers, would become hopelessly muddled.
But let us convert into smaller parts, i. e., convert the eighths into sixteenths.
Then we may write —
JL l •'* 213 5
8i~iO — 1C i Tff — 169
a practical proof of this being easily obtained by an inspection of the common 12-inch rule. It will be noticed that the numbers which compose the fraction -^ (i. e., 5 and 16) are entirely whole numbers.
Example 4. — The tapping size (size cf a hole before being screwed) of a certain Whit worth bolt is given as fo" -f- A"- Express this as a single fraction.
A + A
= i? 4- oU4 converting the &*» into 6yh", ^ 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" bolt» 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 — tsff -f- ^V Express each of these sizes as a single fraction.
Express the following Whitworth bolt-head thicknesses as a single fraction in each case : —
5. (a) J" bolt. Head is | -f- £>• (b} } J" bolt. Head is ft -f ^.
e. (a) r .. - „* + &• (*) 13" .. .. .. H + A.
Express the following Whitworth tapping sizes as a single fraction in each case —
8. (a) i" bolt, tapping size is } $ 4- jJ5. (6) } J" bolt, tapping size is *$ -f ^ .
9. (*) r » » .. » i +A- w r ,. ,. M ., J+A.
10. wr .. M - »«+&• wtr ........ I+A.
A rather different appearance is presented by the case J + -§. We cannot convert the quarters into fifths or the fifths into quarters without complicating the figures ; but we can obviously give each fraction the same denominator by multiplying the 4 by 5, and the 5 by 4, i. e. multiplying the denominators together, giving 20 in each case. Then, remembering that the value of a fraction is unaltered if both top and bottom be multiplied by the same number, the example may be written thus —
J + i = __x-5 + 2_21 4 4 5 4~X 5 5 X 4 = .5 + 8 ^ 13 20 20 20
In Fig. 3 this is shown graphically.
The foregoing method may be applied to any example, but in the more difficult cases the labour and figuring may be reduced by employing the Least Common Denominator (abbreviated to L.C.D.). As an illustration, consider the addition of the fractions ^ and f . Working on the previous method, the common denominator would be 12 X 8 = 96. Now a little thought wall show that the numbers 48 and 24 also contain 12 and 8 an exact number of times, so that for the purpose of addition the fractions may equally well be con- verted into forty-eighths or into twenty-fourths. By using 24 as
14 ARITHMETIC FOR ENGINEERS
the common denominator instead of 96, smaller numbers will be produced, and the calculation simplified in consequence. This 24 is then the smallest possible common denominator, and is the Least Common Multiple (abbreviated to L.C.M.) of the given denomi- nators. Now the L.C.M. of a set of numbers is defined as the least number which will contain each of the given ones an exact number of times, so, keeping this in view, we may deduce the 24 from the given numbers 12 and 8.
Fig. 3.-— Shov/ing that } -f f
The required L.C.M. is to contain the numbers 12 and 8 exactly, and must therefore contain the factors of these numbers (see p. 3). Splitting into prime factors, we have —
12 = 2X2X3 8 = 2X2X2
Now all the factors of 12 will be required, since the 12 itself must divide into the L.C.M. But in the case of the 8, two of the factors already appear in the 12, viz. 2X2, and need not, therefore, be used again. The remaining 2 in the 8 does not appear in the 12, and must therefore be counted. Then the L.C.M. of 12 and 8 = 2x2x3x2 = 24.
The detailed method of finding an L.C.M. is laid out in the following example : —
Example 5. — To find the L.C.M. of 8, 15, 4 and 12.
8 15 \ i*
8-2x2x2 L.C.M. =2x2x2x3x5 I5 = 3X5 =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. 51* 9> 17- 21. 34, 17, 3. 22. 33, 15, 3.
23. 125, 25, 150. 24. 280, 14, 150, 50.
25. 30, i8o> 45> 9, 5- 26. 3, 17, 51, 34, 9.
Applying the method to addition problems, the L.C.M. of the lenominators is first obtained, and each fraction is converted so hat it shall have the L.C.M. for its denominator, i. e. the common lenominator is the Least Common Denominator.
* See Note to Example 6.
16 ARITHMETIC FOR ENGINEERS
Example 7. — Add together f, ^T, J, and -fa.
The L.C.D. has been shown in Example 5 to be 120. Dividing this L.C.D. 120 by each denominator in turn, gives the number 15, which is required to convert the fraction.
Thus, »£ = 15, and therefore J- = 3 X__y _ £. and similarly for the others.
.'. I + A + i + A
-= 3 _X_L5 4- 2 x ^ 4. I x 3-4- 5 * I0 1 These steps may be
8 x 15 15 X 8 4 x 30 12 x 10 I omitted or shortened.
— 45 , 1 6 , 30 , 50 f See a few lines further
I2O I2O I2O I2O J On.
= Hi = If&» or cancelling by 3, = i^
In an actual example the working would not be shown as above, in which there is a considerable amount of repetition for explana- tory purposes. The common denominator need only be written once if a long line be placed above it instead of the several short lines. Also the actual multiplication of numerators may, in most practical cases, be performed mentally. The working would then appear in the following form : —
I + A + i + A
= 45 + 16 -f 30 -f 50 = 141 i 20 i 20
Example 8. — In connection with an electrical resistance problem it was found necessary to evaluate (i. e.t find the value of) the expres- sion i 4- J 4~ iV Calculate the required value.
4, $, 18
4 = 2X2
23 18 = ^ X 3 X 3
.'. L.C.M. = 2x2x3x3
-36.
Explanation. — -The L.C.M. is quite simple. Taking the first frac- tion, 4 into 36 goes 9. Then 9 x 1=9, which is written above the long line. Then 3 into 36 goes 12, and 12 x i = 12, which appears as shown. Lastly, 18 into 36 goes 2, and 2x1 = 2. Addition of the numerators completes the example.
Where the given quantities include whole numbers, then the easier and safer method is to add the whole numbers inde- pendently of the fractions. Usually the whole numbers may be
VULGAR FRACTIONS 17
added mentally. It is unwise to convert the mixed numbers into improper fractions, as unwieldy figures result, and the subsequent labour is greatly increased. Where improper fractions are given (an unlikely condition), they should be converted into mixed numbers.
Example g. — Find the value of3j4-4 + 7i + 6J.
Add the whole numbers mentally and write the result in front of the addition of the fractions, thus —
The expression then =
24
L.C.M. =-8x3
= 24
20 -f ill MaY often be done mentally.
Example 10. — The dimensions of the spindle for a sluice valve are shown in Fig. 4. Determine the overall (i. e., total) length of the spindle.
Fig. 4. — Spindle for a Sluice Valve. Writing down the dimensions in order we have —
= 18 4_±__10 t8 ±JL±_14 16
L.C.M. = 16
Examples like this are of frequent occurrence when dealing with drawings, and with practice they can be done mentally. It is then necessary to be thoroughly familiar with the ordinary measurements of J, T\, etc., and to know at once the simpler con- versions such as f = -{£ = !£» etc- Sucn knowledge is obtained by doing a considerable amount of mechanical drawing. Then the above example would be done mentally as follows : ioj + f is ioj, add 6 gives i6J, add J gives 17$, 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 20TV +
c
TV =
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 + 5J6 + jpa + t- 35. ft + ij + A + 5l-
36. In finding the current required for some electric glow lamps fed in parallel, it was necessary to obtain the value of the expression
ifo H- lib + fs> Find this value-
37. Find the overall length in inches of the lathe mandrel shown at a, Fig. 5. Convert the result into feet and inches.
38. b, Fig. 5, shows an eccentric strap bolt. Find its overall length in inches.
Fig. 5. — Exercises on Addition of Vulgar Fractions.
39. Find the overall length in feet and inches of the piston rod shown at c, Fig. 5.
40. Find the length of the cross-head bolt, shown at d, Fig. 5, from the underside of the head to the point.
41. The vice screw shown at e, Fig. 5, is to be cut from the solid bar. Allowing an extra inch of length for holding, etc., what length of bar must be cut off ?
42. Find the overall length of the armature shaft for an electric motor, shown at/, F*ig. 5.
Subtraction of Vulgar Fractions. — This involves no new principle, the method of finding the L.C.D., and the separate treat- ment of whole numbers, being adopted as in addition. An example is the best explanation.
VULGAR FRACTIONS
Example n. — Find the value of 5^ — 2§.
12 = 2x2x3
_ , 3
-2I - 9 - 3A
24 —
.'. L.C.M. = 2x2x3X2
= 24
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^ + IJ- 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+_L±J = 10J. 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 |
|
= 4«fr °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. 9ft - 6tf .
15. From 3^ take 2j. 16. From i{ j take §^.
17. From ij take ^70. 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. 4ft -f 5j - 7ft.
27. 12 - 3± + Ji - I- 28. M - 2i3o - i + 5i-
29. 3ft - ft + 18 -f 2} - J. 30. i-/a + | — 2J5 -f 3^
31. Find the value of 3ft — ij and take the result from i
32. „ .. ,. * I - I
33. Take the result of ift -f- £ from 3-j%-
34. „ „ „ 9l + ft ., I3S-
ft-
I. A may be irscreased to suih; or- 2. Rackirxg raay be puhal~p.
o
V)
Fig. 7. — Exercises on Subtraction of Vulgar Fractions.
35. From 2ft take the result of f -f f .
36. „ 2i| „ „ „ 3 + 4i 4- 2f.
37. „ 37l .. .. » 3* + 2i| -f- ft.
38. From the result of 2 ft — .p2- take i}J.
39. „ „ „ ft + }" „ &.
40. „ „ „ * - A „ T^J.
41. a, Fig. 7, shows a screw from a lathe head-stock. Find the length of the part C.
42. The valve spindle for a stop valve is shown at b, Fig. 7. Find the length of the plain portion marked A.
43. A milling machine spindle is shown at c, Fig. 7, where by a
22
ARITHMETIC FOR ENGINEERS
draughtsman's oversight the length A of the front bearing has been omitted. Find this value from the other dimensions.
44. Find the value of the missing dimension B from the drawing of the electric motor shaft shown at d in Fig. 7.
45. An electric motor and a centrifugal pump, built by different makers, are to be assembled on a single bedplate as shown at e, Fig. 7, which also gives dimensions from centre line to base.
(a) Supposing a special bedplate is to be made to suit, find what
the dimension A must be.
(b) Supposing that a standard bedplate, in which the dimension
A is ij",be used, find what packing thickness is required under the motor feet.
Multiplication of Vulgar Fractions. — Multiplication is a convenient method of performing a long series of additions of equal quantities. Thus, 5 x 4 is a shorter way of stating both 5+5 + 5 + 5 (*'•*•, 5 added 4 times) and 4 + 44-4 + 4 + 4
B
P
£0
"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 nf 3
~ 4 OI 5"
M L K J H . G
Fig. 8. — Showing that f x | -** TV
(i. e.t 4 added 5 times). The multiplication of a fraction by a whole number may be shown the same way.
Thus, |X5 = | + f + j + f + |
= 3 + 3 + 3 + 3 + 3 = 3 _XJ
4 4
= ¥ = 3l
So that the result is obtained by multiplying the numerator by the whole number.
The multiplication of two fiactions, e. g.t f X f , is very con- veniently shown graphically, as in Fig. 8. The multiplication sign (x) may be replaced by the word " of," for f x f is f taken f of a " time " ; the result will naturally be less than |. In Fig. 8,
VULGAR FRACTIONS 23
the square A F G M represents i unit, which is divided by vertical lines 13 L, C K, D J, and E H, into fifths. Then the area A D J M represents the fraction J. Similarly the horizontal lines R U, Q T, and P S, divide both the unit and the area A D J M into 4 equal parts. Therefore area A D N P is £ of area A D J M, i. e., f of f of a unit. The two sets of lines cut the unit square into 20 small figures such as H G S W, and in the area A D N P there are 9 such figures —
.'.? of 3 =9-
4 5 20
But the 9 is the product of the numerators 3 and 3, and the 20 is
the product of the denominators 4 and 5.
Hence the rule for the multiplication of vulgar fractions — Multiply the numerators together to give the new numerator, and the
denominators together to give the new denominator.
Reduction to lowest terms should follow, but in many cases
the cancelling may be more conveniently performed before multi-
plying up. Thus in the following we first cancel by 3: —
3
$ x 7 = 3 X 7 = 21
16 \\ 16 X 4 64
4
The method is applicable to any number of fractions. When mixed numbers occur, it is usually advisable to convert them into improper fractions.
Example 16. — Find the product of §, if, 2^ and ^. f X If X 2j X A
= I x y x
Cancelling being performed with n, 4 and 3.
Example 17. — If i kilogramme is 2j Ibs., find the number of Ibs. in 235 kilogrammes.
Evidently 235 kilogrammes = 235 times as much as i kilogramme. Then as i kilogramme = 2j Ibs.
235 kilogrammes = 2j x 235 Ibs.
47
= V X W — V X 47 — 517 Ibs.
The cancelling should always be shown neatly, so that the original figures may be easily read. Care in this direction enables the working to be checked through, for possible errors, by any one. Examples involving heavy cancelling are met with when calculating
24 ARITHMETIC FOR ENGINEERS
the speeds of various combinations of gear wheels, such as occur in the feed motions of machine tools, variable speed gears, etc. In examples of this kind, special care must be taken and plenty of space allowed above and below the expression so that the cancelling may not be too confined.
Example 18. — The following figures were obtained when calculating the feed in inches of a drilling machine, for one revolution of the drill. Complete the calculation.
§3 X JO X 7^ X -fo X 16 x J
i i
* i x i
ys/f v ix \/ i \/ i \/ l# \/ £
Ho x ^ x 7* x XV x i x 2
i i
== .}Q x ft = »Jo inch per revolution.
It must be noted that " multiplication " does not necessarily imply an " increase." If the multiplier is itself a fraction, then the resulting product is only a piece of the quantity multiplied, in which case a decrease results.
Exercises 5. On Multiplication of Vulgar Fractions.
Find the value of — •
1.1X7- 2. ^ x 5. 3. .& X 3- 4. 15 x 4-
5. f x j. 6. j x ,V 7. 3? x jj. 8. {? x £,.
9. i? X }£. 10. ft X 2ft. 11. 3-2 X if,. 12. 5* x i ^.
13. y x il xi 14. A x i? x A.
15. 2j X I A X I X J. 16. J X ft X 5t X Ij.
17. Find the value of ij X ij X ij- X ij X il.
The following multiplications refer to certain spur wheel trains. Complete the calculations.
18. Yia X \^ X V- 19. 500 x
20. U x 88 x Sf. 21. V&- x g
The following multiplications give the feed in inches per revolution of the mandrel in an 8" self-acting lathe. Complete the calculation.
22. Greatest surfacing feed = S| X fe X f£ X J.
23. Lowest surfacing feed = §t x 'dV x i« x J-
24. Greatest traversing feed = ^| X ^ X ??, X 14 X f .
25. Lowest traversing feed = gj X ^ X ifg X 14 X |.
26. The speed of a certain lathe spindle, when driving through the back gear, is given by the value of the expression —
140 X ft X jf X Jf r.p.m. Find the value required.
27. The feed of a drilling machine, in inches, for one revolution of the drill spindle, is given by the figures : ? J X JJ X ^ X^jX i6x f. Complete the calculation.
VULGAR FRACTIONS 25
28. (a) The finest measurement that can be made with a certain micrometer is Js 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 I35o 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 * |
5I 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- Tne 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 -fs. 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 TV 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 ~ Iff
3
==17^I7==\\ v $ ^3 12-9 « \\ £
This step usually omitted. Example 20. — Find the value ofijxj-f-2g.
