♦^ •* • % A^ •^ • M" Ur °o /..i^.\ co*.!^^'.^ y,'ii^^\ / ,•* ^0 A^'^ '^^ "^.f. 0^*..i^'>o .4**.--^.\ 00*.-.J^.>o ^:* <.i>^ o^ %^ V^^^^^^' %/-^f.^"V^' V^^V %--^r-^-% '^^4^ A^'^ -j.^ "^^ °o ^..%^' ^ *•«»' -^ . .4°* • A^^ ^ L%v /-^.^^.A .^^\c^,\. /--.^i^ii^A y - 'te.d^ "^ ^V' ♦^M©* ^ 4?W^ *''^ ^«?^ ^^-^^^ ^^^^ ^^-^^^ «*» - o a o °A *'>^' ^ ^ *••« \^ ^ •'■• «^ ... V^ ^^0^ a^^^ "^^'^T^\/ -o^^^--*/ V*^\^^ "^^^^• ate'v *W* .-^^C^-- \./ .-Jfe-t V** r^-t ^^^^' 5^ ^C^ ■ %.*H-^/ V^^*y "^'^-V .... V' O « '^^0^ ^°^ iS^^°- \>/ ^^fe' %/ .-^T: *--*^ -'Jfe: V THE MANUFACTURE OF PULP AND PAPER VOLUME I Pulp and Paper Manufacture IN FIVE VOLUMES An Official Work Prepared under the direction of the Joint Executive Committee of the Vocational Education Committees of the Pulp and Paper Industry of the United States and Canada Vol. I — Mathematics, How to Read Drawings, Physics. II — Mechanics and Hydraulics, Electricity, Chemistry. Ill — Preparation of Pulp. IV, V — Manufacture of Paper. THE MANUFACTURE OF PULP AND PAPER A TEXTBOOK OF MODERN PULP AND PAPER MILL PRACTICE Prepared Under the Direction of the Joint Executive Committee on Vocational Education Representing the Pulp and Paper Industry of the United States and Canada VOLUME I Arithmetic, Elementary Applied Mathematics, How TO Read Drawings, Elements of Physics BY J. J. CLARK, M.E. 5 First Edition Second Impression McGRAW-HILL BOOK COMPANY, Inc. NEW YORK: 370 SEVENTH AVENUE LONDON: 6 & 8 BOUVERIE ST., E. C. 4 1921 \\'>^ ^6 coptbight, 1921, bt the Joint Executive Committee of the Vocational Edtjcation Committees OF THE Pulp and Paper Industry. All Rights Reserved, Including Those of Translation. THE M: A P L E PKESS X O K It FA. PREFACE In numerous communities where night schools and extension classes have been started or planned, or where men wished to study privately, there has been difficulty in finding suitable textbooks. No books were available in English, which brought together the fundamental subjects of mathematics and element- ary science and the principles and practice of pulp and paper manufacture. Books that treated of the processes employed in this industry were too technical, too general, out of date, or so descriptive of European machinery and practice as to be unsuit- able for use on this Continent. Furthermore, a textbook was required that would supply the need of the man who must study at home because he could not or would not attend classes. Successful men are constantly studying; and it is only by studying that they continue to be successful. There are many men, from acid maker and reel-boy to superintendent and mana- ger, who want to learn more about the industry that gives them a livehhood and by study to fit themselves for promotion and in- creased earning power. Pulp and paper makers want to under- stand the work they are doing- — -the how and why of all the various processes. Most operations in this industry are, to some degree, technical, being essentially either mechanical or chemical. It is necessary, therefore, that the person who aspires to under- stand these processes should obtain a knowledge of the under- lying laws of Nature through the study of the elementary sciences and mathematics, and be trained to reason clearly and logically. After considerable study of the situation by the Committee on Education for the Technical Section of the Canadian Pulp and Paper Association and the Committee on Vocational Educa- tion for the Technical Association of the (U. S.) Pulp and Paper Industry, a joint meeting of these committees was held in Buffalo vi PREFACE in September, 1918, and a Joint Executive Committee was ap- pointed to proceed with plans for the preparation of the text, its pubhcation, and the distribution of the books. The scope of the work was defined at this meeting, when it was decided to provide for preHminary instruction in fundamental Mathematics and Elementary Science, as well as in the manufacturing operations involved in modern pulp and paper mill practice. The Joint Educational Committee then chose an Editor, Associate Editor, and Editorial Advisor, and directed the Editor to organize a staff of authors consisting of the best available men in their special lines, each to contribute a section dealing with his specialty. A general outline, with an estimated budget, was presented at the annual meetings in January and February, 1919, of the Canadian Pulp and Paper Association, the Technical Association of the Pulp and Paper Industry and the American Paper and Pulp Association. It received the unanimous approval and hearty support of all, and the budget asked was raised by an appropriation of the Canadian Pulp and Paper Association and contributions from paper and pulp manufacturers and allied industries in the United States, through the efforts of the Technical Association of the Pulp and Paper Industry. To prepare and publish such a work is a large undertaking; its successful accomplishment is unique, as evidenced by these volumes, in that it represents the cooperative effort of the Pulp and Paper Industry of a whole Continent. The work is conveniently divided into sections and bound into volumes for reference purposes; it is also available in pamphlet form for the benefit of students who wish to master one part at a time, and for convenience in the class room. This latter arrangement makes it very easy to select special courses of study; for instance, the man who is specially interested, say, in the manufacture of pulp or in the coloring of paper or in any other special feature of the industry, can select and study the special pamphlets bearing on those subjects and need not study others not relating particularly to the subject in which he is interested, unless he so desires. The scope of the work enables the man with but little education to study in the most efficient manner the preliminary subjects that are necessary to a thorough understanding of the principles involved in the manu- facturing processes and operations; these subjects also afford an exce)lent review and reference textbook to others. The work PREFACE vii is thus especially adapted to the class room, to home study, and for use as a reference book. It is expected that universities and other educational agencies will institute correspondence and class room instruction in Pulp and Paper Technology and Practice with the aid of these volumes. The aim of the Committee is to bring an adequate opportunity for education in his vocation within the reach of every one in the industry. To have a vocational education means to be familiar with the past accomplishments of one's trade and to be able to pass on present experience for the benefit of those who will follow. To obtain the best results, the text must be diligently studied; a few hours of earnest application each week will be well repaid through increased earning power and added interest in the daily work of the mill. To understand a process fully, as in making acid or sizing paper, is like having a light turned on when one has been working in the dark. As a help to the student, many practical examples for practice and study and review questions have been incorporated in the text ; these should be conscientiously answered. The Editor extends his sincere thanks to the Committee and others, who have been a constant support and a source of in- spiration and encouragement; he desires especially to mention Mr. George Carruthers, Chairman, and Mr. R. S. Kellogg, Secretary, of the Joint Executive Committee; Mr. J. J. Clark, Associate Editor, Mr, T. J. Foster, Editorial Advisor, and Mr. John Erhardt of the McGraw-Hill Book Company, Inc. The Committee and the Editor have been generously assisted on every hand ; busy men have written and reviewed manuscript, and equipment firms have contributed drawings of great value and have freely given helpful service and advice. Among these kind and generous friends of the enterprise are: Mr. 0. Bache- Wiig, Mr. James Beveridge, Mr. J. Brooks Beveridge, Mr. H. P. Carruth, Mr. Martin L. Grifiin, Mr. H. R. Harrigan, Mr. Arthur Burgess Larcher, Mr. J. 0. Mason, Mr. Elis Olsson, Mr. George K. Spence, Mr. Edwin Sutermeister, Mr. F. G. Wheeler, and American Writing Paper Co., Dominion Engineering Works, E. I. Dupont de Nemours Co., F. C. Huyck & Sons, Hydraulic Machinery Co., Improved Paper Machinery Co., E. D. Jones & Sons Co., A. D. Little, Inc., National Aniline and Chemical Works, Process Engineers, Pusey & Jones Co., Rice, Barton & viii PREFACE Fales Machine and Iron Works, Ticonderoga Paper Co., Waterous Engine Works Co., and many others. J. Newell Stephenson, Editor For the Joint ExECUTrvE Committee on Vocational Education, George Carruthers, Chairman, R. S. Kellogg, Secretary, T. L. Crossley, G. E. Williamson, C. P. Winslow. Representing the Technical Sec- Representing the Technical As- tion of the Canadian Pulp and Paper sociation of the (U. S.) Pulp and Association Paper Industry. T. L. Crossley, Chairman, George E. Williamson, Chairman, George Carruthers, Hugh P. Baker, A. P. CosTiGANE, Henry J. Guild, Dan Daverin, R. S. Kellogg, C. Nelson Gain, Otto Kress, J. N. Stephenson. W. S. Lucey, C. P. Winslow. CONTENTS Paqh . Preface v SECTION 1 Arithmetic PART I Notation and Enumeration 1-10 Addition and Subtraction . 11-21 Multiplication and Division. . , 22-34 Some Properties op Numbers 34—39 Examination Questions 41-42 PART II Greatest Common Divisor and Least Common Multiple . . . 43-47 Fractions 48-64 Involution 64-67 Decimals and Decimal Fractions 67-76 Signs of Aggregation 76-77 Ratio and Proportion 77-85 Examination Questions 87-88 PART III Square Root 89-98 Percentage 98-106 Compound Numbers 106-113 The Metric System 113-124 The Arithmetical Mean 125-126 Examination Questions 127-128 SECTION 2 Elementary Applied Mathematics PART I Mathematical Formulas 1-4 Algebraic Addition, Subtraction, Multiplication, and Division 4-18 Equations 18-27 Accuracy in Calculation 27-33 Approximate Method for Finding Roots 33-39 Examination Questions 41-42 ix X CONTENTS PART II Page Mensueation of Plane Figtjres 43-47 Triangles 50-60 Quadrilaterals 61-66 Regular Polygons 66-68 The Circle 68-89 Inscribed and Circumscribed Polygons 89-96 Examination Questions 97-100 PART III Mensuration of Solids 101-130 Similar Figures 131-137 Symmetrical Figures 137-143 Examination Questions 145-146 SECTION 3 How to Read Drawings Representing Solids ON Planes 1-11 Special Features Pertaining to Drawings 12-23 Reading Drawings — Visualizing the Object 23-28 Some Examples in Reading Drawing 28 SECTION 4 Elements of Physics PART I Matter and Its Properties 1-6 Motion and Velocity 6-9 Force, Mass and Weight 9-14 Work and Energy 14r-17 Hydrostatics — Pascal's Law 18-30 Buoyancy and Specific Gravity 31-39 Capillarity 39-41 Pneumatics 42-54 Examination Questions 55-56 PART II Properties op Perfect Gases 57-66 Nature op Measurement of Heat 66-80 Change op State 81-90 Light 90-117 Examination Questions 119-120 Index 121-132 SECTION 1 AEITHMETIC PART 1 NOTATION AND ENUMERATION DEFINITIONS 1. A unit is the standard by which anything is measured. For example if it were desired to measure the length of a room, it could be done with a tape line, a rule, a stick of some convenient length, or anything else that would form a basis of comparison. Suppose, for example, a broom- stick were used ; the broomstick would be laid on the floor, with one end a against the wall at one side of the room and with the other end h extending towards the opposite side of the room, as shown in Fig. 1. The point h would be marked and the broom- stick would be shifted so as to occupy the position he, then to the position cd, and so on until the other wall was reached. Each of the lengths ah, he, etc. is equal in length to the broomstick, which is in this case the unit of length. The word that denotes how many times the room is longer than the broomstick is numher. In other words number means an aggregation, or collection, of units; a number may also mean a single unit or a part of a unit — it refers in all cases to the complete measurement. As another example, consider the compensation that a man receives for his work. For working a certain number of hours he receives a certain number of dollars; here two units are involved, §1 1 a ft c d e / 1 s > \ t ! Fig. 1. 2 ARITHMETIC §1 hours and dollars. The unit of hours is established or arrived at by dividing or splitting up a day (another unit) into twenty-four equal parts and calling one of these parts one hour. The unit of money is established by law; it is called in the United States of America one dollar, and is equivalent in value to about twenty- three and one-fourth grains (another unit) of pure gold. Canada has a unit of the same name and of practically the same value. 2. Three different kinds of units, according to their origin, have been mentioned, the broomstick, which may be called an expediency unit; the hour, which is an international unit for measuring time; and the dollar, which is a legal unit, or one estab- lished by law. According to their use, units have many names, and anything that can be expressed as a certain number of units is called a quantity. The science that treats of quantity and its measurement is called mathematics. 3. The difference between quantity and number is this : quantity is a general term and is applied to anything that can be measured or expressed in units; number is simply a term applied to a unit or a collection of units and is usually restricted to expressions containing only figures. 4. The act or process by which numbers are used to reckon, count, estimate, etc. is called computation or calculation. 5. Arithmetic is that branch of mathematics which treats of numbers and their use in computation. Numbers are divided into two general classes, according to their signification: abstract numbers and concrete numbers. 6. An abstract number is one not applied to any object or quantity, as three, twelve, one hundred seventeen, etc. 7. A concrete number is one relating to a particular kind of object or quantity, as five dollars, eleven pounds, seven hours, etc. According to their units, numbers are also divided into two general classes: 8. Like numbers are numbers having the same units: for ex- ample, seven hours, twelve hours, and three hours are like numbers, since they all have the same unit — one hour; fourteen, seventeen and twenty are also like numbers, the unit of each being one. All abstract numbers are like numbers. 9. Unlike numbers are numbers having different units; for example, five dollars, eight hours, and ten pounds are unlike §1 NOTATION AND ENUMERATION 3 numbers since their units are unlike, being respectively one dollar, one hour, and one pound. NOTATION In order that numbers may be recorded, some method of writing and reading them must be devised. 10. Notation is the word used to express the act of and the result obtained by writing numbers. 11. Numeration is the act or process of reading numbers that have been written. 12. Notation is accomplished in three ways: (1) by words; (2) by letters; (3) by figures. The first method, by words, is never used in computation, only the second and third methods having ever been employed for this purpose. The second method, called the Roman notation, is very seldom used at the present time, its only use is for numbering, as the indexes, chapters, etc. of books, and in a few cases for dates. The third method, the Arabic notation, employs certain characters, called figures, and is in universal use today. THE FIRST OR WORD METHOD 13. This is really only the names of the different numbers. The first twelve numbers have distinct names; beyond these all are formed according to a definite system that is readily learned. The names of the first twelve numbers are: one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. The next seven numbers all end in teen, which means and ten, the first part of the word being derived from the words three, four, etc. up to nine; they are thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, and nineteen. Thirteen means three and ten, fourteen means four and ten, etc. The name of the next number is twenty, and the names of the next nine numbers are all compound words, having for their first element the word twenty and for their second element the words one, two, etc. to nine. Thus, twenty-one, twenty-two, etc. to twenty-nine. The word twenty means two tens. The next number is thirty, which means three tens, and the next nine are thirty-one, thirty-two, etc. to thirty-nine. Then follow forty, fifty, sixty, seventy, eighty, and ninety, with their 4 ARITHMETIC §1 combinations of one, two, etc. to nine. The number following ninety-nine is one hundred. The numbers are then repeated, with the element one hundred added, one-hundred-one, one- hundred-two, etc., one-hundred-twenty-one, one-hundred-twenty- two, etc. to one-hundred-ninety-nine, the next number being two-hundred. The notation is continued in this manner to nine- hundred-ninety-nine, the next number being one-thousand; then follow one-thousand-one, one-thousand-two, etc., one-thousand- one-hundred-one, one-thousand-one-hundred-two, etc. to nine- thousand-nine-hundred-ninety-nine, the next number being ten- thousand. Then follow ten-thousand-one, etc. to nine-hundred- thousand-nine-hundred-ninety-nine, the next number being called one-million. The notation is then carried on one-million-one, etc. up to one-thousand-million, which in the United States, France, and the majority of other countries is called a billion; a thousand billion is a trillion, etc. In Great Britain and a few other coun- tries, a billion is a million-million, a trillion a million-billion, etc. THE ROMAN NOTATION 14. The Roman notation uses the letters I, V, X, L, C, D, and M. The values of the letters when standing alone are : I is one, V is five, X is ten, L is fifty, C is one-hundred, D is five-hundred, M is one-thousand. Formerly other letters were used, but these are all that are employed now. Numbers are expressed by letters in accordance with the following principles : I. Repeating a letter repeats its value. Thus, I is one, II is two. III is three, XX is two tens or twenty, XXX is three tens or thirty, CC is two hundred, MM is two- thousand, etc. V, D, and L are never repeated; only I, X, C, and M are ever used more than once in any number, except when they precede a letter of higher value. II. If a letter precedes one of greater value, their difference is denoted; if it follows a letter of greater value, their sum is denoted. Thus, IV is four, VI is six, IX is nine, XII is twelve, XL is forty, LX is sixty, XC is ninety, etc. Only one letter of lower value is allowed to precede one of higher value, and in general, the only letters used are I and X. Further, I precedes only V and X, and X precedes only L and C. Occasionally, however, C is used in §1 NOTATION AND ENUMERATION 5 this manner before M ; as for example, in the date nineteen-hun- dred-four, when instead of writing MDCCCCIV the shorter form MCMIV is used. Pope Leo XIII wrote the date eighteen- hundred ninety-five MDCCCVC instead of MDCCCXCV. On the Columbus memorial in front of the Union Station, Washington, D. C, the year of his birth is expressed by MCDXXXVI, which, of course, is read fourteen-hundred-thirty-six. III. A bar placed over a letter multiples its value by one-thousand. Thus, X is ten-thousand, L is fifty-thousand, XCDXVII is ninety-thousand five-hundred seventeen. In the last expression, it might be thought that D was preceded by X and C; this is not the case, however, as XC is treated as a single letter having a value of ninety-thousand. The following table illustrates the foregoing principles and applications of the Roman notation : VIII is eight XII is twelve XVIII is eighteen XXIX is twenty-nine XXXV is thirty-five XLIV is forty-four LXXVI is seventy-six XCII is ninety-two CXLIV is one-hundred forty-four CCCCL is four-hundred fifty IXLX is nine-thousand sixty C is one-hundred-thousand D is five-hundred-thousand M is one-million THE ARABIC NOTATION 15. The Arabic notation employs ten characters, called figures, to represent numbers; these characters together with their names are : 123456789 naught one two three four five six seven eight nine cipher zero The first figure (0) is called naught, cipher, or zero, and has no value. The other nine figures are called digits, and each has the value denoted by its name. 6 ARITHMETIC §1 16. For numbers higher than nine it is necessary to use two or more figures to express them. Thus ten is expressed by the com- bination 10, the cipher indicating that there are no units in the right-hand figure; 11 is eleven, 12 is twelve, 13 is thirteen, etc. up to 19 or nineteen. All these expressions mean ten and one, ten and two, etc. Twenty is written 20, twenty-one, 21, etc.; thirty is written 30, thirty-three, 33, etc.; and so on up to 99 or ninety- nine. One-hundred is written 100, the right-hand cipher mean- ing that there are no units and the next that there are no tens. Five-hundred-forty-two is written 542, five-hundred-two is writ- ten 502, five-hundred-forty is written 540. In a similar manner one-thousand is written 1000, ten-thousand is written 10000, one-hundred-thousand is written 100000, one-million is written 1000000, etc. 17. As stated before the cipher has no value, but it plays a very important part in expressing numbers by figures ; it not only indi- cates the absence of a digit from the place it occupies, but it also locates the digits with respect to one another. Any number, as for example, 5642, is equivalent to 5000 and 600 and 40 and 2; that is, in this case, it is equivalent to four sepa- rate numbers, three of which are single digits, all followed by one or more ciphers. Such a number might be written 5000 + 600 + 40 + 2, but this is not necessary, since the ciphers can be omitted without impairing the result, with the additional ad- vantage of saving space and figures. 18. The number denoted by a figure in any whole number is determined by writing the figure and then writing after it as many ciphers as there are figures to the right of the one selected. Thus, in the number 2645893, the number denoted by the figure 6 is 600000 or six-hundred-thousand ; the number denoted by the figure 5 is 5000 or five-thousand; etc. NUMERATION 19. For convenience numbers are divided into orders and periods. A figure belongs to the first order when it occupies the units place; it belongs to the second order when it occupies the tens place; etc. In the number 2645893, 2 belongs to the seventh order, 6 to the sixth order, etc. Numbers are divided by commas into ^periods of three figures each, beginning with the first order, §1 NOTATION AND ENUMERATION 7 to assist in reading them; the periods are named according to the name of the lowest order in the periods. Thus, the number 506,273,985,114 contains four full periods; beginning at the right and going to the left, the names of the periods are, respectively, units, thousands, millions, billions. The names of periods and orders represented by the figures in the above number are conveniently shown in the table below. Fourth Period. Billions. Third Period. Millions. Second Period. Thousands. First Period. Units. C ^ d i 6 i a .2 W S u o g ;> TO .2 o m a _2 s •xi c a; o c aj C s 1 el £ o ■+3 -a CO 1 O 1 CO m eS DO o 1 w o m d O d o 02 d o o d T3 M CD o d s O 1 03 1 o 2 6, 3, 9 5, 1 In pointing off these figures, begin at the right-hand figure and count — units, tens, hundreds; the next group of three figures is thousands; therefore, insert a comma (,) before beginning with them. Beginning at the figure 5, say thousands, ten thousands, hundred thousands, and insert another comma; next read millions, ten millions, hundred millions (insert another comma) ; lastly, read billions, ten billions, hundred billions. 20. A number is read by beginning at the left and reading the figures composing each period as though it formed a number by itself and affixing the name of the period, taking each period in succession from left to right. Thus, the above number is read: five-hundred-six billion two-hundred seventy-three million nine- hundred-eighty-five thousand one-hundred-fourteen. Note that the 8 ARITHMETIC §1 name of the units period is not pronounced — it is always understood. Reading a line of figures in this manner is called numeration; and when the numeration is changed back to figures, it is called notation. For instance, the writing of the following figures, 72,584,623 would be the notation, and the numeration would be seventy-two million five-hundred-eighty-four thousand six-hundred-twenty-three. 21. Note. — It is customary to leave the "s" off the words miUions, thou- sands, etc. in cases like the above, both in speaking and writing. 22. Consider the number 500,000. If the 5, which is here a figure of the sixth order, be moved one place to the right, so as to occupy the fifth order, the number becomes 50,000, which is only one-tenth as large as before; while if the 5 be moved one place to the left, so as to occupy the seventh order, the number is ten times larger. In other words, moving a figure to the right de- creases the number it represents ten times for each order it passes into; and moving the figure to the left increases the number represented by the digit ten times for each order passed into. Thus, 5 is one-tenth of 50 and 50 is ten times 5; 50 is one-tenth of 500 and 500 is ten times 50; etc. 23. Integers and Decimals. Any figure, say 5, represents the number of units signified by its name, when standing alone. Moving it one place to the left, it represents five tens or fifty, and is written 50; moving it one place to the right, it represents five- tenths, and is written 0.5. Moving the figure two places to the left of its first position, it represents five hundred, written 500; if moved two places to the right, it represents five hundredths, written 0.05. Now 0.5 means that this number is only one-tenth as large as 5, and 0.05 means that this number is only one-tenth as large as 0.5 or one-hundredth as large as 5. This notation may be continued to the right as far as desired. 24. The dot is called the decimal point ; it is used to point out the unit figure, and the part of the number to the right of the unit figure is called a decimal. All numbers not having any digits to the right of their unit figures (and which do not contain a common fraction) are called whole numbers or integers. The numbers dealt with heretofore have all been integers. The part §1 NOTATION AND ENUMERATION 9 of a number to the left of the decimal point is called the integral part of the number. 25. Decimals are read in much the same manner as integers. The only difference is that after the number has been read as though it were an integer, the name of the lowest (right-hand) order is pronounced with the letters ths added to the name. Thus, 0.52976 is read fifty-two thousand nine-hundred-seventy-six hundred-thousandths; 0.529 is read five-hundred-twenty-nine thousandths. 26. In using decimals be careful to mark the units place by placing the decimal point immediately to the right of the figure in the units place. Then continue the notation to the right of the units place precisely as to the left of it. The resemblance between the names of the places on the two sides of the units place is shown by the following table, in which the differences are marked by italics. -s T) c O ai CO 5 to Co C 03 CO 3 O +1 -o 13 T3 d a m ;=! o m -S -a m o (D O O H Xi t-> .0 +3 +3 el 13 00 03 -fi '3 -f3 g +3 d 3 rd 27. In this (the Arabic) system of notation, each place or order is said to be higher than any place to the right of it, and lower than any place to the left of it. Thus, in the number 43.127, the highest place is tens, and the lowest place is thousandths. 28. In writing and printing, decimals are frequently written .5, .529 etc. instead of 0.5, 0.529, etc. The latter way is the better and more correct, and the student is advised to write the cipher in all cases; for, if the decimal point should fail to print, be indistinct, or through carelessness not written, the cipher will frequently serve to distinguish between integers and decimals. It will be found, however, that the cipher has frequently been omitted throughout this textbook to accustom the reader to both methods of writing decimals. 29. A number, part of which is an integer and part a decimal is called a mixed number. A mixed number is read by reading 10 ARITHMETIC §1 the integral part first, pronouncing the word and, and then reading the decimal part. Thus, the number given in Art. 26, or 543,210.12345 is read five-hundred-forty-three thousand two-hundred-ten and twelve-thousand-three-hundred-forty-five hundred-thousandths. The abbreviation "Art." means Article, and refers to the numbered paragraphs or sections throughout this textbook. The plural is written Arts.; thus, Arts. 1-10, means articles 1 to 10, both inclusive. Decimals are seldom pointed off into periods, especially in English-speaking countries. It may be done, however, (if desired) in the same manner as integers, beginning at the unit figure. Thus, 0.0000000217 would be pointed off 0.00,000,002,17. In reading this decimal, name the periods until the last is reached and then name the order of the right-hand figure; — in this case, units, thousandths, millionths, billionths, ten-billionths; — this determines the name of the lowest order. The number is then read at once as two-hundred-seventeen ten-billionths. EXAMPLES FOR PRACTICE Read the following numbers and write their names: (1) 20103 (2) 1964727 (3) 7926.4867 (4) 0.0078314 Write the following numbers : (5) Eight-hundred-sixty-six million forty-six-thousand seven-hundred- thirty-three. (6) Six-hundred-ninety-one and four-thousand-five-hundred-twenty-three ten-thousandths. (7) Seventeen and eighty-nine-thousand-two-hundred-seventy-six mill- ionths. 30. The Four Fundamental Processes.— The processes of addition, subtraction, multiplication and division are called fundamental because all other processes employed in arithmetical computation are based on them; in other words, no computation can be made without employing one or more of these processes. §1 ADDITION AND SUBTRACTION 11 ADDITION AND SUBTRACTION ADDITION 31. Suppose it is desired to ascertain the length of a room, and that the distance from one end to the edge of a door nearest that end, the width of the door, and the distance from the other edge of the door to the other end of the room are known to be respect- ively, 6 feet, 3 feet, and 8 feet; then without getting a rule or other measure, the length of the room can be found by counting, starting with six, counting three more, getting nine, and then counting eight more after nine, getting seventeen. In other words, the combined lengths of 6 feet, 3 feet, and 8 feet is 17 feet, and the numbers 6, 3, and 8 taken together always give 17. The process of finding a single number equivalent in amount to two or more numbers taken together is called addition, and the single number found as the result of the process is called the sum. Thus, the sum of 6, 3, and 8 is 17. To add two or more numbers is to find their sum. 32. The method just described for finding the sum of several numbers is impracticable except for very small numbers. For example, suppose it were desired to find the sum of $2,904, $976, and $3,520; not only would this take a very long time, but there would also be great danger of making a mistake. A method for finding the sum of a series of numbers will now be described. When it is desired to indicate that several numbers are to be added, they are written in a row and separated by Greek crosses. Thus, the fact that the numbers 6, 3, and 8 are to be added is indicated as follows: 6 + 3 + 8. The cross + is called the sign of addition or plus or plus sign, and the expression would be read, six plus three plus eight. If it is also desired to write the sum as indicating the result of the addition, this is done in the following manner. 6 + 3 + 8 = 17 The sign = is called the equality sign, it is read equals or is equal to, and signifies that the result of the operations indicated on one side is exactly equal to the number (or to the result of a series of operations indicated) on the other side. For example, 5 + 2 + 10 = 17; hence, 6 + 3 + 8 = 5 + 2+ 10 = 17. This means that the result of the operation 6 + 3 + 8 is equal to the result of the operation 5 + 2 + 10 and that either sum is 17. 12 ARITHMETIC il An expression like 6 + 3 + 8= 17 is read, six plus three plus eight equals seventeen. 33. Before one can add several numbers it is necessary that the sum of any two numbers from 1 to 9 be recognized as soon as seen. The following table gives the sum of any two numbers ADDITION TABLE 1 and 1 is 2 2 and 1 is 3 3 and 1 is 4 4 and 1 is 5 1 and 2 is 3 2 and 2 is 4 3 and 2 is 5 4 and 2 is 6 1 and 3 is 4 2 and 3 is 5 3 and 3 is 6 4 and 3 is 7 1 and 4 is 5 2 and 4 is 6 3 and 4 is 7 4 and 4 is 8 1 and 5 is 6 2 and 5 is 7 3 and 5 is 8 4 and 5 is 9 1 and 6 is 7 2 and 6 is 8 3 and 6 is 9 4 and 6 is 10 1 and 7 is 8 2 and 7 is 9 3 and 7 is 10 4 and 7 is 11 1 and 8 is 9 2 and Sis 10 3 and 8 is 11 4 and 8 is 12 1 and 9 is 10 2 and 9 is 11 3 and 9 is 12 4 and 9 is 13 1 and 10 is 11 2 and 10 is 12 3 and 10 is 13 4 and 10 is 14 1 and 11 is 12 2 and 11 is 13 3 and 11 is 14 4 and 11 is 15 1 and 12 is 13 2 and 12 is 14 3 and 12 is 15 4 and 12 is 16 5 and 1 is 6 6 and 1 is 7 7 and 1 is 8 8 and 1 is 9 5 and 2 is 7 6 and 2 is 8 7 and 2 is 9 8 and 2 is 10 5 and 3 is 8 6 and 3 is 9 7 and 3 is 10 8 and 3 is j 1 5 and 4 is 9 6 and 4 is 10 7 and 4 is 11 8 and 4 is 12 5 and 5 is 10 6 and 5 is 11 7 and 5 is 12 8 and 5 is 13 5 and 6 is 11 6 and 6 is 12 7 and 6 is 13 8 and 6 is 14 5 and 7 is 12 6 and 7 is 13 7 and 7 is 14 8 and 7 is 15 5 and 8 is 13 6 and 8 is 14 7 and 8 is 15 8 and 8 is 16 5 and 9 is 14 6 and 9 is 15 7 and 9 is 16 8 and 9 is 17 6 and 10 is 15 6 and 10 is 16 7 and 10 is 17 8 and 10 is 18 5 and 11 is 16 6 and 11 is 17 7 and 11 is 18 8 and 11 is 19 5 and 12 is 17 6 and 12 is 18 7 and 12 is 19 8 and 12 is 20 9 and 1 is 10 10 and 1 is 11 Hand lis 12 12 and 1 is 13 9 and 2 is 11 10 and 2 is 12 11 and 2 is 13 12 and 2 is 14 9 and 3 is 12 10 and 3 is 13 11 and 3 is 14 12 and 3 is 15 9 and 4 is 13 10 and 4 is 14 11 and 4 is 15 12 and 4 is 16 9 and 6 is 14 10 and 5 is 15 11 and 5 is 16 12 and 5 is 17 9 and 6 is 15 10 and 6 is 16 11 and 6 is 17 12 and 6 is 18 9 and 7 is 16 10 and 7 is 17 11 and 7 is 18 12 and 7 is 19 9 and 8 is 17 10 and 8 is 18 11 and 8 is 19 12 and 8 is 20 9 and 9 is 18 10 and 9 is 19 11 and 9 is 20 12 and 9 is 21 9 and 10 is 19 10 and 10 is 20 11 and 10 is 21 12 and 10 is 22 9 and 11 is 20 10 and 11 is 21 11 and 11 is 22 12 and 11 is 23 9 and 12 is 21 10 and 12 is 22 11 and 12 is 23 12 and 12 is 24 Since has no value, the sum of any number and is the number itself; thus, 19 and is 19. §1 ADDITION AND SUBTRACTION 13 from 1 to 12; it should be thoroughly committed to memory, and to do this the reader should propose to himself all kinds of com- binations of two numbers until as soon as the numbers are named he can give the sum almost unconsciously. When at work, when walking, or at any time when he has leisure he should practice by saying 8 and 9 is 17, 9 and 5 is 14, 5 and 11 is 16, 9 and 8 is 17, etc. until he can name the sum instantly and correctly. 34. To Add a Digit and a Number Greater than 12. — The sum of a single digit and a number greater than 12 can always be obtained mentally. For example, consider the numbers 58 and 9; write these numbers, one above the other, so that the unit figures stand in the same column, and draw a line under them. 58 or 9 1- 58 Now add the unit figures, the sum being 17, and write the sum underneath the fine with the unit figure 7 standing in the same column as the unit figures 8 and 9. Now remembering that the 5 in 58 represents 50 add this to the sum of the unit figures, obtain- ing 67 for the sum of 58 and 9 (a) ib) (c) (d) 58 9 58 9 _9 58 _9 58 17 17 67 67 50 50 67 67 Now notice that the same result may be obtained with less figures, in fact mentally, from the following considerations: The sum of any two digits cannot exceed 18, since the largest digit is 9, and 9 + 9 = 18; 18 = 10 + 8 or 1 ten and 8 units. Hence, when two numbers are added as above, and the sum of the unit figures is 10 or a greater number, write the unit figure of the sum of the unit figures and carry the 1 ten to the next left-hand column and add it mentally to the figure or figures in that column. Thus, in (c) above, 9 and 8 is 17; write 7 and carry 1, which added to 5 makes 6, and 58 + 9 = 67. In (d), 8 and 9 is 17; write 7 and carry 1, which added to 5 makes 6. 35. The student will find it greatly to his advantage to practice adding single digits to numbers less than 100 at every convenient opportunity; let him say, for example, 47 and 8 is 55, 23 and 9 is 32, 36 and 4 is 40, etc., the entire operation being performed 14 ARITHMETIC §1 mentally. As soon as proficiency has been obtained in adding simple numbers like these, there will be little trouble in adding any number of figures correctly. Note that if the sum of the figures in the right-hand, or lower, order does not exceed 9, the digit in the next order does not change; but if the sum is 10 or more, the digit is increased by 1. 36. If the larger number contains more than two figures, the operation is exactly the same, except that if the digit in the next to the lowest order is 9, the digit in the next higher order must also be increased by 1. Thus 246 and 5 is 251, since 5 + 6=11, and 4 + 1 = 5; 594 and 9 is 603, since 9 and 4 is 13, 1 and 9 is 10, and 1 and 5 is 6; 1997 and 8 is 2005, since 8 and 7 is 15, 1 and 9 is 10, 1 and 9 is 10, 1 and 1 is 2; etc. 37. To Add any Two Numbers. — The sum of any two numbers is found as follows: Suppose the numbers are 57,962 and 9,458. Write them so that the figures of the same order stand in the same column and draw a line underneath. 57962 9458 67420 Add the figures in the right-hand column first, obtaining 10; write the and carry the 1. Add the 1 to the 5 in the next column (making 6), and then add this result to the 6 aboVe it, obtaining 12; write the 2 and carry the 1. Add 1 to the 4 in the next column (making 5), and add this sum to the 9 above it, obtaining 14; write the 4 and carry the 1. Add 1 to the 9 in the next column (making 10), and add this sum to 7 above it, obtaining 17; write the 7 and carry the 1. The 1 added to the 5 in the next column gives 6, which is written as shown. 38. The procedure is exactly the same when either or both the numbers contain decimals. In this case the easiest way to get the figures of the same order under each other is to place the decimal points under each other in the same column. 26.943 0.434 87.51 5.71 752.933 29.492 32.653 753.367 117.002 In all three of the cases, it will be noted that units are placed under units, tens under tens, tenths under tenths, etc., and that §1 ADDITION AND SUBTRACTION 15 the decimal point in the result is also directly under the decimal points in the numbers added. 39. To Add more than Two Numbers. — Example. — Find the sum of 8, 4, 6, 9, 7 and 3. SOLITTION. — 8 4 6 9 7 _3 37 Ans. Explanation. — Beginning at the bottom of the column, add the first two numbers; to the sum, add the third number; to this sum, add the fourth number; etc. Thus, 3 and 7 is 10; 10 and 9 is 19; 19 and 6 is 25; 25 and 4 is 29; 29 and 8 is 37. After practicing addition in this manner, until a certain degree of proficiency has been attained, the reader should abbreviate the process by naming only the sums; thus, 3, 10, 19, 25, 29, 37. This will greatly increase his speed in adding. EXAMPLES Find the sum of the following sets of numbers. (1) 3+5 + 1+9 + 7 = ? Ans. 25. (2) 4 + 2 + 6+8 = ? Ans. 20. (3) 3+4+6+8 + 7 = ? Ans. 28. (4) 5+8 + 1+6 + 9+4 = ? Ans. 33. 40. When the numbers contain more than one, figure use the following rule: Rule. — I. Write the numbers so that the figures of the same order will form columns having units under units, tens under tens, tenth under tenths, etc., and draw a line under the bottom row of figures. II. Beginning with the right-hand column, add the digits in that column, and write the unit figure of the sum under the line and in the column that was added. Add the tens figure of the sum to the digits in the next column to the left, add this column and write the unit figure of the sum under the line in the column added. Add the tens figure of the sum (if any) to the digits in the third column, proceeding in this manner until the addition is finished, which will be when all the columns have been added. III. If a cipher (0) occurs anywhere, disregard it, since it does not affect the sum. 16 ARITHMETIC §1 Adding the tens figure of the sum of the digits in any column to the digits in the next column to the left is called carrying. Example.— 236 + 109 + 871 + 52 + 467 + 696 = ? Solution. — 236 109 871 52 467 696 2431 Ans. Explanation. — Arrange the numbers as shown, with units under units, tens under tens, etc. Beginning with the right-hand column, say 6, 13, 15, 16, 25, 31; write the 1 under the line and in the column just added, and carry 3 to the next column. Say 3, 12, 18, 23, 30, 33; write 3 and carry 3 to add to the digits in the next column. Say 3, 9, 13, 21, 22, 24. As there are no more columns to add, write the last sum, 24, as shown. The sum of all the numbers is 2431. It is always a good plan to write the abbreviation Ans. (which means answer) after the final result has been obtained. 41. If some of the numbers contain decimals, they are added according to the same rule. Arranging the numbers with units under units, etc., brings the decimal points under one another so that they all stand in the same column ; hence, I of the rule in Art. 40 might be changed to read: arrange the numbers so that the decimal points stand in the same column. The rest of the rule requires no change. Example.— Add the following numbers : 403.7819, 21.875, 5.2742, 369.92, and 7923.917. Solution.— 403.7819 21.875 5.2742 369 . 92 7923.917 8724.7681 Ans. Explanation. — Arranging the numbers so that the decimal points stand in the same column, say 2, 11, and write 1 and carry 1. Say 1, 8, 12, 17, 18; write 8 and carry 1. Say 1, 2, 4, 11, 18, 26; write 6 and carry 2. Say 2, 11, 20, 22, 30, 37; write 7 and carry 3, also writing the decimal point before the 7. Say 3, 6, 15, 20, 21, 24; write 4 and carry 2. Say 2, 4, 10, 12; write 2 and carry 1. Say 1, 10, 13, 17; write 7 and carry 1. Say 1, 8, and write 8. The entire sum is 8724.7681. 42. In bookkeeping, and when making out bills, business state- ments, etc., the decimal point is usually omitted, a vertical line being used in its place. This practice tends to prevent mistakes §1 ADDITION AND SUBTRACTION 17 and saves the writing of decimal points, the vertical line sepa- rating the dollars from the cents. Example.— Add $66.72, $243.27, $127.85, $37.40, and $101.32. Solution.— $ 66 72 243 27 127 85 37 40 10132 $576 56 Ans. Explanation. — Here the vertical line is used instead of the decimal points. The addition is performed in the usual manner, and the sum is found to be $576.56. 43. Checking Results.- — -In business and in engineering, it is of the utmost importance that the final result be correct. While mistakes or blunders are liable to occur, even when the greatest care is taken, they must be overcome by some method of detecting them, and corrections must be made before the final result is accepted. In checking the work of addition, the various num- bers may be added again, but if this is done immediately after the first adding, the same mistake is likely to be made again. A better way is to add the several columns downward; this causes a change in the sequence of the digits added, and tends to prevent a second mistake like the first. If the same result is obtained when adding down as was obtained when adding up, the work is presumed to be correct; but if a different result is obtained, repeat the work until the same result is obtained by adding both ways. This practice of testing the work to see if it is correct is called checking, and the method used in checking is called a check. The reader should apply the check by adding down to the two examples of the preceding article. EXAMPLES (1) 26.48 + 360.72 + 54.068 + 7.205 + 509.045 = ? Ans. 957.518. (2) 19681 + 89320 + 20358 + 33146 + 7508 = ? Ans. 170013. (3) 5818 + 8726 + 6791 + 9809 + 12463 + 752 = ? Ans. 44359. (4) 6.317 + 49.42 + 17.718 + 5.801 + 83.77 = ? Ans. 163.026. (5) The weights of seven bags of alum are 292 pounds, 305 pounds, 301 pounds, 298 pounds, 297 pounds, 307 pounds, and 303 pounds; what is their total weight? Ans. 2103 pounds. (6) What is the weight of six rolls of paper, if the rolls weigh 957 pounds, 1048 pounds, 1075 pounds, 993 pounds, 986 pounds, and 979 pounds? Ans. 6038 pounds. 18 ARITHMETIC §1 (7) Paid the following amounts for bales of rags: $18.82, $18.94, $19.11, $19.20, $18.85, $18.97. What was the total amount paid? Ans. $113.89. (8) Six barrels of rosin were placed on an elevator; if the barrels weighed 389 pounds, 411 pounds, 395 pounds, 399 pounds, 408 pounds, and 406 pounds, how much must the elevator raise, if it carries in addition to the rosin a man weighing 168 pounds? Ans. 2576 pounds. SUBTRACTION 44. Subtraction means taking away. In arithmetic, subtrac- tion is the process of taking one number from another number, or it is the process of finding how much greater one number is than another. If a person has 10 cents and spends 4 cents, he will have 6 cents left. Here 4 cents are taken from 10 cents, and 6 cents are left. The arithmetical operation of taking 4 cents from 10 cents is called subtraction. Again, how much greater is 10 cents than 4 cents? If 6 cents are added to 4 cents, the sum is 10 cents; hence, 10 cents is 6 cents greater than 4 cents. The same result will be obtained by subtracting 4 cents from 10 cents. 45. The sign of subtraction is — , a short horizontal line; it is read minus, and means less; it indicates that the number following it is to be subtracted from the number preceding it. Thus, 10 — 4 = 6; here 4 is to be subtracted from 10, the result being 6; the expression may be read either as 10 minus 4 equals 6 or as 10 less 4 equals 6. The longer expression, 4 subtracted from 10 equals 6, is also correct. 46. The number subtracted is called the subtrahend, and the number from which the subtrahend is subtracted is called the minuend; the result is called the remainder or difference. When it is not desired to be specific regarding the order of the numbers, the word difference is generally used ; thus, the difference between 4 and 10 or the difference between 10 and 4 is 6. But when the order of the numbers is definitely stated, as 10 minus 4 is 6, 6 is called the remainder, it being what is left after taking 4 from 10. In an expression like 15 — 9 = 6, 15 is the minuend, 9 is the subtrahend, and 6 is the remainder. Before explaining the general process of subtraction, the two following principles or laws should be thoroughly understood: 47. Principle I. — If the same number he added to both minuend and subtrahend, the remainder is unchanged. For example, 15 — 9 = 6; if now, 10 be added to both 15 and 9, they become 25 §1 ADDITION AND SUBTRACTION 19 and 19 respectively, and 25 — 19 = 6, the same remainder as before. The reason is evident; the 10 added to the subtrahend is subtracted from the 10 added to the minuend, thus leaving the minuend and subtrahend the same as before the 10 was added. A like result will be obtained, no matter what number is added; thus, adding 7 to 15 and 9, these numbers become 22 and 16, and 22 — 16 = 6, as before. 48. Principle II. — If the remainder he added to the subtrahend, the sum will he the minuend. For instance, if a stick is 13 feet long and 5 feet are cut off-, the length of the stick is then 8 feet. Here 13 — 5 = 8, and 5 is the subtrahend and 8 is the remainder. But when the two parts of the stick are placed together, they must be equal in length to the original stick; that is, 5 + 8 = 13, or subtrahend + remainder = minuend. Hence, instead of saying 5 from 13 leaves 8, it will be equally proper to say "what number added to 5 will make 13?" this number is 8; therefore, 13 - 5 = 8, because 5 + 8 = 13. 49. The reader will find it greatly to his advantage to reverse the addition table of Art. 33. Thus, 9 and 6 is 15; 15 less 9 is 6, and 15 less 6 is 9. Again, 8 and 5 is 13; 13 less 8 is 5, and 13 less 5 is 8. He should practice in this manner both addition and subtraction in spare moments, until the result presents itself instantly as soon as the numbers are named. 50. If the right-hand figure of the minuend is smaller than the digit to be subtracted, add 10 to it, subtract, and then subtract 1 from the next left-hand figure of the minuend; thus, 34 — 8 = 26. Here 10 is added to 4, making 14; 8 from 14 leaves 6; then subtracting 1 from 3, the remainder is 2. This is in accordance with Principle I, Art. 47, since 10 is added to the minuend (really making it 44, but expressed as 30 + 14) and 10 is added to the subtrahend (really making it 18). The result is correct; since, by Principle II, 8 + 26 = 34; also 18 -|- 26 = 44. It is thus seen that 36 - 9 = 27, 43 - 7 = 36, 81 - 4 = 77, etc. The reader should practice these combinations also. 61. Bearing the foregoing in mind, the following is the rule for subtraction : Rule I. — Place the numbers as in addition, with the subtrahend under the minuend, and with units under units, tens under tens, etc. or with the decimal ^points in the same column. II. Beginning with the right-hand digit of the subtrahend, sub- 20 ARITHMETIC §1 tract it from the digit above it in the minuend; do the same with the next digit or figure to the left, proceeding in this manner until all the figures in the subtrahend have been subtracted from the figures above them in the minuend. III. If a figure in the minuend is a cipher or is smaller in value than the figure under it in the subtrahend, add 10 to it before sub- tracting, and then carry 1 to the next figure of the subtrahend before subtracting it from the figure above. IV. Having found the remainder, add it to the subtrahend, figure by figure, and if the sum equals the minuend, the remainder is very probably correct. This last operation is a check. Example 1.— From 93875 take 72032. Solution.— 93875 72032 21843 Ans. Explanation. — Writing the numbers as directed in I of the rule, begin with the right-hand digit and say 2 from 5 is 3, 3 from 7 is 4, write 8 (since from any number leaves the number), 2 from 3 is 1, and 7 from 9 is 2. The remainder, therefore, is 21,833. To check this result, add the remainder to the subtrahend; thus, 3 and 2 is 5, 4 and 3 is 7, 8, 1 and 2 is 3, and 2 and 7 is 9. The sum being the same as the minuend, the work is very probably correct. Example 2.— From $1437.50 take $918.27. Solution.— $1437.50 918.27 $ 519.23 Ans. Explanation. — Since 7 cannot be subtracted from 0, add 10 to 0, obtain- ing 10; then 7 from 10 is 3. Carry 1 to 2 making it 3, and 3 from 5 is 2. Write the decimal point. Then, since 8 cannot be taken from 7, add 10, making 17; and 8 from 17 is 9. Carry 1 to 1 making it 2, and 2 from 3 is 1. Lastly, 9 from 14 is 5. If the remainder be added to the subtrahend, the sum is the minuend. Hence, the correct remainder is $519.23. Example 3. — There are four piles of papers; the first pile contains 986 sheets, the second pile contains 753 sheets, the third pile contains 875 sheets, and the fourth pile contains 1038 sheets; they were all placed in one pile, and 2369 sheets were taken away; how many sheets were left? Solution. — The first step is to find how many sheets there were all together. This found by adding the "°" sheets number of sheets in each pile, the sum or total being '^'^ 3652 sheets. The next step is to subtract from this °'^ sum the number of sheets that were taken away, the l^oo remainder thus found being 1283 sheets. To check, 3652 sheets add down, then add the remainder to the subtrahend, 2369 obtaining 3652, the minuend. Since the work checks 1283 sheets .Ans. in both cases, it is probably correct. §1 ADDITION AND SUBTRACTION 21 Example 4.— From 19.125 take 9.3125. Solution.— 19.1250 9.3125 9.8125 Ans. Explanation. — There is no figure in the minuend above the right-hand figure of the subtrahend; but since the subtrahend contains a decimal, it is allowable (for reasons that will be explained later) to annex ciphers to the minuend until it contains the same number of figures in the decimal part that there are in the subtrahend. The subtraction is then performed in the regular manner. Should there be more figures in the decimal part of the minuend than in the decimal part of the subtrahend, annex ciphers to the subtrahend until the decimal parts of the subtrahend and minuend both contain the same number of figures, before subtracting. 52. Although subtraction is a much simpler and easier opera- tion than addition, it is probable that more mistakes are made in subtraction than in addition, the reason being that the operator fails to check by "adding back" and is more inclined to be care- less. It is well not to attempt to work too fast at first; try to secure accuracy, and speed will come with practice. This advice applies to every process of computation. EXAMPLES (1) From 417.23 take 273.58. Ans. 143.65. (2) From 54.375 take 48.65625. Ans. 5.71875. (3) From 484814 take 258177. . Ans. 226637. (4) From 1002070 take 304098. Ans. 697972. (5) 2033.4238 - 1792.3917 = ? Ans. 241.0321. (6) From a pile of pulp weighing 14,253 tons 9,468 tons were sold; how much pulp remained in the pile? Ans. 4,785 tons. (7) Three cars were receiv^ed containing 58,024 pounds of sulphur; when the stock was cleaned up, mill figures showed that 55,638 pounds had been used; how much had been lost in handling? Ans. 2,386 pounds. (8) To a pile of pulp weighing 4,826 tons, 527 tons were added; how much was left after 3,962 tons had been sold? Ans. 1,391 tons. (9) A bleach liquor tank contained 925 gallons; how much remained after 146 gallons had been used? Ans. 779 gallons. 22 ARITHMETIC §1 MULTIPLICATION AND DIVISION MULTIPLICATION 63. In arithmetic, multiplication may be defined as a short method of adding (or finding the sum of) several equal numbers. For instance, suppose there are 6 books, (") each book containing 1892 pages; how 1892^^^^^ ^^^^ P^^®^ ^^^ ^^ *^^ ^ books? To jgg2 find the total number, 1892 may be 1892 written six times as shown at (a), and the 1892 sum found, which is 11352 pages. By ^^^^ using the process of multiplication, how- 11352 pages Ans. ever, the work will be arranged as shown ,,. at (6), and 1892 is then multiplied by 6, jgg2 the result being the same as before. 6 Note the great saving in figures even in 11352 pages Ans. this simple case. Suppose there had been, say, 576 books each containing 1892 pages; it would then be practically impossible to find the total number of pages by the method shown at (a), but the total can easily be found by multiplication. 54. The number multiplied (1892 in Art. 53) is called the multi- plicand; the number used to multiply the multiplicand (and which shows how many times the multiplicand is to be added) is called the multiplier (this is 6 in Art. 53) ; the result of the opera- tion of multiplication is called the product (in Art. 53, 11352 is the product). When it is not desired to be specific and make a special dis- tinction between the multiplicand and multiplier, these two num- bers are called factors; in Art. 53, 1892 and 6 are factors of 11352, their product. 55. The sign of multiplication is an oblique (St. Andrew's) cross, which is written between the factors; it is read times or multiplied by. Thus, 6 X 8 = 48 is read either as six times eight equals forty-eight or as six multiplied by eight equals forty- eight; here 6 and 8 are factors, 6 being the multiplier in the first case and 8 being the multiplier in the second case. 56. Insofar as the product is concerned, it makes no difference which of two factors is considered as the multiplier, since 6 times MULTIPLICATION AND DIVISION 23 8 is the same, for instance, as 8 times 6, the product in both cases being 48; this may be expressed mathematically as 6 X 8 = 8 X 6 = 48. 57. Before one can multiply, it is necessary that he memorize the multiplication table. This may take a little time, but it is absolutely necessary if the reader is to be successful in this subject. MULTIPLICATION TABLE 1 times 1 s 1 2 times 1 s 2 3 times 1 is 3 4 times 1 is 4 1 times 2 s 2 2 times 2 s 4 3 times 2 is 6 4 times 2 is 8 1 times 3 s 3 2 times 3 s 6 3 times 3 is 9 4 times 3 is 12 1 times 4 s 4 2 times 4 s 8 3 times 4 is 12 4 times 4 is 16 1 times 5 s 5 2 times 5 s 10 3 times 5 is 15 4 times 5 is 20 1 times 6 s 6 2 times 6 s 12 3 times 6 is 18 4 times 6 is 24 1 times 7 s 7 2 times 7 s 14 3 times 7 is 21 4 times 7 is 28 1 times 8 s 8 2 times 8 s 16 3 times 8 is 24 4 times 7 is 32 1 times 9 s 9 2 times 9 s 18 3 times 9 is 27 4 times 9 is 36 1 times 10 s 10 2 times 10 s 20 3 times 10 is 30 4 times 10 is 40 1 times 11 s 11 2 times 11 s 22 3 times 11 is 33 4 times 11 is 44 1 times 12 s 12 2 times 12 s 24 3 times 12 is 36 4 times 12 is 48 5 times 1 s 5 6 times 1 s 6 7 times 1 is 7 8 times 1 is 8 5 times 2 s 10 6 times 2 s 12 7 times 2 is 14 8 times 2 is 16 5 times 3 s 15 6 times 3 s 18 7 times 3 is 21 8 times 3 is 24 5 times 4 s 20 6 times 4 s 24 7 times 4 is 28 8 times 4 is 32 5 times 5 s 25 6 times 5 s 30 7 times 5 is 35 8 times 5 is 40 5 times 6 s 30 6 times 6 s 36 7 times 6 is 42 8 times 6 is 48 5 times 7 s 35 6 times 7 s 42 7 times 7 is 49 8 times 7 is 56 5 times 8 s 40 6 times 8 s 48 7 times 8 is 56 8 times 8 is 64 5 times 9 s 45 6 times 9 s 54 7 times 9 is 63 8 times 9 is 72 5 times 10 s 50 6 times 10 s 60 7 times 10 is 70 8 times 10 is 80 5 times 11 s 55 6 times 11 s 66 7 times 11 is 77 8 times 11 is 88 5 times 12 s 60 6 times 12 s 72 7 times 12 is 84 8 times 12 is 96 9 times 1 s 9 10 times 1 3 10 11 times 1 is 11 12 times 1 is 12 9 times 2 s 18 10 times 2 s 20 11 times 2 is 22 12 times 2 is 24 9 times 3 s 27 10 times 3 s 30 11 times 3 is 33 12 times 3 is 36 9 times 4 s 36 10 times 4 s 40 11 times 4 is 44 12 times 4 is 48 9 times 5 s 45 10 times 5 s 50 11 times 5 is 55 12 times 5 is 60 9 times 6 s 54 10 times 6 s 60 1 1 times 6 is 66 12 times 6 is 72 9 times 7 s 63 10 times 7 s 70 11 times 7 is 77 12 times 7 is 84 9 times 8 s 72 10 times 8 s 80 11 times 8 is 88 12 times 8 is 96 9 times 9 s 81 10 times 9 s 90 11 times 9 is 99 12 times 9 is 108 9 times 10 IS 90 10 times 10 s 100 11 times 10 is 110 12 times 10 is 120 9 times 11 IS 99 10 times 11 s 110 11 times 11 is 121 12 times 11 is 132 9 times 12 s 108 10 times 12 IS 120 11 times 12 is 132 12 times 12 is 144 The product of any number and is (9 X = 0, 167 X = 0, etc.) ; this is evident, since the sum of any number of O's cannot make an integer. It will be noted by reference to the table that the product of any number and 1 is the number itself; thus, 409 X 1 = 409, since 24 ARITHMETIC §1 409 X 1 is 409 I's, which is, of course, 409. Note, also, that the product of any number and 10 is the number itself with a cipher (0) added at the right; thus, 7 X 10 = 70, 526 X 10 = 5260, etc. Note, again, that the product of 11 and any number not greater than 9 is the number repeated; thus, 3X11 = 33, 6X 11 = 66, 9 X 11 = 99, etc. The reader should repeat the different parts of the table to himself at odd times until it becomes so firmly impressed on his memory that as soon as any two numbers are named, their pro- duct will instinctively name itself. 58. To Multiply any Number by a Single Digit. — Multiply 31415927 by 7. Here 31415927 is the _ multiplicand and 7, the multiplier, is omnn^cf^ A written under the right-hand figure of the 219911489 Ans. . . ° . * multiplicand. Draw a line under the factors, as shown, and multiply the right-hand figure of the multiplicand by 7 the multiplier, obtaining 7X7 = 49. Write the unit figure of the product under the unit figure of the multipli- cand, and carry 4, the tens figure of this product. Then say 7 times 2 is 14, to which add 4, the figure carried, making 18; write 8 as shown and carry 1. Say 7 times 9 is 63, add the 1 carried, making 64; write 4 and carry 6. Say 7 times 5 is 35 and 6 (the figure carried) is 41; write 1 and carry 4. Say 7 times 1 is 7 and 4 (the figure carried) is 11; write 1 and carry 1. Say 7 times 4 is 28 and 1 is 29; write 9 and carry 2. Say 7 times 1 is 7 and 2 is 9; write 9 and there is nothing to carry. Lastly, say 7 times 3 is 21, v/hich write. Every figure in the multiplicand has now been multiplied by the multiplier, 7, and the product is 219,911,489. Had the multiplier been 70, 700, 7000, etc., the process would have been exactly the same, except that after the product was found, as many ciphers would have been annexed to the product as there were ciphers to the right of the right-hand digit of the multiplier; thus, 31,415,927 X 7000 = 31415927^^^ 219,911,489,000. The work is shown in 219911489000 Ans. *^^ margin. First multiply by 7 as before; then to the product, annex three ciphers, because there are three ciphers to the right of the right- hand digit of the multiplier . 59. To Multiply When the Multiplier Contains Two or More Digits. — Place the multiplier under the multiplicand, with the §1 MULTIPLICATION AND DIVISION 25 right-hand digit of the multiplier under the right-hand digit of the multiplicand. Multiply by the right-hand digit of the multi- plier and write the product figure by figure under the multiplier, as shown in the margin. This result is 428095 called the first partial product. Note that after multiplying 9 by 2 there is 1 to carry; 1712380 *^®^ ^^y ^ times naught is naught and 1 is 29966650 Ij which write as shown. Now multiply by 1284285 the next figure to the left of 2, in this case 15857494990 Ans. 4. Say 4 times 5 is 20; write the cipher one place to the left of the right-hand figure of the first partial product, thus bringing the cipher under the figure multiphed by. Continue the multiphcation by 4, obtain- ing 1712380, which is called the second partial product. The third figure to be used as a multiplier is 0, and since any number multiphed by is 0, write a cipher one place to the left of the right-hand figure of the second partial product, which brings it directly under the cipher in the multiplier. Now multiply by the next figure, 7, of the multiplier. Say 7 thnes 5 is 35, write 5 alongside the cipher and carry 3; this brings the 5 under the figure used as a multiplier, and makes the third row of figures 29966650, the third partial product, which is equal to 428095 X 70. Finally, multiply by 3, the left-hand digit of the multiplier, and the result is the fourth partial product, the right-hand figure of which is written under 3, the number multiplied by. Now adding the four partial products, the sum is 15,857,494,990, which is the entire product, or the result sought. If there are ciphers to the right of the multiplicand or multiplier or both, pay no attention to them until after the product has been found as just described. Then annex to the entire product as many ciphers as there are ciphers to the right of either or both factors. For instance, to multiply 205000 526700 by 205000, arrange as shown in 26335 — *^^ margin, with the right-hand digits 105340 o^ *^6 multipHcand and multiplier under 107973500000 Ans. ^ach other. Multiply 5267 by 205, the product being 1079735; there are two ciphers to the right of one factor and three to the right of the other factor; hence, annex 2 + 3 = 5 ciphers to the right of the entire product, which is thus found to be 107,973,500,000. 26 ARITHMETIC §1 60. Rule. — Write the multiplier under the multiplicand with the right-hand digits under each other. Beginning with the right-hand digit of the multiplier, and proceeding to the left, multiply the upper factor by each figure of the loiver factor, or multiplier, writing the right-hand figure of each partial product under the figure used as a multiplier. Then add the partial products, and the sum will be the entire product. 61. Check for Multiplication. — The best way to check multi- pHcation is to employ the process called "casting out nines." This consists in dividing (the operation of dividing will be con- sidered in the next article) the two factors by 9, multiplying the remainders, and if the product is greater than 9, divide that by 9; note the remainder. Then divide the entire product by 9, and if the remainder is the same as that first obtained, the work is very probably correct. If the two remainders differ, however, then the work is wrong, some mistake having been made. In- stead of dividing by 9, the remainder may be found by adding the digits; if the sum is greater than 10, add the digits of the sum, pro- ceeding in this manner until a single digit has been found, which will be the remainder when the number is divided by 9. Thus, consider the number 7854; the sum of the digits is 7 + 8 + 5 + 4 = 24; 2 + 4 = 6, and 6 is the remainder when 7854 is divided by 9. Applying this check to the first example of Art. 59, the sum of the digits in the multipHcand is 4 + 2 + 8 + 9 + 5- 28, 2+8 = 10, and 1 + = 1; the sum of the digits in the multiplier is 3 + 7 + 4 + 2 = 16, and 1 + 6 = 7; then 1X7 = 7. The sum of the digits in the entire product is 1 + 5 + 8 + 5 + 7 + 4 + 9 + 4 + 9 + 9 = 61, and 6 + 1=7. Since the remainders are both 7, the work is very probably correct. When adding the digits in this manner, it is not necessary to add any 9's; thus, in the foregoing, the remainder for the multiplicand is 4 + 2 + 8 + 5 = 19 or 1, and the remainder for the product is 1 + 5 + 8 + 5 + 7 + 4 + 4 = 34, and 3 + 4 = 7, both results being the same as obtained before. Applying this check to the second example of Art. 59, 5 + 2 + 6 + 7 = 20, or 2; 2 + 5 = 7; 2 X 7 = 14, and 1 + 4 = 5; 1 + 7 + 7 + 3 + 5 = 23, and 2 + 3 = 5. Since the remain- ders are the same, the work is very probably correct. The reader is strongly advised to apply this check in every case. §1 MULTIPLICATION AND DIVISION 27 EXAMPLES (1) 7854 X 2038 = ? Ans. 16,006,452. (2) 230258 X 90057 = ? Ans. 20,736,344,706. (3) 31831 X 31416 = ? A7is. 1,000,002,696. (4) 543836 X 4688 = ? Ans. 2,549,503,168. (5) 197527 X 98743 = ? Ans. 19,504,408,561. (6) 295369 X 700405 = ? Ans. 206,877,924,445. (7) The average consumption of coal by a mill per year is 28,750 tons; if the average cost per ton is $7.00, what is the annual cost to the mill for coal? Am. $201,250. (8) During a certain period, a mill consumed 686 tons of alum ; if the price paid for the alum was $49.00 per ton, how much was the total amount paid for the alum? Ans. $33,614. (9) What is the value of 13,908 cords of wood at $11.00 per cord? Ans. $152,988. (10) How much must be paid for 12 cans of dye stuff, if each can weighs 47 pounds and the dye stuff is worth $19.00 per pound? Ans. $10,716. (11) At different times a certain mill sold paper as follows: 27,848 pounds at 14 cents per pound; 17,005 pounds at 18 cents per pound; 9,990 pounds at 19 cents per pound; and 36,476 pounds at 15 cents per pound. How much was received from these sales? Ans. $14,329.12. DIVISION 62. Division means a partition, a separating into parts. In arithmetic, division is the process of finding how many times larger one number is than another; thus, since 24 11352 ^^ ^ times 4, 24 is 4 times as large as 6 or 6 times as 1892 large as 4. Division may also be defined as the 9460 process of separating a number into a required 1892 number of equal parts; thus, 24 may be separated 7568 into 6 equal parts of 4 each or 4 equal parts of 6 each. 1^92 Just as multiplication is a short process or 5676 method of adding equal numbers so division is a short process or method of subtracting continuously ^^^ until the remainder is or less than the subtra- — — hend. For example, referring to Art. 53, suppose 1892 there are 11,352 pages in a certain number of books ^^ each containing 1892 pages, and it is required to find the number of books. The work might be ^^<^ done as shown at (a) in the margin, subtracting 11352 ^^^2 from 11352, then subtracting 1892 from the Q remainder, continuing this process until the remainder becomes or less than the subtrahend. 28 ARITHMETIC §1 In this case, the remainder is 0; and as the subtraction was performed 6 times, there are 6 books. By the process of divi- sion, as shown at (b), 1892 is contained 6 times in 11352, because 1892 X 6 = 11352. Note the great saving of j&gures in the second method. Had it been known that the total number of pages was 11,352, the number of books was 6, and it were required to find the number of pages in each book, it would be necessary to subtract 6, by the method shown at (a), 1892 times, which is practically impossible. In multiplication, the object to be attained is to find the product of two numbers (factors); in division, the object is to divide a number into two factors, one of them being given. 63. The number that is to be divided into two factors is called the dividend; the given factor, which is divided into the dividend, is called the divisor; the other factor, which is obtained by divid- ing the dividend by the divisor, is called the quotient; anything that is left over after the division has been performed is called the remainder. In (b), Art. 62, 11,352 is the dividend, 1892 is the divisor, 6 is the quotient, and the remainder is 0. Whenever the remainder is 0, the division is said to be exact. 64. There are several signs for division, the principal one being a colon ( :) or a colon with a short horizontal line between the dots (-^). When either of these two signs occurs between two numbers, it means that the number on the left is to be divided by the number on the right; thus, 24 -^ 6 means that 24 is to be divided by 6, 24 being the dividend and 6 the divisor. In some cases, a vertical line is used in place of the regular sign of division; thus, 24 1 6. The vertical line is seldom used between two numbers; it is most frequently used when the product of several 84 numbers is to be divided by the product of 112 several other numbers; thus, the product of 124, 49, and 75 divided by the product of 84 and 112 may be indicated as shown in the margin, the product of the numbers to the left of the vertical line being the dividend, and the product of the numbers to the right being the divisor. Most commonly, however, in cases of this kind, a horizontal line is used, the number or numbers above the line being divided 24 . by the number or numbers below it; thus, -^ is read 24 over 6, and means 24 divided by 6; also, — qa v 1 1 9 — means that the 124 49 75 §1 MULTIPLICATION AND DIVISION 29 product of 124, 49, and 75 is to be divided by the product of 84 and 112. An inclined line is also frequently used; thus 24/6 means 24 divided by 6. 65. Short Division. — When the divisor is not greater than 12, it is customary to employ what is called short division. The process is best illustrated 8)5 463^863^2 by an example. For instance, 543,832:8 = ? 6 7 9 7 9 Write the divisor to the left of the dividend with a curved line between, as shown in the margin, and draw a straight line under the dividend. Since 8 is greater than 5, the left-hand digit of the dividend, consider the first two figures, 54. Now find what number multiplied by 8 will make 54 or whose product subtracted from 54 will be less than 8; since 8 X 6 = 48 and 8 X 7 = 56, this number is 6, and 54 — 48 = 6. Write 6 under 54 for the first figure of the quotient, and also write 6, the remainder, above and to the right of 4, as shown. This last 6 belongs to the same order as the 4 in 54 and the 3 that follows 4 is of the next lower order; hence, combine the 6 and 3, and call the number 63. Now find what number multiplied by 8 will make 63 or whose product subtracted from 63 will be less than 8; this number is 7, since 8 X 7 = 56 and 63 — 56 = 7. Write 7 under 3 for the second figure of the quotient, and also write 7, the remainder, as shown. The next number to be divided is 78, the 7 being prefixed to the fourth figure of the dividend, which is the figure of the next lower order. Here 8 X 9 = 72, and 78 - 72 = 6; write 9 under 8 for the third figure of the quotient and write 6, the remainder, as shown. Prefixing 6 to 3, the next figure of the dividend, 8 X 7 = 56, 63 — 56 = 7; hence, write 7 for the fourth figure of the quotient and write 7, the remainder, as shown. Finally, 8 X 9 = 72, and there is no remainder; hence, write 9 for the fifth figure of the quotient. As there are no more figures in the dividend, the quo- tient sought is 67,979. That this is correct may be proved by multiplying the quotient by the divisor, the result being 67979 X 8 = 543832, the dividend. In practice, the remainders would not be written in the manner indicated in the foregoing^ — they would be simply carried men- tally. The process would then be as follows : 8 into 54, 6 times and 6 over; 8 into 63, 7 times and 7 over; 8 into 78, 9 times and 6 over; 8 into 63, 7 times and 7 over; 8 into 72, 9 times and no re- 30 ARITHMETIC §1 mainder. Suppose it were required to divide the above number, 543,832 by 9. Say 9 into 54, 6 times and ^ nothing over; 9 into 3 no times; 9 into 38, 60425+ 7 rem. ^ ^.