I*Xi-r2f
_ * x 7 x _5 _ 35 - 2 X 8 X « i - 6^
4
Example 21. — A large pipe-joint is to have a number of bolts evenly spaced on a circle whose length right round is 754"'. The bolts are to be spaced 3$ * from centre to centre. How many bolts must be used ?
A glance at Fig. 10 will show that there are as many bolts as spaces.
VULGAR FRACTIONS 27
Then if we find how many times the 3| is contained in the 75 J we shall have the number of spaces and therefore the number of bolts. Then, 75* ~ 3*
2
*5£ _i_I5 _i5i v i_ ~ ' ~
or, as we cannot use a fraction of a bolt, say 20 bolts.
Example 22. — An engineering drawing is made to a scale of " J* to i ft." (i. £.,
every i ft. on the job is drawn as £" on the Distance round This drawing). Find what fraction of full size "Pitch Circle-11 is 75 2" this represents (i. e., find what fraction of an inch on the drawing represents \" on the job). lg' I0*
The quantities must be changed into the same units before dividing. Now " J" to i ft." means that i ft. is shown as \" , also, changing to inches, i2/r are shown as |*.
/. i* will be J* -f- 12
= f x A = iV'-
4
Thus i" on the job appears as fa" on the drawing, or a scale of |* to i ft. = ^ full size.
Example 23. — It is desired to make measurements from a certain drawing of which the " scale " is not known. A dimension figured as I2J" measures 2J". What is the scale ?
I2J" is shown as 2 1". Then i" will be shown by 2j* -r- 124-"
2 3 Then the drawing is J full size or 2" to i ft.
It should be noted that division does not necessarily produce a decrease, as might at first be inferred from the word " divide/' When we divide by a fraction, i. e., by something less than i, then the result of the division may be considerably greater than the number divided. Thus, if 3 be divided by J the result is 12, the reason being that J is contained 4 times in i, and therefore 12 times in 3 ; and as a proof 12 X J = 3.
28 ARITHMETIC FOR ENGINEERS
Exercises 6. On Division of Vulgar Fractions.
Find the value of the following —
1. I -r ,!-,. 2. fr -r ,V 3. .?., -r- JJ. 4. if -> yV
5. 3 A - y - 6. 1& -r- 3?,- 7. 4J -r i?. 8. i js -- JJ.
9. ^ -r 14- 10. 5t -r 9-
11. The top joint of an oil-separator shell is to have 40 bolts evenly spaced round the pitch circle, and the length of this circle is 123 J*. Find the " pitch " of the bolts (i. e., the distance from, the centre of any bolt to the centre of the next. Fig. 10 will make this clear).
12. The length round a boiler ring on the pitch line of the rivets is 1 8 '-9", and the rivets are to be 2 J" pitch. (Fig. 10 will show what is meant.) How many rivets will be required? (Note. — Work in inches.)
13. The size of the teeth in a spur wheel are often given by the " diametral pitch," i. e., the number of teeth that the wheel has for every inch of its diameter (e. g., a wheel of 50 teeth and 5" diameter has 10 teeth per inch of diameter). Find the diameter of a wheel having 35 teeth and 2j teeth per inch of diameter.
Find the diameters of the following spur wheels — •
14. To have 162 teeth, and 12 teeth per inch of diameter.
15. „ 55 » » 2j „
16. „ 22 „ „ 2| „
The following figures are taken from tests on toothed wheels. Find in each case the " pressure per inch of width/' i. e., pressure ~ width.
17. Pressure on teeth 270 Ib. ; width i \" .
18. „ „ „ 434 » » 1 2"-
19. „ ,, „ 416 „ ,, \\".
20. „ „ „ 1638 „ „ -2\".
What fractions of full size do the following scales represent :—
21. £" to i ft. 22. iV' to i ft. 23. \" to i ft. 24. I" to i ft. 25. ii" to i ft. 26. $\" to i ft. 27. 4" to i ft. 28. TV 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. „ „ „ 7r .. 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* 0—4—1.
==: A - jl __ 3 - _____ 3_ . __ - JL
488 o 8>
which shows a considerable difference.
It should be noted that the bar of division also acts as a bracket.
c \ Q Thus, in the case of - (-> the 6 and the 8 must be considered as
one quantity. Cancelling is not permissible where the terms of the division contain + and — signs, unless all the numbers on either side of the line be divided at the same time. For instance, if it be desired to cancel the above expression by 2, both the 6 and 8 must be
divided by 2, thus —
3 4
8 and not either of the following —
«)$ + ! or
10 x4
8 Thus(i) =" = i& and
Now — 5~ evidently = J-J-, which can be cancelled to |, thus proving the first system of cancelling to be correct.
30 ARITHMETIC FOR ENGINEERS
With compound examples, like the following, it will be found that the working presents a much better appearance, and can easily be re-read at any time, if the various operations are carried out in their proper places in the given expression. The steps in the next examples should be easily followed with but little explanation.
7
Example 24. — Find the value of . 1 |T
1 3 +io
= ? — 49 = -1 v 3° = * 15 ' 30 ^ X ^ 7
1 7 ""
Explanation. — The first operation must be to evaluate the bottom line, then the ^ can be divided by the result. The addition may be done in a separate place and the result collected.
a
Example 25. — Simplify the expression -•.—-*—» .
rs + tr
Expression = ~r^7 inverting the 3 and
- y -^ X4
^ A 3 cancelling by 2
6
9
= t = 3 v ill = 'i7 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+Cxf£ + I5
i 7
» I+I+-9-
2 ^ 7 ^ 16
= 56 + l6 + 63 135 _ 23
112 112 ~~ *!*
Note. — The multiplication must be done first. Example 27. — Simplify the expression (J + §) X & -f ft. The expression = *_+_4 X £ + £ *"*
9 Thcn
. •""" 4 16""" 16
3 Example 28. — Find the v^lue of i — T V~V
The expression = i — --1 <* ^
"To""
i
==I~S==I~^x!7
3 __ 4
ARITHMETIC FOR ENGINEERS
of cuitmq-off tool '1
Example 29. — Cast-iron piston rings, fa" wide, are to be turned from a rough casting. The width of the cutting-off tool is fa*, and 3* is re- quired at one end for holding in the chuck. Allowing y for facing up the end of the pipe, find the least possible length of pipe from which
to cut i dozen rings (see Fig. n).
After facing, which will finish one side of the first ring cut off, each part- ing will produce one ring. Therefore there will be 12 partings.
Total length required == Chuck grip + total width of partings -f- total width of rings -f- allowance for finishing
Clld'_ 3 , 1
Fig. II.
= 3 + VM-Y- + i =-. 13!, say 14 in.
Example 30. — A mortar is to be composed of i part of cement to 3 parts of sand, by weight. Find how much sand and cement arc needed to make 3 cwt. of this mortar.
Using i Ib. of cement, say, we must use 3 Ibs. of sand, as there are to be 3 parts of sand to i of cement ; the total weight of the mortar is then 4 Ibs. Thus the weight of cement is £ of the total weight (i.e., i Ib. cement in 4 Ibs. of mortar), and the weight of the sand is | of the total weight (i. e., 3 Ibs. sand in 4 Ibs. of mortar).
The actual working may be as follows — • Cement = i part vSand = 3 parts
/. Mortar --= 4 parts
Then cement — J- of mortar — J- and sand = i „ „ ^= J
3 cwt. = 3 ,. =
cwt.
Total (as proof) = 3 cwt.
Nute. — The method of working is always the same, no matter how many constituents there may be in the mixture.
Exercises 7. Compound Examples.
Find the value of the following expressions —
1 1H
*• 8| 5' p-^
«.!+-:
2.
6. ^
10. ?
?*
3. A-
s T i
-i9*
2-*V
7.
i V i'
11.
A + iV
2\ —I*
8. 3 s
V
X
VULGAR FRACTIONS
33
Simplify the following expressions : — 12. 2£ - ij x i^. 13. (2j - ij) X i,V
14- iV X | + f X f . 15. A X (I + ft X ».
16. f of | -r * + f . 17. t of f - (* + |).
18. 3i + 2* -^ i - A- 19- (3i + 2i) -r- (J - i
20.
- i*.
21.
22. - x
23. Special set screws are to be cut from bar stock in an automatic machine (see a, Fig. 12). The overall length of the screw is iA*. The width of the cutting-oif tool is J". and the bar stock is in 5 -o*
li^^^.l.-*. __O I Ug_ . ti J If ,i - -, j.,.. tT -jr ___
16^ 1 16 HBr^ ne ^ | ^*"
Lor\c»
Fig. 12. — Illustrating Exercises 7.
lengths. How many screws are obtainable from each bar, and how many bars will be required for an order of 12 gross of screws ? (Note. — I gross — 12 dozen = 144.) (Hint. — Remember that if, say, 20 J screws can be cut from i bar, only 20 screws are really obtainable ; but if, say, 55J bars are required, 56 must be bought.)
24. It is required to turn 9 handles each 5^' long, as shown at 6, Fig. 12, from a mild steel bar. Allowing £-" extra on each handle for cutting-oif, etc., find what length of bar must be cut off for the 9 handles.
25. Some sleeves, as at c, Fig. 12, are to be cut from a piece of wrought-iron pipe, in the lathe. If we allow -^" extra on each sleeve for parting off and i£" extra at the end for holding when cutting off the last one, what length of pipe is required for 16 such sleeves ?
26. Pillars for stop valves, as shown at d, Fig. 12, are to be turned from bars. Allowing an extra \? on each one for cutting off, how many pillars can be cut from a bar 5/-6" long ?
27. Electric contact fingers, J" wide, are to be sawn in a power hack-saw, from extruded bars each 6'-6* long. (See e, Fig. 12.) If the saw-cut is j^" wide, how many fingers can be cut from each bar ?
34 ARITHMETIC FOR ENGINEERS
28. Special nuts arc to be turned from the bar as follows : Seven f * thick, and fifteen £" thick. Allowing an extra fs" on each nut for parting off, find the total length of bar required.
29. Forty-eight angle stiffeners for a plate girder are required 40 J" long, and are sheared from the long angle bar without waste. What length of angle will be required ? If the stock angle bar is in 45 ft. lengths, how many lengths will be needed ?
30. A concrete is to be made of cement i part and coke breeze 4 parts by weight. How much of each constituent is required for 15 cwt. of concrete ?
31. What weight of each metal must be used to make 150 Ibs. of each of the following solders, assuming nothing lost in the melting ?
(a) Lead 3 parts; tin 2 parts.
(b) Tin 4 parts ; lead i part ; bismuth i part.
32. Black gunpowder is composed of nitre 15 parts, charcoal 3 parts, and sulphur 2 parts, by weight. Find the quantities of each material required to make 60 Ibs. of gunpowder.
33. The white metal for fusible plugs is made of varying propor- tions, according to the temperature at which it is required to melt. Find what weight of each metal is required to make i cwt. (112 Ibs.) of each of the following alloys : (a) tin 2 parts, bismuth i part ; (b) lead i part, tin 4 parts, bismuth 5 parts ; (c) lead 5 parts, tin 3 parts, bismuth 8 parts.
34. Find what weight of each material is required to make the following quantities of foundry sand : (a) 2 cwt. of facing sand con- sisting of 6 parts old sand, 4 parts new sand, and i part coal dust ; (b) 10 cwt. of floor sand consisting of 6 parts old sand, 2 parts new sand, and £ part coal dust. (Give results in Ibs.)
35. A filling material, for cast iron, which will expand on solidifying, is composed of lead 9 parts, antimony 2 parts, and bismuth i part. How much of each metal must be used to make 30 Ibs. of filling ?
36. Find what weight of each material is necessary to make 75 tons of concrete, to consist of cement i part, sand 3 parts, and aggregate (broken brick, etc.) 6 parts.
37. When helical springs are used in compression the last three- quarters of the coil at each end is flattened to provide a flat seating; these flattened portions do not form part of the effective coils. Eighteen such springs are required, each having 7 effective coils, the average length of a coil being 3f". Calculate the total length of wire in a complete spring and also the total length of wire required for the 1 8 springs, allowing 2" per spring as wastage in cutting off.
Find the values of the following expressions.
38. I 39. * 40. JJL(L±iL
m _ 140 41. 31 + J> 20 ' 42. (J x &) -i- (f+ 58)- 48.
CHAPTER II DECIMAL FRACTIONS
Decimal Notation. — The ordinary method of numbering is built upon a scale which uses only ten signs, and is therefore called decimal notation. The ten different signs employed, which are called digits or figures, are o, i, 2, 3, 4, 5, 6, 7, 8 and 9; and all numbers are expressed with these only, by employing multiples of the unit in which 10 of each kind make I of the next larger variety. As examples of these multiples we have the ten, which is a set of 10 units ; the hundred, a group of 10 tens, *. e.9 10 X 10 — 100 ; the thousand, a group of 10 hundreds, i. e., 100 X 10 = 1000 ; and so on.
The value of each digit in a number (i. e., the multiple to which that figure refers) is indicated by its position in the number. Thus the extreme right-hand digit is the " units " figure, the next to the left the " tens " figure, the next the " hundreds " figure, and so on.
For example, the number 15 means i set of 10 units + 5 single
units (i.e., i X 10 + 5) the number 746 means (7 X 100) + (4 x 10) + 6, and
5280 „ (5 x 1000) + (2 x 100) + (8 x 10) + o
From the above examples it can be seen that, starting from the extreme left with the most important figure (i. e., the one of the largest multiple) and proceeding to the right, every multiple is jl0 of its left-hand neighbour, until finally we arrive at the simplest unit. Now proceeding to divide the unit, in order to deal with fractional quantities, it is only natural that we should make the first sub-multiple ^0 of the unit, and the next -^ of that (i. e., ^ of jV — j J<j), and so on. Then, if the vulgar system of writing down such fractions were employed, we should have for denomi- nator a number such as 10, 100, 1000, etc., i. e., a number composed
35
ARITHMETIC FOR ENGINEERS
of i with one or more noughts following it.* Fractions having such denominators are called " decimal fractions/' but the denomi- nator is omitted in writing down the fraction. The numerator only is written, the denominator being indicated by the position of the figures in the numerator. A dot (•), called a decimal point, is used to distinguish between whole numbers and the numerators of decimal fractions; figures to the left of the point are whole numbers, and those to the right are fractions. The first place of figures to the right of the point is the T\j- place, the second the y^ place, the third the -nn^ place, and so on.
All the foregoing points are illustrated in the diagram below, which shows the composition of a mixed number when expressed decimally.
Multiples
Sub-multiples
|
o o |
o o |
8 |
0 H |
M |
-13 |
-IS |
E |
g |
|||
|
0 |
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 |
0 |
'S "S |
a |
s |
H a |
||
|
H |
H |
ffi |
H |
£ |
H |
H |
ffi |
H |
H |
||
|
8 |
1 |
5 |
• |
3 |
2 — • — - — -~ |
7 |
Whole Number
Fraction
300
The number 815-327 means 815 + T3^ + T§rr +
But, by the rules of Chap. I, T£ ^ - T|^, and - so that the number 815-327 = SiS^V
Besides omitting the denominator in writing, it is also omitted in reading a decimal fraction : the figures following the point are simply read in order. Thus the above quantity 815*327 is read as eight hundred and fifteen point three two seven. As further examples —
39*37 reads thirty-nine point three seven and — 39 + ^ + ioV
* Numbers such as these (i.e., those composed of i followed by one or more noughts) are called " powers " of 10 : see Chap. III.