^^^ ^^^ 2 over; 9 into 23, 2 times and 5 over; 9 into 52, 5 times and 7 over. Since there are no more figures in the dividend, the quotient is 60,425 and 7 remainder. That by result is correct may be proved by multiplying the quotient by the divisor and adding the remainder to the product; thus, 60425 X 9 = 543825, and 543825 + 7 = 543832. Now note that the remainder is the same as that obtained in Art. 61 by adding the digits; thus 5 + 4 + 3 + 8 + 3 + 2 = 25, and 2 + 5 = 7. For reasons that will be explained later, it is customary to write the remainder over the divisor, with a line between, and annex this expression to the quotient. In the last example, 543832 -j- 9 = 60425^. Ans. This last expression may be read sixty thousand four-hundred-twenty-five and seven over nine. EXAMPLES (1) 197527 -4- 11 = ? Ans. 17,957. (2) 527324 ^ 7 = ? Ans. 75,332. (3) 900725 -H 6 = ? Ans. 150,120f. .^, 49503163 „ A A ioro«Q7 (4) Tg = ? Ans. 4,125,263i^. (5) 1580216/4 = ? Ans 395,054. (6) 4350688 ^ 3 = ? Ans. l,450,229i. (7) 2072623/10 = ? Ans. 207,262i3^. Note. —To divide a number by 10, write all the figures except the last for the quotient; the last figure will be the remainder (see example 7, above) . (8) Since there are 12 inches in one foot, how many feet are equivalent to 237 inches? Ans. l^-^^ feet. (9) How many nickels are equal in value to $3.45, one nickel being equal to 5 cents? Ans. 69 nickels. (10) Twelve of anything make a dozen, and twelve dozen make a gross. How many dozen balls of twine are in a shipment containing 5076 balls? also, how many gross were in the shipment ? Ans. 423 dozen ; 35^ gross. (11) There are 8 pints in one gallon; how many gallons are equivalent to 22,222 pints? Ans. 2777| gallons. (12) One yard is equal to three feet; how many yards are contained in 63360 inches? Ans. 1760 yards. (13) A quire of paper contains 24 sheets. How many quires are in a pile of 1784 sheets? Ans. 74^^ quires. §1 MULTIPLICATION AND DIVISION 31 66. Long Division. — When the divisor is greater than 12, the process called long division is used; this is best explained by an example. For instance, what is 64903358) 386 the quotient when 64,903,358 is 386 168143Hi ^«s. divided by 386 ? Write the divi- 2630 sor to the right of the dividend, 2316 with a line (either straight or 3143 curved) between, and draw a line ^Q^^ under the divisor, as shown. ^^^ Since the divisor contains 3 fig- — ures, compare it with the number Jg^^ made up of the first 3 figures of -r^g the dividend, in this case, 649, jj5g which call the first trial dividend. "^ Note that 649 is larger than 386, which is nearly equal to 400; calling it 400, it is seen that 400 is contained in 649, 1 time, and 1 is thus the first figure of the quotient, which is written under the divisor. Now multiply the divisor by 1, the figure of the quotient just found, and write the product under the first three figures of the dividend, draw a line under it, and subtract, obtain- ing a remainder of 263. Annex to this remainder the next figure of the dividend, in this case 0, and divide 2630, which is the new, or second, trial dividend, by 386, the divisor, which call 400 as be- fore, obtaining 6 for the second figure of the quotient. Multiply the divisor by 6, the figure of the quotient last found, and write the product, 2316, under 2630; subtract as before, obtain- ing 314 as a remainder, to which annex the next figure of the dividend, 3 in this case, making the new, or third, trial dividend 3143. To divide 3143 by 386, note that 8 X 400 = 3200, a number slightly larger than 3143; but as 386 is smaller than 400, try 8 for the third figure of the quotient. Multiplying 386 by 8, the product is 3088, which subtracted from 3143, leaves a remainder of 55, to which annex the next figure of the divi- dend, in this case 3, making the new, or fourth, trial dividend 553. The next figure of the quotient is evidently 1, and the next, or fifth, trial dividend is 1675. Dividing 1675 by 400, the next figure of the quotient is 4; multiplying 386 by 4 and subtracting the product from 1675, the remainder is 131, to which annex the next (in this case, the last) figure of the dividend, obtaining 1318 for the new, or sixth, trial dividend. Dividing 1318 by 400, the 32 ARITHMETIC §1 next figure of the quotient is 3 ; the remainder after multiplying the divisor by 3 is 160. As there are no more figures in the divi- dend, the division of 160 by 386 is indicated by writing 386 under 160 with a line between. The quotient, therefore, is 168,143m. In the foregoing, the number 400, which was used in place of 386 to determine the different figures of the quotient, is called the trial divisor. If the second figure of the divisor is 5 or a larger digit, increase the first figure of the divisor by 1, use the result thus obtained as a trial divisor, and proceed as in short division to obtain the next figure of the quotient. But, if the second figure of the divisor is less than 5, use the first figure of the divisor for a trial divisor. In case there is any doubt as to whether the figure of the quotient so obtained is correct, multiply the second figure of the divisor by the figure thus obtained in the quotient and add the amount to be carried to the product of this figure and the first figure of the divisor, comparing the result with the first two figures of the trial dividend. Thus, in the foregoing ex- ample, to determine whether to try 7 or 8 for the third figure of the quotient 8 X 8 = 64, 8 X 3 = 24, and 24 + 6 = 30; since 30 is smaller than 31, the first two figures of the trial dividend, try 8 for the third figure of the quotient. It may be remarked that the quotient, 168143^|t, is read one hundred sixty-eigh thousand one hundred forty-three and one hundred sixty over three hundred eighty-six. 67. Check for Division. — To check division, cast out 9's from the dividend, divisor, quotient, and remainder; the product of the remainders for the divisor and quotient plus the remainder for the remainder should equal the remainder for the dividend, but if not, a mistake has been made. Thus, in the example of Art. 66, disregarding the 9's, 6 + 4 + 3 -|- 3 + 5 -F 8 = 29, and the remainder is 2 for the dividend. The remainder for the divi- sor is 3 + 8 + 6 = 17, and 1 -|-7 = 8; the remainder for the 1 .9001 S7^r u.r 07 quotient is 1 + 6 + 8 + 1+4 + 3 = 1529918746)43607 _„ j o i o c +u • j Toncoi in -7^0 23, and 2 + 3 = 5; the remamder lo0821 35084 ^-^J--^-— Ans 221708 ^ ^ f°^ ^^® remainder is 1 + 6 = 7; then 218035 8 X 5 = 40, or 4, and 4 + 7 = 11, 367374 1 + 1 = 2, the same remainder as 348856 was found for the dividend; hence, 185186 the work is probably correct. 174428 As another example, divide 10758 1,529,918,746 by 43607. Since §1 MULTIPLICATION AND DIVISION 33 the second figure of the divisor is 3, a digit smaller than 5, use 4 for the trial divisor. Since the first 5 figures of the dividend make a smaller number than the five figures of the divisor, use the first six figures of the dividend for the first trial dividend. Since 4 is contained in 15, 3 times, 3 is the first figure of the quotient. The second figure of the quotient is easily seen to be 5, and the second remainder is 3673; annexing 7, the next figure of the dividend, the third trial dividend, 36737, is smaller than the divisor; hence, write a cipher (0) for the third figure of the quotient, and annex 4, the next figure of the dividend, making the fourth trial dividend 367374. While 36 -^ 4 = 9, 9 is evidently too large, since 43 X 9 = 387; consequently, try 8 for the fourth figure of the quotient. The fifth figure is 4, and the remainder is 10758, which is written over the divisor, as shown. Applying the check, 1 + 5 + 2+1 + 8 + 7 + 4 + 6 = 34, and 3 + 4 = 7; 4 + 3 + 6 + 7 = 20, or 2; 3 + 5 + 8 + 4 = 20, or 2; 1 + 7 + 5 + 8 = 21, and 2 + 1 = 3; then 2X2 = 4, and 4 + 3 = 7, the same remainder as was obtained for the divi- dend; hence, the work is probably correct. 68. Rule I. — Write the divisor to the right of the dividend, with a line between, and draw a line under the divisor. II. Determine the trial divisor as previously described and divide it into the first trial dividend for the first figure of the quotient; multi- ply the divisor by this figure, subtract the product from the trial divi- dend, and annex the next figure of the dividend for a new trial dividend. Divide the second trial dividend by the trial divisor for the second figure of the quotient. Proceed in this manner until all the figures of the dividend have been used. III. If any trial dividend is smaller than the divisor, write a cipher for the corresponding figure of the quotient, and annex the next figure of the dividend for a new trial dividend. IV. If there is a remainder, write it over the divisor with a line between, and annex this expression to the quotient. EXAMPLES 220^ (1) Divide 31415927 by 4726. Ans. 6647^^- 7SQ0 (2) 40073836 ^ 8018 = ? Ans. 4997gQjg- ,„. 712946 ^ , ,o„o359 (3) -^^g- = ? Ans. 13733^. 34 ARITHMETIC §1 (4) Divide 43560 by 209. Ans. 208209- (5) 30159681 ^ 5307 = ? Ans. 5683. 4,396,652,679 _ ^ 51009^^^ ^^^ 86193 ~ • ^''^- ^^""^86193- (7) Divide 2,189,404,900 by 29,950. Ans. 73,102. (8) According to Bessel, the diameter of the earth at the equator is 41,- 847,192 feet; what is the diameter in miles, one mile containing 5280 feet? ,,^^.3192 Ans. 7925^2^ miles. (9) How many reams of 480 sheets each are contained in 75,960 sheets of 120 writing paper? Ans. ISSjg^ reams. (10) How many bales of rags can be made up from 238,996 pounds of rags, if the bales average 596 pounds each? Ans. 401 bales. (11) 12,656 pounds'of paper is to be put up in reams of 25 pounds each; how many reams will it make? Ans. 506 2 5 reams. (12) The freight bill on a shipment of pulpwood called for payment on 246,782 pounds; if the average weight of a cord is 4450 pounds, how many rr2032 cords are there? Ans. 55 . .-^ cords. SOME PROPERTIES OF NUMBERS DIVISIBILITY OF NUMBERS 69. As previously stated, the factors of the product of two numbers are the two numbers which, when multiplied together produce the product. If more than two numbers are multiplied, the product has more than two factors; thus, 4 X 7 X 12 X 25 = 8400, and 8400 may be considered to have as its factors 4, 7, 12, and 25. Since 12 = 3 X 4, and 25 = 5 X 5, 8400 also has as its factors 3, 4, 4, 5, 5, and 7, because 3X4X4X5X5X7 = 8400. These factors may be combined in any way to form other factors; thus, 3 X 5 = 15, 4 X 5 = 20, and 4 X 15 X 20 X 7 = 8400, or 4 X 5 X 5 = 100, and 3 X 4 X 7 X 100 = 8400, etc. 70. A multiple of a number (the given number) is a certain number of times the given number; thus, 24 is a multiple of 6, 6 being the given number, because 4 times 6 is 24 ; it is a multiple of 8, because 3 times 8 is 24; it is a multiple of 12, because 2 times 12 is 24. In other words, any number that can be expressed as the product of two or more factors is a multiple of any one of the §1 SOME PROPERTIES OF NUMBERS 35 factors or of the product of two or more of its factors. For in- stance, 7854 = 2 X 3 X 7 X 11 X 17; it is, therefore, a multi- ple of any one of these numbers and may be exactly divided by any one of them, and when it has been divided by one of them, the quotient may be exactly divided by any one of the remaining factors; thus, 7854 ^ 11 = 714, and 714 may be exactly divided by any of the remaining factors, since 714 -r- 2 = 357; 357 -r- 3 = 119; 119 -^ 7 = 17, the last factor. A multiple of several fac- tors may be exactly divided by the product of any number of those factors; thus, 7854 h- 11 X 17 = 7854 -- 187 = 42; 7854 -f- 3 X 7 X 11 = 7854 -=- 231 = 34; etc. When a number can be exactly divided by another number, the first number is said to be divisible by the second number; if, however, there is a remainder after the division, then the first number is not divisible by the second. For example, 84 is divis- ible by 2, 3, 4, 6, 7, 12, 14, 21, 28, and 42, and by no other num- bers except itself (84) and 1; 84 is also, of course, a multiple of these numbers. 71. An odd number is one whose last (right-hand) figure is 1, 3, 5, 7, or 9; 71, 423, 625, 1007, 1649 are odd numbers. An even number is one whose last figure is 0, 2, 4, 6, or 8; 640, 972, 1774, 31416, and 2008 are even numbers. Any even number is divisible by 2; thig may be considered as another definition of an even number. Thus, any of the above even numbers are divisible by 2. Any number ending in 5 is divisible by 5; thus, 635, 895, etc. are divisible by 5. Any number ending in is divisible by 5 and by 10; thus, 640 = 64 X 10 = 64 X 2 X 5. Since any number ending in is a multiple of 10, it has for two of its factors 2 and 5; it is therefore divisible by 2, 5, and 2 X 5 = 10. See Art. 70. 72. A number that is not divisible by any number except itself and 1 is called a prime number or a prime; all prime numbers except 2 are odd numbers, since any even number is divisible by 2. The prime numbers less than 100 are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. A number that is divisible by some number other than itself and 1 is called a composite number. Composite numbers may be odd or even, and may always be expressed as the product of two or more prime numbers; thus, 84 = 2X2X3X7; 105 = 3X5 36 ARITHMETIC §1 X 7; 69 = 3 X 23; etc. When the factors are prime numbers, they are called prime factors. (a) Any number ending in two ciphers is divisible by 4 and by 25; thus, 62100 = 621 X 100 = 621 X 4 X 25, since 100 = 4 X 25. (6) Any number ending in three ciphers is divisible by 8 and by 125; thus, 621000 = 621 X 1000 = 621 X 8 X 125, since 1000 = 8 X 125. Since 100 = 2 X 50 = 20 X 5, a number ending in two more ciphers is divisible by 2, 4, 5, 10, 20, 25, and 50. Since 1000 = 2 X 500 = 4 X 250 = 5 X 200 = 8 X 125 = 10 X 100 = 20 X 50 = 40 X 25, a number en ding in three or more ciphers is divisible by 2, 4, 5, 8, 10, 20, 40, 50, 100, 125, 200, 250, and 500. (c) If the sum of the digits of any number is a multiple of 3, the number is divisible by 3; when adding the digits, it is not necessary to add 3, 6, or 9. Thus, 31416 is divisible by 3, because 1 + 4+1 = 6, a multiple of 3 ; similarly, 2350976 is not divisible by 3, because 2 + 5 + 7 = 14, which is not a multiple of 3. (d) If the number is even and the sum of the digits is a multi- ple of 3, the number is divisible by 6; thus, 31416 is divisible by 6, because it is even and the sum of the digits is a multiple of 3. Similarly, 790108 is not divisible by 6, because 7 + 1 + 8 = 16, which is not a multiple of 3. Had the number been 780108, 7 + 8 + 1 + 8 = 24, a multiple of 3, and the number is divisible by 6. No odd number is divisible by 6, (e) If the sum of the digits is a multiple of 9, the number is divisible by 9. Thus, 2,072,322 is divisible by 9, because 2+7 + 2 + 3 + 2 + 2 = 18, a multiple of 9; and 3,263,016 is not divisible by 9, because 3 + 2 + 6 + 3 + 1 + 6 = 21, which is not a multiple of 9; it is a multiple of 3, however, and since the number is even, it is divisible by 3 and by 6. (/) A number is divisible by 4 when the last two figuies are divisible by 4; thus, 4692, 543,836, 127,324, etc. are all divisible by 4, because 92, 36, and 24, the last two figures of each number are divisible by 4. But, 1742, 67,583, 98,782 , etc. are not divisible by 4. (g) A number is divisible by 8 when the last three figures are divisible by 8; 1752, 57064, 31416, etc. are divisible by 8, because 752, 064, and 416, the last three figures of each number are divis- ible by 8. But 1852, 57164, and 31426 are not divisible by 8. §1 SOME PROPERTIES OF NUMBERS 37 CANCELATION 73. When one composite number is to be divided by another composite number or when the product of several numbers is to be divided by the product of several other numbers, the work can frequently be shortened by employing the process called cancelation. Cancelation means dividing out or canceling equal factors from the dividend and divisor. For example, suppose 216 were to be divided by 36. According to Art. 72, 216 is divisible by 4 and by 9, and 36 is also divisible by 4 and by 9. Indicating 216 the division by -^, divide both dividend and divisor by 4, and the result is V; again dividing both dividend and divisor, this time by 9, the result is f = 6, and 6 is the quotient of 216 -i- 36. 6 In practice, the work would be performed as follows : -^ = 6. 1 Here 216 is divided by 4, a line is drawn through it, thus striking out or canceling 216, and the quotient, 54, is written above it; 36 is also divided by 4, canceled, and 9, the quotient, is written below it. Since 54 and 9 have a common factor, 9, 54 is divided by 9, canceled, and the quotient, 6, is written above it ; 9 is divided by 9, canceled, and the quotient, 1, is written under it. Since 6 4-1 = 6, the quotient of 216 -^ 36 is 6. The same result might have been obtained by resolving the dividend and divisor into their prime factors and canceling those that are common; thus, 216 _ ^X^X2X^X^X3 _ 2 X 3 - 6 When the auo- W ^X2X^X^ -2X3- b. When the quo tient is 1, it is not customary to write it when canceling. Canceling the same factor in both dividend and divisor does not alter the value of the quotient. 74. If there are several factors in both dividend and divisor, the process is similar; any factor common to both dividend and divisor may be canceled. For example, divide 136 X 54 X 117 by 26 X 51 X 40. Writing the dividend over the divisor, with a line between, 9 ;y ^/ 9 M Xf>^Xm 9X9 81 ^^ X ^I X ^0 5 5 ;^ ;; 5 16i 38 ARITHMETIC §1 By Art. 72, 136 is divisible by 8; 40 is also a multiple of 8; hence, cancel 136 and 40, writing the quotients as shown. Since 26 and 54 are divisible by 2, cancel and write the quotients as shown. By Art. 72, 51 is divisible by 3, and since 27 is a multiple of 3, cancel and write the quotients as shown. Since there is a 17 above the line and another below it, cancel the 17's. By Art 72, 117 is divisible by 9; it is equal to 9 X 13; hence, divide the 13 below the line into 117. As there are no more common factors, 9X9 the original expression is equal to — ^ — > being the product of all the uncanceled factors above the line divided by the product of all the uncanceled factors below the line. As another example, what is the value of 224 X 5700 X 189 X 85 _ ^ 250 X 456 X 21 X 17 ~ • Here, 2 ^ 9 28 ^70 ^f> in m X m^ X if>^ X ^^ _ 28 y 2 X 9 - 504 X m X n xn By Art. 72, 224 and 456 are both divisible by 8; hence, cancel and write the quotients as shown. Cancel the 10 in 250 and 5700, leaving 25 and 570. Cancel the 57 in the divisor into 570 in the dividend. Cancel 5 in 25 and 10, leaving 5 and 2, and cancel the 5 into 85, leaving 17, which cancels 17 in the divisor. 189 and 21 are both divisible by 3, and 63 is a multiple of 7. Canceling as shown, all the factors in the divisor have been canceled, leaving 28 X 2 X 9 = 504 for the quotient. That this is correct may be proved by actual multiplication and division ; thus 224 X 5700 X 189 X 85 = 20,511,792,000; 250 X 456 X 21 X 17 = 40,698,- 000; and 20,511,792,000 -^ 40,698,000 = 504. Rule. — Cancel the factors that are common to both dividend and divisor, and divide the ^product of all the factors that remain in the dividend by the product of all the factors that remain in the divisor. §1 SOME PROPERTIES OF NUMBERS 39 EXAMPLES ,,. 39 X 162 X 87 X 96 _ „ . _„3 ^^^ 19 X 6 X 29 X 42 X 13 " ' ^'''- ^^^• 30 X 40 X 50 X 60 X 70 _ , ^^^ 15 X 25 X 35 X 45 X 55 ~ ^ ^'''- ^^^• , . 24 X 44 X 64 X 84 X 104 ^ *• '^ 33 X 32 X 42 X 52 ... 231 X 328 X 7200 „ a n^ox (^) 21 X 64 X 5 25- = " ^"^- ^^^- 31416 X 55 X 192 _ , ^^^ 121 X 56 X 85 ~ • ^'''- ^^^• ... 5236 X 630 X 128 X 192 „ . _„_, ^^) 196 X 32 X 144 X 90 = ^ ^"^- ^^^*- = ? Ans. 256. AEITHMETIC (PART 1) EXAMINATION QUESTIONS (1) Express in figures the following numbers: (a) ten million nineteen thousand forty-two; (6) seventy thousand six hundred five; (c) five hundred sixty-two hundredths. (2) Express the following numbers in Roman notation: (a) 4068; (b) 44; (c) 657,903; (d) 1920; (e) 1888. (3) What is the sum of $18.04, $1.57, $197.85, $36.43, and $360.52? Ans. $614.41. (4) Seven barrels, with their contents, weigh respectively 297 pounds, 417 pounds, 226 pounds, 388 pounds, 293 pounds, 185 pounds, and 313 pounds; if they are all hoisted in an elevator at one time, and the elevator weighs 2309 pounds, together with two men, one weighing 148 pounds and the other 187 pounds, what is the total weight lifted? Ans. 4763 pounds. (5) From a pile of pulp weighing 12,882 tons, 2057 tons were removed on a certain day and 3836 tons on another day; how many tons remained in the pile? Ans. 6989 tons. (6) If 256,405 be subtracted from a certain number and the remainder is 700,999; (a) what is the number? (6) What num- ber must be subtracted from 1,001,010 to give a remainder of 660,019? ^^ /(a) 957,404 (6) 339,991. (7) What is the product of 461, 217 and 865. Check the re- sult by casting out 9's. Ans. 86,532,005. (8) If the quotient is 5077, the divisor is 6345, and the re- mainder is 2609, what is the dividend? Check by casting out 9's. Ans. 32,216,174. (9) If the dividend is 4,511,856, the quotient is 2803, and the remainder is 1829, what is the divisor? Ans. 1609. (10) There are 5280 feet in a mile; (a) how many miles in 408,806 feet? (6) How many feet in 1894 miles? Ans / ^^) ^^^*^^ '^'^^^' ' 1(6) 10,000,320 feet. 41 42 ARITHMETIC §1 (11) There are five numbers: (1) 840, (2) 231, (3) 1728, (4) 2618, and (5) 1215; (a) which of these numbers are odd? (h) which are even? (c) which are divisible by 3? (d) by 6; (e) by 9; (/) by 12? (12) Referring to the last question, which of the numbers are divisible (a) by 2? (6) by 4? (c) by 8? (d) by 5? /-,o^^.•^.l, 1 , 5236 X 9 X 45 X 26 ^ . ^. (13) Fmd the value of — qi y iqt^ V 88 — '^ cancelation. Ans. 51. (14) A hogshead contains 63 gallons; how many hogsheads are equal to 20,610 gallons? Ans. 327/-3 hogsheads. ^.^^T.■A.^. 1 .1309X1728X448, _. (15) Fmd the value oi o^fi V 88 V 1^4 -^ cancelation. Ans. 291tt. ARITHMETIC (PART 2) GREATEST COMMON DIVISOR AND LEAST COMMON MULTIPLE 75. Greatest Common Divisor. — The greatest common divisor of two or more numbers is the largest number that will exactly divide all of them. Thus, the greatest common divisor (usually- abbreviated G.C.D.) of 36 and 48 is 12, because 12 is the largest number that will divide 36 and 48 — it is the largest common factor of 36 and 48; also, 36 is the G.C.D. of 72, 108, and 144, because it is the largest common factor of those numbers. To find the G.C.D. of two numbers, find the product of their common factors; the work is usually arranged as follows : Find the G.C.D. of 54 and 216. Write the ' numbers as shown in the margin, ' separating them by a comma, and G C D =6 X 9 = 54 Ans ^^^'^ proceed as in short division. 'By Art. 72, 54 and 216 are both di- visible by 9; divide as shown, obtaining 6 and 24 for the quotients. Both 6 and 24 are divisible by 6, the quotients being 1 and 4. The only factor common to 1 and 4 is 1, which has no effect on the G.C.D. Since 54 and 216 have the common factors 9 and 6, they are divisible by the product of these common factors, and the G.C.D. of 54 and 216 is 9 X 6 = 54. To prove the work, 54 ^ 54 = 1, and 216 4- 54 = 4. 76. To find the G.C.D. of more than two numbers, proceed in exactly the same way as in Art. 75, remembering that the divisors must divide all the numbers. Thus, find the G.C.D. of 270, 405, 675, and 945. Arrang- ing the work as shown, it is seen that all the numbers are multiples of 9; hence, divide by 9. The quotients are all divisible by 5; hence, divide 43 9 5 3 270, 405, 675, 945 30, 45, 75, 105 6, 9, 15, 21 G.C.D. = 9 2, 3, X5 X3 5, 7 = 135. Ans 44 ARITHMETIC 391 238 1 238 153 1 153 85 1 85 68 1 68 17 4 68 G.C. D. = 17. Ans. by 5. The quotients in the third line are all multiples of 3; hence, divide by 3. The quotients in the fourth line have no common factor; therefore, the G.C.D. is 9X5X3 = 135. To prove this, divide each of the original numbers by 135; the quotients will be the same as in the bottom line. 77. It frequently happens that the common factors are not readily seen; in such cases, proceed as in the following example. Find the G.C.D. of 238 and 391. Write the larger number, and draw a vertical line on its right; then write the smaller number and draw another vertical line. The first vertical line is the sign of division, but the second vertical line is merely a line of separation. Now divide 391 by 238, and write the quotient to the right of the second vertical line. The remainder is 153, which is divided into 238, the quotient being written under the first quotient. The remainder from the second division is 85, which is divided into 153, giving a remainder of 68, the quo- tient being written under the preceding quotient. Dividing 85 by 68, the remainder is 17. Lastly, 68 divided by 17 gives a quotient of 4 and a remainder 0; and 17, the divisor that gave the remainder 0, is the G.C.D. of 238 and 391. As another example, find the G.C.D. of 31,416 and 1,400,256. Arranging the work as before, the first quotient is 44, and the remainder is 17,952, which is divided into 31,416. The second quotient is 1, and the second remainder is 13,464, which is divided into 17,952. The third quotient is 1, and the third remainder is 4,488, which is contained in 13,464 3 times, and there Therefore, the G.C.D. of 31,416 and 1,400,256 1400256 125664 143616 125664 17952 13464 4488 31416 17952 13464 13464 44 1 1 3 G.C.D. = 4488. Ans. is no remainder is 4488. It is seldom necessary to find the G.C.D. of more than two numbers. The first method should be used whenever possible; but, when the numbers are large or the common factors are not apparent, then use the second method. Observe that the process must be carried out until a remainder of is obtained. If the numbers have no common divisor, the G.C.D. will be 1. In such case, the numbers are said to be prime to each other, though §1 LEAST COMMON MULTIPLE 45 both may be composite numbers. For example, 91 and 136 are both composite, but they are prime to each other, as is evident when the greatest common divisor is found to be 1, in accordance with the work as shown ^ ^ -P, " . in the margin. G.C.D. = 1. Ans. ° 136 91 1 91 90 2 45 1 45 45 = 1. Ans. EXAMPLES (1) Find the G.C.D. of 40, 60, 80, and 100. Ans. 20. (2) Find the G.C.D. of 70, 175, 210, and 245. Ans. 35. (3) Find the G.C.D. of 2387 and 1519. Ans. 217. (4) Find the G.C.D. of 4059 and 6390. Ans. 9. (5) Find the G.C.D. of 11433 and 444. . Ans. 111. (.6) Find the G.C.D. of 364089 and 457368. Ans. 3009. (7) Find the G.C.D. of 1016752 and 991408. Ans. 176. 78. Least Common Multiple. ^ — ^A common multiple of several numbers is a number that is divisible by those numbers; thus, 840 is a common multiple of 40 and 56. The least common multiple of two or more numbers is the smallest number that is divisible by those numbers; thus, 280 is the least common multiple of 40 and 56 ; it is the smallest number that is divisible by both 40 and 56. The product of two or more numbers is a common multiple of the numbers; for the want of a special name, this will be called the prime multiple of the numbers. The prime multiple may be the least common multiple, and will be such if no two of the numbers have a common factor; thus, the prime multiple of 91 and 36 is 91 X 36 = 3276, and this is also the least common multiple of 91 and 36. The words least common multiple are usually abbreviated to L.C.M. If the two numbers have a com- mon factor, it may be canceled from the numbers, and the L.C.M. will then be the product of this common factor and the remaining factors of the numbers. Thus, 56 and 72 have the common factor 8; dividing 56 and 72 by 8, the quotients are 7 and 9, respectively; then the L.C.M. is 8X7X9 = 56 X9 = 7X8X9 = 7X 72 = 504. Note that 504 is equal to 56 X 9 and to 72 X 7; hence, it is divisible by both 56 and 72. Since 8, the common fac- tor, is the greatest common factor, that is, the G.C.D., of 56 and 72, it follows that the L.C.M. of two numbers may be found by 46 ARITHMETIC §1 dividing one of the numbers by their G.C.D. and multiplying the other number by the quotient. For instance, referring to ex- ample (3) at the end of Art. 77, the G.C.D. of 2387 and 1519 is 217; 1519 ^ 217 = 7; then 2387 X 7 = 16,709, the L.C.M. of 1519 and 2387. Or, 2387 ^ 217 = 11, and 1519 X 11 = 16,709. 79. To find the L.C.M. of more than two numbers, proceed in much the same way as in finding the G.C.D. An example will illustrate the process. Find the L.C.M. of 16, 24, 30, and 32. Arrange the work as shown. Since all the numbers contain the common factor 2, divide it out. In the first line of quotients, 2 16, 24, 30, 32 4 8, 12, 15, 16 2 2, 3, 15, 4 3 1, 3, 15, 2 1, 1, 5, 2 L.C.M. =2X4X2X3X5X2= 480. Ans. three of the numbers are multiples of 4(= 2 X 2), but the other number contains no factor in common with 4; hence, divide by 4. The second line of quotients contains two numbers that are multiples of 2; hence, divide by 2. The third line of quotients contains two numbers that are multiples of 3; hence, divide by 3. The numbers in the fourth line of quotients are all prime to each other, and no further division is possible. The product of the divisors and the factors in the last line of quotients is the L.C.M., which is equal to 2X4X2X3X5X2 = 480. When finding the L.C.M., divide out all factors that are com- mon to two or more numbers until a row of quotients is obtained that are prime to one another. As another example, find the L.C.M. of 15, 27, 55, and 99. There is no factor common to all 5 15, 27, 55, 99 11 3, 27, 11, 99 3 3, 27, 1, 9 3 1, 9, 1, 3 1, 3, 1, 1 L.C.M. =5X11X3X3X3 = 1485. Ans. the numbers; but, since 15 and 55 are multiples of 5, divide by 5. Since 11 and 99 are multiples of 11, divide by 11. The remainder of the work is evident. Note that whenever a number is not divisible by a divisor, the number is brought down into the l^ne of quotients. The L.C.M. is5XllX3X3X3 = 1485. §1 LEAST COMMON MULTIPLE 47 If the numbers are such that their factors are not apparent, find the L.C.M. of two of them, and then find the L.C.M. of this result and the third number; then the L.C.M. of the second result and the fourth number, and so on; the last result will be the L.C.M. of all the numbers. For instance, to find the L.C.M. of 893, 1387, and 1121, first find the L.C.M. of 893 and 1387 (or 1121). Pro- ceeding as described in Art. 78, first find the G.C.D. of 893 and 1387. The work is shown in the margin, and the G.C.D. is 19. Then, 893 -^ 19 = 47, and 1387 X 47 = 65189, the L.C.M. of 893 and 1387. Now find in the same way the L.C.M, of 1121 and 65,189. The work for this is also shown in the margin. The G.C.D. of these two numbers is 19; 1121 -^ 19 = 59; and 65189 X 59 = 3,846,141, which is the L.C.M. of 893, 1387, and 1121. It may here be remarked that the greatest common divisor and the least common multiple are of importance in connection with the reduction, addition, and subtraction of fractions, as will shortly appear. 1387 893 893 494 494 399 399 380 95 19 95 65189 5605 9139 8968 171 95 76 76 1121 -- 19 = 59 65189 X 59 = 3846151. A 1121 1026 95 76 19 58 6 1 1 4 (1) (2) (3) (4) (5) (6) (7) Find the L.C.M. Find the L.C.M. Find the L.C.M. Find the L.C.M. Find the L.C.M. Find the L.C.M. Find the L.C.M. EXAMPLES of 28, 49, 63, and 84. of 12, 14, 16, and 18. of 15, 20, 25, and 30. of 4, 8, 12, 16, and 20. of 1955 and 4403. of 119, 204, 272. of 442, 234, and 1001. Ans. 1764. Ans. 1008. Ans. 300. Ans. 240. Ans. 506,345. Ans. 5712. Ans. 306,306. 48 ARITHMETIC §1 FRACTIONS DEFINITIONS 80. When an integer or a unit is divided into equal parts, one or more of these parts is called a fraction of the integer or unit. If, for example, a straight stick be cut into two pieces of the same length, one piece is equal to the other, and either is called one- half of the stick. If the stick is cut into three equal pieces or parts, one of them is called one-third of the stick; if cut into four equal parts, one of them is called one-fourth of the stick; if cut into five equal parts, one of them is called one-jifth of the stick; etc. The expressions one-half, one-third, one-fourth, etc. are fractions. More than one part is denoted by writing the number before the name of the part; thus, two-thirds means two one-thirds, three- fourths means three one-fourths, etc. 81. To express a fraction with figures, it is necessary to write two numbers, one to show into how many parts the integer or unit has been divided, and the other to show how many of these parts are taken or considered. For instance, I means one-half, and indicates one of two equal parts I means one-third, and indicates one of three equal parts I means one-fourth, and indicates one of four equal parts f means two-thirds, and indicates two of three equal parts -f means four-fifths, and indicates four of five equal parts tV means seven-twelfths, and indicates seven of twelve equal parts, etc., etc. 82. It will be observed that a fraction is expressed in one of the ways used to indicate division. The number below the line is called the denominator, because it denominates, or names, the number of parts into which the integer has been divided. The number above the line is called the numerator, because it numerates, or counts, the number of the equal parts that are taken or considered. 83. An expression like f may be interpreted in two ways: First, it is 3 times i of a unit. For example, one dollar is equal to 100 cents; one-fourth of a dollar is 100 ^ 4 = 25 cents; and three-fourths of a dollar is 3 X 25 cents = 75 cents. Here one dollar is the unit. Second, it is one-fourth of three times the unit. If the unit is one dollar, three times the unit is 3 dollars or 300 cents, and one-fourth of 3 times the unit is 300 cents -^ 4 = 75 §1 FRACTIONS 49 cents, the same result as before. In the first case, f is a fraction; in the second case, it is an indication of division. 84. Except when the denominator is 1, 2 or 3, a fraction is read by pronouncing the name of the numerator and then pro- nouncing the name of the denominator after adding ths; thus, f is read three-sevenths, tI is read eleven-sixteenths, tit is read forty-nine one-hundred-twenty-fifths, etc. But if is read thirteen- twenty-firsts; |f is read twenty-nine-forty-seconds; || is read thirty-seven fifty-thirds, etc. 85. If two fractions have the same numerator, but a different denominator, the fraction whose denominator is the smaller is the larger; thus f is larger than f , because one-fifth of anything is smaller than one-fourth of it; hence, 3 one-fifths is smaller than 3 one-fourths For example, three-fourths of a dollar is 75 cents, but three-fifths of a dollar is 60 cents, since one-fifth of a dollar is 100 cents ^ 5 = 20 cents, and 3 X 20 cents = 60 cents. 86. The numerator of a fraction may be greater than the de- nominator, in which case, the value of the fraction is found by dividing the numerator by the denominator; thus, V" = 4, the value of the fraction; ^- = 2, the value of the fraction; etc. If the numerator is equal to the denominator, the value of the fraction is 1; thus, TT = 1, I = 1, etc. This is evident, since if a dollar is divided into, say, 5 equal parts, 5 of these parts make up the dollar. If the numerator is less than the denominator, the value of the fraction is less than 1. 87. A proper fraction is one whose numerator is smaller than its denominator; its value is always less than 1. Thus, f, tV, tI , etc. are proper fractions. An improper fraction is one whose numerator is equal to or greater than its denominator. Thus, f, 1, ||, etc. are impro- per fractions. When it is not desired to specify the numerator and denomina- tor separately, they are called the terms of the fraction; thus, the terms of the fraction |f are 23 and 42. When a fraction is joined to an integer, as in the expression 14|, the expression is called a mixed number. Here 14f means 14 and f more; it has the same meaning as 14 -f f . In reading a mixed number, the word and is used as above, but the plus sign is always understood, though not written or spoken; thus, 5f is read five and two-thirds, and means 5 + f . Mixed numbers 50 ARITHMETIC §1 always occur whenever the dividend is not a multiple of the divi- sor; for instance, the quotients found in examples (3), (4), (6), and (7) of Art. 65 are mixed numbers. 88. In printing and writing, in order to save space, fractions are frequently expressed by using the inclined line instead of the horizontal line; thus, M = f > 113/147 = lif, etc. In such cases, a hyphen is sometimes written between the integer and the fraction of a mixed number; thus, 14-3/8 = 14%; the hyphen shows that the fraction belongs to the integer. When it is desired to indicate that the fraction is to be pro- nounced and it is not desired to write the name in full* the fraction is written as usual and the ending of the name of the denominator is annexed; thus, ^^rds, %2ths, ^%2ds, ^^j^sts, mean two-thirds, seven-twelfths, fifteen twenty-seconds, twenty-three thirty-firsts, etc. The only exception is ^, which is always read one-half, and is always so written and printed. REDUCTION OF FRACTIONS 89. To reduce a fraction is to change its form without changing its value; to change its form means to alter its numerator and de- nominator. It was shown in Arts. 82 and 83 that a fraction may be regarded as an expression of division, the numerator being the dividend and the denominator the divisor. In Art. 73, it was shown that canceling the same factor in both dividend and divisor does not alter the value of the quotient; hence, dividing both numerator and denominator of a fraction by the same number does not alter the value of the fraction. For example, ^f = yV = I- Here both 18 and 24 are first divided by 2, the quo- tients being 9 and 12, respectively. Since 9 and 12 contain the common factor 3, divide both 9 and 12 by 3, the quotients being 3 and 4, respectively. The numerator and denominator of the given fraction might both have been divided by their greatest common divisor, 6, and the same final result, f, obtained. That Hths = fths is easily shown. Thus, a gross is 144; i-|ths of a gross is 18 times ^th of 144; ;2Vth of 144 is 144 -f- 24 = 6, and |-|ths is 18 X 6 = 108. But fths of a gross is 3 times ith of 144; |th of 144 = 36, and fths of 144 is 3 X 36 = 108. In the same way, it is shown that rVths of 144 is 108. Therefore, 1 8 _ 9 _ 3 "5T — TY — 4" §1 FRACTIONS 51 90. Multiplying both numerator and denominator by the same number does not alter (change) the value of the fraction; this is evident, since both terms of the new fraction may be divided by the number used as a multiplier, thus obtaining the original frac- 3X6 tion. For instance, , ^ = |-|, which, as has just been shown 2x7 is equal to |. Similarly, - - = |-f- 91. When the fraction is reduced to lower terms, that is, when the numerator and denominator are made smaller by division, the process is called reduction descending. When the fraction is reduced to higher terms, that is, when the numerator and de- nominator are made larger by multiplication, the process is called reduction ascending. 92. When the fraction has been reduced by division until the numerator and denominator are prime to each other (have no common factor), the fraction is said to be in its lowest terms; thus, f, tV, if? etc. are fractions in their lowest terms. When a fraction in its lowest terms, it is in its simplest form. 93. To reduce a fraction to its lowest terms, cancel all factors that are common to the numerator and denominator. If the factors are not apparent, and it is desired to make sure that the fraction is in its lowest terms, find the greatest common divisor, if any, of the numerator and denominator and divide both terms by it. Example 1. — Reduce ||| to its lowest terms. Solution. — Since both terms are multiples of 4, divide them by 4, and tI"! ~ ii- Both terms of the new fraction are multiples of 7; hence, divid- ing by 7, If = f. Both terms of the new fraction are multiples of 3; hence, dividing by 3, f = f. Since the terms are now prime to each other, the fraction is in its lowest terms. In practice, the work would be arranged as follows : -^-f -f =11=1 = 3. Ans. Example 2. — Reduce f Hit to its simplest form. Solution. — Since both terms are multiples of 8 (see Art. 72), divide them by 8, and f gff f = f f -ff. Apparently, the terms have no common factor, but to make certain, apply the process of finding the G.C.D. of 6409 and 8671; in this case, the G.C.D. is 377. Dividing both terms by 377, ||ff 94. To reduce a fraction to another fraction having a given denominator, divide the given denominator by the denominator of the given fraction and multiply the terms of the fraction by the quotient. For example, to reduce f to a fraction having 96 52 ARITHMETIC §1 3 X 12 36 for its denominator; 96 -^ 8 == 12, and o y io = qa- Arts. The reason for dividing the given denominator by the denomina- tor of the given fraction is evident, since the quotient must be the number by which the denominator of the given fraction must be multipHed in order to equal the given denominator. This opera- tion is of importance in adding and subtracting fractions. 95. To reduce an integer to an improper fraction having a given denominator, multiply the integer by the given denominator, and the product will be the numerator of a fraction having the given denominator. For instance, reduce 7 to a fraction having 12 for its denominator. Here 7 X 12 = 84, and the required fraction is ff . That this result is correct may be proved by dividing the numerator by the denominator, the quotient being 7, the original number. This result may also be obtained in another way; thus, 7 is evidently equal to t, and r -lo — Jo- 96. To reduce a mixed number to an improper fraction, multi- ply the integral part by the denominator of the fraction, add the numerator to the product, and write the sum over the denomina- tor. This is evidently correct, since the mixed number is obtained by dividing the dividend (numerator) by the divisor (denomina- tor), and the remainder is written over the divisor. Thus, reduce 14f to an improper fraction. Here 14 X 8 = 112; 112 + 3 = 115; hence, 14| = ^-^. Regarding ^^ ths as an indication of dividing 115 by 8, 115 ^ 8 = 14f. 97. A common denominator of two or more fractions is a com- mon multiple of the denominators of the fractions; and the least common denominator is the least common multiple of the denomi- nators. For instance, the least common denominator of f , |, and I is 24, because 24 is the L.C.M. of 3, 4, and 8. 98. To reduce two or more fractions to fractions having a least common denominator, find the L.C.M. of the denominators; then, by the method of Art. 94, reduce each fraction to a fraction hav- ing this denominator. Example 1. — Reduce f, j, and f to fractions having a least common denominator. Solution.— The L.C.M. of 3, 4, and 8 is 24; f = if ; i = ^\; and f Example 2. — Reduce ||, f|, and j-f-f to fractions haying a least common denominator. §1 FRACTIONS 53 Solution.— The L.C.M. of 69 and 92 is 276; the L.C.M. of 276 and 161 is 1932, which is therefore the L.C.M. of 69, 92, and 161. Then, 1932 -- 69 = 28, and || X || = ifff; 1932 - 92 = 21, and || X |i = HMj 1932 -i- 161 = 12, and 111 X il = ilM- Therefore, the required fractions are 1232 1197 „riH 1^4 0. 4„„ Example 3. — Which fraction is the larger, jW or Hg ? Solution. — To determine which is the larger, reduce them to a common denominator; then the one that has the greater numerator is the larger. Since 4f § = ||; reduce i%% and |f to a common denominator, preferably, the least common denominator. Since 133 and 39 have no common factor, their L.C.M. is their product, or 133 X 39 = 5187; then y^s = If if' and |5 = I Iff. Therefore, xf§ is the larger fraction. .4ns. Example 4. — There are 1760 yards in a mile; what fraction of a mile is 550 yards? Solution. — Since there are 1760 yards in one mile, the number of miles or parts of a mile in 550 yards is 550 -r- 1760 = tV^tj = tVc" = A- Therefore, 550 yards is j^ths of a mile. Ans. Example 5. — Which is the greater 500 yards or -gV^ths of a mile? Solution. — Since there are 1760 yards in a mile, 500 yards = iWo = y^y**^ - ^%8 mile; -gV^ = if mile. Reducing these fractions to a com- mon denominator, the L.C.M. of 88 and 53 is 88 X 53 = 4664; U = \Ut and II = iff|. Therefore, 500 yards is a little greater than ■^''^^^ths of a mile. Ans. EXAMPLES (1) Reduce to its lowest terms J|-|. Ans. j|. (2) Reduce to its lowest terms ||-J. Ans. |i. (3) Reduce to its lowest terms i|||. Ans. i*^. (4) Express 11 as a fraction having 24 for its denominator. Ans. W*-. (5) Reduce 125.{6 to an improper fraction. Ans. -W- (6) Reduce 852iH to an improper fraction. Ans. ^-^V/^- (7) WhichisthelargeryVs oriff? Ans. ^U- (8) Which is the larger |f or |f ? Ans. ff. (9) Which is the larger ^i or if? Ans. if. (10) Reduce to their least common denominator |, |, /g, and i\. AriQ J8 40 SO SI /I 'to. 4jj, 48, 5 J, -f-g. (11) Reduce to theirleast common denominator |,i, |, H, and H. Amo 2 53 210 lAO 4 95 45.5 ii.nb. ^30) eaiJ' tisd) sso) 'SsiT' (12) A ream of writing paper contains 480 sheets; 328 sheets is what fraction of a ream? Ans. tJth ream. (,13) How many sheets are equal to T^ths of a ream of writing paper? Ans. 280 sheets. (14) Which is the larger, 200 sheets of writing paper or Hth ream? Ans. Hth ream. (15) How many sheets difference are there between 400 sheets of writing paper and fjds ream? Ans. 5 sheets. (16) If a car of sulphur weighs 40,000 pounds, what fraction of a car-load is 26,800 pounds? Ans. iV^th car-load. 54 ARITHMETIC §1 ADDITION OF FRACTIONS 99. The sum of A and yV is rf , because 5 one-sixteenths and 9 one-sixteenths = 14 one-sixteenths = yfths. Like num- bers (see Art. 8) can be added, but unhke numbers cannot be added; thus, 4 feet cannot be added to 6 inches, but 4 feet can be added to 6 feet and 4 inches can be added to 6 inches. In an expression like yV + tV, the denominators show what is to be added, in this case 16ths, and the numerators show how many of the things indicated by the denominator are to be added ; it is exactly the same operation as adding 5 dollars and 9 dollars and obtaining 14 dollars for the sum. In one case, 16ths are added, and in the other case dollars are added. Consequently, if the denominators are alike, the sum of several fractions may be found by adding the numerators and writing the sum over the denomi- 5 + 9 + 13 nator. For instance, A + iV + tI = 7g = f ^ = ill, the last result being obtained by dividing the numerator by the denominator. 100. If the denominators are unlike, the fractions cannot be added until they have been reduced to a common denominator, preferably, the least common denominator; thus, f cannot be added to y, because 2 one-thirds + 4 one-fifths has no particular meaning as it stands — the fractions have no common unit. But, f = H, i = H, and H + H = fl = 1 A. That this result is correct is readily seen. Thus, a bushel of wheat weighs 60 pounds; hence, fds of a bushel weighs 40 pounds; fths of a bushel weighs 48 pounds; and the combined weight is 40 + 48 = 88 pounds. Now yVth of 60 pounds is 4 pounds, and ff ths of 60 pounds is 22 X 4 = 88 pounds, the same result as before. Similarly, to add 4 feet and 6 inches, it is necessary to express the feet in inches or the inches in feet before adding. Since there are 12 inches in 1 foot, there are 12 X 4 = 48 inches in 4 feet; and 48 inches + 6 inches = 54 inches. If it is desired to express the sum in feet, then it is necessary to divide 54 by 12, obtaining 4tV = 4| feet. The same result might have been obtained by dividing the number of inches, 6, by 12, thus obtaining a frac- tion of a foot, which can then be added to the number of feet; thus, 6 -T- 12 = tV = h that is, one-half of a foot, and 4 feet + I foot = 4 + I = 4| feet. §1 FRACTIONS 55 101. Rule. — I. If the fractions have a common denominator, add the numerators and write the sum over the denominator. II. If the fractions do not have a common denominator, reduce them to fractions having a common denominator, -preferably, their least common denominator, and then add. III. If the sum is an improper fraction, reduce it to an integer or mixed number by dividing the numerator by the denominator. IV. // the sum is a proper fraction or a mixed number containing a proper fraction, reduce the fraction to its lowest terms. Example.— Add ^V, H, U, and |f. Solution. — Here the least common denominator is 240. Then, /^ + {^ ^■iz^is _j_4o ,_i65 ^204 I 130- 140 + 165+204+130 639 _,,., -t- 2 -|- 2¥ - 24 + 810 -t- 210 "t" 2?(J 240 "^ 240 ^ ^^^ In practice, the work would be arranged as follows : 7+11:17,18 140 + 165 + 204+130 ^53 ^ Here the denominator of the sum is written only once, the numerators of the fractions being written above it and added. It is seen at a glance that the sum of the numerators is greater than the denominator; hence, they are added separately and the improper fraction reduced to a mixed number, with the fractional part in its lowest terms. 102. The Sum of Two Fractions.— When it is desired to add two fractions whose denominators are unlike and have no common factor, that is, when these are prime to each other, proceed as follows: Add yi and ^. Here the denominators have no com- mon factor; hence, the least common denominator is the pro- duct of the denominators. This product divided by either denominator gives the other denominator, which is used as a multiplier for the numerator. Therefore, multiply the denomina- tors for a new denominator; multiply the numerator of the first fraction by the denominator of the second, and the numerator of the second fraction by the denominator of the first and add theproducts, Thus,-H+ .1,11X^ + 19X13 ^264 + 247 ' ^ ^ 13 X 24 312 — ^^ — 11.9 9. A Even if the denominators have a common factor, this method of adding two fractions is generally to be preferred, because it is quicker than reducing the fractions to their least common denominator. 56 ARITHMETIC §1 103. To add mixed numbers, add the integral and fractional parts separately; if the sum of the fractions is an improper frac- tion, reduce it to a mixed number and add to the sum of the integers. Example.— What is the sum of 23f, SlH, 28M, and 25||? Solution. — Arrange the numbers in the same manner as for addition 233 _ 24 of integers. The least common de- 3111 = If nominator of the fractions is evidently 28|i = 11 64; hence, reduce the fractions to frac- ^^ II tions havmg 64 for a denommator, lUy^ J 6 J" — ■^s¥ writing them as shown. Adding the numerators, the sum is 149, and the sum of the fractions is ^-^ = 2|i. Carry the 2 into the column of integers, which add, obtaining 109, to which annex the fraction, obtaining 109U for the sum of the mixed numbers. Ans. The sum might have been obtained by reducing the mixed numbers to improper fractions and then adding by the regular rule; but this takes longer and there is greater liability of making a mistake. As another example, add df, 5|, 7ri, and 6|. The least common denominator is 120, Reducing the fractions to fractions having 120 for a denominator, the sum of the numerators 24H Ans. is 27.3, which divided by 120 gives a quo- tient of 2^-0^. Carrying the 2 to the sum of the integers and annexing the fraction, the sum of the mixed numbers is 24-f-g-; thus, 2 + 6 -F 7 + 5 + 4 = 24, and 24 -F ii = 24^^. H _*8 ' 1^6 6i = 40 7H = m 6^- 75 l^TJ EXAMPLES (1) Add f, ^j,, A. Ans. IfH. (2) Add H and xV- Ans. lr%. (3) Add I, h I I, jS. Ans. lU- (4) Add III and fff . Ans. 1A¥/y. (5) Add 11 A, 8f, 9if, lOi Ans. 39fi. (6) Add 127f, 85f, 109^, 96|. Ans. 419tVt7. (7) Three kinds of dyes were added to a beater of paper stock, and weighed 9^'i ounces, 8% ounces and 13% ounces; how much dye stuff was used? Ans. 32)^4 ounces. SUBTRACTION OF FRACTIONS 104. Since subtraction is the reverse of addition, fractions must be reduced to a common denominator before the subtraction can 6 — 5 be performed. For example, f — f = — ^ — = I- Here the §1 FRACTIONS 57 common denominator is found as in addition, then the difference of the numerators is taken and the remainder is written over the common denominator. Example. — Find the difference between y^f and ^il"- Solution. — As it is not evident which fraction is the larger, assume that the first fraction is the larger ; if, later, it is found not to be such, the subtra- hend and minuend in the numerator can be transposed. Then, proceeding as in addition of two fractions, 1 fi 17 106 X 268 - 217 X 133 28408 - 28861 _ . m - Ui = 133 X 268 = 58156— • ^^'' expression „ 1 „ . . . 28861 - 28408 shows that the fraction -f^ -J is the larger, and the difference is ^eT^R 105. To subtract one mixed number from another, subtract the integral and fractional parts separately. For example, sub- tract 26f from 42f . The least common de- 42f = if nominator is 24; hence, reduce the fractions 26§ = A to fractions having 24 for a denominator and 16^^ ^\ subtract as shown. The difference is found to be 16^V- Ans. Example.— From 109t\ take 96|. Solution. — Here the fraction in the subtrahend is larger than the fraction lOOj^^ = if lit = ii in the minuend; therefore, proceed exactly 961^ = If ft as in subtraction of integers when the figure in the subtrahend denotes a larger number 12||. Ans. If than the figure over it in the minuend, by- adding 1 to the fraction in the minuend and 1 to the integer in the subtrahend Reducing the mixed number, l^f to an improper fraction, fi — 's?' = If, and 109 - 97 = 12; hence, the difference is 12ff. In a similar manner, if it is desired to subtract a fraction from an integer, reduce 1 to a fraction having the denominator of the given fraction and subtract 1 from the integer. Thus, 15 106. Rule. — Reduce the fractions to fractions having a common denominator, find the difference of the numerators, and write the remainder over the common denominator; then reduce the resulting fraction to its lowest terms. EXAMPLES (1) Find the difference between || and j%\. Ans. tV/^. (2) From 11 A subtract 7§. Ans. 3|i. (3) Subtract 8| from 13. Ans. 4|. (4) From 4:72^^% take 297xV^. Ans. 175^^. 58 ARITHMETIC §1 (5) What number added to 53f will make 751? Ans. 21i|. (6) A package of dye weighs 3ji pounds; after If pounds have been used how much remains? Ans. Ijf pounds. MULTIPLICATION OF FRACTIONS 107. To Find the Product of a Fraction and an Integer. In Art. 53, multiplication was shown to be a short process of addition; hence, 4 X f may be regarded as f + f + f + f 3+3+3+3 4X3 12 - „ .^ . ,^ , -, ,, = y = — ~ — = -;=- = If. Here it is seen that it the numerator of the fraction be multiplied by the integer, the pro- duct is the same as the product of the integer and the fraction. This same result may be arrived at in another way, thus : f ths of 4 is evidently 3 times yth of 4; yth of 4 is 4 -^ 7 = f, and 3 times 4 one-sevenths is 3 X 4 one-sevenths, or 12 one-sevenths = -T- = if ; as before. Again, what is the product of 5 and tt? Here, 5 X tt ~ IT ~ V"= 3f- This same result may be obtained by can- celation; thus, ^ X jk = V" = 3f . As the result of this opera- 3 tion, it is seen that dividing the denominator of the fraction by the integer, multiplies the fraction. This should be evident, since the smaller the denominator the larger the fraction; and if the denominator is 5 times as small, the fraction must be 5 times as large. Therefore, the product of a fraction and an integer may be obtained by multiplying the numerator of the fraction by the integer or by dividing the denominator of the fraction by the integer. If the denominator is not a multiple of the integer, but contains one or more factors common to the integer, these factors maybe canceled before multiplying the numerator. For example, 7 n 4 108. To Multiply a Fraction by a Fraction. — When the word "of" occurs between two fractions or between a fraction and an integer, it has the same meaning as the sign of multiplication; thus, I of f and f of 12 have the same meaning as f X f and §1 FRACTIONS 59 I X 12, respectively. Now f is 2 times ^; | of | is 2V, since 1 X i is I, and ^ of | must be 3 times smaller, ^ ^ = ^V- Since f is 5 times i, | of f must be 5 times as large as ^ of |; hence, | of 8 = T 4 X 5 = A- Since, also, | is 2 times |, and | of | is 3^, f of f is A X 2 = i| = A, or p X ^ = t\- 12 This same result may be obtained in an easier manner by multiplying the numerators of the fractions for the numerator of the product, and multiplying the denominators of the fractions 2X5 for the denominator of the product; thus, f X f = » .. ^ 0X0 _ 1 _ 5 „„ ? V - - ^ - _5 4 The product of any number of fractions may be found in this same way; that is, divide the product of the numerators by the ^ H 7 T2 product of the denominators. For instance, ;s X o X 7^ X 77^ ' p p jip 17 4 ^ = VV- Cancelation should be employed whenever possible. 109. To multiply a mixed number by an integer, multiply the integral and fractional parts separately. For example, 17| X 42 7 '^^ 147 = 17 X 42 + I X 42; g X ^^ = ^ = 36f ; 17 X 42 = 714; 4 and 714 + 36| = 750|. 110. To multiply a mixed number by a mixed number, reduce both to improper fractions, and then multiply in the ordinary 47 25 manner. Thus, 13f X lOH = ^ X ^ = ^ = 146f 111. Rule. — I. The product of two or more fractions is found by dividing the product of the numerators hy the product of the denomina- tors, canceling factors that are common to the numerators and denomi- nators whenever possible before multiplying. II. To find the product of two or more mixed numbers, reduce them to improper fractions before multiplying. 60 ARITHMETIC §1 EXAMPLES (1) § of f of ' = ? Ans. '2 8- (2) II X If = ? Ans. A. (3) 68 X 19-1 = ? Ans. 1348i (4) 33^ X 56| = ? Ans. 1894M. (5) 3| X 6f X T^E = ? Ans. llf. . (6) m X ^11 = ? ■ Ans. x¥A. (7) From a barrel of rosin weighing 404 pounds, xith of the contents were taken; how many pounds were taken? Ans. 277| pounds. (8) A car load of coal weighing 78,500 pounds was received. At the end of a certain period, fths had been used; during a second period, one-half of the remainder was used; and during a third period, fths of what was left at the end of the second period was burned. How many pounds were used during the three periods? Ans. 75,433i| pounds. (9) Referring to the last example, how many pounds were consumed during each of the three periods? ( First period, 29437^ pounds. Ans. \ Second period, 24531| pounds. I Third period, 21464|1 pounds. (10) What is the value of 4| ounces of dye that is worth $1 j^ per ounce? Ans. $5H- (11) From a pile of 11,670 cords of wood, Mtbs was used in one part of a plant; another part of the plant used }^ as much as the first part, and a third part of the plant used K2ths as much as the second part; how many cords were used in the third part? Ans. 1276i%2 cords. (12) Referring to the last example, how many cords were used in the first and second parts of the plant? . f First part, 4376M cords. \ Second part, 21883^ cords. Note.— Observe that the product of two proper fractions is always less than either of the fractions used in finding the product. DIVISION OF FRACTIONS 112. To Divide a Fraction by an Integer. — Suppose it is desired to divide | by 3. Since f is 6 one-sevenths, 6 one-sevenths divided by 3 is 2 one-sevenths = f , in the same way that 6 bushels divided by 3 is 2 bushels. Hence, dividing the numerator by the integer divides the fraction. Again, the quotient of f divided by 3 is evidently only one-third as much as the quotient of f divided by 1, since 3 is 3 times 1; hence, the quotient of f divided by 3 will be ^ of f = i X f = A = f • Note that in the first case, the numerator is divided by the integer, while in the second case, the denominator is multiplied by the integer, §1 FRACTIONS 61 the result being the same in both cases. Therefore, dividing the numerator or multiplying the denominator by an integer divides the fraction. 113. To Divide an Integer by a Fraction.— Suppose it is desired to divide 12 by f; for instance, how many boxes holding f pound each can be filled from a package holding 12 pounds? If 12 be divided by 1, the quotient is 12, but if 12 be divided by a number smaller than 1, the quotient will evidently be greater than 12. Since 1 is 4 times as large as i, 12 divided by f must be 4 times as large as 12 divided by 1; hence, 12 ^ ^ = 12 X 4 = 48. Since f is 3 times i, 12 divided by f will be only one- third as much as 12 divided by i and the quotient will be 48 X i = 16, or 48 4- 3 = 16 = 12 -- f . This same result may be obtained in a much easier manner by turning the fraction upside down (this is called inverting the 4 4 fraction) and multiplying; thus ;^ X h = 16. Consequently, to divide an integer by a fraction, invert the fraction and multiply the integer by the inverted fraction. 114. To Divide a Fraction by a Fraction.— Suppose it is desired to divide f by |. Since the divisor is a fraction, invert it and multiply I by the inverted fraction; thus, f X f = | = If. That this is correct is easily shown. Since f is 3 times i, 3 4- | = 3 X f = f, and one-fourth of f is f -^ 4 = |, or f X i = | ~ Is- Example. — Divide j\% by ^. • 17 ^ SoLUTiON.^Inverting the divisor and multiplying ^ v — = i^ = 1t\. Ans. 115. To divide a mixed number by an integer, divide the inte- gral part by the integer; if there is a remainder, annex the fraction to it, reduce this new mixed number to an improper fraction, and divide it by the integer. Example.— Divide 508| by 8. Solution.— Here 508 -^ 8 = 63 and 4 remainder: annexing the f to 4 it becomes 4f = V ; ¥ - 8 = U- Therefore, 508| ^ 8 = 63f i Ans. The mixed number might have been reduced to an improper fraction and then divided by the integer; the process here given is better, however (particularly, when the divisor is small), and it requires less work. 62 ARITHMETIC §1 116. To divide a mixed number by a mixed number, reduce both to improper fractions, and then proceed in the regular manner. Example. — Divide 15f by 3^%. 41 5 Solution.— 15f = i^^; S/tt = f^; ^ X p = W = 4f|. Ans. 2 21 117. Rule. — I. To divide a fraction hy an integer, divide the numerator or multiply the denominator by the integer. II. To divide an integer or fraction by a fraction, invert the divisor and multiply the integer or fraction by it. m. To divide a mixed number by an integer, divide the integral part by the divisor; if there is a remainder, annex the fractional part to it, reduce the resulting mixed number to an improper fraction, and divide it by the divisor; if there is no remainder, divide the frac- tional part of the mixed number by the divisor. The sum of the two quotients is the entire quotient. Example.— Divide 1296f by 16. 2 11 SoLXTTiON.- 1296 ^ 16 = 81; I "^ 16 = § X jg = 24! hence, 1296f -j- 16 = Slsi- Ans. That this method of dividing a fraction by an integer is correct is easily seen. Since 16 = ^-, inverting the divisor makes it Yt, and multiplying by rg^ multiplies the denominator of the fraction by 16. This method of dividing by an integer is recommended, because it permits of cancelation, when possible. EXAMPLES (1) Divide f^f by 12. Ans. s\. (2) Divide H by 4f . Ans. ^. (3) Divide IM by U. Ans. 2f . (4) Divide 180 by ||. Ans. 328 (5) Divide ^ by f . Ans. |. (6) Divide 19| by 4|. Ans. 4^. (7) Divide 4795i by 21. Ans. 228f. COMPLEX AND COMPOUND FRACTIONS 118. A complex fraction is one that has a fraction for its nu- 8 3 merator or denominator or both terms are fractions; thus, «' f §1 FRACTIONS 63 3 c and ^ are complex fractions . In order to distinguish the numerator 5 from the denominator, the dividing line is made heavier than the dividing line of the fraction or fractions. Thus, in the first complex fraction, the numerator is f and the denominator is 7, the expression meaning also f -^ 7; in the second complex frac- tion, the numerator is 9 and the denominator is f, the expression also denoting 9 -r- 5 ; in the third complex fraction, the numerator is I and the denominator is f , the expression also denoting f -r- |. 119. To simplify a complex fraction, multiply the digits or numbers above and below the heavy line by the denominator of the fraction; the product will be the denominator of the simplified fraction in the first and third cases above, and the numerator in the second case. Or, transpose the deiiominator of the frac- tion fi"om above to helow the line or from below to above the line,as the case may be, and use it as a multiplier. For example, to 3 simplify the first of the above complex fractions, write 8 7 3 3 ■ .u J o = — 5 — = 15 ; in the third = n ^ o = -5% ; m the second case, o S 5 3 case, o = o y o = I- That these results are correct may be 5 proved by performing the operations indicated. Thus, f -r- 7 = §__ _ 3 q ^ 3 _ Q X ^ — if^- onrl 3 _^ 3. _ 3 V 5 7X8~^ ' ^ ~ 3 ~-^' ^"^ 8 ~ 5 — 8 X 3- 3X8 «• Either term or both terms of a complex fraction may consist of two or more fractions connected by one or more of the four signs 11_3 of operation, +, — , X, -^ ; thus, — j — is a complex fraction, , .^ . 1 ^ 112 29 and it IS equal to -j- = ^^- 64 ARITHMETIC §1 120. A compound fraction is a fraction of a fraction. Thus, f of TT, or f X TT, is a compound fraction. Since division of a fraction by a fraction is changed into multipHcation by in- verting the divisor, f -^ yV rnay also be called a compound fraction; and a compound fraction may be defined as an expres- sion containing two or more fractions connected by signs of multipUcation or division. Even the product of one or more integers and one or more fractions may be called a compound fraction; thus, 145 X If , i of f of 147 = | X f X 147, 3| X 288 X tV, etc. may be called compound fractions. 121. If either or both of the terms of a complex fraction is a compound fraction, the best way to simplify it is to use the method 160X^X^^X900X225 of Art. 119. For example, simplify oqnon 4 714 3 9 • ,u , A • . ;0^X38X;WX^00X^^^ Transposmg the two denommators ;^ x lf>m X m^^ ^ m n 100 38X714X9 „,,,33 , = jQQ = 244lf f . Ans. First cancel the ciphers that are common to the numerator and denominator; this is the same as dividing by 10, 100, etc. The cancelation might have been carried farther, since 38 and 714 are both multiples of 2, and 100 is a multiple of 4; but it is easier to divide by 100 than by 25, and the work was left as it stands. 291 As another example, find the value of -3 qi • Reducing the mixed numbers to improper fractions, 4 v 7 -y^g-- = o8y2T,- AUS. H^ 179 X 3 X 7 X ¥ 43 X 3 INVOLUTION 122. The product of several equal factors is called a power of the number used as a factor. Powers are named first, second, third, fourth, fifth, etc. according to the number of equal factors considered or used. Thus, the second power of 7 is 7 X 7 = 49, the third power of 9 is 9 X 9 X 9 = 729, the fifth power of 6 is 6X6X6X6X6 = 7776. §1 INVOLUTION 65 123. Instead of indicating the product of the factors as above, it is customary to shorten the work by writing a small figure above and to the right of the number used as a factor, the small figure indicating the number of times the factor is to be used and cor- responds with the name of the power. Thus, 6^ is read Q fifth or 6 to the fifth power, and means the fifth power of 6, or 6 X 6 X 6 X 6 X 6 = 7776; b"^ is read 5 fourth or 5 to the fourth power, and represents 5X5X5X5 = 625, etc. The second and third powers are usually called the square and cube, respectively; thus, 2V is read 21 square and equals 21 X 21 = 441, 16^ is read 16 cube and equals 16 X 16 X 16 = 4096. The first power of any number is the number itself; thus 217^ = 217, 48^ = 48, etc. The small figure that is written above and to the right of the number is called an exponent. The power itself is the product found by performing the multiplication; thus, the cube of 16 is 4096, the square of 21 is 441, the fifth power of 6 is 7776, the first power of 528 is 528, etc. Involution treats of powers of numbers. 124. To indicate the power of a fraction, enclose the fraction in parenthesis and write the exponent outside the parenthesis; thus, the cube of f is indicated by (f)^, and it equals f X | 3 _ 3 X 3 X 3 2 7 . p . ^ ~ 4:X4:X4: ~ ^^' ^^^^^' ^^ ^01^0, a fraction to any indicated power, raise the numerator and denominator separately to the power indicated. For example, (t^)^ = 9X9X9X9 7461 . 94 16 X 16 X 16 X 16 = 65536 '"^ °*^^^ ^°^^^- ^tV)^ = -^.^ because A X A X tV X A = -A>19X1^X9__ _ 9^ '^ "" 17 16 X 16 X 16 X 16 ~ 16* 125. To raise 10 to any power indicated by the exponent, simply annex to 1 as many ciphers as there are units indicated by the exponent; thus, 10^ = 100,000, 10^= 1000, 10^ = 10,000,000, etc. In the first case, the exponent 5 indicates 5 units and 5 ciphers follow 1; in the second case, 3 ciphers follow 1 because the exponent is 3; in the third case, 7 ciphers follow 1 because the exponent is 7, etc. These results may all be proved by actual multiplication. Conversely, to express any power of 10 as 10 with an exponent, count the number of ciphers, and the number so found will be the exponent. Thus, 100 = 10^; 10000 = lO''; 1000000 = 10«; etc. 66 ARITHMETIC , §1 To raise a mixed number to the power indicated by the ex- ponent, first reduce the mixed number to an improper fraction, and then raise the fraction to the indicated power. Thus, (4i^)^ = (Ur = -\W- = 22tVt. Ans. EXAMPLES (2) 7462 = ? Ans. 556,516. (2) 873 = ? Ans. 658,503. (3) (fi)3 = ? Ans. sSWj. (4) 100,000,000 is what power of 10? Ans. The 8th. (.5) 10^ = ? Ans. 10000. (6) (291)2 = ? Ans. 877|i. 126. To Multiply or Divide by a Power of 10. — Consider what occurs when some integer, as 7035, is multiphed by some power of 10, say 10,000. According to Art. 58, the product is found by multiplying by 1 and then annexing 4 ciphers, the number of ciphers to the right of 1; thus, 7035 This operation is equivalent to the following: to IQOQO multiply any number (integer) by a power of 10, 70350000, simply annex to the number as many ciphers as there are ciphers in the given power of 10 or as many ciphers as are indicated by the exponent of 10. For instance, 10,000 = 10^; hence, to multiply any number by 10,000 or 10^, annex 4 ciphers to the integer. Now any integer, as 7035, may be written 7035., the decimal point being always understood to follow the unit figure (5, in this case) whether written or not; but when it is written, any number of ciphers may be annexed to the number without altering its value. Thus, 7035.0, 7035.0000, and 7035 all have the same value, since the position of the unit figure has not been changed and the addition of the ciphers has not added anything to the number. If, now, 7035.0000 be multiplied by 10,000, the product will be the same as before, or 70350000., the decimal point being moved 4 places to the right. In the product, 5 no longer indicates 5 units, but 5 ten-thousands. From this, it is evident that any number may be multiplied by a power of 10 by moving (shifting) the decimal point as many places to the right as there are ciphers in the power or indicated by the exponent of 10. Thus, 3.1416 X 102 = 3.1416 X 100 = 314.16; 3.1416 X 10* = 3.1416 X §1 DECIMALS AND DECIMAL FRACTIONS 67 10000 = 31416; 3.1416 X 10^ = 3.1416 X 10000000 = 31416000; etc. Rule. — To multiply any number by a power of 10, shift the decimal point as many places to the right as there are ciphers in the power or indicated by the exponent of 10, annexing ciphers, if necessary. 