DECIMAL FRACTIONS 37
But as ^ = £jv, a shorter statement in the vulgar form is 39 $£$ . Similarly, 62-425 reads sixty-two point four two five, and = 62
+ 4 I 2 I 6 A/> 425
TTF -T TTTTF ~T TfftfTT ~ °2 TTJtf tf-
When a whole number alone is written down, the decimal point is omitted, as there is no fraction to consider. If a number such as 15-00 is obtained as the result of any calculation, then the dot and the noughts may be omitted, and the figure written simply as 15, since -oo is, of course, o. But when a proper fraction has to be written down, then the dot is absolutely necessary, as its omission will convert the fraction into a whole number. Thus -305 is the fraction -^nnr* an(^ ^ the dot were omitted the figures would read 305 ! For this reason a nought is sometimes written in front of the point thus, 0-305, so that, should the dot be lost, the appear- ance of the figures o 305 would indicate its intended existence. The number -305 may be read as either " nought point three nought five " or simply " point three nought five.1'
It should also be noted that noughts may be added to the right of the figures of the fraction, and to the left of the figures of the whole number, without affecting the value of the number. Thus 3-1400 is only 3-14, the o's merely stating that no y^ths or ^i^ths exist, which is equally well shown by writing nothing whatever beyond the 4. For similar reasons 0078-54 = 78-54.
Movement of the Decimal Point. — It is a feature of the decimal system that any number may be multiplied or divided by 10, or 100, or 1000, etc., without any actual operation on the figures themselves beyond moving the decimal point. As an example take the number 39-37. This = 39^, which may be expressed in the improper form -^nnr- H the decimal point be moved one place to the right the figures will appear as 393-7, which = 393^ = m*-. Converting to ^ths we have -3-|^, which = Stfp- X 10, so that 393-7 is 10 times 39*37. The converse is, of course, equally true. Thus 39-37 is ^ of 393-7. Hence the rule : to multiply or divide a number by 10 move the decimal point 1 place to the right or left respectively.
Since 100 = 10 x 10, then to multiply or divide by 100 the above rule should be obeyed twice. Similarly, to operate by 1000, which = 10 X 10 x 10, the decimal point must be moved three times. Hence we have a more general statement of the above rule : To multiply or divide a number by any power of 10,* move the decimal point to right or left respectively, 1 place for each 0 in the multiplier.
* See foot-note on p. 36.
38 ARITHMETIC FOR ENGINEERS
As examples : 2-5 X 10 =25 (the dot is not needed final])7).
2-5 ~io = -25
7854 x 100 = 78-54
3-1416 x 1000 = 3141-6.
When a whole number is to be operated upon in this manner, and apparently has no decimal point, e. g., 1760, then it must be remembered that the dot really exists immediately to the right of the units figure, i. e., after the o. Then places may easily be counted off when moving the dot to the left. Thus 1760 -f- 100 = 17-60. When moving the dot to the right the operation is that of adding a o for every time that the number is multiplied by 10. Thus 1760 x 100 = 176,000.
Similarly, when moving the decimal point to the left and no more existing figures are available, noughts may be introduced as necessary. Thus 3-14 -r 100 — "0314, for we may consider 3-14 as 03-14.
Also -0625 -f- 10 = -00625.
Exercises 8. On Movement of Decimal Point.
What is the result of —
1. («) 5'5 X 10. (b) 5-5 — 10. (c) 1-34 X ioo. (d) 1-34 4- ioo. (e) -153-6 x 1000. (/) 453-6 H- 1000.
2. (a) -ooi 1 18 x 1000. (b) -0000087 x 10,000. (c) n>ooWo- (d) 35° x 1,000,000.
3. If i kilowatt = 1000 watts, convert the following into watts — (a) 35 kw. (b) 7-3 kw. (c) -08 kw.
(Hint. — Multiply by 1000.)
4. Convert the following into kilowatts —
(a) 29,500 watts. (b) 231 watts. (c) 305,000 watts.
(Hint. — Divide by 1000.)
5. If i volt — 100,000,000 absolute units, convert the following into absolute units —
(a) 2 -7 volts; (b) -005 volts ; (c) -000003 volts.
(Flint. — Multiply by 100,000,000.)
6. If i Joule = 10,000,000 ergs, convert the following into Joules — (a) 42,600,000 ergs, (b) 550,000 ergs.
Conversion from Decimals to Vulgar Fractions, — This has already appeared in the section on decimal notation, and only a few points remain to be noted. It will be seen, from the examples given in the section referred to, that the denominator consists of i followed by as many o's as there are figures in the numerator ; for example, in the case of 39-37 the denominator (ioo) contains two noughts, one for each of the figures behind the decimal point. Similarly with ^425 there are three noughts, i. e., the denominator is 1000. Hence we have the following rule : For the numerator
DECIMAL FRACTIONS
39
write the figure or figures behind the decimal point. For the denomin- ator write 1 for the decimal point and a 0 for every figure following it. Finally reduce the fraction to its lowest terms.
Example 31. — Convert the following decimals into vulgar fractions — (a) -2; (b) -5; (c) 0-3; (d) -45; W -125-
(a) Numerator = 2. Denominator — i followed by one nought,
i. e., 10, as there is one figure behind the decimal point. /. -2 = j20 ; cancelling by 2 gives £.
(6) -5 =!5o: .. » 5 » £
(c) 0-3 = 1%- ; no reduction possible.
9
(d) -45 = I'\M>; cancelling ^ == JQ.
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 = 1Q0ioV= i wir — iV cancelling by 25 or by 5 and 5 again.
Similarly -0036 = ^VV = roVW = WW cancelling by 4.
(The steps ^Vo anc^ voVoV aDOve arc only introduced for ex- planation and need not be put in the working.)
Since many of the simpler fractions are frequently used in the vulgar form, it is desirable to remember the decimal equivalents of the more common ones. The following table gives the decimal equivalents for every ^ up to i. Certainly the equivalent for every £ should be memorised, and it is advisable to learn also the equivalents for the intermediate sixteenths.
DECIMAL EQUIVALENTS.
|
Fraction. |
Decimal Equivalent. |
Fraction. |
Decimal Equivalent. |
Fraction. |
Dec-mal Equivalent. |
|
T?S ' • ' |
•03125 |
3 B |
•375 |
V?. |
•71875 |
|
,'* - • |
•0()25 |
Ji • • - |
•40625 |
I |
•75 |
|
3», " ' ' |
•09375 |
./,> • • |
•4375 |
¥l - • • |
•78125 |
|
1 8 |
•125 |
M • • • |
•46875 |
U • - |
•8125 |
|
& • • • |
•15625 |
1 li |
•5 |
P • - • |
•84375 |
|
ft • • |
•1875 |
«..- |
•53125 |
7 3 |
•875 |
|
Tts • • • |
•21875 |
10 '"* ' ' |
•5625 |
3S • • • |
•90625 |
|
4 |
•25 |
18 - - • |
'59375 |
8 • • |
'9375 |
|
A ... |
•28125 |
5 y |
•625 |
31 ... |
•96875 |
|
A • • |
•3125 |
;5J • • • |
•65625 |
i |
I '00 |
|
M • • • |
'34375 |
| U - • |
•6875 |
40 ARITHMETIC FOR ENGINEERS
Exercises 9. On Conversion of Decimals to Vulgar
Fractions.
Convert the following decimal fractions into the vulgar form —
1. (a) -65; (b) -45; (c) -425.
2. (a) -028; (b) -875; (c) -3264.
3. (a) 2-205; (b) 1-34; (c) -305.
4. (a) -964; (b) 3-142; (c) -00256.
5. Convert into vulgar fractions : (a) -515625; (b) '390625.
Degree of Accuracy. Significant Figures. — Mathematic- ally there is no limit to the number of figures which may be written on either side of the decimal point, but a few only are required for practical problems. As some operations with decimals produce large numbers of digits in the results, it is desirable to consider how many figures are really necessary in any mixed number. All the quantities with which we have to deal are connected with measurement ; the accuracy with which a quantity can be stated depends upon the nature of the measurements relating to that quantity. Thus we might say with truth that the size of a steel ball, taken from a ball bearing, was 1-2498" ; machines and instru- ments which can measure i^J^n/' (-0001") in a distance of ij* are fairly common, and it is also very desirable to know the size of the ball accurately in order to secure smooth and safe running of the bearing. But to say that the length of a liner's voyage is 2536-4978 miles would be utterly absurd, for the actual distance travelled would depend so much on the state of the weather, ocean currents, etc., etc., that we could not be sure that even the last figure in the whole number (the 6) was correct, much less any of the succeeding figures. We could not give, with confidence, the distance any more accurately than 2540 miles. Evidently, then, there is no use in stating the figures 6-4978, and still less in using them in any calculation in which the above distance is required. A quantity such as this 2540 miles is, then, only an approximate, or nearly correct, measurement. When describing the accuracy to which this distance has been measured we say that it is " correct to 3 significant figures" the digits 2, 5 and 4 being the " significant figures " ; that is to say, it is as near as possible to the truth while only containing 3 significant figures. The o only helps to denote the various multiples of the figures 2, 5 and 4, i. e., it fixes the position of the decimal point and does not itself signify any number.
DECIMAL FRACTIONS 41
All the digits i, 2, 3, 4, 5, 6, 7, 8, and 9 are significant figures in all cases. The digit o is only a significant figure when it has a higher digit on both sides of it, as in the case of the number '305. Then the o is significant, distinguishing the number -305 from '035 or '350. Thus, taking the following numbers : —
In -305 the sig. figs, are 305. In 6o6»5 the sig. figs, are 6065.
„ -035 „ „ „ „ 35. „ 6080 „ „ „ „ 608.
„ -035 „ „ ., » 35- „ -000341 „ „ „ „ 341.
n 2938 „ „ „ „ 2938. „ 30,000,000 „ „ fig. IS 3.
Most of our measurements fall under this " approximate " heading, and consequently are only stated " correct to so many significant figures/' The actual number of significant figures depends upon the nature of the measurement. Thus in giving the horse-power of a ship's engines 2 significant figures are usually all that can be relied upon, e. g., 12,000 h.-p. ; but when stating the length of a base-line used for the large surveys of Great Britain, 7 or more significant figures are given, e. g.> 26405*78 ft. ; and in accurate chemical work it is quite possible to work to 7 significant figures. No definite rules can be given for the number of significant figures desirable in any measurement or result. It is a matter of experience, and the reader must be guided by the style of answer given to the various examples, both worked and set, throughout this book. It may be taken, as a general rule, that 3 or 4 significant figures are quite sufficient for most engineering calculations.
As a matter of interest it may be noted that calculations on the ordinary slide rule (Chap. X) are performed with the " significant figures " only, the position of the decimal point being fixed later.
The adjustment of a quantity to any required number of sig- nificant figures is governed by the idea that any error introduced thereby shall be as small as possible. Thus if the 2536 be required correct to 3 significant figures it should be written 2540, and not 2530. The difference between 2536 and 2540 is 4, whereas between 2536 and 2530 it is 6. Hence less error is introduced by using 2540 rather than 2530. But the number 2533 correct to 3 significant figures is 2530 and not 2540, the difference between 2533 and 2530 being 3, whereas between 2533 and 2540 it is 7.
Hence we have the rule : increase the last figure retained by 1
when the rejected figures are over 5000 but leave it unaltered
when the rejected figures are less than £000 When one figure
only is to be rejected and is 5 exactly, either of the above may be applied.
42 ARITHMETIC FOR ENGINEERS
Example 32. — Express the following numbers correct to 2 significant figures: — (a) -333; (b) 1-666; (c) -01734; (d) -7071; (e) 10-01 ,
(/) 235-5; fe) i'57°8; W °'995r-
(a) *33 as 3 is less than 5; (6) 1-7 as 66 is greater than 50 ; (c) -oi7as34,, „ „ 50; (d) -71 „ 71,, „ „ 50; (e) 10 as 01 „ „ „ 05; (/) 240,, 55,, ,, „ 50; (g) 1-6 as 708 is greater than 500; (h) i-o as 51 is greater than 50. Note. — In case (h) only one sig. figure can be given.
Example 33. — Express the following numbers correct to 3 significant figures:— (a) 3744; (6) 39'37i W 2-7183; (d) 2-3026; (e) 1-897;
(a) 3740; (b) 39-4.' (c) 2-72; (d) 2-30 (i.e.. 2-3); (*) 1-90 (i.e.. 1-9); (/) 4880.
. — In (d), since only 3 figures are required, all figures to the right of the o (the third significant figure) must be eliminated, leaving 2-30. The last o has no use, so that our result is 2-3. This contains only 2 significant figures but cannot be avoided since the third figure is a o.
In (e), since 3 figures are required, the digit to the right of the 9 must go out, but being a 7, i. e., over 5, must be allowed for by adding I to the 9. This gives 1-90, i. e., 1-9, or only 2 figures, which, however, cannot be avoided.
Example 34. — Express the number 3-1415928* correct to (a) 5 significant figures ; (b) 4 significant figures ; (c) 3 significant figures.
(a) 3-1416; (b) 3-142; (c) 3-14-
It should be noted that, after adjusting, the last significant figure is not reliable. Where several numbers which have been thus adjusted are used in a calculation there is a possibility of introducing errors beyond those of measurement. To avoid this it is usually advisable to carry our calculations to i figure more than is required, and to adjust to the required number finally. Thus a problem requiring an answer to 2 significant figures should be worked to 3, and the answer reduced to 2 finally.
Exercises 10. On Significant Figures.
What are the significant figures in the following numbers : — 1- 39-37: 5280; 2-205; 30-45; -0365.
2. 29,850,000; 10,500,000; -0807; 1083; 454.
3. -00000066; 606-5; 'OoniS; 32-09; 1-5708.
4. 10-9; '0000087; -0000105; 600-9; 2000.
* This particular number, which is very important in connection with circles, has been calculated to 700 decimal places !
DECIMAL FRACTIONS 43
Express the following numbers correct to 3 significant figures —
5. 453*6; 30-48; -4343; 62-425; -001118.
6. 14380; 1393; 32-182; 600-9; 3749-9.
Express the following numbers correct to 2 significant figures : — -
7. -0807; ^282; 14-7; -0646; 5-37.
8. 2-205; 606-5; *°i6o4 ; 29,850,000; '1145.
9. The Imperial Gallon contains 277-274 cubic inches. Express this
(a) correct to 4 significant figures ; (b) correct to 3 significant figures.
10. The British pound weight contains 453-5924 grammes. Express this (a) correct to 5 significant figures ; (6) correct to 3 significant figures.
Express the following numbers correct to (a) 5 significant figures ;
(b) 4 significant figures; (c) 2 significant figures —
11. -142857. 12. 2-71828.
Addition and Subtraction of Decimals. — Since ten is the (< rate of exchange " on each side of the decimal point, addition and subtraction with decimal fractions are carried out in exactly the same way as with whole numbers, provided that all the decimal points are kept in a vertical line, and the various decimal figures representing the same decimal place are directly beneath each other. That is to say, tens may only be added to tens and units to units, and similarly tenths may only be added to tenths and hundredths to hundredths, etc.
Thus, finding the total of 25-3, 12-57 an<^ °'^2«
25-3 12-57
O-02
38-49
d c b a
Explanation — Column a, 2nd place of decimals, 2 -{- 7 =9, write 9. Nothing to carry.
b, ISt „ „ 6 -{- 5 -[-3 — 1 4, ,, 4. Carry i.
,, C, units I -f- 2 -[- 5 ~ 8, ,, 8- Nothing to carry.
d,tens i-\-2 = 3, „ 3.
Result = 38-49.