127. Since 3.1416 X 1000 = 3141.6, 3141.6 -^ 1000 must evi- dently equal 3.1416; in other words, to divide by a power of 10, shift the decimal point as many places to the left as there are ciphers in the power or indicated by the exponent of 10. This evidently follows also from the fact that division is the reverse of multiplication; hence, if the decimal point is shifted to the right for multiplication, it must be shifted to the left for division. To divide 3.1416 by 10,000, it is necessary to prefix ciphers to the first figure, 3, in order to make it occupy a position 4 places to the right of unit's place, and 3.1416 -^ 10^ = 3.1416 -4- 10000 = .00031416. In multiplication, the unit figure is removed to the left; in division, it is removed to the right Also, 7035 -^ 10^ = 7035 ^ 10000000 = .0007035. Rule. — To divide any number by a power of 10, shift the decimal point as many places to the left as there are ciphers in the power or indicated by the exponent, prefixing ciphers, if necessary. It will be noted that the 3 in 3.1416 indicates 3 units, while in .00031416, it indicates 3 ten-thousandths, a number ten thou- sand times smaller than 3 units. This is evidently correct, since the divisor is 10,000. DECIMALS AND DECIMAL FRACTIONS DEFINITIONS AND EXPLANATIONS 128. A decimal fraction is one that has a power of 10 for its denominator; yV, tWoT; 3tVVA, etc. are decimal fractions. When the numerator is divided by the denominator, the above fractions become .7, .5236, 3.1416, and are called decimals. 129. Any decimal may be converted into a decimal fraction by writing the decimal (considered as an integer) for the numerator of the fraction and writing for the denominator a power of 10 containing as many ciphers as there are decimal places in the 68 ARITHMETIC §1 decimal. For example, .7854 = Am, .00034 = thUhi^, .052465 ~ yoToooiTj etc. 130. The value of a decimal is not changed by annexing ciphers; thus, .25000 = .25. This is evident, since nothing has been added to the number represented by the decimal, and the position of the first digit relative to the decimal point has not been changed. This may also be shown by converting both decimals into decimal fractions; thus, .25000 = -^f^f^an^^ — i^' and .25 = tVt- Here it is seen that as many ciphers are added to the denominator as are added to the numerator, and these may be canceled in the fraction. 131. If the numerator of a decimal fraction contains the factor 2 or 5, the decimal fraction may be reduced to lower terms, thus becoming an ordinary common fraction. For example, .625 _ 6 25 _ 12 5 _ Si. _ 5. CiOcy — 3 2 _ 4 . 0'7(\A — 2 7 04 = ill; etc. If the decimal does not contain 2 or 5 as a factor, that is,' if it does not end in 5 or is not an even number, the decimal fraction cannot be reduced, because since 2 X 5 = 10 and they are both primes, 2 and 5 are the only factors in any power of 10. Thus, 1000 = 103 = 10X 10 X 10 =2X5X2X5X2X5 = 2X2X2X5X5X5= 23 X 53. 132. Addition and Subtraction of Decimals. — It has already been shown how to add decimals and numbers containing decimals; simply place the decimal points under one another, and add or subtract as in the case of integers, placing the decimal point in the result directly under the decimal point in the num- bers added or subtracted. The only case that need be considered here is that in which the subtrahend contains more decimal places than the minuend. In such a case, annex ciphers to the minuend until it contains as many decimal places as the subtra- hend, and then subtract. Example 1.— From 426.45 take 294.0847. Solution. — Here the minuend contains two decimal places and the sub- trahend contains four; therefore, annexing two ciphers, 426 4500 which does not change the value of the minuend, the sub- 294 0847 traction is performed as in the case of integers. This opera- tion is equivalent to reducing the decimal fractions to a common denominator, which in this case is 10,000. In practice, the ciphers would not be written; they would simply be considered to be added. §1 DECIMALS AND DECIMAL FRACTIONS 69 Example 2.— From 93 take 77.5652. Solution. — The work is arranged as shown in the margin go with the decimal points under each other, and the ciphers «y 5652 that are supposed to be annexed are not written, but are : , understood l^-^^^S Ans. MULTIPLICATION OF DECIMALS 133. The only difference between multiplying decimals and multiplying integers is the locating of the decimal point in the product. The number of decimal places in the product is equal to the sum obtained by adding to the number of decimal places in the multiplicand the number of decimal places in the multipHer. To understand the reason for this, multiply two decimal fractions. Thus, .24 X .637 = j\\ X iVA = tVoVA = -15288. Here it will be noted that the number of ciphers in the denominator of the product is equal to the sum of the number of ciphers in the denominators of the two fractions. But the number of ciphers in the denominators is the same as the number of decimal places .637 in the factors; therefore, multiply the ■ 24 numbers in the same manner as though they 2548 were integers, disregarding the decimal points. 1274 The product is found to be 15288. There are . 15288 Ans. 3 decimal places in the multiplicand and 2 in the multiplier; hence, there are 3+2 = 5 decimal places in the product. Had the numbers been .037 and .24, the product as integers would have been 888; pointing off 5 decimal places, the product as decimals is .00888. 134. Mixed numbers are multiplied in exactly the same way as integers, and the number of decimal places in the product is determined in the same manner as in multiplication of decimals. Example 1.— What is the product of 75.305 and 3.1416? 7 5.305 Solution. — The product of the numbers is found 3.1416 as though they were integers. Since there are 3 4 5 18 3 decimal places in the multiplicand and 4 in the 7 5 3 5 multiplier, 7 decimal places are pointed off in the 3 12 2 product, which is, therefore, 236.578188. 7 5 3 5 2 2 5 9 15 23 6. 5781880 Ans. 70 ARITHMETIC §1 Example 2.— Find the product of 47,082 and .0005073. 4 7 8 2 Solution. — The numbers are multipHed as in . 5 7 3 example 1, as though they were integers. The 14 12 4 6 multiplicand contains no decimal places and the 3 2 9 5 7 4 multiplier contains 7; hence, 7 decimal places are 2 3 5 4 10 pointed off in the product. 2 3.8 8 4 6 9 6 8 Ans. EXAMPLES (1) 47.95 X 126.42 = ? Ans. 6061.839. (2) .0903 X .7854 = ? Ans. .07092162. (3) 730.25 X 36.524 = ? Ans. 26671.651. (4) 1285 X .0001016 = ? Ans. .130556. (6) .00575 X .00036 = ? Ans. .00000207. (6) What will be the cost of 7042 tons of coal at $7.33 per ton? Ans. $51,617.86. (7) If the price of a certain dye is $1.13 per ounce, how much must be paid for 18.625 ounces? Ans. $210.4625, or $210.46. Note. — In all laonetary transactions, if the fractional part of a cent is equal to or greater than J the number of cents is increased by 1; otherwise, the fraction is rejected. In the last example, the fraction of a cent is .25 (found by moving the decimal point so as to follow the cent), and it is therefore rejected. (8) What amount must be paid for 5050 pounds of paper at 12.375 cents per pound? Ans. 62,554 cents', or $625.54. (9) How much freight will have to be paid on 439 tons of limestone at $4.17 per ton? Ans. $1830.63. (10) If a pulp mill uses an average of 9.07 tons of soda ash per month and the cost is $45 per ton, how much will be paid for this material in a year? Ans. $4897.80 DIVISION OF DECIMALS 135. Decimals are divided in exactly the same manner as integers, no attention being paid to the decimal point until all the figures of the dividend have been used. To understand how the decimal point is located in the quotient, consider carefully the following examples and their solutions, paying strict attention to the explanations. Example 1.— Divide 7.092162 by .0903. Solution. — The division is performed as shown in the y 092162( 0903 margin, the quotient being 7854 when the dividend and g 321 tqTka divisor are considered as integers. To locate the decimal zrr .' point, subtract the number of decimal places in the divisor 7004 from the number of decimal places in the dividend, and the remainder will be the number of decimal places in the quo- 4»7d tient. In the present case, the divisor contains 4 decimal '*^^^ places and the dividend 6; hence, there are 6—4 = 2 3612 decimal places in the quotient. 3612 §1 DECIMALS AND DECIMAL FRACTIONS 71 To understand the reason for above proceeding, convert the decimals into decimal fractions and divide by the rule for division of fractions Thus ^^^^^^^ ^ -903_ _ 7092162 100^ 01 iracuons. ±nus, ^qqqqqq • j^qqqq - ioO0000 ^ 903 = ~Tja?r = 78.54. Here 4 ciphers in the numerator of the divisor cancel 4 ciphers in the denominator of the dividend, leaving 6 — 4 = 2 ciphers in the denominator of the quotient. Since the number of ciphers in the denominators of the fractions is the same as the number of decimal places in the corresponding numbers expressed as decimals, the number of decimal places in the quo- tient is always equal to the number of decimal places in the divi- dend minus the number in the divisor. Example 2.— Divide 7.092162 by 903. Solution. — Considering the dividend as an integer, the quotient is 7854, as found in the last example. Since there are no decimal places in the divisor and there are 6 in the dividend, there are 6—0 = 6 decimal places in the quotient, which is therefore equal to .007854. That this is correct may be proved by multiplying the quotient by the divisor. Example 3.— Divide 70921.62 by .0903. Solution. — Considering both numbers as integers, the quotient is 7854. Since the divisor contains a greater number of decimal places than the divi- dend, annex ciphers to the dividend (this does not change its value, see Art. 130) until it contains the same number as the divisor; in this case, 2 ciphers, then since both dividend and divisor contain the same number of decimal places, in this case 4, the quotient is an integer, because 4—4 = 0, and there are no decimal places to be pointed off. Therefore, 70921.62 H- .0903 = 70921.6200 ^ .0903 = 785400, treating the dividend and divisor as integers. The same result may also be arrived at as follows : Expressing the decimals as decimal fractions and dividing as in example 1, 7092162 10000 7092162 ^^ „ , ~~1M~ ^ "903 ^ 9Q3 X 100 = 7854 X 100 = 785400. Or, since multiplying both numerator and denominator by the same number does not alter the value of the fraction, -^^^-^ X T¥¥ = ^ Ull^-^, a fraction whose denominator is the same as the denominator of the divisor, and which equals the decimal 70921.6200, which, in turn, is the dividend with 2 ciphers an- j ^ ^ 709216200 ;0000 „ ^ nexed. But, — JaMa — X qq^ = 785400, the same result as before. 72 ARITHMETIC §1 Example 4.— Divide 4575.12 by 7.854. Solution. — Since the divisor contains one more decimal place than the dividend, 4575 120('7 R'^4 annex a cipher to the dividend. The 3Q27 — 6 2 A quotient is found to be 582, and there 582^19 Am. -g ^ remainder of 4092; hence, the 648 12 Or, 582.521^i9 Ans. q^^tient is 582A|f|. = 582^^, a mixed number composed of an integer and a 19 800 common fraction. It is generally more 15 708 convenient to express the quotient in 4 0920 4It4 = tW such cases as a mixed number composed 3 9270 of an integer and a decimal; in this case, 16500 simply annex one or more ciphers to the 157Q8 remainder and proceed with the di- TQon vision as indicated. In other words, ycr^ tW = .521yt9"' That this is the case 6 6 _ 1 ^^ readily shown. Thus, 62 = 62.000, 66 T8^T - rr? and 62.000 ^ 119 = .521y|9. Note that three ciphers in all were annexed to the remainders, since there were three different remainders used as dividends, and one cipher was annexed to each. Example 5. — Find to seven decimal places the quotient of xfff f -s"- 348.0000000(25575 Solution.— Since there are no decimal 255 75 0136070 Ans. places in either dividend or divisor, annex as g2 250 many ciphers to the dividend following the ^r, ,j2K decimal point as there are decimal places required in the quotient, in this case 7. When all the figures in the dividend have 15 3450 been used, there will be 7 decimal places in 180000 the quotient, because 7-0 = 7. The 179025 quotient to 7 decimal places is .0136070. 9750 136. Whenever the divisor contains a factor other than 2 or 6, there will always be a remainder, no matter to how many decimal places the quotient is carried, and as this is usually the case, it is customary to carry the division one place farther than the num- ber of decimal places specified. If, then the extra figure is 5 or a greater digit, the preceding figure is increased by 1, and a minus sign is written after the quotient to show that it is not quite so large as written or printed; but, if the extra figure is less than 5, it is rejected, and a plus sign is written after the quotient to show that it is a Kttle larger than written or printed. In this example, it is seen that the figure in the 8th decimal place is 3 ; hence, the quotient to 7 decimal places would be written .0136070 +. The quotient to 3 decimal places would be written .014 — . In prac- tice, these signs may be omitted. §1 DECIMALS aWd DECIMAL FRACTIONS 73 Example. — Find the quotient of .0348 4- .00255 to five decimal places. .03480 ( .00255 255 13.647058 930 13.64706— Ans. Solution. — The divisor contains one more 765 decimal place than the dividend ; hence, annex- TT^Q ing one cipher to the dividend, the integral part iKOQ of the quotient is found to be 13. Ciphers are ■ — — now annexed to the different remainders as de- scribed in example 4, and 5 + 1=6 figures of the decimal are found. Since the sixth figure 1°00 Qf ^j^e decimal is 8, the preceding figure, 5, is l^o5 increased by 1, and the quotient to five decimal 1500 places is 13.64706-. 1275 2250 2040 137. Rule — -I. If the divisor does not contain more decimal places than the dividend, divide as though the numbers were integers; subtract the number of decimal places in the divisor from the number in the dividend, and point off in the quotient as many decimal places as are indicated by the remainder. II. If the divisor contains a greater number of decimal places than the dividend, annex ciphers to the dividend until it contains the same number of decimal places as the divisor; if the dividend {with the annexed ciphers) is then larger than the divisor, both being considered as integers, the quotient when all the figures of the dividend (including the annexed ciphers) have been used will be an integer. But, if the dividend is smaller than the divisor, annex ciphers to the dividend and proceed with the division; the quotient will be a decimal, and may be pointed off as in I. III. If there be a remainder, it may be written over the divisor, and the resulting fraction reduced to lower terms, if possible. Or, ciphers may be annexed to the remainders as found and the division continued until the quotient has been determined to any desired number of decimal places. EXAMPLES (1) Divide 268.92 by .007007. Ans. 38378.76+ (2) Divide 304.75 by 3.125. Ans. 97.52. (3) Divide 100.43 by 1000.26. Ans. .100404-. (4) Divide 648.433 by 6509.85. Ans. .099608-. (5) Divide .00218 by 966. Ans. .000002257-. 74 ARITHMETIC §1 (6) The price of a certain dye is $19.25 per pound; if the bill for a purchase of dye amounts to 11398.88, how many pounds were bought? Ans. 72.67— pounds. (7) The amount paid for 3490 tons of coal was $25,642.12; what was the price per ton? Ans. $7.35. (8) Bought press felts at $2.16 per pound. The bill came to $546.48; what was the weight of the felt? Ans. 253 pounds. DECIMALS AND COMMON FRACTIONS 138. To reduce a common fraction to a decimal, divide the numerator by the denominator. If the denominator is a power of 2, as 2, 4, 8, 16, 32, 64, etc., an exact decimal equivalent can be found for the common fraction; this will also occur when the denominator is a power of 5, as 5, 25, 125, 625, 3125, etc. But, if the denominator contains any other factor than 2 or 5, the quotient will never be exact, no matter to how many decimal places the division may be carried. In such cases, find the quotient to one more decimal place than is desired; if the extra figure is 5 or a greater digit, increase the preceding figure by 1 and annex the sign — ; otherwise, reject it and annex the sign +. Examples. — Reduce (a) xi > (b) ii, and (c) ^f y to decimal fractions : Solution. — The work is shown herewith. In (a), the first figure of (b) (c) 17.0 (24 5.000 ( 657 Ans. 16.8 ■ 70833 + Ans. 4 599 .0076103 + Ans. 200 4010 192 3942 ~80 680 72 657 ~80 2300 72 1971 11 9 (a) .0 (16 6 . 6875 1 40 1 28 120 112 80 80 8 329 the quotient evidently denotes .6, since it was necessary to annex one cipher to the dividend, 11, before dividing; and since the divisor (denominator) is a power of 2 (2* = 16), the quotient is exact, the remaining figures being found by annexing ciphers to the different remainders. In (b), the first figure of the quotient denotes .7; the remaining figures are found as in {a). Note tha,t the second remainder, and all succeeding remainders, is 8, and that the third figure of the quotient, and all succeeding figures, is 3. In (c), it is necessary to annex 3 ciphers to the dividend (numerator) before it will contain the divisor (denominator); hence, the first figure §1 DECIMALS AND DECIMAL FRACTIONS 75 of the quotient denotes .007. The division is here carried to 7 decimal places. The next figure will be 5, and the correct value of the quotient to 7 decimal places is .0076104 — . 139. Mixed Fractions. — It sometimes happens that it is desirable to express a quotient exactly, even though it has been reduced to an approximate decimal. For instance, in (6) of the last example, the quotient might have been written .708/i^ = .708^, and the quotient in (c) might have been written .00761 Ay- Expressions of this kind are called mixed fractions. A mixed fraction may be reduced to a common fraction in the same way that a mixed number is reduced to an improper fraction; that is, multiply the decimal by the denominator of the fraction, add the numerator to the product, and write the sum over the denominator. Thus, .708| = •'^Q^ X 3 + -001 2. 124 +.001 2.125 ^, = Q = — o — The numerator, 1, of the fraction be- longs to the same order as the figure of the decimal that precedes it ; in other words, .7081 = jVW + -Em = .708 + '~ To get rid of the decimal in the numerator of the reduced fraction, note 2 125 X 8 that .125 = I; hence, multiply both terms by 8, and ' ^ = XX A.O,-. nn'7a. 2 3_ _ :00761 X 657 + .00023 = U' Again, .00761 /A 657 _ 4.99977 + .00023 _ ^ 657 ~ ^^^" In practice, the work would be performed as .00761 shown in the margin. Here the decimal is 657 multiplied by the denominator, and the num- 5327 erator is added to the product before pointing 3805 off, both the product and the numerator being 4 566 treated as integers. 4 99977 23 140. To Reduce a Decimal to a Fraction 5 00000 Having a Given Denominator. — To reduce a decimal to a fraction having a specified de- nominator, all that is necessary is to multiply the decimal by the specified denominator; the product will be the numerator of the fraction. Thus, to reduce .671875 to a fraction having 64 for its denominator, .671875 X 64 = 43; hence, the fraction is If. 76 ARITHMETIC §1 The reason for this is simple. No number is changed by- multiplying it by 1; 1 = If; and .671875 X fi =- wi = If. Or, reducing the decimal to a decimal fraction, .671875 = iVoVVro; riow multiplying both terms of this fraction by 64, .^ , , . ^ 671875 X 64 43000000 ^^ the specified denommator, j^^^^ ^ ^^ = q^qq^qqq = ^l In the case just given, the product of the decimal and the specified denominator is an integer, 43. This seldom occurs in practice, and what is usually sought is that fraction having the specified denominator that is nearest in value to the decimal. For instance, what fraction having a denominator of 32 is nearest in value to .708? Since .708 X 32 = 22.656, = 23-, .708 = If, approximately. Had the decimal been .703, .703 X 32 = 22.496 = 22 +, and .703 = 11 = ii, approximately. Here, as before, if the first figure of the decimal part of the product is 5 or a greater digit, increase the preceding figure by 1 ; otherwise, reject the decimal. Had it been required to express .703 as a fraction having a denominator of 64, .703 X 64 = 44.992 ., = 45 — , and .703 = ||, approximately. EXAMPLES (1) Reduce f?f to a decimal. Ans. .80859375. (2) Reduce .8035|4 to a common fraction. Ans. fHfH. (3) Reduce ^ffx to a decimal. Ans. .0061597 — . (4) Reduce xflfo to a mixed decimal. Ans. .0569836^^-. (5) Express .8086 inch to the nearest 64th of an inch. Ans. It = rl inch. (6) Express as decimals |, |, |. Ans. .375, .625, .875. (7) Express .3937 to the nearest 12th. Ans. j^^. (8) Express .3937 inch to the nearest 32d inch. Ans. ^ inch. (9) Reduce .05698ff pound to a common fraction. Ans. yHI^ pound. SIGNS OF AGGREGATION 141. When it is desired to indicate that several numbers are to be operated upon as though they were a single number, a sign of aggregation is used to enclose the numbers. The word aggre- gate means to collect, to bring several things together. There are four of these signs : the vinculum , a straight line placed over the numbers, and the parenthesis ( ), brackets [ ], and brace { }, §1 RATIO AND PROPORTION 77 used to enclose the numbers. An expression like 3 X (45 — 28) or 3 X 45 - 28 means that 28 is to be subtracted from 45 and the remainder is to be ' multiplied by 3. In arithmetic, onlj the vinculum and the parenthesis are generally used. 142. Order of Signs of Operation. — When several numbers are connected by signs of operation denoting addition and subtrac- tion, perform the operations indicated in regular order from left to right; thus, 25 - 19 + 47 - 32 - 9 + 50 = 6 + 47 - 32 - 9 + 50 = 53 - 32 - 9 + 50 = 21 - 9 + 50 = 12 + 50 = 62. If, however, a sign of multiphcation or division occurs, the operation indicated by this sign must be performed before adding or subtracting; thus, 25 ~ 3 X 5 + 42 ^ 7 + 38 = 25 - 15 + 6 + 38 = 10 + 6 + 38 = 16 + 38 = 54. If signs of multiphcation and division follow one another with no signs of addition or subtraction between them, perform the operations of multiphcation and division in order from left to right before adding or subtracting; thus, 46 + 54 -^ 6 X 8 - 67 = 46 + 9 X 8 - 67 = 46 + 72 - 67 = 118 - 67 = 51. Now by using signs of aggregation, the order of operations indicated by the signs of operation may be changed; thus, if desired, the last expression may be written (46 + 54 -^ q) X 8 - 67. Here 54 is first divided by 6, the quotient is added to 46, the sum is multiphed by 8, and 67 is then subtracted, the result being (46 + 9) X 8 - 67 = 55 X 8 - 67 = 440 - 67 = 373. When one sign of aggregation includes another, as in this case, always consider the inner sign first. EXAMPLES (1) 159 - (8 + 5 X 16) -=- 11 - 100 = ? Ans 51 ^2) 6 X 13 - 119 ^ 7-50^ 12 = ? Ans. 5QH- (3) 20 + (126 - 4 X 270 -^ 9) X 15 - 35 = ? Ans 75 (4) 96 - 7.3 X 11 ^ 14 + 24 X 8 = ? Ans. 282".2/j" RATIO AND PROPORTION RATIO 143. It is frequently desirable to ascertain the relative sizes of two numbers. For instance, suppose it is desired to know the relative sizes of 28 and 7, that is, how many times 7 is 28 ? Divid- 78 ARITHMETIC §1 ing 28 by 7, 28 -^ 7 = 4; hence, 28 is 4 times 7, or, in other words, 28 is 4 times as large as 7. What has here been done is to compare 28 and 7, the comparison being done by division. If it were desired to compare 7 to 28, that is, to find what part of 28 is 7, divide 7 by 28, obtaining 7 -^ 28 = j; in other words, 7 is 3'^th of 28 or 7 is 3'^th as large as 28. When two numbers are compared in this manner, by dividing the first by the second, the comparison is called a ratio. The language used in the first of the above cases is the ratio of 28 to 7, and in the second case, the ratio of 7 to 28; in either case, the ratio of one number to another is the first number divided by the second. 144. A ratio may be indicated as above by using the sign of division, but it is usual to use the colon (see Art. 64), thus indicat- ing distinctly that a ratio is implied; thus, the ratio of 57 to 36 is written 57 : 36. A ratio is also very frequently indicated in the form of a fraction; thus, the ratio of 128 to 16 is written 128 : 16 128 or -r^. The second, or fractional, form possesses many advan- 1 A tages. The ratio of 16 to 128 is written 16 : 128 or -=-^. The two numbers used in forming or expressing a ratio are called the terms of the ratio. The number to the left of the colon or above the dividing line is called the first term, and the other number is called the second term. In the ratio 32 : 12, 32 or Yn, 32 IS the first term and 12 is the second term. The value of a ratio is the quotient obtained by dividing the first term by the second term; thus, the values of the ratios 57 : 32 36, 16 : 128, j^, etc. are liV, h 2|, etc. respectively. 145. Ratios like the above in which the first term is named first in speaking or writing are called direct ratios. It is frequently convenient to name the second term first, in which case the ratio is called an inverse ratio ; thus, the direct ratio of 24 to 4 is 24 : 4, and its value is 6; but the inverse ratio of 24 to 4 is 4 : 24, and its value is |. The word direct is seldom or never used in connection with ratio; but if the ratio is an inverse one, the word inverse or inversely is always used. If there is nothing to -show that the ratio is inverse, it is always taken to be direct. §1 RATIO AND PROPORTION 79 The best way of writing an inverse ratio is to write it first as if it were direct, and then invert, i.e., transpose, the terms. For example, to write the inverse ratio of 36 to 90, write first 36 : 90 or 7^; now invert the terms and obtain 90 : 36 or 7^. The 90 36 value of the inverse ratio of 36 to 90 is 2| = 2.5, 146. Evidently, only like numbers can be used to make up the terms of a ratio. For instance, 5 dollars cannot be compared with 8 feet; but 5 dollars can be compared with 8 dollars, and 5 feet can be compared with 8 feet. The speed of one shaft can be compared with the speed of another shaft. 147. If both terms of a ratio be multiplied or both be divided by the same number, it will not alter the value of the ratio. Thus, the value of the ratio 32 : 12 is 2f ; multiplying both terms by 3, the ratio becomes 96 : 36, and its value is 2§, as before; dividing both terms by 4, the ratio becomes 8 : 3, and its value is 2f , as before. The reason for this is seen when the ratio is 32 written in the fractional form, j^. Now regarding this ex- pression as a fraction, it has been shown in connection with frac- tions that multiplying or dividing the numerator and denominator by the same number does not alter the value of the fraction. PROPORTION 148. If there are two ratios, each having the same value, and they are written so as to indicate that the ratios are equal, the resulting expression is called a proportion. Thus, the value of the ratio o~i^ is f , and the value of the ratid ^^ifa is | ; placing these two ratios equal to each other, 5-j = ^■,' ^ , an ex- ' 8 hours $1.76' pression that is called a proportion. It is to be noted that the value of any ratio is always an abstract number (Art. 6) ; it is for this reason that the ratio of two concrete numbers can be placed equal to the ratio of two entirely different concrete numbers, as in the above case. Hence, when stating a proportion, it is not customary to write the name of the quanti- ties, and the above proportion would ordinarily be written either as c = YY^ or as 5 : 8 = 1.10 : 1.76. Here 5 is called the first 80 ARITHMETIC §1 term, 8 is called the second term, 1.10 is called the third term, and 1.76 is called the fourth term. When written in the second form above, the two outside terms, 5 and 1.76, are called the extremes, and the two inside terms, 8 and 1.10, are called the means. 149. Law of Proportion. — In any proportion, the product of the extremes is equal to the product of the means; thus, in the propor- tion just given, 5 X 1.76 = 8.8, and 8 X 1.10 = 8.8. To understand the reason for this law, the value of the two ratios in the above proportion is .625, and the proportion may be reduced to the expression .625 = .625. Dividing both numbers by .625, the expression reduces to 1 = 1. Now writing the pro- portion in the form ^ = y^, divide both ratios by %, and the 5 8 1 10 8 8 8 result is g X r = t^ X r, or 1 = ^ = 1. But, 8 X 1.10 is the product of the means and 5 X 1.76 is the product of the extremes, and both are equal to 8.8. There is still another way of writing a proportion that was formerly used by the older writers on mathematical subjects; they employed the double colon in place of the sign of equality when writing the proportion in the second form; they 15 405 would express the proportion y^ = oKt as 15 : 12 ::405 : 324. Here, as before, the product of the extremes equals the product of the means, since 15 X 324 = 12 X 405 = 4860. This last way of writing a proportion is not much used at this time, the sign of equality being preferred. 150. A proportion may be read in two ways. Consider the proportion 16 : 10 = 88 : 55; this may be read 16 is to 10 as 88 is to 55, or it may be read the ratio of 16 to 10 equals the ratio of 88 to 55. Either way is correct, but the second is to be preferred when using the fractional form for expressing the ratios. 151. A proportion, like a ratio, may be direct or inverse; in a direct proportion, both ratios are direct, but in an inverse propor- tion, one of the ratios is inverse. If both ratios were inverse, the proportion would become direct again. For example, the state- ment that the ratio of 16 tolO equals the inverse ratio of 55 to 88 indicates an inverse proportion. To write it, first state it as though it were a direct proportion and then invert one of the §1 RATIO AND PROPORTION 81 ratios; thus, 16 : 10 = 55 : 88. Now inverting the first ratio, 10 : 16 = 55 : 88, whence, 10 X 88 = 16 X 55; or, inverting the second ratio, 16 : 10 = 88 : 55, whence, 16 X 55 = 10 X 88. The proportion as first written was not true, since 16 X 88 = 1408 and 10 X 50 = 550; but by inverting one of the ratios, it became true. The test of any proportion is the law of Art. 149. If the product of the extremes does not equal the 'product of the means, then the expression is not a proportion. 152. To Find the Value of One Unknown Term in a Proportion. The object of any proportion is to find the value of one of the terms when the values of the other three are known; this fact will be made clearer shortly. Let X represent the value of the term that is not known, and suppose the proportion is 14 : 8 = 49 : aj. By the law of pro- portion, 14 X a; = 8 X 49. Now, evidently, if 14 times a; = 8 8 X 49 X 49, 1 times x must equal — zrg — = 28, and the proportion 14 : 8 = 49 : 28 is true, because 14 X 28 = 8 X 49 = 392. Suppose the first term had been unknown; then re : 8 = 49 : 28, 8 X 49 and 1 times x = x = — ^o — = 14. Here it is seen that if the unknown is one of the extremes, its value can be found by dividing the product of the means by the other extreme. Suppose the second term had been unknown; then the propor- tion would have been 14 : re = 49 : 28, from which 14 X 28 14 X 28 = 49 X re, and the value of x is evidently — j^ — = x, or re = 8. Lastly, suppose that the third term had been unknown; then 14: 14 X 28 8 = re : 28, and the value of re is evidently -a = re, or re o = 49. Here it is seen that if the unknown is one of the means, its value can be found by dividing the product of the extremes by the known mean. 153. Proportion is used for solving a great variety of problems. For example, the statement that "the circumferences of any two circles are to each other as their diameters" or that ''the circum- feience of a circle varies directly as its diameter" implies a propor- tion. In the first statement, if the diameter and circumference of any one circle are known and the diameter (or circumference) of some other circle is given, the circumference (or diameter) of that circle may be found by the method explained in Art. 152. 82 ARITHMETIC §1 Thus, the circumference of a circle whose diameter is 5 inches is 15.708 inches, very nearly; hence, to find the circumference of a circle whose diameter is 17.6 inches, first form the proportion 1 7 6 V 1 5 708 5 : 17.6 = 15.708 : x, and x = ' ^, = 55.29216 inches. o Note particularly the way in which the above proportion was formed. The first ratio consists of the two diameters, which are like quantities, and the second ratio consists of the two circum- ferences, which are also like quantities. The first and third terms in every direct proportion must belong or relate to the same thing, and the second and fourth terms must also belong or relate to the same thing; in the present case, 5 and 15.708 are the diame- ter and circumference of one circle, and 17.6 and 55.29216 are the diameter and circumference of the other circle. It will also be noted that the second circle is larger than the first circle ; hence, its circumference must also be larger. But, if x be written for the third term of the proportion, the value that will be found for X will be smaller than the third term, thus showing again that x must be written as the fourth term. If desired, the proportion might have been written 15.708 '.x = b'. 17.6, and the same value will be obtained for x. It is customary, however, to have the first ratio consist of numbers or quantities that are known. Now referring to the second statement above, that the circum- ference of a circle varies directly as its diameter, this means that if the diameter increases, the circumference also increases, and if the diameter decreases, the circumference also decreases, both increasing or decreasing in the same ratio. Therefore, if the circumference of a circle whose diameter is 5 inches is 15.708 inches, and it is desired to find the circumference of the circle when the diameter is increased to 17.6 inches, form the proportion 5 : 17.6 = 15.708 : X. Substituting the value of x in this proportion, 5 : 17.6 = 15.708 : 55.29216. The diameter of the second circle has increased in the ratio 17.6 : 5, whose value is 3.52, and the circumference of the second circle has increased in the ratio 55.29216 : 15.708, whose value is 3.52; hence, the diameter and the circumference increased in the same ratio. From the fore- going, it is seen that the word vary, as here used, always implies a proportion or that a proportion can be formed. In practice, instances frequently occur in which an increase in one of the terms of the first ratio results in a decrease in the value of the corresponding term in the second ratio, and vice versa. §1 RATIO AND PROPORTION 83 Thus, suppose a certain volume of air be confined in a cylinder, and that it is made to keep this volume by means of a piston that is free to move up and down and on which is placed a weight. Now it is evident that if the weight be increased, the piston will move downward and the volume of the air will be less; or, if the weight be decreased, the piston will move upward and the volume of the air will be greater. In other words, the volume decreases as the pressure increases, and vice versa. This fact is expressed more elegantly by saying that the volume varies inversely as the pressure, which, of course, implies an inverse proportion, when a proportion is possible. As an example, it is known that when the temperature remains the same, the volume of a gas (air, for instance) varies inversely as the pressure; if the volume is 12.6 cubic feet when the pressure is 14.7 pounds per square inch, what is the volume when the pressure is 45 pounds per square inch? Forming the proportion as though it were direct, 14.7 : 45 = 12.6 : X. Here it is seen that the value of x obtained from the pro- portion will be greater than 12.6, and it should be less, thus indicating an inverse proportion. Since the proportion is inverse, invert the terms of one of the ratios, say the second, obtaining 14 7 V 19 fi 14.7 :45 = a; : 12.6, from which x - ^^ - 4. 116 cubic feet. Note that the volume is smaller than the original volume, as it ought to be. Example 1. — If 5 men can do a certain piece of work in 22 hours, how long will it take 9 men to do the same work, if they all work at the same rate? Solution. — It is evident that 9 men can do the work in less time than 5 men; hence, the proportion is inverse. Stating as a direct proportion, 5 : 9 = 22: a:; then, inverting the second ratio, 5 : 9 = a; : 22, from which 5 X 22 ^^-. , ^ X = g — = 12^ hours. Ans. Example 2. — If 5 men earn $48.95 in 22 hours, how much will 9 men earn in the same time at the same rate of pay? Solution. — It is here plain that 9 men will receive a greater wage than 5 men; hence, the proportion is direct, and 5:9 = 48.95 : x, from which 48.95 X 9 X = — '—£ = $88.11. Ans. Note that the time, 22 hours, has nothing to do with the proportion, because it is the same in both cases. 154. Compound Proportion. — A simple proportion is one in which all the terms consist of but one number in each term; but when two of the terms or all four of the terms contain more than one number, then the proportion is said to be compounded, or it is called a compound proportion. Suppose the last example had 84 ARITHMETIC §1 been stated thus: if 5 men earn $48.95 in 22 hours, how much will 9 men earn in 15 hours? Here 5 men in 22 hours earn $48.95, and it is required to find how much 9 men will earn in 15 hours, all being paid at the same rate. The proportion in this case should be stated as follows: 5 X 22 : 9 X 15 = 48.95 : x, from which X = ot^ — = $60,073^. The reason for stating the proportion in this manner is that 5 men working 22 hours is the same as one man working 5 X 22 = 110 hours, and during this 110 hours, $48.95 was earned. Also, 9 men working 15 hours is the same as one man working 9 X 15 = 135 hours, and during this 135 hours, a certain amount was earned that is re- presented by X. Instead of multiplying before substituting in the proportion, it is better to indicate the multiplication, so as to take advantage of any opportunity of cancelation; thus, 3 4.45 9^aiXiW_ 2 Example. — If 25 men can dig a ditch 600 feet long, 43^ feet wide, and 3}^ deep in a certain number of days, working 8 hours a day, how long a ditch 4 feet wide and 4 feet deep can 36 men dig when working 10 hours a day? Solution. — If men and hours be considered in forming the first ratio, the first term is compounded of 25 men and 8 hours, representing time; the third term is compounded of 600 feet, 4.5 feet, and 3.5 feet, representing work done; the second term is compounded of 36 men and 10 hours; and the fourth term is compounded of x feet, 4 feet, and 4 feet. Then, 25 X 8 : 36 X 10 = 600 X 4.5 X 3.5 : a; X 4 X 4, from which _ 36X10X600X4.5X3.5 _ "^ ~ 25X8X4X4 ~ ^"^"^^ *^^*- ^''^• EXAMPLES Find the value of x in the following proportions: (1) 6 : 8 = a; : 18. Ans. 13.5. ^2) 20 : x = 8 : 26. Ans. 65. (3) x : 7 = 81 : 91. Ans. 6^3. (4) 11 : 16 = 517 : x. Ans. 752. (5) 5.5 X 18 : 7.2 X 15 = 4.4 X 25 : 10.5 X^Xx. Ans. 7^. '(6) 120 X 64 : 540 X 328 = 110 : a:. Ans. 2536%. (7) If a certain supply of provisions will last 525 men 129 days, how long will it last 603 men?. Ans. 112 + days. §1 RATIO AND PROPORTION 85 (8) Referring to Art. 153, suppose the volume of an air compressor is 1.95 cubic feet. If the cyhnder be filled with air at a pressure of 14.7 pounds per square inch, and the air is compressed until the volume is .31 cubic feet, what is the pressure, the temperature remaining the same? Ans. 92.47— pounds per square inch. (9) Referring to Art. 153/. what is the circumference of a circle whose diameter is 12% inches? Ans. 38.8773 inches. ARITHMETIC (PART 2) EXAMINATION QUESTIONS (1) What is the greatest common divisor of 15,862 and 1309? Ans. 77. (2) Find the G.C.D. of 45, 135, 270 and 405. Ans. 45. (3) Find the L.C.M. of 45, 135, 270 and 405. Ans. 810. (4) Which fraction is the larger, ^\ or ^V^? Ans. ^'j;. (5) What is the value of f + tV + ii + H - W Ans. liW (6) Reduce the following fractions to their lowest terms: (a) 1^1; (&) T2%%', (c) H^; (d) what is the difference between the frac- tions (a) and (6) ? (e) what is the sum of the fractions (6) and (c) ? Ans. (a) f; (6) ff; (c) ^'r, (d) i^r, (e) ffH. (7) What is the value of Ji - ^ + H - li + W Jxrib. 5 4 0- (8) What is the product of the sum and the difference of 4| and3i? Ans. 13AVV. (9) A carload of coal weighing 47,960 pounds was delivered to a paper mill; Aths was used the first week, yoths the second week, and T2ths the third week; how many pounds then remained? Ans. 8502 pound. (10) What must be paid for 8i| ounces of dye if the cost is $1.27 per ounce? Ans. $11.19. (11) Divide (a) 0.67 by 0.0007042; (h) 18| by 653.109. , r (a) 951.43+. ^'''- [{h) .028135-. (12) The distance across the diagonally opposite corners of a hexagonal (six-sided) nut is called the outside diameter, and for an unfinished nut is found as follows: to 1| times the diameter of the bolt add | inch, and multiply the sum by 1.1547. What is the outside diameter of a nut for a 1| inch bolt? Give result to nearest 64th of an inch. Ans. 2^\ = 2^2 inches. 87 88 ARITHMETIC §1 (13) Simplify the complex fraction, — «nQ/i >!/ 29 L 52^^ - DU54 X TTT ^ T2¥ Ans. 6/2. (14) The area of a circle is .7854 timet the square of the di- ameter; what is the area of a circle whose diameter is 5t\? Ans. 21.135+. (15) What is the valu e of 6{9 - [14 - 4(12 - 7) (11 + 5) -^ 48] + 7(45 - 51 - 13)} ? . Ans, 304. (16) Find the value of x in the proportions: (a) 15 : 21 = 48 : a;; (6) 18 : 8 = a; : 54; (c) 12 : x = 128 : 96. ((a) X = 67.2. Ans. \(b) X =^ 121.5. [(c) X = 72. (17) If 5 men can do a certain piece of work in 8 days, how long will it take 10 men to do 3 times as much work? Ans. 12 days. (18) If the pressure of steam in an engine cylinder varies in- versely as the volume, and the pressure is 110 pounds per square inch when the volume is 0.546 cubic feet, what is the pressure when the volume is 2.017 cubic feet? Ans. 29.78— pounds per square inch. AEITHMETIC (PART 3) SQUARE ROOT 155. The power of a number has been previously defined; it is the continued product of as many equal factors as are indicated by the exponent; a very important problem is : given the power, to find what equal factor was used to produce the power. The number so found is called the root of the given number. The roots are named in the same manner as the powers. For example, if two equal numbers are used to form the power, one of them is called the square root of the given number; if three equal factors are used to form the power, one of them is called the cube root; if four equal factors are used to form the power, one of them is called the fourth root; etc. Thus, the square root of 9 is 3, because 3 X 3 = 9; the cube root of 125 is 5, because 5 X 5 X 5 = 125; etc. 156. The root of a number is indicated by writing before it what is called the radical sign \/, and to specify what root, a small figure is written in the opening of the sign; thus, v^, \/, ^, etc. indicate the square root, cube root, and fifth root, respec- tively. The square root is indicated so frequently that it is customary to omit the small figure, which is called the index of the root, using only the sign; hence, \/9, VSl, Vl44, etc. indicate respectively the square root of 9, the square root of 81, the square root of 144, etc. The index cannot be omitted when indicating any other root than the square root. It is also customary to use the vinculum in connection with the radical sign. Thus, ^^1728^ indicates t he cube root of 1728; v^7776 indicates the fifth root of 7776; V24 X 54 indicates the square root of the product of 24 and 54. If it were desired to indicate the product of 54 and the square root of 24, it may be done in two ways, either as 54\/24 or as V24 X 54. It will be observed that in the first form, the sign of multiphcation is omitted between 54 and the radical sign; this is customary, and when so written, multiplication is always understood. 89 90 ARITHMETIC §1 The root of a number is also frequently indicated by using a fractional exponent, the denominator of the fraction indicating the root. For example, 196' has the same meaning as \/l96, 243^ = v^243, 343^ = -^^^343, etc. If the numerator of the frac- tional exponent is some number other than 1, it indicates that the number is to be raised to the power indicated by the numerator and the root taken that is indicated by the denominator. Thus, 243^ indicates that 243 is to be raised to the fourth power and the fifth root is then to be found, or it means that the fifth root of 243 is to be found and the result raised to the fourth power; the final result will be the same. For instance, ■v^243 = 3, and 3* = 81 ; also, 343^ = 3,486,784,401 and -^3,486,784,401 = 81, because 81^ = 3,486,784,401. In other words, 243^ = ( v^243)'or v^243^ and the value of both these expressions is 81. Both expressions may also be written (243*) ^ or (243^)1 157. A number is said to be a perfect square, a perfect cube, a perfect fifth power, etc., when its square root, cube root, fifth root, etc., can be expressed exactly; for instance, 196 is a perfect square, because •\/l96 = 14; 343 is a perfect cube, because -v^343 = 7; 3,486,784,401 is a perfect fifth power, because ^3,486,784,401 = 81. Only comparatively few numbers are perfect powers ; thus, between 1 and 100, the only perfect powers are 1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, and 100, a total of 13 only. Numbers that are not perfect powers are called imperfect powers; thus 7, 12, 26, 47, etc. are imperfect powers. It is to be noted that a number may be perfect for some particular power, but not for others; in fact, this is usually the case. For example, 243 is a perfect fifth power, but it is not perfect for any other power; 16 = 2^ and 4^, and is a perfect fourth power and a perfect square; 64 = 2^ = 4^ = 8^, and is a perfect sixth power, a perfect cube, and a perfect square. Numbers that are perfect for more than one power are rare. The root of an imperfect power is never exact; the root of such a number may be calculated to any number of figures, in the same way that the quotient may be carried to any number of decimal places in division when the divisor contains some factor other than 2 or 5. 158. The square root of a number is very frequently desired; occasionally, some other root is required, but the process of find- ing any root other than the square is so laborious, and such roots §1 SQUARE ROOT 91 are required so seldom, that only the method for finding the square root of any number is described here. When other roots are desired, they are usually found by using tables of logarithms. A method for finding cube and fifth roots is described in Elemen- tary Applied Mathematics. 159. There are many methods of finding the square root of a number, the best general method being that known as Horner's method, which is the one that will be explained here. The first step, no matter what method is used, is to point off the number whose root is to be found into periods of two figures each, beginning at the decimal point and going to the left, and to the right also, in the case of mixed numbers and pure decimals. For this purpose, it is best to use a tick — a mark something like an apostrophe. For example, the number 11,778,624 is pointed off as 1177'86'24; the numbers .7854 and .0003491 would be pointed off as .78'54 and .00'03'49'10; the number 755.29216 would be pointed off as 7'55.29'21'60. When the right hand period of the decimal is not complete, that is, does not contain two figures, add a cipher. The left-hand period of the integral part may contain either one or two figures, according to whether the number of figures in the integral part is even or odd. Suppose the square root of 11,778,624 is required. First point off the number into periods of two figures each. The work is done in two columns, as follows: The first period, 11, of the given number is not a perfect square, and the largest perfect square less than 11 is 9, the square root of which is 3. Write 3 as the first figure of the root, in the same manner as when finding the quotient in division, and also write it at the head of the first column. Now multiply the 3 in the first column by the 3 in the root, and write the product under the first period ; subtract, obtaining a remainder of 2. Add the first figure of the root to the number in the first column, and the sum is 6, to which annex a cipher, making it 60; also annex the second period of the given number to the remainder, 2, in the second 3 3 60 11'77'86'24(3432 9 277 4 256 64 2186 4 680 2049 13724 3 13724 683 3 6860 2 6862 92 ARITHMETIC §1 column, obtaining as a result 277. Divide the last number in the second column, 277, by the last number in the first column, 60, and the quotient is 4, which is probably the next figure of the root. Write 4 for the second figure of the root, add it to 60, the last number in the first column, obtaining 64 ; multiply 64 by 4, the second figure of the root, and subtract the product, 256, from the last number in the second column, 277, obtaining 21 for the remainder. Then add the second figure of the root, 4, to 64, the last number in the first column, and the sum is 68, to which annex a cipher, making it 680. Also annex the third period, 86, to the remainder 21, in the second column, making it 2186. Divide 2186 by 680, and the quotient, 3, is probably the next, or third, figure of the root. Write 3 for the third figure of the root, add it to 680 in the first column, obtaining 683, and multiply 683 by the third figure of the root; the product, 2049, is subtracted from 2186, and the remainder is 137. Add the third figure of the root, 3, to 683, obtaining 686, to which annex a cipher, making it 6860. Annex the fourth (and last) period of the given power to the last remainder, making it 13724; divide this by 6860, and the quotient 2, is the fourth figure of the root. Add the fourth figure of the root to 6860, obtaining 6862 ; multiplying this by the fourth figure of the root, the product is 13724, which subtracted from the last number in the second column gives a remainder of 0. Therefore, \/ll,778.624 = 3432. Ans. / Note that the first column is formed entirely by adding, and that two additions are made for each figure of the root. The above explanation of the method will be made much clearer to the reader if he will take pencil and paper and set down the vari- ous numbers exactly as they occur in the explanation. After a few examples have been solved, the method will be clear, and it will be found easy to remember; in fact, the work will be found to be but little harder than ordinary division. 160. Referring to the last example, the numbers in the first column (the ones to which the ciphers have been annexed) that are used as divisors to determine figures of the root are called trial divisors; the numbers that they divide are called trial dividends, and the numbers that are obtained by multiplying the trial divisors by the root figures may be called, for the want of a better term, the partial squares or partial powers. When the given number is a perfect square, the sum of the partial powers, added as they stand, must equal the given power; thus, §1 SQUARE ROOT 93 the sum of 9, 256, 2049, and 13724, the partial powers, added 9 as they stand, is 11,778,624, the given number, 256 In this example, the first trial divisor is 60, 2049 and the first trial dividend is 277; the second ^^^^ trial divisor is 680, and the second trial divi- 11778624 ^gjjjj ig 2186; etc. The numbers obtained by adding the root figures to the trial divisors are called the complete divisors; 64, 683, and 6862 are the complete divisors. When finding the second figure of the root, it may happen that the product of the complete divisor and the second figure of the root is greater than the partial power; in such case, try a number one unit smaller for the second figure of the root. If the product of the new complete divisor and the new second figure of the root is still greater than the trial divisor, reduce the second figure of the root one unit more. It will never be necessary to make more than two trials, and seldom more than one. Example. — Find the square root of 12,054,784. 3 12'05'47'84 (3472. Ans. Solution.— The work is almost exactly 3 9 the same as in the preceding example. It 60 305 is to be noted, however, that the quotient 4 256 obtained by dividing the first trial dividend 64 ^[947 by the first trial divisor is 5; but when 5 is ^ 4809 added to 60 for the complete divisor, and fisn ~r^S4 *^^^ ^^ multiplied by 5, the product is 7 iQ8«j. greater than the trial dividend; hence, 4 is tried for the second figure of the root. 687 7 6942 161. If the given number whose root is to be found contains a decimal there will be as many integral places in the root as there are periods in the integral part of the given number. Thus, the square root of 1205.4784 contains two integral places; the square root of 43206.231 contains three integral places, because when pointed off, it becomes 4'32'06.23'10, and there are three periods in the integral part; etc. If the given number is a pure decimal, the root will also be a pure decimal; and if there are ciphers between the decimal point and the first digit, there will be as many ciphers between the decimal point and the first digit of the root as there are periods containing no digits. Thus, the square root of .000081 is .009, because when pointed off, the given number becomes .OO'OO'Sl, 94 ARITHMETIC §1 and there are two periods containing no digits; the square root of .000196 is .014, because when pointed off, the given number becomes .00'01'96, and there is one period containing no digit; etc. To prove this statement, square the roots; thus, .009^ = .000081, and .014^ = .000196. Example 1. — What is the square root of 844.4836? Solution. — On dividing the first trial dividend, 444, by the first trial divisor, 40, the quotient is 11; but, the second figure of the root cannot be greater than 9; hence, try 9 for the second 8'44.48'36 (29.06. Ans. 2 40 4 444 9 49 441 34836 9 34836 5800 6 5806 Example 2. — Find 1 3'56. (18. 8( 1 20 1 256 8 224 28 3200 8 2944 360 25600 8 22596 368 300400 8 264089 3760 3631100 6 264089 3766 22596 6 2944 37720 224 7 1 37727 355963689 7 36311 377340 356000000 figure. The second trial divisor is 580, and it is larger than the second trial dividend, which is 348; consequently, annex another cipher to the trial divisor, making it 5800, and annex another period to the trial dividend, making it 34836, which contains the trial divisor 6 times. As there are two periods in the integral part of the given number, there are two figures in the integral part of the root. Find the square root of 356 to three decimal places. Solution. — The first figure of the root is 1, since the first period is less than 4. When dividing the first trial dividend by the first trial divisor, the quotient is 256 -;- 20 = 12 + ; but the second figure of the root cannot exceed 9. Trying 9, the complete divisor is 29, and 29 X 9 = 261, which is greater than the partial power, 256; hence, try 8 for the second figure of the root. The remainder is 256, showing that the given number is an imperfect power. The work is now continued by annexing cipher periods as shown. The sixth figure of the root, the figure in the fourth decimal place, is 9; hence, the root correct to three decimal places is 18.868—. Adding the partial powers, first writing them under the last, 264089, the sum is 355.963689, which is the square of 18.867; if the last remainder, 36311, disregarding the decimal point, be added to this sum, the total is 356, the given number, showing that the work is correct. 162. If the given number is a common fraction, the root may- be found by extracting the square root of the numerator and denominator separately; thus, V^ = I- But, unless both numerator and denominator are perfect squares, it is better to reduce the fraction to a decimal, and find as many figures of the §1 SQUARE ROOT 95 root as are desired; usually, 4 or 5 figures of the root of an im- perfect power (not counting ciphers between the decimal point and the first digit of pure decimals) are suflficient for practical purposes. Example. — Find the square root of | correct to four decimal places. Solution. — Reducing | to a decimal, it 9 .87'50 (.93541 Ans. becomes .875. According to Art 161, the q oj root is a pure decimal and the nrst ngure — — . of it is a digit. After finding the first two 180 650 figures of the root, there are no more periods ^-~. .rir to the given number, and the work is con- 183 10100 tinned by annexing cipher periods to the ^ 9325 different trial divisors, each period con- taining two ciphers. It is evident that the fifth figure of the root is 1; hence, the root correct to 4 decimal places is .9354. To prove that no mistake has been made in the work, add the partial powers, obtaining .87497316 = .9354^; adding the last re- mainder the sum is .875; the given number, which shows that no mistake has been made. Furthermore, when expressed to four figures, .87497316 = .8750, which is .87500000 also correct. It is well to check the work in this manner, as it prevents mistakes. 163. If the reader has followed each step of the foregoing examples with pencil and paper, he should be able to extract the square root of any number. Rule I. — Beginning at the decimal point, point off the given number into periods of two figures each, including the decimal part, if any. II. Arrange the work in two columns, the second column contain- ing the given number. The first figure of the root is the square root of largest square that does not exceed the first period. III. Write the first figure of the root as the first number in the first column, multiply it by the first root figure, and subtract the product from the first period, and annex to the remainder the second period; this is the first trial dividend. Add the first root figure ^ to the number in the first column, and annex a cipher, thus obtaining the first trial divisor. IV. Divide the trial dividend by the trial divisor, and the first figure of the quotient (if less than 10) will probably be the second figure of the root; add this figure to the trial divisor, obtaining the 1860 5 77500 74816 1865 5 2684( .81 18700 4 549 9325 18704 74816 4 187080 .87497316 2684 96 ARITHMETIC §1 complete divisor, which multiply by the second root figure, and sub- tract the product from the trial dividend, to which annex the next period to form the next trial dividend. But if this product is greater than the trial dividend, try a figure one unit less than the quotient just obtained, adding it to the trial divisor, and multiplying as before. If the product is yet greater than the trial divisor, try a root figure one unit less than the last. Having found a satisfactory complete divisor, add to it the root figure last found, and annex a cipher to form a new trial divisor. Divide the second trial dividend by the second trial divisor, and the integral part of the quotient will be the next figure of the root. V. Proceed in this manner until all the periods of the given number have been used. If the number is an imperfect square, limit the root to five figures (counting the first digit as one), unless more figures are especially desired, annexing cipher periods of two figures each, if necessary. VI. // at any time, the trial divisor is larger than the trial dividend, annex a cipher to the trial divisor and annex another period to the trial dividend, placing a cipher in the root. Proceed in this manner until the root has been found or as many figures have been found as are desired. VII. // the quotient obtained by dividing the first trial dividend by the first trial divisor is greater than 9, try 9 for the second figure of the root. The sum of the partial powers added as they stand, must a ways be equal to the square of the root as found, if no mistake has been made in the work, and this sum plus the last remainder must equal the given number. It is a good plan to check the work in every instance by ascertaining if this is the case. EXAMPLES (1) Find the square root of 293,005.69. Ans. 541.3 (2) Find \/lO to four decimal places. Ans. 3.1623 — (3) Find V3.1416 to four decimal places. Ans. 1.7725- (4) Find V36|| to four decimal places. Ans. 6.0557 + (5) Find V.02475 to five decimal places. Ans. .15732 + (6) Find Vf to five decimal places. Ans. .81650 — (7) Find \/.0000000217 to five decimal places. Ans. .00014731- §1 SQUARE ROOT 97 APPLICATIONS OF SQUARE ROOT 164. In problems relating to mensuration, mechanics, strength of materials, and engineering generally, it is required very fre- quently to find the square root of numbers. A few applica- tions will be given here by means of examples. Example 1. — If the area of a circle is known or given, the diameter of the circle can be found by multiplying the square root of the area by 1.1284. What is the diameter of a circle . having an area of 56.28 square inches? Give result to four decimal places. Solution. — Since the result is required to four decimal places, and there are four decimal places in the number 1.1284 and one integral place, making five figures in all, find the square root of 56.28 to 5 + 1 =6 figures, obtain- ing 7.50200 -. Then, the diameter is 1.1284X7.502=8.4652568, or 8.4653 — inches to four decimal places. Ans. Example 2. — When a heavy body falls freely, it strikes the ground with a velocity in feet per second equal to 8.02 times the square root of the height of the fall in feet. If a certain body, say a cannon ball, falls from a height of 750 feet, what will be its velocity when it strikes the ground ? Solution.— The square root of 750 is 27.386+, and 8.02 X 27.386 = 219.64— feet per second, or about 2.5 miles per minute. Ans. Example 3. — The intensity of light varies inversely as the square of the distance of the hght from the object illuminated. If the intensity of illumination of a certain object 5 feet from the source of light is 32 candle- power, at what distance will the intensity of illumination be 18 candlepower? Solution. — Expressed as a direct proportion, 32 : 18 = 5^ : x^. It is necessary to square the distances 5 and x, because the intensity varies in- versely as the square of the distance. Inverting the terms in the second ratio, 32 : 18 = a;2 : 5^, from which x^ = t^ = 44.4444+. Now if lo ^2 = 44.4444, X must equal the square root of 44:A4:4:4:, that is x = a/44.4444 = 6.6667— feet. Ans. This result might also have been obtained as , ., 32 X 25 400 , /400 20 follows: — Ys — "^^^^'\"9~="3" = ^i = 6.6667- feet. The above result means this: if an object is illuminated with an intensity of 32 candlepower when 5 feet from the light, it will be illuminated with an in- tensity of only 18 candlepower when 6| feet from the light. Example 4. — The strength of a simple beam (one that is merely sup- ported at its ends) varies directly as its breadth, directly as the square of its depth, and inversely as its length. If a simple beam made of hemlock is 20 feet long, has a breadth of 3 inches, a depth of 8 inches, and will safely carry a load of 960 pounds, uniformly distributed over the beam, what must be the depth of a similar beam having a length of 16 feet and a breadth of 2 inches to carry safely a uniform load of 1250 pounds? Solution. — This is evidently a problem in compound proportion. Stated as a direct proportion, 3 X 8^ X 20 : 2 X a;^ X 16 = 960 : 1250. It is neces- sary to square 8 and x, because the strength varies as the square of the depth 7 98 ARITHMETIC §1 But the strength varies inversely as the length; hence, transposing 20 and 16, the lengths, 3 X 8^ X 16 : 2 X a:^ X 20 = 960 : 1250, from which , 3X64X16X1250 ,._ -p + vp 2 mn ^TTv^ mi, a;2 = 2 V 20 V QBO — ^ ^ "^ ' ^ ^ V 100 = lOj hence, the depth of the beam must be 10 inches. Ans. The problems just given will suffice to show the importance of knowing some method of extracting the square root of numbers. PERCENTAGE 165. It is frequently desirable to compare numbers or quantities not in relation to each other, but in relation to some fixed number, called the base. For example, the sum of 7, 6, and 12 is 25. Now to compare 7, 6, and 12 with 25, form the ratios 7 : 25, 6 : 25, and 12 : 25. The values of these ratios are found by dividing the first terms by the second, thus obtaining ■2^ = .28, -is = .24, and if = .48. Note that the sum of the fractions is -g^ + A + if = 1, and the sum of the decimal equivalents of the fractions is .28 + .24 + .48 = 1.00 = 1, also, as it must. In the case of the ratios, the base is 25; but in the case of the fractions and decimals, the base is 1, a much more convenient number, and one that does not change, as will continually be the case with ratios. In business transactions, and in many other cases that arise in practice, it is convenient to have T^T = .01 as the base, thus making the values of the ratios 100 times as large as those obtained above. With .01 as the base, the values of the above ratios become .28 -f- ^i-Q = .28 X 100 = 28, .24 X 100 = 24, .48 X 100 = 48, and 28 + 24 + 48 = 100. In the last case, the three numbers added are called 28 per cent, 24 per cent, and 48 per cent, respectively, of 25, which is 100 per cent of 25. The term per cent is an abbreviation of the Latin words per centum, which mean by the hundred, and percentage means a cer- tain number of hundredths of some number. Thus, 28 per cent of 25 is 25 X TW = 25 X .28 = 7, and 7 is the percentage obtained by multiplying 25 by 28 hundredths. Also, 7 per cent of 438 is 438 X t^ = 438 X .07 = 30.66, and 30.66 is the percentage obtained by multiplying 438 by 7 hundredths. 166. The sign or symbol that is used to denote per cent is % ; thus, 7% is read 7 per cent, 113^^% is read 11% per cent, etc. The number that is placed before the symbol and shows how §1 PERCENTAGE 99 many hundredths are to be taken or considered is called the rate per cent; thus, in the last sentence, 7 and ll}i are rates per cent. When the rate per cent is expressed decimally or frac- tionally, thus denoting its actual value, it is called the rate. For instance, if the rate per cent is 5%, the rate is .05 or ^^; if the no 7 KQ7 rate per cent is 58.7%, the rate is .587 or ^ = ^ , etc. The rate, therefore, is always equal to the rate per cent divided by 100; and the rate per cent is 100 times the rate. 167. The rate can frequently be expressed as a simple fraction by reducing the common-fraction form of the rate to its lowest terms. For instance, 6i% = ^ = ^ = -,V, or 6|% = ^ 625 10000 — tV- In the table below, some of the rates per cent commonly used, with their decimal and common fraction equiva- lents, are given. Rate Per Cent Rate Rate Per Cent Rate 11 .015 =^t^ 25 .25 = i 2 •02 =,V 33f .331= i 3i ■03^ =,V 371 .375= f 4 .04 =^V 40 •4=1 41 •04i =A 50 .5 = i 5 .05 =,V 62i .625= 1 6i .0626 =^V 661 .661= 1 8 •08 =A 75 .75 = 1 10 •1 =^^ 83i •831= 1 12i .125 =1 87i .875= 1 161 .16! =i 100 1 =1 20 •2 =i 250 2.5 =2* 1 •00125 = 3.1^ 475 4.75 =4| 168. The majority, if not all, of the fractions in the above table may be used to advantage in computing the percentage instead of their decimal equivalents. For example, to find 12>^% of 35.164, simply divide 35.164 by 8, obtaining 4.3955; this is cor- rect, since 121.^% = |, and 35.164 X i = 35.164 ^ 8. Simi- larly, for the same reasons, 83>^% of 439.2 = 439.2 = ^ 439.2 X 5 = -Q = 366. It is evidently easier to multiply by the 100 ARITHMETIC §1 fractions than it is to multiply by their decimal equivalents, and there is less liability of making mistakes. 169. The base may now be defined as the number that is multiplied by the rate to obtain the percentage. In the first case of Art. 168, 35.164 is the base and 4.3955 is the percentage; in the second case, 439.2 is the base and 366 is the percentage. Consequently, Rule. — To find the percentage, multiply the base by the rate. Example. — Out of a lot of 1500 reams of paper, 37% were sold the first month and 43% were sold the second month; how many were sold in the two months? Solution. — This problem may be solved in two ways : 1st method. The number of reams sold the first month is 1500 X .37 = 555; number sold the second month is 1500 X .43 = 645; number sold in the two months is then 555+645 = 1200. Ans. 2d method. If 37% were sold the first month and 43% were sold the second month, 37% + 43% = 80% were sold in the two months; and 1500 X .80 = 1200 reams were sold in the two months. Ans. 170. If the product of the base and the rate equals the percent- age, the rate must equal the percentage divided by the base, since the product of two factors divided by one of the factors must give the other factor for the quotient. Therefore, Rule. — To find the rate, divide the percentage by the base. The rate per cent is 100 times the rate. Example. — Out of a lot of 1500 reams of paper, 555 were sold; what per cent of the paper was sold ? Solution. — Here 555 is the percentage and 1500 is the base, since it is the number which, multiplied by some rate, gives a product of 555. Hence, 555 -^ 1500 = .37 = the rate, and .37 X 100 = 37, the rate per cent. Therefore, the per cent of paper sold was 37%. Ans. 171. If the rate and percentage are known and it is desired to find the base, divide the percentage by the rate. This is evi- dently correct, since, if the product of the base and rate equals the percentage, the percentage divided by the rate must equal the base. Consequently, Rule. — To find the base, divide the percentage by the rate. Example. — In a certain month, 43% of a certain lot of pulp was sold; if the number of bales sold was 645, how many bales were in the lot? Solution. — Here the rate is .43 and the percentage is 645; hence, the base is 645 -f- .43 = 1500, the number of bales in the lot. Ans. 172. It is to be noted that the rate per cent and the percentage really represent the same thing. The rate per cent is the number of parts in 100 parts of the base, while the percentage is the actual §1 PERCENTAGE 101 number of parts of the base. In other words, the rate per cent is found on the assumption that the base is always 100, while the percentage is computed on the actual value of the base. 173. There are two other terms used in percentage — the amount and the difference. The amount is equal to the sum of the base and percentage, the difference is equal to the remainder obtained by subtracting the percentage from the base. To illustrate the meaning of these terms, suppose that the price of alum is $42.00 per ton, and that the price was increased 123-^%; what is the new price? $42 -T- 8 = $5.25 = 12>^% of $42.00; then the new price is evidently $42.00 + $5.25 = $47.25 = the amount, since it is equal to the sum of the base ($42.00) and the percentage ($5.25). Again, suppose, as before, that the price is $42.00, but that the price is reduced 123^-2%, what is the new price? Since 123-^% of $42.00 is $5.25, the new price is $42.00 - $5.25 = $36.75 = the difference, since it is equal to the base minus the percentage. 174. The amount may be found in an easier way. Since the base alwaj^s represents 1 or 100%, the amount may be represented by 1 + rate or by 100% + rate per cent; hence, if the base and rate are given, the amount may be found by multiplying the base by 1 + rate. Thus, referring to Art. 173, the new price after the increase is $42 X (1 + .125) = $42 X 1.125 = $47.25, since 123^^% = .125. Therefore, Rule. — The amount equals the base multiplied by 1 plus the rate. Example. — If a man's wages is $3.20 per day, and he receives an increase of 15%, how much does he then receive per day? Solution. — The base is $3.20, the rate is .15, and the new rate of wages is the amount, which is $3.20 X 1.15 = $3.68. Ans. 175. If the amount and the rate are known, and it is desired to find the base on which the rate was computed, divide the amount by 1 plus the rate. This is evidently correct, since the amount is equal to the base multiplied by 1 plus the rate. Hence, Rule. — To find the base, divide the amount by 1 plus the rate. Example. — If a man receives $3.68 cents per day in wages, and this is 15% more than he formerly received, how much did he formerly receive? Solution. — Here $3.68 is some number plus 15% of that number; that is, $3.68 is the amount and 15% is the rate per cent. Therefore, he formerly received $3.68 -v- 1.15 = $3.20. Ans. 176. If the base and amount are given and it is desired to know the rate, divide the amount by the base and subtract 1 from the quotient. This is evidently correct, since dividing the amount 102 ARITHMETIC §1 by the base gives a quotient that equals 1 plus the rate, and sub- tracting 1 from this leaves the rate. Rule.— To find the rate when the base and amount are known, divide the amount by the base and subtract 1 from the quotient. Example. — If a man's wages are $3.68 per day and he formerly received $3.20 per day, what rate per cent, increase did he receive? Solution. — Here $3.)68 is the amount, since it equals $3.20 plus a certain percentage of $3.20, the base. Consequently, 3.68 -f- 3.20 = 1.15; 1.15 — 1 = .15, the rate; and .15 X 100 = 15%. Ans. The solution may also be obtained as follows : $3.68 — $3.20 = $0.48, the actual increase per day that he received, and which is the percentage. Hence, the rate is .48 -i- 3.20 = ,15, and the rate per cent is .15 X 100 = 15%. Therefore, also. Rule. — To find the rate when the base and the amount are known, subtract the base from the amount and divide the remainder by the base. The second rule is rather easier to remember and apply than the first rule. 177. The rules just given concerning the amount may also be used when the difference is given, by subtracting the rate instead of adding it. Thus, referring to Art. 173, the new price after the old price had been reduced 12>^% is $42 X (1 - .125) = $42 X .875 = $36.75. Therefore, Rule. — To find the difference, multiply the base by 1 minus the rate. In connection with problems relating to money and prices, the amount deducted from a fixed price or amount (the base) is called the discount. Insofar as percentage is concerned, the words percentage and discount mean the same thing, when finding the difference. Example. — If a certain article is priced at $4.75 and it is sold at a discount of 8%, how much was received for it? Solution. — The base is $4.75, the rate is .08, and the difference is re- quired. Applying the rule, $4.75 X (1 - .08) = $4.75 X .92 = $4.37. Ans. The solution may also be obtained as follows: $4.75 X .08 = $0.38, the discount; $4.75 — $0.38 = $4.37. Ans. Either method may be used, whichever appears to be easier. 178. If the difference and rate are known, and it is desired to find the base, divide the difference by 1 minus the rate. For, since the product of the base and 1 minus the rate equals the difference, the base must equal the difference divided by 1 minus the rate. Consequently, Rule. — To find the base when the rate and difference are known, divide the difference by 1 minus the rate. §1 PERCENTAGE 103 Example. — $4.37 was paid for a certain article that was bought at a dis- count of 8%; what was the original price of the article? Solution. — Here $4.37 is the difference and .08 is the rate; the base = the original price = $4.37 -^ (1 - .08) = $4.37 -i- .92 = $4.75. Ans. 179. If the difference and base are known, and it is desired to find the rate, subtract the difference from the base and the remainder will be the discount; divide the discount by the base, and the quotient will be the rate. Example. — If an article is bought for $4.37 and the original price was $4.75, what was the per cent of discount received? Solution.— The actual discount received is $4.75 - $4.37 = $0.38, and $4.75 is the base; hence, 0.38 -^ 4.75 = .08 = 8% of the original price = per cent of discount. Ans. 180. Chain Discount. — Certain manufacturers that deal in goods that are subject to rapid changes in prices issue cataloges in which the prices are quoted at a much higher figure than they think will ever be asked. They then issue what are called dis- count sheets, which are quickly printed, and are subject to change, in accordance with the state of the market. When more than one discount is quoted on any one article, it is called a chain discount. In computing a chain discount, that quoted first is deducted from the list price, the second discount is deducted from the remainder, the third discount from the last remainder, etc. It is never allowable to add the rates of discount and then deduct. As an example, suppose that the price (catalog) of a certain article is $26.60, and that discounts of 60,10 and 5% are offered; what is the real price of the article? $26.60 X (1 - .60) = $10.64, the price after 60% has been deducted. $10.64 X (1 - .10) = $9,576, or $9.58, the price after deducting 60 and 10%. $9.58 X (1 - .05) = $9,101 or $9.10, the price after all the discounts have been taken out. This same result may be obtained by multiplying $26.60 by the continued product of 1 minus the different discounts, that is by (1 - .60) X (1 - .10) X (1 - .05) = .40 X .90 X .95 = .342; thus, $26.60 X .342 = $9.0972, or $9.10, as before. The real discount in this case is, therefore, 100 — 34.2 = 65.8%, which may be called the equivalent discount, and 1 — .342 = .658 may be called the equivalent discount rate. Rule. — To find the equivalent discount rate for any chain dis- count, subtract all the discount rates from 1 and find their product; then subtract this product from 1. The equivalent discount equals the equivalent discount rate multiplied by 100. The net price or 104 ARITHMETIC §1 net cost may be obtained by subtracting all the discount rates from 1, finding their product, which multiply by the given price or cost. Example. — A bill of goods amounting to $102.04 is subject to a chain discount of 87 J^, 10, 5, and 2}^%; what is the equivalent discount per cent and how much must be paid to settle the bill? Solution. — The equivalent discount rate is 1 — (1. — .875) X (1 — 10) X (1 - .05). X (1 - .025) = 1 - .125 X .9 X .95 X .975 = 1 - .104203125 = .895796875, or 89.5796875%. Ans. The amount to be paid to settle the bill is $102.04 X .104203125 = $10.63+. Ans. 181. Gain or Loss Per Cent. — If an article is bought for a certain amount and is sold for another amount, there is a gain or loss equal to the difference between the two amounts. To find the gain or loss per cent, divide the gain or loss by the cost. Thus, if a manufacturer finds that it costs him $2.43 to turn out a certain article, and he sells it for $3.15, he gains $0.72, and the gain per cent is .72 ^ 2.43 = .2963; .2963 X 100 = 29.63%. But if he had sold it for $2.25, he would have lost $2.43 - $2.25 = $0.18, and the loss per cent. = .18 -^ 2.43 = .0741;. 0741 X 100 = 7.41%. Again, suppose that the transmission system of a certain power plant is found to absorb 4.35 horsepower, but after making certain changes, it absorbs only 3.08 horsepower; what is the saving in per cent of horsepower wasted in transmission? The saving in horsepower is 4.35 — 3.08 = 1.27; the horsepower wasted before the change is the base in computing the per cent; hence, 1.27 -^ 4.35 = .292- ; and .292 X 100 = 29.2%. In other words, 29.2%, of the power formerly wasted is saved by the change. In calculations of this kind, it is sometimes a little difficult to determine which of the two given numbers is the base; this can always be determined correctly by remembering that a gain or loss involves a change of some kind; in the case of buying and selling, the selling is a change (of ownership) ; in the case of the power plant, 4.35 horsepower was changed to 3.08 horsepower. The number used to divide the gain or loss is the number that is changed; in the first illustration above, the number changed was $2.43, and in the second illustration it was 4.35. Rule. — To find the gain or loss per cent, divide the gain or loss by the number that is changed and multiply the quotient by 100. Example 1. — Suppose the indicated horsepower of a certain non-condens- ing steam engine is 168, and that it is 204 after a condenser has been attached. What is the gain per cent in horsepower? Solution. — The horsepower before the change was 168, and after the §1 PERCENTAGE 105 change it was 204; the gain is 204 - 168 = 36; 36 -^ 168 = .2143, and .2143 X 100 = 21.43%, the gain per cent. Ans. Example 2. — A certain business is valued at $234,517.00; three years later, it is valued at $187,250; what was the depreciation in per cent? Solution.— The actual depreciation was $234,517 - $187,250 = $47,267; the value before the change was $234,517; 47,267 -r- 234,517 = .2016-; hence, the depreciation in per cent was .2016 X 100 = 20.16%. Ans. EXAMPLES (1) What is 4%% of $683.32 to the nearest cent? Ans. $32.46. (2) If 475 is 623^% of some number, what is the number? Ans. 760. (3) What is %% of $4521? Ans. $16.95. (4) What per cent of 840 is 119? Ans. 14K %• (5) If an automobile was bought for $1350 and after being used for a year was sold for $925, what was the loss per cent? Ans. 31.48%. (6) Bought some dye stuff for $21.85, receiving a discount of 5% from the selling price; what was the selling price? Ans. $23.00. (7) A machinist turned 47 bolts in one day, which was almost 7% more than he tiirned the day before; how many did he turn the day before? Ans. 44. (9) Bought a line of supplies, the bill amounting to $428.73 and subject to a discount of 333^3, 10, and 3% if settled within ten days; how much is required to settle in ten days? Ans. $249.52. (10) A test showed that the coal consumption in the boiler used to furnish steam for a non-condensing engine was 660 pounds per hour; after attach- ing a condenser, only 541 pounds of coal per hour were required. What was the saving per cent? Ans. 18 + %. (11) The average daily production of a paper mill in 1917 was 94 tons; in 1918, it was 105 tons; what was the per cent of increase? Ans. 11.7%. (12) A mill was using 1.4 cords of wood per ton of paper; as the result of certain changes, a saving of 4% of wood was effected. What amount of wood was then used per ton? Ans. 1.344 cords per ton. (13) A pulp mill increased its production 43%, producing 160 tons per day; how much did it formerly produce? Ans. 112 tons per day. (14) If 165 pounds of clay is used in an order of 2240 pounds of paper, what per cent of the finished product is clay?^ Ans. 7.37%. (15) A sample of paper weighs 9.372 grams, of which 0.621 grams are water and 8.751 grams are fiber; what is the percentage of water and fiber in the paper expressed as per cent? ylns. /6.63% of water. \ 93.37% of fiber. (16) A sample of coal shows 11.3% ash; assuming that all the combustible part of the coal is consumed, how many pounds of ashes must be handled per ton of 2240 pounds of coal? Ans. 253.12 pounds. (17) After having been in use for some time, it was found that a 58-foot belt had stretched 3 % ; what was the length of the belt after being stretched? Ans. 59.74 feet. 106 ARITHMETIC §1 (18) A paper shrinks 1.62% from wet end to calenders; if it was 164 inches wide on the wire, how wide is it when dry V Ans. 161.34 inches. (19) How much bone dry fiber is contained in 793 tons of wet pulp, the moisture content being 43 % ? Ans. 452 tons, (20) The bone dry weight of pulp is 90% of the air dry weight, which is the basis of payment. Referring to example 19, what is the equivalent air dry pulp? Ans. 502.22 tons. COMPOUND NUMBERS 182. A compound number is one that requires more than one unit to express it; thus, if the length of a piece of hose is stated to be 12 feet 7 inches, 12 feet 7 inches is a compound number, because two units — the foot and the inch — are required to express the length. Had the length been stated as 12 %2 feet, the number would be called a simple number, since only one unit — 'the foot — is required to express it. Compound numbers are usually denominate numbers, a denominate number being one having its unit or units de- nominated, or named. 183. A denominate number is always a concrete number (see Art. 7), and when only one unit is named, as 5 yards, 8 pounds, etc., it is a simple number. An abstract number is usually a simple number, but it may be treated as a compound denominate number if desired. Thus, 364 may be regarded as 3 hundreds 6 tens 4 units. 184. In the case of abstract numbers, each figure belongs to a denomination that is ten times as large as the denomination of the next figure on its right and one-tenth as large as the denomi- nation of the next figure on its left. In connection with de- nominate numbers, however, this is seldom or never the case. For instance, in the compound denominate number 4 yards 2 feet 8 inches, it takes 12 inches to make 1 foot, the next higher denomination, and it takes 3 feet to make 1 yard, the next denomination higher than feet. For this reason, it is necessary to memorize certain tables showing the relation of the different denominations for different compound numbers. In order to save time and space in writing, the names of the units are abbreviated, and these abbreviations are given in the tables that follow. 185. The tables that follow should all be thoroughly committed to memory. Only the tables in common use are given here. §1 COMPOUND NUMBERS 107 Under each table is a subsidiary table that shows the relation between the different units; it is not necessary to memorize these, although it will be found convenient to remember some of the principal equivalents, as, for example, that 5280 feet make a mile, that 36 inches make a yard, etc. TABLE I LINEAR MEASURE 12 inches (in.) = 1 foot ft. 3 feet = 1 yard yd. 5)4 yards = 1 rod rd. 40 rods = 1 furlong fur. 8 furlongs (320 rd.) =1 mile mi. mi. fur. rd. yd. ft. in. 1 = 8 = 320 = 1760 = 5280 = 63,360 1 = 40 = 220 = 660 = 7,920 1 = 5.5 = 16.5 = 198 1 = 3 = 36 1 = 12 186. Another abbreviation for feet and inches, much used by draftsmen, mechanics, etc. is (') for feet and (") for inches; hence, 4 feet 8 inches may be written either 4 ft. 8 in. or 4' 8". When the latter form is used, it is advisable to place a dash between the feet and inches; thus, 4' — 8". The subsidiary table is useful in expressing higher units in terms of the lower units, and vice versa. For example, if it were desired to express 37 rd. in inches, the table shows that 1 rd. = 198 inches; hence, 37 rd. = 37 X 198 = 7326 in. Again, what fraction of a mile is 1360 feet? From the table, 1 mi. = 5280 ft.; hence, 1360 ft. = ^^^on ^ ^^ ^^• = .2576 - mi. TABLE II SQUARE MEASURE 144 square inches (sq. in.).. . = 1 square foot sq. ft. 9 square feet = 1 square yard sq. yd. 30 J^ square yards . , = 1 square rod sq. rd. 160 square rods = 1 acre A. 640 acres = 1 square mile sq. mi. sq. mi. A. sq. rd. sq. yd. sq. ft. sq. in. 1 = 640 = 102,400 = 3,097,600 ^ 27,878,400 = 4,014,489,600 1 = 160 = 4,840 = 43,560 = 6,272,640 1 = 30.25= 272.25 = 39,204 1 = 9 = 1,296 1 = 144 108 ARITHMETIC §1 187. A plot of ground in the form of a square, each side of which measures 208.71 feet, say 208 ft. 9 in., contains one acre. TABLE III CUBIC MEASURE 1728 cubic inches (cu. in.).. . . = 1 cubic foot cu. ft. 27 cubic feet = 1 cubic yard cu. yd. 128 cubic feet = 1 cord (wood) cd. 24^ cubic feet = 1 perch (stone, masonry)?. cu. yd. cu. ft. cu. in. 1 = 27 = 46,656 1 = 1,728 188. The cord is used only in measuring wood. A pile of wood 8 ft. long, 4 ft. wide, and 4 ft. high contains one cord. The perch is used in measuring stone and brick walls and other masonry. Some contractors allow 25 cubic feet to the perch, but 24 % cubic feet is the correct value. TABLE IV AVOIRDUPOIS WEIGHT 4373^ grains (gr.) = 1 ounce oz. 16 ounces = 1 pound lb. 100 pounds = 1 hundredweight .... cwt. 20 hundredweight (2000 lb.) = 1 ton T. T. cwt. lb. oz. gr. 1 = 20 = 2000 = 32,000 = 14,000,000 1 = 100 = 1,600 = 700,000 1 = 16 = 7,000 1 = 437.5 189. The ton of 2000 pound, called the short ton, is the one commonly used. The ton of 2240 pounds is called the long ton, and is used to weigh coal, pig iron, and other coarse commodities; it is the basis of freight rates on foreign exports. In connection with the long ton, 14 pounds make a stone, 2 stones make a quarter, 4 quarters make a hundredweight, and 20 hundred- weight make a ton; hence. LONG TON T. cwt. qr. St. lb. 1 = = 20 = 80 = 160 = 2240 1 = 4 = 8 = 112 1 ^ 2 1 28 14 §1 COMPOUND NUMBERS 109 190. For measuring medicines, jewelry, gold, silver, etc., the Troy ounce, which contains 480 grains is used. The Troy pound contains 12 Troy ounces or 5760 grains. The Troy pound is therefore, only |^^t = iflth of an avoirdupois pound. A pound of gold thus weighs 5760 -^ 437.5 = 13tS^ = 13.1657 + avoir- dupois ounces. TABLE V. LIQUID MEASURE 4 gills (gi.) 2 pints 4 quarts 313^ gallons 2 barrels (63 gallons) . hhd. 1 bbl. 2 1 gal. 63 31.5 1 = 1 pint pt. = 1 quart qt. = 1 gallon gal. = 1 barrel bbl. = 1 hogshead hhd. qt. pt. gi. 252 = 504 = 2016 126 = 252 = 1008 4 = 8 = 32 1 = 2 = 8 1 = 4 191. The United States, or wine, gallon contains 231 cubic inches, and a gallon of water weighs very nearly 8^ pounds. A cubic foot contains 1728 ^ 231 = 7.481 gallons, or, roughly, 7| gallons. The British imperial gallon, used in Great Britain and Canada, contains 277.463 cubic inches, and a gallon of water at a temperature of 62 degrees Fahrenheit weighs exactly 10 pounds. The British imperial gallon is equal to 1.2 U. S. gallons, very nearly. What is known as the fluid ounce is the weight of yVth of a pint of water, or listh of a gallon of water. Since a gallon of water weighs S^-i pounds, a fluid ounce weighs 8| X 16 -^ 128 = l-^j = If ounces. 2 pints (.pt.). 8 quarts. . . . 4 pecks TABLE VI DRY MEASURE bu. 1 pk. 1 = 1 quart qt. 1 peck pk. 1 bushel bu. qt. pt. 32 = 64 8 = 16 1=2 192. The unit of dry measure is the Winchester bushel, which contains 2150.42 cubic inches; hence, the dry quart contains 110 ARITHMETIC §1 2150.42 -j- 32 = 67.2 cubic inches, while the liquid quart con- tains 231 -^ 4 = 57 f cubic inches. A box 14 inches square and 11.7 (say llf) deep, inside measurement, holds one bushel. The bushel is used in measuring charcoal and, sometimes, lime and coal. The British bushel is equal to 8 British imperial gallons; it therefore contains 277.463 X 8 = 2219.704 cu. in. and one British bushel equals 2219.704 h- 2150.42 = 1.03222 - Winchester bushels. A cylinder 18| in. in diameter and 8 in. deep holds exactly one bushel. TABLE VII ANGULAR MEASURE 60 seconds (") 60 minutes 90 degrees 4 quadrants (360°) cir. quad. (L) deg. (°) 1 = 4 = 360 1 = 90 1 1 minute ' 1 degree ° 1 quadrant L 1 circle cir. min. (') sec. (") 21,600 = 1,296,000 5,400 = 324,000 60 = 3,600 1 = 60 193. Table VII is very important in connection with all prob- lems in which the measurement of angles is necessary for their solution. The circle is divided into 360 equal parts called degrees; each degree is divided into 60 equal parts called minutes; and each minute is divided into 60 equal parts called seconds. When the circle is divided into four equal parts, each part is called a quadrant, and the 4 equal angles so formed are called right angles. A quadrant and a right angle contain 360° -;- 4 = 90°. 194. Units of Measurement. — It should be evident from Art. 1 that before any measurement can be made it is necessary to establish a unit. For linear measurements (Table I), the fundamental unit, from which the other units are derived, is the yard; the yard is divided into three equal parts each of which is called one foot; the foot is divided into twelve equal parts, each of which is called one inch. The higher units, rods, furlongs, and miles, are obtained by taking a certain number of yards. The units for square and cubic measure are obtained from those for linear measure; thus, the square inch is a square, every side of which measures 1 inch (see Fig. 2). Fig. 2. §1 COMPOUND NUMBERS 111 mp nq Fig. 3. The square foot is a square every edge of which measures 1 foot or 12 inches. If a square foot be divided into 12 equal parts by lines ah, cd, etc., Fig. 3, and again divided into 12 equal parts by lines mn, pq, etc., every one of the little squares that are thus formed will be a square inch. Between a and b, there are 12 inch-squares, between c and d, 12 inch-squares, etc. Conse- quently, the foot-square has been divided into 12 X 12 = 144 inch-squares, and there are 144 square inches in 1 square foot. A square yard may be similarly divided into 9 foot-squares; hence, there are 9 square feet in 1 square yard. Note that 12 X 12 = 122 and 3 X 3 = 3^; there- fore, since there are 5.5 yards in a rod, a square rod contains 5.5^ = 30.25 = 30| square yards; etc. The number of square units in any square may be found by squaring the length of the side of the square. It is for this reason that the second power of a number is usually called the square. A cube measuring 1 inch on every edge is called a cubic inch (see Pig. 4) . A cubic foot is a cube measuring 1 foot on every edge. Such a cube may be divided into 12 equal layers, each of which measures 12 inches on each side and 1 inch high. Each layer may therefore be divided into 12^ equal cubes measuring 1 inch on each edge; that is, each layer may be divided into 12 X 12 inch-cubes; and since there are 12 layers, a cube measuring 12 inches (==1 foot) on every edge may be divided into 12 X 12 X 12 = 12^ little cubes measuring 1 inch on every edge. Similarly, a cubic yard is a cube measuring 1 yard on every edge and may be divided into 3^ = 27 small cubes measuring 1 foot on every edge. Hence, a cubic foot contains 12^ = 1728 cubic inches, and a cubic yard contains 3^ = 27 cubic feet = 36^ = 46,656 cubic inches. The number of cubic •units in any cube may be found by cubing the length of one of the edges. It is for this reason that the third power of a number is called the cube. The fundamental unit of weight is the pound avoirdupois ; the Fig. 4. 112 ARITHMETIC §1 other units of weight are found by dividing or multiplying the pound. The fundamental unit of liquid measure in the United States is the United States or wine gallon, which contains 231 cubic inches. In Canada the law allows the use of the Imperial gallon only for a commercial unit. Since the fundamental unit of dry measure is the Winchester bushel, which contains 2150.42 cubic inches, the units of both liquid and dry measures depend upon the unit of linear measure. The fundamental unit of angular mea- sure is the quadrant or the right angle. The circle is divided into 2 equal parts by a line drawn through the center called a diameter, as ab, Fig. 5. Each half is then divided into 2 equal parts by another diameter cd, thus dividing the entire circle into 4 equal parts. Each of these four equal parts is then divided into 90 equal parts to obtain degrees. TABLE VIII United States money 10 mills (m.) = 1 cent ct. or jif. 10 cents = 1 dime d. 10 dimes (100c.) =1 dollar dol. or $ 10 dollars =1 eagle E. E. $ d. ct. m. 1 = 10 = 100 = 1000 = 10,000 1 = 10 = 100 = 1,000 1 = 10 = 100 1 = 10 Note. — With the exception of the mill and the eagle, the Canadian unit of money have the same names and relative values. 196. The fundamental unit of United States money is the gold dollar, which is defined as 25.8 grains of gold nine-tenths fine; that is, the gold dollar contains 25.8 X .9 = 23.22 grains of pure gold. It will be noted that the scale in the table of United States money is 10, the same as in the Arabic system of notation; it is therefore called a decimal scale, and numbers may be expressed in different units by simply shifting the decimal point. Thus, $12.43 = 1243c, and 978c = $9.78. Since there are lOOj;^ in $1, to change dollars to cents, multiply by 100, that is, shift the decimal point two places to the right; and to change cents to dollars, shift the decimal point two places to the left. §1 COMPOUND NUMBERS 113 TABLE IX TIME 60 seconds (sec.) = 1 minute min. 60 minutes = 1 hour hr. 24 hours = 1 day da. 7 days = 1 week wk. 365 days =1 year yr. yr. da. hr. min. sec. 1 = 365 = 8760 = 525,600 = 31,536,000 1 = 24 = 1,440 = 86,400 1 = 60 = 3,600 1 = 60 197. The fundamental unit of time, universally used, is the second. The year does not contain exactly 365 days, but 365.242216 days = 365K days very nearly; hence, every fourth year, one day is added, making 366 days in what are called leap years. TABLE X MISCELLANEOUS 12 of anything = 1 dozen doz. 12 dozen =1 gross gr. 12 gross =1 great gross g- gr. 24 sheets of paper = 1 quire qr. 20 quires (480 sheets) =1 ream rm. 198. A gross is evidently equal to 12 X 12 = 144. It is now quite common practice to consider a ream as 500 sheets, except for stationery papers. THE METRIC SYSTEM 199. The metric system is thus called because it is based on the meter (also spelled metre), which is equal to 39.370113 inches. By Act of Congress, the meter contains 39.37 inches. Sub- divisions of the meter are the decimeter, centimeter, and millimeter, and multiples of it are dekameter, hektometer, kilometer, and myria- meter. The scale of the system is a decimal one, the same as with ordinary numbers and United States money. The prefixes deci, centi, and milli denote respectively one-tenth, one-hundredth, and one-thousandth of the unit to which they are prefixed, the same as the dime, cent, and mill denote-one-tenth, one-hundredth, and one-thousandth of a dollar. The prefixes deka, hekto, kilo, 8 114 ARITHMETIC §1 and myria, denote respectivly 10, 100, 1000, and 10,000 times the unit to which they are prefixed. TABLE XI LINEAR MEASURE 10 millimeters (mm.) =1 centimeter cm. 10 centimeters = 1 decimeter dm. 10 decimeters = 1 meter m. 10 meters = 1 dekameter Dm. 10 dekameters = 1 hektometer Hm. 10 hektometers = 1 kilometer Km. 10 kilometers = 1 myriameter Mm. 200. The fundamental unit in the foregoing table is the meter; the other units that are principally used are the millimeter and centimeter for short lengths and the kilometer for long dis- tances. The following table gives English equivalents for these units. 1 millimeter = .03937 in. = jf s in. very nearly = .04 inch roughly. 1 centimeter = .3937 in. = |f in. very nearly = .4 in. roughly. 1 meter = 39.37 in. = 39f in. very nearly = 40 in. roughly. 1 kilometer = 39,370 in, = |i mi. very nearly = f mile roughly. 1 meter = 3.281 ft. nearly = 1.094 yd. nearly. It will be observed that the abbreviations for the multiples of the principal unit begin with a capital letter, while those for the sub-multiples and for the principal unit itself begin with a lower case letter. TABLE XII SQUARE MEASURE 100 square millimeters (mm^).. . . = 1 square centimeter cm.^ 100 square centimeters = 1 square decimeter dm.^ 100 square decimeters = 1 square meter m.^ 201. This table does not include measures of land, the principal unit of which is the Are = 100 m.^ The hectare, which equals 100 ares, or 100 X 100 = 10,000 square meters, takes the place of the English acre, and is equivalent to 2.471 acres, say 2| acres, roughly. 1 square meter = 1550 square inches, very nearly = 10.7641 square feet, say lOf square feet, roughly. TABLE Xin CUBIC MEASURE 1000 cubic millimeters (mm. 3).... = 1 cubic centimeter.. . c.c. or cm.' 1000 cubic centimeters = 1 cubic decimeter. . . dm.^ 1000 cubic decimeters =1 cubic meter m.^ 1 cubic meter = 35.3156 cubic feet = 1.308 cubic yards. §1 COMPOUND NUMBERS 115 TABLE XIV LIQTJID MEASURE 10 milliliters (ml.) - 1 centiliter cl. 10 centiliters = 1 deciliter dl. 10 deciliters =1 liter 1- 10 liters = 1 dekaliter Dl. 10 dekaliters = 1 hektoliter. ... HI. 10 hektoliters = 1 kiloliter Kl. 202. The principal units of liquid measure are the liter and the hectoliter. The liter is equal in volume to 1 cubic decimeter = 61.0254 cubic inches = 1.0567 quarts = l^V quarts. The hectoliter = 26.4179 gallons - .9435 barrel. The milliliter is used by physicians, chemists, scientists, and others for measur- ing small amounts. Since a milliliter is xoVo of a hter, and since a liter equals a cubic decimeter = 1000 cubic centimeters, a miUiliter = 1 cubic centimeter, and it is customary to call this unit a cubic centimeter instead of a milliliter. A teaspoon contains }i of a fluid ounce. 1 liter = 1 dm3= 61.0254 cu. in.; 1 quart = 57.75 cu. in. = 16 X 2 = 32 fluid ounces; H of a fluid ounce = 57.75 -^ 32 -^ 8 = .225586 cu. in.; 1 milliliter = 1 cubic centimeter = 61.0254 -^ 1000 = .0610254 cu. in. Therefore, 1 teaspoon = .225586 -^ .0610254 = 3.6966, say 3.7 cubic centimeters, and 1 c.c. = .27052 teaspoon, or a little more than }i teaspoon. Also, 1 fluid ounce = 3.6966 X 8 = 29.5728 cubic centimeters = 30 c.c, nearly. TABLE XV MEASURES OF WEIGHTS 10 milligrams (mg.) =1 centigram eg. 10 centigrams =1 decigram dg. 10 decigrams =1 gram g. 10 grams. 1 dekagram Dg- 10 dekagrams =1 hektogram Hg. 10 hektograms =1 kilogram Kg. 1000 kilograms =1 tonne T. 204. The fundamental unit of weight in the metric system is gram, which is the weight of 1 cubic centimeter of water, weighed under certain conditions, and is equal to 15.432 grains, (more accurately, 15.432356 gr.). The kilogram (frequently called the kilo) = 1000 grams = the weight of 1 liter of water = 2.2046 pounds (more accurately, 2.20462234 pounds). The tonne (also called metric ton) = 1000 kilograms = the weight of 116 ARITHMETIC §1 1 cubic meter of water = 2204.6 pounds. The metric ton is therefore approximately equal to the English long ton. The milligram is used by chemists for weighing minute quantities; it is equal to s\ of a grain approximately, or ^^ gr. very accurately. REDUCTION OF COMPOUND NUMBERS 205. To reduce a compound number is to change it so that the denominations used to express it will be different without chang- ing the value of the number. For example, it may be desirable to express a certain number of miles, rods and feet as feet or inches; or it may be desired to express the number in miles and fraction of a mile. All such changes are called reduction or reducing the number to lower or higher denominations. 206. Reducing to Lower Denominations. — The process is best illustrated by means of an example. Thus, express 3 mi. 126 rd. 9 ft. in feet. Arranging the work as shown, multiply the number of rods in a mi. rd. ft. mile (320) by the number of miles in the 3 126 9 given number, in this case 3, and the ?^ product is the number of rods in 3 miles. ^^^ ^^- To this add the number of rods in the 126 — given number, and the sum, 1086, is the ^„i/' number of rods in 3 mi. 126 rd. Since -rrr there are 163^^ feet in a rod, the product 6516 of 1086 and 16>^ = 17919 is the number 1086 of feet in 1086 rd. To this add the 9 ft. 17919 ft. in the given number, and the sum is 9 ft. 17,928 ft., the number of feet in 3 mi. 236 17928 ft. Ans. rd. 9 ft. Had it been desired to reduce the number to inches, the last result would have been multiplied by 12, and as there are no inches in the given number, the final result would have been 17928 X 12 = 215,136 in. = 3 mi. 126 rd. 9 ft. 207. Rule. — To reduce a compound number to lower denomina- tions, find the product of the number representing the highest denom- ination in the given number and the number of units of the next lower denomination that will make one unit of the next higher denom- ination, and add to this product the number of units of that lower denomination, {if any) in the given number. Repeat this process §1 COMPOUND NUMBERS 117 to reduce to the next lower denomination, continuing in this manner until the required denomination has been used. Example 1. — Reduce 32° 47' 19" to seconds. Solution. — The work is shown in the margin, and should be evident. Since there are 60 minutes in 1 degree, the 1920' product of 60 and 32 is first found, to which is added the 47.' 47 Since there are 60" in 1 minute, the product of 60 and 1967 1967 is then found, to which is added the 19". The num- 60 ber is now reduced to the denomination required. Note 118020" ^^^^ ^i^^ numbers are added— minutes to minutes and IQ" seconds to seconds. It is not customary to write the 1 1sn^Q" Avi abbreviations when reducing. Example 2. — Express 156 m., 5 cm., 2 mm. in milhmeters. Solution. — Write the number without abbreviations, with a period fol- lowing the number of units of the highest denomination, and with ciphers in place of any denomination that is missing. Since there are no decimeters, 156 m. 5 cm. 2 mm. = 156.052 meters. The process is exactly the same as writing a number in the Arabic system of notation. To reduce 156.052 m. to millimeters, multiply by 1000 by shifting the decimal point 3 places to the right, and 156.052 m. = 156.052 mm. Ans. Had it been desired to express the above number in decimeters, multiply it by 10 by moving the decimal point 1 place to the right, obtaining 1560.52 dm.; to express it in centimeters, move the decimal point two places to the right, obtaining 15605.2 cm. 208. Reducing to Higher Denominations. — The process is best illustrated by an example. Thus, reduce 215,136 in. to higher denominations. Since there are 12 inches in 1 215136 in. (12 foot, the number of feet in 215,136 17928.0 ft. (16.5 inches may be found by dividing 1^ 1086 215,136 by 12; there is no remainder, 1428 and the quotient is 17,928 ft. There 1320 • are 16.5 feet in 1 rod; hence, the 1080 QQ^ number of rods in 17,928 feet is -— ^ g A f+ found by dividing 17,928 by 16.5. 1086 rd. (320 "^^^ quotient is 1086 rd., and the 960 3 ^i remainder is apparently 90 ft. In 126 rd. reality, however, it is 9 ft., because 3 mi. 126 rd. 9 ft. Ans. the cipher in 90 is the one following the decimal point in the dividend, and has no value. Since there are 320 rods in 1 mile, divide 118 ARITHMETIC §1 1086 rods by 320 to find how many miles there are in 1086 rd.; the quotient is 3 mi., and the remainder is 126 rd. Therefore, 215,136 in. = 3 mi. 126 rd. 9 ft. Note that after any division has been completed, the remainder is of the same denomination as the dividend; this is necessarily the case, since the remainder is a part of the dividend. The quotient, however, is of a higher denomination than the dividend. 209. Suppose that instead of reducing the inches to a compound number the highest denomination of which, in this case, is a mile, it had been desired to express the number in miles and a fraction (decimal) of a mile. In such case, proceed in exactly the same manner as before, except that the quotient will in all cases be a mixed number (when there is a remainder), the division being carried to as many decimal places as is desired. Thus, in the example just given, 9 ft. =■ 9 ^ 16.5 = .545454+ rd. Adding this to the 126 rd., the sum is 126.545454+ rd. Then, 126.545454 -r- 320 = .39545454+ mile, which added to the 3 mi. makes 3.3954545 mi. 210. To reduce a decimal of one denomination to units of a lower denomination, proceed exactly in accordance with the rule of Art. 207. Thus, .3954545 mi. = .3954545 X 320 = 126.54544 rd.; .54544 X 16.5 = 8.99976, say 9 ft. The result would have been exactly 9 ft. if instead of expressing the rods and feet as a decimal of a mile they are reduced to a fraction of a mile. Thus, 9 -^ 16.5 = j^^=U = t\ rd.; 126 + A = 126^ = ^U^ rd. Iff 2 ^ 320 = illl = 2%\ = .39 Ami. This last expression wUl reduce to 126 rd. 9 ft. exactly. 211. Rule I. — To reduce a denominate number to higher denomi- nations, begin' with the lowest denomination of the given number and divide the number of units of that denomination by the number of units required to make one unit of the next higher denomination, forming either a common fraction {which reduce to its lowest terms) or carrying the quotient to any desired number of decimal places. Add the quotient thus obtained to the number of units in the given number of the same denomination as the quotient (if any), and divide the sum by the number of units required to make one unit of the next higher denomination. Proceed in this manner until the desired denomination is reached. §1 COMPOUND NUMBERS 119 II. //, however, it is desired to reduce a given number of units of a lower denomination to a compound number of higher denomination, divide the given number by the number of units required to make one unit of the next higher denomination; the quotient will be of the next higher denomination, and the remainder {if any) will be of the same denomination as the dividend. If the quotient is larger than the number of units required to make a unit of the next higher denomina- tion, divide it by the number of units required to make one unit of the next higher denomination. Proceed in this manner nutil the highest denomination is reached or a quotient is obtained that is smaller than the number of units required to make a unit of the next higher denomination. Example. — Reduce 123,456" to a compound number; also express it in degrees and decimal of a degree to 5 decimal places. 60) 1 23456" Solution. — Dividing first by 60, tbe quotient 60)2057' + 36" ^^ 205/' and the remainder is 36". Dividing „.o _1_ 17' again by 60, since there are 60' in 1°, the quo- oAo ^'7/ OR" A tient is 34° and the remainder is 17'. Since 34° is less than 90°, the number of degrees necessary fi^^ 12*^4 '>6" "">> to make a quadrant, the work ceases, and 60) 2057.6' 123,456" = 34° 17' 36". The work for reducing 34.29333° + Ans. to degrees and decimal of a degree is evident. Example 2. — Express 3250615 milligrams in kilograms. Solution. — Beginning with the right hand figure, move the decimal point one place to the left for each higher denomination until the desired denomi- nation is reached. Thus, say milligrams, centigrams, decigrams, grams, dekagrams, hektograms, kilograms, placing the pencil point on the position occupied by the decimal point in each case, it being to the right of the milligrams in the beginning. When kilograms is reached, the pencil point will fall between 5 and 3; consequently, 5320615 mg. = 5.320615 Kg. Ans. This same method may be used in reducing metric numbers (except those for square and cubic measure) to lower denomina- tions. For square measure, move the decimal point two places each time the name of a denomination is pronounced, and for cubic measure, move it three places. For instance, 59308726 mm2 = 59.308726 m^, and 607849358 c.c. = 607.849358 m^. Also, 78.06342 m^ = 780634.2 cm^ = 78063420 mm^, and 9.50783 m3 = 9507830 c.c. = 9507830000 mm^, 212. Another way of reducing the lower units of a compound number to a decimal or fraction of higher denomination is the following: To express 23 rd. 3 yd. 1 ft. 8 in. as a decimal of a mile, first reduce the compound number to the lowest denomi- nation in the given number, in this case inches, and 23 rd. 3 120 ARITHMETIC §1 yd. 1 ft. 8 in. = 4682 in. Referring to Table I, and consulting the sub- sidiary table, it is seen that there are 63360 inches in a mile; hence, 4682 in. is /aWirth of a mile. Reducing this fraction to a decimal, W3%%\ = .073895+. Therefore, 23 rd. 3 yd. 1 ft. 8 in. = .073895 mi. This method is to be preferred to the former one, when reducing to a decimal or a fraction of a higher denomination, since it entails no more work, if as much, and is, in general, more accurate. 4682 in. ' rd. yd. 23 3 5.5 ft. 1 115 115 126.5 3 129.5 yd. 3 388.5 1 389.5 ft. 12 4674.0 8 EXAMPLES (1) Reduce 5 mi. 3 fur. 36 rd. 4 yd. 2 ft. 7 in. to inches. Ans. 347,863 in. (2) Reduce the number in (1) to miles and decimal of a mile. Ans. 6.490276 mi. (3) Reduce 13 bbl. 23 gal. 1 pt. to pints. Ans. 3,461 pt. (4) Reduce the number in (3) to barrels and decimal of a barrel. Ans. 13.73413 bbl. (5) Reduce 3 T. 13 cwt. 71 lb. 12 oz. to ounces. Ans. 117,948 oz. (6) Reduce the number in (5) to tons and decimal of a ton. Ans. 3.685875 T. (7) Reduce 14° 9' 54" to seconds. Ans. 50,994". (8) Express 14° 9' 54" in degrees and decimal of degree. Ans. 14.165°. (9) Reduce 75,906" to a compound number. Ans. 21° 5' 6". (10) Reduce 75,906" to degrees. Ans. 21.085°. (11) Reduce 50,000 in. to a compound number. Ans. 6 fur. 12 rd. 2 yd. 2 ft. 8 in. OPERATIONS WITH COMPOUND NUMBERS 213. Addition." — Compound numbers are added in practically the same manner as abstract numbers. The numbers to be added are arranged under one another with like denominations in the same columns. Then, beginning with the lowest denomina- tion, add the numbers in that column; if the sum is greater than the number of units required to make a unit of the next higher denomination, reduce it to the next higher denomination and carry to the next column the number of units so found of that next §1 COMPOUND NUMBERS 121 higher denomination. The process is repeated for the second and subsequent columns. Example L— Add 15° 20' 36", 21° 53' 46", 8° 49' 28", and 76° 51' 17". Solution. — The numbers are arranged as shown in the margin, with 1 ^° 90' ^fi" seconds under seconds, minutes under minutes, etc. _ _„ .„ The sum of the second's column is 127" = 2' 7"; _ .„ <^„ write the 7" and carry the 2' to the next column, „„ _. ^_ adding the 2' to the numbers in that column. The sum thus obtained is 175' = 2° 55'. The 2° is carried rd. yd. ft. in. 11 4 2 4 34 1 9 27 3 1 6 17 1 1 2 90 10 5 21 2 fur. 10 4i 2 9 5yd. = 1 6 122 175 127 ^^ ^j^g j^g^^ column, making the sum 122°, and the sum of all the numbers is 122° 55' 7". It is not customary i.L^ oo / . jxiif,. to write the sums of the columns, as shown here; only the final results are written, unless there are fractions as in the next example. Example 2.— Add 11 rd. 4 yd. 2 ft. 4 in., 34 rd. 1 yd. 9 in., 27 rd. 3 yd. 1 ft. 6 in., and 17 rd. 1 yd. 1 ft. 2 in. Solution. — The numbers are arranged as in example 1, and the sum as found is 2 fur. 10 rd. 4.5 yd. 2 ft. 9 in. Since it is inconvenient to have a fraction in any but the lowest de- nomination, reduce the .5 yd. obtain- ing 1 ft. 6 in., which is added as shown, making the final sum 2 fur. 10 rd. 5 yd. 1 ft. 3 in. 2 fur. 10 rd. 5 yd. 1 ft. 3 in. Ans. Rule. — Flace the numbers to he added under each other, with like denominations in the same columns. Add each column, beginning with that of the lowest denomination. If the sum of the numbers in any column is greater than the number of units required to make a unit of the next higher denomination, reduce the sum to the next higher denomination before adding the next column. When the sum has been found, if the number of units in any denomination contains a fraction, reduce the fraction of a unit to lower denomina- tions and add to the units of lower denomination in the sum previously found. 214. Subtraction.- — The operation of subtraction is practically the same as in subtraction of abstract numbers. Place the sub- trahend under the minuend, with like denominations under each other. If the number of units of any denomination in the sub- trahend is larger than the number above it, reduce one unit of the next higher denomination to the next lower, add it to the number of units of that denomination in the minuend, and then rd. 34 27 yd. ft. 1 3 1 in. 9 6 6 6rd. 2K 2 1 3 yd. ~ 3 6 9 in. 122 ARITHMETIC §1 subtract. Add 1 to the number of units in the next column of the subtrahend before subtracting. This process is exactly similar to that employed in subtracting abstract numbers. Example 1.— Subtract 34° 27' 17" from 90°. SoLTTTiON. — Placing the subtrahend under the minuend, there are no seconds in the minuend; hence, 1' = 60" is added to 90° 00' 00" the minuend, and 60" - 17" = 43". Adding 1' to 27', 34 27 17 the sum is 28', and 60' - 28' = 32'. Adding 1° to 55° 32' 43". Ans. 34°, the sum is 35°, and 90° - 35° = 55°. The re- mainder, therefore, is 55° 32' 43". Example 2. — From 34 rd. 1 yd. 9 in. subtract 27 rd. 3 yd. 1 ft. 6 in. Solution. — Arranging the numbers as in example 1, 1 ft. cannot be sub- tracted from ft.; hence, 1 yd. = 3 ft. is added to the minuend, and 3 ft. — 1 ft. = 2 ft. Adding 1 to the 3 yd. makes 4 yd.; but 4 yd. cannot be subtracted from 1 yd. ; hence, 1 rd. = 5}i yd. is added to the minuend, making 1 + 5K = 6J^ yd., and 6>^ yd. — Ans. 4 yd. = 2H yd. Finally, 27 rd. + 1 rd. - 28 rd., and 34 rd. - 28 rd. = 6 rd. The ^2 yd. is reduced to 1 ft. 6 in. and added, the sum being 3 ft. 9 in. But 3 ft. = 1 yd.; hence, the number of yards is increased by 1, making the final sum 6 rd. 3 yd. 9 in. Rule. — Place the subtrahend under the minuend, with like denominations in the same columns. Beginning with the column of lowest denomination, subtract the numbers in the bottom roiv from those above them. If the number in any column of the subtrahend is larger than the number above it in the minuend, reduce one unit of the next higher denomination to the next lower denomination, add it to the minuend, then subtract, and add 1 to the number in the column of the next higher denomination in the subtrahend. Con- tinue in this manner until the entire remainder has been found. 215. Multiplication. — A compound number may be multiplied by an abstract number in two ways: 1st., by multiplying the units of each denomination separately, and reducing the products to higher denominations; this is advisable when the multiplier is a small number. 2d, reduce the compound number so that it will be expressed in units of one denomination, preferably, the lowest denomination in the compound number, then multiply, and reduce the product to higher denominations; this process is preferred when the multiplier is a large number, say greater than 12, or when it contains a fraction or a decimal. §1 COMPOUND NUMBERS 123 Example 1.— Multiply 56 T. 13 cwt. 72 lb. by 8. Solution.— The product of 8 and 72 lb. is 576 lb. = 5 cwt. 76 lb. Then „ 13 cwt. X 8 = 104 cwt., to which is added the ■ ' 5 cwt. carried from the first product, making ^ 109 cwt. = 5 T. 9 cwt. Lastly, 56 T. X 8 = 448 T., to which is added the 5 T. carried 453 109 576 {Tom the preceding product, making 453 453 T. 9 cwt. 76 lb. Ans. rj. ^j^^ ^^-^ product is 453 T. 9 cwt. 76 lb. Example 2. — Multiply 7 gal. 3 qt. 1 pt. by 53. Solution. — Reducing to pints, 7 gal. 3 qt. 1 pt. = 63 pt.; 63 pt. X53 = 3339 pt. = 6 hhd. 1 bbl. 7 gal. 3 qt. 1 pt. When reducing 3339 pt. to higher denominations, the first result obtained is 6 hhd. 1 bbl. 7J^ gal. 1 qt. 1 pt. Since }i gal. = 2 qt., the final result without fractions is 6 hhd. 1 bbl. 7 gal. 3 qt. 1 pt. Ans. Rule I. — Begin with the loivest denomination in the given num- ber, and multiply the number of units of that denomination by the multiplier; reduce the product to the next higher denomination by dividing by the number of units required to make 1 of the next higher denomination, and add the quotient to the product of the number of units in the next higher denomination and the multiplier. Proceed in this manner until the complete product has been found. If a fraction occurs in connection with any product, reduce it to lower terms. n. // the multiplier is greater than 12, or if it contains a fraction or a decimal, reduce the multiplicand to the lowest denomination given, multiply, and reduce the product to higher denominations. Example.— Multiply 5 yd. 2 ft. 9 in. by 11.7. Solution. — Reducing to inches, 5 yd. 2 ft. 9 in. = 213 in.; 213 in. X 11.7 = 2492.1 in. = 69 yd. 8.1 in., or 12 rd. 3 yd. 8.1 in. Ans. 216. Division. — There are two cases of division: dividing a compound number by an abstract number, in which case, the quotient is a compound number; or dividing a compound number by a compound number of the same kind, in which case, the quotient is an abstract number. The simplest method of per- forming the division (and in most cases, the easiest) is to reduce the dividend to the lowest denomination in the given number, divide, and reduce the quotient to higher denominations. If the divisor is also compound, reduce it to the same denomination as the dividend, and divide as in division of abstract numbers, the quotient being abstract. Example 1.— What is ^th of 143° 25' 41"? Solution.— 143° 25' 41" = 516,341"; 516,341" -=- 7 = 73,763" = 20° 29' 23". Ans. A somewhat easier method of performing this division 124 ARITHMETIC §1 (.which may be used whenever the divisor is small) is shown in the margin. Here proceed as in short division. Then, 143° -f- 7 = 20° + 3° remain- Is! iU der; 3° = 180', which added to the 25', makes 205'. 7)143° 25' 41" 205' 4- 7 = 29' + 2' remainder; 2' = 120", which 20° 29' 23". Ans. added to the 41", makes 161"; and 161" -^ 7 = 23". The work may be arranged as shown or the additions may be performed mentally. The latter practice is not recom- mended, however, as it tends to increase the liability of making a mistake. Example 2. — How many flasks holding 2 gal. 1 qt. 1 pt. may be filled from a cask holding 52 gal. 1 qt.? Solution. — Reducing both numbers to pints, 2 gal. 1 qt. 1 pt. = 19 pt.; 52 gal, 1 qt. = 418 pt.; 418 -r- 19 = 22; hence, 22 flasks may be filled, Ans. Rule. — If the divisor is abstract, reduce the dividend to the lowest denomination in the given number, divide, and reduce the quotient to higher denominations. If both dividend and divisor are com- pound, reduce them to the lowest denomination in either number and divide; the quotient will be abstract. EXAMPLES (1) Find the sum of 17 cwt. 26 lb. 9 oz., 15 cwt. 83 lb. 11 oz., 22 cwt. 55 lb. 6 oz., 24 cwt. 71 lb. 8 oz., and 18 cwt. 48 lb. 5 oz. Ans. 4 T. 18 cwt. 85 lb. 7 oz. (2) What is Ath of 360°? Ans. 32° 43' 38tt". (3) Add 19 rd. 4 yd. 1 ft. 4 in., 31 rd. 2 yd. 9 in., 26 rd. 5 yd. 2 ft. 6 in., and 35 rd. 1 yd. 1 ft. 10 in. Ans. 2 fur. 33 rd. 3 yd. 5 in. (4) The sum of the three angles of any plane triangle is 180°; if two of the angles of a triangle are 36° 13' 26" and 64° 45' 40", what is the other angle? Ans. 79° 0' 54". (5) A piece of tape line 2 rd. 1 yd. 2 ft. 5 in. long was used to measure the distance between two points about one mile apart. The tape was applied 172 times; what was the distance between the points? Ans. 1 mi. 2 fur. 2 yd. 1 ft. 8 in. (6) The distances between the centers of the faces of a number of pulleys on a main shaft were as follows: 5 ft. 8 J in., 8 ft. 6i in., 7 ft. 3| in., 10 ft. 9| in., 9 ft. 4| in., 6 ft. 7i in. What was the distance between the first and last pulleys? Ans. 48 ft. 3| in. (7) A coal hod was found to hold 38 lb. 13 oz. of coal; how many times will a ton of coal fill the hod ? Ans. 51.5+times. (8) Bought a car of pulp weighing 47,526 lb. on which the freight rate was 17 cents per hundredweight; how much was the bill for freight? Remember that all freight rates are computed on the basis of 2240 pounds to the ton, hence, a hundredweight here contains 112 pounds. Ans. $72.14. (9) The weight of a cargo of pulp was 3923 tons 1458 pounds; what was the freight bill, if the rate was $12.75 per ton? Ans. $50,026.55. (10) What was the cost of a belt 95 ft. 7M in. long at 32(^. per foot? Ans. $30.60. (11) A machine makes 1836 pounds of paper per hour; in what time will it make 273^ tons of 2240 pounds. Ans. 33 hr. 33 min. §1 THE ARITHMETICAL MEAN 125 THE ARITHMETICAL MEAN 217. The arithmetical mean of several numbers (or quantities) is the quotient obtained by dividing the sum of the numbers (or quantities) by the number of numbers (or quantities); it is the same as the average of several numbers (or quantities), and is usually called the mean. For example, if goods were bought from three different firms at discounts of 32%, 35%, and SQ}i%, the mean, or average, discount is (32 + 35 + 36.5) -^ 3 = 34.5%. This result presupposes that the same amounts were bought from each firm. Now suppose that from one firm, the bill was $126.40 and the discount was 32%; from another firm, the bill was $87.65 and the discount was 35%: from the third firm, and bill was $68.83 and the discount was 36.5%; what was the mean discount? Here it is necessary to multiply the amount of each bill by the discount 126.40 X 32% (a) = 4044.8% $126.40 X .68 = $ 85.95 87.65 X 35 = 3067.75 87.65 X .65 = 56.97 68.83 X36.5 = 2512.295 68.83 X .635 = 43.71 282.88 Xx =9624.845 $186.63 9624.845 ^ 282.88 = 34.024% = x $282.88 X (1 - .34024) = $186.63. and then divide the sum of the products by the sum of the bills. The result as shown herewith is a mean discount of 34.024%. To prove that this is correct, multiply the amount of each bill by 1 minus the discount and add the products; the sum will evidently be the total amount to be paid, and it must equal the total amount less the mean discount, which is shown to be true, both being equal to $186.63. The process is exactly the same as would be followed in finding the average (mean) price paid for 126.4 lb. of some article at 32 cents per pound, 87.65 lb. at 35 cents per pound, and 68.83 lb. at 36.5 cents per pound. The reason for the process is easily found. If the discount on $1 is 32%, the discount on $126.40 is 126.4 X 32% = 4044.8%. The total discount on $282.88 is 9624.845%; hence, the average, or mean, discount is 9624.845% -^ 282.88 = 34.024% on $1. Example 1. — A machinist receives 52)i per hour in wages. For every hour he works over 8 hours a day, he receives time-and-a-half. His over- time for one week was: Monday, 2}i hr.; Tuesday, 2 hr.; Wednesday, 3 hr.; Thursday, 2 hr.; Friday, IK hr.; Saturday, 1 hr. What was his average rate per hour for that week? 126 ARITHMETIC §1 Solution. — The total number of regular hours of work was 6 X 8 = 48, for which he received 48 X 52 (i = $24.96. The number of hours that he worked overtime was 2.5+2+3+2 + 1.5+1 = 12; his rate for overtime was 1.5 X52(4 = 7^i; and he received for overtime work 12 X 1H = $9.36. The total amount that he received for the week was $24.96 + $9.36 = $34.32. The total number of hours worked was 48 + 12 =60. Therefore, his average, or mean, rate per hour for that week was $34.32 -i- 60 = $.572 = 57.2 cents. Ans. Example 2. — It was desired to measure very accurately the distance between two punch marks. The result of measurements by five different persons was as follows: 10 ft. 8.1 in.; 10 ft. 8.16 in.; 10 ft. 7.97 in.; 10 ft. 8.21 in.; 10 ft. 8.05 in. Which of these measurements is nearest to the mean, or average, of all the measurements? Solution. — The sum of all the measurements is 50 ft. 40.49 in. The number of measurements is 5; hence, the mean is 50 ft. 40.49 in. -i- 5 = 10 ft. 8.098 in. which is very nearly equal to the first measurement, and is nearer that than any of the others. The probable correct value is 10 ft. 8.1 in. Note that the 40 in. was not reduced to feet, because it is to be divided by 5. Example 3. — The floor area of a room in which 36 clerks are employed is 2960 square feet; what is the average number of square feet per clerk? Solution. — Evidently, the average number of square feet per clerk is 2960 -^ 36 = 82f , or a space about 9 ft. square for each clerk, since 9^ = 81. Ans. 10 ft. 8.1 in. 10 8.16 10 7.97 10 8.21 10 8.05 50 40.49 10 ft. 8 . 098 in, Ans. EXAMPLES (1) The daily production of a paper machine for one week was as follows: 97.6 T., 101.2 T., 98.5 T., 90 T., 103.1 T., 96.4 T.; what was the average daily production for the week ? Ans. 97.8 T. (2) The amount of sulphur used in making sulphite pulp was : In January, 149,721 lb. of sulphur for 508 tons of pulp; in February, 141,176 lb. for 476 T.; in March, 152,148 lb. for 519 T.; in April, 148,635 lb. for 493 T.; in May, 147,204 lb. for 527 T.; in June, 153,630 lb. for 468 T.; in July, 152,582 lb. for 483 T.; in August, 151,796 lb. for 479 T.; in September, 154,881 lb. for 492 T.; in October, 150,300 lb. for 512 T.; in November, 153,714 lb. for 483 T.; in December, 149,566 lb. for 506 T. What was the average amount of sulphur used per ton for the year? Ans. 303.625 — lb. per T. (3) Sold 1250 lb. of paper at 22^ per pound; 6700 lb. at 20f5; 10,0001b. at 18ff; 5500 lb. at 21)4; and 15,000 lb. at 17)4. What was the average price received per pound. Ans. 18.5176 — ^. AEITHMETIC (PART 3) EXAMINATION QUESTIONS (1) Eeferring to example 1, Art. 164, what is the diameter to the nearest 64th of an inch of a circle whose area is 218 7 '^- '^-p , . ^ , Ans. 16fi = 16H in. (2) Keferrmg to example 2, Art. 164, with what velocity will a ball of lead strike the ground if it fall from a height of 338 ft.? Ans. 147.45- ft. per sec. {6) Ihe area of a circle is proportional to the square of the diameter. The area of a circle whose diameter is 29| in. is 701 sq. in.; what is the diameter of a circle whose area is 500 sq. in. to the nearest 64th of an inch? Ans. 25^ « in! (4) The price paid for a certain dye was $1.36 per ounce. This represented a dealer's profit of 331%, a wholesaler's profit of 12i%, an importer's profit of 8|%, and a manufacturer's profit of 20%; what was the cost of manufacturing, after allowing 4 cents for handhng, packing, etc.? Give result to the nearest ^®^*: ^ , „, Ans. 66 cents per ounce. (5) A bill for rubber hose amounted to $235.40, hst price- dis- counts of 70, 30, 10 and 5% were allowed; (a) how much was actually paid to settle this bill? (6) what was the equivalent single discount? Ans. (L"^^'-''' ^«^ A 1 r , . ^ ^^) 82.045%. (6) A sample of coal shows 10.83% ash; if the weight of ashes obtained in one day is 1632 lb., actual weight, about how many long tons of coal were burned? Ans. 6 T. 1629 lb. = 6.727 T. (7) A firm desires to pay a special bonus of 2% on the com- missions earned by its salesmen when the sales are in excess of a certain fixed amount; but the 2% is to be computed on the sales (commissions) after deducting the special bonus. What is the actual bonus, expressed as a per cent? Ans. 1.96+%. 127 128 ARITHMETIC §1 (8) In a certain mill 35 men receive 42(/^ per hour, 64 receive IZi per hour, 15 men receive 87|?5 per hour, and 5 men receive $1.12| per hour; (a) what was the average wage per hour per man? (6) if they all received an increase of 12|%, what was the average wage per hour per man? I {a) 67.37^ per hr. I {h) 7b.7H per hr. (9) By making certain changes, a pulp miU increased its daily production from 88 tons to 95 tons. The total cost of operation increased 25% and the price was increased 12|%; what was (a) the profit per cent after the change, and (6) what was the gain or loss per cent in profits, if the profit when the daily production was 88 tons was 18%? a i ^^^ 14.65- % ^'^' I (6) 18.6+ % loss. (10) How many gallons are equivalent to 9.24 cu. ft.? Ans. 69.12 gal. (11) Add 5 yd. 2 ft. 7 in., 3 yd. 1 ft. 8 in., 4 yd. 9 in. and 3 yd. 2 ft. 10 in. Ans. 3 rd. 1 yd. 4 in. (12) The sum of the three angles of any plane triangle is 180°; if two of the angles of a certain triangle are 36° 14' 43" and 65° 27' 13", what is the other angle? Ans. 78° 18' 4". (13) Express 4.807 mi. in miles and lower denominations. Ans. 4 mi. 6 fur. 18 rd. 1 yd. 11.52 in. (14) Express 75° 18' 18" as a decimal part of 360°. Ans. 0.2091805|. (15) What is (a) the weight in ounces of 57 c.c. of water? (6) what is the equivalent volume in cubic inches. Ans I ^""^ 2-^^^^+ °'- I (6) 3.4784+ cu. in. (16) Divide 26 mi. 6 fur. 22 rd. 3 yd. 2 ft. 6 in. by 15. Ans. 1 mi. 6 fur. 12 rd. 2 ft. 11.6 in. (17) Reduce (a) 1 mi. 6 fur. 12 rd. 2 ft. 11.6 in. to inches; (6) express this number in yards. . f (a) 113,291.6 in. ^'''- \ (b) 3146M yd. SECTION 2 ELEMENTARY APPLIED MATHEMATICS (PART 1) MATHEMATICAL FORMULAS g DEFINITIONS 1. A mathematical formula, or, more simply, a formula, is an expression composed of ordinary arithmetical numbers and quantities indicated by letters, which shows at a glance what operations (addition, subtrac- tion, multiplication, division, {^ powers, and roots) are required to be performed in order to obtain a certain desired result. Roughly speaking, a formula is a short, concise expression of a Pj^ j rule, law, or principle. For instance, refer to Fig. 1, which represents a cylinder with a round hole throughout its entire length. The rule for finding the weight of this hollow cylinder may be stated as follows : Rule. — Multiply the sum of the diameters of the cylinder and hole hy their difference; multiply this product by the length of the cylinder, hy the weight in pounds of a cubic inch of the material of which the cylinder is composed, and by .7854, all measurements to he taken in inches. The final product will he the weight of the cylinder in pounds. To express this rule by a formula, let W = weight of cylinder in pounds; w = weight in pounds of a cubic inch of the material composing cylinder; 1 2 ELEMENTARY APPLIED MATHEMATICS §2 D = diameter of cylinder in inches; d = diameter of hole in inches ; I = length of cylinder in inches; then W = .7854wl{D + d){D - d) This last expression is a formula, and it shows at a glance just what operations are required in connection with the quantities tt), I, D, d, and the number .7854 in order to find the value of W, the weight of the cylinder. All that is necessary in order to use the formula is to substitute in it the values of the quantities represented by the letters. Thus, suppose the diameter of the cylinder is 12 in., diameter of hole is 8 in., length of cylinder is 30 in., and that it is composed of cast iron, a cubic inch of which weighs .2604 pounds; what is the weight of the cylinder? Here D — 12, d = S, I = 30, w = .2604; substituting these values for their corresponding letters in the formula, W = .7854 X .2604 X 30(12 + 8) (12 - 8) or, W = .7854 X .2604 X 30 X 20 X 4 = 490.8+ lb. 2. When two or more letters in a formula are written with no sign of addition or subtraction between them or when a letter follows a number, multiplication is understood; multiplication is also understood when a letter precedes or follows a sign of aggre- gation when there is no + or — sign between or when two dif- ferent signs of aggregation follow each other; therefore, .7854wZ (D -{- d)(D — d) has the same meaning as though written .7854 XwXlX {D -{- d) X (D - d). Obviously, the sign of multi- plication cannot be omitted between two numbers, since it would not then be possible to distinguish between the numbers. 3. Positive and Negative Quantities. — Before going further, it is necessary to make a distinction between quantities that indicate exactly opposite directions or meanings. For example, suppose two men start from the same position, one walking due north and the other due south; they are evidently walking in exactly opposite directions . To indicate this fact mathematically, one man is said to walk in a positive direction, while the other is said to walk in a negative direction. Suppose that north is taken as the positive direction; then if one man walks 6 miles due north and the other walks 6 miles due south, the first man is said to walk -f- 6 miles, and the second man is said to walk —6 miles. Positive quantities are always indicated by the plus sign and negative quantities by the minus sign. A negative quantity is §2 MATHEMATICAL FORMULAS 3 just as real as a positive quantity; it simply indicates a direction or meaning exactly opposite in character to that indicated by the positive quantity. A man's income may be considered as positive and his expendi- tures as negative. If going up hill be considered positive, going down hill will be negative. When the mercury in a thermometer is above zero, the reading is considered positive; when it is below zero, the reading is negative; hence, 65 degrees above zero is written + 65°, and 12 degrees below zero is written — 12°. If a push be considered positive, a pull will be negative. Many other instances may be cited; all that is necessary is that if any particular state, meaning, or direction be considered as positive, the state, meaning, or direction that is directly opposite in character will be negative. For instance, the direction in which the hands of a clock move, called clockwise, may be considered as positive; then if the hands are moved backward, or counter- clockwise, this direction will then be negative. 4. In connection with arithmetical problems, it is always pos- sible to ascertain which of two numbers is the larger; or, if several numbers are under consideration, it is always possible to arrange them in their relative order of magnitude. But when quantities are represented by letters, it is seldom possible to arrange them in this manner, and it frequently happens that the subtraction of a larger quantity from a smaller may be indicated in a formula. The distinction between positive and negative quantities becomes of great importance when the quantities are represented by letters. For this reason, it is necessary to know how to add, subtract, multiply, and divide positive and negative quantities that are represented by letters. When quantities are represented by letters, they are called literal quantities, to distinguish them from those represented by figures only or figures combined with a name (concrete numbers) , which are called numerical quantities. The numerical value of any quantity is its value when expressed by figures. Thus, in Art. 1, the numerical value of D was 12; of d, 8; of w, .2604, etc. 5. Coefficients and Exponents. — It was stated in Art. 2 that an expression like 5a means 5 X a. The number 5 which multi- plies a is called the coefficient of a; the coefficient of any literal quantity is always a multiplier of the quantity. In 33-^m, ISapq, 3(a + 76), etc., S}^, 18, 3, etc. are the coefficients of m, apq, (a + 76), etc. 4 ELEMENTARY APPLIED MATHEMATICS §2 It is frequently convenient to represent the coefficients by letters; thus, the area of a ciicle is expressed by the formula A = irr^ in which tt represents 3.141592+ (usually taken as 3.1416), and r represents the radius of the circle. Jn ISapq, 18o may be considered as the coefficient of pq, and 18ap may be considered as the coefficient of q. Here 18 is the numerical coefficient of apq, while a and ap are literal coefficients of pq and q, respectively. In general, however, the word coefficient refers only to the numerical coefficient. If no numerical coefficient is given, it is always understood to be 1. Thus, the coefficient of mn is +1, and mn = -\-lmn; the coefficient of —ay is — 1, and —ay=— lay. Note that when no sign is prefixed to a literal expression, it is always understood to be +; thus, 18 apq — -t- l^apq. The minus sign, however, is never omitted; hence, the numerical coefficient in —22>bx is —23. 6. Exponents have the same meaning in connection with literal quantities that they have when used with numbers. For instance, an"^ = a X n X n; -4:a^¥ = -4XaXaX6X6X6; 8xy = 8 X x^ X y^; etc. Note particularly that any quantity, whether numerical or literal, that has for an exponent is always equal to 1; thus, m" = 1, 56" = 1, etc. The reason for this will be explained in connection with division of literal quantities. Exponents may also be negative, in which case, the quantity affected with a negative exponent is always equal to 1 divided by the quantity when the exponent is the same, but positive; thus, x~^ = —i> — 15ay~^ = ^ — } etc. "When no sign is prefixed to X y a number or exponent, it is always understood to be +; hence, 2 = +2, aj2 = rc+^, etc. 1 divided by a number or quantity is called the reciprocal of that number or quantity; thus, the 1 . .1 reciprocal of 2.5 is t^-^; the reciprocal of x^ is —^> etc. Therefore, any quantity affected with a negative exponent is equal to the reciprocal of that quantity affected with an equal positive exponent. OPERATIONS WITH LITERAL QUANTITIES 7. Addition. — When several literal expressions are connected with one another by the signs + and — , their sum is always Understood. For instance, in the expression 27a^ — 54a^^ -f §2 MATHEMATICAL FORMULAS 5 36at/2 - Sy^, the different parts connected by the signs + and - are called terms. The first term, 27 a^ is understood to be -^27 a^, the second term is -54.a^y, etc. The expression is equivalent to +27 a' + (-54a2y) + SQay^ + {Sy'). 8. Like terms are those having the same literal quantities affected with the same exponents, regardless of the coefficients; thus, 5ay^ and 8ay^ are like terms, and -bm and llhm are also like terms. In the expression of Art. 7, no two of the terms are alike, whence they are called unlike terms. It might be thought at first that the second and third terms were alike, but they are not, since the letters do not have the same exponents. 9. Like terms may be added, but unlike terms cannot be added — the addition of unlike terms can only be indicated as above. I. When adding literal quantities, a slightly different meaning is given to the term addition from that employed in arithmetic, because all numbers used in arithmetic are positive. Referring to Fig. 2, suppose a man desires to walk toward the point B; o -IS + S +10 +15 5 +s +ro +i; A Fig. 2. then any movement that takes him in the direction to the right may be called positive or +, and any movement that takes him in the opposite direction or toward the left will be negative or — , regardless of the point started from. Suppose he starts from A and takes 6 steps toward B; the distance advanced is then +6 steps. If he now takes 9 steps more in the same direction, he will advance +9 steps. The total distance advanced toward B is evidently +15 steps; therefore, the sum of +6 and +9 is +15. This case corresponds to ordinary addition in arithmetic. II. Suppose, however, instead of walking toward B; he had walked toward C, starting from A, as before. If he takes first 6 steps and then 9 steps toward C, his first advance is - 6 steps and his second advance is -9 steps, and his total advance is - 15 steps; from which it is seen that -6 + (-9), or -6 —9, = - 15. That is, to find the sum of two negative quantities, add them and prefix the minus sign. III. Suppose he had walked first 6 steps toward B and then 9 steps toward C; he would evidently stop at -3, since counting 6 ELEMENTARY APPLIED MATHEMATICS §2 9 steps to the left from +6 makes him pause at —3. Therefore, +6-9 = -3 or -9 + 6 = -3. IV. Suppose he had walked 6 steps toward C and then 9 steps toward B; he would evidently stop at +3. Therefore, —6 + 9 = +3 or +9 - 6 = +3. From the foregoing, it is seen that when two numbers have the same sign, their sum is the sum of the two numbers prefixed by the common sign; but when the numbers have unlike signs, the sum is equal to the difference of the two numbers prefixed by the sign of the greater number. 10. To add two like terms having one or more literal quantities in them, all that is necessary is to add their coefficients and prefix to the sum the sign of the greater. For instance, lax + 4:ax = llax; —3ax + 7 ax = 4ax; 12mn — 19mn = —7mn; etc. If there are more than two like terms to be added, find the sum of those having positive and those having negative coefficients separately, and then add the two sums; thus, 3a — 8a — a + 6a — 11a + 4a = —7a, since 3a + 6a + 4a = 13a, —8a — a - 11a = -20a, and 13a - 20a - -7a. To show that the sum may be found by simply adding the coefficients, consider the expression, 7c + 4c = lie. Suppose c represents 5 inches; then 7c = 7 X 5 in. = 35 in., 4c = 4 X 5 in. = 20 in., and 35 in. + 20 in. = 55 in. But 7c + 4c = lie = 11 X 5 in. = 55 in. In other words, 7 of something plus 4 of something is equal to 7 + 4 or 1 1 of something, and it makes no difference what this something is, whether it is 1 in. or 5 in. or 105 in.; all that is necessary is that the terms be alike. 11. Subtraction. — Subtraction may be defined as the difference between two quantities; it may also be defined as that quantity which when added to the subtrahend will give the minuend for the sum. Thus, the difference between 15 and 6 is 9, and 9 is the number that will produce 15 for the sum when it is added to 6; here 6 is the subtrahend and 15 is the minuend. The second definition is the better one to use in connection with subtraction of literal quantities. Referring to Fig. 2, call movement toward B positive or + and movement toward C negative or — , as before. I. Suppose two men to start from A ; let one man walk 9 steps toward B, and the other 15 steps toward B; how far are they §2 MATHEMATICAL FORMULAS 7 apart? The distance between them is evidently 15 — (+9) = 6, since 9 + 6 = 15. 