Apart from mistakes in common addition, the only blunders likely to arise are from not writing the figures and decimal points in distinct separate columns, when figures of the wrong columns may get added together. Therefore great care should be given to the writing down of the figures.
44 ARITHMETIC FOR ENGINEERS
Example 35. — When considering the strength of a certain steel column it was necessary to add together the quantities 6-48, 0-312, and 2 '3 7. Find the value required.
6-48
0-312
2*37
! ! The 2 may be neglected.
9-162 Then result = 9*16.
In subtraction, the number to be taken away is written below the number from which it is to be taken, the subtraction being performed in the ordinary way. The previous remarks on the decimal point and the neat spacing of the figures should be noted.
Thus in taking 1*875 from 2-327 —
2-327 1-875
•452 = Result.
Example 36. — When measuring the difference of pressure between
two points of a pipe in which water is flowing, a mercury U gauge
(see Fig. 13) is sometimes employed. The top surface of the mercury
stands opposite -817 ft. on the scale, and the bottom surface opposite
199 ft. What is the " difference of level/' i. e.t the distance marked " ? "
Top reading = -817
Bottom ,, = -199
Difference of level = -618 ft.
Fig. 13.
Example 37. — In connection with the strength of a cast-iron rail section certain areas had to be measured with a planimeter.* The following three pairs of readings were taken, the area in each case being the difference of the pair of figures. Find these areas (sq. ins.).
(a) 19-45, 12-25; (&) 57'63» 56'52; (c) 7°'76> 69'09-
(a) 19-45 (&) 57*63 (c) 7°76
12-25 56-52 69-09
7-20 sq. ins. i-n sq. in. 1-67 sq. in.
* An instrument for measuring accurately, and with very little calculation, the area of an irregular figure.
DECIMAL FRACTIONS"
45
Example 38. — In finding the bending moment at a certain section of a girder it was necessary to take 62-2 ft. tons from the sum of 3*4, 45-6 and 37'6 ft. tons. Find this value.
3 »4 ft. tons 45*6
Sum
86-6 62-2
Difference = 24-4 ft. tons
Result = 24*4 ft. tons.
Example 39. — A portion of a surveyor's levelling book is shown. The height of the place called G above the place called A is found by subtracting the sum of column 3 from the sum of column 2. Fine? this height, all the figures being in feet.
|
Point. |
Back Sight. |
Fore Sight. |
|
A |
7-52 |
|
|
B |
4-09 |
4-41 |
|
C |
9'75 |
|
|
D |
4-14 |
I «I2 |
|
E |
2-01 |
2'73 |
|
F |
1-99 |
4-69 |
|
G |
8-78 |
|
|
Totals . |
29-50 |
26-89 |
Deduct Total Fore Sights 26-89
(*. e.t Col. 3)
2*61
Result : Place G is 2-61 ft. above place A.
The engineer has seldom to add up long columns of figures. A good deal of the decimal addition will be only with two or three numbers at a time, and hence the method of writing down in vertical columns, shown in the foregoing examples, is often omitted as being laborious. With a little practice one is soon able to sort out the last decimal place (say the fourth) in each number to be added, and to perform the addition of those particular figures mentally; then pass on to the next larger decimal place (say the third) and repeat the process ; and so on until the example is complete ; similarly with subtraction.
46
ARITHMETIC FOR ENGINEERS
The following examples will serve as illustrations —
Example 40 — In a problem on a retaining wall it was necessary to find the sum of the distances 1*19 ft. and 2-33 ft., and then to subtract 2-83 ft. from this sum. Find the resulting measurement.
We have 1-19 + 2-33 — 2-83 = 3-52 - 2-83 = -69 ft.
Example 41. — The expression 92*1 -f '56 — (25 -f- i?'9) relates to the flow of water through a pipe. Find the value required.
92-1 -f- -56 — (25 -f 17-9) = 92-66 — 42-9 = 49-76, say 49-8.
Exercises 11. On Addition and Subtraction of Decimals.
1. Find the sum of 10-5, 63-58 and 25-2.
2. Add together 67-35, 85-62, 105-14 and 5-7.
3. Add together 125-5, 67*825, 1-5625, 13-125 and 12-03125.
4. The total head against which a centrifugal pump has to work is made up of three heads : loss of head in pump 5-5 ft., loss by friction in the pipe 93-6 ft., and height through which water has to be lifted 250-5 ft. Find the total head.
5. In a laboratory experiment the following weights (Ibs.) were placed on a certain hook : -7, 2-9, 3, 3, 2-9, 2-9, 3. 3, 2, 2, 1-5, 1-5, i-i, 2, 3, 1-5. Find the total weight on the hook.
6. The lengths of the lines in a traverse survey were as follows : 13-63, 3-8, 6-97, 9-41, 3*48, 6-1 1, 9-66, 5-08, 6-84, 2-07 chains. Find the total length of line surveyed.
7. In calculating the bending moment on a continuous girder the expression 5 -j- 3-75 -f- 1-06 -f- 1-42 was obtained. Finish the calculation.
8. The following table gives the chemical analysis of some gaseous fuels. The combustible constituents are hydrogen, methane, and carbon monoxide. Determine the total number of parts of combustible matter in each gas.
|
Gas. |
Carbon Monoxide. |
Hydrogen. |
Carbon Dioxide. |
Methane. |
Nitrogen. |
|
parts |
parts |
parts |
par ts |
parts |
|
|
Dowsoii Gas |
25-07 |
i8'73 |
6-57 |
•62 |
49-1 |
|
Mond Gas . . |
10-00 |
23-00 |
I5-00 |
3-00 |
49-00 |
|
Natural Gas |
•41 |
1-64 |
•25 |
93-7° |
3MI |
9. Find the difference between 68-75 an(i 59'&5'
10. Find the result of 125-3 — 68-45 — 26-9.
11. In an experiment on linear expansion, the following were some figures (millimetres) taken for brass and steel —
(a) Brass. Initial reading -5 Final reading 2-3
(b) Steel. ,, ,, 3*215 ,, ,, 4*225
Find the actual expansion in each case (i. e., the difference between each pair of figures) .
DECIMAL FRACTIONS
47
12. The following figures were taken in a delicate experiment on the electrolysis of copper sulphate —
Weight of plate before depositing . . . 63-8203 grammes
,, ,, after ,, ... 64-5091 ,,
Find the weight of copper deposited on the plate.
13. In an experiment on the latent heat of steam the following weights were measured- —
Weight of beaker (empty) 264 -9 grammes
,, ,, „ -V cold water .... 1428-6 ,, ,, ,, ,, -f- cold water -f- steam condensed 1480-5 ,,
Find (a) weight of cold water used; (b) weight of steam condensed.
14. Find the difference (a) between '3575 and -5463, and (b) between "3575 and '6453, figures which occurred in measuring the quantity of water flowing over a V -notch.
15. Kxtreme sizes of a shaft and hole, nominally 3" diameter, when made to limit gauges are given below —
Maximum diameter of hole = 3-0035 in. Minimum ,, ,, ,, — 3-0000 ,,
Maximum ,, ,, shaft = 2-9965 ,,
Minimum ,, ,, ,, = 2-9930 ,,
Find the extreme clearances which can occur, i. e , find the difference in diameter between the largest shaft and the smallest hole; and between the smallest shaft and the largest hole.
16. and 17. The following tables arc from a surveyor's level book. Find the difference between the sums of columns FS and BS in each case (which gives the difference of level between the first and last points).
16. ~~ 17.
|
Pt. |
B.S. |
F.S. |
|
A |
•23 |
|
|
B |
3'4T |
13-96 |
|
C |
4-18 |
9\57 |
|
D |
4'95 |
|
Pt. |
B.S. |
F.S. |
|
A |
2-475 |
|
|
13 |
3-37 |
3*9 I |
|
C |
2-15 |
4") 9 |
|
D |
6-21 |
|
|
E |
2-13 |
5-33 |
|
F |
2-15 |
18. Figures from a surveyor's level book are given. If the levelling is correct the total of column 11 will equal the total of column F. Any difference is an error. Find the totals of columns R and F and find what error exists, if any. (All measurements in feet.)
|
Pt. |
B.S. |
F.S. |
R. |
F. |
|
A |
7*33^ |
— |
||
|
B |
10-265 |
9'455 |
— |
2-119 |
|
C |
3'44i |
5'°55 |
5-210 |
— |
|
D |
4-60 |
3-3^5 |
•076 |
— |
|
E |
6-61 |
9-885 |
— |
5-285 |
|
A |
— |
4-49 |
2'12 |
48 ARITHMETIC FOR ENGINEERS
19. Referring to the last exercise, the sum of column B.S should equal the sum of column F.S. Any difference is an error. Find what error, if any, exists.
20. Find the value of 3*75 — 1-42 + I'5> figures obtained in calcu- lating the bending moment on a beam.
21. One of the reactions of a continuous girder was calculated to be 2 -5 -f- I%5 "h 2*5 — -27, all in tons. Complete the calculation.
22. Find the value of the expression (connected with a bending moment) 12-7 — 47 — 15-55 + 12-3 — 1-75.
Multiplication of Decimals. — Let us consider the multiplica- tion of two decimals, e. g., •& X '2. Convert each decimal into a vulgar fraction, but omit the cancelling. Then '8 becomes ^ and »2 becomes T2^. The expression -8 x *2 may now be written as ^ x ^.
Multiplying (without cancelling) we have —
TV x TV ^ iVV *'• e-> * tenth + 6 hundredths,
or *i6 in the decimal form. It should be noticed that the figures in -16 are simply the product of the figures in -8 and -2 ; also that the total number of decimal places in the quantities multiplied is 2 (i in '8 and I in »2) and there are 2 decimal places in their result -16. Again, let us consider 2-45 X 1-3:
V.AZ _ <7 45 _ 2 iff
* 45 — * T<rff — i<7i7 » also 1-3 = if\ = {£. Hence we have ^g X | J = \\%§ = 3-185.
Again notice that the digits in 3-185 are the product of the digits in 2 '45 and 1-3 ; also the total number of decimal places in the numbers multiplied is 3 (2-45 has 2 decimal places and 1-3 has i), and there are 3 decimal places in the result 3*185.
Any example may be similarly treated and the same points will be seen. Hence we have the rule for the multiplication of decimals —
Multiply the figures together without considering the decimal point. Count up the total number of decimal places in the quantities multiplied, and starting from the right, point off this number of figures in the result obtained.
To illustrate this rule, and also the method of writing down the work, consider the second example just taken, i. e., 2-45 X 1-3. Either number may be taken as the multiplier ; but usually it will be found easier to take the one with the fewer digits by which to multiply, in this case 1-3. Set down the quantities as in ordinary multiplication ; the decimal point need not be placed in any particular position. Multiply, disregarding the decimal point (i. e.t as in ordinary multiplication).
Thus
DECIMAL FRACTIONS 49
Mental work to fix the decimal point
2*45 . , . , .2 decimal places 1-3 . . . . .1 decimal place
735
245 _
3-185 . .... 3 decimal places.
This gives the figures 3185.
Now fix the decimal point thus : Count the number of decimal places in 2*45 (two) and add this to the number of decimal places in 1*3 (one), making 3 decimal places in all. Starting from the right in the product 3185 count off 3 digits and place the decimal point : thus point off the 5, the 8 and the i, and then write the dot. Thus the product of 2-45 and 1*3 is 3-185.
Example 42. — If i cubic foot of water weighs 62-5 Ibs. and cast iron is 7-2 times heavier than water, what is the weight of i cubic foot of cast iron ?
i cu. ft. of cast iron will, of course, weigh 62-5 X 7*2 Ibs. 62-5 i decimal place
1250 4375 450'QO 2 decimal places.
The total number of decimal places in 62-5 and 7-2 being two, two digits are pointed off from the right in the product 450-00. .'. i cu. ft. of cast iron weighs 450 Ibs.
Occasionally, when a large number of decimal places are required to be pointed off in the result, it will be found that there are not sufficient figures available. Then a nought must be supplied for every required figure which does not exist in the product obtained. The following example will illustrate.
Example 43. — In finding the area of a small circle it was necessary to multiply together the numbers -785 and -0039. Find this value.
Again the number with the fewer multiplying figures is taken as the multiplier, i. e.t -0039.
•785 . . 3 decimal places
•0039 4
7065 2355
•0030615 . . 7 decimal places. .*. Result = -00306
50 ARITHMETIC FOR ENGINEERS
The multiplication of 785 by 39 gives the figures 30615 as result. Seven decimal places being required and five figures only existing, two noughts must be inserted to the left of the 3, in order to complete the seven decimal places. The answer then becomes '0030615. The last two figures may be neglected and the result given as '00306.
The foregoing method can be used for the multiplication of more than two numbers. In these cases a large number of figures will be obtained in the result, and many will have to be discarded as useless. Much labour can be saved, and many useless figures avoided, by reducing the result each time to 3 significant figures before using the next multiplier. In the following example the multiplica- tion is worked in ordinary " long hand " manner, and also by the method just suggested. It will be seen that there is a big saving in labour, and the difference in the results is not worth considering. To 3 significant figures the results are the same.
Example 44. — Evaluate 1670 x 2-75 X 3-73 X 6-78.
Long hand working. Working when each result is
reduced to 3 sig. figs.
1670 1670
275 2-75
835° 835°
11690 11690
334° 3340
373 say 4590 to 3 sig. figs.
137775 373
321475
17130-025 1377°
6-78
17120-70
137040200
say
119910175 y
102780150
116141-56950 I368°°
^ 119700
102600
115938-00 to 3 sig. figs. = 116000 to 3 sig. figs. = 116000
DECIMAL FRACTIONS 51
It can be seen from the above that the multiplication of decimals becomes very long and tedious when more than two numbers are multiplied together. There are " contracted " methods of multi- plying which are sometimes advocated, but their value is very doubtful, and they are probably never used in practice. The majority of engineering multiplication (and division) can be carried out with sufficient accuracy by logarithms (see Chap. VI), which are very simply and quickly used and certainly should be known by any one attempting calculation seriously.
Exercises 12. On Multiplication of Decimals.
(Note. — Nos. i to 10 are set to test the student's accuracy of multiplying; the answers should be given both fully and to 3 sig. figs.)
I. Multiply 29-5 by 1-38. 2. Multiply 235 by 1-72. 3. „ -057 „ 13-9- 4. „ 15-95 .. 785- 5. „ 3-14 ,, 19-6. 6. ,, -0087 ,, -054. 7. ,, 13400 ,, -0036. 8. ,, 191*5 ,> '0986. 9. „ -0046 ,, 21-2. 10. ,, 1114 »» 19*25.
II. If a cruiser's speed is 27 knots, what is this in miles per hour? (i knot = I -15 miles per hour.)
12. The speed of a cargo boat is loj knots. The speed of a tramcar is 12 miles per hour. Which vehicle is the faster ?
13. The calculation of the *' moment of inertia" of a cast-iron rail section (a figure used in connection with the strength of the section) gave the figures 2 X 1*78 X 2-7 X 2-7. Calculate the required value.
14. In calculating the weight per foot of a certain angle bar the following figures appeared : 7*11 X 12 X '28 Ib. Finish the calculation.
15. The following calculation was necessary to convert the readings of a water meter into Ibs. per min. : 60 x 62-5 X -026. Find the required value.
16. The calculation for the weight per foot of a large wrought-iron pipe produced 10-25 X '7^5 X 3-4 Ibs. Find the required value.
17. The working volume of an engine cylinder was given by the expression 36 X 3*14 X 24 cu. ins. Find the required value.
18. Find the value of the following figures, which give the amount a certain rail will expand under certain conditions : 360 X 25 X -0000087 inches.