11. Suppose they had walked toward C the same number of steps; then the distance between them is — 15 —( — 9) = —6, since -9 + (-6) = -15. III. Suppose one of the men had walked 9 steps toward B and the other 15 steps toward C; then the distance between them is —15 — (+9) = —24, the —24 meaning that the man at +9 would have to take —24 steps to place him at —15. Also note that 9+ (-24) = -15. IV. Suppose that one of the men had walked 15 steps toward B and the other 9 steps toward C; then the distance between them is 15 — ( — 9) = 24, that is, the man at —9 would have to walk +24 steps to get to +15. Also note that — 9 + 24 = +15. It will be observed in the foiegoing four cases that the difference may be found in each case by changing the sign of the subtrahend and proceeding as in addition. Thus, in I, the sign of the subtra- hend is +; changing it to — and adding, 15 — 9 = +6. In II, -15 + 9 = -6; in III, -15 -9 = -24; and in IV, 15 + 9 = + 24. Therefore, to subtract one term from another, change the sign of the subtrahend and proceed as in addition. For example, 18ar - (7ar) = liar; 12bd - {22bd) = -lObd; Qm^ - (-Slw^) = +37w2; -22bV -(-346V) = + 12&V- 12. Multiplication. — The general rule for the signs in multipli- cation is : If the coefficients of two terms that are multiplied have like signs, the sign of the coefficient in the product will be -{-;if they have unlike signs (one + and the other —), the sign of the coefficient in the product will be —. Thus, +3 X +5 = +15, and —3 X —5 = +15; also -3 X +5 = -15, and +3 X -5 = -15. A little consideration will make this rule clear. Call a man's debts negative and his savings positive; then any tendency to increase the debts will be negative, and any tendency to in- crease the savings or to make the debts less will be positive. Now suppose the man saves 3 dollars each week for 5 weeks; at the end of 5 weeks, he will have saved 3 X 5 = 15 dollars. The 3 dollars and the 15 dollars are necessarily positive, re- presenting savings; the factor 5 is also positive, because it in- dicates that 3 dollars was saved 5 times. Hence, + 3 X + 5 = + 15. In the same way, + 3 X — 5 = — 15, because this 8 ELEMENTARY APPLIED MATHEMATICS §2 operation indicates that 3 dollars was spent 5 times, the total amount spent being 15 dollars. Also, — 3 X + 5 = — 15, because this operation indicates that a debt of 3 dollars was in- creased 5 times, and an increase must be considered as positive, the total increase in debt being 15 dollars. Finally, — 3 X — 5 = + 15; here a debt of 3 dollars is decreased 5 times, because, if + 5 indicates an increase, — 5 must indicate a decrease; hence, if a man's debt decreases, the value of his assets increase, that is, there is a positive change, which amounts to + 15 dollars. A similar line of reasoning may be applied to any similar case involving positive and negative, or opposite, changes or quantities. The product must contain all the literal quantities in both multiplicand and multiplier; thus, 66 X 9ay = 54:aby, the coeffi- cient of the product being equal to the product of the coefficients of the factors. Hence, —laH X 2ax^ = —lAa^axx^ = — 14 a^x*. In other words, if the same literal quantity occurs in both factors, it will have an exponent in the product equal to the sum of the exponents in the two factors. In the last case, the exponent of a in the product is equal to 2 + 1 = 3, and the exponent of x is equal to 1 + 3 = 4. This is always the case, whatever the exponents, and whether positive or negative. Thus, ch^ X c-%-2 = c-^z^, since 2-4= -2, and 5-2 = 3; also, 2.1d-''^ X 3.4(^2 = 7.14^2-79, since 2.1 X 3.4 = 7.14, and 2 + .79 = 2.79. 13. Division. — The general rule for signs in division is very similar to that for multiplication. If the dividend and divisor have like signs, the sign of the quotient is +; if they have unlike signs, the sign of the quotient is — . To find the quotient, divide the coefficient of the dividend by the coefficient of the divisor, and the result will be the coefficient of the quotient; to this annex the literal quantities in both dividend and divisor, but changing the signs of the exponents in the divisor, and then add the ex- ponents of like quantities. If any exponent then becomes 0, the value of the quantity having that exponent is 1, and it thus dis- appears from the quotient. For example, ASa^b^c -r- — 12a% = -462c, since 48 ^ -12 = -4, and a^¥ca-%-^ = a%H = Wc = ¥c. The proof for the law of signs in division is simple: As in arithmetic, the product of the divisor and quotient is the divi- dend (no remainder being considered) . Letting d represent the dividend, p, the divisior, and q the quotient, d ^ p X q. If cZ and p are both positive, q will also be positive, because -|- p §2 MATHEMATICAL FORMULAS 9 X -\- q = -\- pq = -{- d; also, if d and p are both negative, q will be positive, because — p X -\- q= - pq = - d. Further, if d is positive and p negative, q is negative, because -pX-q = -\-pq = + d; also if d is negative and p positive, q is negative, because +pX-q=-pq=-d. Therefore, if the signs of the dividend and divisor are aHke, the quotient is positive; if they are unHke, the quotient is negative. In practice, the exponents of the literal quantities in the divisor are subtracted from the exponents of like quantities in the divi- dend; this produces the same result as changing the signs of the exponents. Thus, in the last example, the exponent of a in the quotient is 2-2 = 0; of &, 3- 1 = 2; and as c does not occur in the divisor, it goes into the quotient unchanged. To prove that any number having for an exponent is equal to 1, let ?i represent any number or quantity, then n -^ n = 1; but n -i- n = n^~^ = n*\ Since the only way that can be obtained for an exponent is to divide a number or quantity by itself, it follows that any number or quantity having for an exponent is equal to 1. Similarly, n^ -^ n^ = n^~^ = w" = 1. 14. Operations with Pol3momials. — An expression consisting of but one term is called a monomial (the prefix mo is a contrac- tion of mon meaning one); an expression consisting of but two terms is called a binomial (hi means two); an expression consisting of more than two terms is called a polynomial {poly means many). Before performing anyof the operations of addition, subtraction, multiplication, or division, it is always advisable to arrange the polynomials according to the descending powers of one of the letters; this is done by writing first that term containing the highest power of the letter chosen, then the term containing the next highest power, etc. until all the terms have been written. For example, arrange 2x^y - 2axy^ + a;^ - ^/^ + x^y^ — aH^ according to descending powers of x. The term containing the highest power of x is x^, that containing the next highest power of X is —aH'^, etc. consequently, the polynomial arranged according to descending powers of x 1% x^ - aH^ + 2x^y + x'^y'^ - 2axy^ — 1/^. If it were desired to arrange the polynomial according to descending powers of y, the result would then be —y^- 2axif + x^y^ + 2xhj — aH'^ + x^. Note that in the first polynomial as arranged, the last term does not contain x, and in the second arrangement, the last two terms do not include y. 10 ELEMENTARY APPLIED MATHEMATICS §2 15. To add two or more polynomials, first arrange the one containing the most terms acording to the descending powers of one of the letters; then place under this the other polynomials, with like terms in the same columns, and add each column separately as in addition of monomials. Example 1. — Find the sum of a;^ — 3xy + y^ + x + y — 1, 2x^ + 4:xy — 3?/2 - 2x - 2y + 3, 3x^ - 5xy - Ay^ + 3x + 4:y - 2, and Qx^ + lOxy + 5y^ + X + y. Solution. — Arrange the first polynomial according to descending powers of X, and then place under it the other polynomials with like terms in the same columns. When adding polynomials, it is customary to begin at the left, instead of at the right, as in arithmetic, and this can be done because there is never an5rthing to carry from one column to the next. The sum of the coefficients in the first column is 12, and the first term of the sum is 12x^. x^ — 3xy + X -{- y^ + y — 1 2x2 + 4a.y _2x -3y^ -2y + 3 3x^ - 5xy + 3x - Ay^ + 4:y - 2 6x^ + lOxy + X + 5y^ + y 12a;2 + 6xy + 3x — y^ + 4y. Ans. The sum of the coefficients in the second column is 10 + 4—5 — 3 = 6, and the second term of the sum is Qxy. The sum of the coeffi- cients in the third column is 1+3 + 1— 2=3, and the third term of the sum is 3x. The sum of the coefficients in the fourth column is 5 + 1 — 4 — 3 = — 1, and the fourth term of the sum is —y^. The sum of the coefficients in the fifth column is 1+4 + 1— 2=4, and the fifth term of the sum is 4y. The sum of the numbers in the sixth column is 3 — 2 — 1=0, and the entire sum is 12^^ + 6xy + 3x — y^ + Ay. Example 2. — Find the sum of 4a — 5& + 3c — 2d, a + & — 4c + 5d, 3a — 7& + 6c + Ad, and a + 4& - c - 7d. 4a — 55 + 3c — 2d Solution. — Since the letters in all these polynom- a + 6 — 4c + 5d ials have the same exponents, 1, arrange them accord- 3a — 76 + 6c + 4d ing to the order of their letters, i.e., alphabetically, a + 46 — c — 7d The sum is then found in the same manner as in 9a - 76 + 4c. Ans. Example 1. 16. To subtract one polynomial from another, arrange the minuend according to descending powers of one of the letters, place the like terms of the subtrahend under it in the same columns, and subtract each term of the subtrahend from the term above it in the minuend, as in subtraction of monomials. If the subtra- hend contains a term not in the minuend, change its sign and write it in the remainder; and if the minuend contains a term not in the subtrahend, write it in the remainder with its sign unchanged. §2 MATHEMATICAL FORMULAS 11 Example 1. — From 4x^ - 2x^ + 3a;^ - 1 + 7a; subtract 6a; + 1 - a;* + 2x3 - 2x\ Solution. — Arranging the minuend according to descending powers of x, and writing the subtrahend under it with like terms in the same columns, begin at 3a;* + Ax^ — 2x2 + 7a; — 1 the left and subtract each term of the —X* + 2x^ — 2x2 _j_ Q-j. _(_ 1 subtrahend from that above it in the 4x* + 2x3 ^ X - 2. ^TOS.niinuend. Since 3 - ( - 1) = 4, the first term in the remainder (difference) is 4x*. The second term of the differ- ence is evidently 2x^, the third term is 0, the fourth term is x, and the coef- ficient of the fifth term is -l-(+l) = -l-l = -2. The final result obtained for the difference is 4x* + 2x^ -\- x — 2. Example 2.— From Sam^ — IZa^m^ + ISa^m^ - 29 subtract Saw^ + 4a2m* + 14a*TO2 - 24:a^m - 38. Solution. — The arrangement according to descending powers of m, and allowing for missing powers in both subtrahend and minuend is shown below. 8am5 - IZa^m^ + ISo^m^ - 29 SawS + 4a2m* + 14a*w2 - 24a% - 38 - 4a2m4 - IZa^m^ + Aa^m^ + 24:a^m + 9. Ans. As there is no term in the minuend over the second term of the subtrahend, change its sign and write it in the difference. There is no term in the subtrahend under the second term of the minuend; hence, bring it down into the difference as it stands. There is no term in the minuend over the fourth term of the subtrahend; hence, change its sign and bring it down into the difference. The rest of the work is evident, and the difference sought is -4a2m4 - IZa^m^ + 4a*m^ +24:a^m + 9. That the result as obtained is correct may be proved by adding the difference to the subtrahend, using the work as it stands; the sum is the minuend. 17. To multiply a polynomial by a monomial, simply multiply separately each term of the polynomial by the multiplier. It is customary to arrange the multiplicand according to the descend- ing powers of one of the letters and begin the multiplication with the term containing the highest power, i.e., begin at the left. Example. — Multiply 5x^ — 2ax^ + Ta^x — 14a3 by AaH^. Solution. — The multiplicand is already arranged according to the descending powers of x. Multiplying each term separately by Aa^x^, the 5x3 _ 2ax2 + 7a^x - Ua^ product is 2Qa^x^ - SaH^ + 28 4(i2^3 a*a;* — SGo^x^. The product is 20a2x« - 8a3x« + 28a%* - 56a«x3. Am. ^^^^^y fo^^d in this case without placmg the multiplier under the multiplicand, but this has been done in the margin to show how the work is arranged. Beginning at the left and taking each term of the multiplicand in succession, 5x3 x 40^x3 = 2QaH^, — 2ax^ X "^aH^ = - Sa^x^, etc. 18. To multiply a polynomial by a binomial or another polyno- mial, arrange both multiplicand and multiplier according to 12 ELEMENTARY APPLIED MATHEMATICS §2 descending powers of one of the letters, placing the first term of the multiplier under the first term of the multiplicand. Then multiply each term of the multiplicand by the first term of the multiplier; the result will be the first partial product. Multiply each term of the multiplicand by the second term of the multiplier; the result will be the second partial product, which is written under the first partial product with like terms under each other. Proceed in this manner until all the terms of the multiplier have been used. Now add the partial products, and the sum will be the entire product. Example 1.— Multiply 4a^ - Sab + &b^ by a - 36. Solution. — Multiplying the multiplicand by a, the product is ^a^ — 5a^b + 6ab^, which is written under the multiplier. Now multiplying by 4q2 _ 5^^ + 6&'' ~ ^^' *^^ product'of 4a2 and — 36 is a — Sb ~ 120^6, which is written in the same . 3 _ — _ 21 I ^~^ column as the — 5a% of the first par- ir, ,1. I ir 19 -I07., tial product. Then — 5ab X — 36 - 12a26 + 15a62 - 186' , _\„ , • , • •,, • ,, , = loab^, which is written in the col- 4a3 - 17a'b + 21ab^ - 186". Ans. ^mn containing 606^. 66^ X - 36 = — 186^, which is written to the right of the preceding product. Add- ing the two partial products, the entire product is 4a^ — 170^6 + 21ab^ — 186'. Example 2. — Multiply D + dhy D — d. Solution. — The work is shown in the margin, and should be evident without any special explanation. The entire pro- duct is D^ — d^. It is to be noted that D and D + d ^ are to be treated as though they were two different D — d letters. This result shows that the product of the D^ -{- Dd sum and difference of two numbers or quantities is — Dd — d^ equal to the difference of their squares, and the for- 2)2 — d^. Ans. niula in Art. 1 may be written ^ = .7854w;Z (Z)2 — d^). In the example that is given in connection with this formula, Z> = 12 and d = 8; D"^ = W- = 144 and ^2 = 8^ = 64; 144 - 64 = 80. But D + d = 12 + 8 = 20; D - d = 12 - 8 = 4; {D -\- d) (D - d) = 20 X 4 = 80, the same result in either case. Example 3.— Multiply x*+ 2x^ + x^ - Ax - U by 5x^ - 2x + 3. Solution. — The work is shown below and requires no special explanation. x^ + 2x^ + x^ - 4a; - 11 5x^ - 2a; + 3 5x6 + 10x5 + 5a;* - 20^' - 55x^ - 2x5 - 4x* - 2x' + 8x2 ^ 22x 3x* + 6x3 4, 3a;2 _ i2x - 33 5x6 ^ 8x5 + 4a;4 _ i5a;3 _ 443.2 _|. jqx - 33. Ans. §2 MATHEMATICAL FORMULAS 13 It will be observed here that when the multiplicand and multiplier are arranged according to the descending powers of some letter, each partial product begins one place farther to the right than the preceding partial product, and that all terms of the partial products follow in regular order. Example 4. — Suppose the multiplier in the last example had been x^ — 2x +3; what would the product have been? Solution. — The work is shown herewith. This example has been selected X* + 2x3 _^ a;2 - 4x - 11 x^ -2x +3 a;8 + 2x5 + X* - 4x3 _ iia;2 - 2x5 _ 4a;4 _ 2x^ + 8x2 ^ 22x 3x4 + 6x3 4, 3a;2 _ i2x - 33 x« + lOx - 33. Ans. to show how terms will sometimes disappear in multiplication, the product in this case being x^ + lOx — 33, the terms containing x^, x\ x^, and x^ having disappeared. 19. To divide a polynomial by a monomial, arrange the polyno- mial according to the descending powers of one of the letters and divide each term separately by the divisor. Thus, to divide 21a4&2 _ I2a%^ + 24:a%^ - d3a¥ by dah^, arrange the work in exactly the same manner as for short division in arithmetic. Sa¥) 21a^b'~ - Ua'^b^ + 24:a^¥ - 33ob^ 7a3 - Aa^b + 8a¥ -lib'. Ans. Beginning with the first term of the dividend and dividing it by 3ab^, the quotient is 7a^, which is the first term of the quotient sought. The other terms are found in the same way, the quotient being 7a^ - 4:a^b + Sah^ - llbK 20. To divide a polynomial by a binomial or by a polynomial, arrange the dividend and divisor according to descending powers of one of the letters, placing them in the same relative positions as for long division in arithmetic. Divide the first term of the dividend by the first term of the divisor, and the result will be the first term of the quotient; multiply the divisor by the first term of the quotient and subtract the product from the dividend. Divide the first term of the remainder just found by the first term of the divisor, and the result will be the second term of the quotient, which multiply into the divisor and subtract the prod- uct from the remainder first found. Proceed in this manner until a remainder of is obtained or until the first term of the last re- mainder is of loioer degree than the first term of the divisor, in which case, write the remainder as the numerator of a fraction whose denominator is the divisor. 14 ELEMENTARY APPLIED MATHEMATICS §2 Example 1. — Divide a^ — 2ab^ + b^ hy a — b. Solution. — Arranging the dividend and divisor according to the descend- ing powers of a, place the divisor to the right of the dividend, with a curved hne between, and draw a hne flS _ 2ab^ + &' (a — b under the divisor to separate it from o3 _ 025 a^ _^ ab - 62. Ans. *^^ quotient. Then, a^ -i- a = a\ ~ iK _ 2 ?)2 -1- 63 (A') *^^ ^^^^ term of the quotient. 27 _ ,2 Multiplying the divisor by a* and — subtracting the product from the div- — ao -t K ) idend, the remainder is shown at (^), JZ - — which is arranged according to de- (C*) scending powers of a. Then, a^b -ir a = ab, the second term of the quotient. Multiplying the divisor by ab and subtracting the product from the remainder (^), the new remainder is shown at (5). Then, —ab^ -^ a = —b^, the third term of the quotient. Multiplying the divisor by —6* and sub- tracting the product from the remainder at (B), the difference is 0, showing that the division is exact. The quotient sought is a^ + ab — b^. Example 2.— Divide 64^5 - 486aS by 2x - 3a. Solution. — The work is shown below and requires no special explanation. 64a;6 - 486o5 ( 2x - 3a 64a:5 - 96ax ^ 32a;4 + 48ax^ + 720^x2 + lOSa^x + 162a*. Ans. 96ax* - 486a5 96aa;4 - 144a2a;3 144a2a;3 - 486a5 144a2a;3 - 21Qa^x^ 21Qa^x^ - 486a« 21&a^x^ - 324:a^x 324a4x - 486as 324a*x - 486a'' That the quotient as obtained is correct may be proved by multi- plying it by the divisor; the product will be the dividend. Example 3.— Divide 9n^ + llcn^ + 17c^n by 3n^ + 4c. Solution. — To obtain the second term of the quotient, note that 9n* + llcn2 + 17c^n (3n^ + 4c ^!^1+12^ 3n2-|c + §f^i^. Ans. - cn^ + 17c^n 9«== + 12c — cre^ — |c^ 17c^n + fc" ^ Slc^r; + 4c2 ^ c'CSln + 4) 3to2 +4c ~ 9^2 + 12c ~ 9tc2 + 12c — cn^ -T- 3n^ = — ic. Note further that the exponent of n in the first term of the second remainder, 170%, is smaller than the exponent of n in the divisor; hence, the remainder is said to be of lower degree than the divisor, and the division ceases, the remainder being written as the numerator of a fraction whose denominator is the divisor. The numerator contains the fraction | ; to get rid of it, multiply both numerator and denominator of the §2 MATHEMATICAL FORMULAS 15 entire fraction by 3 (as shown in arithmetic, this does not alter the value of the fraction), and the result is -q-YaTTo"' ^* ^^ readily seen that Slc^n -f- 4c2 = c^{51n + 4), since if the parenthesis be removed and the terms within it are multiplied by c^, the product will be the binomial Slc^w + 4c2. c^(51n + 4) Therefore, the quotient is Sn* — |c H — q „ , . 2^ ' 21. Signs of Aggregation.' — The signs of aggregation are used much more freely in connection with literal quantities than with numerical quantities, the reason being that numerical quantities can be readily combined to form a single number, which is not possible with literal quantities, unless the terms are alike. The rule given in arithmetic for removing the signs of aggregation applies to both numerical and literal quantities. Thus, a + [b + (c— d)] = a -}- [b + c — d] = a -\- h -\- c — d. Here, as in arithmetic, when one sign of aggregation includes another, the inner one is removed first. This is particularly advisable when some of the signs are negative. Thus, a — [b — {c — d)] = a — [b — c-{-d] = a — b + c — d. Example 1. — Remove the signs of aggregation from 2x — {Sy — [4a; - (52/ -6a;)]}. Solution. — 2x — {3y — [4a; — (5y — 6a;)]} = 2x — {3y — [Ax - by + 6x]} =2x - {3y - [lOx - by]] = 2x - {3y - 10a; + 5y} = 2x - {Sy — lOxj = 2x — Sy + lOx = 12a; — 8?/. First remove the parenthesis; since the quantities enclosed by it are subtracted from 4a;, their signs must be changed, thus making the expression within the brackets 4a; — 5y + 6a;, which is equal to lOx — by. Next remove the brackets, obtaining for the expression within the brace 3?/ — lOx + by, which is equal to Sy — lOx. Now removing the brace, the original expression has reduced to 2x — 8y + 10a;, which is equal to 12a; — 8y. Ans. Example 2. — Remove the signs of aggregation from bm{{a -b)[a^ - 46(a2 - b^)]} Solution.— 5m{ (a - b)[a^ - 4:b{a^ - b^)]} = bm{ (a - b)[a^ - ia% + 463]} = bm{a^ - 4:a^b + Aab^ - a^b + Aa^'b^ - 46^ = bma^ - 20ma% ~bma% + 20ma%^ + 20mab^ — 20mb\ Ans. Having removed the paren- thesis, the polynomial within the brackets is multiplied by the binomial, a — b, and the brackets are removed. Now removing the brace and multiplying the last product by bm, clears the expression of all the signs of aggregation. 22. When two or more terms of an expression have a common factor, the terms may be included in a parenthesis or other sign of aggregation and the common factor placed outside as a multi- plier. For instance, the polynomial x^ + 2ax + a^ may be written x{x -\- 2a) -|- a^ or a;^ + a (2a; + a), if desired. When the 16 ELEMENTARY APPLIED MATHEMATICS §2 first term within the parenthesis has the minus sign before it, as in x^ — 2ax -{- a^ = x^ -\- a{ — 2x -\- a), it is desirable to have the first term positive, and this may be done in the present case by changing the order of terms, writing the expression x^ -\- a(a — 2x) . If, however, it is not desired to change the order of terms or if both terms are negative, place the minus sign outside the parenthesis and change the signs of all the terms within it; in such case, the last expression becomes x^ — a(2x — a). That this is correct will be clear when it is noted that the expression as it stands means that a(2x — a) is to be subtracted from x^, and in subtraction, the signs of the subtrahend are changed, thus making the expression x^ — 2ax + o,^, the original form when the paren- thesis is removed. Suppose it were desired to divide x^ — x^y + xy^ — y^ by x — y. This may be done in the regular way, but a somewhat easier method in this case is the following: x^ — xhj + xy^ —y^ = x^(x — y) -\- y^{x — y) = (x — y){x^ + y^), and this last ex- pression divided by x — y gives x^ + y^ for the quotient. That x^(x — y) -{- y^(x — y) = (x — y)(x^ + y^) is evident, because a; — y is a factor common to both terms, and if the parenthesis be removed from {x^ + y^), the expression reduces to the preceding one. If more than one of the signs of aggregation are used, the inner one is used first and' the terms enclosed by it are treated as a single term when employing the next sign. For example, 3m^ — 12m^n — 6mn^ — 18n^ = 3m^ — 12m^n — Qn^(m + Sri) =3m^ — 6n[2m^ + n(m + 3w)] = 3{m^ — 2n[2m^ + n(m + 3n)]}. Note that the signs of the terms within the parenthesis are not changed when the brackets are written, the entire expression, n(m + 3rz), being treated as though it were a single letter. 23. The Signs of a Fraction.^ — A fraction has three signs, one for the numerator, one for the denominator, and one for the entire fraction, the latter showing whether the fraction is to be added or subtracted. Thus, let a be the numerator and b the denominator, then the fraction may have one of the following . , -\-a a , —a a , -\-a a -{-a a forms: +^ = ^^ + +5 = "6^ + ^ = "6' " +6 = "6' — a a -\-a _a —a a ~ +6"'6' ~ '-b' ^b' ~ ^h^ ~b- To prove that these results are correct, apply the law that if any quantity be multiplied and divided by the same number or §2 MATHEMATICAL FORMULAS • 17 quantity its value is unchanged; thus, 3 X -1 = — 3, and —3 4- — 1 =3. Now remembering that to divide a fraction, its numerator may be divided or its denominator multipHed by the / —a\ . — a ^ — l __«_-, divisor, -IX [+^) -^ -I = ip^ - 5, -L = + — —,-, = +^=y: etc. Now since multiplying by ' +0 0' — 1 changes the sign of the fraction and dividing by — 1 changes the sign of the numerator or denominator, as the case may be, it follows that if two of the three signs of the fraction are changed, the value of the fraction is not altered; hut, if one sign only or all three are changed, the value is altered. Suppose the fraction has the form — ^2_a^ — ) *o ^^^^ *^^ sign before the first term of the numerator + , change the sign of each term of the numerator and the sign of the fraction, obtaining - ^ o f' - The same result might have been ob- ^ x^ — a^ tained by changing the signs of both numerator and denominator, obtaining ^ ?" !^ , but this changes the order of the terms in the ^ a^ — x^ denominator, which is not usually desirable. -- + 24c^ -c + 48c^ _ c-48c^ Similarly, 4^2 + 7^ _ c " Sn^ + 14w - 2c " Sn^ + 14n - 2c* Here, to get rid of the fraction in the numerator, both numer- ator and denominator are multiplied by 2 (this does not alter the value of the fraction); then the sign before the fraction and the signs of the terms in the numerator are changed. If the signs of the numerator and denominator are changed, instead of the sign of the numerator and the sign of the fraction, the result •u u c-48c^ _ c-48c^ will be _g^2 _ j4^ + 2c ~ 2c-14n-8n2- EXAMPLES (1) Find the sum of a^ - b'' + Sa^b - bah\ Za^ + 36^ - Zah^ - ^a% a.3 + 63 4. ^a% 2a^ - 46' - 5ab% Qa^b - Sa^ + lOob^, and 2b3 - Ta^ft + 4ob2. Ans. 4a2 + a% + ah^ +h\ (2) From Sm' - 2m^ - m -1 subtract 2m^ - Sm^ + m + 1. Ans. m^ + m^ — 2m — 8. 2 18 ELEMENTARY APPLIED MATHEMATICS §2 (3) Add a^ - Say + y^ + a + y - 1, Aay - 2a + 2a'' - 3y^ - 2y + S, 3a2 - 5ay - 4y^ + 4?/ - 2, and 5y^ + lOay + 6a^ + y + a. Ans. 12o2 + Gay - y^ + Ay. (4) Subtract a - h - 2{c - d) from 2(a - 6) - c + d. Ans. a — h -\- c — d. (5) Subtract a^ + a;^ — ax from Za^ — 2ax + x^. Ans. 2a^ — ax. (6) Multiply 2r3 + 4r-2 + 8r + 16 by 3r - 6. 4ns. 6r« - 96. (7) Multiply 4m2 — 3mn — n^ by 3m - 2n. Ans. 12m3 - 17m^n + 3mn^ + 2nK (8) Divide a^ - 6a^ + a^ + 4 by a^ - 1. Arts, a^ - 5a^ - 4. (9) Divide 32s5 + t^ by 2s + t. Ans. 16s^ - Ss^i + 4tsH^ - 2st^ + t\ (10) Remove the signs of aggregation from 7a - {3a - 2[4a - 3(5a - 2)]}. Ans. 12 - 18a. (11) Remove the signs of aggregation from 15 - 2y{x +y) {x- 2y) - Ay[2x - 3y(3 - x)] Ans. 16 — Sxy + 3Qy^ — lOxy'^ — 2xhj + 4?/^. (12) Eaclose within parenthesis the third and fourth terms, also the fifth and sixth terms, and then enclose all except the first term in brackets, of the expression 1 — 20w — 36TOn + \2n^ + Sm^n + Sre^. Ans. 1 - 4n[5 + 3(3m - n^) - 2{m^ + n)]. — 7 . (13) Change the signs of the fraction — -r^ without changing its value. , 7 Ans. — c + 1 (14) Change the signs of the fraction 2 _ o — ZTi without changing its , , 2a - 8 2a - 8 value. Ans. ; ^ r or, . , _ -.■ a^ — 2a — 4 ' 4 + 2a — a^ EQUATIONS 24. An equation is an expression of equality between two sets of quantities; thus 4^ = 64, (a + 6)^ = o^ + 2ab + b^, and a^ + 36 = 2a — 7 are equations. It will be noted that an equation consists of two parts, or members, which are separated by the sign of equality. The part on the left of the sign of equality is called the first member, and the part on the right is called the second member. In the first two equations given above, the second member is merely another way of writing the first member, the second member being obtained by performing the operations indicated in the first member. Equations of this kind are not true equations; they are called identical equations. The third equation is a true equation, and since the second member cannot be obtained by rewriting or performing any indicated operations in the first member, it is called an independent equation. Every formula such as was given in Art. 1, is an independent equation. §2 MATHEMATICAL FORMULAS 19 It is to be noted that in every equation, the two members are exactly equal, in the same sense that 2 = 2 or 3 + 4 = 7. Bearing this in mind, the following law will at once be evident: If the same quantity he added to or subtracted from both members, if both members be multiplied or both divided by the same quantity, or if both members be raised to the same power or the same root of both members be taken, the equality of the members is unaltered. For instance, take the identical equation 16 = 16; adding 4 to both members, 16 + 4 = 16 + 4, or 20 = 20; if 4 be-subtracted from both members, 16 - 4 = 16 - 4, or 12 = 12; if both members be multiplied by 4, 16 X 4 = 16 X 4, or 64 = 64; if both members be divided by 4, 16 -^ 4 = 16 -^ 4, or 4 = 4; if both members be squared, 16^ = 16^, or 256 = 256^ finally, if the square root of both members be taken -x/lQ = \/l6, or 4 = 4, What is true of an identical equation, insofar as the above law is concerned, is also true of any independent equation. For ex- ample, if x^ -{- 5x = 16 and 4 be subtracted from both members, the equation becomes x^ -\- 5x - 4: = 16 - 4 = 12; and if 16 be subtracted, the equation becomes x^ -{- 5x — IQ = 0, 25. Transformation of Equations. — To transform an equation is to make alterations in the members without destroying their equality. This is done by applying the law of Art. 24. Case I. — A term may be transferred from one member to the other by changing its sign; this is called transposition. Consider the equation 40 - 6a; -16= 120 - 14rc. Performing the opera- tions indicated in the first member, 24 - 6a; = 120 - 14a;; this operation is called collecting terms. Transposing the 14a; to the first member and the 24 to the second member, 14a; - 6a; = 120 - 24. Collecting terms, 8a; = 96. Dividing both members by S, X = 12. To prove that x = 12, substitute 12 for x in the original equation; if the two members are then equal, 8 is the correct value for x. Thus, 40 - 6 X 12 - 16 = 120 - 14 X 12, or 40 - 72 - 16 = 120 - 168, whence, - 48 = - 48, and the equation is said to be satisfied. To prove the rule for transposition, consider the second equa- tion above, 24 - 6.1; = 120 - 14a;. If 14a; be added to both members (which, by the law of equations, does not destroy the equality), the equation becomes 14a; + 24 - 6a; = 120 - 14a; + 14a; = 120. Note that as the result of this operation, the term 14a; has here been transferred (transposed) to the first member 20 ELEMENTARY APPLIED MATHEMATICS §2 and that its sign has been changed from — to +. Subtracting 24 from both members of the equation 14a: + 24 — 6a; = 120, the equation becomes, 14a; + 24 — 24 — 6a; = 120 — 24, or 14a; — 6a; = 96, from which, 8a; — 96. Note that as the result of this operation, the term 24 has been transposed to the second member and that its sign has been changed from + to — . Evidently, therefore, a term may be transposed from one member to the other by changing its sign. Having reduced the equation to 8a; = 96, both members may be divided by 8 without destroying the equaHty, according to the law of equations. X CO ^ Case II. — Find the value of a; in ^ ~ q ■^~ ^ ^ ^'^' '^^® ^^^* step is to clear the equation of fraction, which may be done by multiplying both members by each denominator in succession. Thus, multiplying first by 2, a; ^ ~^ 'k ~ ^^' multiplying 6a; by 3, 3a; — 2a; H — ^ = 264; multiplying by 5, after first com- bining 3a; and — 2.a;, 5a; + 6a; = 1320; dividing both members by 11, since 5a; + 6a; = 11a;, a; = 120. Substituting this value f • +V, • • 1 ,.120 120 , 120 . . , of X m the origmal equation, -^ ^ — I — r' =44; whence, 60 — 40 + 24 = 44, and the equation is satisfied. Multiplying by 2, 3, and 5 is the same as multiplying by 2 X 3 X 5 = 30. Therefore, instead of multiplying by these numbers separately, both members of the original equation might have 30 X a; been multiplied by 30, and the result would have been — ^ — 30 V r 30 V r ^+ ^-f-^ = 30 X 44; whence, 15a; - 10a; + 6a; = 1320, 1320 or 11a; = 1320, and x = -jr- = 120. Example. — ^Find the value of x in — ^- + ^ = 4 -^ Solution.— The product of the denominators is 2 X 3 X 4 = 24. Mul- . • 1 • u ,, , , „, 24(x + 3) , 24a; _ , 24(5 - x) . tiplymg both members by 24, ^ ■ -\ — ^ = 96 H r , smce T — = H 2; — Since all the numerators are now divisible by the denominators, perform the divisions, thus clearing the equation of fractions, and obtain 12{x + 3) + 8a; = 96 + 6(5 - x), or 12* + 36 + 8x = 96 90 + 30 — 6a;. Transposing and collecting terms, 26x = 90, or a; = ^ = 3x®^. Ans. §2 MATHEMATICAL FORMULAS 21 26. The subject of transformation of equations is of great importance in connection with formulas. For instance, consider the formula, .37wT p = V Suppose that it is desired to find another formula giving the value of w. Multiplying both members of this equation by v, pv = .S7wT, which may be written .37 Tw = pv. Dividing both members by .377^, pv ^ = :37r' a formula that can be used to obtain the value of w. jp As another example, transform the formula t = .53L^/-^ so it can be used to find the value of P. Dividing both members by t IP .53L, the equation becomes —^ry = ^/-^; squaring both members .o6L \ o t^ P P 3.56^2 1 o t:« AT > smce nr.nn = o.oo — . JNow .532L2 S .2809L2 L^ ''"^"^ .2809 multiplying both members by S, L^ ' A numerical factor may be transferred from the denominator to the numerator by multiplying the numerator by the reciprocal 171 of the factor. Thus, let a be the factor and — the fraction; then ' an - X m - Xm ^ — = z = Since -^^^t^ = 3.56 very nearly, aw 1 , _ n .2809 - XaXn a t' _ 3.56^^ t' ^ _±_ y i! ^ o rfi V ^ = .2809L2 ~ L2 ■ ^^' .2809L2 .2809 ^ L^ ^ L^ 3.56^2 27. Prime Marks and Subscripts. — In certain formulas, it is desirable to use the same letter for different measurements or quantities of the same kind. The formula given in Art. 1 is an instance. Here the outside diameter was represented by D and the inside diameter by d. Instead of using a capital letter for one diameter and a lower case letter for the other, the outside diameter might have been represented by d' and the inside 22 ELEMENTARY APPLIED MATHEMATICS §2 diameter by d". Had there been several other diameters used in the formula, they might be indicated by d'", d"" or ^^ ^ d!^ , etc. These marks are called prime marks, and they are read in connec- tion with the letters as follows: d prime, d second, d third, d fourth, etc. In connection with printing, these marks are called superior characters. When exponents are used in connection with letters affected with prime marks, the result is somewhat awkward (thus, d'^, d"^, d'"^, etc.) and it is then customary to use what are called sub- scripts (in printing, called inferior characters) , which are small figures of the same size as those used for exponents and placed slightly below and to the right of the letter; for example, d\ (read d sub-one), d^ (read d sub-two), dz (read d sub-three), etc. Exponents can be used with such letters very readily; thus, d\, d\, dl, etc. Such expressions are read d sub-one square, d sub-two cube, d sub-three fifth, etc. Always remember that letters having different subscripts or prime marks are just as different and are operated on in the same way as though different letters had been used instead. For instance, bd' + M" - 2d' = 3d' + 8d". 28. Application of Formulas. — To apply a formula, substitute in the right-hand member the values of the letters and then per- form the indicated operations. Example 1. — Given the formula / = Ir^ j- + stt ) jr t- find the = (r^ - value of I when r = 12.3 and h = 4.8. Solution. — Substituting in the formula the values of r and h, I = (12.3* 3 X 12.3 X 4.8 , 3 X 4.8^ \ 2 X 4.8 ^ ., oo aa oi, A.-i ^^a\ 4 + —20") 3X12.3 -4.8 = (151.29-44.28+3.456) X .29907 = 110.466 X .29907 = 33.037, very nearly. Ans. ^. , , •, o HA' + 2VA^" + 3A") ' ^ „ Example 2. — Given the formula o = , : — , nnd o 4:{A'.+ VA'A" +A') when h = 14.5, A' = 40.84, and A" = 30.63. Solution. — Substituting in the formula the values given. „ ^ 14.5(40.84 + 2V40.84 X 30.63 + 3 X 30.63) 4(40.84 + \/40.84 X 30.63 + 30.63) ^ 1 4.5(40.84 + 70.737 + 91.89) ^ 14.5 X 203.467 ^ _ . 4(40.84 + 35.3685 + 30.63) 4 X 106.8385 " ' PK Example 3. — Given the formula p = ^Mtr , t^ ' to find D when p = 56, ^ DN + K f ' P = 200, K = 700, and N =150. Solution. — First transform the equation so D will stand alone in the left-hand member. Multiplying both sides of the equation by DN + K, §2 MATHEMATICAL FORMULAS 23 -pND + pK = PK; transposing the term pK to the second member, pND — PK — pK; Dividing both members by pN , n PK-pK _ K{P - p) ^ ~ pN pN ' Now substituting in the formula for D the values of K, P, p, and N, _ 700(200 - 56) _ ^ - 56 X 150 - ^^- '^'''• + f^ + i_" = 7 ^ _ |\ +35. Ans. x = llxV- EXAMPLES (1) Find the value of x in the equation 13 - 5(3 + 4x) = 4x + 20 - 4(x + 20). Ans. x = 2.1 (2) Find the value of x in the equation X 2x , 7x 2"^ 3 (3) Find the value of x in the equation X ^„ ,„ , a(78 - 45a) 15a - 2ax + ^ = 26 - 17a:e. Ans. x = ^ — (4) When w' = 3, w" = 8.5, w'" = 3.6, s' = .0951, s" = 1, s'" = .1138 f' = t" = 60, and t'" = 840, find the value of t in the formula. ^ v^ + w"s^r^+w-s-t"' ^^^ ^ ^ 9,_,,3_^ w's' + w"s" + w"'s"' (5) When H = 178, Ta = 540, and Tc = 1210, find p from the formula p=ff(^^-^) ^^«- P = 1.343. (6) When s = }i{a + b + c), and a = 24.5, h = 37.8, c = 43.3, find r from the formula r = J (s -a)is -b){s -c) _ ^^^ ^ ^ 8.7394+. (7) Given the formula c = V2rh - h^, find c when r = 32| and h = llf . Ans. c = 24.955 -. (8) Given n = —, , find n when L = 1800, tt = 3.1416, d = 8, Vird^ + t^ 3 and ( = o2' ^^*- " ^ 126.94 -. QUADRATIC EQUATIONS 29. To solve an equation is to find the value of the quantity represented by some particular letter; this quantity is called the unknown quantity. In the examples given in Art. 25, the un- known quantity is x. What is termed the degree of an equation is determined by the highest exponent of the unknown quantity. In the equations so far given, the highest exponent of the unknown quantity was 1 ; 24 ELEMENTARY APPLIED MATHEMATICS §2 hence, these equations are said to be of the first degree. An equation of the first degree is also called a linear equation. When the highest exponent of the unknown quantity is 2, as in the equation 3x + Qx^ — 12 = 7(2 -{- x^), which reduces to 2x^ + 3x = 26 after the various transformations have been made, the equation is said to be of the second degree. An equation of the second degree is generally called a quadratic equation. The majority of equations that occur in practice reduce by transformation to linear or quadratic equations. Any linear equation may be represented by ax = b, (1) in which a and h may have any numerical value, and may be positive or negative. If a be negative, it may be rendered posi- tive by multiplying both sides of the equation by — 1 or, what is the same thing, dividing both sides (i.e., both members) by — 1. Thus, —ax = h divided by — 1 becomes ax = —b. Any quadratic equation will reduce to the form ax^ -\- bx = c, (2) in which a, b, and c may have any numerical value, and may be positive or negative. If a be negative, it may be rendered posi- tive by dividing both sides of the equation by — 1. Thus, —ax^ -\- bx = —c divided by — 1 becomes ax^ — bx = c. In equations (1) and (2), x is the unknown quantity whose value is to be found, and which may be represented by any letter whatever instead of x. For instance, the equation might be n^+ 8n = 33; here n is the unknown quantity, a = 1, 6 = 8, and c = 33. In the equation 3s^ — 13s = —14, the unknown quantity is s, a = 3, 6 = — 13, and c = — 14. 30. It is sometimes convenient to use what is called the double sign which is written + or + . The first character is read plus or minus, and the second character is read minus or plus, the name of the upper sign being pronounced first, both characters being a combination of the plus and minus signs. Either charac- ter really represents two signs and is treated as two signs, whence the name, double sign. Thus, 16 ± 5 = 16 + 5 = 21 or 16 — 5 = 11. Since +a X +a = + a^ and —aX—a= -}-a^, it follows that the square root of a^ may be either +a or —a; this fact is indicated by writing -y/a^ = ±a, consequently, 7 + -\/l6 = 11 and 7 — \/lQ = 3 may be represented by 7 + \/l6 = 11 or 3. §2 MATHEMATICAL FORMULAS 25 If the number whose square root is to be extracted is negative, the operation can only be indicated, since no negative number can be squared that will give a positive product. For this reason, such expressions as a/— 16, a/— c, etc. are called imaginary quantities. Since the square root of a negative quantity is in- dicated by V— «j it follows that the square of an imaginary quantity is a negative quantity; thus, (■%/— a)^ = —a, (\/— 16)^ = — 16, etc. 31. Roots of Equations. — Any value of the unknown quantity that satisfies the equation is called a root of the equation. An equation of the first degree, a linear equation, has but one root, while an equation of the second degree, a quadratic equation, has two roots. To solve a linear equation, reduce it to the form ax = h; then, dividing both members by a, a; = -j and the result thus obtained is the root of the equation. To solve a quadratic equation, reduce it to the form ax^ -\- hx = c; then substitute the values of a, h, and c, with their proper signs, in the formula ^ ^ -b ± \/b^ + 4ac 2a This formula gives two values for the unknown quantity; that is, the equation has two roots. To apply the formula, let the equation be Ss^ — 13s = — 14. Substituting 3 for a, — 13 for h, and — 14 for c, _ -(-13) ±V(-13)2+4 X3 X- 14 s — 2X3 13±Vl69 - 168 13 ± 1 = 24 or 2. 6 6 If either of these values be substituted for s in the original equation, it will satisfy the equation. Thus, since 2| = y, 3(1)2 - 13 X i = -^/ - -V- = - ¥- = - 14, and 3 X 2^ - 13 X2=12 — 26=— 14. Since the first member equals the second member in both cases, both of the values obtained for s are said to satisfy the equation, and the roots of the equation are 2f and 2. The above formula giving the value of x is so important that it should be thoroughly committed to memory. 26 ELEMENTARY APPLIED MATHEMATICS §2 7 Example 1. — Find the roots of the equation 2x —-z = 5(21 — x). Solution. — Removing the parenthesis and transposing to the first mem- ber the term containing x, 7x ^ = 105 X + 5 Clearing of fractions, 7x^ + 35x — 7 = 105x + 525 Transposing and collecting terms, 7a;2 — 70x = 532 Dividing by 7, x^ - lOx = 76 -(-10) + V102 +4X1 X 76 Substituting in the formula. 2 X 1 10 ± \/404 10 + 20.1 2 2 = 15.05 or -5.05. Ans. 7 Substituting 15.05 for x in the original equation, 2 X 15.05 — ^ - „_ . _ = 5(21 — 15.05). Performing the operations indicated, 29.75087+ = 29.75, which agrees close enough for practical purposes. Substituting —5.05 7 forx, 2 X —5.05 — _r nc i c = ^[21 — (—5.05)]. Performing the opera- tions indicated, 129.9 = 130.25, which also agrees close enough for most practical purposes. In the first case, the difference between the two values was 00.00087, and in the second case, 000.35. If more accurate results are desired, the square root of 404 must be found to a greater number of deicmal places. In almost every case, only the positive root is desired, and the value found is accurate enough. X 4x^ Example 2. — Find the value of x in 5 — 4 — x^ + 2x ^ = 45 — o o 3x2 4. ^^ SoLXJTioN. — Transposing the 4 to the right-hand member and the two terms containing x to the left-hand member, I + 2x2 - 2x - ^ = 49 (1) Clearing of fractions, 5x + 30x2 _ 30a; _ 12x2 = 735 Collecting terms, 18x2 - 25x = 735 « u +•++••+;, . 1 25 ± \/252 + 4 X 18 X 735 Substitutmg m the formula, x = ^ X lo _ 25 + 231.398- 36 = 7.1221+ or -5.7333-. Ans. Substituting these values in equation (1), which is practically the same as the original equation, Zi^|^ + 2 X 7.12212 - 2 X 7.1221 - ^ ^ 7^12212 = 49; performing the indicated operations, 48.999 = 49; difference = 00.001. Also, -^^333 ^ 2( -5.7333)2 _ 2 x -5.7333 - 4(-5-J333)2 ^ ^^, ^^^^ forming the indicated operations, 49.0004 = 49; difference = 00.0004. Both results are close enough for practical purposes. §2 MATHEMATICAL FORMULAS 27 It is always a good plan to substitute at least one of the roots for the unknown quantity in the original equation in order to make sure that no mistake has been made in the work. EXAMPLES Find the value of x in the following equations: (1) x^ - 16a; = -63. Ans. x = 9 or 7. (2) 3x2 -{- 8x = 56. Ans. x = 3.1882+ or -5.8549-. (.3) 7x2 ^ 20x = 32. Ans. X = U or -4. (4) x2 - §x = li Ans. X = 1| or -|. (5) x3 + (19 - x)3 = 1843. Ans. x = 11 or 8. (6) 24.3x2 - 65.28X = 427.486. Ans. x = 5.74732 or -3.06091. ACCURACY m CALCULATION 32. Significant Figures. — The significant figures of a number are those figures which begin with the first digit and end with the last digitJ For example, in the numbers 2304600 and .00023046, the significant figures are 23046. No attention is paid to the decimal point in connection with significant figures, and ciphers to the left of the first digit and to the right of the last digit are not considered. The significant part of a number is that part which is composed of its significant figures. In the numbers .005236, 52.36, and 5236000, the significant part is 5236, and these numbers all contain four significant figures. In practice, before any formula can be used, it is usually neces- sary to make one or more measurements. Thus, before the formula in Art. 1 can be used, it is necessary to measure the length and the inside and outside diameters of the hollow cylinder. If, however, it is assumed that the inside diameter is, say, 8 in., that the weight is, say, 490 lb., and it is desired to find the outside diameter, it may be readily calculated as follows : W = .7854m;Z (D + d) {D - d) = .7854 wl (D^ - d^), whence, ^ = ^S^l+^- ^/ ^xT604X30 + «' = "•««^^: in. In practice, this would probably be taken as 12 in., but in any case, a measurement would have to be made before this result could be applied. What is true in the case just cited is also true in practically every case that occurs in the application of formulas; measure- ments of some kind — weight, length, area, or volume — must be made either before or after the formula is used. 28 ELEMENTARY APPLIED MATHEMATICS §2 33. Accuracy in Measurements. — In commercial transactions and in ordinary calculations pertaining to engineering matters, measurements are seldom accurate to more than three significant figures. A measurement is said to be correct to n significant figures when, if expressed to n figures, the figures following the nth. figure being considered as a decimal, the difference between the two numbers is greater than —.5 and less than +.5. Thus, the number 3.141593 expressed to four significant figures is 3.142; expressed to three or five significant figures it is 3.14 or 3.1416. Regarding these several numbers as integers and subtracting the original number from them, 3142 — 3141.593 = +.407, which is less than +.5; 314 — 314.1593 = —.1593, which is greater than -.5; and 31416 - 31415.93 = +.07, which is less than +.5. Hence, 3.141593 correct to 3, 4, or 5 significant figures is 3.14, 3.142, or 3.1416. When a number is expressed to a certain number of significant figures, say n, and it is necessary to employ ciphers to make up the required number of n figures, then these ciphers are counted as significant figures. For instance, 4599.996 and 4600.003 when expressed to 6 significant figures are written 4600.00. For any measurement to be correct to n significant figm-es, it is necessary to know that the figures to the right of the nth figure form a number that is greater than —.5 and less than +.5. Suppose that 4 tons of coal (8000 lb.) are bought. The purchaser will not object if there is a pound less and the dealer will not object if there is a pound more; in fact, a difference of 5 or possibly 10 pounds would not be noticed. To obtain 4 significant figures correct in this transaction would require that the amount of coal delivered be not less than 7999.5 lb. and not more than 8000.5 lb. But the consideration of half a pound of coal would be ridiculous in connection with a weight of this amount. To be correct to 3 significant figures, the coal delivered must be not less than 7995 lb. nor more than 8005 lb., and if the amount actually delivered came within these limits, the transaction would be considered quite accurate. What is true of a heavy weight is also true of a light weight. Chemists have balances so accurate that when the proper pre- cautions are taken, a thousandth of a grain may be weighed; but these balances are suitable for weighing only very small amounts, and a weight of even a tenth of a pound might injure them; possibly a hundredth of a pound would be the greatest amount it §2 MATHEMATICAL FORMULAS 29 would be safe to use. Since a pound contains 7000 grains, a hundredth of a pound contains 70 grains, or 70,000 thousandths of a grain, that is, 5 significant figures. To make certain that the fifth figure was correct, it would be necessary that the balances detect a variation of half a thousandth of a grain, which would be beyond the limit of any except very special instruments ; therefore, the fifth figure could not ordinarily be determined, and the weigh- ing would be correct to only four significant figures. The same considerations govern any other kind of measure- ment, and the following law may therefore be laid down: measurements may he considered sufficiently accurate for practical purposes if correct to 3 significant figures; very accurate, if correct to 4 significant figures; and extremely accurate, necessitating special methods and instruments, if correct to 5 or more significant figures, 34. Accuracy in Numerical Operations. — Suppose it is desired to multiply 4343.944819 by 3.141593 and obtain the result correct to 5 significant figures. Using all the figures of both numbers, the work is shown at (a), and the product to 5 significant figures {a) (6) (c) 4342.944819 4342.94 + 4'3'4'2.'9'4 + 3.141593 3.14159 13028.82 3.14159 13028.834457 13028.82 434 2944819 434 29 4 434 29 173 71779276 173 71 76 173 72 4 342944819 4 34 294 4 34 2 1714724095 2 17 1470 2 17 39086503371 39 08646 39 13028834457 13643.73 68746 13643.73 13643.765042756667 | is 13644—. Now limiting both multiplicand and multiplier to one more significant figure than is required in the product, in this case, 5 + 1 = 6 figures, the work is shown at (6) , and the product to 5 significant figures is 13644 as before. Inspecting the work at (6) , it will be observed that if a line be drawn as shown cutting off all figures to the right of the first partial product, the result is 13643.73, which becomes 13644 when reduced to 5 significant figures. Observe that in the multiplication, the left-hand digits of the multiplicand and multiplier are placed under each other and the first partial product is found by multiplying the multiplicand by the first figure of the multiplier. The second partial product is then written one place to the right of the first partial product; the 30 ELEMENTARY APPLIED MATHEMATICS §2 third partial product is written one place to the right of the second partial product; etc. This brings the partial products in the same relative positions insofar as the order of the figures of the product is concerned as would be the case if the multiplying were begun with the right-hand figure of the multiple. The work at (c) is very much the same as at (6) , except that as soon as the first partial product has been obtained, the right- hand figure of the multiplicand is cut off; it is considered, however, in multiplying, in order to ascertain how much to carry. In finding the second partial product, there is nothing to carry in this case because 1 X 4+ = 4+. Having found the second partial product, cut off another figure in the multiplicand; then, since 4 X 9 = 36, call it 40 and carry 4 to the next product, obtaining 4X2 + 4 = 12; write 2 and carry 1, Then, 4X4 + 1 = 17; write 7 and carry 1. 4 X 3 + 1 = 13; write 3 and carry 1. 4 X 4 + 1 = 17, which write. Now cut off another figure, and 1 X 434 = 434, which write. Cut off another figure, and since 5 X 4 = 20, carry 2. Then, 5 X 3 + 2 = 17; write 7 and carry 1. 5X4+1 = 21, which write. Finally, cut off one more figure, and 9X3 = 27, which call 30 and carry 3. Then 9 X 4 + 3 = 39, which write. It will be observed that the first figure obtained in each partial product is written directly under the last figure of the first partial product. Adding the partial products as they stand, the sum is 13643.73; or 13644 when reduced to 5 significant figures. Now counting the number of figures at (a), it will be found that 109 figures were used; at (6), 64 figures were used; at (c), 44 figures were used. The result in all three cases is the same when the product is expressed to 5 significant figures. Not only is there a great saving in the number of figures employed, but there is very much less liability of making mistakes, and it is very much easier to check the work. In order, however, to perform the work as shown at (c), it is necessary to begin multiplying with the first (left-hand) figure of the multiplier. As another example, find the product of 230.2585093 and 15.70796 to 5 significant figures. The work is shown in the margin. Instead of reducing the numbers to 5 + 1 = 6 signifi- cant figures, they are written as they stand and the multiplicand is limited to 6 significant figures by cutting off the remaining figures. The first partial product is 2302.59, because the seventh figure is 5+. It is always best to locate the decimal point in 2302.59 1151 29 161 18 1 61 21 1 §2 MATHEMATICAL FORMULAS 31 the first partial product. To locate it in this case, note that if the 2'3'0 '2'5'8'5093 multiplier were 1.57+ , the first partial 15 70796 product would be 230.259; but the multi- plier is 10 times 1.57+ , and the first partial product is therefore 10 X 230.259 = 2302.59. After multiplying by 9 and obtaining 21 for the fifth product, cut off one more figure of the multiplicand, and og,g go since 6 X 2 = 12, write 1 for the sixth partial product. This method of multi- plying is slightly more accurate than that at (c) in the preceding example, when both factors were reduced to 5 + 1 = 6 figures. The product correct to 5 significant figures is 3616.9. Ans. 35. Division may be performed in a somewhat similar manner. Thus, to find the quotient of 3616.89 -h 230.2585093 to 5 significant figures, arrange the work as shown in the margin. Limit the divisor to 5 + 1 = 6 figures by cutting off all figures beyond the sixth, and then cut off another figure in the divisor every time another remainder is found. There will evidently be two integral places in the quotient, since 3616.890 -=- 230.258+ = 15.7+ . The quotient correct to 5 significant figures is 15.708. As another example, find the quotient of 13643.765 -^ 3.141593 to 5 significant figures. Limiting 13643. 7'6'5(3.'1'4'1'5'9'3 ,, .°. j j- -j j x 5 ^ the divisor and dividend to 6 12566 37 4342.94+ ^ u xt, ■,• -j j -n figures each, the dividend will not contain the divisor; hence, instead of cutting off another figure from the divisor, add another figure to the dividend, and then proceed with the division as shown. The quotient evidently contains 4 in- tegral places, since the quotient of 13643 -T- 3 will contain 4 integral places. The quotient correct to 5 significant figures is 4342.9. Ans. 3616 ,890( 59 '2'3'0. '2'5'8'5093 2302 15, .70795- 1314 30 1151 29 163 01 161 18 1 83 1^ 61 22 21 1 1 1077 39 942 48 134 91 125 66 9 25 6 28 2 97 2 83 14 12 2 32 ELEMENTARY APPLIED MATHEMATICS §2 36. Constants and Variables. — A large majority of the formulas used in practice contain one or more quantities that remain the same no matter what the conditions are that govern the problem. For instance, the formula in Art. 1 contains the quantity .7854. Now, no matter what the values of D, d, I, and w are, the quantity .7854 remains the same; for this reason, it is called a constant. The other quantities in this formula are called variables, because their values vary or change for different cylinders. For all cast- iron cylinders, w = .2604, and, hence, for all cast-iron cyUnders, w is also a constant, its value being .2604. It has been shown that for multiplication and division, the numbers iised may be limited to one more figure than the number of significant figures required in the result without loss of accuracy. What is true in this respect of multiplication and division is also true of addition, subtraction, powers, and roots. Consequently, in all practical applications of formulas, the number of significant figures used in all the quantities contained in the formulas may be limited to one more than is contained in the constant having the smallest number of significant figures, and the result should be limited to the same number of significant figures as this constant contains. For example, the formula A = p ■ o r.^ contains two constants, both having 3 significant figures; hence, the values of G and P may be limited to 3 + 1 = 4 significant figures, and the result when found should be expressed to 3 significant figures. The formula d = L54aD + 2.6 contains two constants, one of which contains but two significant figures; hence, the values of a and D may be limited to 3 significant figures and the value of d when found should be limited to 2 significant figures. 37. In the examples that follow, unless for some reason a special exception is made, aU applications of formulas will be made under the assumption that all the quantities, both constants and variables, have been limited to one more significant figure than the number of significant figures contained in the constant having the least number of significant figures; if there are no constants, then all quantities should ordinarily be limited to 5 significant figures, and the results will be expressed to 4 significant figures. Example. — In the formula s = .„,^y > suppose that W = 1200 pounds, Z = 108 inches, E = 1,500,000 pounds, and / = j^, in which 6 = 3 inches and d = 8 inches; find the value of s. §2 MATHEMATICAL FORMULAS 33 Solution. — In this formula, the constants 48 and 12 are exact, and are therefore correct to any nunaber of significant figures; the number 1,500,000 is correct to only 2 significant figures, and it is doubtful even if the second figure is correct. Consequently, it is useless to employ more than 3 signifi- cant figures in the calculation, 3X8^ First calculate the value of 7, obtaining I = — j^^ = 128. Substituting this value of / and the other values given in the above formula, 9 m P 27 27 _ im X 1083 ^ i^^ X 19^ X m ^ 6561 ^ ^ ~ ^^ X xmm X 128 mm x m 40000 ' ^ • • 4 15000 20000 3^ Ans. 2 The formula just given gives the value of s in inches, s being the deflection of a beam having a certain shape and under certain conditions of loading, etc. Therefore, the deflection in this case would be taken as .16 inch, and this value is as close as can be obtained, although it might be a trifle more or a trifle less. Note that instead of cubing the number 108, the cube was expressed as the product of three factors, in order to employ cancelation. APPROXIMATE METHOD FOR FINDING ROOTS 38. Cube and Fifth Roots of Numbers. — In certain formulas used in engineering, it is sometimes necessary to extract the cube root or, less frequently, the fifth root of a number. These roots may be found by an extension of the method explained in Arith- metic for finding the square root; but it entails a great amount of labor, and for practical purposes an approximate method answers the purpose equally well and is much easier to apply. Of the many approximate methods that have been recommended, the following is, perhaps, the simplest and most accurate. It was discovered by Charles Hutton, a famous English mathematician. Let n = the number whose root is to be found; let r = the index of the root, and let a be a number a little greater or a little less than the exact value of the root, so that a'' is a little greater or a little less than n. Then, r/- r(r + l)n + (r-lKl i /^ V w = 7 :r\ rn — , ^\ A ^^ nearly. (1) L(r — l)n + (r + ijoTA For cube root, r = 3, and formula (1) reduces to 3/- /2n + a\ .^. 34 ELEMENTARY APPLIED MATHEMATICS §2 For fifth root, r = 5, and formula (1) reduces to 5/- /3w + 2a\ The more nearly a*" approaches in value to n the more accurate will be the value obtained for the root. In order to save labor in finding a, the following table has been calculated; it gives the cubes and fifth powers of 1.1, 1.2, 1.3, etc. up to 9.9, andisusedas described below. Suppose it is desired to find the cube root of 34,586. The first step is to point off the number into periods of 3 figures each, 3 being the index, beginning with the decimal point and going to the right and left. If the number contains an integral part, point off that part only; but if it is a pure decimal, point it off to the right. Thus, the above number, when pointed off, becomes 34'586. If it had been .034586, it would become .034'586 when pointed off. The next step is to move the decimal point so that it will follow the first period that contains a digit, and the given number then becomes 34.586. The given number, after shifting the decimal point will be called the altered number. Of course, if the integral part of the given number contains not more than 3 figures, it is not necessary to shift the decimal point. Now referring to the table, and looking in the column headed n^, the number 34.586 is found to He between 32.768 = 3.23 ^nd 35.937 = 3.33; the cube root of 34.586 therefore lies between 3.2 and 3.3, and one of these two numbers is to be selected for a in applying formula (2) . To decide which one, find which of the two cubes just mentioned is nearest in value to the altered number. The easiest way to do this is to add the two numbers and divide the sum by 2 (this is the arithmetical mean of the two numbers). If the altered number is less than the arithmetical mean, use the smaller number, but if it is larger, or if it is near the arithmetical mean in value, use the larger number. In the present case, (32.768 + 35.937) -^ 2 = 34.3525; hence, use the larger number. Then a = 3.3 and a^ = 35.937. The work of applying the formula is shown in the margin. After performing the division, the quotient is 34.586 2 35.937 2 69.172 35.937 71.874 34.586 105.109 ( 106.460 95 814 .98731 9 2950 3.3 8 5168 2.96193 7782 296193 7452 3.258123 330 319 11 §2 MATHEMATICAL FORMULAS CUBES AND FIFTH POWERS 35 n n3 n6 n n3 ns 1.0 1.000 1.00000 5.5 166.375 5032.84375 1.1 1.331 1.61051 5.6 175.616 5507.31776 1,2 1.728 2.48832 5.7 185.193 6016.92057 1.3 2.197 3.71293 5.8 195.112 6563.56768 1.4 2.744 5.37824 5.9 205.379 7149.24299 1.5 3.375 7.59375 6.0 216.000 7776.00000 1.6 4.096 10.48576 6.1 226.981 8445.96301 1.7 4.913 14.19857 6.2 238.328 9161.32832 1.8 5.832 18.89568 6.3 250.047 9924.35643 1.9 6.859 24 . 76099 6.4 262.144 10737.41824 2.0 8.000 32.00000 6.5 274.625 11602.90625 2.1 9.261 40.84101 6.6 287.496 12523.32576 2.2 10.648 51.53632 6.7 300.763 13501.25107 2.3 12.167 64.36343 6.8 314.432 14539.33568 2.4 13.824 79 . 62624 6.9 328.509 15640.31349 2.5 15.625 97.65625 7.0 343.000 16807.00000 2.6 17.576 118.81376 7.1 367.911 18042.29351 2.7 19.683 143.48907 7.2 373.248 19349.17632 2.8 21.952 172.10368 7.3 389.017 20730.71593 2.9 24.389 205.11149 7.4 405.224 22190.06624 3.0 27.000 243.00000 7.5 421.875 23730.46875 3.1 29.791 286.29151 7.6 438.976 25355.25376 3.2 32.768 335.54432 7.7 456.633 27067.84157 3.3 35.937 391.35393 7.8 474.552 28871 . 74368 3.4 39.304 454.35424 7.9 493.039 30770.56399 3.5 42.875 525.21875 8.0 512.000 32768.00000 3.6 46.656 604.66176 8.1 531.441 34867.84401 3.7 50.653 693.43957 8.2 551.368 37073.98432 3.8 54.872 792.35168 8.3 571 . 787 39390.40643 3.9 59.319 902.24199 8.4 592.704 41821.19424 4.0 64.000 1024.00000 8.5 614.125 44370.53125 4.1 68.921 1158.56201 8.6 636.056 47042.70176 4.2 74.088 1306.91232 8.7 658.503 49842 . 09207 4.3 79.507 1470.08443 8.8 681 . 472 52773.19168 4.4 85.184 1649.16224 8.9 704 . 969 55840.59449 4.5 91.125 1845.28125 9.0 729.000 59049.00000 4.6 97.336 2059.62976 9.1 753.571 62403.21451 4.7 103.823 2293.45007 9.2 778.688 65908.15232 4.8 110.592 2548.03968 9.3 804.357 69568.83693 4.9 117.649 2824.75249 9.4 830.584 73390.40224 5.0 125.000 3125.00000 9.5 857.375 77378.09375 5.1 132.651 3450.25251 9.6 884.736 81537.26976 5.2 140.608 3802.04032 9.7 912.673 85873.40257 5.3 148.877 4181.95493 9.8 941.192 90392.07968 5.4 157.464 4591 . 65024 9.9 970.299 95099 . 00499 5.5 166.375 5032.84375 10.0 1000.000 100000.00000 36 ELEMENTARY APPLIED MATHEMATICS §2 multiplied by a = 3.3, and the product is the cube root of 34.586, approximately. Since the decimal point was shifted one period to the left to form the altered number, it must be shifted one place to the right in the root, and v^ 34,586 = 32.58123, approximately. The root to 8 significant figures is 32.581178; therefore, the root as found by the formula was correct to six significant figures. 38. The root when calculated as just described will always be correct to at least five significant figures, except when the altered number is less than 1.331 = l.P and is nearly equal to the arith- metical mean of 1 and 1.331. This case will be discussed in Art. 39. If for any reason, more figures are desired than can be obtained by proceeding as above, express the root as found to 3 or 4 signifi- cant figures, and substitute it for a in this formula. Thus, in the preceding example, the root to 4 figures is 32.58; substituting this f • f 1 ^o^ V^T^^ / 2 X 34586 + 32.58^ \ ,, ,, for a m formula (2), V34,586 = \ 34536 -|- 2 X 32 58V = 32.581177737944— ; the root correct to 14 significant figures is 32.581177737942—; hence, the root as calculated was correct to 13 significant figures. The reason for shifting the decimal point to get the altered number is obvious — to make it easier to use the table. Since ■^34586 = ^1000 X 34.586 = 10^34.586, the decimal point must be moved one place to the right after the root of the altered number has been found. Similarly, -v^. 034586 = \/"17)nn" = TTrV^34.586; consequently, if the decimal point is shifted one period to the right to form the altered number, it must be shifted one place to the left in the root. 39. When the altered number is less than 1.2, the fifth figure of the root as calculated by the foregoing method will usually be incorrect. In such cases, proceed as follows: Suppose the cube root of .001166 be desired. Pointing off, the result is .001 '166. Shifting the decimal point one period to the right, the altered number is 1.166, which is less than 1.2. Now divide the decimal part by the index of the root, and the quotient is .166 -^ 3 = .055+, which is nearly equal to the decimal part of the cube root of 1.166, the integral part being 1; that is, v^l.166 = 1.055, nearly. Substituting this value for a in formula (2), v'Hee = P ^.}''!^l t ! n^S ) 1-055 = .997655025 x 1.055 = V1.I66 + 2 X 1.0553/ 1.052526051. . §2 MATHEMATICAL FORMULAS 37 The root correct to 12 significant figures is 1.05252604197. Hence, ^.001166 = .1052526+, by the formula. 40. The fifth root of a number is found in exactly the same manner, using formula (3). The only difference in the process is that the number must be pointed off into periods of five figures each, because the index of the root is 5. As an example, find the fifth root of 214.83. Here the integral part of the number contains only 3 figures, and it is not necessary to point off the number. Referring to the table, the given number falls between 2.9^ = 205.11149 and 3.0^ = 243; the arithmetical mean of these two numbers is 224.055745, and as this is greater than the given number, use 2.9 for a in formula (3). Substituting in the formula, 5/xrj^ ^ /3 X 214.83 + 2 X 205.11149\ '^^^'^•^^ \2 X 214.83 + 3 X 20531149/ 214.83 214.83 3 2 644.49 429.66 205.11149 205.11149 2 3 410.22298 644.49 1054.71298 1044 99447 615.33447 429.66 (1044.99447 1.00930006 9 71851 2.9 9 40495 31356 2.01860012 908370054 31350 2.926970174 2.9 = 2.92697+. Ans. The whole calculation is shown in the margin. It was really not necessary to use BO many decimal places, but since but very little labor would be saved by using a smaller number, it was not considered worth while to abbreviate the process further. In general, however, only one more signifi- cant figure would be used than was desired in the root. 6 As in the case of cube root, the method will give at least five significant figures correct, except when the altered number lies between 1^ = 1 and 1.1^ = 1.61051 and is near the arithmetical mean of these two numbers. In such cases, proceed as described in connection with cube root, dividing the decimal part of the altered number by the index, in this case 5, to find the decimal part of the root. For example, find the fifth root of 1.308375. Here .308375 -^ 5 = .061675, and -y^l. 308375 = 1.06 nearly. Substituting 1.06 for a in formula (3), .VT^nsW^- / 3Xl.308 3 75+2Xl.06n Vl.dUbd/i) \,2X 1.308375+3X1.06=;' 1 .06 = 1.055197 -, say 1.0552. Ans. 38 ELEMENTARY APPLIED MATHEMATICS §2 The root correct to 9 significant figures is 1.05522833. If more figures of the root are desired, substitute 1.055 for a. 41. The fourth root is very seldom required. In case it should be necessary to find the fourth root, all that need be done is to extract the square root and then exti'act the square root of the result; in other words, -x/n = S-y/n-. For instance, v^97.34 = N/ a/97.34 = ^9.8661 = 3.1410 +• Here the square root of 97.34 = 9.86610+ , and the square root of 9.8661 = 3.1410+ . 42. If the table of cubes and fifth powers is not available, find the first figure of the root by trial; then substitute this value for a in formula (2) or (3) and calculate the root to three figures, calling the result h. Express b to two figures, and substitute it for a in formula (2) or (3) ; the remainder of the work is the same as before, the value of a being calculated instead of being taken from the table. Referring to the example of Art 37, the altered number is 34.586. To find the first figure of the root, note that 3^ = 27 27 -j- 64 and 4^ = 64; the mean of these two powers is ^ = 45.5; hence, use 3 for the first figure of the root. Then, a = 3 and a3 = 33 = 27, and -.y^T^eft - / 2 X 34.586 + 27 \ „ _. ^^^•^^^ - ( 34.586 + 2 X 27 ) = ^-2^- = ^- Expressing this root to two figures, b = 3.3. Substituting this value of 6 for a. in the formula, 3.3^ = 35.937, and It will be observed that insofar as the first five significant figures are concerned, either 3.2 or 3.3 may be substituted for a in the formula; but 3.3 is a slightly better value in this case, since it gives the root correct to 6 figures, while 3.2 gives only 5 figures correct. The same procedure will be followed in the case of fifth roots; thus, referring to Art. 40, the altered number is 214.83. Here 3^ = 243 and 2^ = 32; obviously, 3 is the proper number to sub- stitute for a in formula (3), and Expressing b to two figures, the root is 2.9, which is the same as the value of a used in Art. 40. §2 MATHEMATICAL FORMULAS 39 This method may be appHed to the special cases of Arts. 39 and 40. Thus, to find the value of -v^l.166, the first figure of the root is obviously 1, and L .^^'1 o v 1 ) ^ ^ 1-0524. Now when the altered number does not differ greatly from a^, the value of b may be calculated to 4 or 5 significant figures, and h may be expressed to 3 or even 4 figures, if desired, before sub- stituting for a in the formula; but, unless great accuracy is desired, it is best to express 6 to 3 figures, to save labor in calcu- lation. In the present case, substitute 1.05 for a in the formula. Similarly, in Art. 40, the altered number is 1.308375, and the first figure of the root is 1. Hence, 5/T^7^^5^ / 3 X 1.308375 + 2 X 1 \ . , „_.„ ^1^08^75 = ( 2 X 1.308375 + 3 X 1 ) ^ = ^'^^^^^ In this case, 1.05 or 1.055 may be substituted for a. If desired, this method may be used for finding the square root of numbers instead of employing the exact method de- scribed in Arithmetic. For square root, r = 2, and formula (1) of Art. 36 becomes ^3n + a^\ {n + SaV As an example, find the square root of 3.1416. Here a is evidently 2, and ( 3 1415 _l 3 v^ 4 ) X 2 = 1.773+ . Using the first three figures for a, V3.1416 = 3 ^^^g + 3 x 1 77^ ^ 1.77245594 — . By the exact method, the root to 9 significant figures is 1.77245592; hence, the root as found by the formula was correct to 8 significant figures. The exact method is so easy to apply, that the reader is ad- vised to use it in preference to the formula. /- /3n + a\ Vn = (;^^r^) EXAMPLES Calculate the roots in the following examples to 5 significant figures: (1) v^.04608. Ans. .35851+. (2) ^.86402. Ans. .97119+. (3) v^l.0947. Ans. 1.0306 +. (4) -v/324, 096,815. Ans. 686.90-. (5) -v/4, 063,972. Ans. 20.979+. (6) v^3.1416. Ans. 1.4646 -. ELEMENTAEY APPLIED MATHEMATICS (PART 1) EXAMINATION QUESTIONS (1) Explain the difference between a coefficient and an ex- ponent. (&) What is meant by the reciprocal of a number or 2 , c quantity? (c) What is the reciprocal of a — - ? Ans. _ ^ . (2) Add a — b,b — c, c — d, and d — c. Let a = 20,h = 16, c = 11, and d = 8 and prove the result obtained was correct. Ans. a — c. (3) Find the sum of: a^ - Zax^ - daz^ + xh"" + z^ dax^ - 4a3 + 2a;22 + Qaz^ - x\ 2,a^ + 2a^z^ - 2xh^ + Zah - 5az^, 4:X^ - 6aH - Sa^ - Sz^ + xH\ Ans. Sx^ - ^aH + 2xz^ + Sa^g + 2aH^ - 2az^ - a^ - 2a^ - 2z\ (4) Subtract fa + 36 + io; + ^ax from |a - 76 - 3aa; + ix. Ans. a + ix — iax — 112. (5) Multiply a^ - lOa^b + 40a362 _ gOa^&s _|_ soa?)^ - 326 ^ by a — 26. ^ns. a^ - 12a56 + 60d%'" - lQOa%^ + 2400^6* - 192a65 + 6466. (6) By performing the indicated operations, reduce the follow- ing expression to a simpler form : a2(3a - 5a;) - (2a - x^ia - 2x) + (a^ -\- ax + x^)ia - x) - x\ Ans. ax {7 a — 9x). (7) Divide x^ - Ax^ - Slrc^ - 6x + 120 by a; + 5. Ans. x^ - 9x^ - 6a; + 24. . , P • 3 - 11a; .,, , (8) Change the signs of the fraction ~ ^ i g^ _ q without 3 - 11a; changing the value of the fraction. Ans. g _ g^ _ ^^2 (9) Given the formula v ^ Vo - ^at^, rewrite it so it will ex- press (a) the value of a, and (6) the value of t. 2{vo - v) a= ^, ins. l2{vo - v) 41 12 {vo - \ a 42 ELEMENTARY APPLIED MATHEMATICS §2 (10) Given the formula 23 1 ,00155 _ n s /- ''".552l(23 + ^5l55V^''^'' \ s / r find the value of v when n = .013, s = .005, and r = .559, Ans. V = 3.6+. (11) Remove the signs of aggregation from (27 - 4:x)[40 - 5(3a: + 7)] + 6{24 - (x + 2)(x - 4)[3 - 5x(4 + a;)]} Ans. dOx^ + 60a;3 - 438a;2 - 1349a; + 423. (11) Divide 40x' - 700;^ + Ax^ + 724^3 + 88a;2 _ loSOrc - 3390 by 5x^ - 20a; + 48. Ans. 8x' + 18a;3 - 4a;2 - 44a: - 120 - J'l^\~ ^f '^?q - 5x^ — 20a; + 48 (12) Find the value of y in the equation. o _L 4?/ y c /% \ ^^ + T - 2 = My - ^ + ^j Ans. 7/ = 4.0223+ (13) Find the two values of x in the equation lla:^ — 35a; = 40 Ans. x = 4.0743+ or -.89251- (14) What is the cube root of 705.33> Ans. 8.9015+ (15) What is the fifth root of .76054? Ans. .94673- (16) What is the value of \/26.252 + 17.52? Ans. 31.559+ ELEMENTARY APPLIED MATHEMATICS (PART II) MENSURATION OF PLANE FIGURES DEFINITIONS 43. Mensuration deals with the measurement of the length of lines, the area of surfaces, and the volume of solids; its principles and rules are used in every occupation and industry. The subject is, therefore, of the greatest importance. Every material object occupies space in three directions — in length, in breadth or width, and in thickness or depth. Every magnitude or body is consequently said to have three dimensions — length, breadth, and thickness. 44. A mathematical line indicates direction only; it has only one dimension, length, and has no breadth or thickness. Such a line could not be seen; hence, every visible Hne, no matter how fine it may be, has breadth, but when considered in connection with problems relating to mensuration, the breadth of all lines is disregarded and they are conceived as having only length. 45. A straight Line is one that extends continuously in one unvarying direction. In mathematics, straight lines are assumed to be infinite in length; that is they are assumed to be capable "^ ~ ~ of being extended in either direction without limit. Fig. 3 shows a straight hne, and it is assumed that it can be extended to the right or to the left as far as is desired — one foot, a mile, a hundred thousand miUion miles or farther — without any change in its direction. In practice, only short parts, called segments, of straight lines are con- sidered; and to distinguish one line from another, letters or figures are placed at the ends of the segments. Thus, in Fig. 3, 43 44 ELEMENTARY APPLIED MATHEMATICS §2 the line there shown is called the line AB, when it is considered as extending from A to B, or the line BA, when it is considered as extending from B to A. In mathematics, a straight hne is called a right line, and wiU usually be so designated hereafter. Another definition of a right hne is : a right hne is the shortest path between two points. This definition is obvious, since if the path deviates, even in the sHghtest degree, from a point to another point, the path (hne) will be longer than if it extended straight from one point to the other. 46. A broken line is one that is made up of two or more right line segments; see Fig. 4. A broken Hne is distinguished by placing a letter (or figure) at the ends of each segment. Thus, the broken line in Fig. 4 is called the line ahcdef. A Fig. 4. 47. A curved line or curve is one that has no right hne seg- ment; its direction changes continually throughout its entire length. See Fig. 5. A curve may be finite (hmited) in length or infinite (unhmited) in length, but in either case, no part of it is ever straight. An example of a finite curve is a circle. Fig. 5, and an example of an infinite curve is a parabola, see (a), Fig. 6. Fig. 6. When only a portion of a curve is considered, it is called an arc, the word arc meaning bow, alluding to its shape. Curves are distinguished by placing letters at the ends of the arc and at such other points between as may be deemed advisable. Thus, in Fig. 5, the circle may be designated as ABCD; in (a), §2 MENSURATION OF PLANE FIGURES 45 Fig. 6, the curve may be called abc; in (6), Fig. 6, the curve may be referred to as 12345. 48. When two lines cross or cut each other, they are said to intersect, and the place where they intersect is called the point ot intersection. Thus, in Fig. 4, h, c, d, and e are points of in- tersection of the hnes ah and ch, of he and dc, etc., assuming that these segments of right hnes are prolonged (produced); in Fig. 5, the points A, B,C, and B are the points of intersection of the right hnes AC and DB with the circle. 49. In mathematics, a point indicates position only; it has no dimensions. In practice, a point resembles a very small circle — • more properly, a square — bub is always considered to be without dimension. A point may always be conceived as being the point of intersection of two lines. 50. Parallel lines are right lines that are always the same distance from each other; ^ they never intersect no mat- ^ ^ j^ — __- jj ter to what length they may p^^ ^ be produced. Thus, in Fig. 7, the right hnes AB and CD are parallel. 51. An angle is the difference in direction between two right hnes that intersect. The point of intersection is called the vertex of the angle, and the lines are called the sides. Angles are designated by placing a letter at the vertex and by two other letters, one on each line. In Fig. 8, the right lines AC and DB, inter- secting in the point 0, form four angles, designated as AOB, BOC, COD, and DO A. Two angles on the same side of a hne (which forms one of their sides) and separated by a common side are called adjacent angles. In Fig. 8, AOB and AOD are adjacent angles; AOB and BOC are also adjacent angles; etc. Note that adjacent angles have a common vertex, a common side, and the other side of both angles lies in the same right line. 52. When two right hnes intersect in such a manner that the adjacent angles are equal, the hnes are said to be perpendicular to each other. For example, in Fig. 9, CD has been so drawn 46 ELEMENTARY APPLIED MATHEMATICS §2 FiQ. 9. that all the adjacent angles, COB and CO A, COB and BOD, etc. are equal; hence, AB is said to be perpendicular to CD, and CD fj is said to be perpendicular to AB. All four angles, AOC, COB, BOD, and DO A are equal, and each is called a right angle. _^ 53. A horizontal line is one that is parallel to the horizon or to the water level; with the book held in its usual position for reading, AB, Fig. 9, is a horizontal line. Any line that is perpen- dicular to a horizontal line is a vertical line. In Fig. 9, CD is a vertical line. All vertical Unes have the same direction as a plumb line. 54. An angle that is smaller than a right angle is called an acute angle. In Fig. 10, the angle AOB is evidently smaller than the right angle COB; hence AOB is an acute angle. If an angle is greater than a right angle, it is called an obtuse angle. In Fig. 11, AOB is an obtuse angle, because it is greater than the right angle COB. In Fig. 8, AOB and COD are acute angles, and AOD and BOC are obtuse angles. 65. Angles are measured in several ways, the most common method being to use an arc of a circle. This method is called the angular measure of angles, and the table of angular measure was given in Arithmetic. The entire circle is supposed to be divided into 360 equal parts called degrees, the abbreviation for which is (°) ; each degree is then divided into 60 equal parts called minutes, the abbreviation for which is (') ; and each minute is divided into 60 equal parts called seconds, abbreviation (")• The draftsman measures an angle by means of an instrument, usually made of metal or paper, called a protractor, which is a half circle divided into degrees and half degrees or degrees and quarter degrees. Evidently, the length of the sides of an angle have nothing to do with the size of the angle, since lengthening or shortening the sides does not change the direction of the lines. Consequently, by laying the protractor on the angle in such a man- ner that the center of the half circle coincides with the vertex of -B Fig. 10. Fig. 11. §2 MENSURATION OF PLANE FIGURES 47 the angle and the bottom, or straight, side of the protractor coincides with one side of the angle, the other side will cross the arc of the half circle and the size of the angle can be read on the graduated edge. This is clearly shown in Fig. 12. Here the pro- tractor P is so placed that its center coincides with the vertex of the angle, and the bottom side, or edge, of the protractor coin- cides with the side OB of the angle; the other side, OA, of the Fig. 12. angle crosses the arc at 59", which is the size of the angle. More accurate instruments are in use, but they all are based on the same principle. 55. A surface has no thickness. When a surface is perfectly flat, so that a right line may be drawn on it in any direction and anywhere on it, the surface is called a plane. A plane surface may be tested by laying a straightedge anywhere on it; if the straightedge coincides with the surface throughout the length of the straightedge, no matter where it is placed, the surface is a plane surface. Planes, like right lines, are supposed to be infinite in extent; that is, they are infinite in length and infinite in breadth; but, like lines, only parts of planes are considered in practice. 56. Two planes intersect in a right line. For example, referring to Fig. 13, the planes ABCD and EFGH intersect in the right Hne MN, which is called the line of intersection. If from any point o in the line of intersection MN a line oa be drawn in the plane EFGH, perpendicular to MN, and a line ob be drawn in the plane ABCD, also perpendicular to MN, the angle aob is the angle which the planes make with each other. 48 ELEMENTARY APPLIED MATHEMATICS §2 If this angle is a right angle, in which case, oa and oh are perpen- dicular to each other, the two planes are then said to be perpen- Fig. 13. dicular to each other; otherwise, they make an acute or obtuse angle with each other, according as the angle aoh is acute or obtuse. PLANE FIGURES 57. A plane figure is any outline that can be drawn on a plane surface; and, since a plane is infinite in length and breadth, a plane figure may be of any size and any shape. To be complete, however, the figure must close; that is, if the figure be traced by moving the pencil over it, then, starting from any point on the outline and passing over the entire figure, the pencil must return to the point from which it started. 58. Polygons. — The simplest plane figures are those made up entirely of right lines — called polygons, which means many angles, from 'poly (meaning many) and gonia (meaning angles), and so called because a 'polygon always has as many angles as it has sides. The right fines that form the outline of a polygon are called the sides of the polygon, and the angles included between the sides are called the angles of the polygon. The sum of the lengths of the sides is called the perimeter of the polygon. The perimeter, therefore, is the length of the outline that bounds the polygon; it equals the distance around it. §2 MENSURATION OF PLANE FIGURES 49 59. Polygons are named in accordance with the number of sides that they have. A polygon with three sides is called a triangle; one with four sides is called a quadrilaterial; one of five sides is a pentagon; one of six sides is a hexagon; one of seven sides is a heptagon; one of eight sides is an octagon; one of nine sides is a nonegon; one of ten sides is a decagon; one of eleven sides is an undecagon; one of twelve sides is a dodecagon; etc. In practical work, the triangle, quadrilateral, and hexagon are very freely used, and the pentagon, octagon, and decagon are oc- casionally used; the other polygons are practically never used. 60. A regular polygon is one in which all the sides and all the angles are equal. Fig. 14 shows regular polygons of 3, 4, 5, 6, 8, B C A D Quadrilateral Pentagon Hexagon Octagon Fig. 14. Decagon and 10 sides. In each figure, the sides are all equal in length, and the angles (called the interior angles) are all equal. Thus, in the regular pentagon, AB = BC = CD = DE = EA, and ABC = BCD = CDE = DEA = EAB. 61. To find the number of degrees in one of the equal angles of a regular polygon, let n = the number of sides and a° =the number of degrees in one of the equal angles; then ^o ^ (.n- 2) ^ ^g^o n For instance, in the regular triangle, the interior angles are equal to ^ — X 180° = 60°; in the regular quadrilateral, 4—2 6—2 a° = — J — X 180° = 90°; in the regular hexagon, a° = — ^ — X 180° = 120°; etc. 4 50 ELEMENTARY APPLIED MATHEMATICS §2 TRIANGLES 62. Triangles are classified in two ways: according to their angles, and according to their sides. "When the angles are con- sidered, triangles are called right triangles, when they have one right angle; oblique triangles, when they have no right angle; and equiangular triangles, when all the angles are equal. An equiangular triangle is also an oblique triangle. Fig. 15 shows a right triangle, the right angle being at B. It may here be re- marked that when there is no possibihty of misun- derstanding, angles may be named by a single letter placed at the vertex. Thus, in Fig. 15, angle C is the angle ACB, angle A is the angle CAB, and the angle B is the angle ABC. When the sides are considered, triangles are called isosceles, scalene, or equilateral triangles. An isosceles triangle is one that has two equal sides, see Fig. 16; here CA = CB. A scalene triangle is one in which all the sides have different lengths, see Fig. 17. An equilateral triangle is one in which all the sides are equal, see Fig. 18. A scalene triangle is always an oblique tri- angle. An equilateral triangle is also an isosceles triangle; it Fig. 15. B A B A Fig. 16. Fig. 17. Fig. 18. is likewise an equiangular triangle. A right triangle is usually a scalene triangle, but if the two sides that enclose or form the right angle are equal, it is then an isosceles triangle also. For convenience in mathematical operations, triangles are lettered with capital letters at the vertexes and with small letters (lower case letters) placed at the middle of the sides, the lower case letters being in each instance the same as the capital letters at the angles opposite the sides. Thus, referring to Figs. 15-18, side a is opposite angle A, side 6 is opposite angle B, and side c is opposite angle C. §2 MENSURATION OF PLANE FIGURES 51 The side on which any triangle (or any polygon) is considered as standing, the plane of the figure being supposed to be vertical, is called the base. In Fig. 14, AC is the base of the triangle; AD is the base of the quadrilateral; and AE is the base of the pentagon. In Fig. 15, CB is the base; in Figs. 16-18, AB is the base. 63. Some Properties of Triangles. — (1) In any triangle, the sum of the three angles is always 180°; thus, in Figs. 15-18, A -{- B -\- C = 180°. Therefore, if two of the angles are known, the third can be found by subtracting the sum of the two known angles from 180°. Thus, in Fig. 17, if A = 68° 23' and B =24° 35', C = 180°— (68° 23' + 24° 35') = 87° 2'. (2) In a right triangle, one of the angles being a right angle and therefore equal to 90°, the sum of the other two angles must be 180° — 90° = 90°; consequently, both angles must be acute, since neither can exceed (nor even equal) 90°. Also, this sum must be 90°; because the sum of the three angles is 180°, and since one of the angles if 90°, the sum of the other two angles must be 180° — 90° = 90°. Hence, if one of the acute angles of a right triangle is known, the other can be found by subtracting the known angle from 90°. Thus, in Fig. 15, suppose that the angle A = 28° 42'; then the angle C = 90° —28° 42' = 61° 18'. (3) The longest side is always opposite the greatest angle, and the greatest angle is always opposite the longest side. In Figs. 15 and 17, 6 and c are respectively the longest sides; hence, B and C are respectively the largest angles. (4) If any two sides of a triangle are equal, the angles opposite those sides are also equal. In Fig, 16, a and b are equal; hence A = B. Consequently, in every isosceles triangle, two of the angles are equal. (5) If an isosceles triangle be so placed that the unequal side forms the base, as in Fig. 19, and a perpendicular to the base is drawn from the vertex of the angle opposite, the perpendicular divides the base into two equal parts. The perpendicular is then said to bisect the base, the word bisect meaning to cut in halves. In Fig. 19, AC = CB, and CD is perpendicular to AB; therefore, AD = DB. (6) If two angles of one triangle are equal to two angles of another triangle, the third angle of the first triangle is equal to 52 ELEMENTARY APPLIED MATHEMATICS §2 the third angle of the second, because the sum of the two angles subtracted from 180° is the same in both cases. In Fig. 20, if A = A'andC = C, then B = B'. 64. Similar Triangles. — If the angles of one triangle are equal to the angles of another, the two triangles are said to be similar. If the sides of one triangle are equal to the sides of another, then the triangles are equal. When two triangles are similar, as ABC and A'B'C in Fig. 20, one may be superposed on the other; that is, the vertex of one of Fig. 20. the angles of one triangle may be placed over the vertex of the equal angle of the other triangle, and the sides of the two angles can then be made to coincide, since they both have the same direc- tion from the common vertex, see Fig. 21. Here the vertex C of the angle C in Fig. 20 is placed over C, and the sides CA and CB are made to coincide with the sides C'A' and C'B') then AB is parallel to A'B'. 65. When two triangles are similar, the sides opposite the equal angles are proportional. Thus, in Fig. 20, if the triangles ABC and A'B'C are similar, and angle A = angle A' and angle C = angle C , then AB : A'B' = AC : A'C; also, AB : A'B' = BC: B'C, and BC : B'C = AC : A'C. This is a very important principle, and should be remembered. Had the angle A been superposed on the angle A', Fig. 20, the result would be as shown in Fig. 21, the dotted hne5C being paral- lel to B'C. That AB, Fig. 2 1 , is parallel to A'B' is evident from the fact that since angle A = angle A' and the sides AC and A'C coincide, AB has the same direction as A'B', and when two right §2 MENSURATION OF PLANE FIGURES 53 lines have the same direction, they must either coincide or be parallel. For the same reason, BC is parallel to B'C. Hence, if two sides and an angle of one triangle are equal to two sides and an angle of another triangle, the triangles are equal. 66. The Right Triangle. — The right triangle is so important that it requires special treatment. Referring to Fig. 22, ABC is a right triangle, right-angled at C. The side C, which is opposite the right angle, is called the hypotenuse; the hypotenuse is alwaj^s the longest side of a right triangle, since it is opposite the largest angle. The other two sides are called the short sides or legs. In every right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs; that is, referring to Fig. 22, c^ = a^ -^ 62 (1) This is a very important principle, and should be memorized; it is used more than any other principle in mensuration. By means of it, any side of a right triangle can be found if the lengths of the other two are known. For example, if a and b are known, c = Va^ + fe' (2) If c and a are known, h = sj c"^ — a^ (3) If c and 6 are known, a = -sj c^ — 6^ (4) Formulas (2), (3), and (i) are derived from formula (1) by solving (1) for c, 6, and a. Example 1. — Referring to Fig. 22, suppose the length of AC is 4% in. and the length of BC is 2% in.; what is the length of AB1 SoLTJTiON. — Since 4% = 4.375, and 2% = 2.875, substitute the values for a and 6 in formula (2), obtaining c = V4.3752 + 2.8752 = 5.2351 - in. Ans. It is evident that formula (1) might have been used instead of formula (2); in fact formula (1) may be used in every case of this kind. Example 2. — In Fig. 23, P and P' are two pulleys, whose centers are 10 ft. 8K in. and 4 ft. 6% in., respectively, from the floor AB, which is sup- posed to be level. By dropping plumb hnes from the shafts, the horizontal distance between the pulley centers is found to be 14 ft. 734 in. What is the distance O'O between the pulley centers? 54 ELEMENTARY APPLIED MATHEMATICS §2 Solution. — Since AB is horizontal and OC and O'D are plumb lines, the angle OCD is a right angle. Drawing O'E parallel to OCD, it must be per- pendicular to OC (since AB is perpendicular to OC), and O'EO is also a right angle. Drawing O'O, O'EO is a right triangle, in which the side O'E = 14 ft. 7.25 in. and OE = 10 ft. 8.5 - 4 ft. 6.75 in. = 6 ft. 1.75 in. For con- FiG. 23. venience in calculation, reduce the lengths of these sides to inches, obtain- ing O'E = 175.25 in. and OE = 73.75 in. Substituting these values in formula (2), O'O = V175.252 + 73.752 = 190.14- in. = 15 ft. 10}i in., very nearly (0.14 = }i nearly). Ans. 67. Equality of Angles. — In connection with the last example, it was stated that the angle O'EO was a right angle, because O'E was perpendicular to OC. It might have been stated that O'EO is a right angle because it is equal to OCD, which is a right Fig. 24. angle, in accordance with the following principle: // the two sides of one angle are parallel to the two sides of another angle, both sides of which extend in the same or both in opposite directions, the two angles are equal. Thus, referring to Fig. 24, if A'O' be parallel to AO and B'O' be parallel to BO, both sides extending in the same direction, A'O'B' is equal to AOB. Again, if A"0" be parallel to AO and B"0" be parallel to BO, both sides §2 MENSURATION OF PLANE FIGURES 55 extending in opposite directions to ^0 and BO, A"0"B" is equal to AOB. If one side of an angle coincide with a side of another angle, and the second side of the first angle be parallel to the second side of the other angle, the two angles are equal. Thus, if 0'" B'" and 0""B"" be parallel to OB, Fig. 24, AO"'B"' and 00'" B'" are equal to AOB. This a special case of the above principle, in which two of the parallel hues coincide. For the same reason rOB" = 00""B"" = AOB. Referring to Fig. 23, OE coincides with OC, and since O'E is parallel to CD, O'EO = DCO. ,'''0 A-' Fig. 25. A second principle applying to equahty of angles is: If the two sides of one angle are perpendicular to the two sides of another angle and both angles are acute or both obtuse, the two angles are equal. Referring to Fig. 25, suppose that A'O' and B'O' are per- pendicular to AO and BO, respectively; then AVB' = AOB. Again, suppose that 0"B" and 0"A" are perpendicular to OB and OA, respectively; then B"0"A" = BOA. Finally, suppose that A"'0"' and B'"0"' are respectively perpendicular to ^0 and BO produced (indicated by the dotted hues; then, by the first principle, A""OB"" = AOB, and A"V"'B'" = AOB. These two principles are very important and are frequently used in connection with practical problems. 68. Area of Triangles.— In Fig. 26, let ^C be the base of the triangle ABC. From the vertex of the angle B opposite the base, drop a perpendicular BD; the point where the perpendicular cuts the base is called the foot of the perpendicular, and that part 56 ELEMENTARY APPLIED MATHEMATICS §2 of it included between the vertex B and the foot D is called the altitude of the triangle. The word altitude means height, and it is usually denoted in formulas by the letter h. In Fig. 27, it is necessary to produce the base in order that it may be cut by a perpendicular from B. In both figures, h = BD is the altitude. C A In any triangle, the area is equal to half the product of the base and altitude. Hence, letting A = the area, h = the base, and h = the altitude. Any side may be taken as the base, the altitude being the perpendicular distance between the base and the vertex of the angle opposite the base. Thus, in Fig. 28, AC XBD ^ BC XAF ^ AB X CE 2 ~ 2 2 * Here AF is the altitude when BC is the base, and CE is the altitude when AE is the base. In a right triangle, if one leg is taken as the base, the other leg is the altitude, see Fig. 22. j„- /^\bx'^. j^ Consequently, the area of a right triangle is equal to half the product of its legs. If two triangles have the same base and the same altitude, their areas are equal. Thus, in Fig. 28, if MN is paraUel to AC, the altitudes of the three triangles ABC, ^ilfC, and ylA^C are equal; and since they all have the same base A C, all three triangles have the same area. Example. — If the base of a triangle is 2 ft. 9 in. and the altitude is 1 ft. 4 in., what is the area of the triangle? §2 MENSURATION OF PLANE FIGURES 57 Solution. — The lengths must be expressed in the same single unit, ex- pressing them in feet, 2 ft. 9 in. = 2.75 = 2f ft.; 1 ft. 4 in. = 1^ ft. Then, applying the formula, A = (2.75 X 1§) -h 2 = 1.8| sq. ft. Ans. Or, expressing the lengths in inches, 2 ft. 9 in. = 33 in.; 1 ft. 4 in. = 16 in. Then, A = ^^^^ = 264 sq. in. = ^ = If - 1.8^ sq. ft. Ans. Always remember that feet multiplied by feet give square feet and inches multiplied by inches give square inches. 69. If the triangle is isosceles (an equilateral triangle is also isosceles), and the unequal side is taken as the base, the line repre- senting the altitude bisects the base, see (5) of Art. 63. Hence, referring to Fig. 19, if the three sides a, h, and c are known, a and 6 being equal, the altitude CD = h = -v &^ — (2) = |\/462 — c2, by formula (4) or (3), Art. 66; and the area of the triangle is, A = I X c X iV462 - c2 = |\/462 - c\ Example. — In an equilateral triangle, the length of each side is 73^ in. ; what is the area of the triangle? 7 25 ■ Solution.— In this case, & = c; hence, A = ^V 4 X 7.25" - 7.25^ = -^VB X 7.252 = ^^ V3 = 22.760+ sq. m. Ans. For purposes of reference, let a = one of the equal sides of an isosceles triangle, and let c be the unequal side ; then A = ^V4:a^- c2 (1) For an equilateral triangle, let c = one of the sides; then A = ^Vs = .43301c2 (2) 70. If all three sides of any triangle are known and the altitude is not known and cannot conveniently be measured, let p = A, BD^ = c^ - p^; therefore, BD^ = a^ - p^ + 2hp - &2 = c2 - p2. froni which, a^ = b^ + c^ - 2bp, which is the same as (1). Stated in words, the square of the side opposite an acute angle of any triangle is equal to the sum of the squares of the other two sides minus twice the product ^^\ obtained by multiplying one of ^^^^ / I these sides by the projection of S^"^^^ / I the other side upon that side. \^^ /" ! Thus, in Fig. 31, the projec- ^'-^ \y~^ — H tion of AC upon AB is AD' '^<- J -Afe^ ^ = p\ and a2 = 62 + c2 - 2cp'. \^ h / This, evidently, is the same ^'^N^ y / case as Fig. 32, when the tri- -^jy angle is turned over and AB Pj^j_ 33 of Fig. 31 is made the base. In Fig. 33, DA = p-\-b,p being the projection DC of a upon AC. Then, BD"^ = a^ — p^ = c^ — (& + p)^ = c2 — 62 _ 2bp —p^; from which, c2 = a2 + 62 + 262> (2) Stating formula (2) in words, the square of the side opposite an obtuse angle of any triangle is equal to the sum of the squares of the other two sides plus twice the product obtained by multiplying one of these sides by the projection of the other side upon hat side. Thus, in Fig. 33, the projection of 6 upon BC is CD' = p', and c^ = a^ + 62 + 2ap'. Example. — In the triangle ABC, in which the side A 5 is the longest, AB = 22 in., AC = 13 in., and the projection of AC upon AB measures 9.73 in., what is the length of the side CBl Solution. — The sides AB and AC evidently include an acute angle, since the longest side is always opposite the largest angle; hence, formula (1) applies to the case, substituting c for h, so that the formula becomes a2 = ^2 -f- c2 — 2cp; whence. a = ^h^ + c2 - 2cp = V132 + 222 - 2 X 22 X 9.73 = AB -= 15.029 in. Ans. If the triangle is a right triangle, the projection of one leg upon the other is a point, which has no length and is therefore 0. In this case, c^ = a2 + 62 + 26 X = a^ + 6^, which is the same as formula (1) in Art. 66. §2 MENSURATION OF PLANE FIGURES 61 QUADRILATERALS 73. The Parallelogram. — When the opposite sides of a quadri- lateral are parallel, the quadrilateral is called a parallelogram; Fig. 34 shows two parallelograms ABDC, the sides AB and CD being parallel and equal, and the sides AC and BD being also parallel and equal. When the interior angles are not all equal and the sides are not all equal the parallelogram is called a rhomboid; both parallelo- grams in Fig. 34 are rhomboids. If the angles are not all equal, but the sides are equal, the paral- lelogram is called a rhombus; see Fig. 35, in which AB = BD = DC = CA. The diagonal of a parallelogram is a hne drawn through the figure from the vertex of one acute angle to the vertex of the other acute angle, the line CB in Figs. 34 and 35; this is called the long diagonal. A hne AD, drawn from the vertex of one ob- tuse angle to the vertex of the other obtuse angle, Figs. 34 and 35, is called the short diagonal. The perpendicular distance BE between two parallel sides, in Figs. 34 and 35, is called the altitude of the parallelogram. 74. Some Properties of Paral- lelograms.— (1) The diagonally opposite 'angles of any parallelo- gram are equal; thus, in Figs. 34 and 35,C = B and A = D. This is a consequence of Art. 67, since AB is parallel to CD and AC is parallel to BD. (2) A diagonal divides the parallelogram into two equal tri- angles. Thus, ACB = CDB and ACD = ABD. This is evident since the three sides of one triangle are equal to the three sides of the other, the diagonal being a common side. (3) The diagonals of a parallelogram bisect each other; that Fig. 35. 62 ELEMENTARY APPLIED MATHEMATICS §2 is, P being the point of intersection of the diagonals, PA = PD and PB = PC. (4) The sum of the interior angles of any parallelogram is equal to 4 right angles or 360°. For, referring to Figs. 34 and 35, angle BDE = angle ACD, and CDB + BDE = CDB + ACD = C -\- D = 2 right angles = 180°. Since A = D and B = C, A+B + D-\-C ^2 X 180° = 360°. Therefore, if one angle of a parallelogram is known, all are known. Thus, suppose the angle C in Fig. 35 is 55°; then angle D is 180° - 55° = 125°, A = 125°, and B = 55°. 75. The side on which the parallelogram is supposed to stand is called the base; in Figs. 34 and 35, CD is the base. The area of any parallelogram is equal to the product of the base and altitude; thus, the areas of the parallelograms in Figs, 34 and 35 is equal to CD X BE. Let A = the area, I = the length of the base, and h = the altitude; then, A = Ih. This follows at once from the rule for finding the area of a triangle. Thus, area of triangle CDB, Figs. 34 and 35, equals }i X BE X CD, and since CAB = CDB, the area of the paral- lelogram is2 X}iCD X BE = CD X BE. R c A (a) D A Fig. 36. 76. If one angle of a parallelogram is a right angle, all the angles are right angles, in accordance with Art. 74, and the paral- lelogram is then called a rectangle; see (a). Fig. 36. Here A = B = C = D = a right angle = 90°, and BC = AD and AB = DC. If the four sides of a rectangle are equal, it is called a square. Thus, (6), Fig. 36, is a square, because AB = BC = CD = DA and A=B = C = D^a right angle. §2 MENSURATION OF PLANE FIGURES 63 Referring to (a), Fig. 36, it will be noted that if AD is the base, CD ■= AB is the altitude, and if AB is the base, AD = BC is the altitude. Hence, if one side of a rectangle be called the length, and one of the sides perpendicular to it be called the breadth or depth (according to whether the plane of the rectangle is sup- posed to be horizontal or vertical), the area of the rectangle is equal to the product of the length and breadth or depth; that is, A = lb = Id. in which A = the area, I = AD, Siudb = d = AB = DC. When the rectangle is a square, I = b = d, and A = I X I = l^. 77. Referring to Fig. 37, ABCD is a rhombus. Draw the two diagonals, and represent the long diagonal by d and the short diagonal by d'. The diagonals intersect in P, which is the middle point of BD and AC (see Art. 74), and the four angles about P are right angles. That BPA = BPC is a right triangle is evident from (5) , Art. 63. Here ABC is an isosceles triangle and since P is the middle point of the base, the line from P to 5 is perpen- dicular to the base (see Fig. 19). Con- sidering the triangle DAB {= DCB), AP is the altitude and DB is the base. But d' AP = ^ and DB = d; hence, area oi DAB = }4 x d X %, and area of rhombus ABCD is k Fig. 37. d' A=\XdX~X2 = Ud' (1) sides Fig. 38. then, d"^ That is, the area of a rhombus is equal to one-half the product of its diagonals. Referring to Fig. 38, ABCD is a square— a rhombus whose angles are right angles. The diagonals AC and BD are equal and formula (1) becomes A =\dd = ^d\ (2) If it is desired to find the length of the diagonal of any square, let I = one of the -- l^ -\- l^ = 21', and d = VW = IV2 = 1.4142Z. (3) 64 ELEMENTARY APPLIED MATHEMATICS §2 If the diagonal of a square is given and it is desired to find the length of the sides, l^ -\- l^ = d^ or 21^ = d^, and P = -^j from which I = ^^2 = .lOllOld (4) Example 1. — Fig. 37 is a drawing of a flat, rectangular plate having a hole in it shaped like a rhombus. From the dimensions given, find the area of the plate. Solution. — -The area of the plate is evidently equal to the area of the rectangle minus the area of the rhombus. Area of rectangle = 18.5 X 28.625 = 529.5625 sq. in. Area of rhombus = M X 10.375 X 12.75 = 66.1406 sq. in. area of plate = 463.4219 sq. in. Therefore, area of plate is 463.92 sq. in. Ans. Example 2. — What is the diagonal of a square, one side of which measures 9^6 in.? Solution. — Substituting in formula (3), d = 1.4142 X 9%6 = 13.170- in. Ans. Example 3. — If the diagonal of a square is 11.82 in., what is the length of one of the sides? Solution. — Applying formula (4), I = .707107 X 11.82 = 8.358 in. Ans. 78. Other Quadrilaterals. — If two sides of a quadrilateral are parallel and the other two sides are not parallel, the quadrilateral is called a trapezoid. In Fig. 39, {a) and (5) are trapezoids, the rb) n side BC being parallel to AD. The trapezoid in (6) is peculiar from the fact that it has two right angles, situated at A and B. If no two sides of a quadrilateral are parallel, it is then called a trapezium ; see Fig. 40. To find the area of a trapezoid, divide it into two triangles by- drawing a diagonal AC; then, if BC be taken as the base of one triangle and ^Z> as the base of the other, the altitude of these triangles is^D = CE in (a). Fig. 39, and AB = the perpendicular distance between the parallel sides in (b), Fig. 39. Denoting the §2 MENSURATION OF PLANE FIGURES 65 altitude by h, the side BC by a, and the side AD by h, the area of one triangle is }-'2ah; of the other, 3^^6/i; and the area of the trapezoid is }i ah + K 6/i = }'2h{a + 6); or, This is a very important formula; it is frequently used, and should be carefully committed to memory. Stated in words, the formula becomes the following rule : The area of a trapezoid is FiQ. 40. equal to half the sum of the parallel sides multiplied by the perpen- dicular distance between them. Example. — What is the area of a trapezoid, the lengths of the parallel sides being 6}4 in. and 7M in. and the altitude being 9% in.? Solution. — The sum of the parallel sides is 6.5 + 7.25 = 13.75; hence, ^, . 13.75 X 9.375 „, ^„^ . . the area is ^ = 64.453+ sq. in. Ans. It is better not to divide by 2 until after the multiplication has been performed, unless one of the factors is an even number. In the example just given, both factors were odd numbers. 79. To find the area of a trapezium, draw a diagonal, as BD, Fig. 40; thus dividing the trapezium into two triangles. Using the diagonal as a base, draw the perpendiculars AD and CE from the vertexes opposite the base; then, area of trapezium is equal to ^iBD X EC ^ HBD X AD = ^BD (AD + EC). If through one of the vertexes, say C, a line be drawn parallel to the base (diagonal) BD and a perpendicular be drawn to this line from the other vertex, the length of this perpendicular is 66 ELEMENTARY APPLIED MATHEMATICS §2 equal to AD + EC. Representing this perpendicular by h, the base (diagonal) by b, and the area by A , A = y2bh. 80. The rules and formulas just given for finding the areas of quadrilaterals apply only to what are termed convex polygons. All polygons that have been illustrated up to this point are convex polygons; and in any convex polygon, if any one of their sides be produced at either end of the side, the produced part will lie outside of the bounding line of the polygon. In Fig. 41, (a) is a quadrilateral and (&) is a pentagon. If the sides AB or Fig. 41. CD be produced from the end B, they wiU enter the space included by the bounding Hnes of the polygons. Angles like ABC are called re-entrant angles, and polygons having one or more re- entrant angles are called concave or re-entrant polygons. To find the area of a re-entrant polygon, divide it into triangles, as indicated by the dotted lines, find the area of each triangle, and their sum will be the area of the polygon. Unless otherwise stated, all polygons are supposed to be convex polygons. REGULAR POLYGONS 81. Except in the case of irregular figures bounded by right lines, most of the polygons that occur in practice are regular polygons. If, in any regular polygon having an even number of sides, a line (diagonal) be drawn from any vertex to the vertex opposite that is farthest away, the line will pass through what is called the geometrical center of the polygon; and if two such lines be drawn §2 MENSURATION OF PLANE FIGURES 67 from different vertexes, they will intersect in the geometrical center. Thus, in Fig. 42, which represents a hexagon (a polygon with an even number of sides), AD, BE, and CF all intersect in 0, which is the geometrical center, or, more simply, the center, of the hexagon. If a regular polygon have an odd number of sides, as the pen- tagon. Fig. 43, and a perpendicular be drawn from any vertex to the side opposite, it will pass through the geometrical center of the polygon; and any two such perpendiculars will intersect in the center. Thus, AP, BQ, CR, DS, and ET, which are per- pendicular respectively to CD DE, EA, AB, and BC, all inter- sect in 0, the center of the pentagon. CPU The perpendicular from the center to one of the sides, as OP in Figs. 42 and 43, bisects the side; that is, PC = PD. This perpendicular is called the apothem. The lines AD, BE, etc.. Fig. 42, and AP, BQ, etc.. Fig. 43, divide the polygons into as many equal triangles as the poly- gons have sides and the sum of the areas of these triangles equals the areas of the polygons. The area of one triangle is, letting I = length of one side and a = the apothem, ^^ial. If n = the number of sides of the polygon, the area, A, of the polygon is 3^^ anl. But nl = the perimeter of the polygon = p; hence, A = }ipa. (1) Stated in words, the area of any regular polygon is equal to half its perimeter multiplied by its apothem. 68 ELEMENTARY APPLIED MATHEMATICS §2 TABLE OF REGULAR POLYGONS Number of Apothem Area Number of Apothem Area sides a k sides a k 3 0.28868 0.43301 15 2.3523 17.642 4 0.5 1. 16 2.5137 20.109 5 . 68819 1 . 7205 20 3.1569 31.569 6 0.86603 2.5981 24 3.7979 45.575 7 1.0383 3.6339 25 3.9579 49.474 8 1.2071 .4.8284 30 4.7572 71.358 9 1.3737 6.1818 32 5.0766 81.225 10 1.5388 7.6942 40 6.3531 127.06 11 1.7028 9.3656 48 7.6285 183.08 12 1.8660 11.196 64 10.178 325.69 The area may also be found by means of the above table when the number of sides in the polygon is given in the table. To use the table, let k = the number in the column headed area that coresponds to the given number of sides; let I = the length of the given side; then, A = kl\ (2) For example suppose that the length of a side of a regular octagon is 33-^ in., and it is desired to find the area of the octagon. Referring to the table, when the number of sides is 8, fc = 4.8284; hence, the area is A = 4.8284 X 3.25^ = 51 sq. in. Ans. Example. — One side of a hexagonal bar of iron measures IH in- J what is the area of a cross section of the bar? Solution. — Referring to the table, when the number of sides is 6, k = 2.5981; hence, area = A = 2.5981 X (1^)^ = 7.3985-, say 7.398 sq. in. Ans. The apothem can be used to lay out the polygon ; the manner of doing this will be described later. The apothem as given in the table is for a side equal to 1 ; hence to find the actual length of the apothem when the length of a side of the polygon is given let I = length of side ; then actual length of apothem = la. In the last example, the actual length of the apothem is lii X .86603 = 1.4614 in. THE CIRCLE 82. Definition. — The circle is a curve every point of which is equally distant from a point within it called the center. The §2 MENSURATION OF PLANE FIGURES 69 curve shown in Fig. 44 is a circle, being the center. A circle may be described (drawn) in various ways. Thus, with a pin, punch two holes in a strip of heavy paper; put the pin through one hole and the point of a sharp pencil through the other hole; then, keep- ing the pin stationary, revolve the pencil about the pin, keeping the strip of paper stretched tight, and the pencil will describe a circle. This method is shown in Fig, 45. A better way is to use an instru- ment employed by draftsmen, called compasses. This instru- ment consists of two legs united at one end by a joint, which permits them to open and close to any desired distance apart. One leg has a needle point at one end, Fig. 44. Fig. 45. Fig. 46. and the other leg carries a pencil point or pen. By placing the needle point of the compasses at the center, the leg carrying the 70 ELEMENTARY APPLIED MATHEMATICS §2 pencil point or pen may be revolved about the needle point, thus describing the circle as shown in Fig. 46. 83. Referring to Fig. 44, any part of a circle, as AD, ABC, etc. is called an arc of a circle, a circular arc, or simply, an arc; it is so called from its shape, the word arc meaning how. A right line joining the extremities of an arc is called the chord of the arc or, simply, the chord. Thus, ^C is the chord of the arc ABC, and DE is the chord of the arc DABCE. When the chord passes through the center of the circle, it divides the circle into two equal parts, each of which is called a semicircle, meaning half-circle, and the chord is then called a diameter of the circle or, simply, a diameter. In Fig. 44, DE is a diameter of the circle, because it passes through the center. For the same reason, BF is also a diameter. The arc DBE is equal to the arc DFE, and both are semicircles. The arc FDB is equal to the arc FEB, and both of these arcs are semicircles. A right line drawn from the center to the curve is called a radius of the circle ; thus, OD, Fig. 44, is a radius, and so is OA, OB, etc. The plural of radius is radii; hence, OD, OA, OB, OC, OE, and OF are radii, of the circle DABCEF. All radii of any circle are equal, by definition of the circle, since they equal the distance from the center to the curve. A radius is also equal to one-half the diameter, since the diameter DE = OE + OD, and OE = OD = the radius. Consequently, the diameter equals twice the radius. The perimeter of a circle is commonly called the circumference ; in geometry, it is called the periphery. The word periphery is applied to plane figures having curved outlines, while the word perimeter is applied to plane figures bounded by right lines. 84. The word circle is also appHed to the area contained within the circumference, hence, by area of a circle, the area included by the circumference is always meant. The area included by an arc and two radii drawn to the extremities of the arc is called a sector; in Fig. 44, the area OABC is a sector of the circle DBEF. The area included between an arc and its chord is called a segment; the area ABC A is a segment of the circle DBEF. 85. If a line be drawn from a point without a circle and is terminated by the circumference after passing through the cir- cle, such a line is called a secant. In Fig. 47, PA and PB are se- cants. Evidently, a secant intersects the circumference in two §2 MENSURATION OF PLANE FIGURES 71 Fig. 47. points; thus, PA intersects the circumference in A and D, and PB intersects it in B and E. If, however, the secant just touches the circle, intersecting it in only one point, it is called a tangent ; thus, PC is a tangent, because it intersects the circle in only one point, the point C, which is called the point of tangency. PG is also a tangent, and F is the point of tangency. 86. Some Properties of Circles.^ — (1) If a diameter he drawn perpendicular to any chord, it bisects the chord and also the arc. In Fig. 44, if BF is perpen- dicular to AC, AG = GC, and arc AB = arc BC. (2) Any angle whose vertex is the center of the circle is measured by the arc it inter- cepts; the word intercept here means the part of the circumference cut off by and included between the radii forming the sides of the angle. In Fig. 44, the angle COE is measured by the arc CE that is intercepted by the radii CO and CE. The angle AOC is measured by the intercepted arc ABC) the angle DO A , by the intercepted arc DA ; etc It is here under- stood that the circumference is supposed to be divided into de- grees, minutes, and seconds, as described in Art. 54. Angles whose vertexes are situated at the center of the circle are called central angles or angles at the center. (3) Since central angles are measured by the arcs they inter- cept, it is evident that a diameter perpendicular to the chord of the arc bisects the central angle that is measured by that arc. Thus, by (1), if the diameter BF is perpendicular to AC, arc AB = arc BC, and AOB = BOC, since both angles are measured by equal arcs. (4) If a right line be drawn perpendicular to any chord at its mdddle point, it will pass through the center of the circle having the same arc that is subtended by the chord. In Fig. 44, the arc ABC is subtended by the chord AC. If G is the middle point of the chord and BGF is perpendicular to AC, then BGF must pass through the center of the circle DBEF of which the arc ABC is a part. (5) Two circles are equal when the radius or diameter of one is equal to the radius or diameter of the other; two sectors are equal when the radius and chord of one are equal to the radius and chord 72 ELEMENTARY APPLIED MATHEMATICS §2 of the other; and two segments or two arcs are equal when the radius and chord of one are equal to the radius and chord of the other. (6) // the vertex of an angle lies on the circumference, the angle is measured hy one-half the intercepted arc. In Fig. 48, the vertex of BAC lies on the circumference; it is therefore measured by one- half the arc BC. Angles whose vertexes lie on the circumference Fig. 48. Fig. 49. are called inscribed angles; hence, an inscribed angle is one-half as large as the central angle having the same arc. Thus, BAG = H BOC, and BOG = 2 BAG. (7) // the inscribed angle intercepts a semicircle, the angle is a right angle, since a semicircle contains 360° -f- 2 = 180°, and one- half of 180° is 90°, a right angle. Thus, in Fig. 49, if AG is a diameter, .A 5C, ADC, and AEG are all right angles. Hence, Fig. 50. any angle inscribed in a semicircle is a right angle. This fact is made use of by mechanics to test the roundness of a semicircular hole, as shown in Fig. 50. Here a square is laid across the edges, and is then rotated back and forth. If the sides of the square just touch the edges and the point of the square just touches the bottom, the surface touched is semicircular, since, as shown in §2 MENSURATION OF PLANE FIGURES 73 the figure, ABC is a semicircle and the angle ABC inscribed in it is a right angle. (8) If two chords intersect in a point within a circle, as DE and FG, Fig. 48, which intersect in H, the angle GHE, which equals DHF, is measured by one-half the sum of the arcs intercepted by these equal angles; that is, GHE (or DHF) is measured by one- half of arc GE + arc DF. (9) If two secants are drawn from the same point, the angle between the secants is measured by one-half the difference of the arcs they intercept. In Fig. 47, APB is measured by one- half of arc AB — arc DE. (10) The angle between a secant and a tangent drawn from the same point, also the angle between two tangents drawn from the same point, is measured by one-half the difference of the inter- cepted arcs. In Fig. 47, the angle APF is measured by one-half of arc AF — arc FD, and angle FPC is measured by one-half arc of FBC — arc FEC. (11) The radius drawn to the point of tangency is per- pendicular to the tangent In Fig. 47, if C and F are points of tangency, OC is perpendicular to PC and OF is perpendicular to PF. (12) If two circles in- tersect, the line passing through their centers is perpendicular to their com- mon chord and bisects the chord. In Fig. 51, two circles having the centers O and 0' intersect; AB is their common chord, drawn through the points of intersec- tion; then, the right line passing through the centers and 0' is perpendicular to AB and bisects AB. The circles whose centers are and 0" also intersect, their common chord being CD. Then, the right hne passing through O and 0" is perpendicular to CD and bisects CD. In the first case, the center of the second circle is outside the circumference of the first; but, 74 ELEMENTARY APPLIED MATHEMATICS §2 in the second case, the center of the second circle is within the circumference of the first. (13) From a given point without a circle, two tangents may be drawn to the circle, as PF and PC in Fig. 47. If from the given point, a secant be drawn passing through the center, it bisects the angle formed by the two tangents and is perpendicular to the chord joining the points of tangency. In Fig. 47, if F and C are the points of tangency of the tangents drawn from P, and PB is a secant passing through the center 0, then OPF = OPC = ^^FPC, and PB bisects the chord FC. PB is also perpendicular to FC. It is evident, also, that PB bisects the central angle FOC. 87. Three Important Principles. — (1) // any two chords of a circle intersect, the product of the segments of one line is equal to the product of the segments of the other line, the segments being determined by the point of intersection. In Fig. 52, the chords AB and CD intersect in the point M; then, AM X MB = CM X MD. The chords AB and JK intersect in the point N; then, AN X NB = JN X NK. If EF is a diameter per- pendicular to the chord GH, IG = IH, according to (1), Art. 86, and IE X IF = IGXIH = IG^ = IH^; that is, a diameter per- pendicular to a chord is divided by the chord into two segments whose pro- duct is equal to the square of half the chord. The segment included between the chord and the arc is called the height of the arc or height of the segment; hence if the chord and height of the arc are known, the diameter or radius can be found. For, let c = the chord GH, h = the height IE, and d = c I c\^ the diameter; then IF = d — h, IG = ^, and h (d — h) = i^j ; from which, hd — h^ = or d = 4/i (1) §2 MENSURATION OF PLANE FIGURES 75 Representing the radius by r, r = ^^ ^^'^ If r and c are known and it is desired to find h, it may be found from formula (2), its value being h = r + ^ -v/4r^ — c^ When h is less than r, the minus sign is used, and h = r - W^r^ - C" (3) When h is greater than r, the plus sign is used, and h =r + W^r^ - c^ (4) If r and h are given and c is desired, c = 2V(2r - h)h (5) (2) 7/ from a point without a circle two secants are drawn, the product of the whole secant and the external segment of one line is equal to the product of the whole secant and the external segment of the other line. In Fig. 53, PA and PB are secants drawn from the point P; EP and FP are the external segments; then, PA XPE = PB X PF. (3) If from a point without a circle a secant and a tangent are drawn, the product of the whole secant and its external segment is equal to the square of the tangent. In Fig. 53, let PA be any secant and PC a tangent, both drawn from P; then, PA X PE = PC\ In order to measure PC, it is necessary to know the point of tangency C; and this can be found by drawing from the center a perpendicular to PC, the point where it intersects PC being the point of tangency. Example .^ — ^The chord of an arc has a length of 14% in.; the height of the arc is 3)4 in.; what is the radius? Fig. 53. 76 ELEMENTARY APPLIED MATHEMATICS §2 Solution. — Applying formula (2), c = 14.875, h 14.8752 + 4 X 3.252 3.25, and = 10.135+ in. Ans. 8 X 3.25 88. It is sometimes desirable to know the chord and height of half the arc when the chord and height of the whole arc are given. In such a case, formulas may be found as follows: Referring to Fig. 54, let C = AB3.ndH = CD, the chord and height of the arc ACB; let c = AC and h = EF, the chord and height of the arc AC = half the arc ACB; and let r = the radius of the arc ACB. In the right triangle ADC, AC^= AD^ + CD\ or Q -{■H' C^ + 4:H^ 4 Multiplying and dividing this fraction by 2H, which of course does not alter its value, 2H(C^ + 4H^) the result is C^ + 4g2 8H • = 2rH, since by formula (2), Art. 87, r = c2 = 2rH, and c = \/2rH (.1) To find h, EF XFH = AF^ or h{2r -h)= (|) FH = 2r- h. But c^ = 2rH; hence, h(2r - h) 8H Therefore, 2rH 4 4 rH smce and 2rh rH h^ = ^- Dividing both members of this equation by — 1, to change the sign of h}, h^ - 2rh = - rH Solving this equation by the regular rule for quadratics (Art. 31) 2r + h = V (2r)2 + 4 X 1 X-~ 2 X 1 = r±^r{r-^)^ = r + (2) Vr^ §2 MENSURATION OF PLANE FIGURES 77 To apply either of these formulas, it is first necessary to calcu- late the radius. Example. — Referring to the example in Art. 87, find the chord and height of half the arc. Solution.— The chord AB = 14.875 in., the height CD = 3.25, from which the radius was found to be 10.136 in. By formula (1), the chord of half the arc = c = -\/2 X 10.135 X 3. 25 = 8.1165 -, say 8.116 in. Ans. By formula (2), h = 10.135 ± a/i0.135^ - ^^'^^^ ^ ^'^^ = 10.135 + 9.287 = 19.422 in.; or, 10.135 - 9.287 = .848 in. The smaller value is evidently the one required in this case; hence, h = .848 in. Ans. It may be remarked that the smaller of the two values just ob- tained is the length of EF, while the larger value is the length FH; the sum of these two lengths is 19.422 + .848 = 20.270 = EH = 2r = 2 X 10.135 = 20.270. 89. Circumference and Area of the Circle. — The circumfer- ence of a circle is equal to the diameter multiplied by a number that is universally represented by the Greek letter t (pronounced pi or pe). This number has been calculated to 707 decimal places, and it has been proven that it cannot be expressed by a finite number of figures; its value to 9 significant figures is 3.14159265 + , and is usually expressed as 3.1416; for rough calculations, 3t = --j- is commonly used for tt. Letting c = the circumference, d = the diameter, and A = the area, c = ird = 3.1416 d (1) Smce the diameter equals twice the radius, c = 27rr (2) The area of a circle is equal to x times the square of the radius, or A =Trr^ (3) Since r = |, A = 7r(|) = i7rrf2 = .7854^2 (4) These four formulas are extremely important; they should be carefully committed to memory. From formula (1), <* - ; = due = -^'^'^ (^> From formula (2), r = if = .159155c (6) 78 ELEMENTARY APPLIED MATHEMATICS §2 From formula (3), From formula (5), = J^ = .56419 VA (7) d = J^ = 1.1284 VA (8) Formulas (5) to (8) may be used to calculate the radius or diameter when the circumference or area is known; as a rule, however, these formulas are not used, formulas (1) to (4) being preferred. Example 1. — A pulley has a diameter of 32 in. and makes 175 revolutions per minute; how fast does a point on the rim travel in feet per minute? Solution. — When the pulley has turned around once, the point will have traveled a distance equal to the circumference of the pulley, and since the pulley turns 175 times in one minute, the point will travel 175 times the circumference of the pulley in one minute. Consequently, the distance traveled by the point in one minute is ird X 175 = 3.1416 X 32 X 175 = 17592.96 in. = 1466.08 ft., say 1466 ft. Hence, the speed of the pulley is 1466 ft. per min. Ans. Example 2. — Suppose the speed of a belt is 3160 feet per minute and that it drives a pulley that makes 330 revolutions per minute; what is the diam- eter of the pulley? Solution. — The speed of a point on the circumference of the pulley is the same as the speed of the belt, assuming that there is no slipping of the belt. Consequently, as shown in example (1), ird X 330 = 3160 ft. = 37920 in., or 37920 d = „ „ = 36.5766 = 36f I in., very nearly. Ans. Example 3. — The piston of a steam engine has a diameter of 16 in.; what is the area of the piston surface touched by the steam? Solution. — The area touched by the steam is evidently the area of a circle having a diameter of 16 in. By formula (4), the area is A = .7854 X 162 = 201.0624, say 201 sq. in. Ans. Example .4. — It is desired to bore a hole that shall have an area of 10 sq. in.; what must be the diameter of the hole? Solution. — Either formula (4) or (8) may be used, but formula (8) is rather easier to apply; using it, therefore, d = 1.1284\/l0= 3.5683, say 3.568 in. Ans. Example 5. — The circumference of a flywheel was measured with a tape hne and found to be 35 ft, 103^ in. What is its diameter to the nearest one-eighth inch? Solution. — Either formula (1) or (5) may be used. If (1) be used, it will be necessary to divide the circumference by 3.1416, while if (5) be used, the circumference may be multiplied by .31831. Since most computers §2 MENSURATION OF PLANE FIGURES 79 would rather multiply than divide, use formula (,5). Reducing the feet to inches, 35 ft. 10.25 in. = 430.25 in. Then, d = .31831 X 430.25 = 136.953 in. =11 ft. 5 in. to the nearest one- eighth inch. Ans. Example 6. — When no ambiguity (confusion) is likely, a circle may be designated by referring to its centers only. Fig. 55 shows two pulleys, O and O', driven by a belt. Suppose the diameter of O is 48 in. and it makes 220 revolutions per minute. It is desired to have pulley 0' make 450 revolutions per minute; what must be the diameter of 0'? Solution. — ^Let N and n be the number of revolutions per minute made by O and 0', respectively, being the larger and O' the smaller pulley; let D and d be the respective diameters of and 0'. The speed of the belt Fig. 66. in feet per minute is equal to the circumference of in feet multiplied by the number of times it turns in one minute; hence, if the diameters of the pulleys are given in inches, the speed of the belt is . The speed of the belt is also equal to the circumference in feet of O' multiplied by the number rdn l2" Therefore, tDN 12 of times it turns in one minute; that is, it equals = -^2". Dividing both members of this equation by yx, DN = dn (9) In words, the product of the diameter and number of revolutions of one pulley is equal to the product of the diameter and number of revolutions made in the same time by the other pulley. Provided the unit used to measure D and d is the same, it is immaterial what unit is used; that is, the diameters may be stated in feet, inches, millimeters, etc. Applying the formula to the present case, 48 X 220 = d X 450 48 X 220 from which d = — -^ — - = 22.4| in. = 23H in. very nearly. The diameter of d must be in inches because the diameter of D is in inches, and both diameters must be measured in the same unit. Formula (9) is very important in connection with calculations pertaining to pulleys and belts, and should be carefully memorized. 80 ELEMENTARY APPLIED MATHEMATICS §2 90. Length of Open Belt. — When the belt passes over the pulleys without crossing the line O'O joining the centers, as in Fig. 55, it is called an open belt. Knowing the distance O'O between the centers and the diameters of the pulleys, it is fre- quently desired to know the length of the belt. There is no simple, exact formula that will give the length of the belt, but the formula given below is sufficiently exact for all practical purposes. The lines A' A and C'C are tangent to the pulleys (circles) and 0'. Drawing the radii OA and OC, A and C are the points of tangency. Drawing the radii O'A' and O'C, A' and C are points of tangency. Draw diameters BD and B'D' perpendicu- lar to O'O; then the angles AOB and COD are equal, since the tangents A' A and C'C must intersect in some point, say P (not shown here), and by (13) of Art. 86, arc AE = arc CE; but arc AB = 90° - arc AE = 90° - arc EC = DC, and AOB = COD. Since, O'B' is parallel to OB and O'A' is parahel to OA, B'O'A' = BOA. For the same reason, C'O'D' = COD, and all four angles are equal. The radii of the arcs B'A' and BA are not equal, and the length of the arc AB is not equal to the length of the arc A'B'. Let L = the length of the belt, and let C = the distance O'O between the centers; then it is evident that the length of the belt is L = 2 X A'A -\- semicircle BED + 2 X arc BA + semicircle B'F'D' - 2 X arc B'A'. The difficulty arises in finding an expression for the lengths of the arcs BA and B'A'. Using the same letters as before, the following formula is suf- ficiently exact for all practical purposes: L ^ 2C + |(i) + d) + ^^-^ (1) When using this formula, C, D, and d must all be measured in the same unit. If it is desired to measure L and C in feet and D and d in inches, the formula then reduces to L = 2C'- + .1309(1)" + d") + 576(f -^ (2) by substituting D" = j^ and d" = y^ for D and d, respectively. In formula (2), C means C feet, and D" and d" mean D inches, and d inches. Example. — What length of belt is required when the pulleys have diam- eters of 56 in. and 16 in. and the distance between the centers is 24 ft. 3 in.? Solution. — Substituting in formula (2) the value 24.25 = 24 ft. 3 in., (Ka _ 1 A')2 for C;L =2 X 24.25 + .1309(56 + 16) + 57^ y. 24 25 " ^^.04- ft. Ans. §2 MENSURATION OF PLANE FIGURES 81 Since .04 ft. = .48 in. say ]4 in., the length of the belt may be taken as 58 ft. K in. 91. Length of Crossed Belt. — When the belt passes over the pulleys so as to cross the line O'O joining the centers, as in Fig. 56, it is called a crossed belt. A and Care the points of tangency for the pulley and A' and C are the points of tangency for the pulley 0' . As in the case of the open belt, the angles AOB, COD, A'O'B', and C'O'D' are all equal. The length of the belt is L = 2 X A'A + semicircle BFD + 2 X arc A5 + semicircle Fig. 56. B'F'D' + 2 X arc A'B'. As with the open belt, there is no simple formula giving an exact value for L, but the following is sufficiently exact for all practical purpose : {D + dY Or, L = 2C + ^(Z) + d) + L = 2C' + .1309(i)" + d") + 4C {D" + d"Y (1) (2) 576C' The letters in these formulas have the same values as in the formulas of Art. 90. It will be noted that the only difference between these formulas and those of Art. 90 is the sign of d in the last term. Example. — Using the same values as in the example of Art. 90, find the length of a crossed belt. SoiiUTiON. — Substituting the values given in formula (2), L = 2 X 24.25 + .1309(56 + 16) + 57^ x 24 25 " ^8.296- ft. = 58 ft. 4^^ in., very nearly. Ans. It will be noted that the length of the crossed belt, in this case, is 4M - K = 4M in. longer than the open belt. 92. Concentric and Eccentric Circles. — Two or more circles are said to be concentric when they have the same center. In Fig. 57, the circles ABC and A'B'C have the same center 0; they are, therefore, concentric circles. The distance A' A 82 ELEMENTARY APPLIED MATHEMATICS §2 = B'B = C'C, each distance being that part of the radius included between the circles. If one circle lies within another, but does not have the same center, the circles are said to be eccentric. In Fig. 58, the center of the larger circle is 0, and the center of the smaller circle is 0' ; hence, these circles are eccentric circles. If, in Figs. 57 and 58, the small circle represents a hole, the area of the space included between the hole and the large circle Fig. 57. Fig. 58. is evidently equal to the area of the large circle less the area of the small circle. Let D = diameter of the large circle, usually called the outside diameter, and let d = the diameter of the small circle, usually called the inside diameter; then area of large circle is -[D^ and area of small circle is -rd^. 4 4 surface in either Fig. 57 or Fig. 58 is : A =^2)2 4 4 4^ d^). The area of the flat (1) If the radius is used instead of the diameter, A = t{R^ - r2), (2) in which R is the radius of the outer circle and r is the radius of the inner circle. Example. — A circular disk has a hole through it that is 73^ in. in diam- eter. The outside diameter is 103^ in. What is the area of the flat surface? Solution. — Applying formula (1), remembering that j = .7854. A = .7854 (10.25^ - 7.125^) = 42.645 -, say 42.64 sq. in. Ans. §2 MENSURATION OF PLANE FIGURES 83 EXAMPLES (1) It is desired to bore a hole that shall have an area of exactly 2 sq. in.; what must be its diameter? Ans. 1.596- in. (2) The diameter of a rod is 4% in.; what is its circumference? Ans. 14|| in. (2) A pulley 19 in. in diameter and making 350 revolutions per minute drives a pulley 8 in. in diameter; how many revolutions per minute does the smaller pulley make? Ans. 831}^ r.p.m. (4) The outside diameter of a cross-section of a cyhnder is 27 in., the in- side diameter is 25 in.; what is the area of the cross-section? Ans. 81.68 + sq. in. (5) The circumference of a shaft is 22% in.; what is its diameter? Ans. 7% in. nearly. (6) Two pulleys having diameters of 80 in. and 20 in. are 22 ft. 6 in. be- tween centers; what length of open belt is required to connect these pulleys? Ans. 58 ft. 4i|V in. (7) Two pulleys that are 9 ft. 2 in. between centers are connected by a crossed belt; the diameters of the pulleys being 18 in. and 12 in., what is the length of the belt? Ans. 22 ft. 5i^ in. SECTORS AND SEGMENTS 93. Circular Measure of Angles. — Instead of measuring an angle in degrees, minutes, and seconds, it may be measured by- means of its arc expressed in terms of its radius. For in- stance, referring to Fig. 59, angle AOB = angle DOE. The number of degrees in the angle DOE is to the number of degrees in the semicircle GDEH as arc DFE is to arc GFH. Letting v = angle DOE, v° : 180° = arc DFE : xr, r being the radius OD and irr being >^ X 27rr = length of arc of semicircle. From this pro- portion, v° = arc ACB 180° X arc DFE irr = 57.296° X Fig. 59. arc DFE = 57.296' X R when R = radius OA of arc ACB. Now suppose that the arc DFE = the radius r, then the arc ACB must equal the radius R, and v° = 57.296° X ^ = 57.296° X ;^ = 57.296°; 84 ELEMENTARY APPLIED MATHEMATICS §2 that is, an arc of 57.296° is equal in length to the radius of the arc. Taking this arc as the unit of measure, in which ca,se, it is called a 180° radian, a semicircle is equal to rn oq^o = 3.1416 =7r radians; a quadrant, or 90°, is half a semicircle and is equal to 3^^^ = ^ = 1.5708 radians; any other angle, as ABC, will be equal to the number of degrees in the angle divided by 57.296. An angle measured in this manner is said to be measured in radians or to be measured in circular measure. When an angle is expressed in degrees, it is said to be measured in angular measure. arc DFE From the equation v° = 57.296° , DFE being any arc and r its radius, arc DFE v° , ^^^ . ,. = -- „„^o = angle DFE m radians; r oY.zyb in other words, if the length of an arc he diijided hy its radius, the quotient will be the measure of the angle in radians. Let d (Greek letter, pronounced theta) be the angle in radians, let I = length of the arc, and r — the radius of the arc, then the last equation may be written ^ = e (1) r From (1), I = rd (2) If an angle be expressed in radians, and it be desired to find its equivalent in angular measure, let v be the angle in degrees; y° then, smce ^^ ^qq" = ^' v° = 57.296°^. (3) If the angle is given in angular measure, and it is desired to express it in radians, ^ = .0174533^° (4) 57.296 Example 1. — The length of an arc is 23% in. and the radius of the arc is 32 in.; what is the angle in radians, and what is the angle in angular measure? 23 625 Solution. — From formula (1), d = ' = .73828 radians. Ans. From formula (3), v = 57.296 X .73828 = 42.3005° = 42° 18' 1.8". Ans. Example 2. — A certain angle is equal to 21° 31' 26"; if the radius of an arc having this angle is 153^ in., what is the length of the arc? Solution. — 'Reducing the minutes and seconds to a decimal of a degree, 21° 31' 26' = 21.5239°. By formula (4), 6 = .0174533 X 21.5239; by formula (,2), I = rd = 15.5 X .0174533 X 21.5239 = 5.823- in. Ans. §2 MENSURATION OF PLANE FIGURES 85 94. Length of Circular Arc. — Referring to formula (1), Art. 93, if r is equal to the unit of linear measure (1 foot, 1 inch, etc.), 6 = ^ = I] in other words, the measure of an angle in radians is the length of the arc to a radius I. Thus, in Fig. 59, if OD = OE = 1 inch, and 6 is the circular measure of the angle AOB = v, the arc DFE = 6 inches. Further, if OA = r, the length of the arc ACB = r X 6 in. = rd in. As previously defined, the figure OACB is a sector. By means of the formulas and principles of Art. 93, the length of the arc ACB of any sector can be found when the radius and the central angle (in either angular or circular measure) are known. If the chord AB and the height CI of the arc ACB are given, there is no simple, exact formula for finding the length of the arc ACB, in such cases, the angle AOB would usually be found by means of a table of trigonometric functions; then, knowing the angle, the length I of the arc can be found the formula I = rd ^ .0174533ry (1) If, however, a table of trigonometric functions is not available or if it is not desired to use such a table, the following formula will give results sufficiently exact for all practical purposes ^ 40r^(15 + 16^^) , . 75 + 180^2 + 64^6 ^ ' 1 . ,■ h height of arc , ,, ,. in which t = - = , ■ — 7 and r = the radius. c chord 01 arc To apply the formula, first calculate the value of ^; if r is not given, calculate it by formula (2), Art. 87. If the angle is not greater than a right angle, or 90°, the term 64^^ may be omitted, and the formula then becomes , ^ 40rf(15 + le^'') ^ 8r^(15 + 16^^) , . When the angle is equal to 90°, - = t= ~ ^ = .207107; c ^ hence, if the value of t equals or exceeds .21, use formula (2); but if t is less than .21, formula (3) may be used. Example 1. — What is the length of an arc whose chord is 7 3^ in. and whose height is 2 H^ in.? a a- . 0,1 -r, 75 117 75 ,, 16 75 SoLUTioN.-Smce f = 2H - 7A = 3^ - ^g- = 32 X m ^ 234 = .320513 is greater than .21, use formula (2). By formula (2), Art. 87, „ (M)Ml4(2£i)2 '^ ~ SX2U 86 ELEMENTARY APPLIED MATHEMATICS §2 Substituting in formula (2), , _ 40 X 4.02375 X .320513(15 + 16 X .320513^) ^ ^ .