19. Pressures in connection with compressed air are frequently spoken of as " atmospheres," i atmosphere being 14*7 Ibs, per sq. in. Find the number of Ibs. per sq. in. in 35 atmospheres.
20. Find the number of Ibs. per sq. in. in 32 J atmospheres.
21. The area of the steam cylinder in a steam-driven air compressor
52 ARITHMETIC FOR ENGINEERS
is to be 1*38 times the area of the air cylinder. Find the steam-cyhnder area when the air-cylinder area is 113 sq. ins.
22. Find the total resistance of 2 miles of electric cable whose resistance per yard is -008 ohms (i mile = 1760 yards).
Division of Decimals. — Let us first consider the case where the divisor is a whole number, e. g., 1175 -r 5- Express 1175 as an improper vulgar fraction without cancelling, and it becomes -VoV"' Divide by 5 according to the method stated in Chap. I, for the division of vulgar fractions, i. e.t divide the numerator. In this case divide 1175 by 5 = 235.
Thnn TT-TS ~ < _ IITB • ^ __ 235
men n 75-7-5 — flro- -r 5 — 100*
Expressed decimally the result is 2*35. Now it will be seen that* the digits in 2-35 have been obtained by dividing the digits in 1175 by 5 in the ordinary manner ; also that the whole number of the quotient is complete when the whole number in the 1175 is finished with, i. e., 2 is obtained when u has been divided. Or, when the decimal point, is reached in 1175 it is also reached in the 2-35.
Hence to divide a decimal by a whole number — Divide the digits in the ordinary manner, and place the decimal point In the result when the decimal point is reached in the number being divided.
Then, the working of the above appears thus — •
The decimal point is reached when n has been divided by 5, and therefore comes after the 2 in the result.
Now take the general case of a mixed number divided by a mixed number, e. g.f 8-672 -7- 271. By a simple adjustment the example can be converted into one of the foregoing nature and worked in a similar manner.
8*672 Let us write the example in the form of a fraction, thus ' .
Now on p. 9, when considering cancelling, it was shown that a fraction may be multiplied top and bottom by the same
DECIMAL FRACTIONS 53
number without altering the value of the fraction. Applying this here let us multiply our fraction, both top and bottom, by loo.
8-672 X 100 867-2 (See p. 37, on movement
Then we have — ^ x IOQ = ~^~ of decimal point.)
The example is now similar to the first one taken in this section, with the exception that " long " division is more convenient.
(3»2 Explanation. — Divide in ordi-
813 nary way. Place decimal point in
• - quotient when it is reached in num-
542 her divided, i.e., 271 into 867 goes
542 3, and since decimal point is then
reached in 867-2, it is placed after the 3 in quotient.
A rough test is available: 8-672 is nearly 9, and 2-71 is nearly 3 ; dividing 9 by 3 gives 3, which is in agreement with our result 3-2.
If the divisor contains three decimal places then multiplication by 1000 would be necessary, instead of 100, to convert to a whole number. That is, the decimal point would have to be shifted three places to the right in both numbers, and so on.
From this we have the following rule for the division of decimals : — Convert the divisor into a whole number by shifting th* decimal point to the right. Move the decimal point in the dividend th* same number of places, also to the right, and then perform the division in the ordinary way. The decimal point must be placed in the quotient when it is reached in the number divided. All cases may be dealt with by this rule.
The following examples and remarks serve to illustrate certain special points that arise. When the divisor is already a whole number (a case of frequent occurrence) then no shifting of the decimal point is necessary.
It will frequently be found, when doing the actual dividing, that there are not sufficient figures in the dividend to " bring down." In this case a nought may be brought down each time, since a number like 34-5 may, of course, be written as 34-500. The following example illustrates.
54 ARITHMETIC FOR ENGINEERS
Example 45. — Divide 3-45 by 1*2.
Move the decimal point in 1*2 one place, making 12
3'45 » •• 34'5
Then divide —
I2)34'5(2*875 Note. — 12 into 34 goes 2. Deci-
24 mal point now being reached in
• — 34*5 place it after the 2 in quo-
IO5 tient. Continue as in ordinary
96 division.
90 84
60 Go
Hence 3-45 -r 1-2 = 2-875.
It will soon be found when dividing, that very few examples will work out " exactly,1' as have those taken up to the present. This is especially so with examples of a practical nature, and in these cases the division should not be carried too far. Thus if 2375 be divided by 3*14, the answer is found to be 7*5636942, while yet leaving a remainder ! It would be ridiculous to give a result of anything like this accuracy. In the first number (2375) there are only 4 significant figures, while the second (3*14) only contains 3.
We may therefore reckon with safety that 3 significant figures would be sufficient in the answer ; but the division should be carried to 4 significant figures in order that the third may be as correct as possible. Thus the division can be stopped when the quotient reads 7-563. The answer is then stated as 7-56.
The general rule of 3 or 4 significant figures in the final answer may be followed in all cases, unless other instructions are given.
When shifting the decimal point in the dividend it will some- times be found that sufficient figures do not exist, in which case noughts may be added as desired. This is shown in the following case —
Example 46. — Divide 231*4 by i'938.
The dot has to be moved three decimal places in 1*938, but only one place is available in 231-4. Hence two noughts must be added to
DECIMAL FRACTIONS 55
make up the three places, and the example appears as ^J-j-ijp, the decimal point in the numerator standing after the second o.
1 938)23 1400 (119-40 1938 3760 1938
18220 17442
<— When this point is reached the decimal
778° point is placed in quotient.
280 Result = ii9*4'
Note. — Four sig. figs, are given in result as there are four in each of the given numbers.
When the number divided is a small fraction it is possible that even after shifting the decimal point there will be no whole number. The procedure in this case is shown by the following example —
Example 47. — Divide -01193 by 2*3. Making the divisor 23 the dividend becomes -1193.
"5
43
23
- Result = -00519.
200 — _
184
- (This can be checked roughly by i Co multiplying -005 by 2-3.)
138
Explanation. — The first thing met with in the dividend is the decimal point. Therefore write the dot in the quotient, for if a fraction be divided by a whole number the result must be a fraction. Try if the first figure in the dividend can be divided by the divisor. If so divide, if not put a nought in the quotient and then try the first two figures, and so on. Thus 23 into i won't go ; put a nought after the decimal point. 23 into n won't go; put a second nought. 23 into 119 goes 5 times. Put 5 in the quotient and continue as in ordinary division.
ARITHMETIC FOR ENGINEERS
Exercises 13. On Division of Decimals.
1. Divide 219-5 DV 5- 3. „ 413-8 „ 22-75. 5. „ 1*897 .. 21-35. 7. ,, -0076 ,, -016. 9. „ 621 ,, -31.
2. Divide 32-73 by 7. 4. „ 135°° i. 4'36. 6. ,, -158 ,, -0065. 8. „ -098 „ 11-57. 10. ,, 33000 „ 117.
11. If 14-7 Ibs. per sq. in. = i atmosphere, find how many atmospheres in the following : (a) 165 Ibs. per sq. in. ; (6) 90 Ibs. per sq. in.
12. The following figures are extracts from the results of an experi- ment to determine the coefficient of discharge of a rectangular notch.
Actual discharge
The coefficient is the value coefficient in each case.
Theoretical discharge"
Calculate the
|
Actual discharge. Cu. ft. per sec. . |
•0372 |
•0531 |
•1085 |
|
Theoretical discharge. Cu. ft. per sec. |
•0571 |
•0829 |
•1750 |
|
Coefficient ... .... |
|||
13. If i gallon of water == 10 Ibs., and 2-205 Ibs. = i kilogramme, find how many kilogrammes there are in i gallon of water.
14. If i kilogramme — 1000 grammes, and 453*6 grammes = i lb., find how many Ibs. there are in i kilogramme.
15. If i kilogramme = 2-205 Ibs., and i cwt. = 112 Ibs., find how many kilogrammes go to i cwt.
16. If 746 watts — i horse-power, find how many horse-power = i kilowatt (i. e., 1000 watts).
17. The reaction of a plate girder (i. e,, the upward force at the support) is 97 tons, and i square foot of surface is required for every 20 tons of load. Find how many square feet are required,
18 to 20. In certain experiments the quantity of water delivered by a pipe in a certain time was measured with the results below. Find in each case the quantity ot water passing through in i minute.
18. 100 Ibs. ot water passed in 4-55 minutes.
19. ioo ,, ,, „ 2-37
20. 94 » „ >. 3 '02
21. Calculate the resistance per foot (i. e., ohms for i foot) of an electric cable which is 520 yards long, and has a resistance of 20*7 ohms.
22. If 62-5 Ibs. of water occupy a volume of i cubic foot, find the volume occupied by i Ib. of water.
Conversion of Vulgar Fractions to Decimals. — In Chap. I, p. 7, it is shown that a vulgar fraction is the result of dividing one number by another. If now this division be performed as just described for decimals, a decimal answer will be obtained, and hence our vulgar fraction will have been converted into a
DECIMAL FRACTIONS 57
decimal fraction. Therefore to convert a vulgar into a decimal fraction : Divide the numerator by the denominator.
Example 48. — Convert the following vulgar into decimal fractions : —
(«) i; (^ i; to A
(c) 16)7-0(0-4375 (a) 4)1-00 64^
*25" " 468°
•'• A ^ '4375:
(fc) 8)1-00 r*o ---"*
~~. "'* ~ I2~*
or 8 — J o J 8o
= * of -25 = '125 —
With the more uncommon vulgar fractions that appear, the division may not work right out without remainder. Then the rule of 3 or 4 significant figures may be followed.
Example 49. — The value of tr,* a constant of great use in calculations with circles, may be given by the fraction 3i\V Express this decimally.
113)16-0(0-1415
47° 45*
•*• 3\1A = 3-14 to 3 sig. figs, 1 80 •*"— -
Example 50, — In a certain experiment the following intervals of time were read with a stop-watch — •
(a) i min. 42 sec. ; (b) 2 min. 35 sec. ; (c) i min. 13 sec. For the purposes of calculation it is desired to state these times as minutes only. Find the necessary values (i. e.t convert the seconds to decimals of a minute).
i min. = 60 sec.
(a) i min. 42 sec. = ijft min. = i/^ cancelling by 6 = 1-7 mins.
(b) 2 min. 35 sec. = 2gJ min. = 2^ cancelling by 5
2^ = 2-583 min. or 2*58 to 3 sig. figs.
(c) i min. 13 sec. = ijg min. = 1-217 mm- or I<22 *° 3 sig- fig3-
* Greek letter "pi."
58 ARITHMETIC FOR ENGINEERS
Exercises 14. On Conversion of Vulgar Fractions into
Decimals.
Convert the following into decimal fractions —
1. f. 2. &. 3. V«. 4- i!- 5. J|.
6. &. 7. A. 8. fl. 9. ft. 10. 3l\V
11 to 15. In certain experiments measurement of time had to be taken in minutes and seconds. Convert the seconds into decimals of
a minute in the following cases (i.e., convert the fraction — into
a decimal, since i min. = 60 sec.) —
11. 18 s.ec. 12. 34 sec. 13. n sec. 14. 54 sec. 15. 13 sec.
16 to 20. The " pitch " of a screw thread is > ~, — —j- per inch.
Find the pitch as a decimal for the following cases : —
16. 24 threads per inch. 17. 18 threads per inch.
18. 9 threads per inch. 19. 4-5 threads per inch.
20. 3 '5 threads per inch.
21 to 24. Find the pitch, in decimals, of the following Whitworth Gas Threads : —
21. ii threads per inch. 22. 14 threads per inch. 23. 19 threads per inch. 24. 28 threads per inch.
25. The " go in " end of an internal limit gauge is -0004" smaller than the nominal size, and the " not go in " end is »ooi2" larger than the nominal size, the latter being if". What are the actual sizes of the ends ?
Compound Examples. Approximation for Result.—
Some of the calculations that occur most frequently in engineering problems are the evaluation of expressions containing multiplica-
j i- • • t- 3*28 x 58*7 x 58-7 ™. ,
tion and division, such as ~~r — — L* Ihese may be
32 x 600 J
worked out on the lines of the previous paragraphs. Cancel where possible. Then the product of the numbers above the line may be obtained; also the product of the numbers below the line. Finally the division is performed. The number of figures in the calculations is often very large, and, in consequence, there are plenty of opportunities for making blunders and slips in the working. The most common mistake is that of wrongly placing the decimal point, and in many cases the blunder may pass unnoticed. Some- times the nature of the problem gives a check on the accuracy of the result, e. g., if on calculating the thickness of a boiler plate a result of 7-5* was obtained, then there is evidently a mistake in the working; ^\" boiler plate is obviously absurd, and an examina- tion of the calculations would no doubt show that the result should be *75*> which is reasonable. But in cases where there is no such guide, and, in fact, in all problems, it is advisable to get a
DECIMAL FRACTIONS 59
rough idea of the result before doing the actual detailed working. To those who use a slide rule, or intend to do so, this " approxi- mating for a result " is of great use, for the slide rule really works with the significant figures only, independently of the position of the decimal point.
The general method of approximating is to replace the actual figures by some near convenient round numbers, so that, using these, the approximate calculation becomes a very simple one, and may often be done mentally. The following example will illustrate :—
Example 51. — Find the approximate value of ?53_2?_1!?.
63
The 293 may be called 300^ This is mental work and is » 4*9 » »» 5 r only written here as an
,, 63 ,, ,, ,, 60 j explanation.
Then our approximate expression would be ^~. — ^ and cancel-
ling by 60.
5
I
Hence our result must be about 25 ; it may be anything from, say, 20 to 30, but must have two figures in the whole number.
The actual result is 22-8 to 3 significant figures. Now supposing our detailed working had been bungled, and a result of, say, 2-28 or 228 had been obtained : the mistake would have been shown by our approximation of 25.
While the method just shown is suitable for simple examples, it is not accurate enough for more complicated ones. Also, for those readers who are not accustomed to much mental calculation, a more systematic method is desirable.
Considering the expression -9-- ---- 529 --- 7
b F 31 X 1-8
let us shift the decimal point in each of the compoi_ ..... ________
until a units figure only stands in front of the point (as in the r8) ; at the same time multiplying or dividing by the necessary number of tens to preserve the value of each number. Thus in the top line —
19*5 becomes 1-95 x 10 529 „ 5-29 x 10 x 10
and -67 „ ~7- ' " 10
while in the bottom line —
31 becomes 3*1 X 10 and the r8 remains unchanged, being already in the form required.
60 ARITHMETIC FOR ENGINEERS
Using these " converted numbers/1 we may rewrite the given expression as —
(1-95 x 10) X (5-29 x io X 10) (3-1 x io) x r8
For convenience, let us separate the tens into one expres- sion (b), and the remaining numbers into another (a), thus —
^95_x_5!19 * J?7 x IQ x I0 x IQ
~ 3-1 x 1-8 ~ io x io
(a) (b)
Expression (b), by cancelling, evidently reduces to io. Now replace each number in (a) by its nearest whole number, i. e.t write 2 in place of 1*95 and 3 in place of 3*1, and so on. Then we have —
-^X5*7XIO 3 X X
(a) (b)
- f x io
= 12 (very nearly) x it)
= 120
which is then an approximate answer to the given expression. The actual answer, it may be noted, is 123-9.