^» • 75 + 180 X .3205132 + 64 X .320513^ ^- ' ' ^^- '^'^*- The correct value of Z to 5 significant figures is 9.1748 +in.; hence, the value as calculated is sufiiciently exact for practical purposes. Example 2. — If the radius of the arc is 54 in. and the height of the arc is 8}4 in., what is the length of the arc and what is the central angle? Solution. — Referring to Fig. 59, suppose ACB is the given arc, and let OC be perpendicular to the chord AB; then AI = ^ and CI = h. In the right triangle AIO, 01 = r — h and OA = r. Therefore, ^ = V r^ — {r — h)^ = -^542 - (54 - 8.5)2 = 29.0818 in., and c = 29.0818 X 2 = 58.1636 8 5 in. Consequently, t = _„ " „ . = .14614. Since t is less than .21, formula (3) may be used, and , 8 X 54 X .14614(15 + 16 X .1 4614^) . ^= 15 + 36 X .146142 = 61.422 m. Ans. The correct value of Z to 5 significant figures is 61.422 in. From formulas (1) and (3), Art. 93, = -, and z; = 57.2960 = 57.2-96- 61 422 = 57.296 X ^Ht- = 65.171°- = 65° 10' 15.6". Ans. 04 The exact value is 65° 10' 14.6"; hence, the error is only 1 second, which is too small to be considered in ordinary practice. If the angle is large, that is, if it is greater than 90°, and great accuracy is desired, it will be better to find the ratio of the chord and height of half the arc, which may be designated by t'; then substituting t' for t in formula (3), the length of half the arc will be found very closely. By means of formula (1), Art. 87, and formulas (1) and (2), Art. 88, the value of ^' = - for half the arc can be shown to be , _ VSr/t + c , .V ^ 4l ^^^ Or, t' = ^^^'±^ (5) Having found t', the length of the whole arc will then be given by the formula, l^rf (15 + 16^^^) ^ ~ 15 + 36«'2 y"^) Referring to the first of the two preceding examples, ^ 4^2U ^ ^^^ Substituting this value of t' in formula (6), Z = 9.1749 + in. §2 MENSURATION OF PLANE FIGURES 87 95. Area of Sector and Segment. — The area of any sector of a circle is equal to one-half the product of its radius and arc. Let I = the length of the arc and r = the radius, then A = >^rZ. (1) It will be noted that this formula is the same as for the area of a triangle when I = the base and r = the altitude. Since I = rd, substitute rd for I in (1), and A = yir^d (2) that is, the area of a sector is equal to one-half the product of the angle in radians and the square of the radius. Referring to Fig. 59, the segment ACBA = sector OACB - triangle OAB. Area of triangle OAB = }ixABX 01 ;hut AB is the chord of the arc ACB = c, and 01 = radius OC — minus height of arc = r — h. Consequently, area of segment is A = H^l - Hc(r - h), or . rl — c(r — h) r^d — c(r — h) .„, A = -^ ^ = 2 ^^) Another formula for finding the area of a segment (approxi- mately) is the following, in which D is the diameter; ^ = f \/f - -^^ w This formula will give results sufficiently accurate for most practical purposes, and is rather easier to apply than (3), when it is necessary to calculate both the highest h and the angle d of the arc. Example. — Fig. 60 represents a round tank having an inside diameter of 60 in. The tank lies on a flat surface and is filled with water to a depth of 42 in. If the ends of the tank are flat and the tank is 12 ft. long, how many gallons of water are in the tank? Solution. — The volume of the water is equal to the area of the segment AGB multiplied by the length of the tank. To find the length of the arc AGB, calculate the length of the arc AFB and subtract it from the circum- ference of the circle AFBG. The height of the arc AFB is /i = 60 - 42 = 18 in. To find the chord AB, apply principle (1), Art. 87, and (2)^ = jP^ X H(? = 18 X 42 = 756; from which, c = \/3024 = 54.991 in. Then, h 18 since - = - . „„.. = .327326, use formula (2), Art. 94, and _ 40 X 30 X .327326(15 + 16 X .327326") ^ . 75 + 180 X .3273262 + 64 X .327326« ^"* The circumference of the circle is 3.1416 X 60 = 188.496 in., and the length of the arc AGB is 188.496 - 69.573 = 118.923 in. The area of the 88 ELEMENTARY APPLIED MATHEMATICS segment AGB is evidently equal to the area of the sector AGBO plus the area of the triangle AOB; area of sector = i^ x 30 X 118. 923 = 1783.845 sq. in.; area of triangle = M X 54.991 X (30 - 18) = 329.946 sq. in.; and area of segment = 1783.845 + 329.946 = 2113.791 sq. in. The length of the tank is 12 ft. = 144 in.; hence, the volume of the water is 2113.791 X 144 = 304385.9 cu. in. Since one gallon contains 231 cu. in., the number of gallons in the tank is 304385.9 ^ 231 = 1317.7 -, say 1318 gallons. Ans. Fig. 60. The area of the segment might have been found by finding the area of the small segment AFB and subtracting it from the area of the circle. The flat surface ABCD represents the water level. 96. Area of Fillet. — When two solids intersect so as to form a square corner, as shown in Fig. 61, the strength of the piece can be greatly increased by rounding the corner. This is usiially done by striking an arc of a circle having a small radius. The curved part thus added is called a fillet. While fillets add somewhat to the weight of the piece, it is customary to neglect this in the case of heavy castings; but if an accurate estimate of the weight is desired, it is necessary to include the weight of the fillets. Referring to Fig. 61, ACB is a fillet, and it is assumed that the sides AC and BC form a right angle. Letting OA = OB = r, the area of the fillet is equal to the area of the square AOBC — area of the quadrant AOB; hence, A ^ r^ - ^irr'^ = r^{l — .7854), or A = .2146r2. FiQ. 61. §2 MENSURATION OF PLANE FIGURES 89 Example. — If the radius of a fillet is yi in,, what is the area of the fillet? Solution. — Applying the formula, A = .2146 X {^iY = .01341 sq. in. Ans. It will be noted that the area is quite small; and since the radius is never very large, it is sufficiently exact for practical purposes to take the value of A as one-fifth the square of the radius. EXAMPLES (1) If the length of an arc is 10.26 in. and its radius is 33.14 in., what is the angle in radians and also in angular measure? . f .30960— radians. ^^s- \ 17° 44' 19/'. (2) If a certain angle measures 31° 12' 27", what is the length of the arc having this angle, the radius of the arc being 19.32 in.? Ans. 10.523 in. (3) The chord of an arc is 14^^ in., the height of the arc is 1% in., what is the length of the arc and what is the angle in radians and degrees? [ I = 15.378 in. Ans. \e= .99586 radian. \ v° = 57° 3' 32". (4) Referring to the example of Art. 95, how many gallons are in the tank when the depth of the water is 54 in.? Ans. 1671 gallons. (5) A fillet has a radius of %6 iii-j what is the area of the fillet? Ans. .021 — sq. in. INSCRIBED AND CIRCUMSCRIBED POLYGONS 97. An inscribed polygon is one whose vertexes lie on the circumference of a circle. In Fig. 62 the vertexes of the poly- gon ABCDEF lie on the circumference of the circle ABCDEF, and it is, therefore, an inscribed polygon. A circumsciibed polygon is one whose sides are tangent to a circle. In Fig, 62. the sides of the polygon ABCDEF are all tangent to the circle A'B'C'D'E'F', and it is, therefore, a circumscribed polygon with reference to the circle A'B'C'D'E'F'. If the polygon is a regular polygon, it may always be inscribed in a circle, and it may also be circumscribed about a circle. The polygon in Fig. 62 is a regular hexagon, and it is a property of the regular hexagon that the sides are always equal in length to the radius of the circle in which the hexagon is inscribed; thus AB = BC = OC. Consequently, a regular hexagon may be drawn by describing a circle and then spacing off on its circumference chords equal in length to the radius. The center of the circles within which the regular polygon is inscribed and about which it is circumscribed is the geometrical 90 ELEMENTARY APPLIED MATHEMATICS §2 a side, ^ = half a side= B'B, and (k)^= center of the polygon, and lines drawn from the center to the points of tangency, as OA', OB', etc. are apothems of the circumscribed polygon. Let a = the apothem and r = the radius of the circle within which the regular polygon is in- scribed; then, if s — a^, from which Given the radius r and the length s of the side, describe a circle with the radius r, and then space off on the circumference the length s. The value of a for polygons most commonly used may be obtained from the table given in Art. 81, in which a is a multiple of s. Fig. 62. Fig. 63. Example. — It is desired to inscribe a regular polygon of 11 sides in a circle having a diameter of 2J^ in. Show how this can be done by means of the table in Art. 81. Solution. — The apothems given in the table are multiples of the length of the side; hence, if the side be taken as 1 inch, the apothem for a regular polygon of 11 sides will be 1.7028 in., say 1.7 in. Draw two lines AB and OC at right angles to each other, Fig. 63. Make OC equal to 1.7 in., the apothem, and CB = CA = |s = | in., in this case (taking s as 1 in.), and draw OA and OB. With O as a center and a radius equal to the radius of the given circle = | X 2§ = 1| in., describe a circle; join the points E and F, the points of intersection of OA and OB with the circle, by the line EF, and EF will be one of the sides of the inscribed 11-sided polygon, and may be spaced off around the given circle. §2 MENSURATION OF PLANE FIGURES 91 If it is desired to calculate the length of the side EF, first find the length of OB. In the right triangle OCB, CB = .5, and OC = 1.7028; hence, OB = V-S^ + 1.70282 - 1.7747 in. The triangles AOB and EOF are similar; whence the pro- . EF OF s 1.25 portion -Jd OB' ^^ 1 and s = .70434 in. 1.7747' If the line AB be taken as some length other than 1 inch, it must be multiplied by the value of the apothem as given in the table to find the length of OC. THE ELLIPSE 98. The ellipse is a plain figure so constructed that the sum of the distances from any point on the curve to two fixed points is constant; thus, referring to Fig. 64, F and F' are the two fixed points, and CF + CF' ^ PF + PF' = QF -i- QF' = etc., and the closed curve is an elHpse. The longest line that can be drawn in the figure is the lineal 5, which passes through the two fixed points F and F'; it is called the major axis. From the center of the major axis, draw CD at right angles io AB; then CD is the short- est line that can be drawn in the ellipse, and is called the minor axis. The points A and B at the extremities of the major axis are called the vertexes of the ellipse; the point is called the center, and the fixed points F and F' are called the foci, either 92 ELEMENTARY APPLIED MATHEMATICS §2 point being called a focus of the ellipse. The vertexes, center, and foci all lie on the major axis, and the distance AF must equal the distance F'B; hence, for the point A, FA -\- F'A = F'A + F'B = AB; and the sum of the distances from any point on the curve to the foci is equal to the major axis. For the point C, CF = CF' = lAB = OA = OB. In practice, an ellipse is usually specified by giving the lengths of the major and minor axes, which are commonly called the long and short diameters. An ellipse may be drawn mechanically in the following manner: Lay off the long diameter (major axis), and bisect it, thus locating the center 0; through 0, draw a perpendicular CD, and lay off OC = OD = one-half the short diameter (minor axis); then, with C (or D) as a center and with a radius equal to one-half the long diameter, draw short arcs cutting AB in F and F', thus locating the foci. Stick pins in the paper at F and F' and also at C; tie one end of a piece of thread to one of the pins F orF', and pass the thread around the other two pins, drawing it taut and passing it several times around the pin at the other focus. Now pull out the pin at C, and with a pencil held perpendicular to the plane of the paper and pressing against the thread (but not hard enough to stretch it) , move it so that it keeps the thread tight, thus describing one-half of the el- lipse, say the upper half ACPB; then bring the thread to the other side of the pins, describe the other half of the eUipse or ADQB. 99. Circumference and Area of Ellipse. — In works on mathe- matics, it is customary to denote one-half the major axis by the letter a and one-half the minor axis by the letter 6; thus, in Fig. 64, a = OA = OB, and b = OC = OD. The area of an elhpse is A = irab (1) There is no exact formula giving the circumference (periphery) of an ellipse, but the following formula, in which r = — tj-t and p = the circumference (periphery) , gives results sufficiently exact for all practical purposes: V=Ha-\- b) ^^ ~ ^g^, (2) Example. — The long and short diameters of an ellipse are 163^ in. and 4% in. respectively; what is the area of the ellipse? what is the circumference. §2 MENSURATION OF PLANE FIGURES 93 Solution. — The area, by formula (1), is A - ,ai, - 3.1416 X ^« X f . ^« X 16.25 X 4.5 = .7854 X 16.25 X 4.5 = 57.432+ sq. in. Ans. The circumference (periphery), by formula (2), is p = 3.1416 16.25 , 4.5\ 64 - 3 X .56627^ / 16.25 4^\ 64 - \ 2 "^ 2 / 64 - = 35.264 in. 2 ' 2 / 64 - 16 X .566272 To apply formula (2), first calculate the value of r; thus, r 2a - 26 16.25 - 4.5 Ans. a — b = .56627-. Then, r^ a + h ,320662 -; r« = (r^)^ ~ 2o + 26 16.25 + 4.5 = .3206622 = .102824 + ; the remainder of the work is evident. It may be remarked that the major axis divides the ellipse into two equal parts; the minor axis also divides the eUipse into two equal parts. AREA OF ANY PLANE FIGURE 100. If the figure can be divided into elementary plane figures {i.e. triangles, rectangles, circles, segments, etc.) the area of each of the elementary fig- ures may be calculated, and the sum will be the area of the entire figure. Referring to Fig. 65, the dotted lines show that the figure may be divided into three trape- zoids two of them equal — one rectangle, a seg- ment of a circle, and two equal fillets. From the dimensions marked on the drawing, the areas of all these ele- mentary figures may be found, and their sum will be the area of the entire figure. The work is as follows: There are two trapezoids of the same size as ahdc; the length of the side ac is 2^ + f = 3", and the area of the two trapezoids is 3 + 2i /, B \h _: s" < \^" y < 3?" *- I. -• h ^& ^'4*1 C < -'a ■ d -< 81 b >- Fig. 65. A = X 31 X 2 = 17.72 sq. in. 94 ELEMENTARY APPLIED MATHEMATICS §2 The length of the rectangle is 2| + f + 4| = 7i", and its area is ^ = 7i X If = 12.47 sq. in. The length of the side ih of the trapezoid kjhi is If+lf+ll = 41" and its area is 4f + 2i X 3 = 10.5 sq. in. . The area of the fillets is 2 X i X (f)^ = .06 sq. in. Concerning the segment, the radius and the chord are known and it is necessary to calculate the height. Using formula (3), Art. 87, h = r - ^W^r^ - c" = 1.5 - Ha/4 X 1.5^ - 2.252 = .508". Substituting this value of h in formula (4), Art. 95, A = 4 X 5082 2 X 1.5 - .608 = .79 sq. in. 3 \ .508 The entire area is 17.72 + 12.47 + 10.5 + 0.06 + 0.79 = 41.54 sq. in. 101. If the figure is of such a nature that it cannot be divided into simple elementary figures, proceed as in Fig. 66. Draw a Fig. 66. line AB, either through the figure or outside of it, as in Fig. 66; then draw lines CB and DA tangent to the extreme ends of the given outline and perpendicular to the line AB. The intersec- tion of the tangents with AB locate the points A and B. Divide AB into any convenient even number of equal parts, in the present case 8, and through the points of division, erect perpendiculars (called ordinates) to AB; these intersect the figure in the lines §2 MENSURATION OF PLANE FIGURES 95 a'a, h'h, c'c, etc., which are all parallel to one another, and are equally distant apart, li n = the number of parts into which AB has been divided, and h = the distance between two con- AB secutive ordinates, measured parallel to AB, h = -^• If, now, another series of ordinates are drawn midway between those previously drawn, here indicated by the dotted hnes, and the lengths of these dotted ordinates are measured and added, the sum so obtained multipHed by h will be the approximate area of the figure. This method is called the trapezoidal rule. The greater the number of parts into which AB is divided, that is, the smaller the value of h, the more accurate will be the result obtained for the area. 102. Another rule that is more accurate than the method just described is what is known as Simpson's rule. Having drawn the ordinates dividing the figure into n equal parts (w being an even number), designate the ordinates by yo, yi, y^, etc., as shown in the figure. Then letting i/o and yn be the end ordinates, the area by Simpson's rule is A = I [?/o + i/n + 4(yi + 2/3 + 2/5 + etc.) + 2(i/2 + ?/4 + J/e + etc.)] Expressed in words, Simpson's rule is: Add the end ordinates, four times the odd ordinates, and 2 times the remaining even ordinates, and multiply this sum hy one-third the distance between the ordinates. Applying both methods to the outline given in Fig. 66, the lengths of the dotted ordinates, beginning with the left and pro- ceeding in order to the right, are 0.97", 1.52", 1.51", 1.04", 0.57", 0.67", 0.85", and 0.75"; their sum is 7.88". The number of equal parts is 8 = n, and I = AB = 3.02". Therefore, h = ^-f, and A = ^-f X 7.88 ^^t^^.>f^ = 2.97 sq. in., by the trapezoidal rule. Applying Simpson's rule, ?/o and y^ are both equal to 0; yi = 1.34", yz = 1.34", y^ = 0.64", y^ = 0.85", and the sum of these odd ordinates is 4.17"; y^ = 1-56", y, = 0.67", y, = 0.83", and the sum of these even ordinates is 3.06". Therefore, by the rule, since ^ = -^— '^ 3 "^ .12583, A = .12583 (0 + + 4 X 4.17 + 2 X 3.06) = 2.87 sq. in. 96 ELEMENTARY APPLIED MATHEMATICS §2 103. It was stated in Art., 101 that the greater the number of parts into which AB is divided the greater the accuracy of the result, and this is true whichever method is used. If AB be divided into 16 equal parts ins^tead of 8, the area will be 2.954 sq. in. by the trapezoidal rule and 2.890 by Simpson's rule. For 32 equal divisions, Simpson's rule gives 2.919 sq. in. Tabulat- ing these results, Trapezoidal rule (8 parts) 2.975 sq. in. Error, +0.056 Trapezoidal rule (16 parts) 2.954 sq. in. Error, +0.035 Simpson's rule (8 parts) 2 . 869 sq. in. Error, — . 050 Simpson's rule (16 parts) 2 . 890 sq. in. Error, — . 029 Simpson 's rule (32 parts) 2 . 9 19 sq. in. Error, Assuming that the result obtained by Simpson's rule for 32 parts is correct to all four figures, it will be observed that the error when using the trapezoidal rule is somewhat greater than when using Simpson 's rule. In practice, it is always advis- able to divide AB into at least 10 parts; this not only increases the accuracy, but it makes n a very convenient number to divide by when finding h. If particularly accurate results are desired, it is best to make n = 20, in which case, the result will be as accurate as the limits of measurement permit. ELEMENTARY APPLIED MATHEMATICS (PART 2) EXAMINATION QUESTIONS (1) The length of the diagonal of a square is 5i inches; what is (a) the length of one of the sides, and (6) what is the area? ■ (a) 3.7123 in. ^^^* I (h) 13.781 sq. in. (2) The lengths of the sides of a triangle are 29.1 ft., 21.8 ft., and 36.5 ft.; what is the area of the triangle? Ans. 317.18 sq. ft. (3) The side of a regular dodecagon measures 4| in.; what is the area of the polygon? Ans. 266.08 sq. in. (4) If (a) the diameter of a circle is 35f in., what is its area? (6) What must be the diameter of a circle that will enclose an area of 1800 sq. in.? .^i («) 1003.8 sq. in. ^^^- I (6) 47.873 in. (5) The length of the chord of a circular arc is 46f in., the height of the arc is lOi in., what is (a) the radius? (6) the angle in radians? (c) the angle in degrees? First find length of arc. r (a) 31.636 in. Ans. \ (6) 1.6570 radians. I (c) 94° 56' 27" (6) Referring to the last example, what is (a) the chord of half the arc? (b) what is the height of half the arc? r (a) 25.466 in. ^^^- 1 (&) 2.6756 in. (7) The diameters, of two belt pulleys are 64 in. and 14 in. and the distance between the centers is 18 ft. 9 in. (a) What length of open belt is required? (6) what length of crossed belt? Give lengths to nearest 16th of an inch. Ans /(«)47ft.ll^in. ^''^- I (6) 48 ft. 3-i:V in. T 97 98 ELEMENTARY APPLIED MATHEMATICS §2 (8) If the central angle of a circular arc is 121° 27' 36", (a) what is the angle in radians? (6) if the radius is 241 ft., what is the length of the arc? . f (a) 2.1199 radians. 1(6) 51.938 ft. (9) Find from the dimensions given, the area of the outline shown in Fig. 1. The dotted lines indicate how the figure should be divided. As shown, the figure is divided into two semi-circles, two rectangles, two trapezoids, two sectors, and two fillets. Ans. 28.261 sq. in. Fig. 1. (10) If the chord of a circular arc is 8.5 in. and the height of the arc is li\ in., what is (a) the radius? (6) the length of the arc? (c) the angle in radians? (d) the angle in degrees? (e) the area of the sector? Ans. (a) (h) (c) id) (e) 7.5372 in. 9.0307 in. 1.1981 radians. 68° 38' 56" 34.033 sq. in. §2 EXAMINATION QUESTIONS 99 (11) Referring to the last question, (a) what is the area of the segment? (6) Calculate the area also by formula (4) of Art. 95. ^'^^' 1(6) 7.575 sq. in. (12) If the major and minor axes of an ellipse are 37 in. and 12| in., respectively, (a) what is the area? (6) what is the peri- phery of the elHpse? . / {a) 363.25 sq. in. ^'^^- 1(6) 82.594 in. (13) The outhne shown in Fig. 2 resembles an indicator dia- gram showing the variation of pressure in a steam-engine cylin- der. Find the area of the diagram (a) by the trapezoidal rule, f] \ y c 1 1 D A iin r- jB Fig. 2. and also by (6) Simpson's rule. The end ordinates should be tangent to the curve at C and D and perpendicular to a hne parallel to AB. The value obtained by dividing the area by the perpendicular distance between the end ordinates is called the mean ordinate, and is indicated on the diagram by the dotted line MN; that is, it is equal to the perpendicular distance be- tween MN and AB. Find (c) value of the mean ordinate. Divide diagram into 12 equal parts; for the convenience of the student, the ordinates have been drawn in position. (14) A certain building lot has the shape of a sector of a circle whose radius is 149 ft., the frontage being 117 ft. What is the area of the lot, and what part of an acre is it? Ans. 8716.5 sq. ft. = .20010 A. (15) An opening is to have the shape of an equilateral triangle and is required to have an area of 16 sq. in.; what is the length of one of the sides? Ans. 6.0787 in. 100 ELEMENTARY APPLIED MATHEMATICS §2 (16) Referring to Question 7, suppose the speed of the belt is 2640 ft. per min. and there is no sKp; (a) how many revolutions per minute does the large pulley make? (6) how many does the small pulley make? . ( (a) 157.56 r.p.m. \(b) 720.30 r.p.m. ERRATA Study Paper No. 6 Page 102, Change first two lines of Art. 108, to read as follows: "108. If a prism be intersected by a plane parallel to a lateral edge, the section so formed by the cutting plane is called a" Page 103, Line 20, Change "either base" to read "base of right prism." Page 108, Line 16 from bottom, change "area" to "volume." Page 109, Line 9, strike out " = ah." Page 110, Change answer to Ex. 6 to read "Ans. 13.98, say 14 cu. ft." Art. 121, line 3, change "cd" to "cb." Page 114, Line 5; change "8070 + " to ".8071 -" line 7, change ".807" to ".8071," and "76.61" to 76.6" bottom line, change "r°" to "r\" Page 115, Second line of Solution to Ex. 2; change in two places ".807" to ".8071," and change "7.^63" to "7.2639;" also, in last line, change "7.263" to "7.2639" and change "80.86" to "80.72." Page 116, Line 9, change "wiR + r)iR - r)" to "irl(R + r)iR - r);" also, 4th and 5th lines from bottom, change "201" to "210." Page 117, Line 17 from bottom, change "generatrix" to "directrix." Page 123, Line 8 of Art 133; read "a"' instead of "a". Page 124, Ans. to Ex. 2 should be "593 cu. in." instead of "5930 cu. in.". Page 127, Line 17, the second "area" should be "volume." Page 129, Line 10 from bottom, change "10^" to "11^." Page 130, Ans. to Ex. 5 should be "2368 cu. ft." instead of "339,605 cu. ft." Ans. to Ex. 8 should be "23,513 gal." instead of "23, 692 gal." Page 133, Last line of Solution, change "86.128" to 86.127." Page 136, Line 2, change "5930" to "593;" line 3, change "28497" to "2850." Line 9 from bottom, change "5930" to "593." Line 8 from bottom, change "15394" to "1540.4." Page 145, Ans. to Ex. 7 should be "35.749 cu. yd." instead of "365.88 cu. yd." Page 146, Ans. to Ex. 8 should be "12.179 lb." instead of "12.268 lb." Ans. to Ex. 10 should be "1184.9 gal." instead of "319.17 gal." ELEMENT AEY APPLIED MATHEMATICS (PART 3) MENSURATION OF SOLIDS PRISMS, CYLINDERS, CONES, AND SPHERES POLYEDRONS 104. Every solid has three dimensions — length, breadth, and thickness; see Art. 43. A simple example of a solid is shown in Fig. 67. Here the sides aa'h'h, hh'c'c, cc'd'd, and aa'd'd are called the faces or lateral sides; the end sides abed and a'h'c'd' are called the bases. Both ends and all the faces are j^ai, i.e.^ they form parts of plane surfaces, and in the illustration, the ends are parallel and the opposite faces are also parallel. Fig. 67. The planes of the faces intersect in right lines to form the edges of the soHd; in Fig. 67, the edges formed by the intersec- tion of the sides are a' a, b'h, c'c, and d'd. The planes of the faces also intersect the planes of the bases in right lines to form the edges at the ends of the solid; in Fig. 67, the edges formed by the 101 102 ELEMENTARY APPLIED MATHEMATICS §2 intersection of the faces and ends are ab, he, cd, da, a'h' , h'c', c'd' , and d'a' . 105. Prisms. — Any solid whose ends and sides are made up of plane surfaces is called a polyedron. If the polyedron has two equal and parallel bases, it is called a prism, and if the bases of the prism are rectangles or squares, it is called a parallelopiped. The solid shown in Fig. 67 is a prism and also a parallelopiped. The bases of a prism may be polygons of any shape — regular or otherwise — but they must be parallel and equal. 106. If the planes of the bases are perpendicular to the planes of the sides, the lateral edges are perpendicular to the edges of the bases, and the prism is called a right prism. Fig. 67 shows a right prism, which is also a right parallelopiped. 107. If a prism be intersected by a plane at right angles to the lateral edges, as the plane MN in Fig. 67, the outline of the plane figure so formed will be equal to the bases, and the section is called a right section or a cross section (usually, the latter). In Fig. 67, the plane MN intersects the prism in the cross section a^^h^^ c^^ d^^ , which is equal and parallel to the base ahcd. 108. If a prism be intersected by a plane at right angles, i.e., perpendicular to, the bases, the section so formed is called a longitudinal section; in Fig. 67, the plane PQ intersects the prism in the longitudinal section a"h"h"'a"' , the plane FQ being perpendicular to both bases. Whatever the shape of the prism, every longitudinal section is a rectangle, since it can cut only two faces at one time, or if it cuts three faces, it must pass through one of the edges, thus forming but one side of the rectangle. 109. The altitude of a prism is the perpen- dicular distance between the bases. Thus, in Fig. 68, suppose that the sides are not at right angles to the bases; then In, which represents the perpendicular distance between the planes of the bases, is the altitude. 110. Referring to Figs. 67 and 68, it will be observed that the sides of any prism are parallelograms, and when the prism is a right prism, the sides are rectangles. Since the bases are parallel, all the lateral edges are of equal length. Fig. 68. §2 MENSURATION OF SOLIDS 103 111. Area and Volume of Prism. — By lateral area is meant the area of the outside of the soHd not counting the ends; in the case of a prism, it is equal to the area of the faces. Referring to Fig. 67, note that if the plane MN is perpendicular to one of the faces, it is perpendicular to all of them, and the lines of intersection of the plane with the faces are perpendicular to the edges; thus, a^^h^^ is perpendicular to a' a and h'h, b^^c^^ is per- pendicular to b'b and c'c, etc. Hence, these hues are equal to the altitudes of the parallelograms forming the faces, the lengths being all equal to the lengths of the lateral edges, which are all equal. Consequently, the lateral area of a prism is equivalent to that of a rectangle whose length is one of the lateral edges and whose altitude is equal to the sum of the lengths of the lines formed by the intersection of the plane MN (perpendicular to the lateral edges) with the faces of the prism, and this latter is equal to the perimeter of the polygon formed by the intersection of the plane MN with the faces. Let A = the lateral area of the prism, I = length of a lateral edge, and p = the perimeter of the polygon formed by a right section of the prism = perimeter of either base; then, A =pl (1) If the area of the ends is also included, the result is called the entire area. Let Ae = the entire area and let a = the area of one end; then, Ae = A -{- 2a = pi + 2a (2) 112. By volume of a solid is meant the number of cubic units that it contains; if the unit of measurement is one inch, then the volume is the number of cubic inches that the solid would occupy if made of some soft material that could be formed into a cube. The phrase cubical contents is frequently used instead of the word volume, and the two terms have identically the same meaning. Let V = the volume of a prism, a = the area of one of the bases, and h = the altitude; then, V = ah (1) That is, the volume of a prism is equal to the area of the base multiplied by the altitude. If the base is a rectangle, let I = the length and b = the breadth; then, V = blh (2) 104 ELEMENTARY APPLIED MATHEMATICS §2 If the base is a square, let d = the length of one of the sides; then, V = d% (3) If the base is a square and the altitude is equal to one of the sides of the base, V = h' (4) If the prism is a right prism, the altitude is equal to one of the lateral edges, and if it is also equal to one of the edges of the base, the prism is a cube, and its volume is found by formula (4) . Example 1. — A room is 22 feet long, 14 ft. 8 in. wide, and 9 ft. 10 in. high; how many cubic feet of air does the room contain? Solution. — The cubical contents of the room is, by formula (2), the prod- uct of the length, width, and height. 118 12 ft. Hence, F = 22 X ^ X ^^ 14 ft. 8 in. = 118 456896 144 176 12 = 3172 ft. and 9 ft. 10 S Q cu. ft. Ans. Example 2. — A blowpit 20' X 10' X 14' is filled with stock weighing 65 lb. per cubic foot; what is the weight of the stock? If the stock is 5.7% fiber, what is the weight of the fiber? Solution.— The cubical contents of the blowpit is 20 X 10 X 14 = 2800 cu. ft. Since 1 cu. ft. of stock weighs 65 lb., the weight of the stock is 2800 X 65 = 182,000 lb. Ans. Since 5.7% of the stock is fiber, the weight of the fiber is 182,000 X .057 = 10,374 lb. Ans. 113. Pyramids. — A pyramid is a polyedron having one base, a polygon, and whose sides are triangles meeting at a common point called the vertex. When a pyramid is referred to by letters placed at the vertex and at the corners of the base, the letter at the vertex is written first, followed by a short dash, and then the letters of the base. Thus, the pyramid shown in Fig. 69 would be referred to as the pyramid s-abcd. Note that the pyramid in Fig. 69 has a square base; the sides ash, bsc, csd, and dsa are triangles whose vertexes have the common point s. The polyedron (pyra- mid) in this case is formed by the intersection of five planes — ■ four forming the sides and the fifth forming the base. If, from the vertex s, a perpendicular be drawn to the base, the line so in Fig. 69, this line is the altitude of the pyramid. Fig. 69. §2 MENSURATION OF SOLIDS 105 If the base of a pyramid is a regular polygon and the projection of the vertex upon the base coincides with the geometrical center of the base, the pyramid is called a regular pyramid. Suppose that the base of the pyramid shown in Fig. 69 is a square and that the point o, which is the projection of the vertex s upon the base, is the center of the square, then s-abcd is a regular pjrramid, and the line so is called the axis of the pyramid. The length of so is also equal to altitude of the pyramid. If a line sf be drawn from the vertex s perpendicular to one of the sides of the base of a regular pyramid, this line is called the slant height. There will be as many slant heights as there are sides in the polygon forming the base, and for regular pyramids, all slant heights are equal; if the pyramid is not regular, some or all the slant heights are different. 114. Area and Volume of Pyramid. — The lateral area of any regular pyramid is equal to perimeter of the base multiplied by one-half the slant height. This is evident, since the sides are sf isosceles triangles, and the area of one side is ^ X be, Fig. 69; hence, the lateral area 18 ^ X be X n (n = number of sides in the base). But n X be = p, the perimeter of the base. Let s = the slant height and A. = the lateral area; then, A = ^X p = hp If the pyramid is not regular, calculate the area of each face and find the sum. 115. The volume of any pyramid is equal to the area of the base multiplied by one-third the altitude. Let h = the altitude = so in Figs. 69 and 70, V = the volume, and a = the area of the base; then, h V = ^ X a = Iha Example. — The sides of the base of a regular triangular pyramid are 13K in. long and the altitude is 18% in.; what is the entire area and what is the cubical contents of the pyramid. Solution. — Fig. 70 shows a sketch of the pyramid. Here sf is the slant height and so is the altitude. The base abc is an equilateral triangle, and of is the apothem, the length of which is (see table, Art. 81) .28868 X 13.25 = 3.825 in. The triangle sof is a right triangle, right-angled at o, and sf 106 ELEMENTARY APPLIED MATHEMATICS §2 = -\/3.8252 + I8.3752 = 18.769 in. The lateral area is, by formula of Art. 114, A = 1M69 ^ ^g 25 x 3 = 373.034 sq. in. The area of the base 13.25^ is, by formula (2) of Art. 69, — '-j — VS = 76.021 sq. in. The entire area therefore is 373.03 + 76.02 = 449.05 sq. in. Ans. The volume is given by the formula above, and is 18 ^7^ V = g X 76.02 = 465.62 cu. in. Ans. 116. Frustum of a Pyramid. — If a pyramid be intersected by a plane parallel to the plane of the base and the top part removed, the remaining portion of the pyramid is called a frustum of the pyramid. Thus, referring to Fig. 71, the plane a'b'c'd'e' is « parallel to the base abcde; now removing that ,^|'^ part of the pyramid above the intersecting // I \\\ plane (here indicated by the dotted lines), the „//,_.|-\jf\ remaining part is a frustum of the pyramid // H \)X s-ahcde. The two bases of the frustum are // ' -J— ^ -^^\ similar plane figures (see Art. 139), and they \"l i \ \ ^^ust be parallel. / " \ / ^ good example of a frustum of a pyramid /_ V isthebottomof abin: also, a spout. Here the be , , frustum IS inverted, the large end bemg up, and the ends are open. In such cases, the frustum has the shape of a solid that would exactly fit the opening. The altitude of a frustum is the perpendicular distance between the bases; it is indicated in Fig. 71 by the dotted line o'o, which is a part of the altitude of the pyramid. If the frustum is a part of a regular pyramid, the slant height of the frustum is that part of the slant height of the pyramid that is included between the bases. 117. In problems relating to frustums, the altitude and the length of the sides of both bases are usually given. For a regular pyramid, the sides are equal trapezoids, and the lateral area is n times the area of one side, where n = the number of sides. Let p' = the perimeter of the lower (larger) base and p" = the perimeter of the upper base; then, if s = the slant height of the frustum of a regular pyramid and A = its lateral area, A = Mp' + P") 118. To find the volume of any frustum of a pyramid, whether a regular pyramid or otherwise, let a' = area of lower base, §2 MENSURATION OF SOLIDS 107 a" = area of upper base, h = the altitude, and V = the volume of the frustum; then, V = ih{a' + a" + Va^X^O That is, the volume of a frustum of any pyramid is equal to one-third the altitude multiplied by the sum of the areas of the two bases and the square root of their product. Example. — The lengths of the edges of the lower base of a frustum of a triangular pyramid (one having three sides) are 12 in., 15 in., and 18 in.; the corresponding edges of the upper base are 7 in., 8% in., and lOK in.; if the height of the frustum is 10 inches, what is its volume? Solution. — Using the formula of Art. 70, the area of the upper base is A =i \/26.25(26.25 - 2 X 7) (26.25 - 2 X 8.75) (26.25 - 2 X 10.5) = 30.385 sq. in. since p = 7 + 8.75 + 10.5 = 26.25 in. Using the same formula, area of lower base is 89.294 sq. in. Then, volume of frustum is V = ^ X 10(89.294 + 30.385 + \/89.294 X 30.385) = 572.56 cu. in. Ans. 119. Prismatoids. — A prismatoid is a polyedron whose ends are parallel, but the polygons forming the outline of the ends are not equal and may have a different number of sides. In Fig. 72 is shown a prismatoid, one end of which is a pentagon and the other end a quadrilateral. It will be noted that four of the sides, abgf, aeif, eihd, and cdhg are quadrilaterals, while the side bgc is a triangle. These sides are all plane sur- faces, and are formed by passing planes through the edges of the lower base and the vertexes of the upper base. Thus, planes passed through the vertex g and the edges be and cd intersect to form the lateral edge gc; the intersection of these two planes with planes passing through the edges ab and de -pia. 72. and the vertexes g and / and h and i inter- sect in the lateral edges gb and hd; a fifth plane passed through ae and the vertexes / and i intersects the last two planes in the lateral edges fa and ie. If both ends have the same number of sides and are similarly situated, the prismatoid is called a prismoid. Thus, the frustum of a pyramid is a prismoid, see Fig. 71, since the ends are parallel and both contain the same number of sides. Fig. 73 shows another prismatoid, the sides ade and 6c/ being triangles; one end, abed, is a quadrilateral and the other end is a right line ef. 108 ELEMENTARY APPLIED MATHEMATICS §2 To find the volume of a prismatoid, let h = the altitude = the perpendicular distance between the parallel ends, let' a' = area of lower base, a" = area of upper base, and am = area of the middle cross section; then the volume of the prismatoid is equal to one-sixth the altitude multiplied by the sum of the areas of the upper and lower bases and 4 times the area of the middle cross section; that is. To find the area of the middle section, note that a', Fig. 72, is midway between a and /, b' is midway between 6 and g, etc. ; hence, a'b' fg + ab b'c' = 6c + be Fig. 73. c'd' = gh + cd etc. 2 2' Knowing the lengths of the sides and their position and directions, the area can be found. This formula is called the prismoidal formula; it may be used to calculate the volumes of many solids besides prismatoids. Applying it to the example of the last article, the sides of the 7+12 15 + 8.75 „^_ , = 9.5, 5 = 11.875, and middle section are 18 + 10.5 2 "' 2 14.25; the area of this triangle is 55.964 sq. in. Therefore, the area of the prismoid (frustum) is 7 = >^ X 10 (89.294 + 30.385 + 4 X 55.964) = 572.56 cu. in.. 120. The Wedge.— When the base of the prismatoid is a rectangle and the end parallel to it is a right line parallel to one of the edges of the base, the prismatoid is called a wedge. Referring to Figs. 73 and 74, if the base abed is a rectangle and ef is parallel to ab, the figure represents a wedge. In Fig. 73, the upper base, the line ef, is shorter than ab = dc, and the solid is a prismatoid; in Fig. 74, ef = ab = dc, and the solid is a triangular prism whose bases, eda = fcb, are parallel, and whose altitude is ab. The volume of the prismatoid in Fig. 73 may be found by applying the prisnjoidal formula; but an easier method is to find Fig. 74. §2 MENSURATION OF SOLIDS 109 the sum of parallel sides, divide it by 3, and multiply the quotient by the area of a right section — one taken at right angles to the parallel sides. Thus, letting a = the area of the right section a'e'd', y ^ aX {ah -\- dc + ej) ,^. o When the prismatoid becomes a wedge, ah = dc, and y ^ a X (2ah + ej) ^^. o When a wedge becomes a prism, ah = dc = ef, and V = a X ah = ah The volume of the wedge may also be calculated by the prismoi- dal formula. Thus, let ah = m, be = n, ef = m', and h = the altitude; then, for the middle section, the sides parallel to ah or dc are equal to ^ ; the sides parallel to he or ad are equal to — ^ — = ^; 4 times the area of the middle section is ^ — - X K X 4 = mn + m'n; the area of one end is mn, and the area of the other end is the line e/ = 0; therefore, by the prismoidal formula, ^ = ^ wn + + (mn + m'n) or V = lh(2m + m')n. (3) For the wedge shown in Fig. 74, m = m', and V = ^hmn. (4) EXAMPLES (1) How many cords of wood are contained in three piles having the fol- lowing dimensions: 30' 6" X 7' X 4', 20' X 8' X 4', and 8' X 5' X 4? One cord contains 128 cu. ft. Ans. 12f|, say 13 cords. (2) A spout has the form of a frustum of a square pyramid ; the bases are 18 in. and 10 in. square, and the altitude is 28 in. How many square feet of sheet iron will be required to make this spout? Ans. 11 sq. ft. (3) How many gallons will the spout mentioned in the last example hold when filled? Ans. 24.404 gal. (4) A trench 20 ft. wide and 36 ft. long is dug; the bottom of the trench is level, but the top slopes, me end of the trench being 8 ft. 9 in. deep, the other end 5 ft. 3 in. deep, and the slope gradual from end to end. If the walls and ends are vertical, how many cubic yards of material were removed? Ans. 186f cu. yd. 110 ELEMENTARY APPLIED MATHEMATICS §2 (5) A freight car is loaded with wood in the following manner, all sticks being cut to 4-foot lengths : two piles at each end of the car placed crosswise and one pile in the middle (divided into two equal parts) placed lengthwise of the car. The average height of the piles at one end is 7' 8", at the other end 7' 6", and the height of the pile in the middle is 7'; the inside dimen- sions of the car: length, 36 ft.; width, 8 ft. 6 in.; height 8 ft. If an allow- ance of 18 in. of the car length be made for spacing between the piles, how many cords were placed in the car? Ans. 16.3 cords. Suggestion. — The total length of the piles is 36 — 1.5 = 34.5 ft. placed crosswise (measured lengthwise of the car) is (34.5 — 4) - length of the middle pile is 4 ft. The length of each pile ■ 2 = 153-^ ft.; since the (6) A pedestal has a lower base 28" X 40", and upper base 21" X 30", and the perpendicular distance between the bases (which are parallel) is 28 in. How many cubic feet are in the pedestal? Ans. 14.99, say 15 cu. ft. (7) How many cubic yards of concrete are required to make a foundation wall, the outside dimensions of which are: length, 136 ft.; breadth, 68 ft. 6 in., the height of the wall being uniformly 7 ft. 9 in. and the thickness being 18 in.? Ans. 173.514 cu. yd. THE THREE ROUND BODIES 121. The Cylinder. — If one side ad of a rectangle ahcd, Fig. 75, be fixed in position and the rectangle be revolved around this fixed side, the opposite side cd will generate a curved surface that is called a cylinder. The other two sides of the rec- tangle will generate the circles hh'h"h"' and cc'c"c"', which form the ends or bases, of the cylinder. It will be seen that the bases are parallel and that they are perpendic- ular to the fine ad passing through their centers; the line ad is called the axis of the cylinder. Any line drawn on the cylinder c"' parallel to the axis is called an element of the cylindrical surface, or, simply, an element; in Fig. 75, he, h'c', etc. are ele- ments of the cylinder, and they are evi- dently positions occupied by the line he as it revolves about the axis db while generating the cylinder. 122. A cylinder generated as just described is called a cylinder of revolution, and since the bases are circles and perpendicular to the axis, it is also called a right cylinder with circular base. The same rules and formulas that were given for finding the area and volume of a prism may be used to find the area and vol- FiG. 75. §2 MENSURATION OF SOLIDS 111 ume of a cylinder, but since all right sections are circles, these rules and formulas may be somewhat simplified. Referring to Fig, 76, suppose the cylinder is a cylinder of revolution and that it hes on a plane surface; the line (element) ab will then be the line of contact of plane and the cylinder, and the plane is said to be tangent to the cylinder. Now roll the cylinder on the plane surface until the element ab again comes in contact with the plane; the distance moved through by the cylinder will evidently be equal to the circumference of a circle whose diameter is equal to the diameter of the base. During this movement, every element of ■Trd~ Fig. 76. the cylinder has come into contact with the plane, and the surface thus touched by the cylinder, which is equal to the outside surface of the cylinder, is the rectangle aa'b'b, called the development of the cyHnder. The outside area of a cylinder, corresponding to the lateral area in a prism, is called the convex area, and the sur- face is called the convex surface. The diameter of a cyHnder is the diameter of a right section, and the altitude of a cylinder is the perpendicular distance between the bases ; it is equal to the length of the axis in a cylinder of revolution, and it is represented by h in Fig. 77. Letting d = the diameter of the cylinder, I = the length of the axis, it is plain from Fig. 76 that the convex area of a cylinder of revolution is A = irdl If the cylinder is not a right cylinder, but has parallel bases, as in Fig. 77, it may be made a right cylinder by passing a plane 112 ELEMENTARY APPLIED MATHEMATICS §2 through the cyHnder perpendicular to the axis and just touching one edge of the base at c; the part ebjc thus cut off may be placed on the other end, as indicated by the dotted lines, and the cylinder then becomes a right cylinder e'ecd. The length of the axis is the same as before, that is, o'o = o"'o"', hence, A = irdl, as before. Fig. 77. 123. If one of the bases is perpendicular to the axis and the other is not, for instance, if the cylinder has the form e'bcd, Fig. 77, the area is equal to the sum of the areas of the right cylin- der e'ecd and the wedge-shaped solid cbe. The latter is evidently equal to one-half of a right cylinder whose base is the circle ec (center ©") and whose axis is equal to the element eb; that is, the area of ebc = ^^-2 X ird X eb = ird X oo", since J-^ X e6 e'b -H dc = oo". But, oo" + o"o' 00 = hence, area of cylinder e'bcd is A = irdX o"o"' + Trd X oo" = Trd{oo" + o"o"'), or (e'b -h dc\ A=.d (p^) If both bases are oblique to the axis, cut the cylinder by planes perpendicular to the axis, calculate the areas of both wedge- shaped solids and of the right cylinder included between them, and then find the sum of the three results. §2 MENSURATION OF SOLIDS 113 Example. — The cylindrical part of a plater roll forms a right cylinder 17 in. in diameter and 36 in. long; what is its convex area? Solution. — Using the formula of Art. 122, A = TT X_17 X 36 = 1922 66 sq. in. = 13.352 sq. ft. Ans, 124, If a right cylinder be intersected by a plane oblique to the axis, which cuts the base of the cyUnder, the part thus cut off, ifgk, Fig. 78, is called a cylindrical ungula, the word ungula having reference to an object shaped ^ like a horse's hoof. Let fg be the line in which the cutting plane inter- sects the base. Bisect fg, and draw k'k through the point of bisection o', and perpendicular to fg; then k'k is a diameter of the base. Let o be the middle point of k'k, and ok = ok^ = r, the radius of the base = radius of cylinder. Let o'f = o'g = a, o'k = b, and arc gk = 1/2 arc fkg = ; then, the convex area of the ungula is, when h = the altitude ik, A^'f[ii-r)^ + a] (1) If the line of intersection pass through the center of the base, as indicated by the dotted outline in Fig. 78, 6 and a are both equal to r, and formula (1) then becomes A = ^[^^ ~ ^^r "^ ^1 from which A = 2rh (2) If the cutting plane just touches the other edge of the base, as in the case of ebc, Fig. 77, b = 2r and a = 0. Substituting these 27'/ir irT ~] values in formula (1), ^ = ^[(2r - r)y + Oj, since is then a semicircle and is equal to rr. Reducing this expression, A = ^rh = Yi-Kdh (3) which is the same value as was obtained in Art. 123. Note that - is the angle okg in radians. Example. — Suppose the radius of a right cylinder is 9 in. and that the cylinder is cut by a plane in such manner that the altitude of the ungula is 8 in.; if o = 63^ in., what is the convex area of the ungula? 8 114 ELEMENTARY APPLIED MATHEMATICS §2 Solution. — Referring to Fig. 78, 2a = 2 X 6.5 = IS = fg = chord of arc fkg; b = o'k = height of arc = h = r — )4\^^r^ — c^ = 2.775, by formula (3), Art. 87; — = one-half the length of the arc divided by the radius = ^• Since t = - = ■,„ = .21346, which is very near .21, use formula (3) of C io Art. 94, and- = g" = g ci e i o(if2) = .8070+. Now substituting in formula (1), A = ^ ^ ^-^ ^ [(2.775 - 9)0.807 + 6.5] = 76.61 sq. in. Ans. 2.775 125. The volume of any cylinder whose bases are parallel is equal to the area of the base multiplied by the altitude; it is also equal to the area of a right section multipHed by the length of the axis. If the cyhnder is one of revolution, the base is equal to a right section, and both are equal to the area of a circle whose diameter is equal to the diameter of the cylinder; Hkewise, the altitude is equal to the length of the axis. Let h = the altitude, I = the length of the axis, d — the diameter of the cylinder, a = the area of the base; then, V^ah = ~dH (1) 4 For a cylinder of revolution, V = Tr% = ^ d% (2) 4 When the bases are not perpendicular to the axis, they have the form of an ellipse, and their areas may be calculated by formula (1), Art. 99. Thus, referring to Fig. 77, the lower base bjck is an ellipse, whose major axis (long diameter) is be, and whose minor axis (short diameter) is jk. Knowing these two dimensions (which call D and d, respectively), a in formula (1), Art. 99, is ^ and 6 is ^; hence the area of the ellipse (base) is expressed by Trab = TT X -^ X ^ = jDd, and the volume of the cylinder is V = '^Ddh (3) 126. The volume of an ungula is given by the following formula, in which the letters have the same meaning as in formula (1), Art. 124: V = ^[a(3r° - a') + 3r( b- r)] (1) §2 MENSURATION OF SOLIDS 115 If the line of intersection fg of the cutting plane with the plane of the base passes throu'gh the center, as indicated by the dotted outline i'k'g'f, Fig. 78, a = h = r, and formula (1) reduces to V = |r2/i (2) If the cutting plane just touches the edge of the base, as indi- cated by ebjck, Fig. 77, a = 0, & = 2r, ^ = wr, and formula (1) reduces to (3) V = '^r^h .807, or <^ = .807 X 9 = 7.263. That formula (3) is correct is readily seen, since the ungula ebc is equal to one-half the cylinder whose base is the circle ec and whose altitude is eh = h in formula (1). Example 1. — The shaft of a plater roll is 65 in. long and 8 in. in diameter; if made of steel, one cubic inch of which weighs .2836 lb., what is the weight of the shaft? Solution. — The shaft is a cylinder of revolution, and its volume is V ^'^dH ^j X 82 X 65 = -.7854 X 64 X 65 = 3267.264 cu. in. The weight of the shaft is 3267.264 X .2836 = 926.6- lb. Ans. Example 2. — Referring to the example in Art. 124, what is the volume of the ungula? Solution. — Here r = 9, h = 8, a = 6.5, 6 (found by calculation) = 2.775, and - (found by calculation) = Substituting these values in formula (1), V = 3 ^ ^^^^ [6.5 (3 X 92 - 6.52) + 3 X 9(2.775 -9) 7.263] =80.86 cu. in. Ans. 127. If the cylinder has a hole through it, it is called a hollow cylinder, pipe or tube. A cross section of a hollow cylinder is shown in Fig. 79. It is assumed that the hole is also a cylinder and that its axis coincides with the axis of the cylinder that contains the hole. The volume of the hollow cylinder is evidently equal to the difference of the volumes of the cylinder and hole. If I = the length of the cylinder, R = radius of cylinder, r = radius of hole, V = tRH - irrH = tI(R^ - r') = Trl{R -V r){R - r) Fig. 79. 1{D + d){D - d) (1) 116 ELEMENTARY APPLIED MATHEMATICS §2 Suppose the cross section in Fig. 79 is that of a tube of thickness t; then t = R — r = oa — oh. Suppose further that a circle be drawn midway between the inner and outer circles as indicated by the dotted line. Then the radius oc of this circle is equal to Vm = — ^ — ; its length (circumference) is 2Trm = 27r X — k — = t{R -\- r); multiplying this by the thickness t and length I, the product will be the volume of a fiat plate having the same cubical contents as the tube. Or, since t = R — r, V = 2Trm tl = t{R -hr) {R - r) which is the same as formula (1). Letting 2rm = dm = diameter of middle circle, V = Tvdjl (2) In other words, the cubical contents of a hollow cylinder, pipe, or tube is equal to the continued product of tt = 3 . 1416, the mean diameter, the thickness, and the length of the cylinder. Example 1. — A cylindrical tank made of wrought iron, a cubic inch of which weighs .2778 lb., is to hold 1000 gallons; the tank is to stand with the axis vertical, and the diameter (inside) is to be the same as the height; what is the diameter of the tank? If the thickness of the shell is 3^ in., what is its weight? Solution. — Since the diameter equals the height, d = Z in formula (1), Art. 125; hence, since there are 231 cu. in. in a gallon, = M![^ = ^^ = 1000, or d = ^\^^ = 66.503, say 66J^ in. Ans. 231 924 ' \3.1416 The inside diameter is 66.5 in., and since the thickness is .5 in., the out- side diameter is 66.6 + 2 X .5 = 67.5 in.; then, by formula (1), Art. 127, the cubical contents of the shell is F = ^ X 66.5(67.5 + 66.5) X 1 = 6998.7 cu. in.; this multiplied by .2778, the weight of a cubic inch of wrought iron, is 6998.7 X .2778 = 1944.24, say 1944 lb. Ans. Example 2. — Assuming that the shell in the last example is made by roll- ing a flat sheet, 3^ in. thick, into a cylindrical form, what should be the size of the sheet? Solution. — The shape of the sheet will be that of a rectangle, one side of which is equal to the height of the tank = 66.5 in., and the other side will be equal to the circumference of a circle that is midway between the inside and outside circles of a right section of the tank. The diameter of this circle is evidently equal to the inside diameter plus the thickness, or 66.5 + .5 = 67 in., and its circumference is 3.1416 X 67 = 201.49, say 201^ in. Therefore, the sheet must be 201 K in. long and 66 J^ in. wide. Ans. Note that the mean diameter of the two circles is — '- — ^ ^ = 67 in., the same result as was obtained by adding the thickness to the inside diameter. V = §2 MENSURATION OF SOLIDS 117 128. The term cylinder is not confined to solids having cir- cular bases or whose right sections are circles; it is applied to solids whose bases have any shape whatever, except those classed as prisms, the elements of whose sm'faces are all perpendicular to a right section. Such solids may all be generated by keeping a line in contact with the outline of a plane figure, then moving the line so that it touches every point of the plane figure, and at the same time, always remains parallel to a given right line. The outline of the plane figure is called the directrix, and the moving line is called the gen- eratrix. In Fig. 80, ahcdefgh is the directrix and a'a is the generatri-x. The generatrix moves over the directrix, always remaining parallel to the fixed line AB, and thus generates the cylindrical sur- face shown in the figure. While, strictly speaking, thes directrix should always be a curve in order to apply the term cylinder to the surface generated, it may, nevertheless, be applied to those surfaces in which the generatrix consists of both straight lines and curves. Keeping in mind this definition of a cylindrical surface and call- ing the solid that it bounds a cyhnder, the general formulas previously given for the area and volume of a cylinder will apply in this case also. Example. — What volume of stock is displaced by a beater roll 48 in. wide and 60 in. in diameter that is immersed to a depth of 28 in. in the stock? SoLtTTiON. — The outline of the stock displaced is a cylinder 48 in. long, with equal bases having the shape of a segment of a circle; and a right section of this cylinder is a segment of a circle, the radius of the circle bemg 60 -T- 2 = 30 in., and the height of the segment being 28 in. The volume displaced is equal to the area of the segment multiplied by the length of the cyhnder. To find the area of the segment, first calculate the chord, using formula (5), Art. 87, and c = 2V(2 X 30 - 28)28 = 59.867 in. Since the angle is very large, use formulas (4) and (6) of Art. 94 to find the length of the arc. From formula (4), t' = .19740+, and from formula Fig: 80. 118 ELEMENTARY APPLIED MATHEMATICS §2 (5), I = 90.25 in. The area of the sector is 90.25 X 30 1353.75 sq. in.; , , . , 59.867 X 2 _ _^ area of triangle = ^ = 59.86/ sq. in.; and area of segment = 1353.75 — 59.87 = 1293.88 sq. in. Therefore, volume of cyhnder = volume of stock displaced = 1293 88 X 48 = 62,106 cu. in. Ans. 129. Whenever a cylindrical tank is partly filled with a fluid of some kind and is so placed in position that its axis is neither vertical nor horizontal, an ungula is formed. The upper surface of the fluid is always a horizontal plane, and the intersection of this plane with the cylindrical surface forms either an ungula or what might be called a frustum of an ungula. Thus, referring to Fig. 81, which shows a cylindrical tank having one end higher Fig. 81. than the other and partly filled with some fluid (stock, for ex- ample), the upper surface aa'h'h is a plane, level and parallel to the horizontal line eg. If the tank were extended until the plane of the top surface just touched the bottom of one end, as indicated by the dotted lines, the outline of the fluid contents would be the ungula f-acb. The part that is actually in the tank has, in this case, the outline acbh'c'a' , which may be called a frustum of the ungula f-acb, and its volume is evidently equal to the difference between the volume of the ungula f-acb and the volume of the ungula f-a'c'h'. The point / can be easily found when the length of the tank and the depth of the fluid at each end are known. Thus, let I = length of tank, m = dc = depth at lower end, n = d'c' = depth at upper end, and x = c'f = the additional length of tank required ; then, from the similar triangles fed and fe'd', cf : e'f = ed : c'd'. But e'f = x and ef = I + x; hence, I -]- x: x ^ 7n: n, or nl X = m — n §2 MENSURATION OF SOLIDS 119 Example.— Referring to Fig. 81, suppose that the tank is 42 in. in diam- eter, 12 ft. 6 in. long, and that the upper end is 10 in. higher than the lower end; if partly filled with stock, so that the distance ed, measured along the tank bottom, is 14^ in-, what is the volume of stock in the tank? Solution.— Before the point / can be found, it is necessary to calcula,te the distance c'd' = n in th e formula ab ove. Since c'c = 12 ft. 6 in. = 150 in. and c'g = 10 in., eg = VlSO^ - 10^ = 149.67 in. Draw ck perpendicular to df; then dkc is a right triangle. Since dc is perpendicular to cf, angle dck = angle c'cg, because ck is perpendicular to eg, which is parallel to dj. Con- sequently, triangles dkc and c'gc are similar, and ck : cd = eg : c'c, or ck : 42 - 14.25 = 149.67 : 150, from which ck = 27.689 in. = distance from stock level to horizontal line eg. Draw e'k' perpendicular to df; then, triangle d'k'c' is similar to triangle dkc, since the sides are parallel, and k'g = kc = 27 689 in. But k'c' = 27.689 - 10 = 17.689 in. Then, from the similar triangles d'k'c' and c'gc, e'd' : 150 = 17.689 : 149.67, or c'd' = 17.728 in. In the formula above, I = 150, m = 42 - 14.25 = 27.75, and n = 17.728; hence x = ^^0 X 17-728 ^ 3 j ^ ^j ^^^ ^j ^ 265.34 + 150 = hence, x 27.75-17.728 415.34 in. Calculating first the volume of the ungula f-acb, it is evident that the arc acb is greater than a semicircle, since ed is less than the radius oe = 42 ^ 2 = 21 in. Hence, to find the length of the arc acb, find the length of the arc aeb and subtract it from the circumferen ce of whic h it is a part. According to Art. 87, db^ = ed X dc, or db = Vl4.25 X 27.75 = 19.886 in. = a in formula (1), Art. 126. Since the arc is very large, use formula (2) of Art. 94 to find the length of the arc aeb. Here t = 2 x 19.886 " ■^^^^^> ^"^ 40 X 21 X .35829(15 -H6 X .35829^) _^, ^^^ .^_ ^^e circumfer- * - 75 + 180 X .358292 + 64 X .35829« ence of the circle is 42 X 3.1416 = 131.947 in.; therefore, arc aeb = 131.947 - 52.242 = 79.705 in. = 2 in formula (1), Art. 126, or = 39.8525 in., say 39.853 in. Using this formula, h = 415.34, b = 27.75, a = 19.886, r = 21, <^ = 39.853, and V = -^ ^-^^ [19 886 (3 X 212 _ 19.8862) + 3 x 21(27.75 - 21)39.853 .3X27.75^ = 176,577 cu. in. Next calculate the volume of the ungula f-a'c'b'. The arc a'c'b' is less than a semicircle, because c'd' is less than the radius o'e'. Here d'b' 1 7 7*?^ I = V'r7.728(42 -l7728y = 20.744; t = 2^2077U ^ ■^^^^°' ^""^ '^ ^ 2 40 X 21 X .4273(15 + 16 X .4273^) ^ ^g 711 in - 2 X 75 -M80 X .42732 + 64 X .4273« Now using formula (1), Art. 126, h = 265.34, b = 17.728, a = 20.744, r = 21, <^ = 29.711, and y,/^_ 268. 13 _ [20.744(3 X 21^ - 20.744^) + 3 X 21(17.728 - 21) ^ ^ ^^"^^^ X 29.711] = 61,831 cu. in. Finally, 7 = F' - F" = 176,577 - 61831 = 114,746, say 114,750 cu. in., the volume of the stock in the tank. Ans. 120 ELEMENTARY APPLIED MATHEMATICS §2 The reader will find it excellent practice to work this entire example, performing all the operations herein indicated. 130. The Cone. — If a right triangle be revolved about one of its legs as an axis, the hypotenuse will generate a conical surface, and the triangle as a whole will generate a solid called a cone. Thus, referring to Fig. 82, let soa be a right triangle, right- angled at 0, and suppose the triangle to be revolved about the leg so; then, the hypotenuse sa, and the triangle as a whole will generate the solid s-aa'a"b. The point s is called the vertex of the cone. Any right line drawn from s to the base aa'a"h is called an element of the cone, more properly, an element of the conical surface; thus, sa, sa', sa", etc. are elements of the cone in Fig. 82. The line so is the axis of the cone, and o is the center of the base of the cone. Any cone generated in this manner is a cone of revolution; and because the base is perpendicular to the axis, the cone is also called a right cone. Let a right cone be laid on a plane, and suppose the line of con- tact of the plane and cone to be sa, Fig. 82; now roll the cone on Fig. 83. the plane, the vertex s remaining stationary. Each element of the cone will come into contact with the plane, and since they are all of the same length, the surface of contact thus generated wiU §2 MENSURATION OF SOLIDS 121 be the circular sector saa', Fig. 83, the radius sa being an element of the cone, and the length of the arc aa' wiU be equal to the cir- cumference of the base of the cone, or 2'k X oa, Fig. 82. Since the area of a sector is equal to one-half the product of the radius and the length of the arc, it follows that the convex area of a right cone is equal to one-half the product of the perimeter of base of the cone and the length of an element. The length of an element is called the slant height of the cone; representing this by s, and letting r = the radius of the base, the convex area is A = 3'^s X 2Tr = wrs 131. If a cone be cut by a plane that intersects the element directly opposite the element first cut by the plane, the surface formed by the intersection of the plane and cone is an ellipse, except when the plane is perpendicular to the axis; in this latter case, the section is a circle, which is an ellipse having all its di- ameters equal. Referring to Fig. 84, the section a'c'h', which is formed by the intersection of a plane perpendicular to the axis so of the cone, is a circle; the base, which is « also perpendicular to the axis, is also a /i\ circle. In Fig. 85, the intersecting plane / j \ is not perpendicular to the axis so of / i \ the cone, but it intersects the element /{-— i— -A ah, which is directly (diametrically) op- cit~—'P—Ab' posite the element dc, hence, the section / jc \ aed is an ellipse, and the base is also / | \ an ellipse. / • \ The altitude of a cone is the perpen- /,'-'' — 1 ""'^-^X dicular distance between the vertex and t' \ \ the base. In the case of a right cone, \ ** y the altitude is equal to the length of the ^^-~-~-______---'^ axis; it equals so in Figs. 82 and 84. In ^ the case of an obhque cone, Fig. 85, the ^^^' ^^' base is not perpendicular to the axis, and the altitude is there indicated by H, the perpendicular distance between the base and the vertex. The volume of any cone is equal to one-third the product of area of the base by the altitude. Let a = area of base and h = the altitude; then, V = iah (1) If the cone is a cone of revolution and d = the diameter of 122 ELEMENTARY APPLIED MATHEMATICS §2 the base, the area of the base is o = j^d'^', substituting this value of a in formula (1), V = ^d^h = .2Q18d% (2) If the base is an ellipse, as in Fig. 85, let D = he, the long diam- eter, and d = gf, the short diameter; then, the area of the base is IT jDd = a. Substituting this value of a in formula (1), V = ^Ddh = .2Q18Ddh (3) Example. — Making no allowance for thickness, what is the area of a piece of thin sheet metal that will just cover the convex surface of a cone with a circular base having a diameter of 33 in. and an altitude of 45 in. ? If made of wood weighing 48 lb. per cubic foot, what is the weight of the cone? Solution. — The area of the sheet metal will be the same as the area of the cone; hence, using the formula of Art. 130, A ,xfx V-+(f) = 2484.5 sq. in. Ans. Here the radius is 33 -r- 2, and the slant height is the hypotenuse of a right triangle, one leg of which is the radius and the other leg is the altitude. To find the weight, it is first necessary to calculate the volume. Apply- ing formula (2), above, V = .2618 X 33^ X 45 = 12829.5 cu. in. Since 48 a cubic foot weighs 48 lb., the weight is evidently 12829.5 X Ty^ = 356.4, say 356 lb. Ans. 132. A frustum of a cone is that portion of the cone included between the base and a plane parallel to the base and intersecting a the cone. In Fig. 84, a'ahb' is a frustum of the cone s —ah] in Fig. 85, ahcd is a frustum of the cone s — hgcf. In both cases, the two bases are parallel; but in Fig. 84, both bases are perpendicular to the axis, while in Fig. 85, they are obUque to the axis. It is apparent that a frustum of a right cone is always a right frustum, since if one base is perpendicular to the axis (as it must be in the case of a right cone) the other base must also be perpendicular to the axis.- If a frustum of a right cone be rolled on a plane surface, the area thus generated on the plane, called the development of the Fig. 85. §2 MENSURATION OF SOLIDS 123 frustum, will have the outhne caa'c', Fig. 83, which may be considered to be the development of the frustum in Fig. 84. Here sa corresponds to the element m, Fig. 84, and sc corresponds to sa'. The area caa'c' is equal to the sector saa' — sector sec' . Letting R = radius of lower base and r = radius of upper base, perimeter of lower base = 2TrR = arc aa'; perimeter of upper base = 2Trr = arc cc' ; then area of frustum is equal to one-half the sum of the circumferences of its bases multiplied by the slant height of the frustum. Let s = the slant height = a'a, Fig. 84, then A = K(27ri^ + 2rr)s, from which A = 7rs{R + r) = |s(Z) + d) When D and d are the diameters of the bases. 133. The volume of any frustum of a cone is equal to one-third the altitude multiplied by the sum of the areas of the upper base, the lower base, and the square root of the product of the areas of the bases. Thus, let a' = the area of the lower base, a" = the area of the upper base, and h = the altitude = h in Fig. 85; then, V = Hh{a' + a" + VV^) (1) If the bases are circles, let D = the diameter of the lower TT base and d = the diameter of the upper base; then, a = t dD^, a" = -rd^, and substituting in formula (1) and reducing, V = .2618/i(Z)2 + d^ + Dd) (2) If the bases are not perpendicular to the axis, they are ellipses; letting D' and d' be the long and short diameters of the lower base and D" and d" the long and short diameters of the upper base, a' = jD'd', a" = jD"d", and substituting in formula (1) and reducing, V = .2618h{D'd' + D"d" + VD' D"d'd") (3) Example. — Find the weight of the plater roll shown in Fig. 86; all right sections are circles; the roll is solid; and it is made of cast iron weighing 450 pounds per cubic foot. Solution. — The parts marked A and E are cylinders having the same diameter, 8 in., and are therefore equivalent to a single cylinder having a diameter of 8 in. and a length of 9 + 14.5 = 23.5 in. The part marked C is a cylinder 17 in. in diameter and 36 in. long; the parts marked B and D are frustums of cones whose bases have diameters of 12 in. and 17 in. and an altitude of 3 in. The volume of the roll will be the sum of the volumes of the parts A, B, C, D, and E. The volume of the cylinders may be calcu- 124 ELEMENTARY APPLIED MATHEMATICS §2 lated by formula (2), Art. 125, and the volume of the frustums by formula (2), above. Volume of A and E = .7854 X 8^ X (9 + 14.5) = 1181 cu. in. Volume of C = .7854 X 17^ X 36 = 8171 Volume of 5 and D = .2618 X 3(17^ + 122 + 17 X 12) X 2 = 1001 Total volume =10,353 cu. in. The number of cubic feet is .. ' „ , and the weight of the roll is A„p X 450 = 2696 lb. Ans. Since the constant 450 is given to only three significant figures, the various results were limited to four significant figures. Fig. 86. Example 2. — Referring to Fig. 85, suppose the long and short diameters of the lower base are 16H in. and 11 in., and the long and short diameters of the upper base are 6.6 in. and 4.4 in.; if the altitude is 8 in., what is the volume of the frustum? Solution .— Applying formula (3 ), above, V = .2618 X 8(16.5 X 11 +6.6 X 4.4 + \/l6.5 X 11 X 6.6 X 4.4 ) = 5930 cu. in. Ans. 134. The Sphere. — If a semicircle be revolved about its diameter, the surface thus generated is called a spherical surface; the solid that is bounded by a spherical surface is called a sphere. In Fig. 87(a), suppose the semicircle ras to be revolved about its diameter rs; it wiU then generate a spherical surface, and the line rs is called the axis of the sphere. The middle point o of the axis is the center of the sphere. Any right line drawn from the center o and ending in the spherical surface, as the line og, is the radius of the sphere, and it is evident that all radii of a sphere are equal, since they are equal to the radius of the semicircle that generates the sphere. Any line that passes through the center and is terminated by the spherical surface, as the hne rs, is a diameter of the sphere, and any diameter may be taken as the axis. The points where the- axis intersects the spherical surface are called the poles; in Fig. 87(a); r and s are the poles. §2 MENSURATION OF SOLIDS 125 If a sphere be intersected by a plane, the figure thus formed will be a circle; and if the plane passes through the center of the sphere, the circle is called a great circle; otherwise, it is called a small circle. In (a) Fig. 87, rash, rcsd, resf, etc. are great cir- cles; the circles represented by the lines kj and mn are small circles. All great cir- cles are equal; but a small circle may have any value between that of a great circle and 0. Assuming the earth to be a perfect sphere, the meridians of longitude are all great circles, while all the parallels of latitude (ex- cept the equator) are small circles. 135. The area of a sphere is equal to the area of four great circles. Let r = the radius of a great circle = radius of the sphere, and d = 2r = the diameter; then, ^=47rr2 = 7rd!2 = 3. 1416^2 (1) The volume of a sphere is equal to its area multi- plied hy one-third the radius = 4xr2 X Ir, or y=|7rr3 = i7rd3 = .5236# (2) Example. — A certain manu- facturer states that a can of his paint will cover 100 square feet; how many cans will be required to give two coats of paint to a sphere 100 in. in diameter? If 126 ELEMENTARY APPLIED MATHEMATICS §2 the sphere were hollow and its inside diameter were 100 in. how many- gallons would it hold? Solution. — Applying formula (1) to find the area, which multiply by 2 since two coats are to be applied, A = 3.1416 X 1002 X 2 = 62,832 sq. in. = 62,832 -h 144 = 436+ sq. ft. Therefore, the number of cans of paint required is 436 -H 100 = 4.36, say 4H cans. Ans. To find the number of gallons that the sphere will hold, apply formula (2) to find the volume, which divide by 231, the number of cubic inches in a- gallon; thus, V = .5236 X 1003 = 523,600 cu. in., and 523600 -^ 231 = 2266% gal. Ans. 136. If a sphere be intersected by two parallel planes, the part included between the planes is called a spherical segment. Referring to (a), Fig. 87, kj and mn represent parallel planes, and kamnbj is a spherical segment. It will be noted that a spherical segment is somewhat similar to a frustum of a cone, and has two bases, kj and mn. If, however, one plane is tangent to the sphere, as pq, the spherical segment has but one base. In (a), Fig. 87, rkij and rkmnj are spherical segments of one base. The perpendicular distance between the bases is the altitude of the spherical segment ; thus, il is the altitude of the spherical segment kmnj, ir is the altitude of the spherical segment rkij, etc. The convex surface of a spherical segment is called a zone; thus, the convex surface of kmnj is a zone of two bases, and the convex surface of rkij is a zone of one base. Note that a zone means a surface — not a solid; there is no such thing as the volume of a zone. 137. The area of a zone is equal to the circumference of a great circle of the sphere of which the zone is a part multiplied by the alti- tude of the zone. The altitude of a zone is the same as the altitude of the spherical segment. Let r = the radius of the sphere, d = the diameter of the sphere, and h = the altitude of the zone ; then, A = 27rrh = wdh (1) The volume of a spherical segment is equal to half the sum of its bases multiplied by its altitude plus the volume of a sphere of which that altitude is the diameter. Thus, letting the letters represent the same quantities as before, and letting ri and di be the radius and diameter of one base, and r2 and dz the radius and diameter of the other base, V = }iwiri^ + r2')h + Hrh' = .52S6h[S{ri^ + r^') + h'] (2) Also, V = .ld09h[d{di^ + di^) + 4/i2] (3) §2 - MENSURATION OF SOLIDS 127 If the spherical segment have but one base, as rkij, Fig. 87(a), make r2 = in formula (2), and V = .5236/i(3ri2 + h^) = .lS09h{Sdi^ + 4:h'') (4) 138. Referring to (6) and (c), Fig. 87, suppose a sphere to be generated by revolving the semicircle racs about its diameter rs; then, that part of the sphere that is generated by any sector of the semicircle is called a spherical sector. In (6) , Fig. 87, the spher- ical sector raob is generated by the sector aor of the semicircle; the remainder of the sphere, shoa, is also a spherical sector. In (c). Fig. 87, aohdoc is a spherical sector generated by the sector aoc. The zone forming the convex surface of the spherical sector is called the base of the sector. In (6), Fig. 87, the zone abr is the base of the spherical sector raob; and in (c). Fig. 87, the zone abdc is the base of the spherical sector aobdoc. The volume of a spherical sector is equal to the area of the zone that forms its base multiplied by one-third the radius of the sphere. Since the area of a zone is 2Trrh, the area of a spherical sector is 2xr/i X }ir, and V = ^r% = 2.0944r2/i o Example 1. — Fig. 88 shows a tank 28 in. long, 12 in. wide, and filled with water to a depth of 16 in. A ball 8}i in. in diameter is partly sub- merged in the water, the vertical depth, measured on the axis being 6 in'. To what additional height, x, will the water level be raised by the ball? Solution. — ^Let x = the additional height of the water due to the ball; this height will be the same as though an amount of water equal in volume to that displaced by the ball had been added to the water in the tank, and this volume is equal to the volume of a spherical segment of one base having an altitude of 6 in. the diameter of the sphere being 8.5 in. To find the radius of the base, let CBDA be a section of the ball, Fig. 89, taken through the center; then, CBDA is a great circle, and CD is the axis of the sphere. AB is the water level and is 6 in. from the lowest point D of the ball. AB is also the diameter of the base of the spherical segment ADB, and in Fig. 89, is the chord of the arc ACB. One-half of AB = AE is the radius of the base of the spherical segment. Now applying the principle of Art. 87, AE^ = r-i2 = CE XED = (8.5 - 6) X 6 = 15. Substituting this value of Ti^ in formula (4), Art. 137, V = .5236 X 6(3 X 15 + 6=) = 254.47 cu. in. which is the volume of that part of the ball submerged in the water. The amount of water in the tank is 28 X 12 X 16 = 5376 cu. in. Adding to this the volume of the spherical segment, 5376 + 254.47 = 5630.47 cu. in. Representing the depth of the water after the ball has been placed in it by d, the volume occupied by the water and the submerged part of the ball is 128 ELEMENTARY APPLIED MATHEMATICS §2 28Xl2Xd = 5630.47; from which, d = 5630.47 ^ (28 X 12) = 16.7574- in. Hence, the ball raised the water level 16.7574 - 16 = .7574 in. = x. Ans. If only the value of x had been desired, all that would be necessary to ascertain it, is to divide the volume of the spherical segment by the area of Fig. 88. the bottom of the tank, since the volume of the prism abcdefgh, Fig. 88, must be equal to the volume of the submerged part of the ball, and the area of the base of the prism is the same as the area of the bottom of the tank. Hence, 254.47 -h ( 28 X 12) = .7574 - in., as before. Example 2. — A ball 15 ft. in diameter is to be painted in two colors in such a manner that the top and bottom zones will be of one color and the middle zone of another color, all three zones having the same altitude. If the middle zone is white and the other two black, which Fig. 89. color will require the most paint, and what will be the area of the surfaces covered? SoLTTTiON. — The altitude of each zone is 15 -i- 3 = 5 ft. By formula (1), Art. 137, the area of a zone is irdh; since d and h are the same for all three zones, their areas are equal, and A = 3.1416 X 15 X 5 = 235.62 sq. ft. §2 MENSURATION OF SOLIDS 129 Therefore, the area to be covered with the white paint is 235.62 sq. ft., and the area to be covered with the black paint is 235.62 X 2 = 471.24 sq. ft. Ans. EXAMPLES (1) Fig. 90 shows an iron casting. The part marked C is 8 in. in diam- eter and 1 in. thick; part Z) is 5 in. in diameter; part B is 10 in. square and 1}4 in. thick; there are four strengthening webs marked A, which may be considered as triangular prisms, neglecting the small curve formed where they join the middle cylinder. Fig. 90. Taking the weight of a cubic inch of cast iron as .2604 lb., what is the weight of the casting? The height over all is 103^ in., and a hole 1}4 in. in diameter passes through the center of the casting. Ans. 95.8 lb. (2) A wooden ball 18 in. in diameter has a 9-inch hole through it, the axis of the hole coinciding with the axis of the ball; what was the original volume of the ball, and what is the volume after the hole has been placed in the ball? Note that the part removed when the hole was bored is a cylinder and two spherical segments of one base, both segments being equal. Ans. 3054- cu. in.; 1983+ cu. in. (3) One side of a stock chest 12 ft. high is shaped like a semicircle with a radius of 5 ft. 2 in.; the side opposite is a right line 10 ft. 4 in. long and is 130 ELEMENTARY APPLIED MATHEMATICS §2 joined to the semicircular part by right lines 7 ft. long, the angle between the long side and the two short sides being a right angle. If the chest stands with its axis vertical, what is its capacity in cubic feet when filled to within 8 in. from the top? Ans. 1295 cu. ft. (4) What is the capacity in United States gallons of a vertical cyUndrical stock chest 11 ft. 10 in. high and 14 ft. inside diameter, if it is filled to within one foot of the top, no allowance being made for the displacement of the agitator? Ans. 12,475 gal. (5) If the consistency of the stock is 275 pounds per 1000 U. S. gallons in the last example, what is the dry weight of the contents of the stock chest? Ans. 3431 lb. (6) Find the solid contents in cubic feet of the following logs : 24 logs 16 ft. long, 18 in. at the large end, 15 in. at small end. 36 logs 15 ft. long, 17 in. at the large end, 13 in. at small end. 38 logs 12 ft. long, 15 in. at the large end, 12 in. at small end. 28 logs 20 ft. long, 18 in. at the large end, 14 in. at small end. When calculating the cubical contents of logs, it is customary to take ir = 3, because the logs are seldom, if ever, exactly round, and it is not only useless to take w = 3.1416, but the results will be inaccurate. Ans. 339,605 cu. ft. (7) A cylindrical tank 28 ft. long and 68 in. in diameter lies with its axis horizontal; how many gallons will it hold when filled to within 7.5 in. of the top? Ans. 4969 gal. Fig. 91. (8) A tank reservoir that has the general shape of a frustum of a right cone is 16 ft. in diameter at base, 14 ft. 9 in. in diameter at top, and is 18 ft. high; how many gallons will it hold when filled to within 14 in. of the top? Ans. 23,392 gal. (9) A pipe 3% in. inside diameter discharges water at the rate of 5.62 ft. per sec; how many gallons will it discharge in one hour? The amount discharged is evidently equal to the amount required to fill a pipe of the same cross-sectional area and having a length equal to 5.62 X 60 X 60 ft. Ans. 967.3 gal. (10) Find the volume of a frustum of a cone having the dimensions indi- cated in Fig. 91. Ans. 130.72 cu. ft. §2 MENSURATION OF SOLIDS 131 SIMILAR FIGURES 139. Let ABODE, Fig. 92, be any polygon. From any point within the polygon, draw lines, called rays, to the vertexes of the polygon; and on these rays, lay off distances OA', OB', OC, etc. OA' OB' OC' in such manner that the ratios ^yv^ = jy^ = yy^ = etc. For OA' instance, suppose ^^ =f;then, OA' = f OA, OB' = iOB, OC = ^OC, etc. Joining the points A', B', C, etc. by right lines, another polygon A'B'C'D'E' is obtained that is said to be similar to the polygon ABODE. The angles at the vertexes of the two polygons that are similarly placed are equal and the corresponding sides are proportional; that is, angle A = angle Ar ^ r, IP/. A^'^' B'O' O'D' A , angle B = angle B , etc., and . p = np = ~rPfr = ^tc. Hence, when two polygons are so related that their corresponding angles are equal and their corresponding sides are proportional, they are similar. The point from which the rays are drawn may be taken any- where in plane of the figure — either within or without the polygon; in fact, it may be taken outside the plane of the figure, provided the planes of the two figures are parallel, as in Fig. 93. Here the figure ABODEF is similar to the figure A'B'O'D'E'F', and the planes of these figures are parallel. If the vertexes of the polygons are connected by lines A' A, B'B, etc., they form the edges of a prismoid. If the rays OA, OB, etc. be extended sufficiently far to enable another prismoid to be formed and the edges of the second prismoid are proportional to the first, then the two solids are similar. 132 ELEMENTARY APPLIED MATHEMATICS §2 In the case of similar plane figures or similar solids, lines that are similarly placed are called homologous; thus, in similar poly- gons, the corresponding sides are homologous. For instance, in Fig. 93. Fig. 92, AB and A'B', BC and B'C, etc. are homologous sides; AC and A'C, EB and E'B', etc. are homologous lines. In Fig. 93 if there is another solid similar to the one there shown, as the solid a'h'c'd'e'j'abcdef, then the edges A' A and a' a, B'B and &'6, etc. are homologous; and lines similarly placed, as B'A and Va, A'D and a'd, etc. are homologous lines. 140. Suppose a plane figure to be irregular in shape, as A BCD, Fig. 94; if from a point within it, a sufficient number of rays be drawn, and points on these rays be taken OA' _ OB' _ PC _ sucn xnati /~\ a — /id — r^c* — eijc., then the figure A'B'C'D' will be similar to the figure ABCD. Any lines similarly placed in the two figures will be homologous lines; thus, the fines DA and D'A', AC and A'C, etc. are homologous lines. 141. Given any two similar 'plane figures, their 'perimeters are proportional tQ the lengths of an'y tivo homologous lines; hence, if Fig. 94. §2 MENSURATION OF SOLIDS 133 the perimeter of one of two similar figures is known and the lengths of any two homologous lines are also known or can be measured, the perimenter of the other figure can be found by proportion. Let p and P be the perimeters and I and L be the lengths of two homologous lines; then, P :p =L :Z From which, P = ^ and p = -j-- All regular polygons and circles are similar; two ellipses are similar when their long and short diameters are proportional; two circular segments are similar when the central angles sub- tended by their arcs are equal; etc. Example. — A cylindrical tank lies in a horizontal position and is filled with water to a depth of 31 in. If the inside diameter of the tank is 40 in. and it is known that the wetted perimeter of a tank 6 in. in diameter and similarly filled is 12.919 in., what is the wetted perimeter of the given tank? Solution. — The wetted perimeter is the circular arc touched by the water. Since the two tanks are similarly filled, the depths are homologous lines, the diameters are also homologous lines, and the wetted perimeters are similar. Letting x = the wetted perimeter sought, x : 12.919 = 40 : 6; whence, x = 12.919 X 40 ^ gg^28, say 863^ in. Ans. 142. The areas of any two similar plane figures or solids are proportional to the squares of any two homologous lines. Since all circles are similar, the areas of any two circles are to each other as the squares of their radii, the squares of their diameters, the squares of the chords of equal arcs, etc. In Fig. 94, if the area of ABCD be represented by A and the area of A'B'C'D' by a, and the lengths of the homologous lines AC and A'C are known and are represented by L and I, A :a=U :V LP- V' from which, A = a 'K-r^, and a = A X j-2' Two cylinders are similar when their altitudes are proportional to any two homologous lines and the bases are similar ; two cones are similar when the base of one is similar to the base of the other and the altitudes are proportional to any two homologous lines; two prisms are similar when the bases are similar and the altitudes are proportional to any two homologous edges or lines of the prisms; etc. Therefore, if A is the area of the surface of a solid and a is the area of a similar solid (either the convex area or the entire area of the solids) and L and I are the lengths 134 ELEMENTARY APPLIED MATHEMATICS §2 of any two homologous lines (the diameter or altitude of a cylinder, prism, or cone, or the diameter of either base of a fru- stum of a cone, or the length of any edge of a prism is a homo- logous line when compared with a similar solid), then A:a=L^:l^ Example. — Referring to the example of Art. 131, what is the area of a piece of sheet metal that will exactly cover a cone having a circular base that is 26.4 in. in diameter and whose altitude is 36 in.? Solution. — Since the area of a cone having a diameter of base of 33 in. and an altitude of 45 in. is 2484.5 sq. in., and since -^ = jp = .8, the cones are similar, and letting A = the area of the cone here being con- sidered, A : 2484.5 = 36^: 45^, from which A = 2484.5 X^^ = 2484.5 (^j = 2484.5 (fl^= 2484.5 X .8^ = 1590.1 - sq. in. Ans. If the area had been calculated by the formula of Art. 130, ^ = ^ X ^— - X -vSe^ + ( — ^) = 1590.1 —, the same result as before. 143. The volumes of any two similar solids are to each other as the cubes of their homologous lines. Let V = the volume of one solid and V = the volume of a similar solid; then, if L and I are homologous lines, V :v =L':l' from which, V = v X -j = v ( -yj \ ■ 144. Any cylinder or prism, any pyramid or cone, any frustum of a pyramid or cone, and any sphere or spherical segment may be regarded as a prismoid, and its volume may be calculated by the prismoidal formula. For instance, in the case of a cone, one base of the prismoid is 0; hence, comparing the middle section with the base, a homologous line of the base will be twice as long as the corresponding homologous line of the middle section, and the area of the base will be 2^ = 4 times the area of the middle section. Letting a — area of the base, the area of the middle section is j; and by the prismoidal formula, V = |/i (a + 4 X T + 0) = iO'h, which is the same as formula (1), Art. 131. In the case of a sphere, both ends of the prismoid are points and their areas are 0; the area of the middle section is TT^ and the altitude is 2r. Therefore, by the prismoidal formula, 7 = 1 X 2r(0 + 4 X 7rr2 + 0) = iirr^, which is the same as formula (2) , Art. 135. §2 MENSURATION OF SOLIDS 135 If the cross sections of one prismoid are proportional to similar cross sections of another prismoid, but the altitudes of the prismoids are not in the same proportion, the volumes of the prismoids are proportional to the products of the areas of the cross sections by the altitudes. For instance, suppose that the dimensions of the frustum of two cones are known, that the cross sections of one frustum are proportional to the cross sections similarly placed in the other frustum, but that the ratio of the altitudes (or of any homologous lines not in the section or in planes parallel to the sections) is not the same as the ratio of the cross sections; then letting A and a be the areas of any two homologous sections and H and h the altitudes, V : v = A X H : a X h. Further, letting L and I be any two homologous lines in the sections whose A /L\ 2 areas are A and h,— ^ ly ) • From the foregoing proportion, V = V {—] -r- Substituting the value of — \al h ^ a y-^m and --(1) 2^ H 145. Suppose that through any point in any solid, three planes are passed perpendicular to one another, as the planes AB, CD, and EF, Fig. 95; these planes intersect in the right lines mn, pg, and rs. If now measurements are taken in these planes parallel to the intersecting lines and these measurements are proportional to similar measurements taken in another body, and this proportion is the same, no matter where the point may be situated, the two solids are similar, and their volumes are pro- portional to the products of any set of three homologous lines. For instance, suppose the point be located at a corner of the base of a rectangular parallelopiped; then, a measurement parallel to rs, mn, and pq, will correspond to the length, breadth, and thick- ness of the parallelopiped, which may be represented byL, B, and T, respectively. Homologous lines of a similar parallelopiped may be designated by I, h, and t, and V : V =L XB XT : I X b X t Example. — Referring to example 2, Art. 133, what is the volume of a similar frustum of a cone having an altitude of 13^ in. ? Solution. — Since the frustums are similar, the volumes of the two frus- 136 ELEMENTARY APPLIED MATHEMATICS §2 turns are proportional to the cubes of any two homologous lines, say their (1 ^ ^\ 3 = 28497 cu. in. Ans. Fig. 95. Example 2. — Suppose that the bases of the frustums in the preceding example had been similar, the diameters of the upper bases being 11 in. and 14}i in., while the altitudes are 8 in. and 12% in. respectively; what is the volume of the latter frustum ? Solution. — The ratio of the bases (and of any section parallel to the bases is 14.25 -;- 11 = 1.3 — ; the ratio of the altitudes is 12.375 -^ 8 = 1.5 +; hence, the volumes are proportional to the bases multiplied by the alti- tudes, and V = V (j) ^ ^ = 5930 /14.25\2 12.375 . X i .. -, I X — g — = 15394 cu.m. Ans. Fig. 96. Example 3. — Referring to the ex- ample of Art. 141, suppose the 40-inch tank is 14 ft. 5 in. long and the 6-inch tank is 3 ft. 8 in. long; how many gallons does each hold when filled to the depth specified? Solution. — A cross section of the 40-inch tank is shown in Fig. 96. The wetted perimeter is the length of the arc ACB, which was found to be MENSURATION OF SOLIDS 137 86.127 in. long. The area of the segment ACB is equal to the area of the sector AC BO + area of triangle AOB. Using the principle of Art. 87, AD = s/q X 31 = 16.703 in. Area of sector = 3>^ X 861.27 X 20 = 86.127 sq. in. Area of triangle = 16.703 X (20 - 9) = 183.73 sq. in. Area of segment = 861.27 + 183.73 = 1045 sq. in. Hence, volume of water in 40-inch tank is, since 14 ft. 5 in. = 173 in., 1045 X 173 = 180785 cu. in. = 180785 -7- 231 = 782.6+ gal. Ans. To find the volume of the water in the 6-inch tank, use the proportion / 6\ 2 4 4 given above, and 782.6: w = 40^ X 173 :62 X 44, or w = 782.6 ( j- j X jyo 4.4785— gal. Ans. SYMMETRICAL FIGURES 146. Two figures are said to be symmetrical with respect to a line, called the axis of symmetry, when any perpendicular to the axis that is limited by the outline of the figure is bisected by the axis. Thus, referring to Fig. 97, if the perpendiculars A' A, B'B, CD, etc. are bisected by the line mn, mn is an axis of symmetry, and the two figures ABODE A and A'B'C'D'E'A' are symmetrical. Now suppose that FBCDG and FB'C'D'G are sym- metrical with respect to the axis mn, the sides FG coinciding; then the figure FBCDGD'C'B'F is said to have an axis of symmetry. If two figures are symmetrical, they are evidently equal; thus, in Fig. 97, ABODE - A'B'O'D'E'. Conse- quently, if any figure have an axis of symmetry, this axis bisects the figure ; hence, FBODG = FB'O'D'G' = one- half FBODGD'O'B'F. It is also evident that if the plane of the paper be folded on the line mn, A will fall on A', B will fall on B', will fall on 0', etc., and ABODE will be superposed on and will coincide with A'B^ O'D'E'. Also, the left half of FBODGD'O'B'F will be superposed on and will coincide with the right half; therefore, if a 'plane figure can he so folded that every point on one side of the line of folding will coincide with a point on the other side, the line of folding will he an axis of symmetry and will divide the figure into two equal parts. B A / F \ A> / \ C \J ^ 6 E' / D D' B' C Fig. 97. 138 ELEMENTARY APPLIED MATHEMATICS §2 147. If the figure have two axes of symmetry, as mn and pq, Fig. 98, their point of intersection is called the center of symmetry, and the center of symmetry is the geometrical center of the fi.gure. Every regular polygon has a center of symmetry, which may be Fig. 98. found by drawing lines from the vertexes to the opposite vertexes, when the polygons have an even number of sides, as 4, 6, 8, etc. ; but, if the polygons have an odd number of sides, the axes of symmetry are found by drawing lines from the vertexes perpen- dicular to the sides opposite them. If the number of sides is even. Fig. 99. a line drawn perpendicular to any side at its middle point will bisect the opposite side and be an axis of symmetry. Conse- quently, a regular polygon has as many axes of symmetry as it has sides. See (J.) and (5), Fig. 99. The figure may be folded on any one of these axes and one half will coincide with the other half. §2 MENSURATION OF SOLIDS 139 An isosceles triangle and an arc, sector, or segment of a circle has but one axis of symmetry; see {D), Fig. 99. The figure whose outline is shown at (C) also has but one axis of symmetry. An eUipse has two axes of symmetry, the long and short diameters. 148. A solid has a plane of symmetry when sections equally distant from the plane of symmetry and parallel to it are equal and every point in one section has its symmetrical point in the other section. A frustrum of any cone has at least one plane of symmetry, which includes the axis of the cone and the long diameter of either base. If the frustum is that of a right cone, the bases are perpendicular to the axis, and there are any num- ber of planes of symmetry perpendicular to the bases. If a solid has two planes of symmetry, they intersect in a line of symmetry, and if it has three planes of symmetry, one of which is at right angles to the other two, the three planes intersect in a point of symmetry, which is the center of the solid. Thus, in Fig. 95, if the plane AB is a plane of symmetry, that part of the rectangular parallelepiped in front of the plane is symmetrical to that part behind it; if the plane EF is also a plane of symmetry, that part of the sohd to the right of the plane is symmetrical to that part to the left, and the two planes intersect in the line of symmetry pq. If a third plane AB, perpendicular to the other two is also a plane of symmetry, that part of the solid below the plane is symmetrical to that part above it, and the three planes intersect in the point 0, which is the center of the solid. 149. Now observe that when a body has one plane of sym- metry, the plane divides the body into two equal parts, and the center of the body lies somewhere in this plane. When the body has two planes of symmetry, the center lies in both planes and, hence, lies somewhere in the line of their intersection; the two planes divide the body into four parts, and if they are perpen- dicular to each other, they divide the body into four equal parts. When the body has three planes of symmetry, the center lies in all three planes, which have only one common point^the point of intersection; the three planes divide the body into eight equal parts, and if the planes are perpendicular to one another, the eight parts are all equal. To find the center of a line, bisect the Une and draw a right Hne perpendicular to the given line at the point of bisection; this line will be an axis of symmetry, provided the line is symmetrical 140 ELEMENTARY APPLIED MATHEMATICS §2 (as in the case of a right line or circular arc) ; otherwise, it has no center, unless it is symmetrical with respect to a point. To be syinmetrical with respect to a point, every line drawn through the point and limited by the given line or by the perimeter of the given figure must be bisected by the point, which is called a center of symmetry. Thus, the line shown in Fig. 100 at (a) is symmetrical with respect to the center 0, because OA = OA', Fig. 100. OB = OB', OC = OC, etc., and is the center of the line. For the same reason, the parallelogram at (6) is symmetrical with, respect to the center 0. The line shown at (c) is not symmetrical with respect to a center or to an axis, and therefore has no center. Observe that neither the line at (a) nor the parallelogram at (6) have an axis of symmetry — neither can be folded on any line so one-half can be superposed on the other ; but both have a center of symmetry, which is the center of the figure. SOLIDS OF REVOLUTION 150. Center of Gravity. — The center of gravity of a plane surface or section is that point at which the surface will balance. If the surface have an axis of symmetry, the center of gravity (which may be denoted by the abbreviation c. g.) will lie in that axis; and if it have two axes of symmetry, the c. g. will be their §2 MENSURATION OF SOLIDS 141 point of intersection. The practical way of determining the center of gravity of a section is to make a scale drawing of it on stiff paper, say Bristol board or cardboard, then carefully cut out the section; next suspend the model against a vertical surface from a horizontal pin or needle passed through the model; suspend the line of a plumb bob from the same pin, and where the line crosses the model, draw a line. Now suspend the model from some other point and draw a line where the line of the plumb bob crosses; the intersection of the two lines will be the center of Fig. 101. gravity of the plane surface or section. This method of deter- mining the c. g. is clearly shown in Fig. 101. The model is first drawn and then cut out, including the holes; a pin hole is made at A, so that the heavy part of the model will hang lowest; where the plumb line crosses, the line mn is drawn. The model is then suspended from another point B and another plumb line drawn, which crosses mn at 0, and is the c. g. of the section. If the section have one axis of symmetry, it is necessary to draw but one plumb line, since where this crosses the axis of symmetry will be the c. g. It is advisable to have the plumb lines cross at as nearly a right angle as possible, since the point of intersection will then be easier to determine. If the section (model) be placed on a sharp point directly under 142 ELEMENTARY APPLIED MATHEMATICS §2 the point 0, it will balance; if laid on a knife edge along either of the two plumb lines, it will balance. 151. If any plane section be revolved about a line in that section as an axis, the solid so generated is a solid of revolution. Some examples of solids of revolution are the right cylinder, right cone, and the sphere. The hollow cylinder or tube of Art. 127 is also a solid of revolution, and may be generated by revolving a rectangle about an axis parallel to one of the sides of the rectan- gle; see (a). Fig. 102. If a circle be revolved about an axis, the resulting solid will be what is called a cylindrical ring or torus; see (6), Fig. 102. If the revolving section is a flat ring, as indi- cated by the dotted circle in (&), the torus will be hollow. Fig. 102. Whatever the shape of the revolving section, the volume of the re- sulting solid will be equal to the area of the revolving {generating) section multiplied by the distance passed through by the center of gravity of the section. If r = the perpendicular distance from the center of gravity to the axis, and the section make a complete revolution, the distance passed through by the center of gravity will be the circumference of a circle having a radius r; representing this circumference by c, c = 27rr. If a = the area of the gen- erating section, the volume of the solid is V = 2Trra 152. Referring to (a). Fig. 102, the center of gravity of the section Hes midway between AD and BC, because the rectangle has two axes of symmetry, one of which is parallel to and half way between AD and BC; hence, if the diameter of the circle described by be represented by dm, and AB and AD be repre- sented by t and I, respectively, the area of the rectangle is ^ X Z, and V = irdmtl, which is the same as formula (2), Art. 127. §2 MENSURATION OF SOLIDS 143 Referring to (6), Fig. 102, the c. g. of the section is the center of the generating circle. Let n = the radius of this circle; then the area of the generating section is tTi^, and the volume of the torus is V = 27rViV = 19.7392riV (1) If the torus is hollow, let ri = radius of outer circle of section and ra = radius of inner circle of section; the area of the section is 7r(r2i - rS) = 7r(ri + r^) (r^ - ra), and F = 27rr X 7r(ri + rg) (^1 - ^2), or V = 19.7392(ri + r2)(ri - r2)r (2) Many, if not the majority, of soHds of revolution that occur in practice have generating sections that have an axis of symmetry perpendicular to the axis of revolution; thus, referring to (c). Fig. 102, the sectionA^Ci) has an axis of symmetry O'B perpen- dicular to the axis mn about which the section is revolved. The e.g. must he somewhere on O'B, and may be located by making a model section, on which draw O'B, and then suspending the model from a pin as previously described, indicate where the plumb hne crosses O'B. Or, if preferred, the model may be balanced on a knife edge, and where this edge crosses O'B will be the point 0, the e.g. Methods of finding by calculation the center of gravity of various geometrical figures will be described in the text treating on mechanics. Example. — What is the volume of a torus, if the diameter over all is 13>^ in. and the diameter of a radial section is % in. ? What is its weight if made of cast iron, a cubic inch of which weighs .2604 lb. ? Solution.— Referring to (6), Fig. 102, O'A = 13.5 -^ 2 = 6.75 in.; AB = Ji in.; OA = ]4 -i- 2 = }{e = .4375 in. = ri in formula (1). O'O = 6.75 - .4375 = 6.3125 in. = r; then, V = 19.7392 X .43752 X 6.3125 = 23.85 - cu. in. Ans. The weight = 23.85 X .2604 = 6.211- lb. Ans. By radial section is meant a section made by a plane passing through the center of the body; such a section includes the axis of the body. All the sections shown in Fig. 102 are radial sections. ELEMENTAEY APPLIED MATHEMATICS (PART 3) EXAMINATION QUESTIONS (1) Find (a) the convex area and (6) the entire area of a cone of revolution whose base is 22 in. in diameter and whose altitude is 38 in. ,^, /(«) 1367.1 sq. in. ^^^- I (b) 1747.2 sq. in. (2) A hopper having somewhat the shape of a frustum of a pyramid has the following dimensions: upper end, a rectangle 44 in. by 28 in.; lower end, a square 8 in. by 8 in.; altitude 36 in. How many gallons will it take to fill the hopper? Ans. 82.286 gal. (3) A piece of cast iron has the shape of a pentagonal prism; the ends are regular pentagons, each edge measuring 2j in., and the altitude is 4| in. What is its weight, a cubic inch of cast iron weighing .2604 lb.? Ans. 10.206 lb. (4) The base of a triangular pyramid is an equilateral triangle, one edge which measures 5| in. ; the altitude is 9| in. ; what is its volume? Ans. 43.116 cu. in. (5) A shallow pan has the shape of a frustum of a cone, and its inside dimensions are: upper end 14 in. diameter, lower end 13 in. diameter, and distance between ends 2| in. ; what arc the cubical contents of the pan? Ans. 304.31 cu. in. (6) The base of a wrought-iron wedge is 10 J in. by 3i in.; the upper edge is 7| in. and the altitude is 15 in. Taking the weight of a cubic inch of wrought iron as .2778 lb., what is the weight of the wedge? Ans. 64.328 lb. (7) How many cubic yards are contained in a foundation wall whose length is 24 ft. 3 in., breadth is 15 ft. 8 in., inside measurements, and thickness is 16 in.? The height of the wall is 8 ft. 6 in. Ans. 365.88 cu. yd. 10 145 146 ELEMENTARY APPLIED MATHEMATICS §2 (8) A wooden ball has a cylindrical hole extending through it, the axis of the hole coinciding with the axis of the ball. If the diameter of the ball is 11 in. and of the hole is 4| in., what is the weight of the ball, if a cubic inch of the wood weighs .023 lb. ? Ans. 12.268 1b. (9) A cylindrical tank lies with its axis in a horizontal position; it is filled with stock to within 8 in. of the top; its diameter is 60 in., and its length is 21 ft. 6 in., inside measurements. How many gallons of stock are in the tank? Ans. 2907.6 gal. (10) A cylindrical tank 48 in. in diameter and 16 ft. long is partly filled with stock. The tank lies in such position that the level of the liquid just touches the upper end of the dia- meter at one end, and cuts the diameter at the other end in its middle point. How many gallons of stock are in the tank? Ans. 319.17 gal. (11) It is desired to make a cylindrical tank to hold 800 gal., the height to be 1| times the diameter; what should be the in- side diameter and height to the nearest 16th of an inch? . ( Diameter, 53^1 in. '^''^' I Height, 80 1 in. (12) A wrought-iron sheU 9 ft. 7^ in. long is open at both ends; it is 54 in. outside diameter and | in. thick. Taking the weight of a cubic inch of wrought iron as .2778 lb., what is the weight of the shell? Ans. 3362.7 lb. (13) If the periphery of an ellipse whose diameters are 37 in. and 12| in. is 73.22 in., what is the periphery of an ellipse whose diameters are 9i in. and 3| in.? Ans. 18.305 in. (14) The area of a certain figure is 430.6 sq. in., and the length of a line drawn through the figure is 21 j in.; what is the area of a similar figure, if the length of a line similarly placed is ]9| in.? Ans. 357.96 sq. in. (15) The volume of a frustum of a cone having an altitude of 8 J in. is 6473 cu. in.; what is the volume of a similar cone having an altitude of 6| in.? Ans. 3352 cu. in. (16) What is the volume generated by revolving an isosceles triangle about an axis parallel to the base, under the following conditions? base is 6^ in., the other two sides are each 4| in., distance from axis to base is 7 in. The distance from the center of gravity of any triangle to the base is one-third the altitude. Ans. 513.84 cu. in. SECTION 3 HOW TO EEAD DRAWINGS REPRESENTING SOLIDS ON PLANES PRELIMINARY EXPLANATIONS 1. The Picture Plane. — When viewing an object, it is essential that the object be illuminated in some manner, either by light originating in or on the object itself or by light coming from an- other source and being reflected from the object. Light travels in right (straight) Hues, called rays, and no matter what their length, whether a fraction of an inch or milHons of mUes, every ray of light is absolutely straight. The number of rays of light from any particular object is infinite, and they extend in every direc- tion (in right lines) from the object unless they are stopped by some opaque substance that light cannot penetrate. A certain number of these rays enter the eye, wherein an image or picture of the object is formed and the object is then said to be seen. If, when viewing an object, a sheet of paper (transparent so light can pass through it) be held between the eye and the object and the various lines seen on the object are traced on the paper with a pen or pencil, the result will be a drawing or picture of the object viewed from the position occupied by the eye relative to the object. For every change in the position of the eye, there will be a change in the shape of the drawing, and for every change in distance between the paper and the eye, there will be a change in the size of the drawing, these two facts are clearly shown in Fig. 1. Here ABCDEFG is a rectangular prismoid- — in this case, a frustum of a rectangular pyramid — which contains a hole pasing part way through it; 5 is the eye; and P is the plane of the paper, called the picture plane. Now imagine lines drawn from the points A, B, C, etc. of the object to the §3 1 HOW TO READ DRAWINGS §3 eye; these lines correspoDd to the rays of Hght from the object to the eye, and they pierce the picture plane P in the points a, b, c, etc. Joining these points by lines as shown, the result is the outline abcdefg, which is a drawing or picture of the object ABCDEFG. It is evident that if the point S be moved up or down the line S'S" or be moved perpendicular to this line in a plane perpendicular to the plane of the paper, say in a plane parallel to the picture plane, the rays wUl make a different angle with the picture plane and the shape of the drawing will Fig. 1. be different. It is further evident that as the picture plane is brought nearer to the eye, the drawing will be smaller, and if moved farther fiom the eye, the drawing will be larger, owing to the convergence of the rays. 2. Note that every ray makes an angle with the picture plane that is different from that made by any other ray; thus, the angle made by the ray AS is different from that made by BS, FS, etc. Now assume the picture plane to be so located that it is perpen- dicular to the parallel planes of the bases of the frustum and is parallel to the parallel edges CB and FG; assume further that the point S, called the point of sight, Ues on a ray passing through the center of the face CBGF and perpendicular to the picture plane. Then, if the picture plane be very near the object and the point of sight be a great distance from the object, all the rays will make §3 REPRESENTING SOLIDS ON PLANES angles with the picture plane that are very nearly equal to a right angle; and, if the point of sight be assumed to be situated at an infinite distance from the object, the rays will all be parallel and will all be perpendicular to the picture plane. The resulting drawing will then be a projection of the object on the picture plane, as shown in Fig. 2. Here only one side of the object is represented on the drawing, the side CBGF; the entire line AB is projected in the single point a, h; the lines DC, EF, and the line representing the edge running back from G are also projected in single points, since they are all perpendicular to the picture plane. It will be noted that the hole does not appear at all. 3. WhUe the drawing in Fig. 2 shows only one side of the object and gives scarcely any intima- tion as to the shape of the object, it nevertheless possesses several advantages over that shown in Fig. 1. On the frustum, the lines AB and CD, also AD and CB, are equal and parallel, but in Fig. 1, these lines are neither equal nor parallel, and it would be very difficult to determine from the lengths of the lines ah, he, etc. the true lengths of the edges AB, BC, etc. In Fig. 2, ch and fg are parallel, as they should be; ch = CB and fg = FG. Further, by using dotted lines to indicate lines that are hidden, the depth and diameter of the hole can also be shown. 4. Projection Drawings.^ — The drawing shown in Fig. 1 is called a scenographic projection drawing, a perspective drawing, or, simply, a perspective ; the drawing shown in Fig. 2 is called an orthographic projection drawing or, simply, a projection drawing. In the majority of cases that arise in practice, two views are given of the object, one view being taken parallel to the horizontal plane and the other parallel to the vertical plane; and when both these views are projection drawings and are fully dimensioned, they will usually suffice for reproducing the object in its exact size. If the object is a complicated machine or machine part, three views may be necessary. If, in addition, the object has a Fig. 2. 4 HOW TO READ DRAWINGS §3 complicated interior arrangement that is hidden by the outside surface of the object, interior views (sections) are frequently necessary or desirable, in order to obviate the use of too many lines, which may make the drawing very hard to read. These features will be made clear in what follows. Suppose the frustum of Fig. 1 be enclosed in a glass box, as illustrated in Fig. 3, and that the front and right-hand sides and the top be treated as picture planes, the glass being assumed to be transparent. Proj ecting the object on the front plane ilf, the result is the outline d"c"J''e", which may be regarded as a drawing of the frustum on the picture plane M when the eye is at an infinite distance from it. The hnes Dd" , Cc", etc., are the light rays from the object to the eye; as applied to drawings, these lines are called projectors. In a similar manner, Cc'", Bh'", etc. are projectors from the object to the plane side P, and c"'h"'g"'j"' is the projection of the frustum on the plane P, considered as a picture plane. Likewise, Ee' , Dd' , etc. are projectors from the object to the top plane N , and the outline there shown is a pro- jection drawing on the plane N, considered as a picture plane. Note that the hole is shown in all three views, being represented as a circle in the top plane, and as a dotted rectangle in the front and side planes. The line of intersection made on a plane A by the intersection with it of another plane B is called the trace of the plane B; Thus, in Fig. 3, the intersection of plane M with plane N is the line JK, and JK is the trace of plane M on plane N; likewise, JK is the trace of plane N on plane M. Similarly, KL is the trace of plane P on plane N, and KI is the trace of plane P on plane M. The lines JK, KL, and KI are so important that they have received special names : JK is called the front trace ; KL is called the side trace; and KI is called the vertical side trace. If a plane be passed through the line EF (the lower front edge of the frustum) parallel to the plane M, the trace of this plane on N will be the line e'/'", and the trace on plane P will be the line f^f". (A plane is of infinite extent.) It will be observed that e'f" is parallel to the projector Ff"; in fact, it is a projec- tor from /' to the side trace, which it intersects in /^". Also, f^f" is parallel to the projector Ee' and is a projector from/"' to the side trace KL, which it intersects in /*" also. Similarly, e'' is the projector of e' and e" on the front trace JK, and is the §3 REPRESENTING SOLIDS ON PLANES point of intersection of the traces e"e^ and e^/i' of the plane h'e''e"H, which has been passed through EH parallel to plane P. Consequently, if the position of the projection of any point on the planes N and ilf, iV and P, or M and P is known, the position of the projection of the point on the third plane can easily be found. For example, suppose the positions of c' and c" are known, and it is desired to find the position of this point on plane P. Draw c'c^" perpendicular to KL, which it intersects in c^"; through c'", draw c^^'c'" perpendicular to KL; also Fig. 3. 0"^^ perpendicular to KI, which it intersects in c"^; draw c^^c"' perpendicular to KI (and parallel to KL) ; C^'c'" intersects c^^c'" in c'", which is the position of the projection of point C of the object on the side plane P. If the projections c" and c'" had been given and the projection c' on the plane N had been desired, project c" on the front trace in c^; project c'" on the side trace in c'"", draw C'c' and c'^c' perpendicular respectively to JK and KL, and they intersect in c', which is the required projection. The outlines a'h'c'd'e'f'g'h', d"c^Te",2.ndic"h"'g"'f"^rQ projec- tion drawings of the rectangular frustum; they are not suitable as they stand for use as working drawings or for any other purpose. To make them suitable, imagine the side M of the glass box to be hinged to the topside N, and the side P to be hinged to the top 6 HOW TO READ DRAWINGS §3 also. Now lift M until its plane coincides with the plane of N, and lift P until its plane coincides with the plane of N also ; the result will be as shown in Fig. 4. Note that the projectors gr V and gi^g'"^ 6'6^^ and ¥"}}'", etc. become a single right line in Fig. 4; and the same thing is true of the projectors e'e'" and e'^e", d'd" and d^d", etc. In other words, g' may be projected directly to g'", h' to 6'", etc.; and e' may be projected directly to e', d' to d", etc., the only requisite being that the distance between g"'j"' and h"'c"' or the distance between e"j" and d"c" must be known, and this will be given by measurements made on the object. Instead of the arrangement of views shown in Fig. 4, another arrangement equally correct may be and is frequently employed. Referring to Fig. 3, suppose that instead of hinging the side P to the top N, the side P is hinged to the front M; then pulling out the side P until its plane coincides with the plane of M, lift both planes until the common plane of M and P coincides with the plane of N. The result will be the arrangement shown in Fig. 5. The arrangement shown in Fig. 4 is the more natural, but that shown in Fig. 5 may sometimes be more convenient. Either Fig. 4 or Fig. 5 is a correct projection drawing of the frustrum, and when properly dimensioned, may be used as a working drawing. 5. Names of, and Number of, Views. — Referring to Figs. 4 and 5, view N is called either a top view or a plan; it is the view obtained when the eye is directly over the object. View M is called either a front view or a front elevation or, simply, the eleva- tion ; it is the view obtained when the eye is directly in front of the object. View P is called a side view or a side elevation; it is the view obtained when the eye is so situated as to see the object from the side, the picture plane being at right angles to the front and top planes. In both Figs. 4 and 5, three views are given, and, theoretically, three views on planes perpendicular to one another are sufficient to give the size and shape of any object, since no solid has more than three dimensions^ — length, breadth, and thickness. In practice, however, it is sometimes advisable to show more than three views, because the number of lines on the drawing would otherwise be so numerous that the drawing would be very difficult to read. Since the glass box in Fig. 3 has six sides, it is possible to get six views without modifying the shape of the box. Thus §3 REPRESENTING SOLIDS ON PLANES what is termed a bottom view or inverted plan is obtained by using TQRI as a picture plane, another side view by using TQSJ as a picture plane, and a back view or rear elevation by using RQSL as a picture plane. To get these views into proper posi- tion, imagine the bottom plane to be hinged to plane M, the side planes to be hinged to plane N, and the back plane to be hinged to plane N also. Now revolve the bottom plane TQRI downward 8 L h P h ' 9 ff \ V »'/ ] 9'" b'" f^^ y n ■•* '^i e 1 d c C" c L___^ f" / 1 \ , fl V C f /'" J i" c r K M ir J.. J II I Fig. 4. until it coincides with plane M, and then revolve both planes up- ward until they coincide with plane iV, likewise, revolve planes TQSJ, IRLK, and RQSL upward until they also coincide with plane N; the result will be as shown in Fig. 6. Except for the letters at the corners, the side views are alike and the front and back views are also alike, since the object is symmetrical; the bottom view, however, is somewhat different from the top view by reason of the dotted lines, which are used because the base of the frustum hides everything above it when the frustum is viewed from below. 8 HOW TO READ DRAWINGS §3 When all six views are shown, the arrangement in Fig. 6 is to be preferred. However, the bottom and sides can be imagined to be hinged in any other manner desired that will permit all the planes to be brought into coincidence with the top plane. Conse- quently, the inverted plan may be placed to the right of the right side elevation c"'b"'g"'f", to the left of the left side eleva- tion d^"a*"A''"e''", or above the rear elevation a"'h'''-g'"%'"\ and the M o d \ I \ c IL K f. n Fig. 5. two side elevations may be placed to the right and left of the front elevation, as noted in Fig. 5, if desired. It will be noted that the letters at the outside corners on the inverted plan are capital letters, corresponding to those at the corners of the base of the frustum, because as shown in Fig. 3 the frustum is sup- posed to rest on the bottom plane and there are, consequently, no projectors from the base to the picture plane. 6. Working Drawings. — A working drawing is a projection drawing on which all necessary dimensions are marked and on which all necessary notes are written or printed that are required §3 REPRESENTING SOLIDS ON PLANES 9 [-- E M f a b Vtll j J/tl^ p G Fig. 6. 10 HOW TO READ DRAWINGS §3 in order that the object represented may be made or reproduced. A working drawing of the frustum of Fig. 1 is shown in Fig. 7. It will be observed that only two views are given; two views are all that are required in this case, as the following considerations will show: The lines mn and jyq are called center lines; in the present case, they are axes of symmetry in the plan and Tnn is an axis of sym- metry in the front elevation. It is plainly evident that the point of intersection of mn and 'pq is the center of the circle that repre- sents the plan of the hole. The plan shows that there are two ^ b.a/l 1 p / \ ^ \ 4^' pjg^^" -t^^H^ .i~.^^^^^^ u. -4f- ^ Fig. 7. Pig. 8. surfaces, ABCD and EFGH, whose projections are rectangles, and the elevation shows that these surfaces are planes (j&at surfaces) and that they are parallel. If they were not plane surfaces, the lines DC and EF in Fig. 7 would not be straight, and if they were not parallel, the lines DC and EF would not be parallel. It is seen from the dimensions that EF and HG are both 4 in. from the center line pq, and that EH and FG are both 2j in. from the center line mn; also, AB and DC are both 2| in. from pq, and DA and CB are both if^ in. from mn. Since these four dimen- sions are given only once, it is inferred that HG, AB, DC, and EF are parallel to pq, and that EH, DA, CB, and FG are parallel to §3 REPRESENTING SOLIDS ON PLANES 11 mn, thus making HGFE and ABCD rectangles, since the main center lines are always drawn perpendicular to each other. The two planes are parallel and are located symmetrically with respect to the center lines mn and pq; their distance apart is given in the elevation, and is found to be 6^ in. The general shape is, consequently, a prismoid, since if the sides were not plane surfaces, some of the lines connecting the top and bottom, as DE, CF, BG, and AH (in both plan and elevation) would not be straight. The hole is IJ in. in diameter and 3 in. deep, and extends down from the top. To make the object from say, a piece of wood, a rectangular prism having a cross section of 4^" X 8" and a length of 6|" would be made. On one end, lines would be drawn parallel to the long sides and 21 — If^ = /^ in. from them; also, lines would be drawn parallel to the short sides and 4 — 2f = If in. from them, as indicated in the elevation, Fig. 8. Then the material included between the outer edges and the dotted lines would be planed off, the result being the frustum. The center of the hole could then be located by drawing the diagonals, as indi- cated in the plan, or by laying off ah and cd, as indicated, both being equal to If^^", and drawing M; then lay off ho or do equal to 2f", thus locating the center o. Having found the position of o, drill or bore a hole l\" in diameter and 3" deep, the axis of the hole being perpendicular to the bases, and the work is completed. It will be noticed that no projectors are shown in Figs. 7 or 8; they are never shown on finished drawings or on working draw- ings. -When reading a drawing, that is, when studying it to find out the shape of the object and the details of its parts, the pro- jectors can always be imagined to be present; and if the eye does not readily perceive the connection between different parts of two views, this may usually be found by means of a straightedge assisted, perhaps, with dividers to set off distances. This feature will be dwelt on more fully later. SPECIAL FEATURES PERTAINING TO DRAWINGS 7. Different Kinds of Lines Used on Drawings. — In Fig. 9, are shown five different lines, which are used in the following manner: Line I.— This is a full line and may be of any convenient weight (by weight is here meant thickness). This line should have the same weight wherever used on any particular drawing; 12 HOW TO READ DRAWINGS §3 it is employed in all cases when the outline of the object can be seen with the eye in the position it is assumed to occupy when the view is drawn. This line is used more than any of the others and is, consequently, the most important. Line II. — This line, called the dotted line, consists of a succes- sion of dots or very short dashes; it is used to show the outline of parts that cannot be seen by the eye when in its assumed position relative to the view being drawn. The inverted plan, or bottom view, in Fig. 6 is a good example of the use of the dotted line. Viewing the frustum from the bottom all that can be seen is the outline of the base, which is drawn with full lines. The top and the edges connecting the top and bottom are then represented by dotted lines, and likewise the hole. This line is never used for any purpose other than to represent invisible outlines. I Fig. 9. Line III. — The broken and dotted line consists of a series of long dashes with a single dot or very short dash between the long dashes; it is used for center Unes, and is also frequently employed to denote the traces of planes, showing where a section has been taken. Its use as a center line is shown in Figs. 6 and 7. Its use to indicate where a section is taken will be discussed later. Line IV. — This broken and dotted liiie consists of a succession of long dashes and two dots; it is used for the same purpose as line III, and both rarely appear on the same drawing, but when they do, line III is used for center lines and line IV to indicate where a section has been taken. Line V. — The broken line consists of a series of long dashes; it is generally used for the same purpose as projectors, to indicate the extension or prolongation of lines, but its principal use is for dimension lines. It is also sometimes used to indicate the extension of lines that are not actually a part of the view being drawn. Fig. 7 shows how it is employed in extending lines and in dimension lines. While the foregoing specifies the manner in which these lines §3 REPRESENTING SOLIDS ON PLANES 13 are most generally used, there are occasional exceptions, particu- larly in the case of working drawings. In some drafting offices, the regular full line is made very heavy and a light full line is used in the same manner as line III; this practice, however, is not recommended. Again, it is quite common practice to use the light full line for dimension lines, it being broken only where the dimension is placed. But, when lines III, IV, and V are used, they are usually employed as specified above. 8. Center Lines. — Center lines are used for two purposes: for what may be termed base lines, from which to take measurements; and to indicate curved surfaces, particularly circles and cylindrical surfaces. They are also used as axes of symmetry. Referring to Fig. 6, the center line pq serves two purposes: it is an axis of symmetry for the plan and the two side elevations; it shows, in connection with the circle in the plan, that the hole is round — a right circular cylinder, in fact. The center line mn also serves the same two purposes, being an axis of symmetry for the plan, the front and rear elevations, and the inverted plan. The point of intersection of these two center lines locates the center of the circle that represents the top and bottom views of the hole. On working drawings, center lines are invariably drawn to define the axis of cylindrical surfaces, and their presence on a drawing passing through a rectangle is sufficient in itself, as a rule, to show that the rectangle is the projection of a cylinder. In the case of a circle, two center lines are always drawn at right angles to each other, thus locating the center of the circle. The use of center lines as base lines is illustrated in Fig. 7; here the dimensions 2j", 2j", 2f", 2f" etc. which are measured from the center lines outward, serve to locate the position of the hole, and they also show by their equality that the center lines are axes of symmetry. 9. Dimension Lines. — Whenever it is possible to avoid it, dimension lines are seldom drawn across the face of any view; this is to prevent confusion in reading the drawing, and also to make the dimensions more prominent. By the use of extension lines, the dimension lines and dimensions are very largely kept off the different views. Practically the only exception to this rule is when giving the diameters of circles and the radii of cir- cular arcs. The use of extension lines is shown in Figs. 6 and 7. In Fig. 7 it will be noted that the diameter of the hole is printed on the front elevation, thus indicating the diameter of the cylin- 14 HOW TO READ DRAWINGS §3 drical hole instead of the diameter of the circle in the plan; this is done because of the crossing of the center lines on the circle. When there is room enough, the dimension lines have arrow- heads, one at each end, with the dimension written or printed about midway between the ends; but, when the space is limited, as in the case of the diameter of the hole in Fig. 7, short arrows with their heads pointing toward each other are used, the dimen- sion being placed between the arrowheads. In some cases, even this will not suffice, the dimension being placed alongside the arrow, as illustrated in Fig. 8 in connection with the dimen- sion -33- . What are termed the general or over all dimensions are those relating to the extreme or outer bounding lines of the figure. In Fig. 7, the over all dimensions are 8", 4|", and 6j", the first two giving the size of the base and the third the height of the frustum. The general dimensions include these and also 5|" and 3t1", which give the size of the top. It might be argued that both of the two dimensions, which added together make the over all dimension 8" are not necessary, since if one is given the other can be found by subtracting from 8; this is true, but the idea of giving both is to save subtracting, which is sometimes awkward. For instance, the distance AB, Fig. 7, is Syf"; if the distance from A to the center line were given as Ifi", it is not readily apparent that the dis- tance from B to the center line is ifi", and this can be found only by subtracting ifl" from Syf", a somewhat awkward operation and one that takes a certain amount of time, together with the possibility of making a mistake. It is for the same reason that the over all dimensions are given — ^to obviate the necessity of adding the intermediate dimensions. Over all dimensions are placed outside of, that is, beyond, all shorter measurements. 10. Scales. — Up to this point, drawings have been described as though every line on them was of the same length as the corre- sponding measurements taken on the object. It is obvious that this is not feasible in many cases, for not only would it be extremely difficult to make the drawing but it would also be very difficult if not impossible to read it. The matter of storing and preserving drawings must also be considered. Consequently, most of the drawings in actual use are made to scale, as it is termed; by this is meant that the drawing is smaller than if all the dimensions corresponded in length with those of the object. In, such cases, §3 REPRESENTING SOLIDS ON PLANES 15 lines on the drawing are only one-half, one-third, one-fourth, etc. as long as those they represent on the object. Sometimes, when the object is small and it is desired to bring out certain details prominently, the drawing may be made two, three, four, etc., times as large as the object. Any drawing that has been made accurately, every line being carefully measured with a scale, is called a scale drawing. If every line on the drawing is of the same length as the line it represents on the object, the scale is full size or 12" = 1 ft. If a line on the drawing is only half as long as the corresponding line on the object, the scale is half size or 6" = 1 ft. if the lines on the drawing are only a quarter, an eighth, etc. as long as the corre- sponding lines on the object, the scale is quarter size, eighth size etc. A quarter-size scale is 3" = 1 ft. because one-fourth of 12" is 3"; hence, 3" measured on the drawing equal 1 ft. measured on the object. If 3" be divided into 12 equal parts, each part will represent 1 inch on the object when the drawing is made to a scale of 3" = 1 ft. Dividing each of these parts into halves, quarters, eighths, etc., these smaller divisions represent halves, quarters, eighths, etc. of an inch. The scales most commonly used are: full size (12" = 1 ft.), half size (6" = 1 ft.), 3" = 1 ft. (quarter size), 1^" = 1 ft. (eighth size), and f" = 1 ft. (sixteenth size). The first three scales are the most used, and a scale smaller than |" = 1 ft. is very seldom needed in drawings of machines. Architects and civil engineers use scales very much smaller, a favorite scale with architects being 1" = 4 ft. (or |" = 1 ft.). Any drawing that has been made to scale should always have the scale used marked on it. It sometimes happens that a part of the drawing may be made to one scale and a part to another scale; in every such case, the scale used should always appear on the drawing in connection with the part to which it applies. To use a scale to find the actual length of a particular line on the object, proceed as follows: suppose the scale is 3" = 1 ft., and that an ordinary 12-inch scale, with inches divided into halves, quarters, eighths, sixteenths, and thirty-seconds, is used. Suppose further that the actual length of the line on the drawing that is being measured is 3if in. Since the scale is 3" = 1 ft., 1 in. on the drawing represents 12 -r- 3 = 4 in. on the object; I in. on the drawing represents 1 in. on the object; and sV in- on the drawing represents -gV X 4 = |in. on the object. Con- 16 HOW TO READ DRAWINGS §3 sequently, 3if in. on the drawing represent 3X4+13X1 = 12 + If = 13f in. on the object. Had the scale been 1|" = 1 ft., 1 in. on the drawing = 12 -^ 1| = 8 in. on the object; 3V in. on the drawing = tV X 8 = Jin. on the object; and 3|^f in. on the drawing represent 3 X 8 + 13 X i = 24 + 3i = 27i in. on the object. Observe that since the second scale is twice as small as the first, the length of the line when measured to the second scale should be twice as long as when measured to the first scale; and this is the case, since 13f X 2 = 27|. The process of determining the length of a line on the object by measuring the corresponding line on the drawing is called scaling the drawing, and when a measurement has thus been taken, the drawing is said to be scaled. Anj?- drawing may be scaled pro- vided the scale to which the drawing was made is known, pro- vided further that the drawing has been made accurately and the scale used for measuring is accurate. Special rules (scales) are made for making drawings to scale and measuring them. 11. Abbreviations and Notes on Drawings. — When a line is drawn from the center to the arc to indicate a radius, the abbre- viation r. or rad. is almost invariably written after the dimension, thus making it clear that the dimension is the radius of the arc. An arrowhead is placed at the end of the dimension line touching the arc, but not at the center. The abbreviation D., Dia., Diam., d., dia., or diam. is used to follow the dimension indicating the diameter of a circle or a cylindrical surface. Thds or ihds means threads; thus, 8 thds means 8 threads to the inch, and refers to screw threads. The letter/, or fin. uiesms finish, and indicates that the surface on which it is written in the drawing is to be finished. If only a part of the surface is to be finished, this is frequently indicated by drawing a line near to and parallel to the line representing the surface, but extending only as far as the sur- face is to be finished, and writing on it /. or fin. In addition to the dimensions, working drawings usually carry a number of "notes," which are written or printed on the drawings for the benefit of the workmen. In connection with these notes, certain terms are employed that deserve explanation. Thus, the word cored means that the hole is to be made by means of a core when casting and is not to be finished; hore or hored means that the hole is to be cored and finished by boring; drill means that the hole is to be made by drilling; ream or reamed means that after the hole has been bored or drilled is to be finished by reaming; the §3 REPRESENTING SOLIDS ON PLANES 17 word tap means that the hole is to be threaded, a tap being used for this purpose; faced means that the surface is to be finished in a lathe or boring mill, or other machine tool that will make the surface flat— usually by revolving it against the cutting tool; planed means that the surface is to be finished by planing, the tool used being a planer, shaper, or a milling machine; grind means that the surface is to be finished by grinding; scraped means that the surface is to be finished by scraping, that is, hand finished with a scraper; tool finish means that after the sur- face has been finished on the machine, nothing further is to be done to it. Other terms are sometimes used, but they are usually self evident and need not be referred to here. SECTIONS AND SECTIONAL VIEWS 12. Why Sections are Used.— While all hidden surfaces and parts may be represented on a drawing by dotted lines, it is nevertheless frequently advisable to show what are called sec- tional views or sections instead of making the drawing in the regular manner. The cutting plane may be considered to pass through every part of the object that it can touch or its trace may be limited to only a short length relating merely to a single detail. Sections are usually so taken that lines which would appear dotted in a regular projection drawing appear as full lines in the sectional view, thus making the drawing easier to read. This fact is brought out in Fig. 10, which is a drawing of a clamp- ing ring. The center hues mn and pq are axes of symmetry. The view in the middle may here be considered as a front eleva- tion, the view on the right being a side elevation and that on the left a sectional elevation. Note how much clearer the sectional elevation is than the side elevation. Considering the sectional view, the section is taken along the center line mn, the cutting plane being perpendicular to the flat surface included between the circles e and g, that is, perpendicular to the plane of the paper; that part of the ring to the left of mn is then imagined to be removed, and a drawing is made of the re- maining part, the eye being situated to the left of mn. Now note particularly that whenever a section is taken, any surface touched by the cutting plane is indicated on the drawing by cross hatch- ing, as it is termed, the cross hatching consisting of parallel hues, drawn at an angle, usually 45°, to the horizontal. By using 18 HOW TO READ DRAWINGS §3 different kinds of cross hatching, different materials may be in- dicated, so that an inspection of the section lines (cross hatch- ing) will show at once what material is to be used in making the object represented by the drawing. Thus, in Fig. 10, the cross hatching there used represents steel, as will be explained pres- ently; hence, the clamping ring is to be made of steel. Supposing the right-hand view to be omitted, the drawing may then be read as follows: All dimension lines extending up and down and parallel to the center line mn are evidently diameters of circles. The dimension 9fi" is the diameter of the circle marked a; diameter of circle h is 9f|", of circle c, 9if"; of circle d, 9i"; of circle e, 8", and of circle/, Q%". Circles h and c are dotted, because, imagining the sectional view to be a full view, when the ring is looked at from a point to the right of the sectional view, all that part of it to the left of CC is hidden. Further, circles h and c are imaginary, because the surface they are supposed to indicate is a curved surface; still, they are useful in helping to understand the draw- ing. The fact that the lines CD and CD' slope shows that the ring is a frustum of a cone, or that part of it is which is included between DD' and CC That part included between circles e and g is flat, and extends downward to form a cylindrical ring, whose inside diameter is 5|", outside diameter is 8", and altitude is tV + tI = Ig"- Evidently, there is a groove between circles e and d; there is also another groove on the under side, as in- dicated at E and E', which are at the bottom of another conical surface. According to the note, there are 8 holes spaced equally, the dotted circle showing that they are all situated at the same distance from the center o; and the sectional view shows that they pass clear through the ring. The note states that these holes are of such size that they may be threaded by tapping with a T^th inch tap having 12 threads per inch. The note at the bottom of the drawing, which reads "/" all over, means that the ring is to be finished all over, that is, no part of it is to be left rough. 13. Standard Sections. — The sections shown in Fig. 11 are practically in universal use. The first form, shown at (a), con- sists of parallel lines spaced at equal distances apart, all lines being of the same weight; this form is always used for cast iron. The sectioning for wrought iron is shown at (&) ; it is made by drawing light and heavy lines alternately, parallel and equally dis- REPRESENTING SOLIDS ON PLANES 19 Si M P^W" ^ ^• fS t--=:D pR 20 HOW TO READ DRAWINGS §3 tant apart. The sectioning for steel is shown at (c) ; it is made by drawing pairs of parallel lines, the distance between two consecu- tive pairs being about twice the distance between the lines form- ing the pairs. The sectioning for brass is shown at (d); it is drawn in the same manner as the sectioning for cast iron, except that every other line is broken, being made up of short dashes. The sectioning for wood is shown at (e) ; the upper half shows the Cast Iron Steel Brass ■ TFood (d) form used when the section is taken across the grain, and the lower half is used when the section is taken lengthwise, or with the grain. There are other forms for other materials, but they are not much used and there are no other recognized standards, different forms being used to indicate the same material in different draft- ing rooms. The ones shown in Fig. 11 are well recognized and are in nearly universal use. 14. When the surfaces of two different parts come together in a sectional view, the section lines of the two parts are made to slope in different directions whenever possible. Thus, referring to Fig. 12, which represents a section taken through a bracket, which holds up a Une shaft, and its bearing P is the bracket,