In practice this method may be considerably abridged. Thus in expression (a) the " nearest whole numbers " may be obtained at one operation from the values in the given expression, the inter- mediate step being a mental one. For instance, taking the 19*5, the btep 1*95 is performed mentally, the 2 only being written down ; again, taking the "67. the step 6'7 is mental, 7 only being written. Also, the result of expression (a) is not required with great accuracy ; thus in the foregoing -./- is called "12 very nearly/' since the ex- pression -3/ would give 12. A certain amount of caution in this direction is necessary, however, otherwise some of the " approxi- mate " results may be too far from the truth. In expression (b) a stroke (i) may very conveniently be used to represent each io, while the X signs may be omitted, since the only operations in- volved are those of multiplication and division.* These strokes may be cancelled in the usual way, and every stroke remaining
* It must be clearly understood that this method of approximating cannot be applied to expressions involving -f and — signs.
DECIMAL FRACTIONS 61
after the cancelling will, of course, signify a 10. Instead of expres- sion (b) in the foregoing we should then have-^-, the remaining
stroke on the top line indicating that the result of expression (a) is to be multiplied by 10; similarly a stroke beneath the line would indicate division by 10. These " strokes " are perhaps best placed at the right-hand side of the sheet, and will be shown in this position in the succeeding examples, under the heading of " Tens/'
In connection with these " strokes " (i. e., tens), it is seen from the detailed working already shown, that when a " converted num- ber " contains a multiplying 10 the stroke must appear on the same side of the division bar as the number itself; but when con- taining a dividing 10, the stroke must appear on the other side of the
bar. Thus, in the case of the -67 ( which = — ~ \ the stroke must
be under the bar, while the number itself is on top, i. e.> the stroke is on the other side. Note also that in a number containing a mul- tiplying 10 the decimal point is shifted to the left when obtaining the " nearest whole number/' and in one containing a dividing 10, to the right. Considering all these points, we obtain the following rules for approximation —
Taking each number in turn, shift the decimal point mentally until there is a units figure only in front of the point. Find the nearest whole number to the quantity so obtained, and write it in a new expression (called (a) in the foregoing working) ; at the same time place a stroke in an auxiliary expression (called (b) in the foregoing work- ing) for every place that the decimal point has been shifted, according to the following rules —
When the point is shifted to the left place the stroke (or strokes) on the same side of the division bar as the number occupies in the original expression; when shifted to the right place the stroke (or strokes) on the opposite side of the line.
Work out the simple expression (a) and cancel the " strokes " so far as possible. For every remaining stroke, multiply or divide the result of (a) by 10, according to whether the strokes are above or below the line. This gives an approximate result.
// is important to notice that this method, although lengthy in description, is very quick and easy in actual use.
62 ARITHMETIC FOR ENGINEERS
Example 52. — The expression ^ amperes had to be
' J * -000328 X 1200 *
evaluated in connection with an experiment on electrolysis. Find the required value.
Approximation —
____4 __ 3 X 1-2
4
= I very nearly.
Tens —
no strokes remaining,
/. no multiplication or
divisions by 10. Therefore the result is about i ampere.
Explanation of Approximation.
Top line. — Convert -39 to 3-9 and write 4. Decimal point having been moved one place to right, place a stoke below the line.
Bottom line. — Convert -000328 to 3-28 and write 3. Decimal point moved 4 places to right, place 4 strokes above the line.
Convert 1200 to 1-2 and write i.
The strokes all cancel out, so that the approximate answer is the result of the first expression, i. e.t about i.
Working in full — •
•000328 -39 _ 390
1200 -394 ~ 394
394)3900(0-9898 3546
•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&« fiSs- are *3-
Approximation — Tens.
* XJL - i _ .2<
'A v fc — * — -5 4 A ^
•25 = approximate result .'. Actual result must be -23, this being the nearest number to •25 that can be made from the sig. figs. 23.
DECIMAL FRACTIONS
Approximation —
5 X 33°°° X 12 c. -.
— ^ ^ ~ Sig. figs, are 209.
35 X 144 X 3-14 X 6 b b J
Tens —
_ 5 ""
= about J = -2 .". an approx. result = -2 X 100 = 20
.'. Actual result must be 20-9.
\\\
Two strokes above.
/• Multiply by 10 X 10
= IOO.
Approximation —
'- — = about 5
i X 1-5 D
/. an approx. result = — - — = 11 jooo
.'. Actual result = -00434.
arc 434.
Tens—
Tlf /. divide by 10 X 10 X 10
= 1000.
Note. — When a number, after being "converted," is nearly 1-5, as with the 144 at (b) above, it is best to call it 1-5 rather than i or 2, to avoid an answer which is too approximate (see also c above) .
Example 54. — Determine, without actually working out the expres- sions, whether the answers given below are reasonably correct. , N -000267 -
-^0852
= '°314
Tens —
J1L.
= J = -3 UU
.*. divide by 10.
/. Result is about ~ = -03, and answer given is reasonably correct.
Approximating —
4 X 7-14 X 60 "4rx~6"28 X 2-1
Approximating
Tens— 1
.'. multiply by 10.
/. Result is approximately 2 X 10 = 20, and answer given is reasonably correct.
(c) 1-41 x 182 x -00034 x i -08 x 187 x 187 = 33200.
Approximating —
1X2X3x1x2x2= about 24.
.*. Result is about 24 x 100 = 2400.
Tens— 11UU
UU /. multiply by 100.
64 ARITHMETIC FOR ENGINEERS
.*. Given result is incorrect, being 10 times too large, and should be 3320.
(j\ '00307 X '0123 x 8 02 x ro2
'
Approximating —
3X1 X 8xi
I
24 about
= -26
Tens — • \\
\\rn~
.". divide by 1000. /. Given result is incorrect, and should be '026.
Example 55. — A calculation for the indicated horse-power (I.H.P.) of a petrol engine reduced to the figures
.'. Result is about — -— — -024.
T«r»n •
Tens
„ + ^ ,03 ^ _sj>* -iA_^_^. Find 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 'I4°
mate for the answer, and, given that the significant figures are 6009, state the calorific value (in B.T.U. per cu. ft.) correct to the units figure.
Approximation —
Tens—
U
.*. multiply by 1000. .*. Approximate result = -6 X 1000 = 600 and actual result — 600-9. .'. Calorific value to units fig. = 601 B.T.U. per cu. ft.
Exercises 15. On Approximating.
1 to 8. Determine, without actually working out the expressions, whether the answers given in the following are reasonably correct.
2. — ^£6_ = 6o6oo.
. . . _
•00136 -00341 x -7
X 9 X
32 X
9*87
3 I'3I x 33QQQ X 150 = . 4 X 9 X 38 x 3-14 =
6400 X 32-2 J 5< " 32 X 144 X 6 55'
K
5- - = '494'
< 3-aa X 900
7. 2x-785x 6-25x140=1375. s. *'°75 x x 8o'4
DECIMAL FRACTIONS 65
9 to 18. Find an approximate answer to each of the following expressions. Given also the significant figures in the results (obtained, say, from a slide rule), state, from an inspection of your approximate answers, what the actual results are (i. e.t put the decimal point in its correct place among the given figures).
9« v.jT-^TTg SiS- fi6s- in answer are 478.
10. I000° x "5 X -5
3-25 X 12 " " " ^'
11 10° X IIP X_2640
33000 " " "
_ __
30 x 1-06 X 10-56 " " " 7I4*
?i2<_3-i4_x 1-5 X 24
- - 4 x-3l „ „ ., 214.
3-14 X 24 X 1000
- ~~~~
15 ? x goo x 32-2
* 9:67rx^7~x~:6 " " " 44'
16 4 X 3-14 X -18 X_7£\5 X_72_'5
•342
17. -5 35-_.— 5 ~~ ., „ „ 245.
12
7 x -0625 x 2240 6 X 232
18.- z-TT^-n „ „ „ 705.
In some examples expressions appear with the four simple arithmetical rules in various combinations. Then the best method of working is to preserve the general form of the expression throughout the various steps, as was adopted in Chap. I, p. 30. The following example illustrates : —
Example 57. — In calculating the safe eccentric load on a stanchion, the following expression was arrived at : —
-- 6- — x~6- — tons- Complete the calculation. 1 + ^6~x~ir36
The first step necessary is to evaluate the right-hand term under the line.
tons («) 6'5 X 6-5 = 42-3
- - 6*5 X 6-5
5-36 ~
. + ~6 <*> 5-36 X 5-36 = 28-7
tons = 96*4 tons
66 ARITHMETIC FOR ENGINEERS
Exercises 16. Compound Examples.
1. The height of a mercury barometer is 30". The height of the water barometer in feet is found by multiplying this figure by -yj»
Find the required height.
2 and 3. In finding the centre of gravity of two rolled sections the following expressions had to be evaluated. Find the values required.
2. I^LX 57 inches< 3. 1*7*. 5L5_inches.
7-19 3*6
4. The current to pass through an electro-magnet was calculated
to be ^ - %—~ amperes. Complete the calculation.
5. The resistances in an electric circuit, which is to carry a current of 4-5 amperes, are 3-5 ohms, '75 ohms, and 2-38 ohms. The voltage (volts) required is equal to " current (amperes) X total resistance (ohms)." Find (a) total resistances; (b) voltage required.
6. The bending moment at a point on a continuous girder was
given by the expression — -- - — . Find the required value.
7. A stream is found to deliver 48 cu. ft. of water per sec. with a total fall of 1 20 ft. Find the horse-power of the stream, which is
it. 48 X 62-5 X 120 given thus : - --- J -- . b 550
8. Find the thickness of plate for a steam boiler 6'-6" diameter, to work at 100 Ibs. per sq. in. pressure, from the following expression :
100 X 78 . , ----- / - inches. 2 x 9000 x -7
9. A dynamo is to supply 600 .lamps, each of 16 candle-power. Each lamp takes 1*2 watts for every candle-power, (a) How many watts must the dynamo supply? If the voltage is no, the current
watt s (amperes) supplied is --- -- - (b), what current is supplied?
10. The tensile test of an iron wire showed that the " Modulus of
Elasticity "= 3 4 - Ibs. per sq. in. Complete the calculation
J '105 X -00104 r *i r
(3 sig. figs, will do).
11. The following figures were obtained in a torsional test on
wrought iron : IO'2 X ^8 X 5Jl3 X 5I'8 Ibs. per sq. in. Find the •300 X *3oo
required value.
12. A coal gas was tested for its " calorific " or " heating " value. When -146 of a cubic foot of gas was burnt, the temperature of 4-9 Ibs. of water rose from 51° F. to 69° F. Calculate the calorific value (in B.T.U.) which equals Lbg^of.wateLX rigejntemperature
' ^ cu. ft. of gas burnt
13. Find the value of the expression 6-48 -j- '3125 -f- 4*68 x '506, figures obtained when calculating the strength of a stanchion.
14. Calculation in connection with a column gave the expression —
- ------ tons. Complete the calculation.
DECIMAL FRACTIONS 67
15. Evaluate the expression 2 x (21*6 + 246*5) + 427, which refers to the strength of a built-up girder.
16. The bending moment on a girder, at a certain point, was found to be given by the expression 2S^±J°^±± «» + 26 + 32) tons ft.
4O
Complete the working.
17. Find the value of the expression 5 x (9*5 + 8-8 + 5*6) + 15 X (7-8 -f 6-7) tons ft., which is the bending moment on a bridge truss.
18. One of the reactions of a continuous girder was found to be
•5 X 20 , '5 X IO . 16-45 , 16-45 — !V75 ± t^- • t_ AI_ i •
_J_^ -- ^ J> ----- -- ±2 ^ -- 5^ --- 2-L2 tons. Finish the working.
2 2 20 10 b
19. The weight of a proposed plate girder was estimated to be
_85L2LZ2_x_i4_ tons. Find its amount. 1400 x 7 — 70 x 14
20. In calculating the safe eccentric load on a column the following expression was arrived at: —
— ___4 ---- tons. Complete the calculation.
T + 3'15_X_5_ P "*" 2-82 x 2-87
21. Find the value of the expression 130 -f- '85 (606-5 — -7 x 170).
22. Evaluate _ 4 lbs per sq in ^ which occurred
in a steam engine calculation.
23. The following figures were obtained when calculating the theoretical amount of air required to burn i Ib. of oil- —
4*35(2-26 -f 1*2) Ibs. Find the required value.
24. Complete the following calculation, which refers to the mean effective pressure in a steam engine —
75 X -67 X i'4°5 — 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 -S48~ = 9^ tons. This supposed constant consumption is called the average value or mean value. A common way of making the statement would be, " On the average the consumption is 9^ tons per week/' meaning that the consumption per week is in the neighbourhood of g\ tons. From the above we obtain the rule : —
To find the average of a series of quantities, find the sum of all the quantities and divide by the number of quantities.
There are many applications of the use of the average, as will be seen from the examples and exercises in this section. Particularly, when testing or experimenting to find the value of a certain quantity, the experiment is repeated three or four or more times and slightly different results will be obtained owing to errors of instruments, etc. Then the average of all the measured results is found, and we may safely assume that this average is nearer the actual truth than any of the measured values, as the errors have been spread over the whole set of numbers.
Example 58. — A wrought-iron shaft under a torsion (twisting) test had its diameter measured in several places, the values being •625", '62I", -626", -623", -624", •622"', '623', -62I*. Find the average diameter of the shaft.
•625- •621 •626 •623
•624
•622 Average diameter = •623"'
•623 """""
•621
8)4-985 •6231
It should be noticed that the accuracy of the average should not exceed the accuracy of the given figures.
DECIMAL FRACTIONS
69
A case of common occurrence to the engineer is the " averaging " of an indicator diagram from a steam or gas engine for the purpose of finding the indicated horse-power. Roughly an indicator dia- gram is of the form shown on the left of Fig. 14, and as will be seen in Chap. VII, the idea of " averaging," i. e.t finding the average height, is to get a figure equal in area to the diagram, but with a constant or unvarying height as on the right of Fig, 14. Certain
This hei
lines »n
HV is t"He of Irfce dotted Irhe. diareuvt.
Pig. of same avca ID of consVanV' and avea
diagram
The figures ave e^ual m Fig. 14. — "Averaging " an Indicator Diagram.
lines (called ordinates) are drawn, as indicated by the dotted lines, and measured, and the average of these heights is found.
Example 59. — The ordinates, in inches, of an indicator diagram from a steam engine, were as follows : -47, '72, '82, '83, '76, '60, ^44, •28, '2O, and -08. Find (a) the average ordinate. One inch of ordinate represents 70 Ibs. per sq. in. Find (6) the average pressure in Ibs. per sq. in. (called the mean effective pressure, or M.E.P.).
Oj
.S
*s (
'47 •72 •82
•83 •76 •60
'44 •28
•20
(a) Average ordinate = '52*
i* of ordinate = 70 Ibs. per sq. in.
/. •52// of ordinate = 70 x '52 = 36-4 Ibs. per sq, in.
(b) M.E.P. = 36*4 Ibs. per sq. in.
10)5*20 •52
Example 60. — The readings of temperature in the table on p. 70 were made in an experiment on the calorific value of coal gas. Find (a) the average inlet temperature, (b) the average outlet temperature, and.
7o
ARITHMETIC FOR ENGINEERS
(c) the rise in temperature (i. e.t average outlet temperature, — average inlet temperature).
|
Inlet temp. "C. |
Outlet temp. CC. |
|
9H7 |
30-62 |
|
9-46 |
31-^3 3I-80 3I-90 |
|
32-6I 32-82 |
|
|
9-42 |
33-13 33-22 |
|
3)28-35 12)385-88 |
|
|
9-45 32T57 |
Exercises 17. On Averages.
1. Three separate experiments were made to determine the " re- fractive index " of a certain kind of glass, the results being 1-52, i'5Q5 and 1-49. Find the average value.
2. Experiments to determine the "coefficient of discharge" of a V -notch gave the following results ; -575, -574, '580, -581, -581, '581. Find the average value.
3. In measuring the internal resistance of an accumulator the experiment was performed five times with the following results : -025, •028, -03, -0286, -0275 ohms. Find the average resistance.
4. The diameter of an iron wire measured in several places for a tensile test was as follows : -0365, '0346, -0365, '0364, -0358, -0368, •0365, '0348 in. Find the average diameter.
5. The following values of the coefficient of friction between two wood surfaces were obtained in an experiment : '417, '4°9» '367, 395, •37, -389, -403 and -415. Find the average value.
6. During a boiler test the following readings were taken on the water gauge measuring the supply of feed water to the boiler : 31*6, 3i-7. 3i-7. 3i-6, 33, 34'5. 32'4> 26-5, 20-0, 23-8, 24-0, 24-0, 24-5, 247, 24-5 Ibs. per minute. Find the average value of the supply.
7 to 9. The following figures were taken during the test of a steam electric plant ; the steam pressure, voltage, and current being kept as steady as possible.
Steam pressure (Ibs. per sq. in.) : 96, 100, 99, 100, 100, 98, 102, 102, 102, 103. Voltage : no, in, 111-5, IO9» I09'5' I09'5» IIJ» II2, in, no. Current (amperes) : 220, 223, 223, 230, 229, 224, 225, 217, 220, 200.
7. Find the average steam pressure.
8. Find the average voltage.
9. Find the average current.
DECIMAL FRACTIONS 71
10. The following readings were taken on an anemometer at differ- ent parts of the ashpit, in measuring the air supplied to a furnace : 57°» 580, 620, 650, 660, 600, 600, 460 cu. ft. per min. Find the average value.
11. In an experiment to find the specific resistance of a piece of wire the following readings of diameter were taken : '465, '463, -460, •460, '461, '460, -462, '464 millimetres. Find the average diameter.
12. Tests on a number of materials, to determine the ratio of shear to tensile strength gave the following results : -67, '64, -66, -72, -81, •82, -88, -85, -85, -87, -83, -73, -91, -59, -90, -98. Find the average value.
Percentage. — The words " Percentage " and " so much per cent." are frequently met with, a percentage being another way of expressing a fraction (or division) whose denominator is 100. Thus the statement " 5 per cent/' (written as 5%) means 5 parts out of 100 parts, *'. e.t T^ in the vulgar form. Cancelling reduces this to -£$ ; or it may be expressed in the decimal form -05, shifting the point two places to the left to divide by 100. Both forms are useful. From the above we obtain the following rules : —
1. To convert a percentage into a decimal fraction, shift the decimal point two places to the left.
2. To convert a percentage into a vulgar fraction, write, the percentage as numerator, and write 100 as denominator ; then reduce to the lowest terms.
Example 61. — Convert the following percentages into decimal fractions: (a) 30%, (b) 37'5% W 1-13% (<*) -028%.
(a) Shifting decimal point two places to left (to divide by 100), 30% = -3
(b) Similarly 37-5% = -375
(c) „ 1-13% = 'OII3
(d) „ -028% = -00028
Example 62. — Convert the following percentages into vulgar frac- tions : (a) 45%, (6) 12*%, (c) 33*% (d) 26-25%, (e) -oz5%.
(«) 45% = J4050' Cance»ing by 5 =• A
i
(6) «4% = £j - «| -=- 100 = » X JL ^
4
(c) 331% =fg = 33} ,. I00 = \Si x _1_ =i
26*2^ 262^
(d) 26-25%= -j^ = — ^Q- Cancelling by 25 and 5 = fj
(e) -025% = —5. == — 25 - • Cancelling by 25 = ,A6
v ' J /0 IOO IOOOOO o / O ^gOQ
72 ARITHMETIC FOR ENGINEERS
It is often required to find a certain percentage of a quantity, meaning that we are to find a certain fraction of the quantity. To do this, express the percentage as a fraction and multiply by the quantity.
Example 63. — An electric motor is to work at 30 h.-p. under ordinary full load conditions. It is required, however, to deliver 25% overload (i. e.t 25% extra) for a certain period. Find the greatest h.-p. the motor must be capable of giving.
Full load = 30 h.-p.
Overload = 25% = -^ x 3Q = 7-5
2
.'. Greatest h.-p. = 30 4- 7*5 = 37'5 h.-p.
The vulgar form is the more useful when the percentage is equiva- lent to a simple fraction with I as numerator; e. g., 20% = £, I2i% = J, etc. In these cases, to multiply as per rule we merely have to divide by the denominator of the fraction ; thus the above can be done mentally : 25% = J, therefore divide the 30 h.-p. by denominator 4. For other cases the decimal form is better, as multiplying by a decimal is easier than by a complicated vulgar fraction. It is useful to memorise the equivalents of the simpler percentages, which are given below.
i% = r™ = -oi 25% -J--25
5% = A = '05 33i% = i = '3333
10% =TV =-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 {£ 5s- = & °f £8 55- Dividing £8 by 5 gives i and £3 over. £3 ~ 605. and adding in the 5 gives 655. This divided by 5 = 135. Thus 20% of £8 55. = £i 135.
A common error in such examples as the last is to say that " 10% advance with 20% discount is equal to 10% discount." This is not so, since the 10% and the 20% refer to different quantities : the 10% to the old price, and the 20% to the new price : the 20% cannot be used until the old price has been found.
Example 65. — In steel-makers' catalogues it is stated that the actual weights of rolled steel section bars may vary 2j% on either side of the listed weights. Calculate the greatest and least weights of a channel bar listed at 19-3 Ibs. per foot.
Listed weight = 19-3 Ibs. per ft.
2. *>
2}% allowance = — x 19-3
= -482 Ibs. per ft. /. Greatest weight = 19*3 + '48 = 19*78 Ibs. per ft.
and least „ = 19-3 — '48 = 18-82 Ibs. per ft.
Conversion to a Percentage. — It is frequently necessary to express some quantity as a fraction of another, and the most usual way is to express it as a percentage. Suppose that the nominal full load of an alternator is 60 kilowatts, but at some par- ticular time it is called upon to deliver 75 kw., an increase of 15. Now this in itself is not much guide as to how much the machine has been overloaded, as the overload allowable depends on the size of the machine. Thus an alternator designed for 6000 kw. could easily deliver another 15 kw. ; but it would be impossible to get another 15 kw. out of a machine designed for, say, 20 kw. In each case the actual increase is the same, but as a fraction of the full load is very different ; this fraction is the important figure. Taking the big machine 15 kw. expressed as a fraction of 6000 kw. is yJo^ ; with the small machine 15 as a fraction of 20 is i£. Now to avoid the confusion of having vulgar fractions with different denominators, the fractions are expressed as a percentage, i. e.t they are converted so that the denominator is always 100.
From Rule 2 on p. 71, it follows that to convert a vulgar fraction into a percentage we must multiply by 100. Converting the above fractions to percentages we have —
i 5
_il x iqq = i% or -25% overload, and ^ x ^ = 75% overload.
4 *
74 ARITHMETIC FOR ENGINEERS
Then 15 kw. increase on 20 kw. is an overload of 75% but an in- crease of 15 on 6000 kw. is only *25% overload. Thus we have the rule : To express one quantity (A) as a percentage ol another quantity
(B) write (A) over (B) as a vulgar fraction, multiply by 100, and evaluate.
^ Thus A expressed as a percentage of B is ^ X 100.
Examples where this occurs are many and include : efficiencies of all kinds, errors of instruments and experiments, changes of speed and load, extensions of test bars, the composition of alloys, etc., etc.
Example 66. — An ammeter, when tested, was found to read 4-63 amperes, the true value being 4-5 amperes. Find the percentage error (i. e.t express the error as a percentage of the true value).
Apparent value == 4-63 True value = 4*50
/. Error = -13 too high
.\ % error = ~3 X I0° = 2-89% too high, 4-5 _____
i. e.t for every 100 amperes indicated the error is 2-9 amperes.
Example 67. — The following figures were obtained in a tensile test of a piece of steel —
Original length 5*. Original area of cross section -994 sq. in.
Final length 6'O$". Final area of cross section -407 sq. in.
(a) Express the total extension as a percentage of the original length, (b) Express the contraction in area as a percentage of the original area.
(a) Total extension = Final length — original length.
= 6-05 — 5 = i -05"
/. Extension as % of original length = --°5 *— *?_° = 21%
(b) Contraction in area = Original — Final area = -994 — '407 = •587 sq. in.
.'. Contraction as % of original area = --5..Z. x IQO = 59-05%,
•994 say, 59%
Actually, the statements made are not as complete as in the above two examples. Thus in Example 66, the usual statement is " find the percentage error/1 and the figure of which the error is to be a percentage would not be stated. Then the student may be in doubt as to whether he ought to express the -13 amperes as a
DECIMAL FRACTIONS 75
percentage of the true value 4-5, or the apparent value 4*63. A slight knowledge of the particular work with which the example is connected is useful in this respect, but the following will help —
Instrument and experimental errors are stated as a percentage of the true value.
With changes of load and speed, the error is stated as a percentage of the normal or full load, or speed.
Extensions arid^contractions of test bars are stated as percentages of the original length or area.
With composition of alloys, etc., the quantity of each material is given as a percentage of the total quantity.
Example 68. — A sample of coal weighing 3-35 Ibs. was found on analysis to contain : Carbon 2-52 Ibs., hydrogen -234 lb., oxygen -197 lb., nitrogen -133 lb., and the remainder ash. Find the percentage com- position of the coal (i. e.t express the weight of each constituent as a percentage of the total weight).
Carbon . . . . 2-52 Ibs. Total coal ... 3-35 Ibs.
Hydrogen .... -234 lb. Total (less ash) . 3*084 ,,
Oxygen . . . . -197 ,, /. Ash = difference = *266 lb.
Nitrogen . . . . -133 ,,
Total (without ash) . 3*084 Ibs.
|
Carbon Hydrogen Oxygen Nitrogen Ash |
2-52 X ioo ~ TaJ •234 x ioo |
3'35 ' _ 23*4 |
75-2% 5-88% 3*97% 7'95% |
|
3'35 •197 X ioo |
3'35 19-7 |
||
|
3'35 •133 X ioo |
3'35 13*3 |
||
|
3'35 •266 x ioo |
3'35 26-6 |
||
|
3*35 |
3'35 |
Total 99-99%
The sum of the percentages of all the constituents should be 100%. Any serious difference would indicate a mistake some- where. Due to slight approximations, etc., small differences may arise, but if, as in the above, the percentages total 99-99%, the figures can quite well be accepted. The exact difference allowable depends on the delicacy of the experiments, etc.
Example 69. — The chemical analysis of a " Self-hardening " steel gave the following results : Carbon -63%, chromium 1-04%, manganese •05%, silicon -15%. Find the percentage of iron " by difference " (t. e., subtract the total % of matter other than iron, from 100%).
76 ARITHMETIC FOR ENGINEERS
This example does not involve any finding of percentages, but is merely an addition dealing with percentages.
Carbon . . . '63% 100-00
Chromium . . 1*04,, 1-87
Manganese . . -05 ,, - -
Silicon . . . -15 ,, 98-13% Iron
•'. Iron " by difference " = 98-13
Exercises 18. On Percentages.
1. A firm engaged in the manufacture of valves, makes the following extra charges for drilling holes in the flanges of cast-iron bodies : —
i y holes 45. per dozen holes ; i" holes 25. 6d. per dozen. For cast-steel bodies they charge 50% extra. Find the rates for the above holes in cast steel.
2. If 5% extra is charged for packing a job for export, find the charge for packing a job costing £50.
3. If the estimated cost of a certain job is 26s. and a profit of 12^% is to be added, find the selling price of the job.
4. The ordinary price of a certain clutch is £4. Due to an increase in cost of material the price is increased 5%. Find (a) the new charge. A customer is now allowed a discount of 25%. Find (b) the actual amount he is to pay.
5. A wrought-iron shaft will only transmit 70% of the h.-p. that the same-sized steel shaft will transmit at the same speed. If the h.-p. of a certain steel shaft is 256, what is that of the same-sized wrought- iron shaft ?
6. A square foot of wrought-iron plate i" thick weighs 40 Ibs. If steel is 2% heavier than wrought iron, find the weight of the same- sized sheet of steel.
7. The " working volume " of a steam engine cylinder is 5*96 cu. ft. The " clearance " volume is 8% of the working volume. Find (a) this clearance volume in cubic feet; (b) the total volume in cubic feet (i. e., clearance -f working).
8. A variation in weight of 2% either side of the standard weight is allowed in electric conductors. The standard weight of 1000 yds. of 37/13 conductor is 2900 Ibs. Find the maximum and minimum weights allowable.
9. The Admiralty test load on a \" crane chain is 3 tons. At the Elswick Works the test load is 10% higher. Find the Elswick test load.
10. It is found that an engine governor keeps the speed between 486 and 474 revolutions per minute. Find the change of speed and express it as a percentage of the average speed.
11. At 40° F., i cu. ft. of water weighs 62*43 Ibs., and at 400° F. weighs 53*63 Ibs. Find the decrease in weight, and express it as a percentage of the greatest weight.
12. The correct value of " g " (the acceleration due to gravity) in the metric system at London is 981 units. An experiment gave the value 992. Calculate the percentage error (i. e,, express difference as a percentage of the true value).
13. A pump was rated by its makers at 300 gallons per minute. On test it filled a tank, holding 750 gallons, 4 times in 9 min. 27 sec.
DECIMAL FRACTIONS
77
Find (a) the test rating of the pump (i. e., the gallons delivered in i min. on test), and (b) the difference between the makers' and the test ratings. Express this as a percentage of the makers' rating, saying whether it is above or below.
14 to 16. The composition of some alloys is given below. Find in each case the percentage composition.
14. Bell metal : 16 parts copper, 5 parts tin.
15. Soft gun metal : 16 parts copper, i part tin.
16. Ajax plastic bronze : 13 parts copper, i part tin, 6 parts lead.
17. A solution of electrolyte for copper plating is composed of I Ib. copper sulphate, i Ib. sulphuric acid, and 10 Ibs. water. Calculate the percentage composition of the solution.
18. A quantity of exhaust gas is known to consist of the following : Water -18 cu. ft., carbon dioxide -24 cu. ft., nitrogen '852 cu ft., oxygen •08 cu. ft. Express the quantity of each constituent as a percentage of the total quantity.
19. The following figures were calculated from an analysis of a sample of flue gas : Carbon dioxide 1-54, oxygen 1-81, nitrogen 11-43. Express these figures as percentages of the total.
20 to 23. The chemical analyses of various irons and steels are given in the accompanying table. Find the percentage of iron in each case, by difference.
|
No. |
MATERIAL |
Carbon % |
Silicon % |
Sulphur % |
Phos- phorus % |
Manga- nese % |
Iron % |
|
20 |
Locomotive Boiler |
||||||
|
Plate (Mild Steel) |
•I48 |
•024 |
•039 |
•022 |
•562 |
||
|
21 |
Tram Rail (Mild |
||||||
|
Steel) .... |
•739 |
•347 |
•O^O |
•028 |
•720 |
||
|
22 |
Cast Iron |
3-23 |
1-46 |
•I4 |
•638 |
•5<> |
|
|
23 |
Wrought Iron . |
•04 |
•17 |
•013 |
•31 |
•OQ |
24 to 26. Engine governors are designed to prevent the engine speed from changing more than a certain amount when the load is removed or applied suddenly. The variation allowed depends upon the kind of work —
For cotton spinning the change is not to be more than 2% above and below normal speed.
For electric traction the change is not to be more than 1*5% above and below normal speed.
For machine shop work the change is not to be more than 3% above and below normal speed.
Find the greatest and least speeds in the following cases —
24. Normal speed 200 revs, per min., for cotton spinning.
25. Normal speed 180 revs, per min., for electric traction.
26. Normal speed 85 revs, per min., for machine shop drive.
27. A salt solution is made by adding 65 grammes of salt to 250 grammes of water. Find the percentage of salt in the solution (i. e.f salt as a % of the total).
28. A mixture for " blueing " steel is composed of water 10 Ibs., hypo 2 oz., sugar of lead 2 oz. Calculate the percentage composition by weight of the mixture. (Note. — Work in ounces.)
78 ARITHMETIC FOR ENGINEERS
Ratio. — It is often necessary to state the number of times that one quantity contains another, and this is usually given in the form of a ratio. Thus, if the diameter of a steam engine cylinder is 12" and the stroke 18", the stroke is evidently ij times the dia- meter, since 18" = 1-5 X 12". Now this fact of the stroke being equal to i J diameters is expressed by the words " the Ratio of stroke
to diameter is 1*5 to 1." The 1*5 is evidently the value of — = S-j-— °
i. e., — j-S - -v-5-. Thus, the ratio of one quantity (A) to another 2nd quantity ^ j \ i
(B) is the number of times that (A) contains (B).
It should be clear that we cannot give a ratio between two quantities of different measures. Thus, we cannot say that the ratio of 27 tons to 9 miles is 3 to i ; certainly 27 tons is not 3 times 9 miles. Hence the measures must be alike. Further, the value of the ratio cannot be worked out without bringing both quantities to the same unit. Thus, consider the ratio of 2 ft. to 6 in. Obviously this is not £ or J or i to 3, for 2 ft. is not one-third of 6 in. But,
converting the 2 ft. into inches (24*) the ratio is ^ or- or 4 to i,
meaning that 2 ft. is 4 times 6 in., which is certainly true.
Thus we obtain the rule : To find the ratio of one quantity (A) to another quantity (B), reduce both to the same units and divide the first
A quantity by the second, i. e. ^.
Example 70. — The " grate area " of a boiler is 22 sq. ft., and the heating surface is 1365 sq. ft. Find the ratio of heating surface to grate area.
The first-named quantity in the ratio required is heating surface, so that heating surface is to be divided by grate area.
Then, ratio of heating surface to grate area = ^--^ = 62*1
or, say, 62 to i.
Example 71. — A locomotive crank is 13" long and the connecting rod is 6 ft. 2 in. long. Find the ratio of the length of connecting rod to length of crank.
Convert to the same units, preferably inches. Then the rod is 74* long. The required ratio is " connecting rod to crank," so that con- necting rod is to be divided by crank.
Then ratio of connecting rod to crank = J J = 5-69, or, say, 5-7 to i.
Slight variations in the manner of stating a ratio are met with. The word " to " is frequently replaced by two dots, thus (:), so that
DECIMAL FRACTIONS 79
the statement " 4 : I " is read as " 4 to i." Sometimes only the result of the division is stated, the words " to I " being under- stood. Thus a ratio 57 means a ratio of 57 to i.
It should be noticed that two ratios always exist between two quantities. Thus, considering our original case of stroke and diameter we can have —
(a) The ratio of stroke to diameter, which = — ^ — = - x ' dia. 12
or i -5 to i i. e., stroke = i*5 times diameter.
(b) The ratio of diameter to stroke, which = — — r- =— 0 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 twro ratios are "reciprocal," i. e.t either one is the other inverted.
Thus stroke to diameter = — = — -5 and diameter to stroke =
12 I
—$ = — = '667 ; one can always be obtained from the other. 18 1*5
Example 72. — The following figures refer to a steam-driven air com- pressor : Area of steam cylinder =113 sq. ins.; area of air cylinder = 93*5 S(l' ms> Calculate the ratio of air cylinder area to steam cylinder area and give also the reverse ratio of steam cylinder to air cylinder.
Ratio air cylinder area to steam cylinder area = 93_5
= '827, i. e.t -827 to i Then the ratio steam cylinder to air cylinder = i to '827
Example 73. — The efficiency of any machine is the ratio of the out- put to the input. A generator gives an output of 27*6 kilowatts for an input of 51 h.-p. Find the efficiency.
(Note — i kilowatt = 1*34 h.-p.)
8o ARITHMETIC FOR ENGINEERS
Reduce to same units, preferably horse-power.
Then, since i kw. = 1-34 h.-p.
27-6 kw. =s 27-6 x 1-34 = 37 h.-p.
Efficiency = Ratio of output to input = f£ = '725
Note. — In the case of efficiency it is usual to state only the decimal result as shown, omitting the words " to i." In many cases also this result is converted to a percentage, which for the above would be 72-5%.
Ratios are used when speaking of the relative sizes of the cylinders in a compound or multi-stage expansion steam engine. The case of the compound engine, with only 2 cylinders, may be similarly treated, but with triple and quadruple expansion 3 and 4 cylinders have to be dealt with, and 2 or 3 ratios are combined into one short statement.
Example 74. — The triple-expansion engines of the White Star liner Olympic have the following cylinder areas : High Pressure (H.P.) 2290 sq. ins.; Intermediate Pressure (I. P.) 5542 sq. ins., and Low Pres- sure (L.P.) 7390 sq. ins. Find the "cylinder ratios " (viz. ratios of the cylinder areas), taking H.P. as i.
There will be two ratios ; and H.P. area will be the 2nd quantity in each case.
The area ratios are : 1. I. P. to H.P. 2. L.P. to H.P.
1. Ratio of I.P. to H.P. = |^ - 2-42 : i
2. Ratio of L.P. to H.P. - ~ = 3^3 • i
Instead of writing these results as two, they are combined into one by giving " the ratio of L.P. to I.P. to H.P/' ; thus 3-23 : 2-42 : i, meaning that if the H.P. area be i, then the I.P. area is 2-42 and the L.P. area 3*23. Then the ratio of trie L.P. to the I.P. should be
according to this, ^—^ = 1*335. This is proved by taking the actual
areas given. Then ratio of L.P. to I.P. = ^ 1-335 as before.
The dots (:) are invariably used when giving cylinder ratios.
Ratios connected with certain triangles are very important, and are the foundation of a special branch of mathematics known as " Trigonometry/1 (See Chap. XI.)
DECIMAL FRACTIONS Si
Exercises 19. On Ratios.
Find the ratio of connecting rod length to crank length in the follow- ing engines : —
1. (a) A petrol motor ; crank 2 J", connecting rod n*.
(b) A marine engine ; crank 30", connecting rod 5'-!*.
2. (a) A high-speed electric-lighting engine; crank 4", connecting rod i'-S\
(b) A power-driven air-compressor; crank 2\", connecting rod 14". The " diagram factor " of a steam engine is the ratio of the actual mean pressure to the theoretical mean pressure. Find the diagram factor for the following cases : —
3. A marine engine; theor. mean press. 36 Ibs. per sq. in.; actual mean press. 23 Ibs. per sq. in.
4. Corliss engine ; theor. mean press. 29 Ibs. per sq. in. ; actual mean press. 25 Ibs. per sq. in.
5. The " buckling factor " of a column is the ratio of its length (inches) to a measurement known as its "least radius of gyration" (inches). Find the buckling factor : (a) when length — 14 ft., and least radius of gyration = 4*13"; (b) when length = 38/-6/y, and least radius of gyration 4-1.*
6. Find the ratio of length to diameter of a Lancashire boiler : (a) when length = 24 ft. and diameter 7/-6y/; (b) when length = 21 ft. and diameter = 6 '-6*. Note — work in feet.
7. Find in each of the following boilers the ratio of heating surface to grate area : —
(a) Heating surface 720 sq. ft. ; grate area 24-75 sq. ft.
(b) Heating surface 1616 sq. ft. ; grate area 53-6 sq. ft.
8. Values for the tensile and shearing strength of two metals are given below. Find in each case the ratio of shear strength to tensile stress : —
(a) Copper: Tensile 14 tons per sq. in.; shear n-6 tons per sq. in.
(b) Hard rolled bronze: Tensile 26-9 tons per sq. in.; shear 16-06 tons per sq. in.
9. The results of tests giving the elastic limit stress and breaking stress (in tons per sq. in.) of various metals are given below. Find for the two materials ratio of the elastic stress to breaking stress in each case.
(a) Gun steel. Elastic stress 24-8. Breaking stress 46-9.
(b) Lowmcor iron. Elastic stress 12-5. Breaking stress 22*1.
10. At atmospheric pressure *oi6 cu. ft. of water, when evaporated, becomes 26-37 cu. ft. of steam. Find the " relative volume " (i. e.t the ratio of steam volume to water volume).
11. A two-stage air-compressor has the following dimensions : — Area of H.P. steam cylinder, 314 sq. ins. ; Area of H.P. air cylinder,
380 sq. ins.
Area of L.P. steam cylinder, 1385 sq. ins. Area of L.P. air cylinder, 1018 sq. ins.
Find the ratio of L.P. to H.P. (a) for steam end ; (b) for air end.
12. The areas of the cylinders in a triple-expansion marine engine are as follows : H.P. 962 sq. ins ; I. P. 2290 sq. ins. ; L.P. 6082 sq. ins. Find the ratio of L.P. : I.P : H.P.
13. The quadruple expansion engines of an Atlantic liner have the G
82 ARITHMETIC FOR ENGINEERS
following areas : H.P. 1097; I.P. No. I, 1905; I. P. No. 2, 4390; and L.P. 9852 sq. ins. Find the cylinder ratios (i.e., L,P. : 2nd I.P. : ist I.P. : H.P.).
14. The following figures refer to two American locomotives : —
(a) Firebox heating surface 231 sq. ft.; tubes heating surface 3193 sq. ft.
(b) Firebox heating surface 294-5 S(l- **•; tubes heating surface 3625 sq. ft.
Find for each case the ratio of total heating surface to the fire- box heating surface. (Note. — Total heating surface covers tubes and firebox.)
15. The efficiency of any simple lifting machine is the ratio of the theoretical effort to the actual effort. Find the efficiency if the theoretical effort is -59 Ib. and the actual 1-43 Ibs.
16. and 17. The " velocity ratio " of any machine is the ratio of the movement of the effort to the movement of the load. Find this ratio in each of the following cases : —
16. Weston pulley block. Effort moves 30"; load moves 2-5".
17. Geared capstan. Effort moves 31 '-5"; load moves 18".
18. The "expansion ratio" in a simple steam engine is approxim- ately the ratio of the initial pressure to the final pressure. Find the expansion ratio in the following cases : —
(a) Initial pressure 72 Ibs. per sq. in. ; final pressure 18 Ibs. per sq. in.
(b) Initial pressure 43 Ibs. per sq. in ; final pressure 17-5 Ibs. per sq. in.
19. If a cast-iron propeller costs ^24, then the same sized propeller would cost : (a) in steel, ^38, (b) in Delta metal £115, (c) in gun metal £130, (d) in manganese bronze £135, (e) in aluminium bronze £145, (/) in phosphor bronze ^170. Find the ratio of the cost of each of these materials to the cost of the cast-iron propeller.
Proportion. — When two ratios have the same value, the four quantities composing the ratios are said to be " proportionals/' or " in proportion." Thus the ratio of 3 to 2 is f or 1-5. Similarly the ratio of 12 to 8 is -1/ which also equals 1-5. Then evidently the ratio of 3 : 2 = the ratio of 12 : 8 and the numbers 3, 2, 12 and 8 are said to be in proportion. Thus 4 quantities are in proportion when the ratio of the ist to the 2nd is the same as the ratio of the 3rd to the 4th. The equals sign is commonly replaced by four dots, thus (: :), and the above result is stated as 3 : 2 : :i2 : 8, and is read shortly as "3 is to 2 as 12 is to 8." A more useful form is obtained by stating the ratios as vulgar fractions, thus f = -\2. Although each ratio must have the same units and measures, they need not be the same in the two ratios, since each ratio is only a number.
Proportion has its use when, knowing the ratio between two quantities, and the value of one of them, we wish to find the value of the other quantity ; that is, knowing three of the quantities in the proportion we wish to find the fourth.
DECIMAL FRACTIONS 83
Thus, supposing the ratio of the H.P. diameter to the L.P. diameter in a compound steam engine is to be 3 : 5, and the L.P. diameter is 32" ; and it is desired to find the H.P. diameter.
Then the statement H.P. : L.P. : : 3 : 5 is a proportion, meaning that if the H.P. is, say, 3" dia. the L.P. is 5" dia., and if the H.P. is, say, 6" dia. the L.P. is 10" dia., or, whatever the actual size the H.P. is f of the L.P.
Then since in this case L.P. = 32"
H.P. = £ of 32 = 19*2" or, say, 19^" dia.
Now if our answer is correct the ratio of the calculated H.P. to the L.P. should be 3 : 5 = £ or -6 to i.
Ratio of 19^ (H.P.) to 32 (L.P.) = -^~ = -6 to i, which proves
the work.
Proportion is best treated by the simple equation (see p. 164), when any one of the quantities can be easily obtained. But from the foregoing calculation we can deduce a rule which will suit all ordinary cases. It is seen in the above that H.P. = L.P. x £. Now numbering the terms in the proportion H.P. : L.P. 1:3:5,
from left to right as No. i, 2, 3, and 4, then No. i == ^^ No> 3.
Should the given ratio be written so that the required quantity is No. 2, then the ratio should be reversed as in Ex. 77.
Example 75. — Brass is an alloy of copper and zinc; how much copper must be mixed with 80 Ibs. of zinc when the ratio of copper to
zinc is 7:3?
Copper : Zinc : : 7 : 3
-, Zinc x 7 Then copper = -
• gg-x-Z = ,867 Ibs.
When the given ratio is stated as " so much to i," then the i will have no effect on the result, so that only one operation, instead of two as above, is required.
Example. 76. — The ratio of the lengths of connecting rod to crank in a steam engine is to be 4-5 (i. e., 4-5 : i). Find the length of the connecting rod if the crank is 8".
Connecting rod : Crank : : 4*5 : i
Then connecting rod =
84 ARITHMETIC FOR ENGINEERS
Example 77. — The efficiency (i. e., the ratio of the output to the input) of a transmission gear is '87. Find the input if the output is required to be 35 horse-power.
output : input : : -87 : i
Input (which is here the 2nd quantity) is required ; then the state- ment must be changed to —
input : output : : i : '87. Then input = <"*P^ X i __3S = 40-3 h.-p.
Example 78. — In a triple expansion engine the ratios of the cylinder areas (H.P. : I.P. : L.P.) are to be as 2 : 5 : 13. The L.P. area is 2463 sq. ins. Find the areas of the other two cylinders.
Considering first the I. P., we have the proportion —
I.P. : L.P. : : 5 : 13 Then I.R- k*2L3 = ^XJ, = ^ sq> ^
Now take the H.P. ; we have another proportion — H.P. : L.P. : : 2 : 13
TT _, L.P. X 2 2463 X 2 /. H.P.= —— =— — = 379 sq. ins.
Exercises 20. Proportion.
1. Pewter consists of tin and lead, the ratio of tin to lead by weight being 4:1. If a founder only has in stock ij cwt. (168 Ibs.) of tin and desires to use the whole of it in making pewter, how much lead will be required ?
2. A quantity of sulphuric acid is to be diluted so that the final solution contains water and acid in the ratio of 9 : i. How much water must be mixed with £ gallon of acid ?
3. The ratio of the L.P. cylinder area